Aufgaben:Exercise 3.7Z: Rectangular Signal with Echo - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:54:18Z

Fix interlanguage link: resolve redirect chain

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	[[Category:Signal Representation: Exercises\|^3.4 The Convolution Theorem^]]		[[Category:Signal Representation: Exercises\|^3.4 The Convolution Theorem^]]
	[[de:Aufgaben:3.~~7Z_Rechtecksignal_mit_Echo~~]]		[[de:Aufgaben:Aufgabe 3.7Z: Rechtecksignal mit Echo]]

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2026-03-16T12:32:38Z

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	[[Category:Signal Representation: Exercises\|^3.4 The Convolution Theorem^]]		[[Category:Signal Representation: Exercises\|^3.4 The Convolution Theorem^]]
			[[de:Aufgaben:3.7Z_Rechtecksignal_mit_Echo]]

Javier: Text replacement - "Dirac impulse" to "Dirac delta"

2021-05-26T07:05:07Z

Text replacement - "Dirac impulse" to "Dirac delta"

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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The <u>second suggested solution</u> is correct:		'''(1)'''  The <u>second suggested solution</u> is correct:
	*The impulse response is equal to the received signal  $r(t)$, if a single Dirac ~~impulse~~ is present at the input at time  $t = 0$ :		*The impulse response is equal to the received signal  $r(t)$, if a single Dirac delta is present at the input at time  $t = 0$ :
	:$$h(t) = \delta (t) + \alpha \cdot \delta( {t - \tau } ).$$		:$$h(t) = \delta (t) + \alpha \cdot \delta( {t - \tau } ).$$

Guenter at 12:24, 28 April 2021

2021-04-28T12:24:59Z

← Older revision		Revision as of 14:24, 28 April 2021
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	[[File:P_ID532__Sig_Z_3_7_b_neu.png\|right\|frame\|Convolution of square wave signal  $s(t)$  and impulse response  $h(t)$]]		[[File:P_ID532__Sig_Z_3_7_b_neu.png\|right\|frame\|Convolution of square wave signal  $s(t)$  and impulse response  $h(t)$]]
	'''(2)'''  It holds ~~that~~  $r(t) = s(t) ∗ h(t)$. This convolution operation is most easily performed graphically:		'''(2)'''  It holds  $r(t) = s(t) ∗ h(t)$.  This convolution operation is most easily performed graphically:

	The values of the received signal are generally:		The values of the received signal are generally:
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	'''(3)'''  Using a similar procedure as in '''(2)''' , a DC ~~signal~~ of $2\hspace{0.02cm}\text{ V}$ is obtained for $r(t)$ :		'''(3)'''  Using a similar procedure as in  '''(2)''',  a direct signal  $\rm (DC)$  of  $2\hspace{0.02cm}\text{ V}$  is obtained for  $r(t)$ :
	*The gaps in the signal $s(t)$ are completely filled by the echo $s(t - T/2)$.		*The gaps in the signal  $s(t)$  are completely filled by the echo  $s(t - T/2)$.
	*This result can also be derived in the frequency domain.		*This result can also be derived in the frequency domain.  The channel frequency response is with  $\alpha = 1$  and  $\tau = T/2$:
	*The channel frequency response is with $\alpha = 1$ and $\tau = T/2$:
	:$$H( f ) = 1 + 1 \cdot {\rm{e}}^{ - {\rm{j\pi }}fT} = 1 + \cos ( {{\rm{\pi }}fT} ) - {\rm{j}} \cdot {\rm{sin}}( {{\rm{\pi }}fT} ).$$		:$$H( f ) = 1 + 1 \cdot {\rm{e}}^{ - {\rm{j\pi }}fT} = 1 + \cos ( {{\rm{\pi }}fT} ) - {\rm{j}} \cdot {\rm{sin}}( {{\rm{\pi }}fT} ).$$
	*Apart from the DC component, the input signal ${s(t)}$ only has components at $f = f_0 = 1/T$, $f = 3 \cdot f_0$, $f = 5 \cdot f_0$ etc..		*Apart from the DC component, the input signal  ${s(t)}$  only has components at  $f = f_0 = 1/T$,  $f = 3 \cdot f_0$,  $f = 5 \cdot f_0$,  etc..
	*At these frequencies, however, both the real– and the imaginary part of ${H(f)}$ are equal to zero.		*At these frequencies, however, both the real– and the imaginary part of  ${H(f)}$  are equal to zero.
	*Thus, for the output spectrum with $A_0 = 1 \text{ V}$ and $H(f = 0) = 2$ we obtain:		*Thus, for the output spectrum with  $A_0 = 1 \text{ V}$  and  $H(f = 0) = 2$  we obtain:
	:$$R(f) = A_0 \cdot H(f = 0) \cdot \delta (f) = 2\;{\rm{V}} \cdot \delta (f).$$		:$$R(f) = A_0 \cdot H(f = 0) \cdot \delta (f) = 2\;{\rm{V}} \cdot \delta (f).$$
	The inverse Fourier transformation thus also yields $r(t) \underline{= 2 \text{ V= const}}$.		The inverse Fourier transformation thus also yields  $r(t) \underline{= 2 \text{ V= const}}$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 12:15, 28 April 2021

2021-04-28T12:15:51Z

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	{Which statements are true regarding the impulse response  $h(t)$?		{Which statements are true regarding the impulse response  $h(t)$?
	\|type="[]"}		\|type="[]"}
	- For  $0 ≤ t < \tau$:   $h(t) = 1$ ~~is true~~, for  $t > \tau$:    $h(t) = 1 + \alpha$.		- For  $0 ≤ t < \tau$  holds  $h(t) = 1$, and for  $t > \tau$  holds   $h(t) = 1 + \alpha$.
	+ It holds that  $h(t) = \delta (t) + \alpha \cdot \delta(t - \tau)$.		+ It holds that  $h(t) = \delta (t) + \alpha \cdot \delta(t - \tau)$.
	- $h(t)$  has a Gaussian shape.		- $h(t)$  has a Gaussian shape.

Guenter: Guenter moved page Aufgaben:Exercise 3.7Z: Rectangular signal with Echo to Aufgaben:Exercise 3.7Z: Rectangular Signal with Echo

2021-04-28T12:14:02Z

Guenter moved page Aufgaben:Exercise 3.7Z: Rectangular signal with Echo to Aufgaben:Exercise 3.7Z: Rectangular Signal with Echo

← Older revision		Revision as of 14:14, 28 April 2021
(No difference)

Guenter at 12:13, 28 April 2021

2021-04-28T12:13:45Z

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	[[File:EN_Sig_Z_3_7.png\|right\|frame\|~~Transmit~~ signal $s(t)$ & signal $r(t)$ with echo]]		[[File:EN_Sig_Z_3_7.png\|right\|frame\|Transmission signal  $s(t)$, and  <br>reception signal  $r(t)$  with echo]]
	We consider a periodic ~~square wave~~ signal  $s(t)$  with the possible amplitude values  $0\text{ V}$  and  $2\text{ V}$  and the period duration  $T_0 = T = 1 \text{ ms}$. At the jump points, ~~for example~~ at  $t = T/4$, the signal value is  $1\text{ V}$. The DC component $($~~also~~ i.e. the Fourier coefficient  $A_0)$  of the signal is ~~also~~  $1\text{ V}$.		We consider a periodic rectangular signal  $s(t)$  with the possible amplitude values  $0\text{ V}$  and  $2\text{ V}$  and the period duration  $T_0 = T = 1 \text{ ms}$.  At the jump points, e.g. at  $t = T/4$, the signal value are  $1\text{ V}$.  The DC component  $($i.e. the Fourier coefficient  $A_0)$  of the signal is  $1\text{ V}$, too.

	Further applies:		Further applies:
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	* The coefficients  $A_n$  with even  $n$  are also zero.		* The coefficients  $A_n$  with even  $n$  are also zero.

	* For odd values of  $n$  ~~on the other hand,~~ the following applies:		* For odd values of  $n$,  the following applies:
	:$$A_n = ( { - 1} )^{\left( {n - 1} \right)/2} \cdot \frac{{4\;{\rm{V}}}}{{n \cdot {\rm{\pi }}}}.$$		:$$A_n = ( { - 1} )^{\left( {n - 1} \right)/2} \cdot \frac{{4\;{\rm{V}}}}{{n \cdot {\rm{\pi }}}}.$$

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	*The latter is characterised by the attenuation factor  $\alpha$  and the transit time  $\tau$ .		*The latter is characterised by the attenuation factor  $\alpha$  and the transit time  $\tau$ .
	*Therefore, the following applies to the received signal:		*Therefore, the following applies to the received signal:
	$$r(t) = s(t) + \alpha \cdot s( {t - \tau } ).$$		:$$r(t) = s(t) + \alpha \cdot s( {t - \tau } ).$$
	The frequency response of the channel is  $H(f) = R(f)/S(f)$, the impulse response is denoted by  $h(t)$ .		The frequency response of the channel is  $H(f) = R(f)/S(f)$,  the impulse response is denoted by  $h(t)$ .


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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|The Convolution Theorem and Operation]].		*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|The Convolution Theorem and Operation]].
	*Important information can be found ~~in particular~~ on the page  [[Signal_Representation/The_Convolution_Theorem_and_Operation#~~Convolution_of_a_Function_With_a_Dirac_Function~~\|Convolution of a ~~Function With~~ a Dirac ~~Function~~]].		*Important information can be found on the page  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_of_a_function_with_a_Dirac_function\|Convolution of a function with a Dirac function]].


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	<quiz display=simple>		<quiz display=simple>
	{Which statements are true regarding the impulse response  $h(t)$~~ ~~?		{Which statements are true regarding the impulse response  $h(t)$?
	\|type="[]"}		\|type="[]"}
	- For  $0 ≤ t < \tau$   $h(t) = 1$ is true, for  $t > \tau$    $h(t) = 1 + \alpha$.		- For  $0 ≤ t < \tau$:   $h(t) = 1$ is true, for  $t > \tau$:    $h(t) = 1 + \alpha$.
	+ It holds that  $h(t) = \delta (t) + \alpha \cdot \delta(t - \tau)$.		+ It holds that  $h(t) = \delta (t) + \alpha \cdot \delta(t - \tau)$.
	- $h(t)$  has a ~~gaussian~~ shape.		- $h(t)$  has a Gaussian shape.


	{Calculate the signal  $r(t)$  for the channel parameters  $\alpha = -0.5$  and  $\tau = T/4$. <br>What values result at the given times?		{Calculate the reception signal  $r(t)$  for the channel parameters  $\alpha = -0.5$  and  $\tau = T/4$. <br>What values result at the given times?
	\|type="{}"}		\|type="{}"}
	$r(t = 0.2 \cdot T)\ = \ $ { 1 3% }  $\text{V}$		$r(t = 0.2 \cdot T)\ = \ $ { 1 3% }  $\text{V}$
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	{Calculate the signal  $r(t)$  with  $\alpha = 1$  and  $\tau = T/2$. Interpret the result in the frequency domain. <br>What value results for  $t = T/2$?		{Calculate the reception signal  $r(t)$  with  $\alpha = 1$  and  $\tau = T/2$.  Interpret the result in the frequency domain. <br>What value results for  $t = T/2$?
	\|type="{}"}		\|type="{}"}
	$r(t = T/2)\ = \ $ { 2 3% }  $\text{V}$		$r(t = T/2)\ = \ $ { 2 3% }  $\text{V}$

Guenter: Guenter moved page Aufgaben:Exercise 3.7Z: Square Wave With Echo to Aufgaben:Exercise 3.7Z: Rectangular signal with Echo

2021-04-28T12:03:19Z

Guenter moved page Aufgaben:Exercise 3.7Z: Square Wave With Echo to Aufgaben:Exercise 3.7Z: Rectangular signal with Echo

← Older revision		Revision as of 14:03, 28 April 2021
(No difference)

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:39:08Z

Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

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	[[Category:~~Exercises for~~ Signal Representation\|^3.4 The Convolution Theorem^]]		[[Category:Signal Representation: Exercises\|^3.4 The Convolution Theorem^]]

Noah at 21:17, 28 January 2021

2021-01-28T21:17:57Z

← Older revision		Revision as of 23:17, 28 January 2021
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	*Thus, for the output spectrum with $A_0 = 1 \text{ V}$ and $H(f = 0) = 2$ we obtain:		*Thus, for the output spectrum with $A_0 = 1 \text{ V}$ and $H(f = 0) = 2$ we obtain:
	:$$R(f) = A_0 \cdot H(f = 0) \cdot \delta (f) = 2\;{\rm{V}} \cdot \delta (f).$$		:$$R(f) = A_0 \cdot H(f = 0) \cdot \delta (f) = 2\;{\rm{V}} \cdot \delta (f).$$
	The Fourier ~~inverse~~ transformation thus also yields $r(t) \underline{= 2 \text{ V= const}}$.		The inverse Fourier transformation thus also yields $r(t) \underline{= 2 \text{ V= const}}$.
	{{ML-Fuß}}		{{ML-Fuß}}