Aufgaben:Exercise 1.3: Rectangular Functions for Transmitter and Receiver - Revision history

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	'''2.'''  The variance of the noise component of the detection signal – referred to as the  "detection noise power" – can be calculated in both the time and frequency domains.		'''2.'''  The variance of the noise component of the detection signal – referred to as the  "detection noise power" – can be calculated in both the time and frequency domains.
	*For the present rectangular waveform,  calculation in the time domain yields faster results:		*For the present rectangular waveform,  calculation in the time domain yields faster results:
	:$$\sigma _d ^2 \ = \ \frac{N_0 }{2} \cdot \int_{ -		:$$\sigma _d ^2 \ = \ \frac{N_0 }{2} \cdot \int_{ -\infty }^{ + \infty } {\left\| {h_{\rm E}( t )} \right\|^2\hspace{0.1cm}{\rm{d}}t} =\frac{N_0 }{2} \cdot \int_{ -T/2 }^{ + T/2 } {\left\| {h_{\rm E}( t )} \right\|^2\hspace{0.1cm}{\rm{d}}t} = \ \frac{N_0 }{2} \cdot\frac{1}{T^2} \cdot T = \frac{N_0 }{2T} = \frac{2 \cdot 10^{-5}\,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\rmW}}\hspace{0.05cm}.$$
	\infty }^{ + \infty } {\left\| {h_{\rm E}( t )} \right\|^2
	\hspace{0.1cm}{\rm{d}}t} =\frac{N_0 }{2} \cdot \int_{ -
	T/2 }^{ + T/2 } {\left\| {h_{\rm E}( t )} \right\|^2
	\hspace{0.1cm}{\rm{d}}t} = \ \frac{N_0 }{2} \cdot\frac{1
	}{T^2} \cdot T = \frac{N_0 }{2T} = \frac{2 \cdot 10^{-5}
	\,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\rm
	W}}\hspace{0.05cm}.$$
	*The frequency domain calculation would be as follows with  $H_{\rm E}(f) = {\rm sinc}(fT)$:		*The frequency domain calculation would be as follows with  $H_{\rm E}(f) = {\rm sinc}(fT)$:
	:$$\sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{+ \infty } {\left\| {H_{\rm E}( f )} \right\|^2\hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{-\infty }^{ \infty } {\rm sinc}^2(f T)\hspace{0.1cm}{\rm{d}}f =\frac{N_0 }{2T} \hspace{0.05cm}.$$		:$$\sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{+ \infty } {\left\| {H_{\rm E}( f )} \right\|^2\hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{-\infty }^{ \infty } {\rm sinc}^2(f T)\hspace{0.1cm}{\rm{d}}f =\frac{N_0 }{2T} \hspace{0.05cm}.$$

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	*You can use the interactive applet  [[Applets:Complementary_Gaussian_Error_Functions\|"Complementary Gaussian Error Functions"]]  to determine error probabilities.		*You can use the interactive applet  [[Applets:Complementary_Gaussian_Error_Functions\|"Complementary Gaussian Error Functions"]]  to determine error probabilities.
	*Consider  [[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem\|"Wiener-Khintchine's theorem"]]  when calculating the detection noise power:		*Consider  [[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem\|"Wiener-Khintchine's theorem"]]  when calculating the detection noise power:
	:$$ \sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{		:$$ \sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{+ \infty } {\left\| {H_{\rm E}( f )} \right\|^2\hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{ -\infty }^{ + \infty } {\left\| {h_{\rm E}( t )} \right\|^2\hspace{0.1cm}{\rm{d}}t}\hspace{0.05cm}.$$
	+ \infty } {\left\| {H_{\rm E}( f )} \right\|^2
	\hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{ -
	\infty }^{ + \infty } {\left\| {h_{\rm E}( t )} \right\|^2
	\hspace{0.1cm}{\rm{d}}t}\hspace{0.05cm}.$$


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	{{ML-Kopf}}		{{ML-Kopf}}
	'''1.'''  For '''System A''',  the convolution of the two equal-width rectangular functions  $g_{s}(t)$  and  $h_{\rm E}(t)$  leads to a triangular detection pulse with the maximum at  $t = 0$:		'''1.'''  For '''System A''',  the convolution of the two equal-width rectangular functions  $g_{s}(t)$  and  $h_{\rm E}(t)$  leads to a triangular detection pulse with the maximum at  $t = 0$:
	:$$g_d (t = 0) = \int_{ - T/2}^{		:$$g_d (t = 0) = \int_{ - T/2}^{+ T/2} { g_s(t) \cdot h_{\rm E}( t )} \hspace{0.1cm}{\rm{d}}t =s_0\cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rmW}}}\hspace{0.05cm}.$$
	+ T/2} { g_s(t) \cdot h_{\rm E}( t )} \hspace{0.1cm}{\rm{d}}t =s_0
	\cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm
	W}}}\hspace{0.05cm}.$$
	There is no intersymbol interfering because for  $\| t \|\ge T$  the detection pulse is  $g_{d}(t) = 0$.		There is no intersymbol interfering because for  $\| t \|\ge T$  the detection pulse is  $g_{d}(t) = 0$.

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	W}}\hspace{0.05cm}.$$		W}}\hspace{0.05cm}.$$
	*The frequency domain calculation would be as follows with  $H_{\rm E}(f) = {\rm sinc}(fT)$:		*The frequency domain calculation would be as follows with  $H_{\rm E}(f) = {\rm sinc}(fT)$:
	:$$\sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{		:$$\sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{+ \infty } {\left\| {H_{\rm E}( f )} \right\|^2\hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{-\infty }^{ \infty } {\rm sinc}^2(f T)\hspace{0.1cm}{\rm{d}}f =\frac{N_0 }{2T} \hspace{0.05cm}.$$
	+ \infty } {\left\| {H_{\rm E}( f )} \right\|^2
	\hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{-
	\infty }^{ \infty } {\rm sinc}^2(f T)\hspace{0.1cm}{\rm{d}}f =
	\frac{N_0 }{2T} \hspace{0.05cm}.$$


	'''3.'''  Due to the time-limited pulse shape  (this means:  no intersymbol interfering!),  the bipolar approach assumed here yields:		'''3.'''  Due to the time-limited pulse shape  (this means:  no intersymbol interfering!),  the bipolar approach assumed here yields:
	:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {\rm Q} \left( \frac{ 6 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)		:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {\rm Q} \left( \frac{ 6 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)= {\rm Q}(6) \hspace{0.1cm}\underline {= 0.987 \cdot 10^{-9}} \hspace{0.05cm}.$$
	= {\rm Q}(6) \hspace{0.1cm}\underline {= 0.987 \cdot 10^{-9}} \hspace{0.05cm}.$$
	*'''System A'''  represents the matched filter realization of the optimal binary receiver,  so the following equations would also be applicable:		*'''System A'''  represents the matched filter realization of the optimal binary receiver,  so the following equations would also be applicable:
	:$$E_{\rm B} = s_0^2 \cdot T = 36\, {\rm W} \cdot 10^{-5} {\rm s}\hspace{0.3cm}		:$$E_{\rm B} = s_0^2 \cdot T = 36\, {\rm W} \cdot 10^{-5} {\rm s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)={\rm Q} \left( \sqrt{\frac{2 \cdot 36 \cdot 10^{-5}\,\, {\rm Ws}}{2 \cdot 10^{-5} \,\, {\rmWs}}}\right)={\rm Q}(6)\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)
	={\rm Q} \left( \sqrt{\frac{2 \cdot 36 \cdot 10^{-5}\,\, {\rm Ws}}{2 \cdot 10^{-5} \,\, {\rm
	Ws}}}\right)={\rm Q}(6)
	\hspace{0.05cm}.$$


	'''4.'''  Since  '''System B'''  uses the same receiver filter as  '''System A''',  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.		'''4.'''  Since  '''System B'''  uses the same receiver filter as  '''System A''',  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.
	*However,  the basic detection pulse is now no longer triangular,  but has a more pointed shape.  At time $t = 0$ applies:		*However,  the basic detection pulse is now no longer triangular,  but has a more pointed shape.  At time $t = 0$ applies:
	:$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{		:$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot\frac{s_0 }{2} \cdot T = \frac{s_0 }{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rmW}}\hspace{0.05cm}.$$
	+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot
	\frac{s_0 }{2} \cdot T = \frac{s_0 }{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm
	W}}\hspace{0.05cm}.$$
	*'''System B'''  is also free of intersymbol interfering.  Therefore,  one obtains for the bit error probability:		*'''System B'''  is also free of intersymbol interfering.  Therefore,  one obtains for the bit error probability:
	:$$p_{\rm B} = {\rm Q} \left( \frac{g_d (t = 0)}{\sigma_d}\right)= {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)		:$$p_{\rm B} = {\rm Q} \left( \frac{g_d (t = 0)}{\sigma_d}\right)= {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)= {\rm Q}(3) \hspace{0.1cm}\underline {= 0.135 \cdot 10^{-2}} \hspace{0.05cm}.$$
	= {\rm Q}(3) \hspace{0.1cm}\underline {= 0.135 \cdot 10^{-2}} \hspace{0.05cm}.$$
	*On the other hand,  the following calculation is  '''not'''  applicable here:		*On the other hand,  the following calculation is  '''not'''  applicable here:
	:$$E_{\rm B} = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm		:$$E_{\rm B} = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rmd}t = 2\cdot s_0^2 \cdot \int ^{+T/2} _{0} \left( 1- \frac{2t}{T}\right)^2\,{\rmd}t = \frac{s_0^2 \cdot T }{3} = 12 \cdot 10^{-5} \,{\rm Ws}$$
	d}t = 2\cdot s_0^2 \cdot \int ^{+T/2} _{0} \left( 1- \frac{2t}{T}\right)^2\,{\rm		:$$\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4}\hspace{0.05cm}.$$
	d}t = \frac{s_0^2 \cdot T }{3} = 12 \cdot 10^{-5} \,{\rm Ws}$$
	:$$\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)
	={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4}
	\hspace{0.05cm}.$$
	*One would thus compute a bit error probability that is too low,  since the implicit assumption of a matched filter does not hold.		*One would thus compute a bit error probability that is too low,  since the implicit assumption of a matched filter does not hold.

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	'''5.'''  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   '''System C''', <br>the same basic detection pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.		'''5.'''  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   '''System C''', <br>the same basic detection pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
	*Therefore,  as in  '''System B''':		*Therefore,  as in  '''System B''':
	:$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm		:$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rmW}}\hspace{0.05cm}.$$
	W}}\hspace{0.05cm}.$$
	*In contrast,  the detection noise power is now smaller than in systems  '''A'''  and  '''B''':		*In contrast,  the detection noise power is now smaller than in systems  '''A'''  and  '''B''':
	:$$\sigma _d ^2 = \frac{N_0}{2} \cdot \frac{1}{T^2} \cdot \int^{+T/2} _{-T/2} \left( 1- \frac{2t}{T}\right)^2\,{\rm		:$$\sigma _d ^2 = \frac{N_0}{2} \cdot \frac{1}{T^2} \cdot \int^{+T/2} _{-T/2} \left( 1- \frac{2t}{T}\right)^2\,{\rmd}t = \frac{N_0}{6T}\hspace{0.1cm}\underline { = 0.333 \,{\rm W}}.$$
	d}t = \frac{N_0}{6T}\hspace{0.1cm}\underline { = 0.333 \,{\rm W}}.$$
	*This now gives us for the bit error probability:		*This now gives us for the bit error probability:
	:$$p_{\rm B} = {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{0.577 \,\sqrt{\rm W}}\right)		:$$p_{\rm B} = {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{0.577 \,\sqrt{\rm W}}\right)\approx {\rm Q}(5.2)\hspace{0.1cm}\underline { \approx 10^{-7} } \hspace{0.05cm}.$$
	\approx {\rm Q}(5.2)\hspace{0.1cm}\underline { \approx 10^{-7} } \hspace{0.05cm}.$$
	*The apparent increase in error probability by a factor of about  $100$  compared to subtask  '''(3)'''  is due to the severe mismatch compared to the matched filter.		*The apparent increase in error probability by a factor of about  $100$  compared to subtask  '''(3)'''  is due to the severe mismatch compared to the matched filter.
	*The improvement over subtask  '''(4)'''  is due to the higher signal energy.		*The improvement over subtask  '''(4)'''  is due to the higher signal energy.

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	[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]
			[[de:Aufgaben:1.3_Einfluss_von_g(s)_von_(t)_und_h(E)_von_(t)]]

Guenter at 07:07, 20 May 2022

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	'''4.'''  Since  '''System B'''  uses the same receiver filter as  '''System A''',  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.		'''4.'''  Since  '''System B'''  uses the same receiver filter as  '''System A''',  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.
	*However,  the basic detection pulse ~~'''KORREKTUR: basic transmitter pulse'''~~ is now no longer triangular,  but has a more pointed shape.  At time $t = 0$ applies:		*However,  the basic detection pulse is now no longer triangular,  but has a more pointed shape.  At time $t = 0$ applies:
	:$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{		:$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{
	+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot		+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot

Guenter at 07:06, 20 May 2022

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	'''5.'''  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   '''System C''', <br>the same basic detection pulse ~~'''KORREKTUR: basic transmitter pulse'''~~ is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.		'''5.'''  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   '''System C''', <br>the same basic detection pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
	*Therefore,  as in  '''System B''':		*Therefore,  as in  '''System B''':
	:$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm		:$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm

Guenter at 07:05, 20 May 2022

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	<quiz display=simple>		<quiz display=simple>

	{Calculate for  $\text{System A}$  the basic detection pulse ~~'''KORREKTUR: basic transmitter pulse'''~~  $g_{d}(t) = g_{ s}(t) \star h_{\rm E}(t)$.  What value  $g_0 = g_{d}(t=0)$  results at time  $t = 0$?		{Calculate for  $\text{System A}$  the basic detection pulse  $g_{d}(t) = g_{ s}(t) \star h_{\rm E}(t)$.  What value  $g_0 = g_{d}(t=0)$  results at time  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$g_0 \hspace{0.28cm} = \ $ { 6 3% } $\ \rm W^{1/2}$		$g_0 \hspace{0.28cm} = \ $ { 6 3% } $\ \rm W^{1/2}$

Hwang at 15:06, 9 May 2022

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Hwang at 15:04, 9 May 2022

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	'''4.'''  Since  '''System B'''  uses the same receiver filter as  '''System A''',  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.		'''4.'''  Since  '''System B'''  uses the same receiver filter as  '''System A''',  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.
	*However,  the basic ~~detection~~ pulse is now no longer triangular,  but has a more pointed shape.  At time $t = 0$ applies:		*However,  the basic transmitter pulse is now no longer triangular,  but has a more pointed shape.  At time $t = 0$ applies:
	:$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{		:$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{
	+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot		+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot
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	'''5.'''  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   '''System C''', <br>the same basic ~~detection~~ pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.		'''5.'''  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   '''System C''', <br>the same basic transmitter pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
	*Therefore,  as in  '''System B''':		*Therefore,  as in  '''System B''':
	:$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm		:$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm

Hwang at 15:00, 9 May 2022

2022-05-09T15:00:26Z

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	\cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm		\cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm
	W}}}\hspace{0.05cm}.$$		W}}}\hspace{0.05cm}.$$
	There is no intersymbol interfering because for  $\| t \|\ge T$  the detection pulse is'  $g_{d}(t) = 0$.		There is no intersymbol interfering because for  $\| t \|\ge T$  the detection pulse is  $g_{d}(t) = 0$.

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	{{ML-Kopf}}		{{ML-Kopf}}
	'''1.'''  For '''System A''',  the convolution of the two equal-width rectangular functions  $g_{s}(t)$  and  $h_{\rm E}(t)$  leads to a triangular detection pulse with the maximum at  $t = 0$:		'''1.'''  For '''System A''',  the convolution of the two equal-width rectangular functions  $g_{s}(t)$  and  $h_{\rm E}(t)$  leads to a triangular detection pulse with the maximum at  $t = 0$:
	:$$g_d (t = 0) = \int_{ - T/2}^{+ T/2} { g_s(t) \cdot h_{\rm E}( t )} \hspace{0.1cm}{\rm{d}}t =s_0\cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\~~rmW~~}}}\hspace{0.05cm}.$$		:$$g_d (t = 0) = \int_{ - T/2}^{+ T/2} { g_s(t) \cdot h_{\rm E}( t )} \hspace{0.1cm}{\rm{d}}t =s_0\cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm W}}}\hspace{0.05cm}.$$
	There is no intersymbol interfering because for  $\| t \|\ge T$  the detection pulse is  $g_{d}(t) = 0$.		There is no intersymbol interfering because for  $\| t \|\ge T$  the detection pulse is  $g_{d}(t) = 0$.

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	'''2.'''  The variance of the noise component of the detection signal – referred to as the  "detection noise power" – can be calculated in both the time and frequency domains.		'''2.'''  The variance of the noise component of the detection signal – referred to as the  "detection noise power" – can be calculated in both the time and frequency domains.
	*For the present rectangular waveform,  calculation in the time domain yields faster results:		*For the present rectangular waveform,  calculation in the time domain yields faster results:
	:$$\sigma _d ^2 \ = \ \frac{N_0 }{2} \cdot \int_{ -\infty }^{ + \infty } {\left\| {h_{\rm E}( t )} \right\|^2\hspace{0.1cm}{\rm{d}}t} =\frac{N_0 }{2} \cdot \int_{ -T/2 }^{ + T/2 } {\left\| {h_{\rm E}( t )} \right\|^2\hspace{0.1cm}{\rm{d}}t} = \ \frac{N_0 }{2} \cdot\frac{1}{T^2} \cdot T = \frac{N_0 }{2T} = \frac{2 \cdot 10^{-5}\,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\~~rmW~~}}\hspace{0.05cm}.$$		:$$\sigma _d ^2 \ = \ \frac{N_0 }{2} \cdot \int_{ -\infty }^{ + \infty } {\left\| {h_{\rm E}( t )} \right\|^2\hspace{0.1cm}{\rm{d}}t} =\frac{N_0 }{2} \cdot \int_{ -T/2 }^{ + T/2 } {\left\| {h_{\rm E}( t )} \right\|^2\hspace{0.1cm}{\rm{d}}t} = \ \frac{N_0 }{2} \cdot\frac{1}{T^2} \cdot T = \frac{N_0 }{2T} = \frac{2 \cdot 10^{-5}\,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\rm W}}\hspace{0.05cm}.$$
	*The frequency domain calculation would be as follows with  $H_{\rm E}(f) = {\rm sinc}(fT)$:		*The frequency domain calculation would be as follows with  $H_{\rm E}(f) = {\rm sinc}(fT)$:
	:$$\sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{+ \infty } {\left\| {H_{\rm E}( f )} \right\|^2\hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{-\infty }^{ \infty } {\rm sinc}^2(f T)\hspace{0.1cm}{\rm{d}}f =\frac{N_0 }{2T} \hspace{0.05cm}.$$		:$$\sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{+ \infty } {\left\| {H_{\rm E}( f )} \right\|^2\hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{-\infty }^{ \infty } {\rm sinc}^2(f T)\hspace{0.1cm}{\rm{d}}f =\frac{N_0 }{2T} \hspace{0.05cm}.$$
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	:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {\rm Q} \left( \frac{ 6 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)= {\rm Q}(6) \hspace{0.1cm}\underline {= 0.987 \cdot 10^{-9}} \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {\rm Q} \left( \frac{ 6 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)= {\rm Q}(6) \hspace{0.1cm}\underline {= 0.987 \cdot 10^{-9}} \hspace{0.05cm}.$$
	*'''System A'''  represents the matched filter realization of the optimal binary receiver,  so the following equations would also be applicable:		*'''System A'''  represents the matched filter realization of the optimal binary receiver,  so the following equations would also be applicable:
	:$$E_{\rm B} = s_0^2 \cdot T = 36\, {\rm W} \cdot 10^{-5} {\rm s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)={\rm Q} \left( \sqrt{\frac{2 \cdot 36 \cdot 10^{-5}\,\, {\rm Ws}}{2 \cdot 10^{-5} \,\, {\~~rmWs~~}}}\right)={\rm Q}(6)\hspace{0.05cm}.$$		:$$E_{\rm B} = s_0^2 \cdot T = 36\, {\rm W} \cdot 10^{-5} {\rm s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)={\rm Q} \left( \sqrt{\frac{2 \cdot 36 \cdot 10^{-5}\,\, {\rm Ws}}{2 \cdot 10^{-5} \,\, {\rm Ws}}}\right)={\rm Q}(6)\hspace{0.05cm}.$$


	'''4.'''  Since  '''System B'''  uses the same receiver filter as  '''System A''',  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.		'''4.'''  Since  '''System B'''  uses the same receiver filter as  '''System A''',  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.
	*However,  the basic detection pulse is now no longer triangular,  but has a more pointed shape.  At time $t = 0$ applies:		*However,  the basic detection pulse is now no longer triangular,  but has a more pointed shape.  At time $t = 0$ applies:
	:$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot\frac{s_0 }{2} \cdot T = \frac{s_0 }{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\~~rmW~~}}\hspace{0.05cm}.$$		:$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot\frac{s_0 }{2} \cdot T = \frac{s_0 }{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm W}}\hspace{0.05cm}.$$
	*'''System B'''  is also free of intersymbol interfering.  Therefore,  one obtains for the bit error probability:		*'''System B'''  is also free of intersymbol interfering.  Therefore,  one obtains for the bit error probability:
	:$$p_{\rm B} = {\rm Q} \left( \frac{g_d (t = 0)}{\sigma_d}\right)= {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)= {\rm Q}(3) \hspace{0.1cm}\underline {= 0.135 \cdot 10^{-2}} \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left( \frac{g_d (t = 0)}{\sigma_d}\right)= {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)= {\rm Q}(3) \hspace{0.1cm}\underline {= 0.135 \cdot 10^{-2}} \hspace{0.05cm}.$$
	*On the other hand,  the following calculation is  '''not'''  applicable here:		*On the other hand,  the following calculation is  '''not'''  applicable here:
	:$$E_{\rm B} = \int^{+\infty} _{-\infty} g_s^2(t)\,{\~~rmd~~}t = 2\cdot s_0^2 \cdot \int ^{+T/2} _{0} \left( 1- \frac{2t}{T}\right)^2\,{\~~rmd~~}t = \frac{s_0^2 \cdot T }{3} = 12 \cdot 10^{-5} \,{\rm Ws}$$		:$$E_{\rm B} = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm d}t = 2\cdot s_0^2 \cdot \int ^{+T/2} _{0} \left( 1- \frac{2t}{T}\right)^2\,{\rm d}t = \frac{s_0^2 \cdot T }{3} = 12 \cdot 10^{-5} \,{\rm Ws}$$
	:$$\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4}\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4}\hspace{0.05cm}.$$
	*One would thus compute a bit error probability that is too low,  since the implicit assumption of a matched filter does not hold.		*One would thus compute a bit error probability that is too low,  since the implicit assumption of a matched filter does not hold.
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	'''5.'''  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   '''System C''', <br>the same basic detection pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.		'''5.'''  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   '''System C''', <br>the same basic detection pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
	*Therefore,  as in  '''System B''':		*Therefore,  as in  '''System B''':
	:$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\~~rmW~~}}\hspace{0.05cm}.$$		:$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm W}}\hspace{0.05cm}.$$
	*In contrast,  the detection noise power is now smaller than in systems  '''A'''  and  '''B''':		*In contrast,  the detection noise power is now smaller than in systems  '''A'''  and  '''B''':
	:$$\sigma _d ^2 = \frac{N_0}{2} \cdot \frac{1}{T^2} \cdot \int^{+T/2} _{-T/2} \left( 1- \frac{2t}{T}\right)^2\,{\~~rmd~~}t = \frac{N_0}{6T}\hspace{0.1cm}\underline { = 0.333 \,{\rm W}}.$$		:$$\sigma _d ^2 = \frac{N_0}{2} \cdot \frac{1}{T^2} \cdot \int^{+T/2} _{-T/2} \left( 1- \frac{2t}{T}\right)^2\,{\rm d}t = \frac{N_0}{6T}\hspace{0.1cm}\underline { = 0.333 \,{\rm W}}.$$
	*This now gives us for the bit error probability:		*This now gives us for the bit error probability:
	:$$p_{\rm B} = {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{0.577 \,\sqrt{\rm W}}\right)\approx {\rm Q}(5.2)\hspace{0.1cm}\underline { \approx 10^{-7} } \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{0.577 \,\sqrt{\rm W}}\right)\approx {\rm Q}(5.2)\hspace{0.1cm}\underline { \approx 10^{-7} } \hspace{0.05cm}.$$
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	[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]
	[[de:Aufgaben:1.~~3_Einfluss_von_g(s)_von_(t)_und_h(E)_von_(t)~~]]		[[de:Aufgaben:Aufgabe 1.3: Rechteckfunktionen für Sender und Empfänger]]

← Older revision		Revision as of 17:06, 9 May 2022
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	<quiz display=simple>		<quiz display=simple>

	{Calculate for  $\text{System A}$  the basic transmitter pulse  $g_{d}(t) = g_{ s}(t) \star h_{\rm E}(t)$.  What value  $g_0 = g_{d}(t=0)$  results at time  $t = 0$?		{Calculate for  $\text{System A}$  the basic detection pulse '''KORREKTUR: basic transmitter pulse'''  $g_{d}(t) = g_{ s}(t) \star h_{\rm E}(t)$.  What value  $g_0 = g_{d}(t=0)$  results at time  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$g_0 \hspace{0.28cm} = \ $ { 6 3% } $\ \rm W^{1/2}$		$g_0 \hspace{0.28cm} = \ $ { 6 3% } $\ \rm W^{1/2}$
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	'''4.'''  Since  '''System B'''  uses the same receiver filter as  '''System A''',  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.		'''4.'''  Since  '''System B'''  uses the same receiver filter as  '''System A''',  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.
	*However,  the basic transmitter pulse is now no longer triangular,  but has a more pointed shape.  At time $t = 0$ applies:		*However,  the basic detection pulse '''KORREKTUR: basic transmitter pulse''' is now no longer triangular,  but has a more pointed shape.  At time $t = 0$ applies:
	:$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{		:$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{
	+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot		+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot
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	'''5.'''  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   '''System C''', <br>the same basic transmitter pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.		'''5.'''  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   '''System C''', <br>the same basic detection pulse '''KORREKTUR: basic transmitter pulse''' is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
	*Therefore,  as in  '''System B''':		*Therefore,  as in  '''System B''':
	:$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm		:$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm