Aufgaben:Exercise 2.5Z: Linear Distortions with DSB-AM - Revision history

Guenter at 15:57, 8 December 2021

2021-12-08T15:57:45Z

Guenter at 15:33, 8 December 2021

2021-12-08T15:33:59Z

Guenter: Guenter moved page Aufgaben:Exercise 2.5Z: Distortions with DSB-AM to Aufgaben:Exercise 2.5Z: Linear Distortions with DSB-AM

2021-12-08T15:19:27Z

Guenter moved page Exercise 2.5Z: Distortions with DSB-AM to Exercise 2.5Z: Linear Distortions with DSB-AM

Guenter at 15:19, 8 December 2021

2021-12-08T15:19:12Z

Reed: /* Solution */

2021-12-03T15:51:50Z

Solution

Reed at 15:50, 3 December 2021

2021-12-03T15:50:01Z

Reed at 15:34, 3 December 2021

2021-12-03T15:34:29Z

Reed at 15:33, 3 December 2021

2021-12-03T15:33:58Z

Reed at 15:27, 3 December 2021

2021-12-03T15:27:42Z

Reed at 15:27, 3 December 2021

2021-12-03T15:27:10Z

@@ Line 65: / Line 65: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID1014__Mod_Z_2_5_a.png|right|frame|Receiver spectrum]]
+[[File:P_ID1014__Mod_Z_2_5_a.png|right|frame|Spectrum&nbsp; $R(f)$&nbsp; of the received signal]]
-'''(1)'''&nbsp; In general,&nbsp; $R(f) = S(f) · H_K(f)$.&nbsp; This gives the line spectrum as shown in the adjacent sketch (all weights still have to be supplemented by the unit "V").
+'''(1)'''&nbsp; In general,&nbsp; $R(f) = S(f) · H_K(f)$.&nbsp; This gives the line spectrum as shown in the adjacent sketch&nbsp; (all weights still have to be supplemented by the unit "Volt").
-*When the weight of the spectral line is &nbsp; $f = -f_{\rm O}$, then:
+*For the weight of the spectral line at&nbsp; $f = -f_{\rm O}$&nbsp; it is valid:
 :$${\rm Re}[R(-f_{\rm O})]\hspace{0.15cm}\underline{=0.2 \ \rm V},$$
 :$${\rm Im}[R(-f_{\rm O})]\hspace{0.15cm}\underline{=0.1 \ \rm V}.$$
@@ Line 84: / Line 83: @@
    A_{\rm N}\cdot \frac{I_{\rm U} - I_{\rm O}}{2}\cdot \frac{\rm j}{2} \cdot \left[-\delta(f - f_{\rm N}) + \delta(f + f_{\rm N}) \right]$$
 :$$ \Rightarrow \hspace{0.3cm}v(t) = A_{\rm N}\cdot \frac{R_{\rm U} +R_{\rm O}}{2}\cdot\cos (\omega_{\rm N}\cdot t)+ A_{\rm N}\cdot \frac{I_{\rm U} -I_{\rm O}}{2}\cdot\sin (\omega_{\rm N}\cdot t)\hspace{0.05cm}.$$
-*When &nbsp; $R_{\rm U} = 0.8, I_{\rm U} = -0.2, R_{\rm O} = 0.4, I_{\rm O} = -0.2$&nbsp; it follows:
+*When &nbsp; $R_{\rm U} = 0.8,\ I_{\rm U} = -0.2,\ R_{\rm O} = 0.4,\ I_{\rm O} = -0.2$&nbsp; it follows:
 :$$v(t) = 0.6 \cdot A_{\rm N}\cdot \cos (\omega_{\rm N}\cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} v(t=0) = 0.6 \cdot A_{\rm N}\hspace{0.15cm}\underline {= 1.2\,{\rm V}}\hspace{0.05cm}.$$
-*There is attenuation by a factor of &nbsp; $0.6$ compared to &nbsp; $q(t)$&nbsp;.
+*There is attenuation by a factor of &nbsp; $0.6$ compared to&nbsp; $q(t)$.
 *The synchronous demodulator receives more information about the source signal through the lower sideband than through the upper one.
-*Because of the property&nbsp; $I_{\rm O} = I_{\rm U}$&nbsp;, &nbsp; $v(t)$&nbsp; is also cosine-shaped.
+*Because of the property&nbsp; $I_{\rm O} = I_{\rm U}$,&nbsp; $v(t)$&nbsp; is also cosine-shaped.
-*Accordingly, either no delay occurs or the delay is an even multiple of the period.
+*Accordingly,&nbsp; it is valid either no delay occurs or the delay is an even multiple of the period.
@@ Line 95: / Line 94: @@
 '''(3)'''&nbsp; The following equations apply here:
-:$$ H_{\rm K}(f_{\rm N}+ f_{\rm T})  =  R_{\rm O} + {\rm j} \cdot I_{\rm O} \hspace{0.05cm},\hspace{0.5cm}
+:$$ H_{\rm K}(f_{\rm N}+ f_{\rm T})  =  R_{\rm O} + {\rm j} \cdot I_{\rm O} \hspace{0.05cm}, $$
-H_{\rm K}(f_{\rm N}- f_{\rm T})  =  H_{\rm K}^{\star}(f_{\rm T}- f_{\rm N}) = R_{\rm U} - {\rm j} \cdot I_{\rm U} $$
+:$$ H_{\rm K}(f_{\rm N}- f_{\rm T})  =  H_{\rm K}^{\star}(f_{\rm T}- f_{\rm N}) = R_{\rm U} - {\rm j} \cdot I_{\rm U} $$
 :$$\Rightarrow \hspace{0.2cm} H_{\rm MKD}(f_{\rm N})  =  {1}/{2} \cdot \big[(R_{\rm O} +R_{\rm U}) + {\rm j} \cdot (I_{\rm O} -I_{\rm U}) \big]\hspace{0.05cm},\hspace{0.2cm}
 H_{\rm MKD}(-f_{\rm N})  =  H_{\rm MKD}^\star(f_{\rm N}) = {1}/{2} \cdot \big[(R_{\rm O} +R_{\rm U}) - {\rm j} \cdot (I_{\rm O} -I_{\rm U}) \big]\hspace{0.05cm}.$$
-*Thus, one obtains the same result as in (2), but faster ⇒ <u>Answer 2</u>.
+*Thus,&nbsp; one obtains the same result as in&nbsp; '''(2)''',&nbsp; but faster &nbsp; ⇒ &nbsp; <u>Answer 2</u>.
-'''(4)'''&nbsp; For &nbsp; $f > 0$&nbsp; the resulting frequency response is:
+'''(4)'''&nbsp; For&nbsp; $f > 0$&nbsp; the resulting frequency response is:
 :$$H_{\rm MKD}(f)  =  {1}/{2} \cdot \left[ H_{\rm K}(f_{\rm T}+ f) + H_{\rm K}^\star(f_{\rm T}-f)\right]=  {1}/{2} \cdot \left[ \frac{1}{1 + 3{\rm j} \cdot (\frac{f_{\rm T}+f}{f_{\rm T}} - 1)} + \frac{1}{1 - 3{\rm j} \cdot (\frac{f_{\rm T}-f}{f_{\rm T}} - 1)}\right] $$
 :$$ \Rightarrow \hspace{0.3cm} H_{\rm MKD}(f) = \frac{1}{1 + {\rm j} \cdot {3f}/{f_{\rm T}} } \hspace{0.05cm}.$$
-*Inserted at the point where &nbsp; $f = f_{\rm N}$&nbsp; this leads to the result:
+*Inserted at the point where&nbsp; $f = f_{\rm N}$&nbsp; this leads to the result:
 :$$H_{\rm MKD}(f_{\rm N}) = \frac{1}{1 + {\rm j} \cdot {3f_{\rm N}}/{f_{\rm T}} } \hspace{1.0cm}
   \Rightarrow \hspace{0.3cm}{\rm magnitude} = \frac{1}{\sqrt{1 + ({3f_{\rm N}}/{f_{\rm T}} )^2}} \hspace{0.05cm}, \hspace{0.3cm} {\rm phase} = {\rm arctan}\hspace{0.1cm}({3f_{\rm N}}/{f_{\rm T}}) \hspace{0.05cm}.$$

@@ Line 5: / Line 5: @@
 [[File:P_ID1013__Mod_Z_2_5.png|right|frame|Considered system model]]
 As in &nbsp;[[Aufgaben:Exercise_2.5:_DSB-AM_via_a_Gaussian_Channel|Exercise 2.5]]&nbsp; here we will also examine:
-*the DSB–AM/synchronous demodulator combination
+*the DSB–AM/synchronous demodulator combination,
 *considerations involving a linear distorting channel.
@@ Line 11: / Line 11: @@
 Let the source signal &nbsp;$q(t)$&nbsp; be a cosine signal with amplitude &nbsp;$A_{\rm N}$&nbsp; and frequency &nbsp;$f_{\rm N}$, such that the spectrum of the modulated signal is as follows:
 :$$S(f)= \frac{A_{\rm N}}{4} \cdot \big[\delta(f + f_{\rm O}) + \delta(f + f_{\rm U}) + \delta(f - f_{\rm U}) + \delta(f - f_{\rm O}) \big]\hspace{0.05cm}.$$
-The abbreviations stand for &nbsp;$f_{\rm O} = f_{\rm T} + f_{\rm N}$&nbsp; ('''O''' for German "Oberes Seitenband" i.e. Upper Sideband) and &nbsp;$f_{\rm U} = f_{\rm T} - f_{\rm N}$&nbsp; ('''U''' for "Unteres Seitenband" i.e. Lower Sideband).
+The abbreviations stand for &nbsp;
 The channel frequency response is only given for these two frequencies and is:
 :$$ H_{\rm K}(f_{\rm O}) = R_{\rm O} + {\rm j} \cdot I_{\rm O},\hspace{0.2cm}H_{\rm K}(f_{\rm U}) = R_{\rm U} + {\rm j} \cdot I_{\rm U} \hspace{0.05cm}.$$
-For negative frequencies, &nbsp;$H_{\rm K}(– f) = H_{\rm K}^*(f)$ always holds.
+For negative frequencies,&nbsp; $H_{\rm K}(– f) = H_{\rm K}^*(f)$ always holds.
 Use the following values for numerical calculations :
@@ Line 22: / Line 25: @@
 In subtask&nbsp; '''(3)'''&nbsp; the solution should be found from the resulting frequency response of modulator, channel and demodulator:
 :$$H_{\rm MKD}(f) = {1}/{2} \cdot \big[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\big]\hspace{0.05cm}.$$
-Finally, in subtask&nbsp; '''(4)'''&nbsp; the following channel frequency response is considered (the plot is only valid for positive frequencies):
+Finally,&nbsp; in subtask&nbsp; '''(4)'''&nbsp; the following channel frequency response is considered&nbsp; (this equationt is only valid for positive frequencies):
 :$$ H_{\rm K}(f) = H_{\rm(4)}(f) = \frac{1}{1 + 3{\rm j} \cdot ({f}/{f_{\rm T}} - 1)}\hspace{0.05cm}.$$
@@ Line 29: / Line 32: @@
 Hints:
 *This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]].
 *Particular reference is made to the page&nbsp;  [[Modulation_Methods/Synchronous_Demodulation#Influence_of_linear_channel_distortions|Influence of linear channel distortions]].
@@ Line 43: / Line 43: @@
 <quiz display=simple>
-{Let &nbsp;$R_{\rm U} = 0.8, \ I_{\rm U} = -0.2, \ R_{\rm O} = 0.4, I_{\rm O} = -0.2.$&nbsp; Calculate and sketch the spectrum &nbsp;$R(f)$&nbsp; at the channel output. <br>What is the spectral line at&nbsp; $-f_{\rm O}$?
+{Let &nbsp;$R_{\rm U} = 0.8, \ I_{\rm U} = -0.2, \ R_{\rm O} = 0.4, \ I_{\rm O} = -0.2.$&nbsp; Calculate and sketch the spectrum &nbsp;$R(f)$&nbsp; at the channel output. <br>What is the spectral line at&nbsp; $-f_{\rm O}$?
 |type="{}"}
 ${\rm Re}[R(-f_{\rm O})] \ = \ $ { 0.2 3% } $\ \text{V}$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1013__Mod_Z_2_5.png|right|frame|Considering the channel frequency response]]
+[[File:P_ID1013__Mod_Z_2_5.png|right|frame|Considered system model]]
 As in &nbsp;[[Aufgaben:Exercise_2.5:_DSB-AM_via_a_Gaussian_Channel|Exercise 2.5]]&nbsp; here we will also examine:
 *the DSB–AM/synchronous demodulator combination
-*considerations involving a linear distorting channel .
+*considerations involving a linear distorting channel.

@@ Line 107: / Line 107: @@
 *Inserted at the point where &nbsp; $f = f_{\rm N}$&nbsp; this leads to the result:
 :$$H_{\rm MKD}(f_{\rm N}) = \frac{1}{1 + {\rm j} \cdot {3f_{\rm N}}/{f_{\rm T}} } \hspace{1.0cm}
-  \Rightarrow \hspace{0.3cm}{\rm magnitude} = \frac{1}{\sqrt{1 + ({3f_{\rm N}}/{f_{\rm T}} )^2}} \hspace{0.05cm}, \hspace{0.3cm} {\rm Phase} = {\rm arctan}\hspace{0.1cm}({3f_{\rm N}}/{f_{\rm T}}) \hspace{0.05cm}.$$
+  \Rightarrow \hspace{0.3cm}{\rm magnitude} = \frac{1}{\sqrt{1 + ({3f_{\rm N}}/{f_{\rm T}} )^2}} \hspace{0.05cm}, \hspace{0.3cm} {\rm phase} = {\rm arctan}\hspace{0.1cm}({3f_{\rm N}}/{f_{\rm T}}) \hspace{0.05cm}.$$
 *When&nbsp; $f_{\rm N}/f_{\rm T} = 0.1$&nbsp; we get the magnitude&nbsp; $0.958$&nbsp; and the phase&nbsp; $16.7^\circ$.&nbsp; Thus, the sink signal is:
 :$$v(t) = 0.958 \cdot 2\,{\rm V}\cdot \cos (\omega_{\rm N}\cdot t + 16.7^\circ) \hspace{0.3cm}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1013__Mod_Z_2_5.png|right|frame|Berücksichtigung des Kanalfrequenzgangs]]
+[[File:P_ID1013__Mod_Z_2_5.png|right|frame|Considering the channel frequency response]]
 As in &nbsp;[[Aufgaben:Exercise_2.5:_DSB-AM_via_a_Gaussian_Channel|Exercise 2.5]]&nbsp; here we will also examine:
 *the DSB–AM/synchronous demodulator combination
@@ Line 65: / Line 65: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID1014__Mod_Z_2_5_a.png|right|frame|Empfangsspektrum]]
+[[File:P_ID1014__Mod_Z_2_5_a.png|right|frame|Receiver spectrum]]
-'''(1)'''&nbsp; Allgemein gilt&nbsp; $R(f) = S(f) · H_K(f)$.&nbsp; Damit erhält man das Linienspektrum gemäß nebenstehender Skizze&nbsp; (alle Gewichte sind noch um die Einheit „V” zu ergänzen).
+'''(1)'''&nbsp; In general,&nbsp; $R(f) = S(f) · H_K(f)$.&nbsp; This gives the line spectrum as shown in the adjacent sketch (all weights still have to be supplemented by the unit "V").
-*Für das Gewicht der Spektrallinie bei&nbsp; $f = -f_{\rm O}$ gilt:
+*When the weight of the spectral line is &nbsp; $f = -f_{\rm O}$, then:
 :$${\rm Re}[R(-f_{\rm O})]\hspace{0.15cm}\underline{=0.2 \ \rm V},$$
 :$${\rm Im}[R(-f_{\rm O})]\hspace{0.15cm}\underline{=0.1 \ \rm V}.$$
@@ Line 74: / Line 74: @@
-'''(2)'''&nbsp; Die Spektralfunktion&nbsp; $V(f)$&nbsp; des Sinkensignals&nbsp; $v(t)$&nbsp; lautet:
+'''(2)'''&nbsp; The spectral function &nbsp; $V(f)$&nbsp; of the sink signal&nbsp; $v(t)$&nbsp; is:
 :$$V(f)  =  \big[ R(f) \star \left[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \right]\big]\cdot H_{\rm E}(f).$$
-*Nach den Gesetzmäßigkeiten der Fouriertransformation kann hierfür auch geschrieben werden:
+*According to the laws of the Fourier transform, this can also be written as:
 :$$V(f)  =   \frac{A_{\rm N}}{4} \cdot (R_{\rm O} + {\rm j} \cdot I_{\rm O}) \cdot \delta(f - f_{\rm N}) + \frac{A_{\rm N}}{4} \cdot (R_{\rm U} + {\rm j} \cdot I_{\rm U}) \cdot \delta(f + f_{\rm N})+$$
 :$$\hspace{2.25cm}+ \frac{A_{\rm N}}{4} \cdot (R_{\rm O} - {\rm j} \cdot I_{\rm O}) \cdot \delta(f + f_{\rm N})+ \frac{A_{\rm N}}{4} \cdot (R_{\rm U} - {\rm j} \cdot I_{\rm U}) \cdot \delta(f - f_{\rm N}) \hspace{0.05cm}.$$
-*Alle anderen Terme liegen um die doppelte Trägerfrequenz und werden durch den Tiefpass eliminiert.
+*All other terms are around twice the carrier frequency and are eliminated by the low-pass filter.
-* Umsortieren und Zusammenfassen der Terme führt zu:
+* Rearranging and combining the terms results in:
 :$$V(f)  =  A_{\rm N}\cdot \frac{R_{\rm U} +R_{\rm O}}{2}\cdot \frac{1}{2} \cdot \left[\delta(f - f_{\rm N}) + \delta(f + f_{\rm N}) \right] +
    A_{\rm N}\cdot \frac{I_{\rm U} - I_{\rm O}}{2}\cdot \frac{\rm j}{2} \cdot \left[-\delta(f - f_{\rm N}) + \delta(f + f_{\rm N}) \right]$$
 :$$ \Rightarrow \hspace{0.3cm}v(t) = A_{\rm N}\cdot \frac{R_{\rm U} +R_{\rm O}}{2}\cdot\cos (\omega_{\rm N}\cdot t)+ A_{\rm N}\cdot \frac{I_{\rm U} -I_{\rm O}}{2}\cdot\sin (\omega_{\rm N}\cdot t)\hspace{0.05cm}.$$
-*Mit&nbsp; $R_{\rm U} = 0.8, I_{\rm U} = -0.2, R_{\rm O} = 0.4, I_{\rm O} = -0.2$&nbsp; folgt daraus:
+*When &nbsp; $R_{\rm U} = 0.8, I_{\rm U} = -0.2, R_{\rm O} = 0.4, I_{\rm O} = -0.2$&nbsp; it follows:
 :$$v(t) = 0.6 \cdot A_{\rm N}\cdot \cos (\omega_{\rm N}\cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} v(t=0) = 0.6 \cdot A_{\rm N}\hspace{0.15cm}\underline {= 1.2\,{\rm V}}\hspace{0.05cm}.$$
-*Es ergibt sich gegenüber&nbsp; $q(t)$&nbsp; eine Dämpfung um den Faktor&nbsp; $0.6$.
+*There is attenuation by a factor of &nbsp; $0.6$ compared to &nbsp; $q(t)$&nbsp;.
-*Der Synchrondemodulator bekommt durch das untere Seitenband mehr Information über das Quellensignal als über das obere.
+*The synchronous demodulator receives more information about the source signal through the lower sideband than through the upper one.
-*Wegen der Eigenschaft&nbsp; $I_{\rm O} = I_{\rm U}$&nbsp; ist&nbsp; $v(t)$&nbsp; ebenfalls cosinusförmig.
+*Because of the property&nbsp; $I_{\rm O} = I_{\rm U}$&nbsp;, &nbsp; $v(t)$&nbsp; is also cosine-shaped.
-*Es tritt demnach keine Laufzeit auf bzw. die Laufzeit ist ein geradzahliges Vielfaches der Periodendauer.
+*Accordingly, either no delay occurs or the delay is an even multiple of the period.
-'''(3)'''&nbsp; Hier gelten folgende Gleichungen:
+'''(3)'''&nbsp; The following equations apply here:
 :$$ H_{\rm K}(f_{\rm N}+ f_{\rm T})  =  R_{\rm O} + {\rm j} \cdot I_{\rm O} \hspace{0.05cm},\hspace{0.5cm}
 H_{\rm K}(f_{\rm N}- f_{\rm T})  =  H_{\rm K}^{\star}(f_{\rm T}- f_{\rm N}) = R_{\rm U} - {\rm j} \cdot I_{\rm U} $$
 :$$\Rightarrow \hspace{0.2cm} H_{\rm MKD}(f_{\rm N})  =  {1}/{2} \cdot \big[(R_{\rm O} +R_{\rm U}) + {\rm j} \cdot (I_{\rm O} -I_{\rm U}) \big]\hspace{0.05cm},\hspace{0.2cm}
 H_{\rm MKD}(-f_{\rm N})  =  H_{\rm MKD}^\star(f_{\rm N}) = {1}/{2} \cdot \big[(R_{\rm O} +R_{\rm U}) - {\rm j} \cdot (I_{\rm O} -I_{\rm U}) \big]\hspace{0.05cm}.$$
-*Man erhält somit das gleiche Ergebnis wie unter&nbsp; (2), aber schneller   ⇒   <u>Lösungsvorschlag 2</u>.
+*Thus, one obtains the same result as in (2), but faster ⇒ <u>Answer 2</u>.
+'''(4)'''&nbsp; For &nbsp; $f > 0$&nbsp; the resulting frequency response is:
 :$$H_{\rm MKD}(f)  =  {1}/{2} \cdot \left[ H_{\rm K}(f_{\rm T}+ f) + H_{\rm K}^\star(f_{\rm T}-f)\right]=  {1}/{2} \cdot \left[ \frac{1}{1 + 3{\rm j} \cdot (\frac{f_{\rm T}+f}{f_{\rm T}} - 1)} + \frac{1}{1 - 3{\rm j} \cdot (\frac{f_{\rm T}-f}{f_{\rm T}} - 1)}\right] $$
 :$$ \Rightarrow \hspace{0.3cm} H_{\rm MKD}(f) = \frac{1}{1 + {\rm j} \cdot {3f}/{f_{\rm T}} } \hspace{0.05cm}.$$
-*Eingesetzt an der Stelle&nbsp; $f = f_{\rm N}$&nbsp; führt dies zum Ergebnis:
+*Inserted at the point where &nbsp; $f = f_{\rm N}$&nbsp; this leads to the result:
 :$$H_{\rm MKD}(f_{\rm N}) = \frac{1}{1 + {\rm j} \cdot {3f_{\rm N}}/{f_{\rm T}} } \hspace{1.0cm}
-  \Rightarrow \hspace{0.3cm}{\rm Betrag} = \frac{1}{\sqrt{1 + ({3f_{\rm N}}/{f_{\rm T}} )^2}} \hspace{0.05cm}, \hspace{0.3cm} {\rm Phase} = {\rm arctan}\hspace{0.1cm}({3f_{\rm N}}/{f_{\rm T}}) \hspace{0.05cm}.$$
+  \Rightarrow \hspace{0.3cm}{\rm magnitude} = \frac{1}{\sqrt{1 + ({3f_{\rm N}}/{f_{\rm T}} )^2}} \hspace{0.05cm}, \hspace{0.3cm} {\rm Phase} = {\rm arctan}\hspace{0.1cm}({3f_{\rm N}}/{f_{\rm T}}) \hspace{0.05cm}.$$
-*Mit&nbsp; $f_{\rm N}/f_{\rm T} = 0.1$&nbsp; erhält man den Betrag&nbsp; $0.958$&nbsp; und die Phase&nbsp; $16.7^\circ$.&nbsp; Damit lautet das Sinkensignal:
+*When&nbsp; $f_{\rm N}/f_{\rm T} = 0.1$&nbsp; we get the magnitude&nbsp; $0.958$&nbsp; and the phase&nbsp; $16.7^\circ$.&nbsp; Thus, the sink signal is:
 :$$v(t) = 0.958 \cdot 2\,{\rm V}\cdot \cos (\omega_{\rm N}\cdot t + 16.7^\circ) \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} v(t=0)= 1.916\,{\rm V}\cdot \cos ( 16.7^\circ)\hspace{0.15cm}\underline { = 1.835\,{\rm V}}\hspace{0.05cm}.$$

@@ Line 43: / Line 43: @@
 <quiz display=simple>
-{Es gelte &nbsp;$R_{\rm U} = 0.8, \ I_{\rm U} = -0.2, \ R_{\rm O} = 0.4, I_{\rm O} = -0.2.$&nbsp; Berechnen und skizzieren Sie das Spektrum &nbsp;$R(f)$&nbsp; am Kanalausgang. <br>Wie lautet die Spektrallinie bei&nbsp; $-f_{\rm O}$?
+{Let &nbsp;$R_{\rm U} = 0.8, \ I_{\rm U} = -0.2, \ R_{\rm O} = 0.4, I_{\rm O} = -0.2.$&nbsp; Calculate and sketch the spectrum &nbsp;$R(f)$&nbsp; at the channel output. <br>What is the spectral line at&nbsp; $-f_{\rm O}$?
 |type="{}"}
 ${\rm Re}[R(-f_{\rm O})] \ = \ $ { 0.2 3% } $\ \text{V}$
@@ Line 49: / Line 49: @@
-{Wie lautet das Sinkensignal &nbsp;$v(t)$?&nbsp; Berücksichtigen Sie bei der Berechnung auch den Tiefpass des Synchrondemodulators. <br>Wie groß ist der Signalwert bei &nbsp;$t = 0$?
+{What is the sink signal &nbsp;$v(t)$?&nbsp; Take the synchronous demodulator's low-pass into account during calculation. <br>What is the signal value when&nbsp;$t = 0$?
 |type="{}"}
 $ v(t = 0) \ = \ $ { 1.2 3% } $\ \text{V}$
-{Berechnen Sie nun das Sinkensignal &nbsp;$v(t)$&nbsp; über den resultierenden Frequenzgang &nbsp;$H_{\rm MKD}(f)$&nbsp; und bewerten Sie den Rechengang.
+{Now calculate the sink signal &nbsp;$v(t)$&nbsp; over the resulting frequency response &nbsp;$H_{\rm MKD}(f)$&nbsp; and evaluate the calculation process.
 |type="()"}
-- Die Berechnung gemäß Teilaufgabe&nbsp; '''(2)'''&nbsp; führt schneller zum Erfolg.
+- The calculation according to subtask '''(2)'''&nbsp; leads to faster success.
-+ Die Berechnung gemäß Teilaufgabe&nbsp; '''(3)'''&nbsp; führt schneller zum Erfolg.
++ The calculation according to subtask '''(3)'''&nbsp; leads to faster success.
-{Berechnen Sie &nbsp;$v(t)$&nbsp; für den Kanalfrequenzgang &nbsp;$ H_{\rm K}(f) = H_{\rm(4)}(f)$.&nbsp; Wie groß ist der Signalwert bei &nbsp;$t = 0$?
+{Calculate &nbsp;$v(t)$&nbsp; for the channel frequency response &nbsp;$ H_{\rm K}(f) = H_{\rm(4)}(f)$.&nbsp; What is the signal value when &nbsp;$t = 0$?
 |type="{}"}
 $ v(t = 0) \ = \ $ { 1.835 3% }  $\ \text{V}$
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 [[File:P_ID1014__Mod_Z_2_5_a.png|right|frame|Empfangsspektrum]]