Aufgaben:Exercise 2.7: Two-Way Channel once more - Revision history

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	:$$Y_1(f) = X_1(f) \cdot H(f) = T \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$		:$$Y_1(f) = X_1(f) \cdot H(f) = T \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$

	*The complex exponential functions can be converted using the  [[Signal_Representation/~~Calculating_With_Complex_Numbers~~#Representation_by_magnitude_and_phase\|Euler theorem]]  as follows:		*The complex exponential functions can be converted using the  [[Signal_Representation/Calculating_with_Complex_Numbers#Representation_by_magnitude_and_phase\|Euler theorem]]  as follows:
	:$${\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T}\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}\cdot \big[ {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$		:$${\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T}\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}\cdot \big[ {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]
	[[de:Aufgaben:2.~~7_Nochmals_Zweiwegekanal~~]]		[[de:Aufgaben:Aufgabe 2.7: Nochmals Zweiwegekanal]]

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 *a rectangular pulse &nbsp;$x_2(t)$&nbsp; with height&nbsp; $1$&nbsp; in the range of&nbsp; $0 \ \text{...} \  2T$;
 *a periodic rectangular signal &nbsp;$x_3(t)$&nbsp; with period duration &nbsp;$T = T_0$:
-:$$x_3(t) = \left\{ \begin{array}{c} 1 \\
+:$$x_3(t) = \left\{ \begin{array}{c} 1 \\0 \\  \end{array} \right.\quad \quad\begin{array}{c}   {\rm{for}}  \\ {\rm{for}}\\    \end{array}\begin{array}{*{20}c}{ 0 < t  < T/2,}  \\{ T/2 < t  < T,}  \\\end{array}$$*a periodic rectangular signal &nbsp;$x_4(t)$&nbsp; with period duration &nbsp;$T = 2T_0$::$$x_4(t) = \left\{ \begin{array}{c} 1 \\0 \\  \end{array} \right.\quad \quad\begin{array}{c}  {\rm{for}}  \\ {\rm{for}}\\    \end{array}\begin{array}{*{20}c}{ 0 < t  < T,}  \\{ T < t  < 2T.}  \\\end{array}$$

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	Using the transit times $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$  considered in the subtasks  '''(1)''' ... '''(4)''',  the following is obtained for the frequency response of the two-way channel,  whose magnitude is depicted in the upper graph:		Using the transit times $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$  considered in the subtasks  '''(1)''' ... '''(4)''',  the following is obtained for the frequency response of the two-way channel,  whose magnitude is depicted in the upper graph:
	:$$H(f) = 1 + {\rm e}^{-{\rm j}\hspace{0.04cm}2 \pi f T} = 1 +		:$$H(f) = 1 + {\rm e}^{-{\rm j}\hspace{0.04cm}2 \pi f T} = 1 +\cos(2 \pi f T) - {\rm j} \cdot \sin(2 \pi f T)$$
	\cos(2 \pi f T) - {\rm j} \cdot \sin(2 \pi f T)$$		:$$\Rightarrow \hspace{0.4cm}\|H(f)\| = \sqrt{2\left(1 + \cos(2 \pi fT)\right)}= 2 \cdot \|\cos(\pi f T)\|.$$
	:$$\Rightarrow \hspace{0.4cm}\|H(f)\| = \sqrt{2\left(1 + \cos(2 \pi f
	T)\right)}= 2 \cdot \|\cos(\pi f T)\|.$$

	The bottom graph shows the phase function for  $T_1 = 0$  and   $T_2 = T = 4 \ \rm ms$:		The bottom graph shows the phase function for  $T_1 = 0$  and   $T_2 = T = 4 \ \rm ms$:
	:$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) = \arctan \frac{\sin(2 \pi f		:$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) = \arctan \frac{\sin(2 \pi fT)}{1 + \cos(2 \pi f T)} = \arctan \big[\tan(\pi f T)\big].$$
	T)}{1 + \cos(2 \pi f T)} = \arctan \big[\tan(\pi f T)\big].$$

	*In the frequency region  $\|f\| < 1/(2T)$,    $b(f)$  increases linearly:   $b(f) = \pi \cdot f \cdot T.$		*In the frequency region  $\|f\| < 1/(2T)$,    $b(f)$  increases linearly:   $b(f) = \pi \cdot f \cdot T.$
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	{ 0 < t < T/2,} \\		{ 0 < t < T/2,} \\
	{ T/2 < t < T,} \\		{ T/2 < t < T,} \\
	\end{array}$$		\end{array}$$*a periodic rectangular signal  $x_4(t)$  with period duration  $T = 2T_0$::$$x_4(t) = \left\{ \begin{array}{c} 1 \\
	*a periodic rectangular signal  $x_4(t)$  with period duration  $T = 2T_0$:
	:$$x_4(t) = \left\{ \begin{array}{c} 1 \\
	0 \\ \end{array} \right.\quad \quad		0 \\ \end{array} \right.\quad \quad
	\begin{array}{c} {\rm{for}} \\ {\rm{for}}		\begin{array}{c} {\rm{for}} \\ {\rm{for}}
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The solution in the time domain leads faster to the final result:		'''(1)'''  The solution in the time domain leads faster to the final result:
	:$$y_1(t) = x_1(t) \star h(t) = x_1(t) \star \delta (t)		:$$y_1(t) = x_1(t) \star h(t) = x_1(t) \star \delta (t)+ x_1(t) \star \delta (t - T) = x_1(t) + x_1(t-T).$$
	+ x_1(t) \star \delta (t - T) = x_1(t) + x_1(t-T).$$

	*Thus,  $y_1(t)$  is a rectangular pulse of height  $1$  and width  $2T$.		*Thus,  $y_1(t)$  is a rectangular pulse of height  $1$  and width  $2T$.

	*The same result – but in a more time-consuming way – is obtained by computing in the frequency domain:		*The same result – but in a more time-consuming way – is obtained by computing in the frequency domain:
	:$$Y_1(f) = X_1(f) \cdot H(f) = T \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot		:$$Y_1(f) = X_1(f) \cdot H(f) = T \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$
	\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$

	*The complex exponential functions can be converted using the  [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_magnitude_and_phase\|Euler theorem]]  as follows:		*The complex exponential functions can be converted using the  [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_magnitude_and_phase\|Euler theorem]]  as follows:
	:$${\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T}		:$${\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T}\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}\cdot \big[ {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$
	\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}
	\cdot \big[ {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$

	*Hence,  the following can be formulated for the output spectrum:		*Hence,  the following can be formulated for the output spectrum:
	:$$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}		:$$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}, \; \; {\rm mit } \; \;Y_{11}(f) = 2T \cdot \frac {\sin(\pi f T) \cdot \cos(\pi f T)}{\pif T} = 2T \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$
	, \; \; {\rm mit } \; \;
	Y_{11}(f) = 2T \cdot \frac {\sin(\pi f T) \cdot \cos(\pi f T)}{\pi
	f T} = 2T \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$

	[[File:P_ID925__LZI_A_2_7_a.png\|right\|frame\|Solutions  '''(1)'''  and  '''(2)''']]		[[File:P_ID925__LZI_A_2_7_a.png\|right\|frame\|Solutions  '''(1)'''  and  '''(2)''']]
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	*$X_3(f)$  is a line spectrum with components at frequencies  $f = 0$,  $f = \pm f_0 = \pm 1/T$,  $f = \pm 3f_0$,  etc..		*$X_3(f)$  is a line spectrum with components at frequencies  $f = 0$,  $f = \pm f_0 = \pm 1/T$,  $f = \pm 3f_0$,  etc..
	*However,  the following holds at these discrete frequencies:		*However,  the following holds at these discrete frequencies:
	:$$\|H(f)\| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow		:$$\|H(f)\| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm}\tau_{\rm P}(f) = 0.$$
	\hspace{0.3cm}\tau_{\rm P}(f) = 0.$$

	*From this it follows again that  $y_3(t) = 2 \cdot x_3(t) $.		*From this it follows again that  $y_3(t) = 2 \cdot x_3(t) $.
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	*The frequency response thus has zeros for all odd multiples of  $f_0$.		*The frequency response thus has zeros for all odd multiples of  $f_0$.
	*The only remaining spectral line of  $Y_4(f)$  is at  $f = 0$  where the following holds:		*The only remaining spectral line of  $Y_4(f)$  is at  $f = 0$  where the following holds:
	:$$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f)		:$$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f)\hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$
	\hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$



	'''(5)'''  Now the frequency response is with  $T_1 = 1 \ \rm ms$,  $T_2 = 5 \ \rm ms$  and  $T = T_2 -T_1 = 4 \ \rm ms$:		'''(5)'''  Now the frequency response is with  $T_1 = 1 \ \rm ms$,  $T_2 = 5 \ \rm ms$  and  $T = T_2 -T_1 = 4 \ \rm ms$:
	:$$H(f) = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f		:$$H(f) = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi fT_1}+ {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi fT_2}= \big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}\big]\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi fT_1}.$$
	~~T_1~~}+ {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f
	~~T_2~~}= \big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}
	\big]\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f
	~~T_1~~}.$$

	*The expression in parentheses describes the frequency response already considered so far.		*The expression in parentheses describes the frequency response already considered so far.

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]
			[[de:Aufgaben:2.7_Nochmals_Zweiwegekanal]]

Guenter at 15:11, 6 October 2021

2021-10-06T15:11:54Z

Guenter at 15:00, 6 October 2021

2021-10-06T15:00:22Z

Guenter at 14:51, 6 October 2021

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	In contrast to the general representation in exercise 2.6, the two attenuation factors are equal here:   $z_1 = z_2 = 1$.		In contrast to the general representation in exercise 2.6, the two attenuation factors are equal here:   $z_1 = z_2 = 1$.
	*For mobile communications, this corresponds to an echo at a distance of  $T_2 - T_1$  and of the same strength as the signal on the main path~~, for example~~.		For mobile communications, this corresponds for example to an echo at a distance of  $T_2 - T_1$  and of the same strength as the signal on the main path.  For this, the transit time  $T_1$  is assumed.
	*For this, the transit time  $T_1$  is assumed.


	Using the transit times $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$  considered in the subtasks  '''(1)''' ... '''(4)''',  the following is obtained for the frequency response of the two-way channel, whose magnitude is depicted in the upper graph:		Using the transit times $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$  considered in the subtasks  '''(1)''' ... '''(4)''',  the following is obtained for the frequency response of the two-way channel,  whose magnitude is depicted in the upper graph:
	:$$H(f) = 1 + {\rm e}^{-{\rm j}\hspace{0.04cm}2 \pi f T} = 1 +		:$$H(f) = 1 + {\rm e}^{-{\rm j}\hspace{0.04cm}2 \pi f T} = 1 +
	\cos(2 \pi f T) - {\rm j} \cdot \sin(2 \pi f T)$$		\cos(2 \pi f T) - {\rm j} \cdot \sin(2 \pi f T)$$
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	T)\right)}= 2 \cdot \|\cos(\pi f T)\|.$$		T)\right)}= 2 \cdot \|\cos(\pi f T)\|.$$

	The bottom graph shows the phase function:		The bottom graph shows the phase function for  $T_1 = 0$  and   $T_2 = T = 4 \ \rm ms$:
	:$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) = \arctan \frac{\sin(2 \pi f		:$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) = \arctan \frac{\sin(2 \pi f
	T)}{1 + \cos(2 \pi f T)} = \arctan \big[\tan(\pi f T)\big].$$		T)}{1 + \cos(2 \pi f T)} = \arctan \big[\tan(\pi f T)\big].$$

	The following trigonometric transformation was used here:		*In the frequency region  $\|f\| < 1/(2T)$,    $b(f)$  increases linearly:   $b(f) = \pi \cdot f \cdot T.$
			*Also in further sections of the phase function,  the phase always increases linearly from  $-\pi/2$  to  $+\pi/2$ .
			*The following trigonometric transformation was used here:
	:$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$		:$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$

	~~The bottom graph shows the phase function for $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$:~~
	*In the frequency range  $\|f\| < 1/(2T)$,    $b(f)$  increases linearly:   $b(f) = \pi \cdot f \cdot T.$
	*Also in further sections of the phase function, the phase always increases linearly from  $-\pi/2$  to  $+\pi/2$ .

			In the questions,  $y_i(t)$  denotes the signal at the output of the two-way channel if the signal  $x_i(t)$  is applied to the input  $(i = 1, 2, 3, 4)$.
	In the questions, $y_i(t)$  denotes the signal at the output of the two-way channel if the signal $x_i(t)$  is applied to the input  $( i = 1, 2, 3, 4)$.

	These signals are examined as input signals:		These signals are examined as input signals:
	*a rectangular pulse  $x_1(t)$  with height  $1$  between  $t= 0$  and  $t= T$; ~~<br>~~ $x_1(t) = 0$  holds for  $t < 0$  and  $t > T$  $($the value $0.5$ occurs at the two jump discontinuities~~, respectively~~$)$;		*a rectangular pulse  $x_1(t)$  with height  $1$  between  $t= 0$  and  $t= T$;  $x_1(t) = 0$  holds for  $t < 0$  and  $t > T$  $($the value $0.5$ occurs at the two jump discontinuities$)$;
	*a rectangular pulse  $x_2(t)$  with height  $1$  in the range of  $0 \ \text{...} \ 2T$;		*a rectangular pulse  $x_2(t)$  with height  $1$  in the range of  $0 \ \text{...} \ 2T$;
	*a periodic rectangular ~~pulse~~  $x_3(t)$  with period  $T = T_0$:		*a periodic rectangular signal  $x_3(t)$  with period duration  $T = T_0$:
	:$$x_3(t) = \left\{ \begin{array}{c} 1 \\		:$$x_3(t) = \left\{ \begin{array}{c} 1 \\
	0 \\ \end{array} \right.\quad \quad		0 \\ \end{array} \right.\quad \quad
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	{ T/2 < t < T,} \\		{ T/2 < t < T,} \\
	\end{array}$$		\end{array}$$
	*a periodic rectangular ~~pulse~~  $x_4(t)$  with period  $T = 2T_0$:		*a periodic rectangular signal  $x_4(t)$  with period duration  $T = 2T_0$:
	:$$x_4(t) = \left\{ \begin{array}{c} 1 \\		:$$x_4(t) = \left\{ \begin{array}{c} 1 \\
	0 \\ \end{array} \right.\quad \quad		0 \\ \end{array} \right.\quad \quad
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			Please note:
			*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Linear_Distortions\|Linear Distortions]].
			*For subtasks  '''(1)''' to '''(4)''',  $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$  hold.
			*In subtask  '''(5)''',  the case  $T_1 = 1 \ \rm ms$  and  $T_2 = 5 \ \rm ms$  is considered.
	''Please note:''
	*The ~~task~~ belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Linear_Distortions\|Linear Distortions]].
	*For subtasks '''(1)''' to '''(4)''',  $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$ hold.
	*In subtask '''(5)''', the case  $T_1 = 1 \ \rm ms$  and  $T_2 = 5 \ \rm ms$  is considered.

Oezer at 21:37, 20 September 2021

2021-09-20T21:37:45Z

← Older revision		Revision as of 23:37, 20 September 2021
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	*The expression in parentheses describes the frequency response already considered so far.		*The expression in parentheses describes the frequency response already considered so far.
	*The second term causes an additional runtime around  $ \tau = T_1$ and it holds for all signals  $(i = 1, 2, 3, 4)$:		*The second term causes an additional runtime around  $ \tau = T_1$ and the following holds for all signals  $(i = 1, 2, 3, 4)$:
	:$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$		:$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$

	All statements regarding the distortions are still valid. This corresponds to <u>proposed solution 1</u>.		All statements regarding the distortions are still valid. This corresponds to <u>proposed solution 1</u>.
	{{ML-Fuß}}		{{ML-Fuß}}

Oezer at 21:32, 20 September 2021

2021-09-20T21:32:58Z

← Older revision		Revision as of 23:32, 20 September 2021
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	These signals are examined as input signals:		These signals are examined as input signals:
	*a rectangular pulse  $x_1(t)$  with height  $1$  between  $t= 0$  and  $t= T$; <br> $x_1(t) = 0$  holds for  $t < 0$  and  $t > T$  $($the value$0.5$ occurs at the two jump discontinuities, respectively$)$;		*a rectangular pulse  $x_1(t)$  with height  $1$  between  $t= 0$  and  $t= T$; <br> $x_1(t) = 0$  holds for  $t < 0$  and  $t > T$  $($the value $0.5$ occurs at the two jump discontinuities, respectively$)$;
	*a rectangular pulse  $x_2(t)$  with height  $1$  in the range of  $0 \ \text{...} \ 2T$;		*a rectangular pulse  $x_2(t)$  with height  $1$  in the range of  $0 \ \text{...} \ 2T$;
	*a periodic rectangular pulse  $x_3(t)$  with period  $T = T_0$:		*a periodic rectangular pulse  $x_3(t)$  with period  $T = T_0$:

Oezer at 21:31, 20 September 2021

2021-09-20T21:31:56Z

← Older revision		Revision as of 23:31, 20 September 2021
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	:$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$		:$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$

	The bottom graph shows ~~thep hase~~ function for $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$:		The bottom graph shows the phase function for $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$:
	*In the frequency range  $\|f\| < 1/(2T)$,    $b(f)$  increases linearly:   $b(f) = \pi \cdot f \cdot T.$		*In the frequency range  $\|f\| < 1/(2T)$,    $b(f)$  increases linearly:   $b(f) = \pi \cdot f \cdot T.$
	*Also in further sections of the phase function, the phase always increases linearly from  $-\pi/2$  to  $+\pi/2$ .		*Also in further sections of the phase function, the phase always increases linearly from  $-\pi/2$  to  $+\pi/2$ .

← Older revision		Revision as of 17:11, 6 October 2021
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	*Thus,  $y_1(t)$  is a rectangular pulse of height  $1$  and width  $2T$.		*Thus,  $y_1(t)$  is a rectangular pulse of height  $1$  and width  $2T$.

	*The same result – but in a more time-consuming way – is obtained by computing in the ~~spectral~~ domain:		*The same result – but in a more time-consuming way – is obtained by computing in the frequency domain:
	:$$Y_1(f) = X_1(f) \cdot H(f) = T \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot		:$$Y_1(f) = X_1(f) \cdot H(f) = T \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot
	\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$		\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$
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	\cdot \big[ {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$		\cdot \big[ {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$

	*Hence, the following can be formulated for the output spectrum:		*Hence,  the following can be formulated for the output spectrum:
	:$$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}		:$$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}
	, \; \; {\rm mit } \; \;		, \; \; {\rm mit } \; \;
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	f T} = 2T \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$		f T} = 2T \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$

	[[File:P_ID925__LZI_A_2_7_a.png\|right\|frame\|~~Input~~ and ~~output signals~~]]		[[File:P_ID925__LZI_A_2_7_a.png\|right\|frame\|Solutions  '''(1)'''  and  '''(2)''']]
	Here, the relation  $\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$  is used.		Here,  the relation  $\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$  is used.

	*The inverse Fourier transform of  $Y_{11}$  results in a rectangle of width  $2T$, that is symmetric about  $t = 0$ .		*The inverse Fourier transform of  $Y_{11}$  results in a rectangle of width  $2T$,  that is symmetric about  $t = 0$ .
	*Due to the phase function this is shifted into the range  $0$ ... $2T$  and the result of the time domain computation is confirmed.		*Due to the phase function this is shifted into the range  $0$ ... $2T$  and the result of the time domain computation is confirmed.


	Despite the fact that $y_1(t)$ is rectangular just as $x_1(t)$, there are distortions:		Despite the fact that  $y_1(t)$  is rectangular just as  $x_1(t)$,  there are distortions:
	*These are linear because of  $T_y > T_x$ . In the frequency range of interest  $($that is all frequencies for a si–shaped spectrum$)$,    $\|H(f)\|$  is not constant. So, there are attenuation distortions.		*These are linear because of  $T_y > T_x$.  In the frequency range of interest  $($that is all frequencies for a sinc–shaped spectrum$)$,    $\|H(f)\|$  is not constant.  So, there are attenuation distortions.
	*In addition, there are also phase distortions since the phase does not increase linearly with  $f$ in the whole range   ⇒   The <u>proposed solutions 1, 3, 4 and 5</u> are correct.		*In addition, there are also phase distortions since the phase does not increase linearly with  $f$  in the whole range   ⇒   The  <u>proposed solutions 1, 3, 4 and 5</u>  are correct.


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	'''(3)'''  The period  $T_0 = T$  of the periodic signal  $x_3(t)$  is exactly as large as the delay on the second path. Therefore,  $y_3(t) = 2 \cdot x_3(t) $  holds and no distortions are observable.		'''(3)'''  The period  $T_0 = T$  of the periodic signal  $x_3(t)$  is exactly as large as the delay on the second path. Therefore,  $y_3(t) = 2 \cdot x_3(t) $  holds and no distortions are observable.

	The ~~spectral~~ domain computation leads to the same result.		The frequency domain computation leads to the same result.
	*$X_3(f)$  is a line spectrum with components at frequencies  $f = 0$,  $f = \pm f_0 = \pm 1/T$,  $f = \pm 3f_0$,  etc..		*$X_3(f)$  is a line spectrum with components at frequencies  $f = 0$,  $f = \pm f_0 = \pm 1/T$,  $f = \pm 3f_0$,  etc..
	*However, the following holds at these discrete frequencies:		*However,  the following holds at these discrete frequencies:
	:$$\|H(f)\| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow		:$$\|H(f)\| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow
	\hspace{0.3cm}\tau_{\rm P}(f) = 0.$$		\hspace{0.3cm}\tau_{\rm P}(f) = 0.$$

	*From this it follows again that  $y_3(t) = 2 \cdot x_3(t) $.		*From this it follows again that  $y_3(t) = 2 \cdot x_3(t) $.
	*Thus, only <u>proposed solution 1</u> is correct.		*Thus,  only  <u>proposed solution 1</u>  is correct.




	'''(4)'''  It can be seen that  $y_4(t) = 1$  is distorted with respect to  $x_4(t)$  from the lower sketch of the second graph. These are attenuation distortions  ⇒  <u>Proposed solution 2</u> is correct as the following consideration shows.		'''(4)'''  It can be seen that  $y_4(t) = 1$  is distorted with respect to  $x_4(t)$  from the lower sketch of the second graph. These are attenuation distortions  ⇒  <u>Proposed solution 2</u>  is correct as the following consideration shows.
	*Due to  $T_0 = 2T$  the signal  $x_4(t)$  has the ~~fundamental~~ frequency  $f_0 = 1/(2T)$.		*Due to  $T_0 = 2T$  the signal  $x_4(t)$  has the basic frequency  $f_0 = 1/(2T)$.
	*The frequency response thus has zeros for all odd multiples of  $f_0$~~ ~~.		*The frequency response thus has zeros for all odd multiples of  $f_0$.
	*The only remaining spectral line of  $Y_4(f)$  is at  $f = 0$ where the following holds:		*The only remaining spectral line of  $Y_4(f)$  is at  $f = 0$  where the following holds:
	:$$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f)		:$$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f)
	\hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$		\hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$
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	*The expression in parentheses describes the frequency response already considered so far.		*The expression in parentheses describes the frequency response already considered so far.
	*The second term causes an additional runtime around  $ \tau = T_1$ and the following holds for all signals  $(i = 1, 2, 3, 4)$:		*The second term causes an additional runtime around  $ \tau = T_1$  and the following holds for all signals  $(i = 1, 2, 3, 4)$:
	:$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$		:$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$

	All statements regarding the distortions are still valid. This corresponds to <u>proposed solution 1</u>.		All statements regarding the distortions are still valid.  This corresponds to  <u>proposed solution 1</u>.
	{{ML-Fuß}}		{{ML-Fuß}}

← Older revision		Revision as of 17:00, 6 October 2021
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	<quiz display=simple>		<quiz display=simple>
	{Compute the output signal  $y_1(t)$  for the input signal  $x_1(t)$. Which of the statements are true?		{Compute the output signal  $y_1(t)$  for the input signal  $x_1(t)$.  Which of the statements are true?
	\|type="[]"}		\|type="[]"}
	+ $y_1(t)$  is rectangular like  $x_1(t)$~~ ~~.		+ $y_1(t)$  is rectangular like  $x_1(t)$.
	- $y_1(t)$  is triangular in shape.		- $y_1(t)$  is triangular in shape.
	+ The absolute pulse duration is  $2T$.		+ The absolute pulse duration is  $2T$.
	+ $y_1(t)$  exhibits attenuation distortions with respect to  $x_1(t)$~~ ~~.		+ $y_1(t)$  exhibits attenuation distortions with respect to  $x_1(t)$.
	+ $y_1(t)$  exhibits phase distortions with respect to  $x_1(t)$~~ ~~.		+ $y_1(t)$  exhibits phase distortions with respect to  $x_1(t)$.


	{Compute the signal  $y_2(t)$. What values are obtained at times  $t= 0.5 T$,  $t= 1.5 T$  and  $t= 2.5 T$?		{Compute the signal  $y_2(t)$.  What values are obtained at times  $t= 0.5 T$,  $t= 1.5 T$  and  $t= 2.5 T$?
	\|type="{}"}		\|type="{}"}
	$y_2(t = 0.5T) \ = \ $ { 1 3% }		$y_2(t = 0.5T) \ = \ $ { 1 3% }
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	{Compute the signal  $y_3(t)$. Check which statements are true.		{Compute the signal  $y_3(t)$.  Check which statements are true.
	\|type="[]"}		\|type="[]"}
	+ $y_3(t)$  is undistorted with respect to  $x_3(t)$~~ ~~.		+ $y_3(t)$  is undistorted with respect to  $x_3(t)$.
	- $y_3(t)$  exhibits attenuation distortions with respect to  $x_3(t)$~~ ~~.		- $y_3(t)$  exhibits attenuation distortions with respect to  $x_3(t)$.
	- $y_3(t)$  exhibits phase distortions with respect to  $x_3(t)$~~ ~~.		- $y_3(t)$  exhibits phase distortions with respect to  $x_3(t)$.


	{Which statements are true for the output signal  $y_4(t)$~~ ~~?		{Which statements are true for the output signal  $y_4(t)$?
	\|type="[]"}		\|type="[]"}
	- $y_4(t)$  is undistorted with respect to  $x_4(t)$~~ ~~.		- $y_4(t)$  is undistorted with respect to  $x_4(t)$.
	+ $y_4(t)$  exhibits attenuation distortions with respect to  $x_4(t)$~~ ~~.		+ $y_4(t)$  exhibits attenuation distortions with respect to  $x_4(t)$.
	- $y_4(t)$  exhibits phase distortions with respect to  $x_4(t)$~~ ~~.		- $y_4(t)$  exhibits phase distortions with respect to  $x_4(t)$.


	{Now  $T_1 = 1 \ \rm ms$  and  $T_2 = 5 \ \rm ms$ hold. What are the changes compared to the previous results?		{Now  $T_1 = 1 \ \rm ms$  and  $T_2 = 5 \ \rm ms$ hold.  What are the changes compared to the previous results?
	\|type="[]"}		\|type="[]"}
	+ The above statements regarding distortions are still valid.		+ The above statements regarding distortions are still valid.