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

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

2021-10-06T14:51:04Z

Oezer at 21:37, 20 September 2021

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Oezer at 21:32, 20 September 2021

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Oezer at 21:31, 20 September 2021

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Oezer at 21:30, 20 September 2021

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Oezer at 21:30, 20 September 2021

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Oezer at 21:09, 20 September 2021

2021-09-20T21:09:32Z

Oezer at 21:00, 20 September 2021

2021-09-20T21:00:55Z

@@ Line 111: / Line 111: @@
 *Thus,&nbsp; $y_1(t)$&nbsp; is a rectangular pulse of height&nbsp; $1$&nbsp; and width&nbsp; $2T$.
-*The same result &ndash; but in a more time-consuming way &ndash; is obtained by computing in the spectral domain:
+*The same result &ndash; but in a more time-consuming way &ndash; is obtained by computing in the frequency domain:
 :$$Y_1(f) = X_1(f) \cdot H(f) = T  \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot
   \big[  1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$
@@ Line 120: / Line 120: @@
   \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) .$$
-*Hence, the following can be formulated for the output spectrum:
+*Hence,&nbsp; the following can be formulated for the output spectrum:
 :$$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}
   , \; \; {\rm mit }  \; \;
@@ Line 126: / Line 126: @@
 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&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)''']]
-Here, the relation &nbsp;$\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$&nbsp; is used.
+Here,&nbsp; the relation &nbsp;$\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$&nbsp; is used.
-*The inverse Fourier transform of&nbsp; $Y_{11}$&nbsp; results in a rectangle of width&nbsp; $2T$, that is symmetric about&nbsp;  $t = 0$&nbsp;.
+*The inverse Fourier transform of&nbsp; $Y_{11}$&nbsp; results in a rectangle of width&nbsp; $2T$,&nbsp; that is symmetric about&nbsp;  $t = 0$&nbsp;.
 *Due to the phase function this is shifted into the range&nbsp; $0$ ... $2T$&nbsp; 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&nbsp; $y_1(t)$&nbsp; is rectangular just as&nbsp; $x_1(t)$,&nbsp; there are distortions:
-*These are linear because of&nbsp; $T_y > T_x$&nbsp;. In the frequency range of interest&nbsp; $($that is all frequencies for a si&ndash;shaped spectrum$)$,&nbsp; &nbsp; $|H(f)|$&nbsp; is not constant. So, there are attenuation distortions.
+*These are linear because of&nbsp; $T_y > T_x$.&nbsp; In the frequency range of interest&nbsp; $($that is all frequencies for a sinc&ndash;shaped spectrum$)$,&nbsp; &nbsp; $|H(f)|$&nbsp; is not constant.&nbsp; So, there are attenuation distortions.
-*In addition, there are also phase distortions since the phase does not increase linearly with&nbsp; $f$ in the whole range &nbsp; &rArr; &nbsp; 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&nbsp; $f$&nbsp; in the whole range &nbsp; &rArr; &nbsp; The&nbsp; <u>proposed solutions 1, 3, 4 and 5</u>&nbsp; are correct.
@@ Line 151: / Line 151: @@
 '''(3)'''&nbsp; The period&nbsp; $T_0 = T$&nbsp; of the periodic signal &nbsp;$x_3(t)$&nbsp; is exactly as large as the delay on the second path. Therefore, &nbsp;$y_3(t) = 2 \cdot x_3(t) $&nbsp; 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)$&nbsp; is a line spectrum with components at frequencies&nbsp; $f  = 0$,&nbsp; $f  =  \pm f_0 = \pm 1/T$,&nbsp; $f  =  \pm 3f_0$,&nbsp; etc..
-*However, the following holds at these discrete frequencies:
+*However,&nbsp; the following holds at these discrete frequencies:
 :$$|H(f)| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow
   \hspace{0.3cm}\tau_{\rm P}(f) = 0.$$
 *From this it follows again that&nbsp; $y_3(t) = 2 \cdot x_3(t) $.
-*Thus, only <u>proposed solution 1</u> is correct.
+*Thus,&nbsp; only&nbsp; <u>proposed solution 1</u>&nbsp; is correct.
-'''(4)'''&nbsp;  It can be seen that &nbsp;$y_4(t) = 1$&nbsp; is distorted with respect to &nbsp;$x_4(t)$&nbsp; from the lower sketch of the second graph. These are attenuation distortions &nbsp;&#8658;&nbsp; <u>Proposed solution 2</u> is correct as the following consideration shows.
+'''(4)'''&nbsp;  It can be seen that &nbsp;$y_4(t) = 1$&nbsp; is distorted with respect to &nbsp;$x_4(t)$&nbsp; from the lower sketch of the second graph. These are attenuation distortions &nbsp;&#8658;&nbsp; <u>Proposed solution 2</u>&nbsp; is correct as the following consideration shows.
-*Due to&nbsp; $T_0 = 2T$&nbsp; the signal &nbsp;$x_4(t)$&nbsp; has the fundamental frequency &nbsp;$f_0 = 1/(2T)$.
+*Due to&nbsp; $T_0 = 2T$&nbsp; the signal &nbsp;$x_4(t)$&nbsp; has the basic frequency &nbsp;$f_0 = 1/(2T)$.
-*The frequency response thus has zeros for all odd multiples of &nbsp;$f_0$&nbsp;.
+*The frequency response thus has zeros for all odd multiples of &nbsp;$f_0$.
-*The only remaining spectral line of &nbsp;$Y_4(f)$&nbsp; is at &nbsp;$f = 0$ where the following holds:
+*The only remaining spectral line of &nbsp;$Y_4(f)$&nbsp; is at &nbsp;$f = 0$&nbsp; where the following holds:
 :$$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.$$
@@ Line 180: / Line 180: @@
 *The expression in parentheses describes the frequency response already considered so far.
-*The second term causes an additional runtime around &nbsp;$ \tau = T_1$ and the following holds for all signals&nbsp; $(i = 1, 2, 3, 4)$:
+*The second term causes an additional runtime around &nbsp;$ \tau = T_1$&nbsp; and the following holds for all signals&nbsp; $(i = 1, 2, 3, 4)$:
 :$$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.&nbsp; This corresponds to&nbsp; <u>proposed solution 1</u>.
 {{ML-Fuß}}

@@ Line 63: / Line 63: @@
 <quiz display=simple>
-{Compute the output signal&nbsp; $y_1(t)$&nbsp; for the input signal&nbsp; $x_1(t)$. Which of the statements are true?
+{Compute the output signal&nbsp; $y_1(t)$&nbsp; for the input signal&nbsp; $x_1(t)$.&nbsp; Which of the statements are true?
 |type="[]"}
-+ $y_1(t)$&nbsp; is rectangular like&nbsp; $x_1(t)$&nbsp;.
++ $y_1(t)$&nbsp; is rectangular like&nbsp; $x_1(t)$.
 - $y_1(t)$&nbsp; is triangular in shape.
 + The absolute pulse duration is&nbsp; $2T$.
-+ $y_1(t)$&nbsp; exhibits attenuation distortions with respect to&nbsp; $x_1(t)$&nbsp;.
++ $y_1(t)$&nbsp; exhibits attenuation distortions with respect to&nbsp; $x_1(t)$.
-+ $y_1(t)$&nbsp; exhibits phase distortions with respect to&nbsp; $x_1(t)$&nbsp;.
++ $y_1(t)$&nbsp; exhibits phase distortions with respect to&nbsp; $x_1(t)$.
-{Compute the signal&nbsp; $y_2(t)$. What values are obtained at times&nbsp; $t= 0.5 T$,&nbsp; $t= 1.5 T$&nbsp; and&nbsp; $t= 2.5 T$?
+{Compute the signal&nbsp; $y_2(t)$.&nbsp; What values are obtained at times&nbsp; $t= 0.5 T$,&nbsp; $t= 1.5 T$&nbsp; and&nbsp; $t= 2.5 T$?
 |type="{}"}
 $y_2(t = 0.5T) \ = \ $  { 1 3% }
@@ Line 79: / Line 79: @@
-{Compute the signal&nbsp; $y_3(t)$. Check which statements are true.
+{Compute the signal&nbsp; $y_3(t)$.&nbsp; Check which statements are true.
 |type="[]"}
-+ $y_3(t)$&nbsp; is undistorted with respect to&nbsp; $x_3(t)$&nbsp;.
++ $y_3(t)$&nbsp; is undistorted with respect to&nbsp; $x_3(t)$.
-- $y_3(t)$&nbsp; exhibits attenuation distortions with respect to&nbsp; $x_3(t)$&nbsp;.
+- $y_3(t)$&nbsp; exhibits attenuation distortions with respect to&nbsp; $x_3(t)$.
-- $y_3(t)$&nbsp; exhibits phase distortions with respect to&nbsp; $x_3(t)$&nbsp;.
+- $y_3(t)$&nbsp; exhibits phase distortions with respect to&nbsp; $x_3(t)$.
-{Which statements are true for the output signal&nbsp; $y_4(t)$&nbsp;?
+{Which statements are true for the output signal&nbsp; $y_4(t)$?
 |type="[]"}
-- $y_4(t)$&nbsp; is undistorted with respect to&nbsp; $x_4(t)$&nbsp;.
+- $y_4(t)$&nbsp; is undistorted with respect to&nbsp; $x_4(t)$.
-+ $y_4(t)$&nbsp; exhibits attenuation distortions with respect to&nbsp; $x_4(t)$&nbsp;.
++ $y_4(t)$&nbsp; exhibits attenuation distortions with respect to&nbsp; $x_4(t)$.
-- $y_4(t)$&nbsp; exhibits phase distortions with respect to&nbsp; $x_4(t)$&nbsp;.
+- $y_4(t)$&nbsp; exhibits phase distortions with respect to&nbsp; $x_4(t)$.
-{Now &nbsp;$T_1 = 1 \ \rm ms$&nbsp; and &nbsp;$T_2  = 5 \ \rm ms$ hold. What are the changes compared to the previous results?
+{Now &nbsp;$T_1 = 1 \ \rm ms$&nbsp; and &nbsp;$T_2  = 5 \ \rm ms$ hold.&nbsp; What are the changes compared to the previous results?
 |type="[]"}
 + The above statements regarding distortions are still valid.

@@ Line 8: / Line 8: @@
 In contrast to the general representation in exercise 2.6, the two attenuation factors are equal here: &nbsp; $z_1 = z_2 = 1$.
-*For mobile communications, this corresponds to an echo at a distance of&nbsp; $T_2 - T_1$&nbsp; 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&nbsp; $T_2 - T_1$&nbsp; and of the same strength as the signal on the main path.&nbsp; For this, the transit time&nbsp; $T_1$&nbsp; is assumed.
-Using the transit times&nbsp;$T_1 = 0$&nbsp; and &nbsp;$T_2 = T = 4 \ \rm ms$&nbsp; considered in the subtasks&nbsp; '''(1)''' ... '''(4)''',&nbsp; 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&nbsp;$T_1 = 0$&nbsp; and &nbsp;$T_2 = T = 4 \ \rm ms$&nbsp; considered in the subtasks&nbsp; '''(1)''' ... '''(4)''',&nbsp; the following is obtained for the frequency response of the two-way channel,&nbsp; whose magnitude is depicted in the upper graph:
 :$$H(f) = 1 +  {\rm e}^{-{\rm j}\hspace{0.04cm}2 \pi f T} = 1 +
 \cos(2 \pi f T) - {\rm j} \cdot \sin(2 \pi f T)$$
@@ Line 18: / Line 17: @@
 T)\right)}= 2 \cdot |\cos(\pi f T)|.$$
-The bottom graph shows the phase function:
+The bottom graph shows the phase function  for&nbsp; $T_1 = 0$&nbsp; and &nbsp; $T_2 = T = 4 \ \rm ms$:
 :$$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].$$
-The following trigonometric transformation was used here:
+*In the frequency region&nbsp; $|f| < 1/(2T)$,&nbsp; &nbsp; $b(f)$&nbsp; increases linearly: &nbsp; $b(f) = \pi \cdot f \cdot T.$
 :$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$
 In the questions,&nbsp; $y_i(t)$&nbsp; denotes the signal at the output of the two-way channel if the signal&nbsp; $x_i(t)$&nbsp; is applied to the input&nbsp; $(i = 1, 2, 3, 4)$.
 These signals are examined as input signals:
-*a rectangular pulse &nbsp;$x_1(t)$&nbsp; with height&nbsp; $1$&nbsp; between &nbsp;$t= 0$&nbsp; and &nbsp;$t= T$; <br>&nbsp;$x_1(t) = 0$&nbsp; holds for &nbsp;$t < 0$&nbsp; and &nbsp;$t > T$&nbsp; $($the value $0.5$ occurs at the two jump discontinuities, respectively$)$;
+*a rectangular pulse &nbsp;$x_1(t)$&nbsp; with height&nbsp; $1$&nbsp; between &nbsp;$t= 0$&nbsp; and &nbsp;$t= T$;&nbsp; $x_1(t) = 0$&nbsp; holds for &nbsp;$t < 0$&nbsp; and &nbsp;$t > T$&nbsp; $($the value $0.5$ occurs at the two jump discontinuities$)$;
 *a rectangular pulse &nbsp;$x_2(t)$&nbsp; with height&nbsp; $1$&nbsp; in the range of&nbsp; $0 \ \text{...} \  2T$;
-*a periodic rectangular pulse &nbsp;$x_3(t)$&nbsp; with period &nbsp;$T = T_0$:
+*a periodic rectangular signal &nbsp;$x_3(t)$&nbsp; with period duration &nbsp;$T = T_0$:
 :$$x_3(t) = \left\{ \begin{array}{c} 1 \\
 \\  \end{array} \right.\quad \quad
@@ Line 43: / Line 40: @@
 { T/2 < t  < T,}  \\
 \end{array}$$
-*a periodic rectangular pulse &nbsp;$x_4(t)$&nbsp; with period &nbsp;$T = 2T_0$:
+*a periodic rectangular signal &nbsp;$x_4(t)$&nbsp; with period duration &nbsp;$T = 2T_0$:
 :$$x_4(t) = \left\{ \begin{array}{c} 1 \\
 \\  \end{array} \right.\quad \quad
@@ Line 55: / Line 52: @@
 Please note:
+*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]].
+*For subtasks&nbsp; '''(1)''' to '''(4)''', &nbsp;$T_1 = 0$&nbsp; and &nbsp;$T_2 = T = 4 \ \rm ms$&nbsp; hold.
+*In subtask&nbsp; '''(5)''',&nbsp; the case &nbsp;$T_1 = 1 \ \rm ms$&nbsp; and &nbsp;$T_2  = 5 \ \rm ms$&nbsp; is considered.

@@ Line 187: / Line 187: @@
 *The expression in parentheses describes the frequency response already considered so far.
-*The second term causes an additional runtime around &nbsp;$ \tau = T_1$ and it holds for all signals&nbsp; $(i = 1, 2, 3, 4)$:
+*The second term causes an additional runtime around &nbsp;$ \tau = T_1$ and the following holds for all signals&nbsp; $(i = 1, 2, 3, 4)$:
 :$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$
-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}}

@@ Line 33: / Line 33: @@
 These signals are examined as input signals:
-*a rectangular pulse &nbsp;$x_1(t)$&nbsp; with height&nbsp; $1$&nbsp; between &nbsp;$t= 0$&nbsp; and &nbsp;$t= T$; <br>&nbsp;$x_1(t) = 0$&nbsp; holds for &nbsp;$t < 0$&nbsp; and &nbsp;$t > T$&nbsp; $($the value$0.5$ occurs at the two jump discontinuities, respectively$)$;
+*a rectangular pulse &nbsp;$x_1(t)$&nbsp; with height&nbsp; $1$&nbsp; between &nbsp;$t= 0$&nbsp; and &nbsp;$t= T$; <br>&nbsp;$x_1(t) = 0$&nbsp; holds for &nbsp;$t < 0$&nbsp; and &nbsp;$t > T$&nbsp; $($the value $0.5$ occurs at the two jump discontinuities, respectively$)$;
 *a rectangular pulse &nbsp;$x_2(t)$&nbsp; with height&nbsp; $1$&nbsp; in the range of&nbsp; $0 \ \text{...} \  2T$;
 *a periodic rectangular pulse &nbsp;$x_3(t)$&nbsp; with period &nbsp;$T = T_0$:

@@ Line 25: / Line 25: @@
 :$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$
-The bottom graph shows thep hase function for&nbsp;$T_1 = 0$&nbsp; and &nbsp;$T_2 = T = 4 \ \rm ms$:
+The bottom graph shows the phase function for&nbsp;$T_1 = 0$&nbsp; and &nbsp;$T_2 = T = 4 \ \rm ms$:
 *In the frequency range&nbsp; $|f| < 1/(2T)$,&nbsp; &nbsp; $b(f)$&nbsp; increases linearly: &nbsp; $b(f) = \pi \cdot f \cdot T.$
 *Also in further sections of the phase function, the phase always increases linearly from&nbsp; $-\pi/2$&nbsp; to&nbsp; $+\pi/2$&nbsp;.

@@ Line 7: / Line 7: @@
 :$$h(t) = \delta ( t - T_1) + \delta ( t - T_2).$$
-In contrast to the general representation in task 2.6, the two attenuation factors are equal here: &nbsp; $z_1 = z_2 = 1$.
+In contrast to the general representation in exercise 2.6, the two attenuation factors are equal here: &nbsp; $z_1 = z_2 = 1$.
 *For mobile communications, this corresponds to an echo at a distance of&nbsp; $T_2 - T_1$&nbsp; and of the same strength as the signal on the main path, for example.
 *For this, the transit time&nbsp; $T_1$&nbsp; is assumed.

@@ Line 146: / Line 146: @@
-'''(2)'''&nbsp; Aufgrund der bereits in&nbsp; '''(1)'''&nbsp; angegebenen Gleichung
+'''(2)'''&nbsp; Due to the equation
 :$$y_2(t)  = x_2(t) + x_2(t-T)$$
-erhält man einen stufenförmigen Verlauf entsprechend dem unteren Diagramm der obere Grafik.
+already specified in&nbsp; '''(1)'''&nbsp; a step-like curve shape corresponding to the lower diagram of the upper graph is obtained.
-Die gesuchten Zahlenwerte sind: &nbsp; $y_2(t = 0.5 T)   \hspace{0.15cm}\underline{= 1}, \hspace{0.3cm} y_2(t = 1.5 T)   \hspace{0.15cm}\underline{= 2},
+The numerical values that are looked for are: &nbsp; $y_2(t = 0.5 T)   \hspace{0.15cm}\underline{= 1}, \hspace{0.3cm} y_2(t = 1.5 T)   \hspace{0.15cm}\underline{= 2},
 \hspace{0.3cm}y_2(t = 2.5 T)  \hspace{0.15cm}\underline{ = 1}.$
 [[File:P_ID927__LZI_A_2_7_c.png|right|frame|Solutions&nbsp; '''(3)'''&nbsp; and&nbsp; '''(4)''']]
-'''(3)'''&nbsp; Die Periodendauer&nbsp; $T_0 = T$&nbsp; des periodischen Signals &nbsp;$x_3(t)$&nbsp; ist genau so groß wie die Verzögerung auf dem zweiten Pfad. Deshalb ist &nbsp;$y_3(t) = 2 \cdot x_3(t) $&nbsp; und es sind keine Verzerrungen feststellbar.
+'''(3)'''&nbsp; The period&nbsp; $T_0 = T$&nbsp; of the periodic signal &nbsp;$x_3(t)$&nbsp; is exactly as large as the delay on the second path. Therefore, &nbsp;$y_3(t) = 2 \cdot x_3(t) $&nbsp; holds and no distortions are observable.
-Die Spektralbereichsberechnung führt zum gleichen Ergebnis.
+The spectral domain computation leads to the same result.
-*$X_3(f)$&nbsp; ist ein Linienspektrum mit Anteilen bei den Frequenzen&nbsp; $f  = 0$,&nbsp; $f  =  \pm f_0 = \pm 1/T$,&nbsp; $f  =  \pm 3f_0$,&nbsp; usw..
+*$X_3(f)$&nbsp; is a line spectrum with components at frequencies&nbsp; $f  = 0$,&nbsp; $f  =  \pm f_0 = \pm 1/T$,&nbsp; $f  =  \pm 3f_0$,&nbsp; etc..
-*Bei diesen diskreten Frequenzen gilt aber exakt:
+*However, the following holds at these discrete frequencies:
 :$$|H(f)| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow
   \hspace{0.3cm}\tau_{\rm P}(f) = 0.$$
-*Auch daraus folgt wieder&nbsp; $y_3(t) = 2 \cdot x_3(t) $.
+*From this it follows again that&nbsp; $y_3(t) = 2 \cdot x_3(t) $.
-*Richtig ist somit nur der <u>Lösungsvorschlag 1</u>.
+*Thus, only <u>proposed solution 1</u> is correct.
-'''(4)'''&nbsp; Aus der unteren Skizze der zweiten Grafik geht hervor, dass &nbsp;$y_4(t) = 1$&nbsp; gegenüber &nbsp;$x_4(t)$&nbsp; verzerrt ist. Dabei handelt es sich um Dämpfungsverzerrungen &nbsp;&#8658;&nbsp; <u>Lösungsvorschlag 2</u>, wie die folgende Überlegung zeigt.
+'''(4)'''&nbsp;  It can be seen that &nbsp;$y_4(t) = 1$&nbsp; is distorted with respect to &nbsp;$x_4(t)$&nbsp; from the lower sketch of the second graph. These are attenuation distortions &nbsp;&#8658;&nbsp; <u>Proposed solution 2</u> is correct as the following consideration shows.
-*Wegen&nbsp; $T_0 = 2T$&nbsp; weist das Signal &nbsp;$x_4(t)$&nbsp; die Grundfrequenz &nbsp;$f_0 = 1/(2T)$ auf.
+*Due to&nbsp; $T_0 = 2T$&nbsp; the signal &nbsp;$x_4(t)$&nbsp; has the fundamental frequency &nbsp;$f_0 = 1/(2T)$.
-*Bei allen ungeraden Vielfachen von &nbsp;$f_0$&nbsp; hat somit der Frequenzgang Nullstellen.
+*The frequency response thus has zeros for all odd multiples of &nbsp;$f_0$&nbsp;.
-*Die einzige verbleibende Spektrallinie von  &nbsp;$Y_4(f)$&nbsp; liegt bei &nbsp;$f = 0$, wobei gilt:
+*The only remaining spectral line of &nbsp;$Y_4(f)$&nbsp; is at &nbsp;$f = 0$ where the following holds:
 :$$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f)
 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$
@@ Line 179: / Line 179: @@
-'''(5)'''&nbsp; Der Frequenzgang lautet nun mit &nbsp;$T_1 = 1 \ \rm ms$, &nbsp;$T_2  = 5 \ \rm ms$&nbsp; und &nbsp;$T = T_2 -T_1 = 4 \ \rm ms$:
+'''(5)'''&nbsp; Now the frequency response is with &nbsp;$T_1 = 1 \ \rm ms$, &nbsp;$T_2  = 5 \ \rm ms$&nbsp; and &nbsp;$T = T_2 -T_1 = 4 \ \rm ms$:
 :$$H(f) =  {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f
   T_1}+ {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f
@@ Line 186: / Line 186: @@
   T_1}.$$
-*Der Klammerausdruck beschreibt den bereits bisher betrachteten Frequenzgang.
+*The expression in parentheses describes the frequency response already considered so far.
-*Der zweite Term bewirkt eine zusätzliche Laufzeit um &nbsp;$ \tau = T_1$, und es gilt für alle Signale&nbsp; $(i = 1, 2, 3, 4)$:
+*The second term causes an additional runtime around &nbsp;$ \tau = T_1$ and it holds for all signals&nbsp; $(i = 1, 2, 3, 4)$:
 :$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$
-Alle Aussagen hinsichtlich der Verzerrungen sind weiter gültig. Dies entspricht dem <u>Lösungsvorschlag 1</u>.
+All statements regarding the distortions are still valid. This corresponds to  <u>proposed solution 1</u>.
 {{ML-Fuß}}

@@ Line 136: / Line 136: @@
 Here, the relation &nbsp;$\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$&nbsp; is used.
-*The inverse Fourier transform of&nbsp; $Y_{11}$&nbsp; results in a rectangle symmetric about&nbsp;  $t = 0$&nbsp; and of width&nbsp; $2T$.
+*The inverse Fourier transform of&nbsp; $Y_{11}$&nbsp; results in a rectangle of width&nbsp; $2T$, that is symmetric about&nbsp;  $t = 0$&nbsp;.
 *Durch die Phasenfunktion wird dieser in den Bereich&nbsp; $0$ ... $2T$&nbsp; verschoben und das Ergebnis der Zeitbereichsberechnung bestätigt.