Aufgaben:Exercise 1.6Z: Two Optimal Systems - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:54:47Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:54, 16 March 2026
Line 86:		Line 86:
	:$$E_{\rm B} =\int_{-\infty}^{+\infty}\|G_s(f)\|^2 \,{\rm d} f = G_0^2\cdot 2 f_0\hspace{0.05cm}.$$		:$$E_{\rm B} =\int_{-\infty}^{+\infty}\|G_s(f)\|^2 \,{\rm d} f = G_0^2\cdot 2 f_0\hspace{0.05cm}.$$
	*Using the results from  '''(2)'''  and  '''(4)''',  it follows:		*Using the results from  '''(2)'''  and  '''(4)''',  it follows:
	:$$G_0^2 = \frac{E_{\rm B}}{2 f_0} = \frac{5 \cdot 10^{-7}\,{\rm V^2/Hz}}{2 \cdot 10^{6}\,{\~~rmHz~~}}= 2.5 \cdot 10^{-13}\,{\rm V^2/Hz^2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_0 \hspace{0.1cm}\underline {= 0.5 \cdot 10^{-6}\,{\rm V/Hz}}\hspace{0.05cm}.$$		:$$G_0^2 = \frac{E_{\rm B}}{2 f_0} = \frac{5 \cdot 10^{-7}\,{\rm V^2/Hz}}{2 \cdot 10^{6}\,{\rm Hz}}= 2.5 \cdot 10^{-13}\,{\rm V^2/Hz^2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_0 \hspace{0.1cm}\underline {= 0.5 \cdot 10^{-6}\,{\rm V/Hz}}\hspace{0.05cm}.$$


Line 98:		Line 98:

	[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimization of Baseband Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimization of Baseband Systems^]]
	[[de:Aufgaben:1.~~6Z_Zwei_Optimalsysteme~~]]		[[de:Aufgaben:Aufgabe 1.6Z: Zwei Optimalsysteme]]

Maintenance script: Join 1 split formula blocks

2026-03-16T14:16:04Z

Join 1 split formula blocks

← Older revision		Revision as of 16:16, 16 March 2026
Line 9:		Line 9:
	*System  $\rm A$  uses the NRZ basic transmission pulse  $g_{s}(t)$  according to the upper sketch with amplitude  $s_{0} = 1 \ \rm V$  and duration  $T = 0.5\ \rm µ s$.		*System  $\rm A$  uses the NRZ basic transmission pulse  $g_{s}(t)$  according to the upper sketch with amplitude  $s_{0} = 1 \ \rm V$  and duration  $T = 0.5\ \rm µ s$.
	*In contrast,  system  $\rm B$,  which is to operate at the same bit rate as system  $\rm A$,  has a rectangular basic transmission pulse spectrum:		*In contrast,  system  $\rm B$,  which is to operate at the same bit rate as system  $\rm A$,  has a rectangular basic transmission pulse spectrum:
	:$$G_s(f) = \left\{ \begin{array}{c} G_0 \\		:$$G_s(f) = \left\{ \begin{array}{c} G_0 \\0 \\ \end{array} \right.\quad\begin{array}{{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array}\begin{array}{{20}c}\|f\| < f_0 \hspace{0.05cm}, \\\|f\| > f_0 \hspace{0.05cm}.\\\end{array}$$
	0 \\ \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm{for}}
	\\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c}
	\|f\| < f_0 \hspace{0.05cm}, \\
	\|f\| > f_0 \hspace{0.05cm}.\\
	\end{array}$$

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:30:29Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:30, 16 March 2026
Line 70:		Line 70:

	'''(2)'''  The energy of the NRZ basic transmission pulse of system  $\rm A$  is given by		'''(2)'''  The energy of the NRZ basic transmission pulse of system  $\rm A$  is given by
	:$$E_{\rm B} =		:$$E_{\rm B} =\int_{-\infty}^{+\infty}g_s^2 (t)\,{\rm d} t =s_0^2 \cdot T = {1\,{\rm V^2}}\cdot {0.5 \cdot 10^{-6}\,{\rm s}}\hspace{0.1cm}\underline { = 0.5 \cdot 10^{-6}\,{\rm V^2/Hz}}\hspace{0.05cm}.$$
	\int_{-\infty}^{+\infty}g_s^2 (t)\,{\rm d} t =
	s_0^2 \cdot T = {1\,{\rm V^2}}\cdot {0.5 \cdot 10^{-6}\,{\rm s}}\hspace{0.1cm}\underline { = 0.5 \cdot 10^{-6}\,{\rm V^2/Hz}}\hspace{0.05cm}.$$


Line 92:		Line 90:

	'''(5)'''  For the energy of the basic transmission pulse can also be written:		'''(5)'''  For the energy of the basic transmission pulse can also be written:
	:$$E_{\rm B} =		:$$E_{\rm B} =\int_{-\infty}^{+\infty}\|G_s(f)\|^2 \,{\rm d} f = G_0^2\cdot 2 f_0\hspace{0.05cm}.$$
	\int_{-\infty}^{+\infty}\|G_s(f)\|^2 \,{\rm d} f = G_0^2
	\cdot 2 f_0\hspace{0.05cm}.$$
	*Using the results from  '''(2)'''  and  '''(4)''',  it follows:		*Using the results from  '''(2)'''  and  '''(4)''',  it follows:
	:$$G_0^2 = \frac{E_{\rm B}}{2 f_0} = \frac{5 \cdot 10^{-7}\,{\rm V^2/Hz}}{2 \cdot 10^{6}\,{\rm		:$$G_0^2 = \frac{E_{\rm B}}{2 f_0} = \frac{5 \cdot 10^{-7}\,{\rm V^2/Hz}}{2 \cdot 10^{6}\,{\rmHz}}= 2.5 \cdot 10^{-13}\,{\rm V^2/Hz^2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_0 \hspace{0.1cm}\underline {= 0.5 \cdot 10^{-6}\,{\rm V/Hz}}\hspace{0.05cm}.$$
	Hz}}= 2.5 \cdot 10^{-13}\,{\rm V^2/Hz^2}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_0 \hspace{0.1cm}\underline {= 0.5 \cdot 10^{-6}\,{\rm V/Hz}}
	\hspace{0.05cm}.$$

Maintenance script: Add German interlanguage link

2026-03-16T12:30:34Z

Add German interlanguage link

← Older revision		Revision as of 14:30, 16 March 2026
Line 111:		Line 111:

	[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimization of Baseband Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimization of Baseband Systems^]]
			[[de:Aufgaben:1.6Z_Zwei_Optimalsysteme]]

Guenter at 09:24, 4 May 2022

2022-05-04T09:24:33Z

← Older revision		Revision as of 11:24, 4 May 2022
Line 79:		Line 79:
	*Thus, for system  $\rm A$,  the impulse response  $h_{\rm E}(t)$  is rectangular and the frequency response  $H_{\rm E}(f)$  is sinc-shaped.		*Thus, for system  $\rm A$,  the impulse response  $h_{\rm E}(t)$  is rectangular and the frequency response  $H_{\rm E}(f)$  is sinc-shaped.
	*For system  $\rm B$,  $H_{\rm E}(f)$  is rectangular like  $G_{s}(f)$  and thus the impulse response  $h_{\rm E}(t)$  is an sinc-function.		*For system  $\rm B$,  $H_{\rm E}(f)$  is rectangular like  $G_{s}(f)$  and thus the impulse response  $h_{\rm E}(t)$  is an sinc-function.
	*Statement 3 is false:   An in  $\rm A$,  ~~'''~~B~~'''~~.		*Statement 3 is false:   An integrator has a rectangular impulse response and would be suitable for the realization of system  $\rm A$,  but not for system  $\rm B$.


	'''(4)'''  For system ~~'''~~B~~'''~~ $G_{d}(f)$ nearly coincides with $G_{s}(f)$.		'''(4)'''  For system  $\rm B$   ⇒   $G_{d}(f)$  nearly coincides with  $G_{s}(f)$.
	*There is only a difference in the Nyquist frequency, but this does not affect the considerations here:		*There is only a difference in the Nyquist frequency,  but this does not affect the considerations here:
	*While $G_{s}(f_{\rm Nyq}) = 1/2$, $G_{d}(f_{\rm Nyq}) = 1/4$.		*While  $G_{s}(f_{\rm Nyq}) = 1/2$,  $G_{d}(f_{\rm Nyq}) = 1/4$.

	*This results in a Nyquist system with rolloff factor $r = 0$.		*This results in a Nyquist system with rolloff factor  $r = 0$.
	*From this follows for the Nyquist frequency from the condition that the symbol duration should also be $T = 0.5\ \rm µ s$:		*From this follows for the Nyquist frequency from the condition that the symbol duration should also be  $T = 0.5\ \rm µ s$:
	:$$f_{\rm 0} = f_{\rm Nyq} = \frac{1 } {2 \cdot T} = \frac{1 } {2 \cdot 0.5 \cdot 10^{-6}\,{\rm s}}\hspace{0.1cm}\underline {= 1\,{\rm MHz}}\hspace{0.05cm}.$$		:$$f_{\rm 0} = f_{\rm Nyq} = \frac{1 } {2 \cdot T} = \frac{1 } {2 \cdot 0.5 \cdot 10^{-6}\,{\rm s}}\hspace{0.1cm}\underline {= 1\,{\rm MHz}}\hspace{0.05cm}.$$

Line 95:		Line 95:
	\int_{-\infty}^{+\infty}\|G_s(f)\|^2 \,{\rm d} f = G_0^2		\int_{-\infty}^{+\infty}\|G_s(f)\|^2 \,{\rm d} f = G_0^2
	\cdot 2 f_0\hspace{0.05cm}.$$		\cdot 2 f_0\hspace{0.05cm}.$$
	*Using the results from '''(2)''' and '''(4)''', it follows:		*Using the results from  '''(2)'''  and  '''(4)''',  it follows:
	:$$G_0^2 = \frac{E_{\rm B}}{2 f_0} = \frac{5 \cdot 10^{-7}\,{\rm V^2/Hz}}{2 \cdot 10^{6}\,{\rm		:$$G_0^2 = \frac{E_{\rm B}}{2 f_0} = \frac{5 \cdot 10^{-7}\,{\rm V^2/Hz}}{2 \cdot 10^{6}\,{\rm
	Hz}}= 2.5 \cdot 10^{-13}\,{\rm V^2/Hz^2}		Hz}}= 2.5 \cdot 10^{-13}\,{\rm V^2/Hz^2}
Line 102:		Line 102:


	'''(6)'''  <u>Solution 1</u> is correct:		'''(6)'''  <u>Solution 1</u>  is correct:
	*System ~~'''~~A~~'''~~ represents the optimal system even with peak limitation.		*System  $\rm A$  represents the optimal system even with peak limitation.
	*On the other hand, system ~~'''~~B~~'''~~ would be unsuitable due to the extremely unfavorable crest factor.		*On the other hand,  system  $\rm B$  would be unsuitable due to the extremely unfavorable crest factor.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 09:15, 4 May 2022

2022-05-04T09:15:24Z

← Older revision		Revision as of 11:15, 4 May 2022
Line 65:		Line 65:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Both systems operate according to the specification with the same bit rate.		'''(1)'''  Both systems operate according to the specification with the same bit rate.
	*The NRZ basic transmission pulse of system ~~'''~~A~~'''~~ has the symbol duration $T = 0.5\ \rm µ s$.		*The NRZ basic transmission pulse of system  $\rm A$  has the symbol duration $T = 0.5\ \rm µ s$.
	*This results in the bit rate $R = 1/T$ $ \underline{= 2\ \rm Mbit/s}$.		*This results in the bit rate  $R = 1/T$ $ \underline{= 2\ \rm Mbit/s}$.


	'''(2)'''  The energy of the NRZ basic transmission pulse of system ~~'''~~A~~'''~~ is given by		'''(2)'''  The energy of the NRZ basic transmission pulse of system  $\rm A$  is given by
	:$$E_{\rm B} =		:$$E_{\rm B} =
	\int_{-\infty}^{+\infty}g_s^2 (t)\,{\rm d} t =		\int_{-\infty}^{+\infty}g_s^2 (t)\,{\rm d} t =
Line 75:		Line 75:


	'''(3)'''  The <u>first two statements are true</u>:		'''(3)'''  The  <u>first two statements are true</u>:
	*In both cases $h_{\rm E}(t)$ must be equal in form to $g_{s}(t)$ and $H_{\rm E}(f)$ must be equal in form to $G_{s}(f)$.		*In both cases  $h_{\rm E}(t)$  must be equal in form to  $g_{s}(t)$  and  $H_{\rm E}(f)$  must be equal in form to  $G_{s}(f)$.
	*Thus, for system ~~'''~~A~~'''~~, the impulse response $h_{\rm E}(t)$ is rectangular and the frequency response $H_{\rm E}(f)$ is si-shaped.		*Thus, for system  $\rm A$,  the impulse response  $h_{\rm E}(t)$  is rectangular and the frequency response  $H_{\rm E}(f)$  is sinc-shaped.
	*For system ~~'''~~B~~'''~~, $H_{\rm E}(f)$ is rectangular like $G_{s}(f)$ and thus the impulse response $h_{\rm E}(t)$ is an si-function.		*For system  $\rm B$,  $H_{\rm E}(f)$  is rectangular like  $G_{s}(f)$  and thus the impulse response  $h_{\rm E}(t)$  is an sinc-function.
	*Statement 3 is false:   An ~~integrator has a rectangular impulse response and would be suitable for the realization of system '''~~A~~'''~~, ~~but not for system~~ '''B'''.		*Statement 3 is false:   An in  $\rm A$,  '''B'''.

Guenter at 09:08, 4 May 2022

2022-05-04T09:08:11Z

← Older revision		Revision as of 11:08, 4 May 2022
Line 4:		Line 4:


	[[File:P_ID1293__Dig_Z_1_6.png\|right\|frame\|Optimal systems in ~~the <br>~~time and frequency domain]]		[[File:P_ID1293__Dig_Z_1_6.png\|right\|frame\|Optimal systems in time and frequency domain]]
	Consider two binary transmission systems  $\rm A$  and  $\rm B$ , which have the same error behavior for an AWGN channel with noise power density  $N_{0}$.  In both cases, the bit error probability is:		Consider two binary transmission systems  $\rm A$  and  $\rm B$,  which have the same error behavior for an AWGN channel with noise power density  $N_{0}$.  In both cases,  the bit error probability is:
	:$$p_{\rm B} = {\rm Q} \left( \sqrt{{2 \cdot E_{\rm B}}/{N_0}}\right)\hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left( \sqrt{{2 \cdot E_{\rm B}}/{N_0}}\right)\hspace{0.05cm}.$$
	*System  $\rm A$  uses the NRZ basic transmission pulse  $g_{s}(t)$  according to the upper sketch with amplitude  $s_{0} = 1 \ \rm V$  and duration  $T = 0.5\ \rm µ s$.		*System  $\rm A$  uses the NRZ basic transmission pulse  $g_{s}(t)$  according to the upper sketch with amplitude  $s_{0} = 1 \ \rm V$  and duration  $T = 0.5\ \rm µ s$.
	*In contrast, system  $\rm B$ , which is to operate at the same bit rate as system  $\rm A$,  has a rectangular basic transmission pulse spectrum:		*In contrast,  system  $\rm B$,  which is to operate at the same bit rate as system  $\rm A$,  has a rectangular basic transmission pulse spectrum:
	:$$G_s(f) = \left\{ \begin{array}{c} G_0 \\		:$$G_s(f) = \left\{ \begin{array}{c} G_0 \\
	0 \\ \end{array} \right.\quad		0 \\ \end{array} \right.\quad
Line 21:		Line 21:


			Notes:
	''Notes:''		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems\|"Optimization of Baseband Transmission Systems"]].
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems\|Optimization of Baseband Transmission Systems]].

	*~~Note that here~~ the pulse amplitude is given in "volts", so that the average energy per bit  $(E_{\rm B})$  has the unit  $\rm V^{2}/Hz$.		*Here,  the pulse amplitude is given in  "volts",  so that the average energy per bit  $(E_{\rm B})$  has the unit  $\rm V^{2}/Hz$.


Line 43:		Line 42:
	{Which statements are true for the receiver filters of systems  $\rm A$  and  $\rm B$?		{Which statements are true for the receiver filters of systems  $\rm A$  and  $\rm B$?
	\|type="[]"}		\|type="[]"}
	+For system  $\rm A$,   $H_{\rm E}(f)$  has a si-shaped curve.		+For system  $\rm A$,   $H_{\rm E}(f)$  has a sinc-shaped curve.
	+For system  $\rm B$,   $H_{\rm E}(f)$  is an ideal rectangular ~~lowpass~~ filter.		+For system  $\rm B$,   $H_{\rm E}(f)$  is an ideal rectangular low-pass filter.
	-$H_{\rm E}(f)$  can be realized by an integrator in system  $\rm B$.		-$H_{\rm E}(f)$  can be realized by an integrator in system  $\rm B$.

	{For which cutoff frequency  $f_{0}$  does system  $\rm B$  have symbol duration  $T$?		{For which cutoff frequency  $f_{0}$  does system  $\rm B$  have the symbol duration  $T$?
	\|type="{}"}		\|type="{}"}
	$f_{0} \ = \ ${ 1 3% } $\ \rm MHz$		$f_{0} \ = \ ${ 1 3% } $\ \rm MHz$

	{How large should the constant height  $G_{0}$  of the spectrum of  $\rm B$  be chosen so that the same energy per bit results as for system  $\rm A$?		{How large should the constant height  $G_{0}$  of the spectrum  $\rm B$  be chosen so that the same energy per bit results as for system   $\rm A$?
	\|type="{}"}		\|type="{}"}
	$G_{0} \ = \ $ { 0.5 3% } $\ \cdot 10^{-6} \ \rm V/Hz$		$G_{0} \ = \ $ { 0.5 3% } $\ \cdot 10^{-6} \ \rm V/Hz$

Guenter at 13:58, 19 March 2022

2022-03-19T13:58:23Z

← Older revision		Revision as of 15:58, 19 March 2022
Line 111:		Line 111:


	[[Category:Digital Signal Transmission: Exercises\|^1.4 ~~Optimierung der Basisbandsysteme~~^]]		[[Category:Digital Signal Transmission: Exercises\|^1.4 Optimization of Baseband Systems^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 1.6Z: Zwei Optimalsysteme to Aufgaben:Exercise 1.6Z: Two Optimal Systems

2022-03-19T13:55:02Z

Guenter moved page Aufgaben:Aufgabe 1.6Z: Zwei Optimalsysteme to Aufgaben:Exercise 1.6Z: Two Optimal Systems

← Older revision	Revision as of 15:55, 19 March 2022
(No difference)

Hwang at 13:26, 18 March 2022

2022-03-18T13:26:53Z

Show changes