Aufgaben:Exercise 1.3: Measured Step Response - Revision history

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

Oezer at 00:11, 1 July 2021

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Oezer at 00:10, 1 July 2021

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Oezer at 00:09, 1 July 2021

2021-07-01T00:09:06Z

Oezer at 23:48, 30 June 2021

2021-06-30T23:48:20Z

Oezer at 23:47, 30 June 2021

2021-06-30T23:47:42Z

Oezer at 15:47, 29 June 2021

2021-06-29T15:47:57Z

Oezer at 15:42, 29 June 2021

2021-06-29T15:42:40Z

@@ Line 72: / Line 72: @@
 </quiz>
-===Sample solution===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>Approaches 2 und 3</u> are correct:
-*The output signal is&nbsp; $y_1(t)=0$ as long as the input signal is&nbsp; $x_1(t) = 0$&nbsp;. This means that there is a causal system on hand.
+*The output signal is&nbsp; $y_1(t)=0$&nbsp; as long as the input signal is&nbsp; $x_1(t) = 0$.&nbsp; 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".&nbsp; Only causal systems are realisable and only in realisable systems something can be measured.
-*The input signal&nbsp; $x_1(t)$&nbsp; can be interpreted as a direct signal for very large times&nbsp; $(t \gg 0)$&nbsp;. If&nbsp; $H(f)$&nbsp; was a high-pass filter, then&nbsp; $y_1(t)$&nbsp; would have to go towards zero for&nbsp; $t → ∞$&nbsp;. This means: &nbsp; $H(f)$&nbsp; represents a low-pass filter.
+*The input signal&nbsp; $x_1(t)$&nbsp; can be interpreted as a direct&nbsp; (DC)&nbsp; signal for very large times&nbsp; $(t \gg 0)$.&nbsp; If&nbsp; $H(f)$&nbsp; was a high-pass filter, then&nbsp; $y_1(t)$&nbsp; would have to go towards zero for&nbsp; $t → ∞$.&nbsp; This means: &nbsp; $H(f)$&nbsp; represents a low-pass filter.
@@ Line 86: / Line 86: @@
-'''(3)'''&nbsp; The step response&nbsp; $σ(t)$&nbsp; is equal to the output signal&nbsp; $y(t)$ if&nbsp; $x(t) = γ(t)$&nbsp; is applied to the input.
+'''(3)'''&nbsp; The step response&nbsp; $σ(t)$&nbsp; is equal to the output signal&nbsp; $y(t)$,&nbsp; if&nbsp; $x(t) = γ(t)$&nbsp; is applied to the input.
 *Because of&nbsp; $x_1(t) = 4 \hspace{0.05cm} \rm {V} · γ(t)$&nbsp; the following holds in the range from&nbsp; $0$&nbsp; to&nbsp; $T = 2 \ \rm ms$:
 :$$\sigma(t) = \frac{y_1(t)}{ {\rm 4\, V} } = 0.5 \cdot\big( {t}/{T} - 0.5 ({t}/{T})^2\big).$$
 *At time&nbsp; $t = T = 2 \ \rm ms$&nbsp; the step response reaches its final value&nbsp; $0.25$.
-*For&nbsp; $t = T/2 = 1 \ \rm ms$&nbsp; the numerical value&nbsp; $3/16  \; \underline{\: = \: 0.1875}$ is obtained.
+*For&nbsp; $t = T/2 = 1 \hspace{0.15cm} \rm ms$&nbsp; the numerical value&nbsp; $3/16  \; \underline{\: = \: 0.1875}$ is obtained.
 *Please note that the step response&nbsp; $σ(t)$&nbsp; as well as the step function&nbsp; $γ(t)$&nbsp; 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)'''&nbsp; The step response&nbsp; $σ(t)$&nbsp;  is the integral over the impulse response&nbsp; $h(t)$.
 *Thus,&nbsp; $h(t)$&nbsp; is obtained from&nbsp; $σ(t)$&nbsp; by differentiation with respect to time.
@@ Line 102: / Line 102: @@
 :$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$
 :$$\Rightarrow \hspace{0.2cm} h(t = {\rm 2\, ms}) = h(t = T) \hspace{0.15cm}\underline{= 0}.$$
-*For&nbsp; $t < 0$&nbsp; and&nbsp; $t ≥ T$&nbsp; this,&nbsp; $h(t)=0$, always holds.
+*For&nbsp; $t < 0$&nbsp; and&nbsp; $t ≥ T$,&nbsp; &nbsp; $h(t)=0$&nbsp; always holds.
 *The value&nbsp; $h(t = 0)$&nbsp; at exactly&nbsp; $t = 0$&nbsp; must be determined from the mean value between the left-hand and right-hand limits:
 :$$h(t=0) = {1}/{2} \cdot \left[ \lim_{\varepsilon
@@ Line 110: / Line 110: @@
-[[File:P_ID829__LZI_A_1_3e.png | computed rectangular response| rechts|frame]]
+[[File:P_ID829__LZI_A_1_3e.png | rechts|frame| Rectangular response]]
-'''(5)'''&nbsp; The rectangular pulse&nbsp; $x_2(t)$&nbsp; can also be represented as the difference of two steps shifted by&nbsp; $±T/2$&nbsp;:
+'''(5)'''&nbsp; The symmetrical rectangular pulse&nbsp; $x_2(t)$&nbsp; can also be represented as the difference of two steps shifted by&nbsp; $±T/2$:&nbsp;
 :$$x_2(t) = A \cdot \big[\gamma(t + {T}/{2}) - \gamma(t - {T}/{2})\big].$$
-*Hence, the output signal is equal to the difference of two step responses shifted by&nbsp; $±T/2$&nbsp;:
+*Hence, the output signal is equal to the difference of two step responses shifted by&nbsp; $±T/2$:
 :$$y_2(t) = A \cdot \big[\sigma(t + {T}/{2}) - \sigma(t - {T}/{2})\big].$$
-*For&nbsp; $t = \: -T/2 = -1\ \rm ms$&nbsp; the following holds:&nbsp; $y_2(t) \;\underline{ = 0}$.
+*For&nbsp; $t = \: -T/2 = -1\ \rm ms$&nbsp; the following holds: &nbsp; $y_2(t) \;\underline{ = 0}$.
 *For the other time points considered the following is obtained as indicated in the graph:

@@ Line 3: / Line 3: @@
 [[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 $$x_1(t) = 4\hspace{0.05cm} {\rm V} \cdot \gamma(t)$$&nbsp; (blue curve in the upper sketch)&nbsp; is applied to the input of a linear time-invariant (LTI) transmission system
 *with the frequency response&nbsp;  $H(f)$
 *and the impulse response&nbsp; $h(t)$.
@@ Line 15: / Line 15: @@
-In the last subtask&nbsp; '''(5)'''&nbsp; the output signal&nbsp; $y_2(t)$&nbsp; is to be determined if a symmetrical rectangular pulse&nbsp; $x_2(t)$&nbsp; of duration&nbsp; $T = 2 \hspace{0.05cm} {\rm ms}$&nbsp;  is applied to the input (see red curve in the upper graph).
+In the last subtask&nbsp; '''(5)'''&nbsp; the output signal&nbsp; $y_2(t)$&nbsp; is to be determined if a symmetrical rectangular pulse&nbsp; $x_2(t)$&nbsp; of duration&nbsp; $T = 2 \hspace{0.15cm} {\rm ms}$&nbsp;  is applied to the input (see red curve in the upper graph).
@@ Line 26: / Line 26: @@
 ''Please note:''
-*The task belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain|System Description in Time Domain]].
+*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain|System Description in Time Domain]].
 *The rectangular pulse&nbsp; $x_2(t)$&nbsp; can also be written as follows with &nbsp; $A = 2 \hspace{0.05cm} \text{V}$&nbsp;:
 :$$x_2(t) = A \cdot \big [\gamma(t + {T}/{2}) - \gamma(t - {T}/{2})\big ].$$
-*The frequency response&nbsp; $H(f)$&nbsp; of the LTI system considered here can be taken from the exercise description of&nbsp; [[Aufgaben:Aufgabe_3.8:_Dreimal_Faltung%3F|Exercise 3.8]]&nbsp; in the book "Signal Representation”. However, the abscissa and ordinate parameters have to be adjusted accordingly.
+*The frequency response&nbsp; $H(f)$&nbsp; of the LTI system considered here can be taken from the exercise description of&nbsp; [[Aufgaben:Exercise_3.8:_Triple_Convolution%3F|Exercise 3.8]]&nbsp; in the book&nbsp; "Signal Representation”.&nbsp; However, the abscissa and ordinate parameters have to be adjusted accordingly.
 *For the solution of the problem on hand, though,&nbsp; $H(f)$&nbsp; is not explicitly required.
@@ Line 52: / Line 52: @@
-{What is the step response&nbsp; $σ(t)$? What is its value at&nbsp; $t = T/2$&nbsp;?
+{What is the step response&nbsp; $σ(t)$?&nbsp; What is its value at&nbsp; $t = T/2$&nbsp;?
 |type="{}"}
 $σ(t = \rm 1 \: ms) \ = \ $ { 0.1875 5%  }
-{Compute the impulse response&nbsp; $h(t)$&nbsp; of the system. What values does it have at the times&nbsp; $t = T/2$&nbsp; and&nbsp; $t = T$?
+{Compute the impulse response&nbsp; $h(t)$&nbsp; of the system.&nbsp; What values does it have at times&nbsp; $t = T/2$&nbsp; and&nbsp; $t = T$?
 |type="{}"}
 $h(t = \rm 1 \: ms)  \ =\ $ { 125 } &nbsp;$\text {1/s}$
 $h(t = \rm 2 \: ms) \ =\ ${ 0. } &nbsp;$\text {1/s}$
-{The rectangular pulse&nbsp; $x_2(t)$&nbsp; is applied to the input. What is the output signal&nbsp; $y_2(t)$&nbsp; at the times&nbsp; $t_1 = -1  \text { ms}$,&nbsp; $t_2 = 0$ ,&nbsp; $t_3 = +1   \text { ms}$ &nbsp;and&nbsp; $t_4 = +2  \text { ms}$?
+{The rectangular pulse&nbsp; $x_2(t)$&nbsp; is applied to the input.&nbsp; What is the output signal&nbsp; $y_2(t)$&nbsp; at the times&nbsp; $t_1 = -1  \text { ms}$,&nbsp; $t_2 = 0$,&nbsp; $t_3 = +1   \text { ms}$ &nbsp;and&nbsp; $t_4 = +2  \text { ms}$?
 |type="{}"}
 $y_2(t = t_1) \ =\ $ { 0. } &nbsp;$\text {V}$

@@ Line 98: / Line 98: @@
 '''(4)'''&nbsp; The step response&nbsp; $σ(t)$&nbsp;  is the integral over the impulse response&nbsp; $h(t)$.
 *Thus,&nbsp; $h(t)$&nbsp; is obtained from&nbsp; $σ(t)$&nbsp; by differentiation with respect to time.
-*In the range&nbsp; $0 < t < T$&nbsp; therefore the following is valid:
+*Therefore, in the range&nbsp; $0 < t < T$&nbsp; the following is valid:
 :$$h(t) = \frac{{\rm d}\hspace{0.1cm}\sigma(t)}{{\rm d}t}=  0.5 \cdot\left( \frac{1}{T} - 0.5 (\frac{2t}{T^2})\right)  = \frac{0.5}{T} \cdot (1- \frac{t}{T})$$
 :$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$

@@ Line 98: / Line 98: @@
 '''(4)'''&nbsp; The step response&nbsp; $σ(t)$&nbsp;  is the integral over the impulse response&nbsp; $h(t)$.
 *Thus,&nbsp; $h(t)$&nbsp; is obtained from&nbsp; $σ(t)$&nbsp; by differentiation with respect to time.
-*n the range&nbsp; $0 < t < T$&nbsp; therefore the following is valid:
+*In the range&nbsp; $0 < t < T$&nbsp; therefore the following is valid:
 :$$h(t) = \frac{{\rm d}\hspace{0.1cm}\sigma(t)}{{\rm d}t}=  0.5 \cdot\left( \frac{1}{T} - 0.5 (\frac{2t}{T^2})\right)  = \frac{0.5}{T} \cdot (1- \frac{t}{T})$$
 :$$\Rightarrow \hspace{0.2cm} h(t = {\rm 1\, ms}) = h(t = T/2) = \frac{0.25}{T} \hspace{0.15cm}\underline{= 125 \cdot{1}/{ {\rm s} } },$$

@@ Line 91: / Line 91: @@
 *At time&nbsp; $t = T = 2 \ \rm ms$&nbsp; the step response reaches its final value&nbsp; $0.25$.
 *For&nbsp; $t = T/2 = 1 \ \rm ms$&nbsp; the numerical value&nbsp; $3/16  \; \underline{\: = \: 0.1875}$ is obtained.
-*Please note that the step response&nbsp; $σ(t)$&nbsp; as well as the step function&nbsp; $γ(t)$&nbsp; has no unit.
+*Please note that the step response&nbsp; $σ(t)$&nbsp; as well as the step function&nbsp; $γ(t)$&nbsp; have no unit.

@@ Line 75: / Line 75: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>Approaches 2 und 3</u> are correct:
-*The output signal is&nbsp; $y_1(t)=0$ as long as the input signal is&nbsp; $x_1(t) = 0$&nbsp;.  This means that there is a causal system on hand.
+*The output signal is&nbsp; $y_1(t)=0$ as long as the input signal is&nbsp; $x_1(t) = 0$&nbsp;. This means that there is a causal system on hand.
-*Zum gleichen Ergebnis hätte man allein durch die Aussage „das Ausgangssignal wurde gemessen” kommen können. Nur kausale Systeme sind realisierbar und nur bei realisierbaren Systemen kann etwas gemessen werden.
+*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.
-*Das Eingangssignal&nbsp; $x_1(t)$&nbsp; kann für sehr große Zeiten&nbsp; $(t \gg 0)$&nbsp; als Gleichsignal interpretiert werden. Wäre&nbsp; $H(f)$&nbsp; ein Hochpass, dann müsste&nbsp; $y_1(t)$&nbsp; für&nbsp; $t → ∞$&nbsp; gegen Null gehen. Das heißt: &nbsp; $H(f)$&nbsp; stellt einen Tiefpass dar.
+*The input signal&nbsp; $x_1(t)$&nbsp; can be interpreted as a direct signal for very large times&nbsp; $(t \gg 0)$&nbsp;. If&nbsp; $H(f)$&nbsp; was a high-pass filter, then&nbsp; $y_1(t)$&nbsp; would have to go towards zero for&nbsp; $t → ∞$&nbsp;. This means: &nbsp; $H(f)$&nbsp; represents a low-pass filter.
-'''(2)'''&nbsp; Der Gleichsignalübertragungsfaktor kann aus&nbsp; $x_1(t)$&nbsp; und&nbsp; $y_1(t)$&nbsp; abgelesen werden, wenn der Einschwingvorgang abgeklungen ist:
+'''(2)'''&nbsp; The direct signal transmission factor can be read from&nbsp; $x_1(t)$&nbsp; and&nbsp; $y_1(t)$&nbsp; when the transient has decayed:
 :$$H(f =0) = \frac{y_1(t \rightarrow \infty)}{x_1(t \rightarrow \infty)}=
   \frac{ {\rm 1\, V} }{ {\rm 4\, V} } \hspace{0.15cm}\underline{= 0.25}.$$
-'''(3)'''&nbsp; Die Sprungantwort&nbsp; $σ(t)$&nbsp; ist gleich dem Ausgangssignal&nbsp; $y(t)$, wenn am Eingang&nbsp; $x(t) = γ(t)$&nbsp; anliegen würde.
+'''(3)'''&nbsp; The step response&nbsp; $σ(t)$&nbsp; is equal to the output signal&nbsp; $y(t)$, if&nbsp; $x(t) = γ(t)$&nbsp; is applied to the input.
 *Wegen&nbsp; $x_1(t) = 4 \hspace{0.05cm} \rm {V} · γ(t)$&nbsp; gilt somit im Bereich von&nbsp; $0$&nbsp; bis&nbsp; $T = 2 \ \rm ms$:
 :$$\sigma(t) = \frac{y_1(t)}{ {\rm 4\, V} } = 0.5 \cdot\big( {t}/{T} - 0.5 ({t}/{T})^2\big).$$