Aufgaben:Exercise 1.3: Measured Step Response - Revision history

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^1.2 System Description in Time Domain^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^1.2 System Description in Time Domain^]]
	[[de:Aufgaben:1.~~3_Gemessene_Sprungantwort~~]]		[[de:Aufgaben:Aufgabe 1.3: Gemessene Sprungantwort]]

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	'''(2)'''  The direct signal transmission factor can be read from  $x_1(t)$  and  $y_1(t)$  when the transient has decayed:		'''(2)'''  The direct signal transmission factor can be read from  $x_1(t)$  and  $y_1(t)$  when the transient has decayed:
	:$$H(f =0) = \frac{y_1(t \rightarrow \infty)}{x_1(t \rightarrow \infty)}=		:$$H(f =0) = \frac{y_1(t \rightarrow \infty)}{x_1(t \rightarrow \infty)}=\frac{ {\rm 1\, V} }{ {\rm 4\, V} } \hspace{0.15cm}\underline{= 0.25}.$$
	\frac{ {\rm 1\, V} }{ {\rm 4\, V} } \hspace{0.15cm}\underline{= 0.25}.$$


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	*For  $t < 0$  and  $t ≥ T$,    $h(t)=0$  always holds.		*For  $t < 0$  and  $t ≥ T$,    $h(t)=0$  always holds.
	*The value  $h(t = 0)$  at exactly  $t = 0$  must be determined from the mean value between the left-hand and right-hand limits:		*The value  $h(t = 0)$  at exactly  $t = 0$  must be determined from the mean value between the left-hand and right-hand limits:
	:$$h(t=0) = {1}/{2} \cdot \left[ \lim_{\varepsilon		:$$h(t=0) = {1}/{2} \cdot \left[ \lim_{\varepsilon\hspace{0.03cm} \to \hspace{0.03cm}0} h(- \varepsilon)+ \lim_{\varepsilon\hspace{0.03cm} \to \hspace{0.03cm} 0} h(+ \varepsilon)\right] = \left[ 0 + {0.5}/{T}\right] = {0.25}/{T}= 250 \cdot{1}/{ {\rm s} }.$$
	\hspace{0.03cm} \to \hspace{0.03cm}0} h(- \varepsilon)+ \lim_{\varepsilon
	\hspace{0.03cm} \to \hspace{0.03cm} 0} h(+ \varepsilon)\right] = \left[ 0 + {0.5}/{T}\right] = {0.25}/{T}= 250 \cdot{1}/{ {\rm s} }.$$


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	*For the other time points considered the following is obtained as indicated in the graph:		*For the other time points considered the following is obtained as indicated in the graph:
	:$$y_2(t = 0) = A \cdot \big[\sigma(0.5 \cdot T) - \sigma(-0.5 \cdot T)\big] =		:$$y_2(t = 0) = A \cdot \big[\sigma(0.5 \cdot T) - \sigma(-0.5 \cdot T)\big] ={\rm 2\, V}\cdot \left[0.1875 - 0\right] \hspace{0.15cm}\underline{= {\rm 0.375\, V}},$$
	{\rm 2\, V}\cdot \left[0.1875 - 0\right] \hspace{0.15cm}\underline{= {\rm 0.375\, V}},$$		:$$y_2(t = T/2) = y_2(t = 1\,{\rm ms}) =A \cdot \big[\sigma( T) - \sigma(0)\big] ={\rm 2\, V}\cdot \left[0.25 - 0\right] \hspace{0.15cm}\underline{= {\rm 0.5\, V}},$$
	:$$y_2(t = T/2) = y_2(t = 1\,{\rm ms}) =A \cdot \big[\sigma( T) - \sigma(0)\big] =		:$$y_2(t = T) = A \cdot \big[\sigma(1.5 \cdot T) - \sigma(0.5 \cdot T)\big] ={\rm 2\, V}\cdot \big[0.25 - 0.1875\big] \hspace{0.15cm}\underline{= {\rm 0.125\, V}}.$$
	{\rm 2\, V}\cdot \left[0.25 - 0\right] \hspace{0.15cm}\underline{= {\rm 0.5\, V}},$$
	:$$y_2(t = T) = A \cdot \big[\sigma(1.5 \cdot T) - \sigma(0.5 \cdot T)\big] =
	{\rm 2\, V}\cdot \big[0.25 - 0.1875\big] \hspace{0.15cm}\underline{= {\rm 0.125\, V}}.$$

	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^1.2 System Description in Time Domain^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^1.2 System Description in Time Domain^]]
			[[de:Aufgaben:1.3_Gemessene_Sprungantwort]]

Guenter at 10:21, 12 July 2021

2021-07-12T10:21:31Z

Guenter at 10:05, 12 July 2021

2021-07-12T10:05:04Z

Oezer at 00:12, 1 July 2021

2021-07-01T00:12:42Z

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	'''(4)'''  The step response  $σ(t)$  is the integral over the impulse response  $h(t)$.		'''(4)'''  The step response  $σ(t)$  is the integral over the impulse response  $h(t)$.
	*Thus,  $h(t)$  is obtained from  $σ(t)$  by differentiation with respect to time.		*Thus,  $h(t)$  is obtained from  $σ(t)$  by differentiation with respect to time.
	*In the range  $0 < t < T$  ~~therefore~~ the following is valid:		*Therefore, in the range  $0 < t < T$  the following is valid:
	:$$h(t) = \frac{{\rm d}\hspace{0.1cm}\sigma(t)}{{\rm d}t}= 0.5 \cdot\left( \frac{1}{T} - 0.5 (\frac{2t}{T^2})\right) = \frac{0.5}{T} \cdot (1- \frac{t}{T})$$		:$$h(t) = \frac{{\rm d}\hspace{0.1cm}\sigma(t)}{{\rm d}t}= 0.5 \cdot\left( \frac{1}{T} - 0.5 (\frac{2t}{T^2})\right) = \frac{0.5}{T} \cdot (1- \frac{t}{T})$$
	:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$		:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$

Oezer at 00:11, 1 July 2021

2021-07-01T00:11:16Z

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	'''(4)'''  The step response  $σ(t)$  is the integral over the impulse response  $h(t)$.		'''(4)'''  The step response  $σ(t)$  is the integral over the impulse response  $h(t)$.
	*Thus,  $h(t)$  is obtained from  $σ(t)$  by differentiation with respect to time.		*Thus,  $h(t)$  is obtained from  $σ(t)$  by differentiation with respect to time.
	*n the range  $0 < t < T$  therefore the following is valid:		*In the range  $0 < t < T$  therefore the following is valid:
	:$$h(t) = \frac{{\rm d}\hspace{0.1cm}\sigma(t)}{{\rm d}t}= 0.5 \cdot\left( \frac{1}{T} - 0.5 (\frac{2t}{T^2})\right) = \frac{0.5}{T} \cdot (1- \frac{t}{T})$$		:$$h(t) = \frac{{\rm d}\hspace{0.1cm}\sigma(t)}{{\rm d}t}= 0.5 \cdot\left( \frac{1}{T} - 0.5 (\frac{2t}{T^2})\right) = \frac{0.5}{T} \cdot (1- \frac{t}{T})$$
	:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$		:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$

Oezer at 00:10, 1 July 2021

2021-07-01T00:10:50Z

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	*At time  $t = T = 2 \ \rm ms$  the step response reaches its final value  $0.25$.		*At time  $t = T = 2 \ \rm ms$  the step response reaches its final value  $0.25$.
	*For  $t = T/2 = 1 \ \rm ms$  the numerical value  $3/16 \; \underline{\: = \: 0.1875}$ is obtained.		*For  $t = T/2 = 1 \ \rm ms$  the numerical value  $3/16 \; \underline{\: = \: 0.1875}$ is obtained.
	*Please note that the step response  $σ(t)$  as well as the step function  $γ(t)$  ~~has~~ no unit.		*Please note that the step response  $σ(t)$  as well as the step function  $γ(t)$  have no unit.

Oezer at 00:09, 1 July 2021

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	'''(3)'''  The step response  $σ(t)$  is equal to the output signal  $y(t)$ if  $x(t) = γ(t)$  is applied to the input.		'''(3)'''  The step response  $σ(t)$  is equal to the output signal  $y(t)$ if  $x(t) = γ(t)$  is applied to the input.
	*~~Wegen~~  $x_1(t) = 4 \hspace{0.05cm} \rm {V} · γ(t)$  ~~gilt somit im Bereich von~~  $0$  ~~bis~~  $T = 2 \ \rm ms$:		*Because of  $x_1(t) = 4 \hspace{0.05cm} \rm {V} · γ(t)$  the following holds in the range from  $0$  to  $T = 2 \ \rm ms$:
	:$$\sigma(t) = \frac{y_1(t)}{ {\rm 4\, V} } = 0.5 \cdot\big( {t}/{T} - 0.5 ({t}/{T})^2\big).$$		:$$\sigma(t) = \frac{y_1(t)}{ {\rm 4\, V} } = 0.5 \cdot\big( {t}/{T} - 0.5 ({t}/{T})^2\big).$$
	*~~Zum Zeitpunkt~~  $t = T = 2 \ \rm ms$  ~~erreicht die Sprungantwort ihren Endwert~~  $0.25$.		*At time  $t = T = 2 \ \rm ms$  the step response reaches its final value  $0.25$.
	*~~Für~~  $t = T/2 = 1 \ \rm ms$  ~~ergibt sich der Zahlenwert~~  $3/16 \; \underline{\: = \: 0.1875}$.		*For  $t = T/2 = 1 \ \rm ms$  the numerical value  $3/16 \; \underline{\: = \: 0.1875}$ is obtained.
	*~~Beachten Sie bitte, dass die Sprungantwort~~  $σ(t)$  ~~ebenso wie die Sprungfunktion~~  $γ(t)$  ~~keine Einheit besitzt~~.		*Please note that the step response  $σ(t)$  as well as the step function  $γ(t)$  has no unit.



	[[File:P_ID840__LZI_A_1_3d.png \|~~Berechnete Impulsantwort~~ \| rechts\|frame]]		[[File:P_ID840__LZI_A_1_3d.png \|Computed impulse response \| rechts\|frame]]
	'''(4)'''  ~~Die Sprungantwort~~  $σ(t)$  ~~ist das Integral über die Impulsantwort~~  $h(t)$.		'''(4)'''  The step response  $σ(t)$  is the integral over the impulse response  $h(t)$.
	*~~Damit ergibt sich~~  $h(t)$  ~~aus~~  $σ(t)$  ~~durch Differentiation nach der Zeit~~.		*Thus,  $h(t)$  is obtained from  $σ(t)$  by differentiation with respect to time.
	*~~Im Bereich~~  $0 < t < T$  ~~gilt deshalb~~:		*n the range  $0 < t < T$  therefore the following is valid:
	:$$h(t) = \frac{{\rm d}\hspace{0.1cm}\sigma(t)}{{\rm d}t}= 0.5 \cdot\left( \frac{1}{T} - 0.5 (\frac{2t}{T^2})\right) = \frac{0.5}{T} \cdot (1- \frac{t}{T})$$		:$$h(t) = \frac{{\rm d}\hspace{0.1cm}\sigma(t)}{{\rm d}t}= 0.5 \cdot\left( \frac{1}{T} - 0.5 (\frac{2t}{T^2})\right) = \frac{0.5}{T} \cdot (1- \frac{t}{T})$$
	:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$		:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$
	:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 2\, ms}) = h(t = T) \hspace{0.15cm}\underline{= 0}.$$		:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 2\, ms}) = h(t = T) \hspace{0.15cm}\underline{= 0}.$$
	*~~Für~~  $t < 0$  ~~und~~  $t ≥ T$  ~~gilt stets~~  $h(t)=0$.		*For  $t < 0$  and  $t ≥ T$  this,  $h(t)=0$, always holds.
	*~~Der Wert~~  $h(t = 0)$  ~~bei exakt~~  $t = 0$  ~~muss aus dem Mittelwert zwischen links~~- ~~und rechtsseitigem Grenzwert ermittelt werden~~:		*The value  $h(t = 0)$  at exactly  $t = 0$  must be determined from the mean value between the left-hand and right-hand limits:
	:$$h(t=0) = {1}/{2} \cdot \left[ \lim_{\varepsilon		:$$h(t=0) = {1}/{2} \cdot \left[ \lim_{\varepsilon
	\hspace{0.03cm} \to \hspace{0.03cm}0} h(- \varepsilon)+ \lim_{\varepsilon		\hspace{0.03cm} \to \hspace{0.03cm}0} h(- \varepsilon)+ \lim_{\varepsilon
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	[[File:P_ID829__LZI_A_1_3e.png \| ~~Berechnete Rechteckantwort~~\| rechts\|frame]]		[[File:P_ID829__LZI_A_1_3e.png \| computed rectangular response\| rechts\|frame]]
	'''(5)'''  ~~Der Rechteckimpuls~~  $x_2(t)$  ~~kann auch als die Differenz zweier um~~  $±T/2$  ~~verschobener Sprünge dargestellt werden~~:		'''(5)'''  The rectangular pulse  $x_2(t)$  can also be represented as the difference of two steps shifted by  $±T/2$ :
	:$$x_2(t) = A \cdot \big[\gamma(t + {T}/{2}) - \gamma(t - {T}/{2})\big].$$		:$$x_2(t) = A \cdot \big[\gamma(t + {T}/{2}) - \gamma(t - {T}/{2})\big].$$
	*~~Damit ist das Ausgangssignal gleich der Differenz zweier um~~  $±T/2$  ~~verschobener Sprungantworten~~:		*Hence, the output signal is equal to the difference of two step responses shifted by  $±T/2$ :
	:$$y_2(t) = A \cdot \big[\sigma(t + {T}/{2}) - \sigma(t - {T}/{2})\big].$$		:$$y_2(t) = A \cdot \big[\sigma(t + {T}/{2}) - \sigma(t - {T}/{2})\big].$$
	*~~Für~~  $t = \: -T/2 = -1\ \rm ms$  ~~gilt~~  $y_2(t) \;\underline{ = 0}$.		*For  $t = \: -T/2 = -1\ \rm ms$  the following holds:  $y_2(t) \;\underline{ = 0}$.

	*~~Für die weiteren betrachteten Zeitpunkte erhält man wie~~ in ~~der Grafik angegeben~~:		*For the other time points considered the following is obtained as indicated in the graph:
	:$$y_2(t = 0) = A \cdot \big[\sigma(0.5 \cdot T) - \sigma(-0.5 \cdot T)\big] =		:$$y_2(t = 0) = A \cdot \big[\sigma(0.5 \cdot T) - \sigma(-0.5 \cdot T)\big] =
	{\rm 2\, V}\cdot \left[0.1875 - 0\right] \hspace{0.15cm}\underline{= {\rm 0.375\, V}},$$		{\rm 2\, V}\cdot \left[0.1875 - 0\right] \hspace{0.15cm}\underline{= {\rm 0.375\, V}},$$

Oezer at 23:48, 30 June 2021

2021-06-30T23:48:20Z

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	'''(3)'''  The step response  $σ(t)$  is equal to the output signal  $y(t)$, if  $x(t) = γ(t)$  is applied to the input.		'''(3)'''  The step response  $σ(t)$  is equal to the output signal  $y(t)$ if  $x(t) = γ(t)$  is applied to the input.
	*Wegen  $x_1(t) = 4 \hspace{0.05cm} \rm {V} · γ(t)$  gilt somit im Bereich von  $0$  bis  $T = 2 \ \rm ms$:		*Wegen  $x_1(t) = 4 \hspace{0.05cm} \rm {V} · γ(t)$  gilt somit im Bereich von  $0$  bis  $T = 2 \ \rm ms$:
	:$$\sigma(t) = \frac{y_1(t)}{ {\rm 4\, V} } = 0.5 \cdot\big( {t}/{T} - 0.5 ({t}/{T})^2\big).$$		:$$\sigma(t) = \frac{y_1(t)}{ {\rm 4\, V} } = 0.5 \cdot\big( {t}/{T} - 0.5 ({t}/{T})^2\big).$$

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	</quiz>		</quiz>

	===~~Sample solution~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Approaches 2 und 3</u> are correct:		'''(1)'''  <u>Approaches 2 und 3</u> are correct:
	*The output signal is  $y_1(t)=0$ as long as the input signal is  $x_1(t) = 0$ . This means that there is a causal system on hand.		*The output signal is  $y_1(t)=0$  as long as the input signal is  $x_1(t) = 0$.  This means that there is a causal system on hand.
	*One could have arrived at the same result just by considering the statement "the output signal was measured". Only causal systems are realisable and only in realisable systems something can be measured.		*One could have arrived at the same result just by considering the statement "the output signal was measured".  Only causal systems are realisable and only in realisable systems something can be measured.
	*The input signal  $x_1(t)$  can be interpreted as a direct signal for very large times  $(t \gg 0)$ . If  $H(f)$  was a high-pass filter, then  $y_1(t)$  would have to go towards zero for  $t → ∞$ . This means:   $H(f)$  represents a low-pass filter.		*The input signal  $x_1(t)$  can be interpreted as a direct  (DC)  signal for very large times  $(t \gg 0)$.  If  $H(f)$  was a high-pass filter, then  $y_1(t)$  would have to go towards zero for  $t → ∞$.  This means:   $H(f)$  represents a low-pass filter.


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	'''(3)'''  The step response  $σ(t)$  is equal to the output signal  $y(t)$ if  $x(t) = γ(t)$  is applied to the input.		'''(3)'''  The step response  $σ(t)$  is equal to the output signal  $y(t)$,  if  $x(t) = γ(t)$  is applied to the input.
	*Because of  $x_1(t) = 4 \hspace{0.05cm} \rm {V} · γ(t)$  the following holds in the range from  $0$  to  $T = 2 \ \rm ms$:		*Because of  $x_1(t) = 4 \hspace{0.05cm} \rm {V} · γ(t)$  the following holds in the range from  $0$  to  $T = 2 \ \rm ms$:
	:$$\sigma(t) = \frac{y_1(t)}{ {\rm 4\, V} } = 0.5 \cdot\big( {t}/{T} - 0.5 ({t}/{T})^2\big).$$		:$$\sigma(t) = \frac{y_1(t)}{ {\rm 4\, V} } = 0.5 \cdot\big( {t}/{T} - 0.5 ({t}/{T})^2\big).$$
	*At time  $t = T = 2 \ \rm ms$  the step response reaches its final value  $0.25$.		*At time  $t = T = 2 \ \rm ms$  the step response reaches its final value  $0.25$.
	*For  $t = T/2 = 1 \ \rm ms$  the numerical value  $3/16 \; \underline{\: = \: 0.1875}$ is obtained.		*For  $t = T/2 = 1 \hspace{0.15cm} \rm ms$  the numerical value  $3/16 \; \underline{\: = \: 0.1875}$ is obtained.
	*Please note that the step response  $σ(t)$  as well as the step function  $γ(t)$  have no unit.		*Please note that the step response  $σ(t)$  as well as the step function  $γ(t)$  have no unit.



	[[File:P_ID840__LZI_A_1_3d.png ~~\|Computed impulse response~~ \| rechts\|frame]]		[[File:P_ID840__LZI_A_1_3d.png\|rechts\|frame \|Impulse response]]
	'''(4)'''  The step response  $σ(t)$  is the integral over the impulse response  $h(t)$.		'''(4)'''  The step response  $σ(t)$  is the integral over the impulse response  $h(t)$.
	*Thus,  $h(t)$  is obtained from  $σ(t)$  by differentiation with respect to time.		*Thus,  $h(t)$  is obtained from  $σ(t)$  by differentiation with respect to time.
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	:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$		:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$
	:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 2\, ms}) = h(t = T) \hspace{0.15cm}\underline{= 0}.$$		:$$\Rightarrow \hspace{0.2cm} h(t = {\rm 2\, ms}) = h(t = T) \hspace{0.15cm}\underline{= 0}.$$
	*For  $t < 0$  and  $t ≥ T$  ~~this,~~  $h(t)=0$, always holds.		*For  $t < 0$  and  $t ≥ T$,    $h(t)=0$  always holds.
	*The value  $h(t = 0)$  at exactly  $t = 0$  must be determined from the mean value between the left-hand and right-hand limits:		*The value  $h(t = 0)$  at exactly  $t = 0$  must be determined from the mean value between the left-hand and right-hand limits:
	:$$h(t=0) = {1}/{2} \cdot \left[ \lim_{\varepsilon		:$$h(t=0) = {1}/{2} \cdot \left[ \lim_{\varepsilon
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	[[File:P_ID829__LZI_A_1_3e.png ~~\| computed rectangular response~~\| rechts\|frame]]		[[File:P_ID829__LZI_A_1_3e.png \| rechts\|frame\| Rectangular response]]
	'''(5)'''  The rectangular pulse  $x_2(t)$  can also be represented as the difference of two steps shifted by  $±T/2$ :		'''(5)'''  The symmetrical rectangular pulse  $x_2(t)$  can also be represented as the difference of two steps shifted by  $±T/2$:
	:$$x_2(t) = A \cdot \big[\gamma(t + {T}/{2}) - \gamma(t - {T}/{2})\big].$$		:$$x_2(t) = A \cdot \big[\gamma(t + {T}/{2}) - \gamma(t - {T}/{2})\big].$$
	*Hence, the output signal is equal to the difference of two step responses shifted by  $±T/2$~~ ~~:		*Hence, the output signal is equal to the difference of two step responses shifted by  $±T/2$:
	:$$y_2(t) = A \cdot \big[\sigma(t + {T}/{2}) - \sigma(t - {T}/{2})\big].$$		:$$y_2(t) = A \cdot \big[\sigma(t + {T}/{2}) - \sigma(t - {T}/{2})\big].$$
	*For  $t = \: -T/2 = -1\ \rm ms$  the following holds:  $y_2(t) \;\underline{ = 0}$.		*For  $t = \: -T/2 = -1\ \rm ms$  the following holds:   $y_2(t) \;\underline{ = 0}$.

	*For the other time points considered the following is obtained as indicated in the graph:		*For the other time points considered the following is obtained as indicated in the graph:

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	[[File:P_ID817__LZI_A_1_3.png \|right\|frame\|Measured step response]]		[[File:P_ID817__LZI_A_1_3.png \|right\|frame\|Measured step response]]
	A step-shaped signal $$x_1(t) = 4\hspace{0.05cm} {\rm V} \cdot \gamma(t)$$ (blue curve) is applied to the input of a linear time-invariant (LTI) transmission system		A step-shaped signal $$x_1(t) = 4\hspace{0.05cm} {\rm V} \cdot \gamma(t)$$  (blue curve in the upper sketch)  is applied to the input of a linear time-invariant (LTI) transmission system
	*with the frequency response  $H(f)$		*with the frequency response  $H(f)$
	*and the impulse response  $h(t)$.		*and the impulse response  $h(t)$.
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	In the last subtask  '''(5)'''  the output signal  $y_2(t)$  is to be determined if a symmetrical rectangular pulse  $x_2(t)$  of duration  $T = 2 \hspace{0.~~05cm~~} {\rm ms}$  is applied to the input (see red curve in the upper graph).		In the last subtask  '''(5)'''  the output signal  $y_2(t)$  is to be determined if a symmetrical rectangular pulse  $x_2(t)$  of duration  $T = 2 \hspace{0.15cm} {\rm ms}$  is applied to the input (see red curve in the upper graph).


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	''Please note:''		''Please note:''
	*The ~~task~~ belongs to the chapter  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain\|System Description in Time Domain]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain\|System Description in Time Domain]].
	*The rectangular pulse  $x_2(t)$  can also be written as follows with   $A = 2 \hspace{0.05cm} \text{V}$ :		*The rectangular pulse  $x_2(t)$  can also be written as follows with   $A = 2 \hspace{0.05cm} \text{V}$ :
	:$$x_2(t) = A \cdot \big [\gamma(t + {T}/{2}) - \gamma(t - {T}/{2})\big ].$$		:$$x_2(t) = A \cdot \big [\gamma(t + {T}/{2}) - \gamma(t - {T}/{2})\big ].$$
	*The frequency response  $H(f)$  of the LTI system considered here can be taken from the exercise description of  [[Aufgaben:~~Aufgabe_3~~.8:~~_Dreimal_Faltung~~%3F\|Exercise 3.8]]  in the book "Signal Representation”. However, the abscissa and ordinate parameters have to be adjusted accordingly.		*The frequency response  $H(f)$  of the LTI system considered here can be taken from the exercise description of  [[Aufgaben:Exercise_3.8:_Triple_Convolution%3F\|Exercise 3.8]]  in the book  "Signal Representation”.  However, the abscissa and ordinate parameters have to be adjusted accordingly.
	*For the solution of the problem on hand, though,  $H(f)$  is not explicitly required.		*For the solution of the problem on hand, though,  $H(f)$  is not explicitly required.

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	{What is the step response  $σ(t)$? What is its value at  $t = T/2$ ?		{What is the step response  $σ(t)$?  What is its value at  $t = T/2$ ?
	\|type="{}"}		\|type="{}"}
	$σ(t = \rm 1 \: ms) \ = \ $ { 0.1875 5% }		$σ(t = \rm 1 \: ms) \ = \ $ { 0.1875 5% }


	{Compute the impulse response  $h(t)$  of the system. What values does it have at ~~the~~ times  $t = T/2$  and  $t = T$?		{Compute the impulse response  $h(t)$  of the system.  What values does it have at times  $t = T/2$  and  $t = T$?
	\|type="{}"}		\|type="{}"}
	$h(t = \rm 1 \: ms) \ =\ $ { 125 }  $\text {1/s}$		$h(t = \rm 1 \: ms) \ =\ $ { 125 }  $\text {1/s}$
	$h(t = \rm 2 \: ms) \ =\ ${ 0. }  $\text {1/s}$		$h(t = \rm 2 \: ms) \ =\ ${ 0. }  $\text {1/s}$

	{The rectangular pulse  $x_2(t)$  is applied to the input. What is the output signal  $y_2(t)$  at the times  $t_1 = -1 \text { ms}$,  $t_2 = 0$ ,  $t_3 = +1 \text { ms}$  and  $t_4 = +2 \text { ms}$?		{The rectangular pulse  $x_2(t)$  is applied to the input.  What is the output signal  $y_2(t)$  at the times  $t_1 = -1 \text { ms}$,  $t_2 = 0$,  $t_3 = +1 \text { ms}$  and  $t_4 = +2 \text { ms}$?
	\|type="{}"}		\|type="{}"}
	$y_2(t = t_1) \ =\ $ { 0. }  $\text {V}$		$y_2(t = t_1) \ =\ $ { 0. }  $\text {V}$