LNTwww - User contributions [en]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T10:02:24Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which two-port network has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right $p$–half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.
*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*The double zero is at  $p_{\rm o} = 0$ and the double pole at  $p_{\rm x} = -A = -2$.

'''(3)'''  The following holds for configuration  $(1)$ :
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  Similarly, the following is obtained for configuration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Thus, the suggested solutions 2 and 3 are correct in contrast to statement 1:
* While  $H_{\rm L}(p)$  has two conjugate complex zeros,
*  $H_{\rm L}\hspace{0.01cm}'(p)$  only has a single zero at  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  The following applies for configuration  $(3)$ :
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*The zero of  $H_{\rm L}\hspace{0.01cm}'(p)$  is now at  $p_{\rm o}\hspace{0.01cm}' = -2$.
*The constant is  $K\hspace{0.01cm}' = 4$   ⇒   only suggested solution 2 is correct here.

'''(6)'''  Finally, the following holds for configuration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Suggested solution 2 is also correct here. In general, it can be said that:
*The partial fraction decomposition changes the number and position of the zeros.
* On the contrary, the poles of $H_{\rm L}\hspace{0.01cm}'(p)$ are always identical to those of $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T10:00:35Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which two-port network has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right $p$–half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.
*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*The double zero is at  $p_{\rm o} = 0$ and the double pole at  $p_{\rm x} = -A = -2$.

'''(3)'''  The following holds for configuration  $(1)$ :
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  Similarly, the following is obtained for configuration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Thus, the suggested solutions 2 and 3 are correct in contrast to statement 1:
* While  $H_{\rm L}(p)$  has two conjugate complex zeros,
*  $H_{\rm L}\hspace{0.01cm}'(p)$  only has a single zero at  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  The following applies for Konfiguration  $(3)$ :
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*The zero of  $H_{\rm L}\hspace{0.01cm}'(p)$  is now at  $p_{\rm o}\hspace{0.01cm}' = -2$.
*The constant is  $K\hspace{0.01cm}' = 4$   ⇒   only suggested solution 2 is correct here.

'''(6)'''  Finally, the following holds for configuration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Suggested solution 2 is also correct here. In general, it can be said that:
*The partial fraction decomposition changes the number and position of the zeros.
* On the contrary, the poles of $H_{\rm L}\hspace{0.01cm}'(p)$ are always identical to those of $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T09:59:33Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which two-port network has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right $p$–half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.
*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*The double zero is at  $p_{\rm o} = 0$ and the double pole at  $p_{\rm x} = -A = -2$.

'''(3)'''  The following holds for configuration  $(1)$ :
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  Similarly, the following is obtained for configuration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Thus, the suggested solutions 2 and 3 are correct in contrast to statement 1:
* While  $H_{\rm L}(p)$  has two conjugate complex zeros,
*  $H_{\rm L}\hspace{0.01cm}'(p)$  only has a single zero at  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  The following applies for Konfiguration  $(3)$ :
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*The zero of  $H_{\rm L}\hspace{0.01cm}'(p)$  is now at  $p_{\rm o}\hspace{0.01cm}' = -2$.
*The constant is  $K\hspace{0.01cm}' = 4$   ⇒   only suggested solution 2 is correct here.

'''(6)'''  Finally, the following holds for configuration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Suggested solution 2 is also correct here. In general, it can be said that:
*The partial fraction decomposition changes the number and position of the zeros.
* On the contrary, the poles of $H_{\rm L}\hspace{0.01cm}'(p)$ are always identical to those of $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T09:57:00Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which quadripole has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right $p$–half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.
*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*The double zero is at  $p_{\rm o} = 0$ and the double pole at  $p_{\rm x} = -A = -2$.

'''(3)'''  The following holds for configuration  $(1)$ :
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  Similarly, the following is obtained for configuration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Thus, the suggested solutions 2 and 3 are correct in contrast to statement 1:
* While  $H_{\rm L}(p)$  has two conjugate complex zeros,
*  $H_{\rm L}\hspace{0.01cm}'(p)$  only has a single zero at  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  The following applies for Konfiguration  $(3)$ :
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*The zero of  $H_{\rm L}\hspace{0.01cm}'(p)$  is now at  $p_{\rm o}\hspace{0.01cm}' = -2$.
*The constant is  $K\hspace{0.01cm}' = 4$   ⇒   only suggested solution 2 is correct here.

'''(6)'''  Finally, the following holds for configuration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Suggested solution 2 is also correct here. In general, it can be said that:
*The partial fraction decomposition changes the number and position of the zeros.
* On the contrary, the poles of $H_{\rm L}\hspace{0.01cm}'(p)$ are always identical to those of $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T07:30:18Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which quadripole has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right $p$–half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.
*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*The double zero is at  $p_{\rm o} = 0$ and the double pole at  $p_{\rm x} = -A = -2$.

'''(3)'''  The following holds for configuration  $(1)$ :
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T07:30:00Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which quadripole has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right $p$–half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.
*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*The double zero is at  $p_{\rm o} = 0$ and the double pole at  $p_{\rm x} = -A = -2$.

'''(3)'''  The following holds for the configuration  $(1)$ :
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T07:27:59Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which quadripole has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right $p$–half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.
*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*Die doppelte Nullstelle liegt bei  $p_{\rm o} = 0$, der doppelte Pol bei  $p_{\rm x} = -A = -2$.

'''(3)'''  Für die Konfiguration  $(1)$  gilt:
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T07:27:17Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which quadripole has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right half-plane$p$– for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.
*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*Die doppelte Nullstelle liegt bei  $p_{\rm o} = 0$, der doppelte Pol bei  $p_{\rm x} = -A = -2$.

'''(3)'''  Für die Konfiguration  $(1)$  gilt:
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T07:26:45Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which quadripole has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left $p$–half-plane.
*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*Die doppelte Nullstelle liegt bei  $p_{\rm o} = 0$, der doppelte Pol bei  $p_{\rm x} = -A = -2$.

'''(3)'''  Für die Konfiguration  $(1)$  gilt:
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T07:25:46Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which quadripole has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left $p$–half-plane.
*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*Die doppelte Nullstelle liegt bei  $p_{\rm o} = 0$, der doppelte Pol bei  $p_{\rm x} = -A = -2$.

'''(3)'''  Für die Konfiguration  $(1)$  gilt:
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T07:22:54Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore, in this exercise the  $p$–transfer function
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Which of the sketched two-port networks are all-pass filters?
|type="[]"}
+ Configuration  $(1)$,
+ configuration  $(2)$,
- configuration  $(3)$,
- configuration  $(4)$.

{Which quadripole has the transfer function  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Configuration  $(1)$,
- configuration  $(2)$,
- configuration  $(3)$,
+ configuration  $(4)$.

{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. Enter the function value for  $p = 0$ .
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(2)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
+ The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(3)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for the configuration  $(4)$.  Which statements are true here?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
- The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1 and 2 are correct:
*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left $p$–half-plane.
*Mit  $K = 1$  ist dann die Dämpfungsfunktion  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*Aus der Grafik auf der Angabenseite erkennt man:   Die Konfigurationen  $(1)$ und  $(2)$ erfüllen genau diese Symmetrieeigenschaften.

'''(2)'''  The suggested solution 4 is correct:
*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*Die doppelte Nullstelle liegt bei  $p_{\rm o} = 0$, der doppelte Pol bei  $p_{\rm x} = -A = -2$.

'''(3)'''  Für die Konfiguration  $(1)$  gilt:
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T06:58:33Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pole-zero diagrams]]
In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$ .
* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
*The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$ .

Rather, a partial fraction decomposition corresponding to
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
must be made beforehand. Then,
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}$$ holds for the impulse response.
 $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called ''all-pass filters''.
*This refers to two-port networks for which the Fourier spectral function satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] ist angegeben, wie die Pole und Nullstelle eines solchen Allpasses liegen müssen.

Weiterhin soll in dieser Aufgabe die  $p$–Übertragungsfunktion
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "Konfiguration $(5)$" näher untersucht werden, die bei richtiger Wahl des Parameters  $A$  durch eines der vier in der Grafik vorgegebenen Pol–Nullstellen–Diagramme dargestellt werden kann.

''Hinweise:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Bei welchen der skizzierten Vierpole handelt es sich um Allpässe?
|type="[]"}
+ Konfiguration  $(1)$,
+ Konfiguration  $(2)$,
- Konfiguration  $(3)$,
- Konfiguration  $(4)$.

{Welcher Vierpol hat die Übertragungsfunktion  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Konfiguration  $(1)$,
- Konfiguration  $(2)$,
- Konfiguration  $(3)$,
+ Konfiguration  $(4)$.

{Berechnen Sie die Funktion  $H_{\rm L}\hspace{0.01cm}'(p)$  nach einer Partialbruchzerlegung für die Konfiguration  '''(1)'''. Geben Sie den Funktionswert für  $p = 0$  ein.
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(2)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
+ Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(3)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
- Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(4)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
- Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Richtig sind die Lösungsvorschläge 1 und 2:
*Nach den in der Aufgabe 3.4Z angegebenen Kriterien liegt immer dann ein Allpass vor, wenn es zu jeder Polstelle  $p_{\rm x} = - A + {\rm j} \cdot B$  in der linken $p$–Halbebene eine entsprechende Nullstelle  $p_{\rm o} = + A + {\rm j} \cdot B$  in der rechten Halbebene gibt.
*Mit  $K = 1$  ist dann die Dämpfungsfunktion  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*Aus der Grafik auf der Angabenseite erkennt man:   Die Konfigurationen  $(1)$ und  $(2)$ erfüllen genau diese Symmetrieeigenschaften.

'''(2)'''  Richtig ist der Lösungsvorschlag 4:
*Die Übertragungsfunktion  $H_{\rm L}^{(5)}(p)$  wird ebenso durch die Konfiguration  $(4)$  beschrieben, wie die nachfolgende Rechnung zeigt:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*Die doppelte Nullstelle liegt bei  $p_{\rm o} = 0$, der doppelte Pol bei  $p_{\rm x} = -A = -2$.

'''(3)'''  Für die Konfiguration  $(1)$  gilt:
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T06:48:24Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pol–Nullstellen–Diagramme]]
In der Grafik sind vier Vierpole durch ihre Pol–Nullstellen–Diagramme  $H_{\rm L}(p)$  gegeben.
* Sie alle haben gemein, dass die Anzahl  $Z$  der Nullstellen gleich der Anzahl  $N$  der Polstellen ist.
*Der konstante Faktor ist jeweils  $K=1$.

Im Sonderfall  $Z = N$  kann zur Berechnung der Impulsantwort  $h(t)$  der Residuensatz nicht direkt angewendet werden.

Vielmehr muss vorher eine Partialbruchzerlegung entsprechend
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
vorgenommen werden. Für die Impulsantwort gilt dann
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}.$$
 $h\hspace{0.03cm}'(t)$  ist die Laplace–Rücktransformierte von  $H_{\rm L}\hspace{0.05cm}'(p)$ , bei der die Bedingung  $Z' < N'$  erfüllt ist.

Bei zwei der vier angegebenen Konfigurationen handelt es sich um so genannte ''Allpässe''.
*Darunter versteht man Vierpole, bei denen die Fourier–Spektralfunktion die Bedingung  $|H(f)| = 1$   ⇒   $a(f) = 0$  erfüllt.
*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters|Exercise 3.4Z]] ist angegeben, wie die Pole und Nullstelle eines solchen Allpasses liegen müssen.

Weiterhin soll in dieser Aufgabe die  $p$–Übertragungsfunktion
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "Konfiguration $(5)$" näher untersucht werden, die bei richtiger Wahl des Parameters  $A$  durch eines der vier in der Grafik vorgegebenen Pol–Nullstellen–Diagramme dargestellt werden kann.

''Hinweise:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Bei welchen der skizzierten Vierpole handelt es sich um Allpässe?
|type="[]"}
+ Konfiguration  $(1)$,
+ Konfiguration  $(2)$,
- Konfiguration  $(3)$,
- Konfiguration  $(4)$.

{Welcher Vierpol hat die Übertragungsfunktion  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Konfiguration  $(1)$,
- Konfiguration  $(2)$,
- Konfiguration  $(3)$,
+ Konfiguration  $(4)$.

{Berechnen Sie die Funktion  $H_{\rm L}\hspace{0.01cm}'(p)$  nach einer Partialbruchzerlegung für die Konfiguration  '''(1)'''. Geben Sie den Funktionswert für  $p = 0$  ein.
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(2)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
+ Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(3)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
- Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(4)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
- Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Richtig sind die Lösungsvorschläge 1 und 2:
*Nach den in der Aufgabe 3.4Z angegebenen Kriterien liegt immer dann ein Allpass vor, wenn es zu jeder Polstelle  $p_{\rm x} = - A + {\rm j} \cdot B$  in der linken $p$–Halbebene eine entsprechende Nullstelle  $p_{\rm o} = + A + {\rm j} \cdot B$  in der rechten Halbebene gibt.
*Mit  $K = 1$  ist dann die Dämpfungsfunktion  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*Aus der Grafik auf der Angabenseite erkennt man:   Die Konfigurationen  $(1)$ und  $(2)$ erfüllen genau diese Symmetrieeigenschaften.

'''(2)'''  Richtig ist der Lösungsvorschlag 4:
*Die Übertragungsfunktion  $H_{\rm L}^{(5)}(p)$  wird ebenso durch die Konfiguration  $(4)$  beschrieben, wie die nachfolgende Rechnung zeigt:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*Die doppelte Nullstelle liegt bei  $p_{\rm o} = 0$, der doppelte Pol bei  $p_{\rm x} = -A = -2$.

'''(3)'''  Für die Konfiguration  $(1)$  gilt:
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T06:46:35Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pol–Nullstellen–Diagramme]]
In der Grafik sind vier Vierpole durch ihre Pol–Nullstellen–Diagramme  $H_{\rm L}(p)$  gegeben.
* Sie alle haben gemein, dass die Anzahl  $Z$  der Nullstellen gleich der Anzahl  $N$  der Polstellen ist.
*Der konstante Faktor ist jeweils  $K=1$.

Im Sonderfall  $Z = N$  kann zur Berechnung der Impulsantwort  $h(t)$  der Residuensatz nicht direkt angewendet werden.

Vielmehr muss vorher eine Partialbruchzerlegung entsprechend
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
vorgenommen werden. Für die Impulsantwort gilt dann
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}.$$
 $h\hspace{0.03cm}'(t)$  ist die Laplace–Rücktransformierte von  $H_{\rm L}\hspace{0.05cm}'(p)$ , bei der die Bedingung  $Z' < N'$  erfüllt ist.

Bei zwei der vier angegebenen Konfigurationen handelt es sich um so genannte ''Allpässe''.
*Darunter versteht man Vierpole, bei denen die Fourier–Spektralfunktion die Bedingung  $|H(f)| = 1$   ⇒   $a(f) = 0$  erfüllt.
*In [[Aufgaben:3.4Z_Verschiedene_Allpässe|Aufgabe 3.4Z]] ist angegeben, wie die Pole und Nullstelle eines solchen Allpasses liegen müssen.

Weiterhin soll in dieser Aufgabe die  $p$–Übertragungsfunktion
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "Konfiguration $(5)$" näher untersucht werden, die bei richtiger Wahl des Parameters  $A$  durch eines der vier in der Grafik vorgegebenen Pol–Nullstellen–Diagramme dargestellt werden kann.

''Hinweise:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition|Partial fraction decomposition]].

===Questions===

<quiz display=simple>
{Bei welchen der skizzierten Vierpole handelt es sich um Allpässe?
|type="[]"}
+ Konfiguration  $(1)$,
+ Konfiguration  $(2)$,
- Konfiguration  $(3)$,
- Konfiguration  $(4)$.

{Welcher Vierpol hat die Übertragungsfunktion  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Konfiguration  $(1)$,
- Konfiguration  $(2)$,
- Konfiguration  $(3)$,
+ Konfiguration  $(4)$.

{Berechnen Sie die Funktion  $H_{\rm L}\hspace{0.01cm}'(p)$  nach einer Partialbruchzerlegung für die Konfiguration  '''(1)'''. Geben Sie den Funktionswert für  $p = 0$  ein.
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(2)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
+ Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(3)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
- Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(4)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
- Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Richtig sind die Lösungsvorschläge 1 und 2:
*Nach den in der Aufgabe 3.4Z angegebenen Kriterien liegt immer dann ein Allpass vor, wenn es zu jeder Polstelle  $p_{\rm x} = - A + {\rm j} \cdot B$  in der linken $p$–Halbebene eine entsprechende Nullstelle  $p_{\rm o} = + A + {\rm j} \cdot B$  in der rechten Halbebene gibt.
*Mit  $K = 1$  ist dann die Dämpfungsfunktion  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*Aus der Grafik auf der Angabenseite erkennt man:   Die Konfigurationen  $(1)$ und  $(2)$ erfüllen genau diese Symmetrieeigenschaften.

'''(2)'''  Richtig ist der Lösungsvorschlag 4:
*Die Übertragungsfunktion  $H_{\rm L}^{(5)}(p)$  wird ebenso durch die Konfiguration  $(4)$  beschrieben, wie die nachfolgende Rechnung zeigt:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*Die doppelte Nullstelle liegt bei  $p_{\rm o} = 0$, der doppelte Pol bei  $p_{\rm x} = -A = -2$.

'''(3)'''  Für die Konfiguration  $(1)$  gilt:
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

2021-10-26T06:43:53Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1789__LZI_Z_3_7.png|right|frame|Pol–Nullstellen–Diagramme]]
In der Grafik sind vier Vierpole durch ihre Pol–Nullstellen–Diagramme  $H_{\rm L}(p)$  gegeben.
* Sie alle haben gemein, dass die Anzahl  $Z$  der Nullstellen gleich der Anzahl  $N$  der Polstellen ist.
*Der konstante Faktor ist jeweils  $K=1$.

Im Sonderfall  $Z = N$  kann zur Berechnung der Impulsantwort  $h(t)$  der Residuensatz nicht direkt angewendet werden.

Vielmehr muss vorher eine Partialbruchzerlegung entsprechend
:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
\hspace{0.05cm}$$
vorgenommen werden. Für die Impulsantwort gilt dann
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}.$$
 $h\hspace{0.03cm}'(t)$  ist die Laplace–Rücktransformierte von  $H_{\rm L}\hspace{0.05cm}'(p)$ , bei der die Bedingung  $Z' < N'$  erfüllt ist.

Bei zwei der vier angegebenen Konfigurationen handelt es sich um so genannte ''Allpässe''.
*Darunter versteht man Vierpole, bei denen die Fourier–Spektralfunktion die Bedingung  $|H(f)| = 1$   ⇒   $a(f) = 0$  erfüllt.
*In [[Aufgaben:3.4Z_Verschiedene_Allpässe|Aufgabe 3.4Z]] ist angegeben, wie die Pole und Nullstelle eines solchen Allpasses liegen müssen.

Weiterhin soll in dieser Aufgabe die  $p$–Übertragungsfunktion
:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
\hspace{0.05cm}$$
⇒   "Konfiguration $(5)$" näher untersucht werden, die bei richtiger Wahl des Parameters  $A$  durch eines der vier in der Grafik vorgegebenen Pol–Nullstellen–Diagramme dargestellt werden kann.

''Hinweise:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*Bezug genommen wird insbesondere auf die Seite  [[Linear_and_Time_Invariant_Systems/Laplace–Rücktransformation#Partialbruchzerlegung|Partialbruchzerlegung]].

===Questions===

<quiz display=simple>
{Bei welchen der skizzierten Vierpole handelt es sich um Allpässe?
|type="[]"}
+ Konfiguration  $(1)$,
+ Konfiguration  $(2)$,
- Konfiguration  $(3)$,
- Konfiguration  $(4)$.

{Welcher Vierpol hat die Übertragungsfunktion  $H_{\rm L}^{(5)}(p)$?
|type="()"}
- Konfiguration  $(1)$,
- Konfiguration  $(2)$,
- Konfiguration  $(3)$,
+ Konfiguration  $(4)$.

{Berechnen Sie die Funktion  $H_{\rm L}\hspace{0.01cm}'(p)$  nach einer Partialbruchzerlegung für die Konfiguration  '''(1)'''. Geben Sie den Funktionswert für  $p = 0$  ein.
|type="{}"}
$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(2)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
+ Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(3)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
- Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

{Berechnen Sie  $H_{\rm L}\hspace{0.01cm}'(p)$  für die Konfiguration  $(4)$.  Welche Aussagen treffen hier zu?
|type="[]"}
- $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Nullstellen wie  $H_{\rm L}(p)$.
+ $H_{\rm L}\hspace{0.01cm}'(p)$  besitzt die gleichen Polstellen wie  $H_{\rm L}(p)$.
- Der konstante Faktor von  $H_{\rm L}\hspace{0.01cm}'(p)$  ist  $K' = 8$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Richtig sind die Lösungsvorschläge 1 und 2:
*Nach den in der Aufgabe 3.4Z angegebenen Kriterien liegt immer dann ein Allpass vor, wenn es zu jeder Polstelle  $p_{\rm x} = - A + {\rm j} \cdot B$  in der linken $p$–Halbebene eine entsprechende Nullstelle  $p_{\rm o} = + A + {\rm j} \cdot B$  in der rechten Halbebene gibt.
*Mit  $K = 1$  ist dann die Dämpfungsfunktion  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
*Aus der Grafik auf der Angabenseite erkennt man:   Die Konfigurationen  $(1)$ und  $(2)$ erfüllen genau diese Symmetrieeigenschaften.

'''(2)'''  Richtig ist der Lösungsvorschlag 4:
*Die Übertragungsfunktion  $H_{\rm L}^{(5)}(p)$  wird ebenso durch die Konfiguration  $(4)$  beschrieben, wie die nachfolgende Rechnung zeigt:
:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
=\frac{p/A}{{p/A}+2+ {A/p}}
= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$
*Die doppelte Nullstelle liegt bei  $p_{\rm o} = 0$, der doppelte Pol bei  $p_{\rm x} = -A = -2$.

'''(3)'''  Für die Konfiguration  $(1)$  gilt:
:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

'''(4)'''  In gleicher Weise ergibt sich für die Konfiguration  $(2)$:
:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Richtig sind also die Lösungsvorschläge 2 und 3 im Gegensatz zur Aussage 1:
* Während  $H_{\rm L}(p)$  zwei konjugiert–komplexe Nullstellen aufweist,
*besitzt  $H_{\rm L}\hspace{0.01cm}'(p)$  nur eine einzige Nullstelle bei  $p_{\rm o}\hspace{0.01cm}' = 0$.

'''(5)'''  Für die Konfiguration  $(3)$  gilt:
:$$H_{\rm L}(p) =
\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$
*Die Nullstelle von  $H_{\rm L}\hspace{0.01cm}'(p)$  liegt nun bei  $p_{\rm o}\hspace{0.01cm}' = -2$.
*Die Konstante ist  $K\hspace{0.01cm}' = 4$   ⇒   richtig ist hier nur der Lösungsvorschlag 2.

'''(6)'''  Schließlich gilt für die Konfiguration  $(4)$:
:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$
Richtig ist auch hier der Lösungsvorschlag 2. Allgemein lässt sich sagen:
*Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert.
*Die Pole von $H_{\rm L}\hspace{0.01cm}'(p)$ sind dagegen stets identisch mit denen von $H_{\rm L}(p)$.
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7: Impulse Response of a High-Pass Filter

2021-10-26T06:38:12Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:EN_LZI_A_3_7.png|right|frame|High-pass filter of second-order]]
We assume the sketched arrangement.  The transfer functions of the two identical high-pass filters are:
:$$H_{\rm L}^{(1)}(p) = H_{\rm L}^{(2)}(p) =\frac{p}{p+A}
\hspace{0.05cm} .$$
Since the two-port networks are decoupled in terms of resistance by an isolation amplifier, the total transfer function can be written as follows:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p)
\hspace{0.05cm} .$$
At the same time it is known that the following equation is valid:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}
\hspace{0.05cm} .$$
If this function is represented in pole–zero notation, it will turn out that here the number of zeros  $(Z)$  is equal to the number of poles  $(N)$ .  Therefore, a direct application of the residue theorem is not possible here.

Instead, in order to compute the time function  $h(t)$  a partial fraction decomposition corresponding to 
$H_{\rm L}(p) =1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.05cm}$ 
must be made. Thus, the following holds for the impulse response:
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}.$$
$Z' < N'$ holds with respect to  $H_{\rm L}'(p)$ . Thus, the continuous component  $h'(t)$  of the impulse response can be determined using the residue theorem.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The residual of an  $l$–fold pole  $p_{\rm x}$  within the function  $H_{\rm L}(p)$  is:
:$${\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.03cm}\{H_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p t}\}=
\frac{1}{(l-1)!}\cdot \frac{{\rm d}^{\hspace{0.05cm}l-1}}{{\rm d}p^{\hspace{0.05cm}l-1}}\hspace{0.15cm}
\left \{H_{\rm L}(p)\cdot (p - p_{\rm x})^{\hspace{0.05cm}l} \cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.05cm} .$$
*The derivative of the product  $y(x) = f(x) \cdot g(x)$  is given as follows:
:$$\frac{{\rm d}{\hspace{0.05cm}y(x)}}{{\rm d}\hspace{0.05cm}x}= \frac{{\rm d}{\hspace{0.05cm}f(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot g(x) + \frac{{\rm d}{\hspace{0.05cm}g(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot f(x)
\hspace{0.05cm} .$$

===Questions===

<quiz display=simple>
{Represent  $H_{\rm L}(p)$  in pole–zero notation.  How many zeros  $(Z)$  and poles  $(N)$  are there?  What is the constant factor  $K$?
|type="{}"}
$Z \hspace{0.28cm} = \ $ { 2 }
$N \hspace{0.2cm} = \ $ { 2 }
$K \hspace{0.2cm} = \ $ { 1 }

{What is the parameter $A$  of the two partial two-port networks?
|type="{}"}
$A \ = \ $ { 0.5 3% }

{Convert  $H_{\rm L}(p) = 1 - H_{\rm L}'(p)$ .  What result is obtained for  $H_{\rm L}'(p)$?
|type="[]"}
- $H_{\rm L}'(p) = p^2/(p+0.5)^2$,
- $H_{\rm L}'(p) = p/(p+0.5)^2$,
+ $H_{\rm L}'(p) = (p+0.25)/(p+0.5)^2$.

{Compute the time function  $h'(t)$.  What are the numerical values for the given times?
|type="{}"}
$h'(t = 0) \ = \ $ { 1 3% }
$h'(t = 1) \ = \ $ { 0.455 3% }
$h'(t → ∞)\ = \ $ { 0. }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Starting from the given equation,  $H_{\rm L}(p)$  can be transformed as follows:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}=\frac{p^2}{p^2 + p +1/4}=\frac{p^2}{(p +1/2)^2}
\hspace{0.3cm}
\Rightarrow \hspace{0.3cm} \hspace{0.15cm}\underline{ Z = 2\hspace{0.05cm} , \hspace{0.2cm}N = 2\hspace{0.05cm} ,
\hspace{0.2cm}K = 1}
\hspace{0.05cm} .$$

'''(2)'''  The total transfer function is as follows according to the information page:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p) =\frac{p^2}{(p+A)^2}
\hspace{0.05cm} .$$
A comparison with the result of subtask  '''(1)'''  shows that  $\underline{A = 0.5}$  must hold.

'''(3)'''  The last suggested solution is correct:
*The following is obtained based on the equation computed in subtask  '''(1)''' :
:$$H_{\rm L}(p) =\frac{p^2}{p^2 + p +0.25}= \frac{p^2 + p +0.25}{p^2 + p
+0.25}- \frac{p +0.25}{p^2 + p
+0.25}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{p +0.25}{p^2 + p
+0.25}= \frac{p +0.25}{(p
+0.5)^2}
\hspace{0.05cm} .$$

'''(4)'''  Concerning the function  $H_{\rm L}'(p)$,   $Z' = 1$,  $N' = 2$  and  $K' = 1$ hold.

The two poles at  $p_{\rm x} = -0.5$  coincide such that only one residual needs to be determined:
:$$h\hspace{0.03cm}'(t) \hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.7cm}\{H_{\rm L}\hspace{-0.05cm}'(p)\cdot {\rm e}^{p t}\}=
\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ \frac{p +0.25}{(p
+0.5)^2} \cdot (p
+0.5)^2 \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.2cm}\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ (p
+0.25) \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
\hspace{0.05cm} .$$
[[File:P_ID1788__LZI_A_3_7_d.png|right|frame|Impulse response of the high-pass filter (red); continuous component $h\hspace{0.03cm}'(t)$ (blue)]]
The following is obtained using the product rule of differential calculus:
$$h\hspace{0.03cm}'(t) \hspace{0.15cm} = \hspace{0.15cm}
{\rm e}^{p \hspace{0.05cm}t} + ( p + 0.25) \cdot t \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.15cm} (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm} $$
$$\Rightarrow \hspace{0.3cm}h\hspace{0.03cm}'(t = 0) \hspace{0.15cm} = \underline{1}\hspace{0.05cm} ,\hspace{0.3cm} h\hspace{0.03cm}'(t = 1) \hspace{0.15cm} = \underline {0.455}\hspace{0.05cm} \hspace{0.05cm} ,\hspace{0.3cm}
h\hspace{0.03cm}'(t \rightarrow \infty) \hspace{0.15cm} = \underline {= 0}\hspace{0.05cm} .$$
The graph shows in each case for non–negative times
*the impulse response  $h'(t)$  of the equivalent low-pass filter as a blue curve,
*the total impulse response of the considered high-pass filter as a red curve:
:$$h(t) =
\delta (t) - (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7: Impulse Response of a High-Pass Filter

2021-10-26T06:32:03Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:EN_LZI_A_3_7.png|right|frame|High-pass filter of second-order]]
We assume the sketched arrangement.  The transfer functions of the two identical high-pass filters are:
:$$H_{\rm L}^{(1)}(p) = H_{\rm L}^{(2)}(p) =\frac{p}{p+A}
\hspace{0.05cm} .$$
Since the two-port networks are decoupled in terms of resistance by an isolation amplifier, the total transfer function can be written as follows:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p)
\hspace{0.05cm} .$$
Gleichzeitig ist bekannt, dass folgende Gleichung gültig ist:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}
\hspace{0.05cm} .$$
If this function is represented in pole–zero notation, it will turn out that here the number of zeros  $(Z)$  is equal to the number of poles  $(N)$ .  Therefore, a direct application of the residue theorem is not possible here.

Instead, in order to compute the time function  $h(t)$  a partial fraction decomposition corresponding to 
$H_{\rm L}(p) =1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.05cm}$ 
must be made. Thus, the following holds for the impulse response:
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}.$$
$Z' < N'$ holds with respect to  $H_{\rm L}'(p)$ . Thus, the continuous component  $h'(t)$  of the impulse response can be determined using the residue theorem.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The residual of an  $l$–fold pole  $p_{\rm x}$  within the function  $H_{\rm L}(p)$  is:
:$${\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.03cm}\{H_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p t}\}=
\frac{1}{(l-1)!}\cdot \frac{{\rm d}^{\hspace{0.05cm}l-1}}{{\rm d}p^{\hspace{0.05cm}l-1}}\hspace{0.15cm}
\left \{H_{\rm L}(p)\cdot (p - p_{\rm x})^{\hspace{0.05cm}l} \cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.05cm} .$$
*The derivative of the product  $y(x) = f(x) \cdot g(x)$  is given as follows:
:$$\frac{{\rm d}{\hspace{0.05cm}y(x)}}{{\rm d}\hspace{0.05cm}x}= \frac{{\rm d}{\hspace{0.05cm}f(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot g(x) + \frac{{\rm d}{\hspace{0.05cm}g(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot f(x)
\hspace{0.05cm} .$$

===Questions===

<quiz display=simple>
{Represent  $H_{\rm L}(p)$  in pole–zero notation.  How many zeros  $(Z)$  and poles  $(N)$  are there?  What is the constant factor  $K$?
|type="{}"}
$Z \hspace{0.28cm} = \ $ { 2 }
$N \hspace{0.2cm} = \ $ { 2 }
$K \hspace{0.2cm} = \ $ { 1 }

{What is the parameter $A$  of the two partial two-port networks?
|type="{}"}
$A \ = \ $ { 0.5 3% }

{Convert  $H_{\rm L}(p) = 1 - H_{\rm L}'(p)$ .  What result is obtained for  $H_{\rm L}'(p)$?
|type="[]"}
- $H_{\rm L}'(p) = p^2/(p+0.5)^2$,
- $H_{\rm L}'(p) = p/(p+0.5)^2$,
+ $H_{\rm L}'(p) = (p+0.25)/(p+0.5)^2$.

{Compute the time function  $h'(t)$.  What are the numerical values for the given times?
|type="{}"}
$h'(t = 0) \ = \ $ { 1 3% }
$h'(t = 1) \ = \ $ { 0.455 3% }
$h'(t → ∞)\ = \ $ { 0. }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Starting from the given equation,  $H_{\rm L}(p)$  can be transformed as follows:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}=\frac{p^2}{p^2 + p +1/4}=\frac{p^2}{(p +1/2)^2}
\hspace{0.3cm}
\Rightarrow \hspace{0.3cm} \hspace{0.15cm}\underline{ Z = 2\hspace{0.05cm} , \hspace{0.2cm}N = 2\hspace{0.05cm} ,
\hspace{0.2cm}K = 1}
\hspace{0.05cm} .$$

'''(2)'''  The total transfer function is as follows according to the information page:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p) =\frac{p^2}{(p+A)^2}
\hspace{0.05cm} .$$
A comparison with the result of subtask  '''(1)'''  shows that  $\underline{A = 0.5}$  must hold.

'''(3)'''  The last suggested solution is correct:
*The following is obtained based on the equation computed in subtask  '''(1)''' :
:$$H_{\rm L}(p) =\frac{p^2}{p^2 + p +0.25}= \frac{p^2 + p +0.25}{p^2 + p
+0.25}- \frac{p +0.25}{p^2 + p
+0.25}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{p +0.25}{p^2 + p
+0.25}= \frac{p +0.25}{(p
+0.5)^2}
\hspace{0.05cm} .$$

'''(4)'''  Concerning the function  $H_{\rm L}'(p)$,   $Z' = 1$,  $N' = 2$  and  $K' = 1$ hold.

The two poles at  $p_{\rm x} = -0.5$  coincide such that only one residual needs to be determined:
:$$h\hspace{0.03cm}'(t) \hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.7cm}\{H_{\rm L}\hspace{-0.05cm}'(p)\cdot {\rm e}^{p t}\}=
\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ \frac{p +0.25}{(p
+0.5)^2} \cdot (p
+0.5)^2 \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.2cm}\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ (p
+0.25) \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
\hspace{0.05cm} .$$
[[File:P_ID1788__LZI_A_3_7_d.png|right|frame|Impulse response of the high-pass filter (red); continuous component $h\hspace{0.03cm}'(t)$ (blue)]]
The following is obtained using the product rule of differential calculus:
$$h\hspace{0.03cm}'(t) \hspace{0.15cm} = \hspace{0.15cm}
{\rm e}^{p \hspace{0.05cm}t} + ( p + 0.25) \cdot t \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.15cm} (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm} $$
$$\Rightarrow \hspace{0.3cm}h\hspace{0.03cm}'(t = 0) \hspace{0.15cm} = \underline{1}\hspace{0.05cm} ,\hspace{0.3cm} h\hspace{0.03cm}'(t = 1) \hspace{0.15cm} = \underline {0.455}\hspace{0.05cm} \hspace{0.05cm} ,\hspace{0.3cm}
h\hspace{0.03cm}'(t \rightarrow \infty) \hspace{0.15cm} = \underline {= 0}\hspace{0.05cm} .$$
The graph shows in each case for non–negative times
*the impulse response  $h'(t)$  of the equivalent low-pass filter as a blue curve,
*the total impulse response of the considered high-pass filter as a red curve:
:$$h(t) =
\delta (t) - (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7: Impulse Response of a High-Pass Filter

2021-10-26T06:26:51Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:EN_LZI_A_3_7.png|right|frame|High-pass filter of second-order]]
We assume the sketched arrangement.  The transfer functions of the two identical high-pass filters are:
:$$H_{\rm L}^{(1)}(p) = H_{\rm L}^{(2)}(p) =\frac{p}{p+A}
\hspace{0.05cm} .$$
Since the two-port networks are decoupled in terms of resistance by an isolation amplifier, the total transfer function can be written as follows:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p)
\hspace{0.05cm} .$$
Gleichzeitig ist bekannt, dass folgende Gleichung gültig ist:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}
\hspace{0.05cm} .$$
If this function is represented in pole–zero notation, it will turn out that here the number of zeros  $(Z)$  is equal to the number of poles  $(N)$ .  Therefore, a direct application of the residue theorem is not possible here.

Instead, in order to compute the time function  $h(t)$  a partial fraction decomposition corresponding to 
$H_{\rm L}(p) =1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.05cm}$ 
must be made. Thus, the following holds for the impulse response:
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}.$$
$Z' < N'$ holds with respect to  $H_{\rm L}'(p)$ . Thus, the continuous component  $h'(t)$  of the impulse response can be determined using the residue theorem.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The residual of an  $l$–fold pole  $p_{\rm x}$  within the function  $H_{\rm L}(p)$  is:
:$${\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.03cm}\{H_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p t}\}=
\frac{1}{(l-1)!}\cdot \frac{{\rm d}^{\hspace{0.05cm}l-1}}{{\rm d}p^{\hspace{0.05cm}l-1}}\hspace{0.15cm}
\left \{H_{\rm L}(p)\cdot (p - p_{\rm x})^{\hspace{0.05cm}l} \cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.05cm} .$$
*The derivative of the product  $y(x) = f(x) \cdot g(x)$  is given as follows:
:$$\frac{{\rm d}{\hspace{0.05cm}y(x)}}{{\rm d}\hspace{0.05cm}x}= \frac{{\rm d}{\hspace{0.05cm}f(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot g(x) + \frac{{\rm d}{\hspace{0.05cm}g(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot f(x)
\hspace{0.05cm} .$$

===Questions===

<quiz display=simple>
{Represent  $H_{\rm L}(p)$  in pole–zero notation.  How many zeros  $(Z)$  and poles  $(N)$  are there?  What is the constant factor  $K$?
|type="{}"}
$Z \hspace{0.28cm} = \ $ { 2 }
$N \hspace{0.2cm} = \ $ { 2 }
$K \hspace{0.2cm} = \ $ { 1 }

{What is the parameter $A$  of the two partial two-port networks?
|type="{}"}
$A \ = \ $ { 0.5 3% }

{Convert  $H_{\rm L}(p) = 1 - H_{\rm L}'(p)$ .  What result is obtained for  $H_{\rm L}'(p)$?
|type="[]"}
- $H_{\rm L}'(p) = p^2/(p+0.5)^2$,
- $H_{\rm L}'(p) = p/(p+0.5)^2$,
+ $H_{\rm L}'(p) = (p+0.25)/(p+0.5)^2$.

{Compute the time function  $h'(t)$.  What are the numerical values for the given times?
|type="{}"}
$h'(t = 0) \ = \ $ { 1 3% }
$h'(t = 1) \ = \ $ { 0.455 3% }
$h'(t → ∞)\ = \ $ { 0. }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Starting from the given equation,  $H_{\rm L}(p)$  can be transformed as follows:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}=\frac{p^2}{p^2 + p +1/4}=\frac{p^2}{(p +1/2)^2}
\hspace{0.3cm}
\Rightarrow \hspace{0.3cm} \hspace{0.15cm}\underline{ Z = 2\hspace{0.05cm} , \hspace{0.2cm}N = 2\hspace{0.05cm} ,
\hspace{0.2cm}K = 1}
\hspace{0.05cm} .$$

'''(2)'''  The total transfer function is as follows according to the information page:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p) =\frac{p^2}{(p+A)^2}
\hspace{0.05cm} .$$
A comparison with the result of subtask  '''(1)'''  shows that  $\underline{A = 0.5}$  must hold.

'''(3)'''  The last suggested solution is correct:
*The following is obtained based on the equation computed in subtask  '''(1)''' :
:$$H_{\rm L}(p) =\frac{p^2}{p^2 + p +0.25}= \frac{p^2 + p +0.25}{p^2 + p
+0.25}- \frac{p +0.25}{p^2 + p
+0.25}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{p +0.25}{p^2 + p
+0.25}= \frac{p +0.25}{(p
+0.5)^2}
\hspace{0.05cm} .$$

'''(4)'''  Bezüglich der Funktion  $H_{\rm L}'(p)$  gilt  $Z' = 1$,  $N' = 2$  und  $K' = 1$.

Die beiden Pole bei  $p_{\rm x} = -0.5$  fallen zusammen, so dass nur ein Residium ermittelt werden muss:
:$$h\hspace{0.03cm}'(t) \hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.7cm}\{H_{\rm L}\hspace{-0.05cm}'(p)\cdot {\rm e}^{p t}\}=
\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ \frac{p +0.25}{(p
+0.5)^2} \cdot (p
+0.5)^2 \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.2cm}\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ (p
+0.25) \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
\hspace{0.05cm} .$$
[[File:P_ID1788__LZI_A_3_7_d.png|right|frame|Impulsantwort des Hochpasses (rot); kontinuierlicher Anteil $h\hspace{0.03cm}'(t)$ (blau)]]
Mit der Produktregel der Differentialrechnung erhält man:
$$h\hspace{0.03cm}'(t) \hspace{0.15cm} = \hspace{0.15cm}
{\rm e}^{p \hspace{0.05cm}t} + ( p + 0.25) \cdot t \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.15cm} (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm} $$
$$\Rightarrow \hspace{0.3cm}h\hspace{0.03cm}'(t = 0) \hspace{0.15cm} = \underline{1}\hspace{0.05cm} ,\hspace{0.3cm} h\hspace{0.03cm}'(t = 1) \hspace{0.15cm} = \underline {0.455}\hspace{0.05cm} \hspace{0.05cm} ,\hspace{0.3cm}
h\hspace{0.03cm}'(t \rightarrow \infty) \hspace{0.15cm} = \underline {= 0}\hspace{0.05cm} .$$
Die Grafik zeigt jeweils für nicht–negative Zeiten
*als blaue Kurve die Impulsantwort  $h'(t)$  des äquivalenten Tiefpasses,
*als rote Kurve die gesamte Impulsantwort des betrachteten Hochpasses:
:$$h(t) =
\delta (t) - (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7: Impulse Response of a High-Pass Filter

2021-10-26T06:22:21Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:EN_LZI_A_3_7.png|right|frame|High-pass filter of second-order]]
We assume the sketched arrangement.  The transfer functions of the two identical high-pass filters are:
:$$H_{\rm L}^{(1)}(p) = H_{\rm L}^{(2)}(p) =\frac{p}{p+A}
\hspace{0.05cm} .$$
Since the two-port networks are decoupled in terms of resistance by an isolation amplifier, the total transfer function can be written as follows:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p)
\hspace{0.05cm} .$$
Gleichzeitig ist bekannt, dass folgende Gleichung gültig ist:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}
\hspace{0.05cm} .$$
If this function is represented in pole–zero notation, it will turn out that here the number of zeros  $(Z)$  is equal to the number of poles  $(N)$ .  Therefore, a direct application of the residue theorem is not possible here.

Instead, in order to compute the time function  $h(t)$  a partial fraction decomposition corresponding to 
$H_{\rm L}(p) =1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.05cm}$ 
must be made. Thus, the following holds for the impulse response:
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}.$$
$Z' < N'$ holds with respect to  $H_{\rm L}'(p)$ . Thus, the continuous component  $h'(t)$  of the impulse response can be determined using the residue theorem.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The residual of an  $l$–fold pole  $p_{\rm x}$  within the function  $H_{\rm L}(p)$  is:
:$${\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.03cm}\{H_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p t}\}=
\frac{1}{(l-1)!}\cdot \frac{{\rm d}^{\hspace{0.05cm}l-1}}{{\rm d}p^{\hspace{0.05cm}l-1}}\hspace{0.15cm}
\left \{H_{\rm L}(p)\cdot (p - p_{\rm x})^{\hspace{0.05cm}l} \cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.05cm} .$$
*The derivative of the product  $y(x) = f(x) \cdot g(x)$  is given as follows:
:$$\frac{{\rm d}{\hspace{0.05cm}y(x)}}{{\rm d}\hspace{0.05cm}x}= \frac{{\rm d}{\hspace{0.05cm}f(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot g(x) + \frac{{\rm d}{\hspace{0.05cm}g(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot f(x)
\hspace{0.05cm} .$$

===Questions===

<quiz display=simple>
{Represent  $H_{\rm L}(p)$  in pole–zero notation.  How many zeros  $(Z)$  and poles  $(N)$  are there?  What is the constant factor  $K$?
|type="{}"}
$Z \hspace{0.28cm} = \ $ { 2 }
$N \hspace{0.2cm} = \ $ { 2 }
$K \hspace{0.2cm} = \ $ { 1 }

{What is the parameter $A$  of the two partial two-port networks?
|type="{}"}
$A \ = \ $ { 0.5 3% }

{Convert  $H_{\rm L}(p) = 1 - H_{\rm L}'(p)$ .  What result is obtained for  $H_{\rm L}'(p)$?
|type="[]"}
- $H_{\rm L}'(p) = p^2/(p+0.5)^2$,
- $H_{\rm L}'(p) = p/(p+0.5)^2$,
+ $H_{\rm L}'(p) = (p+0.25)/(p+0.5)^2$.

{Compute the time function  $h'(t)$.  What are the numerical values for the given times?
|type="{}"}
$h'(t = 0) \ = \ $ { 1 3% }
$h'(t = 1) \ = \ $ { 0.455 3% }
$h'(t → ∞)\ = \ $ { 0. }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Starting from the given equation,  $H_{\rm L}(p)$  can be transformed as follows:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}=\frac{p^2}{p^2 + p +1/4}=\frac{p^2}{(p +1/2)^2}
\hspace{0.3cm}
\Rightarrow \hspace{0.3cm} \hspace{0.15cm}\underline{ Z = 2\hspace{0.05cm} , \hspace{0.2cm}N = 2\hspace{0.05cm} ,
\hspace{0.2cm}K = 1}
\hspace{0.05cm} .$$

'''(2)'''  Die Gesamtübertragungsfunktion lautet entsprechend der Angabe:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p) =\frac{p^2}{(p+A)^2}
\hspace{0.05cm} .$$
Ein Vergleich mit dem Ergebnis der Teilaufgabe  '''(1)'''  zeigt, dass  $\underline{A = 0.5}$  sein muss.

'''(3)'''  Richtig ist der letzte Lösungsvorschlag:
*Ausgehend von der in der Teilaufgabe  '''(1)'''  berechneten Gleichung erhält man
:$$H_{\rm L}(p) =\frac{p^2}{p^2 + p +0.25}= \frac{p^2 + p +0.25}{p^2 + p
+0.25}- \frac{p +0.25}{p^2 + p
+0.25}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{p +0.25}{p^2 + p
+0.25}= \frac{p +0.25}{(p
+0.5)^2}
\hspace{0.05cm} .$$

'''(4)'''  Bezüglich der Funktion  $H_{\rm L}'(p)$  gilt  $Z' = 1$,  $N' = 2$  und  $K' = 1$.

Die beiden Pole bei  $p_{\rm x} = -0.5$  fallen zusammen, so dass nur ein Residium ermittelt werden muss:
:$$h\hspace{0.03cm}'(t) \hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.7cm}\{H_{\rm L}\hspace{-0.05cm}'(p)\cdot {\rm e}^{p t}\}=
\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ \frac{p +0.25}{(p
+0.5)^2} \cdot (p
+0.5)^2 \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.2cm}\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ (p
+0.25) \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
\hspace{0.05cm} .$$
[[File:P_ID1788__LZI_A_3_7_d.png|right|frame|Impulsantwort des Hochpasses (rot); kontinuierlicher Anteil $h\hspace{0.03cm}'(t)$ (blau)]]
Mit der Produktregel der Differentialrechnung erhält man:
$$h\hspace{0.03cm}'(t) \hspace{0.15cm} = \hspace{0.15cm}
{\rm e}^{p \hspace{0.05cm}t} + ( p + 0.25) \cdot t \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.15cm} (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm} $$
$$\Rightarrow \hspace{0.3cm}h\hspace{0.03cm}'(t = 0) \hspace{0.15cm} = \underline{1}\hspace{0.05cm} ,\hspace{0.3cm} h\hspace{0.03cm}'(t = 1) \hspace{0.15cm} = \underline {0.455}\hspace{0.05cm} \hspace{0.05cm} ,\hspace{0.3cm}
h\hspace{0.03cm}'(t \rightarrow \infty) \hspace{0.15cm} = \underline {= 0}\hspace{0.05cm} .$$
Die Grafik zeigt jeweils für nicht–negative Zeiten
*als blaue Kurve die Impulsantwort  $h'(t)$  des äquivalenten Tiefpasses,
*als rote Kurve die gesamte Impulsantwort des betrachteten Hochpasses:
:$$h(t) =
\delta (t) - (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7: Impulse Response of a High-Pass Filter

2021-10-25T08:19:34Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:EN_LZI_A_3_7.png|right|frame|High-pass filter of second-order]]
Wir gehen von der skizzierten Anordnung aus.  Die Übertragungsfunktionen der beiden identischen Hochpässe lauten:
:$$H_{\rm L}^{(1)}(p) = H_{\rm L}^{(2)}(p) =\frac{p}{p+A}
\hspace{0.05cm} .$$
Da die Vierpole durch einen Trennverstärker widerstandsmäßig entkoppelt sind, lässt sich für die Gesamtübertragungsfunktion schreiben:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p)
\hspace{0.05cm} .$$
Gleichzeitig ist bekannt, dass folgende Gleichung gültig ist:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}
\hspace{0.05cm} .$$
Stellt man diese Funktion in Pol–Nullstellen–Form dar, so wird sich herausstellen, dass hier die Anzahl der Nullstellen  $(Z)$  gleich der Anzahl der Pole  $(N)$  ist.  Eine direkte Anwendung des Residuensatzes ist hier deshalb nicht möglich.

Um die Zeitfunktion  $h(t)$  berechnen zu können, muss vielmehr eine Partialbruchzerlegung entsprechend 
$H_{\rm L}(p) =1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.05cm}$ 
vorgenommen werden. Damit gilt für die Impulsantwort:
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}.$$
Bezüglich  $H_{\rm L}'(p)$  gilt  $Z' < N'$. Somit kann der kontinuierliche Anteil  $h'(t)$  der Impulsantwort mit dem Residuensatz ermittelt werden.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*Das Residium eines  $l$–fachen Pols  $p_{\rm x}$  innerhalb der Funktion  $H_{\rm L}(p)$  lautet:
:$${\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.03cm}\{H_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p t}\}=
\frac{1}{(l-1)!}\cdot \frac{{\rm d}^{\hspace{0.05cm}l-1}}{{\rm d}p^{\hspace{0.05cm}l-1}}\hspace{0.15cm}
\left \{H_{\rm L}(p)\cdot (p - p_{\rm x})^{\hspace{0.05cm}l} \cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.05cm} .$$
*Die Ableitung des Produkts  $y(x) = f(x) \cdot g(x)$  ist wie folgt gegeben:
:$$\frac{{\rm d}{\hspace{0.05cm}y(x)}}{{\rm d}\hspace{0.05cm}x}= \frac{{\rm d}{\hspace{0.05cm}f(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot g(x) + \frac{{\rm d}{\hspace{0.05cm}g(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot f(x)
\hspace{0.05cm} .$$

===Questions===

<quiz display=simple>
{Stellen Sie  $H_{\rm L}(p)$  in Pol–Nullstellen–Form dar.  Wieviele Nullstellen  $(Z)$  und Pole  $(N)$  gibt es?  Wie groß ist der konstante Faktor  $K$?
|type="{}"}
$Z \hspace{0.28cm} = \ $ { 2 }
$N \hspace{0.2cm} = \ $ { 2 }
$K \hspace{0.2cm} = \ $ { 1 }

{Wie groß ist der Parameter $A$  der beiden Teilvierpolen?
|type="{}"}
$A \ = \ $ { 0.5 3% }

{Wandeln Sie  $H_{\rm L}(p) = 1 - H_{\rm L}'(p)$  um.  Welches Ergebnis erhält man für  $H_{\rm L}'(p)$?
|type="[]"}
- $H_{\rm L}'(p) = p^2/(p+0.5)^2$,
- $H_{\rm L}'(p) = p/(p+0.5)^2$,
+ $H_{\rm L}'(p) = (p+0.25)/(p+0.5)^2$.

{Berechnen Sie die Zeitfunktion  $h'(t)$.  Welche Zahlenwerte ergeben sich für die angegebenen Zeitpunkte?
|type="{}"}
$h'(t = 0) \ = \ $ { 1 3% }
$h'(t = 1) \ = \ $ { 0.455 3% }
$h'(t → ∞)\ = \ $ { 0. }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Ausgehend von der vorgegebenen Gleichung kann  $H_{\rm L}(p)$  wie folgt umgeformt werden:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}=\frac{p^2}{p^2 + p +1/4}=\frac{p^2}{(p +1/2)^2}
\hspace{0.3cm}
\Rightarrow \hspace{0.3cm} \hspace{0.15cm}\underline{ Z = 2\hspace{0.05cm} , \hspace{0.2cm}N = 2\hspace{0.05cm} ,
\hspace{0.2cm}K = 1}
\hspace{0.05cm} .$$

'''(2)'''  Die Gesamtübertragungsfunktion lautet entsprechend der Angabe:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p) =\frac{p^2}{(p+A)^2}
\hspace{0.05cm} .$$
Ein Vergleich mit dem Ergebnis der Teilaufgabe  '''(1)'''  zeigt, dass  $\underline{A = 0.5}$  sein muss.

'''(3)'''  Richtig ist der letzte Lösungsvorschlag:
*Ausgehend von der in der Teilaufgabe  '''(1)'''  berechneten Gleichung erhält man
:$$H_{\rm L}(p) =\frac{p^2}{p^2 + p +0.25}= \frac{p^2 + p +0.25}{p^2 + p
+0.25}- \frac{p +0.25}{p^2 + p
+0.25}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{p +0.25}{p^2 + p
+0.25}= \frac{p +0.25}{(p
+0.5)^2}
\hspace{0.05cm} .$$

'''(4)'''  Bezüglich der Funktion  $H_{\rm L}'(p)$  gilt  $Z' = 1$,  $N' = 2$  und  $K' = 1$.

Die beiden Pole bei  $p_{\rm x} = -0.5$  fallen zusammen, so dass nur ein Residium ermittelt werden muss:
:$$h\hspace{0.03cm}'(t) \hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.7cm}\{H_{\rm L}\hspace{-0.05cm}'(p)\cdot {\rm e}^{p t}\}=
\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ \frac{p +0.25}{(p
+0.5)^2} \cdot (p
+0.5)^2 \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.2cm}\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ (p
+0.25) \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
\hspace{0.05cm} .$$
[[File:P_ID1788__LZI_A_3_7_d.png|right|frame|Impulsantwort des Hochpasses (rot); kontinuierlicher Anteil $h\hspace{0.03cm}'(t)$ (blau)]]
Mit der Produktregel der Differentialrechnung erhält man:
$$h\hspace{0.03cm}'(t) \hspace{0.15cm} = \hspace{0.15cm}
{\rm e}^{p \hspace{0.05cm}t} + ( p + 0.25) \cdot t \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.15cm} (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm} $$
$$\Rightarrow \hspace{0.3cm}h\hspace{0.03cm}'(t = 0) \hspace{0.15cm} = \underline{1}\hspace{0.05cm} ,\hspace{0.3cm} h\hspace{0.03cm}'(t = 1) \hspace{0.15cm} = \underline {0.455}\hspace{0.05cm} \hspace{0.05cm} ,\hspace{0.3cm}
h\hspace{0.03cm}'(t \rightarrow \infty) \hspace{0.15cm} = \underline {= 0}\hspace{0.05cm} .$$
Die Grafik zeigt jeweils für nicht–negative Zeiten
*als blaue Kurve die Impulsantwort  $h'(t)$  des äquivalenten Tiefpasses,
*als rote Kurve die gesamte Impulsantwort des betrachteten Hochpasses:
:$$h(t) =
\delta (t) - (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.7: Impulse Response of a High-Pass Filter

2021-10-25T08:18:06Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:EN_LZI_A_3_7.png|right|frame|Hochpass zweiter Ordnung]]
Wir gehen von der skizzierten Anordnung aus.  Die Übertragungsfunktionen der beiden identischen Hochpässe lauten:
:$$H_{\rm L}^{(1)}(p) = H_{\rm L}^{(2)}(p) =\frac{p}{p+A}
\hspace{0.05cm} .$$
Da die Vierpole durch einen Trennverstärker widerstandsmäßig entkoppelt sind, lässt sich für die Gesamtübertragungsfunktion schreiben:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p)
\hspace{0.05cm} .$$
Gleichzeitig ist bekannt, dass folgende Gleichung gültig ist:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}
\hspace{0.05cm} .$$
Stellt man diese Funktion in Pol–Nullstellen–Form dar, so wird sich herausstellen, dass hier die Anzahl der Nullstellen  $(Z)$  gleich der Anzahl der Pole  $(N)$  ist.  Eine direkte Anwendung des Residuensatzes ist hier deshalb nicht möglich.

Um die Zeitfunktion  $h(t)$  berechnen zu können, muss vielmehr eine Partialbruchzerlegung entsprechend 
$H_{\rm L}(p) =1- H_{\rm L}\hspace{-0.05cm}'(p)
\hspace{0.05cm}$ 
vorgenommen werden. Damit gilt für die Impulsantwort:
:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
\hspace{0.05cm}.$$
Bezüglich  $H_{\rm L}'(p)$  gilt  $Z' < N'$. Somit kann der kontinuierliche Anteil  $h'(t)$  der Impulsantwort mit dem Residuensatz ermittelt werden.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*Das Residium eines  $l$–fachen Pols  $p_{\rm x}$  innerhalb der Funktion  $H_{\rm L}(p)$  lautet:
:$${\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.03cm}\{H_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p t}\}=
\frac{1}{(l-1)!}\cdot \frac{{\rm d}^{\hspace{0.05cm}l-1}}{{\rm d}p^{\hspace{0.05cm}l-1}}\hspace{0.15cm}
\left \{H_{\rm L}(p)\cdot (p - p_{\rm x})^{\hspace{0.05cm}l} \cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.05cm} .$$
*Die Ableitung des Produkts  $y(x) = f(x) \cdot g(x)$  ist wie folgt gegeben:
:$$\frac{{\rm d}{\hspace{0.05cm}y(x)}}{{\rm d}\hspace{0.05cm}x}= \frac{{\rm d}{\hspace{0.05cm}f(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot g(x) + \frac{{\rm d}{\hspace{0.05cm}g(x)}}{{\rm
d}\hspace{0.05cm}x}\cdot f(x)
\hspace{0.05cm} .$$

===Questions===

<quiz display=simple>
{Stellen Sie  $H_{\rm L}(p)$  in Pol–Nullstellen–Form dar.  Wieviele Nullstellen  $(Z)$  und Pole  $(N)$  gibt es?  Wie groß ist der konstante Faktor  $K$?
|type="{}"}
$Z \hspace{0.28cm} = \ $ { 2 }
$N \hspace{0.2cm} = \ $ { 2 }
$K \hspace{0.2cm} = \ $ { 1 }

{Wie groß ist der Parameter $A$  der beiden Teilvierpolen?
|type="{}"}
$A \ = \ $ { 0.5 3% }

{Wandeln Sie  $H_{\rm L}(p) = 1 - H_{\rm L}'(p)$  um.  Welches Ergebnis erhält man für  $H_{\rm L}'(p)$?
|type="[]"}
- $H_{\rm L}'(p) = p^2/(p+0.5)^2$,
- $H_{\rm L}'(p) = p/(p+0.5)^2$,
+ $H_{\rm L}'(p) = (p+0.25)/(p+0.5)^2$.

{Berechnen Sie die Zeitfunktion  $h'(t)$.  Welche Zahlenwerte ergeben sich für die angegebenen Zeitpunkte?
|type="{}"}
$h'(t = 0) \ = \ $ { 1 3% }
$h'(t = 1) \ = \ $ { 0.455 3% }
$h'(t → ∞)\ = \ $ { 0. }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Ausgehend von der vorgegebenen Gleichung kann  $H_{\rm L}(p)$  wie folgt umgeformt werden:
:$$H_{\rm L}(p) =\frac{4}{1/p^2 + 4/p +4}=\frac{p^2}{p^2 + p +1/4}=\frac{p^2}{(p +1/2)^2}
\hspace{0.3cm}
\Rightarrow \hspace{0.3cm} \hspace{0.15cm}\underline{ Z = 2\hspace{0.05cm} , \hspace{0.2cm}N = 2\hspace{0.05cm} ,
\hspace{0.2cm}K = 1}
\hspace{0.05cm} .$$

'''(2)'''  Die Gesamtübertragungsfunktion lautet entsprechend der Angabe:
:$$H_{\rm L}(p) = H_{\rm L}^{(1)}(p) \cdot H_{\rm L}^{(2)}(p) =\frac{p^2}{(p+A)^2}
\hspace{0.05cm} .$$
Ein Vergleich mit dem Ergebnis der Teilaufgabe  '''(1)'''  zeigt, dass  $\underline{A = 0.5}$  sein muss.

'''(3)'''  Richtig ist der letzte Lösungsvorschlag:
*Ausgehend von der in der Teilaufgabe  '''(1)'''  berechneten Gleichung erhält man
:$$H_{\rm L}(p) =\frac{p^2}{p^2 + p +0.25}= \frac{p^2 + p +0.25}{p^2 + p
+0.25}- \frac{p +0.25}{p^2 + p
+0.25}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{p +0.25}{p^2 + p
+0.25}= \frac{p +0.25}{(p
+0.5)^2}
\hspace{0.05cm} .$$

'''(4)'''  Bezüglich der Funktion  $H_{\rm L}'(p)$  gilt  $Z' = 1$,  $N' = 2$  und  $K' = 1$.

Die beiden Pole bei  $p_{\rm x} = -0.5$  fallen zusammen, so dass nur ein Residium ermittelt werden muss:
:$$h\hspace{0.03cm}'(t) \hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{\rm x}}
\hspace{0.7cm}\{H_{\rm L}\hspace{-0.05cm}'(p)\cdot {\rm e}^{p t}\}=
\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ \frac{p +0.25}{(p
+0.5)^2} \cdot (p
+0.5)^2 \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.2cm}\frac{\rm d}{{\rm d}p}\hspace{0.15cm}
\left \{ (p
+0.25) \cdot {\rm e}^{p
\hspace{0.05cm}t}\right\}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
\hspace{0.05cm} .$$
[[File:P_ID1788__LZI_A_3_7_d.png|right|frame|Impulsantwort des Hochpasses (rot); kontinuierlicher Anteil $h\hspace{0.03cm}'(t)$ (blau)]]
Mit der Produktregel der Differentialrechnung erhält man:
$$h\hspace{0.03cm}'(t) \hspace{0.15cm} = \hspace{0.15cm}
{\rm e}^{p \hspace{0.05cm}t} + ( p + 0.25) \cdot t \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.5}
= \hspace{0.15cm} (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm} $$
$$\Rightarrow \hspace{0.3cm}h\hspace{0.03cm}'(t = 0) \hspace{0.15cm} = \underline{1}\hspace{0.05cm} ,\hspace{0.3cm} h\hspace{0.03cm}'(t = 1) \hspace{0.15cm} = \underline {0.455}\hspace{0.05cm} \hspace{0.05cm} ,\hspace{0.3cm}
h\hspace{0.03cm}'(t \rightarrow \infty) \hspace{0.15cm} = \underline {= 0}\hspace{0.05cm} .$$
Die Grafik zeigt jeweils für nicht–negative Zeiten
*als blaue Kurve die Impulsantwort  $h'(t)$  des äquivalenten Tiefpasses,
*als rote Kurve die gesamte Impulsantwort des betrachteten Hochpasses:
:$$h(t) =
\delta (t) - (1- {t}/{4})
\cdot{\rm e}^{-t/2}
\hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-25T08:15:01Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
In this exercise, we consider a causal signal  $x(t)$  with the Laplace transform
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
corresponding to the graph  (one red zero and two green poles).

In contrast, the signal  $y(t)$  has the Laplace spectral function
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Thus, the red zero does not belong to  $Y_{\rm L}(p)$.

Finally, the signal  $z(t)$  with the Laplace tansform
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
is considered, in particular the limiting case for  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*The frequency variable  $p$  is normalized such that time  $t$  is in microseconds after applying the residue theorem.
*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .
*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Compute the signal  $x(t)$. Which of the following statements are correct?
|type="[]"}
+ $x(t)$  is a causal cosine signal.
- $x(t)$  is a causal sinusoidal signal.
+ The amplitude of  $x(t)$  is  $1$.
+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.

{Compute the signal  $y(t)$. Which of the following statements are correct?
|type="[]"}
- $y(t)$  is a causal cosine signal.
+ $y(t)$  is a causal sinusoidal signal.
- The amplitude of  $y(t)$  is  $1$.
+ The period of  $y(t)$  is  $T = 1 \ \rm µ s$.

{Which statements are true for the signal  $z(t)$ ?
|type="[]"}
+ For  $ \beta > 0$,   $z(t)$  is cosine-shaped.
- For  $ \beta > 0$,   $z(t)$  is sinusoidal.
+ The limiting case  $\beta → 0$  results in the step function  $\gamma(t)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1, 3 and 4 are correct:
*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  The suggested solutions 2 and 4 are correct:
*In principle, this subtask could be solved in the same way as subtask  '''(1)'''.
*However, the integration theorem can also be used.
*This says among other things that multiplication by  $1/p$  in the spectral domain corresponds to integration in the time domain:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Please note:''     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior|Exercise 3.6]]  as  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.

'''(3)'''  The suggested solutions 1 and 3 are correct:
*A comparison with the computation of  $x(t)$  shows that  $z(t) = \cos (\beta \cdot t)$  holds for  $t \ge 0$  and  $z(t) = 0$  for  $t < 0$ .
*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.
*The same result is obtained by consideration in the spectral domain:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-25T08:14:17Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
In this exercise, we consider a causal signal  $x(t)$  with the Laplace transform
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
corresponding to the graph  (one red zero and two green poles).

In contrast, the signal  $y(t)$  has the Laplace spectral function
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Thus, the red zero does not belong to  $Y_{\rm L}(p)$.

Finally, the signal  $z(t)$  with the Laplace tansform
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
is considered, in particular the limiting case for  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*The frequency variable  $p$  is normalized such that time  $t$  is in microseconds after applying the residue theorem.
*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .
*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Compute the signal  $x(t)$. Which of the following statements are correct?
|type="[]"}
+ $x(t)$  is a causal cosine signal.
- $x(t)$  is a causal sinusoidal signal.
+ The amplitude of  $x(t)$  is  $1$.
+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.

{Compute the signal  $y(t)$. Which of the following statements are correct?
|type="[]"}
- $y(t)$  is a causal cosine signal.
+ $y(t)$  is a causal sinusoidal signal.
- The amplitude of  $y(t)$  is  $1$.
+ The period of  $y(t)$  is  $T = 1 \ \rm µ s$.

{Which statements are true for the signal  $z(t)$ ?
|type="[]"}
+ For  $ \beta > 0$,   $z(t)$  is cosine-shaped.
- For  $ \beta > 0$,   $z(t)$  is sinusoidal.
+ The limiting case  $\beta → 0$  results in the step function  $\gamma(t)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1, 3 and 4 are correct:
*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  The suggested solutions 2 and 4 are correct:
*In principle, this subtask could be solved in the same way as subtask  '''(1)'''.
*However, the integration theorem can also be used.
*This says among other things that multiplication by  $1/p$  in the spectral domain corresponds to integration in the time domain:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Please note:''     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior|Exercise 3.6]]  als  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.

'''(3)'''  The suggested solutions 1 and 3 are correct:
*A comparison with the computation of  $x(t)$  shows that  $z(t) = \cos (\beta \cdot t)$  holds for  $t \ge 0$  and  $z(t) = 0$  for  $t < 0$ .
*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.
*The same result is obtained by consideration in the spectral domain:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-25T08:12:44Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
In this exercise, we consider a causal signal  $x(t)$  with the Laplace transform
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
corresponding to the graph  (one red zero and two green poles).

In contrast, the signal  $y(t)$  has the Laplace spectral function
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Thus, the red zero does not belong to  $Y_{\rm L}(p)$.

Finally, the signal  $z(t)$  with the Laplace tansform
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
is considered, in particular the limiting case for  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*The frequency variable  $p$  is normalized such that time  $t$  is in microseconds after applying the residue theorem.
*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .
*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Compute the signal  $x(t)$. Which of the following statements are correct?
|type="[]"}
+ $x(t)$  is a causal cosine signal.
- $x(t)$  is a causal sinusoidal signal.
+ The amplitude of  $x(t)$  is  $1$.
+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.

{Compute the signal  $y(t)$. Which of the following statements are correct?
|type="[]"}
- $y(t)$  is a causal cosine signal.
+ $y(t)$  is a causal sinusoidal signal.
- The amplitude of  $y(t)$  is  $1$.
+ The period of  $y(t)$  is  $T = 1 \ \rm µ s$.

{Which statements are true for the signal  $z(t)$ ?
|type="[]"}
+ For  $ \beta > 0$,   $z(t)$  is cosine-shaped.
- For  $ \beta > 0$,   $z(t)$  is sinusoidal.
+ The limiting case  $\beta → 0$  results in the step function  $\gamma(t)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1, 3 and 4 are correct:
*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  The suggested solutions 2 and 4 are correct:
*In principle, this subtask could be solved in the same way as subtask  '''(1)'''.
*However, the integration theorem can also be used.
*This says among other things that multiplication by  $1/p$  in the spectral domain corresponds to integration in the time domain:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Please note'':     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior|Exercise 3.6]]  als  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.

'''(3)'''  The suggested solutions 1 and 3 are correct:
*A comparison with the computation of  $x(t)$  shows that  $z(t) = \cos (\beta \cdot t)$  holds for  $t \ge 0$  and  $z(t) = 0$  for  $t < 0$ .
*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.
*The same result is obtained by consideration in the spectral domain:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-25T08:11:12Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
In this exercise, we consider a causal signal  $x(t)$  with the Laplace transform
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
corresponding to the graph  (one red zero and two green poles).

In contrast, the signal  $y(t)$  has the Laplace spectral function
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Thus, the red zero does not belong to  $Y_{\rm L}(p)$.

Finally, the signal  $z(t)$  with the Laplace tansform
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
is considered, in particular the limiting case for  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*The frequency variable  $p$  is normalized such that time  $t$  is in microseconds after applying the residue theorem.
*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .
*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Compute the signal  $x(t)$. Which of the following statements are correct?
|type="[]"}
+ $x(t)$  is a causal cosine signal.
- $x(t)$  is a causal sinusoidal signal.
+ The amplitude ofn  $x(t)$  is  $1$.
+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.

{Compute the signal  $y(t)$. Which of the following statements are correct?
|type="[]"}
- $y(t)$  is a causal cosine signal.
+ $y(t)$  is a causal sinusoidal signal.
- The amplitude of  $y(t)$  is  $1$.
+ The period of  $y(t)$  is  $T = 1 \ \rm µ s$.

{Which statements are true for the signal  $z(t)$ ?
|type="[]"}
+ For  $ \beta > 0$,   $z(t)$  is cosine-shaped.
- For  $ \beta > 0$,   $z(t)$  is sinusoidal.
+ The limiting case  $\beta → 0$  results in the step function  $\gamma(t)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1, 3 and 4 are correct:
*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  The suggested solutions 2 and 4 are correct:
*In principle, this subtask could be solved in the same way as subtask  '''(1)'''.
*However, the integration theorem can also be used.
*This says among other things that multiplication by  $1/p$  in the spectral domain corresponds to integration in the time domain:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Please note'':     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior|Exercise 3.6]]  als  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.

'''(3)'''  The suggested solutions 1 and 3 are correct:
*A comparison with the computation of  $x(t)$  shows that  $z(t) = \cos (\beta \cdot t)$  holds for  $t \ge 0$  and  $z(t) = 0$  for  $t < 0$ .
*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.
*The same result is obtained by consideration in the spectral domain:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-25T08:10:12Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
In this exercise, we consider a causal signal  $x(t)$  with the Laplace transform
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
corresponding to the graph  (one red zero and two green poles).

In contrast, the signal  $y(t)$  has the Laplace spectral function
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Thus, the red zero does not belong to  $Y_{\rm L}(p)$.

Finally, the signal  $z(t)$  with the Laplace tansform
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
is considered, in particular the limiting case for  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*The frequency variable  $p$  is normalized such that time  $t$  is in microseconds after applying the residue theorem.
*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .
*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Compute the signal  $x(t)$. Which of the following statements are correct?
|type="[]"}
+ $x(t)$  is a causal cosine signal.
- $x(t)$  is a causal sinusoidal signal.
+ The amplitude ofn  $x(t)$  is  $1$.
+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.

{Compute the signal  $y(t)$. Which of the following statements are correct?
|type="[]"}
- $y(t)$  is a causal cosine signal.
+ $y(t)$  is a causal sinusoidal signal.
- The amplitude of  $y(t)$  is  $1$.
+ The period of  $y(t)$  is  $T = 1 \ \rm µ s$.

{Which statements are true for the signal  $z(t)$ ?
|type="[]"}
+ For  $ \beta > 0$,   $z(t)$  is cosine-shaped.
- For  $ \beta > 0$,   $z(t)$  is sinusoidal.
+ The limiting case  $\beta → 0$  results in the step function  $\gamma(t)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1, 3 and 4 are correct:
*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  The suggested solutions 2 and 4 are correct:
*In principle, this subtask could be solved in the same way as subtask  '''(1)'''.
*However, the integration theorem can also be used.
*This says among other things that multiplication by  $1/p$  in the spectral domain corresponds to integration in the time domain:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Please note'':     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior|Exercise 3.6]]  als  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.

'''(3)'''  The suggested solutions 1 and 3 are correct:
*A comparison with the computation of  $x(t)$  shows that  $z(t) = \cos (\beta \cdot t)$  holds for  $t \ge 0$  and  $z(t) = 0$  holds for  $t < 0$ .
*The limit process for  $\beta → 0$  thus results in the step function  $\gamma(t)$.
*The same result is obtained by consideration in the spectral domain:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Linear and Time Invariant Systems/Inverse Laplace Transform

2021-10-25T08:04:30Z

Oezer:

{{Header
|Untermenü=Beschreibung kausaler realisierbarer Systeme
|Vorherige Seite=Laplace_Transform_and_p-Transfer_Function
|Nächste Seite=Some_Results_from_Transmission_Line_Theory
}}
==Problem formulation and prerequisites==
 
{{BlaueBox|TEXT=
$\text{Task:}$  This chapter deals with the following problem:
*The  $p$–spectral function  $Y_{\rm L}(p)$  is given in  "pole-zero"  notation.
*The  '''inverse Laplace transform''', i.e. the associated time function  $y(t)$  is searched-for,  where the following notation should hold:
:$$y(t) = {\rm L}^{-1}\{Y_{\rm L}(p)\}\hspace{0.05cm} , \hspace{0.3cm}{\rm briefly}\hspace{0.3cm}
y(t) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\bullet\quad Y_{\rm L}(p)\hspace{0.05cm} .$$}}

[[File:EN_LZI_T_3_3_S1.png |right|frame|Prerequisites for the chapter "Inverse Laplace Transform"]]
 
The graph summarizes the prerequisites for this task.

*$H_{\rm L}(p)$  describes the transfer function of the causal system and  $Y_{\rm L}(p)$  specifies the Laplace transform of the output signal  $y(t)$  considering the input signal  $x(t)$ .  $Y_{\rm L}(p)$  is characterized by  $N$  poles,  by  $Z ≤ N$  zeros and by the constant  $K.$
*Poles and zeros exhibit the properties mentioned in the  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros|last chapter]]:  Poles are only allowed in the left  $p$–half plane or on the imaginary axis;  zeros are also allowed in the right  $p$–half plane.
*All singularities – this is the generic term for poles and zeros – are either real or exist as pairs of conjugate-complex singularities.  Multiple poles and zeros are also allowed.
*With the input  $x(t) = δ(t)$   ⇒   $X_{\rm L}(p) = 1$   ⇒   $Y_{\rm L}(p) = H_{\rm L}(p)$, the output signal  $y(t)$  then describes the [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|impulse response]]  $h(t)$  of the transmission system.  For this purpose, only the singularities drawn in green in the graph may be used for computation.
*A step function  $x(t) = γ(t)$   ⇒   $ X_{\rm L} = 1/p$  at the input causes the output signal  $y(t)$  to be equal to the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Step_response|step response]]   $σ(t)$ of $H_{\rm L}(p)$ .  In addition to the singularities of  $H_{\rm L}(p)$,  the pole (shown in red in the graph) at  $p = 0$  must now also be taken into account for computation.
*Only signals for which  $X_{ \rm L}(p)$  can be expressed in pole-zero notation are possible as input  $x(t)$  (see the  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Some_important_Laplace_correspondences|table]]  in the chapter "Laplace Transform and $p$–Transfer Function"), for example a cosine or sine signal switched on at time  $t = 0$ .
*So, a rectangle as input signal  $x(t)\ \ ⇒ \ \ X_{\rm L}(p) = (1 - {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot \hspace{0.05cm} T})/p$  is not possible in the approach described here.  However, the rectangular response  $y(t)$  can be computed indirectly as the difference of two step responses.

==Some results of the theory of functions==
 
In contrast to the  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_first_Fourier_integral|Fourier integrals]],  which differ only slightly in the two directions of transformation,  for  "Laplace"  the computation of  $y(t)$  from  $Y_{\rm L}(p)$ – that is the inverse transformation – is
*much more difficult than computing  $Y_{\rm L}(p)$  from  $y(t)$,
*unresolvable or solvable only very laboriously by elementary means.

{{BlaueBox|TEXT=
$\text{Definition:}$ 
In general, the following holds for the  '''inverse Laplace trans''':
:$$y(t) = {\rm L}^{-1}\{Y_{\rm L}(p)\}= \lim_{\beta \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \hspace{0.15cm} \frac{1}{ {\rm j} \cdot 2 \pi}\cdot \int_{ \alpha - {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \beta } ^{\alpha+{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \beta} Y_{\rm L}(p) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\hspace{0.1cm}{\rm
d}p \hspace{0.05cm} .$$
*The integration is parallel to the imaginary axis.
*The real part  $α$  is to be chosen such that all poles are located to the left of the integration path.}}

[[File:EN_LZI_T_3_3_S2.png |right|frame|Line integral together with left and right circular integral]]

The left graph illustrates this line integral along the red dotted vertical  ${\rm Re}\{p\}= α$.  This integral is solvable using  [https://en.wikipedia.org/wiki/Jordan%27s_lemma Jordan's lemma of complex analysis].  In this tutorial only a very short and simple summary of the approach is depicted:
*The line integral can be divided into two circular integrals so that all poles are located in the left circular integral while the right circular integral may only contain zeros.
*According to the theory of functions, the right circular integral yields the time function  $y(t)$  for negative times.  Due to causality,  $y(t < 0)$  must be identical to zero,  but according to the fundamental theorem of the theory of functions this is only true if there are no poles in the right  $p$–half-plane.
*In contrast,  the integral over the left semicircle yields the time function for  $t ≥ 0$.  This encloses all poles and can be computed using the  "residue theorem"  in a (relatively) simple way,  as it will be shown on the next pages.

==Formulation of the residue theorem==
 
It is further assumed that the transfer function  $Y_{\rm L}(p)$  can be expressed in pole-zero notation by
*the constant factor  $K$,
*the  $Z$  "zeros"  $p_{{\rm o}i}$  $(i = 1$, ... , $Z)$  and
*the  $N$  "poles"  $p_{{\rm x}i}$  $(i = 1$, ... , $N$).

We also assume  $Z < N$.

The number of distinguishable poles is denoted by  $I$.  Multiple poles are counted only once to determine  $I$.  Thus, the following holds for the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Some_results_of_the_theory_of_functions|sketch]]  in the last section considering the double pole:   $N = 5$  and  $I = 4$.

{{BlaueBox|TEXT=
$\text{Residue Theorem:}$ 
Considering the above conditions, the  '''inverse Laplace transform'''  of  $Y_{\rm L}(p)$  for times  $t ≥ 0$  is obtained as the sum of  $I$  natural oscillations of the poles,  which are called the  "residuals"  – abbreviated as "Res":
:$$y(t) = \sum_{i=1}^{I}{\rm Res} \bigg \vert _{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}_i}} \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot {\rm e}^{p \hspace{0.05cm}t}\} \hspace{0.05cm} .$$

Since  $Y_{\rm L}(p)$  is only specifiable for causal signals,  $y(t < 0) = 0$  always holds for negative times.

*In general,  the following holds for a pole of multiplicity  $l$ :
:$${\rm Res} \bigg \vert _{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot {\rm e}^{p t}\}= \frac{1}{(l-1)!}\cdot \frac{ {\rm d}^{\hspace{0.05cm}l-1} }{ {\rm d}p^{\hspace{0.05cm}l-1} }\hspace{0.15cm} \left \{Y_{\rm L}(p)\cdot (p - p_{ {\rm x}_i})^{\hspace{0.05cm}l}\cdot {\rm e}^{p \hspace{0.05cm}t}\right\} \bigg \vert_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{0.05cm} .$$
*The following is obtained out of it with  $l = 1$  for a simple pole as a special case:
:$${\rm Res} \bigg\vert_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot {\rm e}^{p t}\}= Y_{\rm L}(p)\cdot (p - p_{ {\rm x}_i} )\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg \vert _{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{0.05cm} .$$}}

On the next pages,  the  "residue theorem"  is illustrated by three detailed examples corresponding to the three constellations in  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros| $\text{Example 3}$]]  in the chapter  "Laplace Transform and p-Transfer Function":
*So, we consider again the two-port network with an inductance  $L = 25 \ \rm µH$  in the longitudinal branch as well as the the series connection of an ohmic resistance  $R = 50 \ \rm Ω$  and a capacitance  $C$  in the transverse branch.
*For the latter,  we again consider three different values,  namely  $C = 62.5 \ \rm nF$,  $C = 8 \ \rm nF$  and  $C = 40 \ \rm nF$.
*The following is always assumed:  $x(t) = δ(t) \; ⇒ \; X_{\rm L}(p) = 1$   ⇒   $Y_{\rm L}(p) = H_{\rm L}(p)$   ⇒   the output signal  $y(t)$  is equal to the impulse response  $h(t)$.

==Aperiodically decaying impulse response==
 
The following is obtained for the  $p$–transfer function computed on the page  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Pole-zero_representation_of_circuits|pole-zero representation of circuits]]  with the capacitance  $C = 62.5 \ \rm nF$  and the other numerical values given in the graph below:
[[File: EN_LZI_T_3_3_S3a.png|right|frame|Aperiodically decaying impulse response]]

:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x 1})(p - p_{\rm x 2})}= 2 \cdot \frac {p + 0.32 }
{(p +0.4)(p +1.6 )} \hspace{0.05cm} .$$

Note the normalization of  $p$,  $K$ and also of all poles and zeros by the factor  ${\rm 10^6} · 1/\rm s$.

The impulse response is composed of  $I = N = 2$  natural oscillations. For $t < 0$,  these are equal to zero.
*The residual of the pole at  $p_{{\rm x}1} =\ –0.4$  yields the following time function:
:$$h_1(t) = {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}} \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= H_{\rm L}(p)\cdot (p - p_{{\rm x}1})\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}$$
: $$\Rightarrow \hspace{0.3cm}h_1(t) = 2 \cdot \frac {p + 0.32 } {p +0.4}\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.4}= - \frac {2 } {15}\cdot {\rm e}^{-0.4 \hspace{0.05cm} t} \hspace{0.05cm}. $$
*In the same way, the residual of the second pole at  $p_{{\rm x}2} = \ –1.6$  can be computed:
:$$h_2(t) = {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}} \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= H_{\rm L}(p)\cdot (p - p_{{\rm x}2})\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}$$
:$$\Rightarrow \hspace{0.3cm}h_2(t) = 2 \cdot \frac {p + 0.32 } {p +1.6}\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-1.6}= \frac {32 } {15}\cdot {\rm e}^{-1.6 \hspace{0.05cm} t} \hspace{0.05cm}. $$

The graph shows  $h_1(t)$  and  $h_2(t)$  as well as the sum signal  $h(t)$.
*The normalization factor  $1/T = 10^6 · \rm 1/s$ is taken into account here so that the time is normalized to  $T = 1 \ \rm µ s$ .
*For  $t =0$,  $T \cdot h(t=0) = {32 }/ {15} -{2 }/ {15}= 2 \hspace{0.05cm}$  is obtained as a result.
*For times  $t > 2 \ \rm µ s$,  the impulse response is negative  (although only slightly and difficult to see in the graph).

==Attenuated-oscillatory impulse response==
 
The component values  $R = 50 \ \rm Ω$,  $L = 25 \ \rm µ H$  and  $C = 8 \ \rm nF$ result in two conjugate complex poles at  $p_{{\rm x}1} = \ –1 + {\rm j} · 2$  and  $p_{{\rm x}2} = \ –1 - {\rm j} · 2$. 
[[File:EN_LZI_T_3_3_S3b.png|right|frame| Attenuated-oscillatory impulse response]]

*The zero is located at  $p_{\rm o} = \ –2.5$.
*$K = 2$  holds and all numerical values are to be multiplied again by the factor  $1/T$  $(T = 1\ \rm µ s$).

When applying the residue theorem to this configuration the following is obtained:
:$$h_1(t) = K \cdot \frac {p_{\rm x 1} - p_{\rm o }} {p_{\rm x 1} - p_{\rm x 2}} \cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 1} \cdot \hspace{0.05cm}t} $$
:$$\Rightarrow \hspace{0.3cm}h_1(t) = 2 \cdot \frac {-1 + {\rm j}\cdot 2 +2.5} {(-1 + {\rm j}\cdot 2) - (-1 - {\rm j}\cdot 2)}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 1}
\cdot\hspace{0.05cm}t}$$
:$$\Rightarrow \hspace{0.3cm}h_1(t) = 2 \cdot \frac {1.5 + {\rm j}\cdot 2} {{\rm j}\cdot 4}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 1} \cdot\hspace{0.05cm}t}= (1 - {\rm j}\cdot 0.75)\cdot {\rm
e}^{-t}\cdot {\rm e}^{\hspace{0.03cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 2t} \hspace{0.05cm} ,$$

:$$ h_2(t) = K \cdot \frac {p_{\rm x 2} - p_{\rm o }} {p_{\rm x 2} - p_{\rm x 1}}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 2} \hspace{0.03cm}\cdot \hspace{0.05cm}t} $$
:$$\Rightarrow \hspace{0.3cm} h_2(t) = 2 \cdot \frac {-1 - {\rm j}\cdot 2 +2.5} {(-1 - {\rm j}\cdot 2) - (-1 + {\rm j}\cdot 2)}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 2} \hspace{0.03cm}\cdot\hspace{0.05cm}t} $$
:$$\Rightarrow \hspace{0.3cm}h_2(t) =2 \cdot \frac {1.5 - {\rm j}\cdot 2} {-{\rm j}\cdot 4}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 2} \hspace{0.03cm}\cdot\hspace{0.05cm}t}= (1 + {\rm j}\cdot 0.75)\cdot {\rm e}^{-t}\cdot {\rm e}^{\hspace{0.03cm}-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 2t} \hspace{0.05cm} . $$

Using  [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_magnitude_and_phase|Euler's theorem]]  the following is obtained for the sum signal:
:$$h(t) = h_1(t) + h_2(t)\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}h(t) = {\rm e}^{-t}\cdot \big [ (1 - {\rm j}\cdot 0.75)\cdot (\cos() + {\rm j}\cdot \sin())+
+ (1 + {\rm j}\cdot 0.75)\cdot (\cos() - {\rm j}\cdot \sin())\big ]$$
:$$\Rightarrow \hspace{0.3cm}h(t) ={\rm e}^{-t}\cdot \big [ 2\cdot \cos(2t) + 1.5 \cdot \sin(2t)\big ]\hspace{0.05cm} . $$

The graph shows the attenuated-oscillatory impulse response  $h(t)$  attenuated by  ${\rm e}^{–t}$  for this pole–zero configuration.

==Critically-damped case==
 
With  $R = 50 \ \rm Ω$,  $L = 25 \ \rm µ H$  and   $C = 40 \ \rm nF$  we get the so-called "critically-damped case":
:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x })^2}= 2 \cdot \frac {p + 0.5 } {(p +1)^2} \hspace{0.05cm} .$$

The capacitance value  $C = 40 \ \rm nF$  is the smallest possible value for which there are just real pole places. 

These coincide, that is  $p_{\rm x} = \ -1$  is a double pole place.  The time function is thus according to the residue theorem with  $l = 2$:
:$$h(t) = {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= \frac{ {\rm d} }{ {\rm d}p}\hspace{0.15cm}
\left \{H_{\rm L}(p)\cdot (p - p_{ {\rm x} })^2\cdot {\rm e}^{p \hspace{0.05cm}t}\right\} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } = K \cdot \frac{ {\rm d} }{ {\rm d}p}\hspace{0.15cm}\left \{ (p - p_{ {\rm o} })\cdot {\rm e}^{p \hspace{0.05cm}t}\right\} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } \hspace{0.05cm} .$$

[[File:EN_LZI_T_3_3_S3c.png |right|frame| Impulse response and step response of the critically-damped case]]
Using the  "product rule"  of differential calculus, this gives:
:$$h(t) = K \cdot \left [ {\rm e}^{p \hspace{0.05cm}t} + (p - p_{ {\rm o} })\cdot t \cdot {\rm e}^{p \hspace{0.05cm}t} \right ] \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-1} = {\rm e}^{-t}\cdot \left ( 2 - t \right)
\hspace{0.05cm} .$$

The graph shows this impulse response (green curve) in normalized representation.  It differs only slightly from the one with the two different poles at  $-0.4$  and  $-1.6$ .

The signal drawn in red   ⇒   $y(t) = 1 - {\rm e}^{-t} + t \cdot {\rm e}^{-t}$  results when a step function is also considered at the input   ⇒   "step response".

To calculate the step response  $\sigma(t) = y(t)$  one can alternatively
*consider an additional pole at  $p = 0$   in the residual calculation (marked in red), or
*form the integral over the impulse response  $h(t)$.

==Partial fraction decomposition==
 
Prerequisite for the application of the residue theorem is that there are less zeros than poles   ⇒   $Z$  must always be smaller than  $N$ .

If,  on the other hand,  as in the case of a high-pass filter  $Z = N$,  then
*is the limit of the spectral function  $H_{\rm L}(p)$  for large  $p$  not equal to zero,
*if the associated time signal  $y(t)$  also contains a  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain|Dirac delta functions]],
*the residue theorem fails and a  [https://en.wikipedia.org/wiki/Partial_fraction_decomposition '''Partial fraction decomposition''']  must be performed.

The procedure is to be clarified exemplarily for a high pass of first order.

[[File:P_ID1775__LZI_T_3_3_S5_neu.png |right|frame| Impulse response of low-pass (blue) and high-pass (red)]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The  $p$-transfer function of a  "first-order RC high-pass filter"  can be transformed by splitting off a constant as follows:
:$$\frac{p}{p + RC} = 1- \frac{RC}{p + RC}\hspace{0.05cm} .$$
Thus, the high-pass impulse response is:
:$$h_{\rm HP}(t) = \delta(t) - h_{\rm TP}(t) \hspace{0.05cm} .$$

The graph shows
*as a red curve the high pass–impulse response  $h_{\rm HP}(t)$,
*as a blue curve the impulse response  $h_{\rm TP}(t)$  of the equivalent low pass.

The Dirac delta function is the Laplace transform of the constant value  $1$,  while the second function to be subtracted gives the impulse response of the equivalent low-pass filter, which is given by the residue theorem with  
:$$Z = 0,\hspace{0.2cm} N =1,\hspace{0.2cm} K = RC.$$ }}

==Exercises for the chapter==

[[Aufgaben:Exercise_3.5:_Circuit_with_R,_L_and_C|Exercise 3.5: Circuit with R, L and C]]

[[Aufgaben:Exercise_3.5Z:_Application_of_the_Residue_Theorem|Exercise 3.5Z: Application of the Residue Theorem]]

[[Aufgaben:Exercise_3.6:_Transient_Behavior|Exercise 3.6: Transient Behavior]]

[[Aufgaben:Exercise_3.6Z:_Two_Imaginary_Poles|Exercise 3.6Z: Two Imaginary Poles]]

[[Aufgaben:Exercise_3.7:_Impulse_Response_of_a_High-Pass_Filter|Exercise 3.7: Impulse Response of a High-Pass Filter]]

[[Aufgaben:Exercise_3.7Z:_Partial_Fraction_Decomposition|Exercise 3.7Z: Partial Fraction Decomposition]]

{{Display}}

Aufgaben:Exercise 3.6: Transient Behaviour

2021-10-25T08:03:38Z

Oezer: Oezer moved page Aufgaben:Exercise 3.6: Transient Behaviour to Aufgaben:Exercise 3.6: Transient Behavior

#REDIRECT [[Aufgaben:Exercise 3.6: Transient Behavior]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-25T08:03:38Z

Oezer: Oezer moved page Aufgaben:Exercise 3.6: Transient Behaviour to Aufgaben:Exercise 3.6: Transient Behavior

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  by magnitude and phase using  $H_{\rm L}(p)$ .   What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm deg$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm deg$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than $1$ to achieve the correct phase position of  $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signal  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$   at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-25T08:02:43Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
In this exercise, we consider a causal signal  $x(t)$  with the Laplace transform
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
corresponding to the graph  (one red zero and two green poles).

In contrast, the signal  $y(t)$  has the Laplace spectral function
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Thus, the red zero does not belong to  $Y_{\rm L}(p)$.

Finally, the signal  $z(t)$  with the Laplace tansform
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
is considered, in particular the limiting case for  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*The frequency variable  $p$  is normalized such that time  $t$  is in microseconds after applying the residue theorem.
*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .
*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Compute the signal  $x(t)$. Which of the following statements are correct?
|type="[]"}
+ $x(t)$  is a causal cosine signal.
- $x(t)$  is a causal sinusoidal signal.
+ The amplitude ofn  $x(t)$  is  $1$.
+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.

{Compute the signal  $y(t)$. Which of the following statements are correct?
|type="[]"}
- $y(t)$  is a causal cosine signal.
+ $y(t)$  is a causal sinusoidal signal.
- The amplitude of  $y(t)$  is  $1$.
+ The period of  $y(t)$  is  $T = 1 \ \rm µ s$.

{Which statements are true for the signal  $z(t)$ ?
|type="[]"}
+ For  $ \beta > 0$,   $z(t)$  is cosine-shaped.
- For  $ \beta > 0$,   $z(t)$  is sinusoidal.
+ The limiting case  $\beta → 0$  results in the step function  $\gamma(t)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1, 3 and 4 are correct:
*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  The suggested solutions 2 and 4 are correct:
*In principle, this subtask could be solved in the same way as subtask  '''(1)'''.
*However, the integration theorem can also be used.
*This says among other things that multiplication by  $1/p$  in the spectral domain corresponds to integration in the time domain:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Please note'':     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  sind auf dem Angabenblatt zu  [[Aufgaben:3.6_Einschwingverhalten|Exercise 3.6]]  als  $c_{\rm K}(t)$  bzw.  $s_{\rm K}(t)$  dargestellt.

'''(3)'''  Richtig sind die Lösungsvorschläge 1 und 3:
*Ein Vergleich mit der Berechnung von  $x(t)$  zeigt, dass  $z(t) = \cos (\beta \cdot t)$  für  $t \ge 0$  und  $z(t) = 0$  für  $t < 0$  gilt.
*Der Grenzübergang für  $\beta → 0$  führt damit zur Sprungfunktion  $\gamma(t)$.
*Zum gleichen Ergebnis kommt man durch die Betrachtung im Spektralbereich:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-25T07:54:32Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
In this exercise, we consider a causal signal  $x(t)$  with the Laplace transform
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
corresponding to the graph  (one red zero and two green poles).

In contrast, the signal  $y(t)$  has the Laplace spectral function
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Thus, the red zero does not belong to  $Y_{\rm L}(p)$.

Finally, the signal  $z(t)$  with the Laplace tansform
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
is considered, in particular the limiting case for  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*The frequency variable  $p$  is normalized such that time  $t$  is in microseconds after applying the residue theorem.
*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .
*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Compute the signal  $x(t)$. Which of the following statements are correct?
|type="[]"}
+ $x(t)$  is a causal cosine signal.
- $x(t)$  is a causal sinusoidal signal.
+ The amplitude ofn  $x(t)$  is  $1$.
+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.

{Compute the signal  $y(t)$. Which of the following statements are correct?
|type="[]"}
- $y(t)$  is a causal cosine signal.
+ $y(t)$  is a causal sinusoidal signal.
- The amplitude of  $y(t)$  is  $1$.
+ The period of  $y(t)$  is  $T = 1 \ \rm µ s$.

{Which statements are true for the signal  $z(t)$ ?
|type="[]"}
+ For  $ \beta > 0$,   $z(t)$  is cosine-shaped.
- For  $ \beta > 0$,   $z(t)$  is sinusoidal.
+ The limiting case  $\beta → 0$  results in the step function  $\gamma(t)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The suggested solutions 1, 3 and 4 are correct:
*The following is obtained for signal  $x(t)$  for positive times by applying the residue theorem:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  Richtig sind die Lösungsvorschläge 2 und 4:
*Prinzipiell könnte diese Teilaufgabe in gleicher Weise gelöst werden wie die Teilaufgabe  '''(1)'''.
*Man kann aber auch den Integrationssatz heranziehen.
*Dieser besagt unter anderem, dass die Multiplikation mit  $1/p$  im Spektralbereich der Integration im Zeitbereich entspricht:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Hinweis'':     Das kausale Cosinussignal  $x(t)$  sowie das kausale Sinussignal  $y(t)$  sind auf dem Angabenblatt zu  [[Aufgaben:3.6_Einschwingverhalten|Aufgabe 3.6]]  als  $c_{\rm K}(t)$  bzw.  $s_{\rm K}(t)$  dargestellt.

'''(3)'''  Richtig sind die Lösungsvorschläge 1 und 3:
*Ein Vergleich mit der Berechnung von  $x(t)$  zeigt, dass  $z(t) = \cos (\beta \cdot t)$  für  $t \ge 0$  und  $z(t) = 0$  für  $t < 0$  gilt.
*Der Grenzübergang für  $\beta → 0$  führt damit zur Sprungfunktion  $\gamma(t)$.
*Zum gleichen Ergebnis kommt man durch die Betrachtung im Spektralbereich:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-25T00:21:33Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
In this exercise, we consider a causal signal  $x(t)$  with the Laplace transform
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
corresponding to the graph  (one red zero and two green poles).

In contrast, the signal  $y(t)$  has the Laplace spectral function
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Thus, the red zero does not belong to  $Y_{\rm L}(p)$.

Finally, the signal  $z(t)$  with the Laplace tansform
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
is considered, in particular the limiting case for  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*The frequency variable  $p$  is normalized such that time  $t$  is in microseconds after applying the residue theorem.
*A result  $t = 1$  is thus to be interpreted as  $t = T$  with  $T = 1 \ \rm µ s$ .
*The  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  is as follows using the example of the function  $X_{\rm L}(p)$  with two simple poles at  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Compute the signal  $x(t)$. Which of the following statements are correct?
|type="[]"}
+ $x(t)$  is a causal cosine signal.
- $x(t)$  is a causal sinusoidal signal.
+ The amplitude ofn  $x(t)$  is  $1$.
+ The period of $x(t)$ is  $T = 1 \ \rm µ s$.

{Compute the signal  $y(t)$. Which of the following statements are correct?
|type="[]"}
- $y(t)$  is a causal cosine signal.
+ $y(t)$  is a causal sinusoidal signal.
- The amplitude of  $y(t)$  is  $1$.
+ The period of  $y(t)$  is  $T = 1 \ \rm µ s$.

{Which statements are true for the signal  $z(t)$ ?
|type="[]"}
+ For  $ \beta > 0$,   $z(t)$  is cosine-shaped.
- For  $ \beta > 0$,   $z(t)$  is sinusoidal.
+ The limiting case  $\beta → 0$  results in the step function  $\gamma(t)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Richtig sind die Lösungsvorschläge 1, 3 und 4:
*Durch Anwendung des Residuensatzes erhält man für das Signal  $x(t)$  bei positiven Zeiten:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  Richtig sind die Lösungsvorschläge 2 und 4:
*Prinzipiell könnte diese Teilaufgabe in gleicher Weise gelöst werden wie die Teilaufgabe  '''(1)'''.
*Man kann aber auch den Integrationssatz heranziehen.
*Dieser besagt unter anderem, dass die Multiplikation mit  $1/p$  im Spektralbereich der Integration im Zeitbereich entspricht:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Hinweis'':     Das kausale Cosinussignal  $x(t)$  sowie das kausale Sinussignal  $y(t)$  sind auf dem Angabenblatt zu  [[Aufgaben:3.6_Einschwingverhalten|Aufgabe 3.6]]  als  $c_{\rm K}(t)$  bzw.  $s_{\rm K}(t)$  dargestellt.

'''(3)'''  Richtig sind die Lösungsvorschläge 1 und 3:
*Ein Vergleich mit der Berechnung von  $x(t)$  zeigt, dass  $z(t) = \cos (\beta \cdot t)$  für  $t \ge 0$  und  $z(t) = 0$  für  $t < 0$  gilt.
*Der Grenzübergang für  $\beta → 0$  führt damit zur Sprungfunktion  $\gamma(t)$.
*Zum gleichen Ergebnis kommt man durch die Betrachtung im Spektralbereich:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-24T14:45:15Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
In this exercise, we consider a causal signal  $x(t)$  with the Laplace transform
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
corresponding to the graph  (one red zero and two green poles).

In contrast, the signal  $y(t)$  has the Laplace spectral function
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Thus, the red zero does not belong to  $Y_{\rm L}(p)$.

Abschließend wird noch das Signal  $z(t)$  mit der Laplace–Transformierten
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
betrachtet, insbesondere der Grenzfall für  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*Die Frequenzvariable  $p$  ist so normiert, dass nach Anwendung des Residuensatzes die Zeit  $t$  in Mikrosekunden angegeben ist.
*Ein Ergebnis  $t = 1$  ist somit als  $t = T$  mit  $T = 1 \ \rm µ s$  zu interpretieren.
*Der  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  lautet am Beispiel der Funktion  $X_{\rm L}(p)$  mit zwei einfachen Polstellen bei  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Berechnen Sie das Signal  $x(t)$. Welche der folgenden Aussagen sind richtig?
|type="[]"}
+ $x(t)$  ist ein kausales Cosinussignal.
- $x(t)$  ist ein kausales Sinussignal.
+ Die Amplitude von  $x(t)$  ist  $1$.
+ Die Periodendauer von $x(t)$ ist  $T = 1 \ \rm µ s$.

{Berechnen Sie das Signal  $y(t)$. Welche der folgenden Aussagen sind richtig?
|type="[]"}
- $y(t)$  ist ein kausales Cosinussignal.
+ $y(t)$  ist ein kausales Sinussignal.
- Die Amplitude von  $y(t)$  ist  $1$.
+ Die Periodendauer von  $y(t)$  ist  $T = 1 \ \rm µ s$.

{Welche Aussagen treffen für das Signal  $z(t)$  zu?
|type="[]"}
+ Für  $ \beta > 0$  verläuft  $z(t)$  cosinusförmig.
- Für  $ \beta > 0$  verläuft  $z(t)$  sinusförmig.
+ Der Grenzfall  $\beta → 0$  führt zur Sprungfunktion  $\gamma(t)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Richtig sind die Lösungsvorschläge 1, 3 und 4:
*Durch Anwendung des Residuensatzes erhält man für das Signal  $x(t)$  bei positiven Zeiten:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  Richtig sind die Lösungsvorschläge 2 und 4:
*Prinzipiell könnte diese Teilaufgabe in gleicher Weise gelöst werden wie die Teilaufgabe  '''(1)'''.
*Man kann aber auch den Integrationssatz heranziehen.
*Dieser besagt unter anderem, dass die Multiplikation mit  $1/p$  im Spektralbereich der Integration im Zeitbereich entspricht:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Hinweis'':     Das kausale Cosinussignal  $x(t)$  sowie das kausale Sinussignal  $y(t)$  sind auf dem Angabenblatt zu  [[Aufgaben:3.6_Einschwingverhalten|Aufgabe 3.6]]  als  $c_{\rm K}(t)$  bzw.  $s_{\rm K}(t)$  dargestellt.

'''(3)'''  Richtig sind die Lösungsvorschläge 1 und 3:
*Ein Vergleich mit der Berechnung von  $x(t)$  zeigt, dass  $z(t) = \cos (\beta \cdot t)$  für  $t \ge 0$  und  $z(t) = 0$  für  $t < 0$  gilt.
*Der Grenzübergang für  $\beta → 0$  führt damit zur Sprungfunktion  $\gamma(t)$.
*Zum gleichen Ergebnis kommt man durch die Betrachtung im Spektralbereich:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Linear and Time Invariant Systems/Inverse Laplace Transform

2021-10-24T14:34:01Z

Oezer:

{{Header
|Untermenü=Beschreibung kausaler realisierbarer Systeme
|Vorherige Seite=Laplace_Transform_and_p-Transfer_Function
|Nächste Seite=Some_Results_from_Transmission_Line_Theory
}}
==Problem formulation and prerequisites==
 
{{BlaueBox|TEXT=
$\text{Task:}$  This chapter deals with the following problem:
*The  $p$–spectral function  $Y_{\rm L}(p)$  is given in  "pole-zero"  notation.
*The  '''inverse Laplace transform''', i.e. the associated time function  $y(t)$  is searched-for,  where the following notation should hold:
:$$y(t) = {\rm L}^{-1}\{Y_{\rm L}(p)\}\hspace{0.05cm} , \hspace{0.3cm}{\rm briefly}\hspace{0.3cm}
y(t) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\bullet\quad Y_{\rm L}(p)\hspace{0.05cm} .$$}}

[[File:EN_LZI_T_3_3_S1.png |right|frame|Prerequisites for the chapter "Inverse Laplace Transform"]]
 
The graph summarizes the prerequisites for this task.

*$H_{\rm L}(p)$  describes the transfer function of the causal system and  $Y_{\rm L}(p)$  specifies the Laplace transform of the output signal  $y(t)$  considering the input signal  $x(t)$ .  $Y_{\rm L}(p)$  is characterized by  $N$  poles,  by  $Z ≤ N$  zeros and by the constant  $K.$
*Poles and zeros exhibit the properties mentioned in the  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros|last chapter]]:  Poles are only allowed in the left  $p$–half plane or on the imaginary axis;  zeros are also allowed in the right  $p$–half plane.
*All singularities – this is the generic term for poles and zeros – are either real or exist as pairs of conjugate-complex singularities.  Multiple poles and zeros are also allowed.
*With the input  $x(t) = δ(t)$   ⇒   $X_{\rm L}(p) = 1$   ⇒   $Y_{\rm L}(p) = H_{\rm L}(p)$, the output signal  $y(t)$  then describes the [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|impulse response]]  $h(t)$  of the transmission system.  For this purpose, only the singularities drawn in green in the graph may be used for computation.
*A step function  $x(t) = γ(t)$   ⇒   $ X_{\rm L} = 1/p$  at the input causes the output signal  $y(t)$  to be equal to the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Step_response|step response]]   $σ(t)$ of $H_{\rm L}(p)$ .  In addition to the singularities of  $H_{\rm L}(p)$,  the pole (shown in red in the graph) at  $p = 0$  must now also be taken into account for computation.
*Only signals for which  $X_{ \rm L}(p)$  can be expressed in pole-zero notation are possible as input  $x(t)$  (see the  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Some_important_Laplace_correspondences|table]]  in the chapter "Laplace Transform and $p$–Transfer Function"), for example a cosine or sine signal switched on at time  $t = 0$ .
*So, a rectangle as input signal  $x(t)\ \ ⇒ \ \ X_{\rm L}(p) = (1 - {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot \hspace{0.05cm} T})/p$  is not possible in the approach described here.  However, the rectangular response  $y(t)$  can be computed indirectly as the difference of two step responses.

==Some results of the theory of functions==
 
In contrast to the  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_first_Fourier_integral|Fourier integrals]],  which differ only slightly in the two directions of transformation,  for  "Laplace"  the computation of  $y(t)$  from  $Y_{\rm L}(p)$ – that is the inverse transformation – is
*much more difficult than computing  $Y_{\rm L}(p)$  from  $y(t)$,
*unresolvable or solvable only very laboriously by elementary means.

{{BlaueBox|TEXT=
$\text{Definition:}$ 
In general, the following holds for the  '''inverse Laplace trans''':
:$$y(t) = {\rm L}^{-1}\{Y_{\rm L}(p)\}= \lim_{\beta \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \hspace{0.15cm} \frac{1}{ {\rm j} \cdot 2 \pi}\cdot \int_{ \alpha - {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \beta } ^{\alpha+{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \beta} Y_{\rm L}(p) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\hspace{0.1cm}{\rm
d}p \hspace{0.05cm} .$$
*The integration is parallel to the imaginary axis.
*The real part  $α$  is to be chosen such that all poles are located to the left of the integration path.}}

[[File:EN_LZI_T_3_3_S2.png |right|frame|Line integral together with left and right circular integral]]

The left graph illustrates this line integral along the red dotted vertical  ${\rm Re}\{p\}= α$.  This integral is solvable using  [https://en.wikipedia.org/wiki/Jordan%27s_lemma Jordan's lemma of complex analysis].  In this tutorial only a very short and simple summary of the approach is depicted:
*The line integral can be divided into two circular integrals so that all poles are located in the left circular integral while the right circular integral may only contain zeros.
*According to the theory of functions, the right circular integral yields the time function  $y(t)$  for negative times.  Due to causality,  $y(t < 0)$  must be identical to zero,  but according to the fundamental theorem of the theory of functions this is only true if there are no poles in the right  $p$–half-plane.
*In contrast,  the integral over the left semicircle yields the time function for  $t ≥ 0$.  This encloses all poles and can be computed using the  "residue theorem"  in a (relatively) simple way,  as it will be shown on the next pages.

==Formulation of the residue theorem==
 
It is further assumed that the transfer function  $Y_{\rm L}(p)$  can be expressed in pole-zero notation by
*the constant factor  $K$,
*the  $Z$  "zeros"  $p_{{\rm o}i}$  $(i = 1$, ... , $Z)$  and
*the  $N$  "poles"  $p_{{\rm x}i}$  $(i = 1$, ... , $N$).

We also assume  $Z < N$.

The number of distinguishable poles is denoted by  $I$.  Multiple poles are counted only once to determine  $I$.  Thus, the following holds for the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Some_results_of_the_theory_of_functions|sketch]]  in the last section considering the double pole:   $N = 5$  and  $I = 4$.

{{BlaueBox|TEXT=
$\text{Residue Theorem:}$ 
Considering the above conditions, the  '''inverse Laplace transform'''  of  $Y_{\rm L}(p)$  for times  $t ≥ 0$  is obtained as the sum of  $I$  natural oscillations of the poles,  which are called the  "residuals"  – abbreviated as "Res":
:$$y(t) = \sum_{i=1}^{I}{\rm Res} \bigg \vert _{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}_i}} \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot {\rm e}^{p \hspace{0.05cm}t}\} \hspace{0.05cm} .$$

Since  $Y_{\rm L}(p)$  is only specifiable for causal signals,  $y(t < 0) = 0$  always holds for negative times.

*In general,  the following holds for a pole of multiplicity  $l$ :
:$${\rm Res} \bigg \vert _{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot {\rm e}^{p t}\}= \frac{1}{(l-1)!}\cdot \frac{ {\rm d}^{\hspace{0.05cm}l-1} }{ {\rm d}p^{\hspace{0.05cm}l-1} }\hspace{0.15cm} \left \{Y_{\rm L}(p)\cdot (p - p_{ {\rm x}_i})^{\hspace{0.05cm}l}\cdot {\rm e}^{p \hspace{0.05cm}t}\right\} \bigg \vert_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{0.05cm} .$$
*The following is obtained out of it with  $l = 1$  for a simple pole as a special case:
:$${\rm Res} \bigg\vert_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot {\rm e}^{p t}\}= Y_{\rm L}(p)\cdot (p - p_{ {\rm x}_i} )\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg \vert _{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{0.05cm} .$$}}

On the next pages,  the  "residue theorem"  is illustrated by three detailed examples corresponding to the three constellations in  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros| $\text{Example 3}$]]  in the chapter  "Laplace Transform and p-Transfer Function":
*So, we consider again the two-port network with an inductance  $L = 25 \ \rm µH$  in the longitudinal branch as well as the the series connection of an ohmic resistance  $R = 50 \ \rm Ω$  and a capacitance  $C$  in the transverse branch.
*For the latter,  we again consider three different values,  namely  $C = 62.5 \ \rm nF$,  $C = 8 \ \rm nF$  and  $C = 40 \ \rm nF$.
*The following is always assumed:  $x(t) = δ(t) \; ⇒ \; X_{\rm L}(p) = 1$   ⇒   $Y_{\rm L}(p) = H_{\rm L}(p)$   ⇒   the output signal  $y(t)$  is equal to the impulse response  $h(t)$.

==Aperiodically decaying impulse response==
 
The following is obtained for the  $p$–transfer function computed on the page  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Pole-zero_representation_of_circuits|pole-zero representation of circuits]]  with the capacitance  $C = 62.5 \ \rm nF$  and the other numerical values given in the graph below:
[[File: EN_LZI_T_3_3_S3a.png|right|frame|Aperiodically decaying impulse response]]

:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x 1})(p - p_{\rm x 2})}= 2 \cdot \frac {p + 0.32 }
{(p +0.4)(p +1.6 )} \hspace{0.05cm} .$$

Note the normalization of  $p$,  $K$ and also of all poles and zeros by the factor  ${\rm 10^6} · 1/\rm s$.

The impulse response is composed of  $I = N = 2$  natural oscillations. For $t < 0$,  these are equal to zero.
*The residual of the pole at  $p_{{\rm x}1} =\ –0.4$  yields the following time function:
:$$h_1(t) = {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}} \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= H_{\rm L}(p)\cdot (p - p_{{\rm x}1})\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}$$
: $$\Rightarrow \hspace{0.3cm}h_1(t) = 2 \cdot \frac {p + 0.32 } {p +0.4}\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.4}= - \frac {2 } {15}\cdot {\rm e}^{-0.4 \hspace{0.05cm} t} \hspace{0.05cm}. $$
*In the same way, the residual of the second pole at  $p_{{\rm x}2} = \ –1.6$  can be computed:
:$$h_2(t) = {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}} \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= H_{\rm L}(p)\cdot (p - p_{{\rm x}2})\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}$$
:$$\Rightarrow \hspace{0.3cm}h_2(t) = 2 \cdot \frac {p + 0.32 } {p +1.6}\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-1.6}= \frac {32 } {15}\cdot {\rm e}^{-1.6 \hspace{0.05cm} t} \hspace{0.05cm}. $$

The graph shows  $h_1(t)$  and  $h_2(t)$  as well as the sum signal  $h(t)$.
*The normalization factor  $1/T = 10^6 · \rm 1/s$ is taken into account here so that the time is normalized to  $T = 1 \ \rm µ s$ .
*For  $t =0$,  $T \cdot h(t=0) = {32 }/ {15} -{2 }/ {15}= 2 \hspace{0.05cm}$  is obtained as a result.
*For times  $t > 2 \ \rm µ s$,  the impulse response is negative  (although only slightly and difficult to see in the graph).

==Attenuated-oscillatory impulse response==
 
The component values  $R = 50 \ \rm Ω$,  $L = 25 \ \rm µ H$  and  $C = 8 \ \rm nF$ result in two conjugate complex poles at  $p_{{\rm x}1} = \ –1 + {\rm j} · 2$  and  $p_{{\rm x}2} = \ –1 - {\rm j} · 2$. 
[[File:EN_LZI_T_3_3_S3b.png|right|frame| Attenuated-oscillatory impulse response]]

*The zero is located at  $p_{\rm o} = \ –2.5$.
*$K = 2$  holds and all numerical values are to be multiplied again by the factor  $1/T$  $(T = 1\ \rm µ s$).

When applying the residue theorem to this configuration the following is obtained:
:$$h_1(t) = K \cdot \frac {p_{\rm x 1} - p_{\rm o }} {p_{\rm x 1} - p_{\rm x 2}} \cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 1} \cdot \hspace{0.05cm}t} $$
:$$\Rightarrow \hspace{0.3cm}h_1(t) = 2 \cdot \frac {-1 + {\rm j}\cdot 2 +2.5} {(-1 + {\rm j}\cdot 2) - (-1 - {\rm j}\cdot 2)}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 1}
\cdot\hspace{0.05cm}t}$$
:$$\Rightarrow \hspace{0.3cm}h_1(t) = 2 \cdot \frac {1.5 + {\rm j}\cdot 2} {{\rm j}\cdot 4}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 1} \cdot\hspace{0.05cm}t}= (1 - {\rm j}\cdot 0.75)\cdot {\rm
e}^{-t}\cdot {\rm e}^{\hspace{0.03cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 2t} \hspace{0.05cm} ,$$

:$$ h_2(t) = K \cdot \frac {p_{\rm x 2} - p_{\rm o }} {p_{\rm x 2} - p_{\rm x 1}}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 2} \hspace{0.03cm}\cdot \hspace{0.05cm}t} $$
:$$\Rightarrow \hspace{0.3cm} h_2(t) = 2 \cdot \frac {-1 - {\rm j}\cdot 2 +2.5} {(-1 - {\rm j}\cdot 2) - (-1 + {\rm j}\cdot 2)}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 2} \hspace{0.03cm}\cdot\hspace{0.05cm}t} $$
:$$\Rightarrow \hspace{0.3cm}h_2(t) =2 \cdot \frac {1.5 - {\rm j}\cdot 2} {-{\rm j}\cdot 4}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 2} \hspace{0.03cm}\cdot\hspace{0.05cm}t}= (1 + {\rm j}\cdot 0.75)\cdot {\rm e}^{-t}\cdot {\rm e}^{\hspace{0.03cm}-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 2t} \hspace{0.05cm} . $$

Using  [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_magnitude_and_phase|Euler's theorem]]  the following is obtained for the sum signal:
:$$h(t) = h_1(t) + h_2(t)\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}h(t) = {\rm e}^{-t}\cdot \big [ (1 - {\rm j}\cdot 0.75)\cdot (\cos() + {\rm j}\cdot \sin())+
+ (1 + {\rm j}\cdot 0.75)\cdot (\cos() - {\rm j}\cdot \sin())\big ]$$
:$$\Rightarrow \hspace{0.3cm}h(t) ={\rm e}^{-t}\cdot \big [ 2\cdot \cos(2t) + 1.5 \cdot \sin(2t)\big ]\hspace{0.05cm} . $$

The graph shows the attenuated-oscillatory impulse response  $h(t)$  attenuated by  ${\rm e}^{–t}$  for this pole–zero configuration.

==Critically-damped case==
 
With  $R = 50 \ \rm Ω$,  $L = 25 \ \rm µ H$  and   $C = 40 \ \rm nF$  we get the so-called "critically-damped case":
:$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x })^2}= 2 \cdot \frac {p + 0.5 } {(p +1)^2} \hspace{0.05cm} .$$

The capacitance value  $C = 40 \ \rm nF$  is the smallest possible value for which there are just real pole places. 

These coincide, that is  $p_{\rm x} = \ -1$  is a double pole place.  The time function is thus according to the residue theorem with  $l = 2$:
:$$h(t) = {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= \frac{ {\rm d} }{ {\rm d}p}\hspace{0.15cm}
\left \{H_{\rm L}(p)\cdot (p - p_{ {\rm x} })^2\cdot {\rm e}^{p \hspace{0.05cm}t}\right\} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } = K \cdot \frac{ {\rm d} }{ {\rm d}p}\hspace{0.15cm}\left \{ (p - p_{ {\rm o} })\cdot {\rm e}^{p \hspace{0.05cm}t}\right\} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } \hspace{0.05cm} .$$

[[File:EN_LZI_T_3_3_S3c.png |right|frame| Impulse response and step response of the critically-damped case]]
Using the  "product rule"  of differential calculus, this gives:
:$$h(t) = K \cdot \left [ {\rm e}^{p \hspace{0.05cm}t} + (p - p_{ {\rm o} })\cdot t \cdot {\rm e}^{p \hspace{0.05cm}t} \right ] \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-1} = {\rm e}^{-t}\cdot \left ( 2 - t \right)
\hspace{0.05cm} .$$

The graph shows this impulse response (green curve) in normalized representation.  It differs only slightly from the one with the two different poles at  $-0.4$  and  $-1.6$ .

The signal drawn in red   ⇒   $y(t) = 1 - {\rm e}^{-t} + t \cdot {\rm e}^{-t}$  results when a step function is also considered at the input   ⇒   "step response".

To calculate the step response  $\sigma(t) = y(t)$  one can alternatively
*consider an additional pole at  $p = 0$   in the residual calculation (marked in red), or
*form the integral over the impulse response  $h(t)$.

==Partial fraction decomposition==
 
Prerequisite for the application of the residue theorem is that there are less zeros than poles   ⇒   $Z$  must always be smaller than  $N$ .

If,  on the other hand,  as in the case of a high-pass filter  $Z = N$,  then
*is the limit of the spectral function  $H_{\rm L}(p)$  for large  $p$  not equal to zero,
*if the associated time signal  $y(t)$  also contains a  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain|Dirac delta functions]],
*the residue theorem fails and a  [https://en.wikipedia.org/wiki/Partial_fraction_decomposition '''Partial fraction decomposition''']  must be performed.

The procedure is to be clarified exemplarily for a high pass of first order.

[[File:P_ID1775__LZI_T_3_3_S5_neu.png |right|frame| Impulse response of low-pass (blue) and high-pass (red)]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The  $p$-transfer function of a  "first-order RC high-pass filter"  can be transformed by splitting off a constant as follows:
:$$\frac{p}{p + RC} = 1- \frac{RC}{p + RC}\hspace{0.05cm} .$$
Thus, the high-pass impulse response is:
:$$h_{\rm HP}(t) = \delta(t) - h_{\rm TP}(t) \hspace{0.05cm} .$$

The graph shows
*as a red curve the high pass–impulse response  $h_{\rm HP}(t)$,
*as a blue curve the impulse response  $h_{\rm TP}(t)$  of the equivalent low pass.

The Dirac delta function is the Laplace transform of the constant value  $1$,  while the second function to be subtracted gives the impulse response of the equivalent low-pass filter, which is given by the residue theorem with  
:$$Z = 0,\hspace{0.2cm} N =1,\hspace{0.2cm} K = RC.$$ }}

==Exercises for the chapter==

[[Aufgaben:Exercise_3.5:_Circuit_with_R,_L_and_C|Exercise 3.5: Circuit with R, L and C]]

[[Aufgaben:Exercise_3.5Z:_Application_of_the_Residue_Theorem|Exercise 3.5Z: Application of the Residue Theorem]]

[[Aufgaben:Exercise_3.6:_Transient_Behaviour|Exercise 3.6: Transient Behaviour]]

[[Aufgaben:Exercise_3.6Z:_Two_Imaginary_Poles|Exercise 3.6Z: Two Imaginary Poles]]

[[Aufgaben:Exercise_3.7:_Impulse_Response_of_a_High-Pass_Filter|Exercise 3.7: Impulse Response of a High-Pass Filter]]

[[Aufgaben:Exercise_3.7Z:_Partial_Fraction_Decomposition|Exercise 3.7Z: Partial Fraction Decomposition]]

{{Display}}

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-24T14:32:34Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Zwei imaginäre Polstellen und eine Nullstelle ]]
In dieser Aufgabe betrachten wir ein kausales Signal  $x(t)$  mit der Laplace–Transformierten
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
entsprechend der Grafik  (eine rote Nullstelle und zwei grüne Pole).

Das Signal  $y(t)$  besitze dagegen die Laplace–Spektralfunktion
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Die rote Nullstelle gehört somit nicht zu  $Y_{\rm L}(p)$.

Abschließend wird noch das Signal  $z(t)$  mit der Laplace–Transformierten
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
betrachtet, insbesondere der Grenzfall für  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*Die Frequenzvariable  $p$  ist so normiert, dass nach Anwendung des Residuensatzes die Zeit  $t$  in Mikrosekunden angegeben ist.
*Ein Ergebnis  $t = 1$  ist somit als  $t = T$  mit  $T = 1 \ \rm µ s$  zu interpretieren.
*Der  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  lautet am Beispiel der Funktion  $X_{\rm L}(p)$  mit zwei einfachen Polstellen bei  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Berechnen Sie das Signal  $x(t)$. Welche der folgenden Aussagen sind richtig?
|type="[]"}
+ $x(t)$  ist ein kausales Cosinussignal.
- $x(t)$  ist ein kausales Sinussignal.
+ Die Amplitude von  $x(t)$  ist  $1$.
+ Die Periodendauer von $x(t)$ ist  $T = 1 \ \rm µ s$.

{Berechnen Sie das Signal  $y(t)$. Welche der folgenden Aussagen sind richtig?
|type="[]"}
- $y(t)$  ist ein kausales Cosinussignal.
+ $y(t)$  ist ein kausales Sinussignal.
- Die Amplitude von  $y(t)$  ist  $1$.
+ Die Periodendauer von  $y(t)$  ist  $T = 1 \ \rm µ s$.

{Welche Aussagen treffen für das Signal  $z(t)$  zu?
|type="[]"}
+ Für  $ \beta > 0$  verläuft  $z(t)$  cosinusförmig.
- Für  $ \beta > 0$  verläuft  $z(t)$  sinusförmig.
+ Der Grenzfall  $\beta → 0$  führt zur Sprungfunktion  $\gamma(t)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Richtig sind die Lösungsvorschläge 1, 3 und 4:
*Durch Anwendung des Residuensatzes erhält man für das Signal  $x(t)$  bei positiven Zeiten:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  Richtig sind die Lösungsvorschläge 2 und 4:
*Prinzipiell könnte diese Teilaufgabe in gleicher Weise gelöst werden wie die Teilaufgabe  '''(1)'''.
*Man kann aber auch den Integrationssatz heranziehen.
*Dieser besagt unter anderem, dass die Multiplikation mit  $1/p$  im Spektralbereich der Integration im Zeitbereich entspricht:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Hinweis'':     Das kausale Cosinussignal  $x(t)$  sowie das kausale Sinussignal  $y(t)$  sind auf dem Angabenblatt zu  [[Aufgaben:3.6_Einschwingverhalten|Aufgabe 3.6]]  als  $c_{\rm K}(t)$  bzw.  $s_{\rm K}(t)$  dargestellt.

'''(3)'''  Richtig sind die Lösungsvorschläge 1 und 3:
*Ein Vergleich mit der Berechnung von  $x(t)$  zeigt, dass  $z(t) = \cos (\beta \cdot t)$  für  $t \ge 0$  und  $z(t) = 0$  für  $t < 0$  gilt.
*Der Grenzübergang für  $\beta → 0$  führt damit zur Sprungfunktion  $\gamma(t)$.
*Zum gleichen Ergebnis kommt man durch die Betrachtung im Spektralbereich:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6Z: Two Imaginary Poles

2021-10-24T14:27:21Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Zwei imaginäre Polstellen und eine Nullstelle ]]
In dieser Aufgabe betrachten wir ein kausales Signal  $x(t)$  mit der Laplace–Transformierten
:$$X_{\rm L}(p) =
\frac { p} { p^2 + 4 \pi^2}=
\frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
\hspace{0.05cm}$$
entsprechend der Grafik  (eine rote Nullstelle und zwei grüne Pole).

Das Signal  $y(t)$  besitze dagegen die Laplace–Spektralfunktion
:$$Y_{\rm L}(p) =
\frac { 1} { p^2 + 4 \pi^2}
\hspace{0.05cm}.$$
Die rote Nullstelle gehört somit nicht zu  $Y_{\rm L}(p)$.

Abschließend wird noch das Signal  $z(t)$  mit der Laplace–Transformierten
:$$Z_{\rm L}(p) =
\frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
\hspace{0.05cm}$$
betrachtet, insbesondere der Grenzfall für  $\beta → 0$.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*Die Frequenzvariable  $p$  ist so normiert, dass nach Anwendung des Residuensatzes die Zeit  $t$  in Mikrosekunden angegeben ist.
*Ein Ergebnis  $t = 1$  ist somit als  $t = T$  mit  $T = 1 \ \rm µ s$  zu interpretieren.
*Der  [[Linear_and_Time_Invariant_Systems/Laplace–Rücktransformation#Formulierung_des_Residuensatzes|Residuensatz]]  lautet am Beispiel der Funktion  $X_{\rm L}(p)$  mit zwei einfachen Polstellen bei  $ \pm {\rm j} \cdot \beta$:
:$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
\beta}}
\hspace{0.05cm}.$$

===Fragebogen===

<quiz display=simple>
{Berechnen Sie das Signal  $x(t)$. Welche der folgenden Aussagen sind richtig?
|type="[]"}
+ $x(t)$  ist ein kausales Cosinussignal.
- $x(t)$  ist ein kausales Sinussignal.
+ Die Amplitude von  $x(t)$  ist  $1$.
+ Die Periodendauer von $x(t)$ ist  $T = 1 \ \rm µ s$.

{Berechnen Sie das Signal  $y(t)$. Welche der folgenden Aussagen sind richtig?
|type="[]"}
- $y(t)$  ist ein kausales Cosinussignal.
+ $y(t)$  ist ein kausales Sinussignal.
- Die Amplitude von  $y(t)$  ist  $1$.
+ Die Periodendauer von  $y(t)$  ist  $T = 1 \ \rm µ s$.

{Welche Aussagen treffen für das Signal  $z(t)$  zu?
|type="[]"}
+ Für  $ \beta > 0$  verläuft  $z(t)$  cosinusförmig.
- Für  $ \beta > 0$  verläuft  $z(t)$  sinusförmig.
+ Der Grenzfall  $\beta → 0$  führt zur Sprungfunktion  $\gamma(t)$.

</quiz>

===Musterlösung===
{{ML-Kopf}}
'''(1)'''  Richtig sind die Lösungsvorschläge 1, 3 und 4:
*Durch Anwendung des Residuensatzes erhält man für das Signal  $x(t)$  bei positiven Zeiten:
:$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
\frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
\frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p
\hspace{0.05cm}\cdot \hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
\frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
\hspace{0.05cm} .$$
:$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
{1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
t}\right ] = \cos(2\pi t)
\hspace{0.05cm} .$$

'''(2)'''  Richtig sind die Lösungsvorschläge 2 und 4:
*Prinzipiell könnte diese Teilaufgabe in gleicher Weise gelöst werden wie die Teilaufgabe  '''(1)'''.
*Man kann aber auch den Integrationssatz heranziehen.
*Dieser besagt unter anderem, dass die Multiplikation mit  $1/p$  im Spektralbereich der Integration im Zeitbereich entspricht:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
\tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
\hspace{0.05cm} .$$

''Hinweis'':     Das kausale Cosinussignal  $x(t)$  sowie das kausale Sinussignal  $y(t)$  sind auf dem Angabenblatt zu  [[Aufgaben:3.6_Einschwingverhalten|Aufgabe 3.6]]  als  $c_{\rm K}(t)$  bzw.  $s_{\rm K}(t)$  dargestellt.

'''(3)'''  Richtig sind die Lösungsvorschläge 1 und 3:
*Ein Vergleich mit der Berechnung von  $x(t)$  zeigt, dass  $z(t) = \cos (\beta \cdot t)$  für  $t \ge 0$  und  $z(t) = 0$  für  $t < 0$  gilt.
*Der Grenzübergang für  $\beta → 0$  führt damit zur Sprungfunktion  $\gamma(t)$.
*Zum gleichen Ergebnis kommt man durch die Betrachtung im Spektralbereich:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
z(t) = \gamma(t)
\hspace{0.05cm} .$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T14:24:59Z

Oezer:

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T14:21:53Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  by magnitude and phase using  $H_{\rm L}(p)$ .   What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm deg$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm deg$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than $1$ to achieve the correct phase position of  $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signal  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T14:20:21Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  by magnitude and phase using  $H_{\rm L}(p)$ .   What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm deg$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm deg$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than $1$ to achieve the correct phase position of  $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T14:18:56Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  by magnitude and phase using  $H_{\rm L}(p)$ .   What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm deg$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm deg$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   to achieve the correct phase position of  $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$   to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T14:16:15Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  by magnitude and phase using  $H_{\rm L}(p)$ .   What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm deg$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm deg$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   to achieve the correct phase position of  $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T13:56:06Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  by magnitude and phase using  $H_{\rm L}(p)$ .   What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm deg$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm deg$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T13:51:53Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  by magnitude and phase using  $H_{\rm L}(p)$ .   What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm deg$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm deg$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for sample solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for sample solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T13:49:13Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  by magnitude and phase using  $H_{\rm L}(p)$ .   What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm Grad$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm Grad$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for sample solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for sample solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T13:48:16Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  by magnitude and phase using  $H_{\rm L}(p)$  using  $H_{\rm L}(p)$ .   What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm Grad$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm Grad$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for sample solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for sample solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T13:46:42Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  using  $H_{\rm L}(p)$  by magnitude and phase.  What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm Grad$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm Grad$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for sample solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for sample solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-24T13:42:36Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{for}} \\ {\rm{for}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the diagram below.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  using  $H_{\rm L}(p)$  by magnitude and phase.  What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm Grad$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm Grad$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for sample solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for sample solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-23T21:50:48Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the diagram below.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  using  $H_{\rm L}(p)$  by magnitude and phase.  What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm Grad$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm Grad$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for sample solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for sample solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Finally, the following is obtained for the last residual:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Thus, the output signal is as follows for a causal cosine signal applied to the input:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$ at this time.

* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-23T21:42:03Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the diagram below.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  using  $H_{\rm L}(p)$  by magnitude and phase.  What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm Grad$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm Grad$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for sample solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for sample solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Using Euler's theorem this can also be expressed as follows:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .

* Schließlich erhält man für das letzte Residuum:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Damit lautet das Ausgangssignal bei kausalem Cosinussignal am Eingang:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:Zum Vergleich:   Das Signal  $y_{\rm C}(t)$  hat zu diesem Zeitpunkt den Wert  $0.303$.

* Dagegen ergibt sich beim kausalen Sinussignal am Eingang allgemein und speziell zum Zeitpunkt des ersten Maximums bei  $t = 0.45 \cdot T$:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]

Aufgaben:Exercise 3.6: Transient Behavior

2021-10-23T21:38:10Z

Oezer:

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves, each causal]]
In this exercise, we consider a cosine signal  $c(t)$  with amplitude  $1$  and period $T = 1 \ \rm µ s$, which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})
\hspace{0.05cm} .$$
In contrast, the causal cosine signal  $c_{\rm K}(t)$  (red curve) starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
0 \end{array} \right.
\begin{array}{c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}}
\end{array}\begin{array}{*{20}c}
{ t \ge 0\hspace{0.05cm},} \\
{ t < 0\hspace{0.05cm}.}
\end{array}$$
Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$ :
:$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$
On the contrary, for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified:
:$$C_{\rm L}(p) =
\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
Accordingly, the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$:
:$$S_{\rm L}(p) =
\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the diagram below.

The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
:$$H_{\rm L}(p) =
\frac {2 /T} { p + 2 /T} \quad
\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad
h(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

''Please note:''
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].

*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  using  $H_{\rm L}(p)$  by magnitude and phase.  What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm Grad$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm Grad$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$ if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the length of influence  $T_h$  of the filter impulse response, that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value. Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and  $p$  by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is thus obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
\frac {f_0} { {\rm j} \cdot \pi f + f_0}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2
}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm
ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm
Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm
arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=
-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
This signal is shown dotted in blue in the left graph for sample solution  '''(5)''' .

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2
}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
This signal is sketched dotted in blue in the right graph for sample solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value. Thus, the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) =
{2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
{0.02}/{T}\hspace{0.05cm}.$$

'''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However, since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

[[File:P_ID1785__LZI_A_3_6_e.png|center|frame|Transient behavior of a causal cosine and a causal sine signal]]

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right. Note the transit time of  $T/5$  $($corresponding to the phase  $72^\circ)$  in both cases.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$   $y_{\rm C}(t)$ .
*In contrast, for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$ to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signals  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}
\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=
\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm
x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Similarly, the following is obtained for the second part:
:$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p
\hspace{0.05cm}t}
\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdot
p_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm
x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot
\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm
x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot
\hspace{0.03cm}t}
\hspace{0.05cm} .$$

* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$,
:$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}$$
:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }
\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
\hspace{0.03cm}t/T}
\hspace{0.05cm} $$ is obtained.

* Mit Hilfe des Eulerschen Satzes kann hierfür auch geschrieben werden:
:$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2
}}= y_{\rm C}(t)\hspace{0.05cm}.$$
Man erkennt, dass  $y_{1\hspace{0.03cm}+2}(t)$  gleich dem in der Teilaufgabe  '''(2)'''  berechneten Signal  $y_{\rm C}(t)$  ist.

* Schließlich erhält man für das letzte Residuum:
:$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot
\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot
\pi)} =\frac {- {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { 1+\pi^2}
\hspace{0.05cm} .$$

* Damit lautet das Ausgangssignal bei kausalem Cosinussignal am Eingang:
:$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm
e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
:Zum Vergleich:   Das Signal  $y_{\rm C}(t)$  hat zu diesem Zeitpunkt den Wert  $0.303$.

* Dagegen ergibt sich beim kausalen Sinussignal am Eingang allgemein und speziell zum Zeitpunkt des ersten Maximums bei  $t = 0.45 \cdot T$:
:$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm
e}^{\hspace{0.05cm}-2
\hspace{0.03cm}t/T}} { {1 + \pi^2
}}$$
:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm
e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2
}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]