Modulation Methods/Synchronous Demodulation - Revision history

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	In the books  "Signal Representation"  and  "Theory of Stochastic Signals”,  it was shown that		In the books  "Signal Representation"  and  "Theory of Stochastic Signals”,  it was shown that
	*the [[Signal_Representation/~~Fourier_Transform_and_its_Inverse~~#The_first_Fourier_integral\|$\text{spectrum}$]]  of a cosine signal  $x(t) = A · \cos(2πf_{\rm T}t)$  is given by:		*the [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|$\text{spectrum}$]]  of a cosine signal  $x(t) = A · \cos(2πf_{\rm T}t)$  is given by:
	:$$X(f) = \frac{A}{2}\cdot \delta(f + f_{\rm T}) + \frac{A}{2}\cdot \delta(f - f_{\rm T}) \hspace{0.05cm}, $$		:$$X(f) = \frac{A}{2}\cdot \delta(f + f_{\rm T}) + \frac{A}{2}\cdot \delta(f - f_{\rm T}) \hspace{0.05cm}, $$

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	*The power-spectral density  $\rm (PSD)$  after the low-pass filter is exactly as large for  $ \vert f \vert < B_{\rm NF}$  and zero beyond this range.		*The power-spectral density  $\rm (PSD)$  after the low-pass filter is exactly as large for  $ \vert f \vert < B_{\rm NF}$  and zero beyond this range.
	:$${\it \Phi}_\varepsilon (f) = \left\{ \begin{array}{c} N_0 \\ 0 \\ \end{array} \right.\quad		:$${\it \Phi}_\varepsilon (f) = \left\{ \begin{array}{c} N_0 \\ 0 \\ \end{array} \right.\quad\begin{array}{{10}c} {\rm{for} }\\ \\ \end{array}\begin{array}{{20}c}\vert f \vert < B_{\rm NF} \hspace{0.05cm}, \\ {\rm otherwise} \hspace{0.05cm}. \\ \end{array}$$
	\begin{array}{{10}c} {\rm{for} }\\ \\ \end{array}\begin{array}{{20}c}
	\vert f \vert < B_{\rm NF} \hspace{0.05cm}, \\ {\rm otherwise} \hspace{0.05cm}. \\ \end{array}$$

	*By integration,  we obtain the power  $P_ε = 2N_0 · B_{\rm NF}$.   Thus,  with this intermediate result,  the  "sink SNR"  can be given as:		*By integration,  we obtain the power  $P_ε = 2N_0 · B_{\rm NF}$.   Thus,  with this intermediate result,  the  "sink SNR"  can be given as:
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	:$$s(t) = \big[ q(t) + A_{\rm T}\big] \cdot \cos(\omega_{\rm T} \cdot t ) = A_{\rm T} \cdot \cos(\omega_{\rm T} \cdot t ) + {A_{\rm N} }/{2}\cdot \cos\big [(\omega_{\rm T}+ \omega_{\rm N}) \cdot t \big] + {A_{\rm N} }/{2}\cdot \cos\big[(\omega_{\rm T}- \omega_{\rm N}) \cdot t \big]\hspace{0.05cm}.$$		:$$s(t) = \big[ q(t) + A_{\rm T}\big] \cdot \cos(\omega_{\rm T} \cdot t ) = A_{\rm T} \cdot \cos(\omega_{\rm T} \cdot t ) + {A_{\rm N} }/{2}\cdot \cos\big [(\omega_{\rm T}+ \omega_{\rm N}) \cdot t \big] + {A_{\rm N} }/{2}\cdot \cos\big[(\omega_{\rm T}- \omega_{\rm N}) \cdot t \big]\hspace{0.05cm}.$$
	*The power of the source signal,  in relation to a resistance of  $1 \ \rm Ω$,  with the period  $T_{\rm N}$,  is:		*The power of the source signal,  in relation to a resistance of  $1 \ \rm Ω$,  with the period  $T_{\rm N}$,  is:
	:$$P_{q} = \frac{1}{T_{\rm N} }\cdot\int_{0}^{ T_{\rm N} }		:$$P_{q} = \frac{1}{T_{\rm N} }\cdot\int_{0}^{ T_{\rm N} }{q^2(t)}\hspace{0.1cm}{\rm d}t = \frac{A_{\rm N}^2}{T_{\rm N} }\cdot\int_{0}^{T_{\rm N} }{\cos^2(2 \pi\cdot{t}/{T_{\rm N} })}\hspace{0.1cm}{\rm d}t = \frac{A_{\rm N}^2}{2}\hspace{0.05cm}.$$
	{q^2(t)}\hspace{0.1cm}{\rm d}t = \frac{A_{\rm N}^2}{T_{\rm N} }\cdot\int_{0}^{T_{\rm N} }
	{\cos^2(2 \pi\cdot{t}/{T_{\rm N} })}\hspace{0.1cm}{\rm d}t = \frac{A_{\rm N}^2}{2}\hspace{0.05cm}.$$

	*Accordingly,  for the power of the transmitted signal,  one obtains:		*Accordingly,  for the power of the transmitted signal,  one obtains:
	:$$P_{\rm S} = \frac{A_{\rm T}^2}{2} + \frac{(A_{\rm N}/2)^2}{2} + \frac{(A_{\rm N}/2)^2}{2} =		:$$P_{\rm S} = \frac{A_{\rm T}^2}{2} + \frac{(A_{\rm N}/2)^2}{2} + \frac{(A_{\rm N}/2)^2}{2} =\frac{A_{\rm T}^2}{2} + \frac{A_{\rm N}^2}{4}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_{\rm S} = {1}/{2} \cdot \left( P_q + A_{\rm T}^2 \right)\hspace{0.05cm}.$$
	\frac{A_{\rm T}^2}{2} + \frac{A_{\rm N}^2}{4}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_{\rm S} = {1}/{2} \cdot \left( P_q + A_{\rm T}^2 \right)
	\hspace{0.05cm}.$$

	*This equation is valid for both  »DSB–AM without carrier«  $(A_{\rm T} = 0)$  and  »DSB–AM with carrier«.  Since  $q(t)$  was assumed to be a harmonic oscillation,  »DSB–AM without carrier«  given a modulation depth  $m = A_{\rm N}/A_{\rm T}$,  this can also be written as:		*This equation is valid for both  »DSB–AM without carrier«  $(A_{\rm T} = 0)$  and  »DSB–AM with carrier«.  Since  $q(t)$  was assumed to be a harmonic oscillation,  »DSB–AM without carrier«  given a modulation depth  $m = A_{\rm N}/A_{\rm T}$,  this can also be written as:
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	[[File: EN_Mod_T_2_2_S7b.png\|right\|frame\|Linear und double–logarithmic representation of sink SNR]]		[[File: EN_Mod_T_2_2_S7b.png\|right\|frame\|Linear und double–logarithmic representation of sink SNR]]
	The following is a detailed discussion of this equation.  Already in the chapter  [[Modulation_Methods/Quality_Criteria#Investigations_at_the_AWGN_channel\|"Investigations at the AWGN channel"]]  it was justified why it makes sense to state the sink SNR  $ρ_v$  as a function of the performance parameter  $ξ$  named below:		The following is a detailed discussion of this equation.  Already in the chapter  [[Modulation_Methods/Quality_Criteria#Investigations_at_the_AWGN_channel\|"Investigations at the AWGN channel"]]  it was justified why it makes sense to state the sink SNR  $ρ_v$  as a function of the performance parameter  $ξ$  named below:
	:$$\xi = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}}		:$$\xi = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}}\hspace{0.5cm}\Rightarrow \hspace{0.5cm} \rho_v = \frac{\xi}{1 + {2}/{m^2}}\hspace{0.05cm}.$$
	\hspace{0.5cm}\Rightarrow \hspace{0.5cm} \rho_v = \frac{\xi}{1 + {2}/{m^2}}
	\hspace{0.05cm}.$$

	The two graphs show the corresponding curves		The two graphs show the corresponding curves
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	*For  »DSB–AM with carrier«  with a modulation depth  $m$:		*For  »DSB–AM with carrier«  with a modulation depth  $m$:
	:$$\rho_v = \xi \cdot \frac{1}{1 + {2}/{m^2}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$\rho_v = \xi \cdot \frac{1}{1 + {2}/{m^2}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg }\hspace{0.1cm}\rho_v = 10 \cdot {\rm lg }\hspace{0.1cm}\xi - 10 \cdot {\rm lg }\hspace{0.1cm} \left({1 + {2}/{m^2}}\right)\hspace{0.05cm}.$$
	10 \cdot {\rm lg }\hspace{0.1cm}\rho_v = 10 \cdot {\rm lg }\hspace{0.1cm}\xi - 10 \cdot {\rm lg }\hspace{0.1cm} \left({1 + {2}/{m^2}}\right)\hspace{0.05cm}.$$

	*In the double logarithmic representation  (see right graph),  this leads to a downward parallel shift of the curves,  <br>e.g. at  $m = 1$  by  $4.77$ dB  and at  $m = 0.5$  by  $9.54$ dB.		*In the double logarithmic representation  (see right graph),  this leads to a downward parallel shift of the curves,  <br>e.g. at  $m = 1$  by  $4.77$ dB  and at  $m = 0.5$  by  $9.54$ dB.

Maintenance script: Add German interlanguage link

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	{{Display}}		{{Display}}
			[[de:Modulationsverfahren/Synchrondemodulation]]

Guenter at 09:32, 13 January 2023

2023-01-13T09:32:42Z

Show changes

Guenter at 09:24, 13 January 2023

2023-01-13T09:24:05Z

Hwang at 20:56, 2 January 2023

2023-01-02T20:56:07Z

← Older revision		Revision as of 22:56, 2 January 2023
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	[[Aufgaben:Aufgabe_2.4Z:_Tiefpass-Einfluss_bei_Synchrondemodulation\|Exercise 2.4Z: Low-pass influence with Synchronous Demodulation]]		[[Aufgaben:Aufgabe_2.4Z:_Tiefpass-Einfluss_bei_Synchrondemodulation\|Exercise 2.4Z: Low-pass influence with Synchronous Demodulation]]

	[[Aufgaben:~~Aufgabe_2~~.5:~~_ZSB–AM_über_einen_Gaußkanal~~\|Exercise 2.5: DSB-AM via a Gaussian Channel]]		[[Aufgaben:Exercise_2.5:_DSB-AM_via_a_Gaussian_channel\|Exercise 2.5: DSB-AM via a Gaussian Channel]]

	[[Aufgaben:Exercise_2.5Z:_Linear_Distortions_with_DSB-AM\|Exercise 2.5Z: Linear Distortions with DSB-AM]]		[[Aufgaben:Exercise_2.5Z:_Linear_Distortions_with_DSB-AM\|Exercise 2.5Z: Linear Distortions with DSB-AM]]

Hwang at 20:54, 2 January 2023

2023-01-02T20:54:46Z

← Older revision		Revision as of 22:54, 2 January 2023
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	*When  $m = 3$,  it is only an insignificant deterioration of less than one dB compared to  »DSB–AM without carrier«.		*When  $m = 3$,  it is only an insignificant deterioration of less than one dB compared to  »DSB–AM without carrier«.

	==Exercises for the ~~Chapter~~==		==Exercises for the chapter==
	<br>		<br>
	[[Aufgaben:Aufgabe_2.4:_Frequenz–_und_Phasenversatz\|Exercise 2.4: Frequency and Phase Offset]]		[[Aufgaben:Aufgabe_2.4:_Frequenz–_und_Phasenversatz\|Exercise 2.4: Frequency and Phase Offset]]

Hwang at 20:54, 2 January 2023

2023-01-02T20:54:30Z

← Older revision		Revision as of 22:54, 2 January 2023
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	*All statements are valid under the assumption of an ideal synchronous demodulator.  In this case,  the  »DSB–AM with carrier«  method actually makes no sense.  The added carrier would only lead to an unnecessarily large transmit power and could not be used for demodulation.		*All statements are valid under the assumption of an ideal synchronous demodulator.  In this case,  the  »DSB–AM with carrier«  method actually makes no sense.  The added carrier would only lead to an unnecessarily large transmit power and could not be used for demodulation.
	*The curves are valid for perfect frequency and phase synchronization.  However,  in order to be able to determine the parameters  $f_{\rm T}$  and  $\mathbf{ϕ_{\rm T} }$  from the received signal  $r(t)$  with less effort,  a small carrier component in the transmitted signal makes sense.		*The curves are valid for perfect frequency and phase synchronization.  However,  in order to be able to determine the parameters  $f_{\rm T}$  and  $\mathbf{ϕ_{\rm T} }$  from the received signal  $r(t)$  with less effort,  a small carrier component in the transmitted signal makes sense.
	*When  $m = 3$,  it is only an ~~insignifican~~ deterioration of less than one dB compared to  »DSB–AM without carrier«.		*When  $m = 3$,  it is only an insignificant deterioration of less than one dB compared to  »DSB–AM without carrier«.

	==Exercises for the Chapter==		==Exercises for the Chapter==

Hwang at 20:45, 2 January 2023

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	==Relationship between the powers from source signal and transmitted signal==		==Relationship between the powers from source signal and transmitted signal==
	<br>		<br>
	To characterise the relationship between the sink–SNR  $\rho_v$  and the transmission power  $P_{\rm S}$,  we first need the relationships between		To characterise the relationship between the sink–SNR  $\rho_v$  and the transmission '''KORREKTUR: transmit?''' power  $P_{\rm S}$,  we first need the relationships between
	*source signal  $q(t)$   ⇒   power  $P_q$,  and		*source signal  $q(t)$   ⇒   power  $P_q$,  and
	*transmitted signal  $s(t)$   ⇒   transmission power  $P_{\rm S}$.		*transmitted signal  $s(t)$   ⇒   transmission '''KORREKTUR: transmit?''' power  $P_{\rm S}$.


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	The curves are to be interpreted as follows:		The curves are to be interpreted as follows:
	*For the system variant  »DSB–AM without carrier«  with  $m → ∞$  the above equation yields the simple relationship  $ρ_v = ξ$.  This gives the bisector of the angle for both the linear and double logarithmic plots.		*For the system variant  »DSB–AM without carrier«  with  $m → ∞$  the above equation yields the simple relationship  $ρ_v = ξ$.  This gives the bisector of the angle for both the linear and double logarithmic plots.
	*A greater transmission power  $P_{\rm S}$  leads to a better sink SNR,  as does a larger transmission factor   $α_{\rm K}$  (⇒ lower attenuation).  However,   $10 · \lg ρ_v$  is also increased by a smaller noise power density  $N_0$  and a smaller bandwidth  $B_{\rm NF}$,  other things being equal.		*A greater transmission '''KORREKTUR: transmit?''' power  $P_{\rm S}$  leads to a better sink SNR,  as does a larger transmission factor   $α_{\rm K}$  (⇒ lower attenuation).  However,   $10 · \lg ρ_v$  is also increased by a smaller noise power density  $N_0$  and a smaller bandwidth  $B_{\rm NF}$,  other things being equal.
	*For  »DSB–AM with carrier«  with a modulation depth  $m$:		*For  »DSB–AM with carrier«  with a modulation depth  $m$:
	:$$\rho_v = \xi \cdot \frac{1}{1 + {2}/{m^2}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$\rho_v = \xi \cdot \frac{1}{1 + {2}/{m^2}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}

Hwang at 20:44, 2 January 2023

2023-01-02T20:44:25Z

← Older revision		Revision as of 22:44, 2 January 2023
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	\vert f \vert < B_{\rm NF} \hspace{0.05cm}, \\ {\rm otherwise} \hspace{0.05cm}. \\ \end{array}$$		\vert f \vert < B_{\rm NF} \hspace{0.05cm}, \\ {\rm otherwise} \hspace{0.05cm}. \\ \end{array}$$

	*By integration,  we obtain the power  $P_ε = 2N_0 · B_{\rm NF}$.   Thus,  with this intermediate result,  the sinking '''KORREKTUR: sink'''SNR can be given as:		*By integration,  we obtain the power  $P_ε = 2N_0 · B_{\rm NF}$.   Thus,  with this intermediate result,  the sinking '''KORREKTUR: sink''' SNR can be given as:
	:$$\rho_v = \frac{\alpha_{\rm K}^2 \cdot P_q}{P_\varepsilon} = \frac{\alpha_{\rm K}^2 \cdot P_q}{N_0 \cdot B_{\rm NF} } \hspace{0.05cm}.$$		:$$\rho_v = \frac{\alpha_{\rm K}^2 \cdot P_q}{P_\varepsilon} = \frac{\alpha_{\rm K}^2 \cdot P_q}{N_0 \cdot B_{\rm NF} } \hspace{0.05cm}.$$

Hwang at 20:44, 2 January 2023

2023-01-02T20:44:08Z

← Older revision		Revision as of 22:44, 2 January 2023
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	\vert f \vert < B_{\rm NF} \hspace{0.05cm}, \\ {\rm otherwise} \hspace{0.05cm}. \\ \end{array}$$		\vert f \vert < B_{\rm NF} \hspace{0.05cm}, \\ {\rm otherwise} \hspace{0.05cm}. \\ \end{array}$$

	*By integration,  we obtain the power  $P_ε = 2N_0 · B_{\rm NF}$.   Thus,  with this intermediate result,  the sinking SNR can be given as:		*By integration,  we obtain the power  $P_ε = 2N_0 · B_{\rm NF}$.   Thus,  with this intermediate result,  the sinking '''KORREKTUR: sink'''SNR can be given as:
	:$$\rho_v = \frac{\alpha_{\rm K}^2 \cdot P_q}{P_\varepsilon} = \frac{\alpha_{\rm K}^2 \cdot P_q}{N_0 \cdot B_{\rm NF} } \hspace{0.05cm}.$$		:$$\rho_v = \frac{\alpha_{\rm K}^2 \cdot P_q}{P_\varepsilon} = \frac{\alpha_{\rm K}^2 \cdot P_q}{N_0 \cdot B_{\rm NF} } \hspace{0.05cm}.$$

← Older revision		Revision as of 22:56, 2 January 2023
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	[[Aufgaben:Aufgabe_2.4Z:_Tiefpass-Einfluss_bei_Synchrondemodulation\|Exercise 2.4Z: Low-pass influence with Synchronous Demodulation]]		[[Aufgaben:Aufgabe_2.4Z:_Tiefpass-Einfluss_bei_Synchrondemodulation\|Exercise 2.4Z: Low-pass influence with Synchronous Demodulation]]

	[[Aufgaben:~~Aufgabe_2~~.5:~~_ZSB–AM_über_einen_Gaußkanal~~\|Exercise 2.5: DSB-AM via a Gaussian Channel]]		[[Aufgaben:Exercise_2.5:_DSB-AM_via_a_Gaussian_channel\|Exercise 2.5: DSB-AM via a Gaussian Channel]]

	[[Aufgaben:Exercise_2.5Z:_Linear_Distortions_with_DSB-AM\|Exercise 2.5Z: Linear Distortions with DSB-AM]]		[[Aufgaben:Exercise_2.5Z:_Linear_Distortions_with_DSB-AM\|Exercise 2.5Z: Linear Distortions with DSB-AM]]

← Older revision		Revision as of 11:24, 13 January 2023
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	:$$B(f) = R(f) \star Z_{\rm E}(f)\hspace{0.05cm}.$$		:$$B(f) = R(f) \star Z_{\rm E}(f)\hspace{0.05cm}.$$

	*Convolution of the Dirac delta function  $ - {\rm j} \cdot δ(f – f_{\rm T})$  with the ~~wholly '''KORREKTUR:~~ purely~~'''~~ imaginary spectrum  $R(f)$  leads to completely real spectral components around  $f = 0$  and  $f = 2f_{\rm T}$.  These components are marked with a  "plus"  on the lower graph.		*Convolution of the Dirac delta function  $ - {\rm j} \cdot δ(f – f_{\rm T})$  with the purely imaginary spectrum  $R(f)$  leads to completely real spectral components around  $f = 0$  and  $f = 2f_{\rm T}$.  These components are marked with a  "plus"  on the lower graph.

	*The second convolution product  ${\rm j} · δ(f + f_{\rm T}) \star R(f)$  yields a low-frequency spectral component around $f = 0$  in addition to a component at  $–2f_{\rm T}$.  These spectral components are marked with  "minus".		*The second convolution product  ${\rm j} · δ(f + f_{\rm T}) \star R(f)$  yields a low-frequency spectral component around $f = 0$  in addition to a component at  $–2f_{\rm T}$.  These spectral components are marked with  "minus".
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	In particular,  we make the following assumptions:		In particular,  we make the following assumptions:
	*Only double-sideband amplitude modulation with modulation depth  $m$  and an ideal synchronous demodulator without phase and frequency offsets are considered.		*Only double-sideband amplitude modulation with modulation depth  $m$  and an ideal synchronous demodulator without phase and frequency offsets are considered.
	*According to the extended AWGN channel model,  where  $α_{\rm K}$  is a frequency-independent transmission factor of the channel and the interference ~~'''KORREKTUR:~~ noise~~'''~~ signal  $n(t)$  models white noise with two-sided noise power density  $N_0/2$,  the following holds for the received signal:		*According to the extended AWGN channel model,  where  $α_{\rm K}$  is a frequency-independent transmission factor of the channel and the interference  $($noise$)$  signal  $n(t)$  models white noise with two-sided noise power density  $N_0/2$,  the following holds for the received signal:
	:$$r(t) = \alpha_{\rm K} \cdot s(t) + n(t) \hspace{0.05cm}.$$		:$$r(t) = \alpha_{\rm K} \cdot s(t) + n(t) \hspace{0.05cm}.$$
	*Representing a source signal  $q(t)$  of bandwidth  $B_{\rm NF}$ , a cosine-shaped message signal of frequency  $B_{\rm NF}$  is assumed here:		*Representing a source signal  $q(t)$  of bandwidth  $B_{\rm NF}$ , a cosine-shaped message signal of frequency  $B_{\rm NF}$  is assumed here:
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	\vert f \vert < B_{\rm NF} \hspace{0.05cm}, \\ {\rm otherwise} \hspace{0.05cm}. \\ \end{array}$$		\vert f \vert < B_{\rm NF} \hspace{0.05cm}, \\ {\rm otherwise} \hspace{0.05cm}. \\ \end{array}$$

	*By integration,  we obtain the power  $P_ε = 2N_0 · B_{\rm NF}$.   Thus,  with this intermediate result,  the ~~sinking '''KORREKTUR:~~ sink~~'''~~ SNR can be given as:		*By integration,  we obtain the power  $P_ε = 2N_0 · B_{\rm NF}$.   Thus,  with this intermediate result,  the  "sink SNR"  can be given as:
	:$$\rho_v = \frac{\alpha_{\rm K}^2 \cdot P_q}{P_\varepsilon} = \frac{\alpha_{\rm K}^2 \cdot P_q}{N_0 \cdot B_{\rm NF} } \hspace{0.05cm}.$$		:$$\rho_v = \frac{\alpha_{\rm K}^2 \cdot P_q}{P_\varepsilon} = \frac{\alpha_{\rm K}^2 \cdot P_q}{N_0 \cdot B_{\rm NF} } \hspace{0.05cm}.$$

	In the next section,  we still establish the relationship between the power  $P_q$  of the source signal and the ~~transmission '''KORREKTUR:~~ transmit~~?'''~~ power  $P_{\rm S}$.}}		In the next section,  we still establish the relationship between the power  $P_q$  of the source signal and the transmit power  $P_{\rm S}$.}}

	==Relationship between the powers from source signal and transmitted signal==		==Relationship between the powers from source signal and transmitted signal==
	<br>		<br>
	To characterise the relationship between the sink–SNR  $\rho_v$  and the ~~transmission '''KORREKTUR:~~ transmit~~?'''~~ power  $P_{\rm S}$,  we first need the relationships between		To characterise the relationship between the sink–SNR  $\rho_v$  and the transmit power  $P_{\rm S}$,  we first need the relationships between
	*source signal  $q(t)$   ⇒   power  $P_q$,  and		*source signal  $q(t)$   ⇒   power  $P_q$,  and
	*transmitted signal  $s(t)$   ⇒   ~~transmission '''KORREKTUR:~~ transmit~~?'''~~ power  $P_{\rm S}$.		*transmitted signal  $s(t)$   ⇒   transmit power  $P_{\rm S}$.


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	The curves are to be interpreted as follows:		The curves are to be interpreted as follows:
	*For the system variant  »DSB–AM without carrier«  with  $m → ∞$  the above equation yields the simple relationship  $ρ_v = ξ$.  This gives the bisector of the angle for both the linear and double logarithmic plots.		*For the system variant  »DSB–AM without carrier«  with  $m → ∞$  the above equation yields the simple relationship  $ρ_v = ξ$.  This gives the bisector of the angle for both the linear and double logarithmic plots.
	*A greater ~~transmission '''KORREKTUR:~~ transmit~~?'''~~ power  $P_{\rm S}$  leads to a better sink SNR,  as does a larger transmission factor   $α_{\rm K}$  (⇒ lower attenuation).  However,   $10 · \lg ρ_v$  is also increased by a smaller noise power density  $N_0$  and a smaller bandwidth  $B_{\rm NF}$,  other things being equal.		*A greater transmit power  $P_{\rm S}$  leads to a better sink SNR,  as does a larger transmission factor   $α_{\rm K}$  (⇒ lower attenuation).  However,   $10 · \lg ρ_v$  is also increased by a smaller noise power density  $N_0$  and a smaller bandwidth  $B_{\rm NF}$,  other things being equal.
	*For  »DSB–AM with carrier«  with a modulation depth  $m$:		*For  »DSB–AM with carrier«  with a modulation depth  $m$:
	:$$\rho_v = \xi \cdot \frac{1}{1 + {2}/{m^2}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$\rho_v = \xi \cdot \frac{1}{1 + {2}/{m^2}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}