Aufgaben:Exercise 2.5: DSB-AM via a Gaussian channel - Revision history

Guenter at 12:26, 6 December 2021

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Guenter at 11:59, 6 December 2021

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Guenter at 10:56, 6 December 2021

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Guenter: Guenter moved page Aufgaben:Exercise 2.5: DSB-AM via a Gaussian Channel to Aufgaben:Exercise 2.5: DSB-AM via a Gaussian channel

2021-12-06T10:47:54Z

Guenter moved page Exercise 2.5: DSB-AM via a Gaussian Channel to Exercise 2.5: DSB-AM via a Gaussian channel

Reed at 15:07, 3 December 2021

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Reed at 18:09, 1 December 2021

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@@ Line 13: / Line 13: @@
 On the page &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Influence_of_linear_channel_distortions| Influence of linear channel distortions]]&nbsp; it was shown that the entire system is is sufficiently accurately characterized by the resulting frequency response
 :$$H_{\rm MKD}(f) = {1}/{2} \cdot \big[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\big]$$
-Here the subscript stands for '''M'''odulator–'''K''' (for German "Kanal" i.e. channel) –'''D'''emodulator.
+Here the subscript stands for&nbsp; $\rm M$odulator &ndash; $\rm K$&nbsp; (for German "Kanal" i.e. channel) &ndash; $\rm D$emodulator.
 The source signal &nbsp;$q(t)$&nbsp; is composed of two cosine oscillations:
@@ Line 66: / Line 66: @@
 {{ML-Kopf}}
 [[File:P_ID1011__Mod_A_2_5_a.png|right|frame|Resulting baseband frequency response for&nbsp; $f_{\rm T} = f_{\rm M}$]]
-'''(1)'''&nbsp; The equation given states that the BP frequency response &nbsp; $H_{\rm K}(f)$&nbsp; has to be shifted left and right by the carrier frequency &nbsp; $f_{\rm T}$&nbsp;, respectively, and the two components have to be added up.
+'''(1)'''&nbsp; The equation given states that the band-pass frequency response &nbsp; $H_{\rm K}(f)$&nbsp; has to be shifted to left and to right by the carrier frequency $f_{\rm T}$,&nbsp; and the two components have to be added up.
-*The factor &nbsp; $1/2$&nbsp; must still be taken into account (see plot).
+*The factor&nbsp; $1/2$&nbsp; must still be taken into account&nbsp; (see plot).
-*At low frequencies, this results in a Gaussian function around the center frequency  "0":
+*At low frequencies,&nbsp; this results in a Gaussian function around the center frequency&nbsp;  "0":
 :$$H_{\rm MKD}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left ({f}/{\Delta f_{\rm K}}\right)^2} \hspace{0.05cm}.$$
-*The two components at &nbsp; $±2f_{\rm T}$&nbsp; need not be considered further.  For the two frequencies we are looking for &nbsp; $f_1 = 1\ \rm  kHz$&nbsp; and &nbsp; $f_5 = 5 \ \rm kHz$&nbsp;, we obtain:
+*The two components at &nbsp; $±2f_{\rm T}$&nbsp; need not be considered further.&nbsp;  For the two frequencies we are looking for &nbsp; $f_1 = 1\ \rm  kHz$&nbsp; and &nbsp; $f_5 = 5 \ \rm kHz$,&nbsp; we obtain:
 :$$ H_{\rm MKD}(f = f_1) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{1\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/100}\hspace{0.15cm}\underline {\approx 0.969} \hspace{0.05cm},$$
 :$$H_{\rm MKD}(f = f_5) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/4} \hspace{0.3cm}\hspace{0.15cm}\underline {\approx 0.456} \hspace{0.05cm}.$$
@@ Line 77: / Line 77: @@
-'''(2)'''&nbsp; With&nbsp; $ω_1 = 2π · 1\ \rm   kHz$&nbsp; and&nbsp; $ω_5 = 2π · 5 \ \rm  kHz$&nbsp;, it holds that:
+'''(2)'''&nbsp; With&nbsp; $ω_1 = 2π · 1\ \rm   kHz$&nbsp; and&nbsp; $ω_5 = 2π · 5 \ \rm  kHz$,&nbsp; it holds that:
 :$$ v(t)  =  0.969 \cdot 2\,{\rm V}\cdot \cos (\omega_1 \cdot t)+ 0.456 \cdot 3\,{\rm V}\cdot \cos (\omega_5 \cdot t) =  \underline { 1.938\,{\rm V}}\cdot \cos (\omega_1 \cdot t) + \hspace{0.15cm}\underline {1.368\,{\rm V}}\cdot \cos (\omega_5 \cdot t) \hspace{0.05cm}.$$
-*It can be seen that now, unlike the source signal &nbsp; $q(t)$&nbsp;, the component at &nbsp; $1 \ \rm kHz$ &nbsp; &rArr; &nbsp; $A_1 = 1.938 \ \rm V$&nbsp; is larger than the &nbsp;  $5 \ \rm kHz$ component &nbsp; &rArr; &nbsp; $A_5 = 1.368 \ \rm V$, because the channel attenuates the &nbsp; $49 \ \rm kHz$&nbsp; and&nbsp; $51 \ \rm kHz$&nbsp; frequencies less than the spectral components at &nbsp; $45 \ \rm kHz$&nbsp; und&nbsp; $55 \ \rm kHz$.
+*It can be seen that now&nbsp; &ndash; unlike the source signal &nbsp; $q(t)$ &ndash; &nbsp; the component at &nbsp; $1 \ \rm kHz$ &nbsp; &rArr; &nbsp; $A_1 = 1.938 \ \rm V$&nbsp; is larger than the &nbsp;  $5 \ \rm kHz$ component &nbsp; &rArr; &nbsp; $A_5 = 1.368 \ \rm V$, because the channel attenuates the &nbsp; $49 \ \rm kHz$&nbsp; and&nbsp; $51 \ \rm kHz$&nbsp; frequencies less than the spectral components at &nbsp; $45 \ \rm kHz$&nbsp; and&nbsp; $55 \ \rm kHz$.
-'''(3)'''&nbsp; The two spectral functions shifted by &nbsp; $±f_{\rm T}$&nbsp; now no longer come to lie directly on top of each other, but are offset from each other by &nbsp; $10 \ \rm kHz$&nbsp;.
+'''(3)'''&nbsp; The two spectral functions shifted by &nbsp; $±f_{\rm T}$&nbsp; now no longer come to lie directly on top of each other,&nbsp; but there are an offset from each other by&nbsp; $10 \ \rm kHz$&nbsp;.
-*The resulting frequency response &nbsp; $H_{\rm MKD}(f)$&nbsp; is thus no longer Gaussian, but characterized according to the sketch below:
+*The resulting frequency response&nbsp; $H_{\rm MKD}(f)$&nbsp; is thus no longer Gaussian,&nbsp; but characterized according to the sketch below:
 :$$H_{\rm MKD}(f ) = {1}/{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f - 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f + 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right]\hspace{0.05cm}.$$
@@ Line 94: / Line 94: @@
 :$$\hspace{1.25cm}= \frac{1}{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{60\, {\rm kHz}- 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{-50\, {\rm kHz}+ 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right] = 0.022 + 0.500 \hspace{0.15cm}\underline {= 0.521}\hspace{0.05cm}.$$
 [[File:P_ID1012__Mod_A_2_5_c.png|right|frame|Resulting baseband frequency response for $f_{\rm T} \ne f_{\rm M}$]]
-*While the synchronous demodulator extracts information about the message signal from both sidebands in the same way at &nbsp; $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$&nbsp;, the lower sideband (LSB) provides the larger contribution at&nbsp; $f_{\rm T} = 55\ \rm  kHz$&nbsp;.
+*While the synchronous demodulator extracts information about the message signal from both sidebands in the same way at &nbsp; $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$,&nbsp; the lower sideband (LSB) provides the larger contribution at&nbsp; $f_{\rm T} = 55\ \rm  kHz$.
@@ Line 101: / Line 101: @@
-'''(4)'''&nbsp; With the result of the previous subtask, one obtains:
+'''(4)'''&nbsp; With the result of the previous subtask,&nbsp; one obtains:
 :$$ A_1  = 0.463 \cdot 2\,{\rm V}\hspace{0.15cm}\underline { = 0.926\,{\rm V}}\hspace{0.05cm},$$
 :$$A_5  = 0.521 \cdot 3\,{\rm V} \hspace{0.15cm}\underline {= 1.563\,{\rm V}}\hspace{0.05cm}.$$
-*In diesem Fall sind die linearen Verzerrungen sogar weniger stark, da nun auch der&nbsp; $1 \ \rm kHz$–Anteil stärker gedämpft wird.
+*In this case,&nbsp; the linear distortions are even less strong,&nbsp; since now also the&nbsp; $1 \ \rm kHz$ portion is attenuated more.
 '''(5)'''&nbsp; YES is correct:
-*With the carrier frequency&nbsp; $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$&nbsp;, the &nbsp; $5 \ \rm kHz$component is more attenuated than the &nbsp; $1 \ \rm kHz$ component, while at&nbsp; $f_{\rm T}  = 55 \ {\rm kHz} \ne f_{\rm M}$&nbsp;, the &nbsp; $1 \ \rm kHz$ component is slightly more attenuated.
+*With the carrier frequency&nbsp; $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$,&nbsp; the&nbsp; $5 \ \rm kHz$&nbsp; component is more attenuated than the&nbsp; $1 \ \rm kHz$&nbsp; component,&nbsp; while at&nbsp; $f_{\rm T}  = 55 \ {\rm kHz} \ne f_{\rm M}$,&nbsp; the&nbsp; $1 \ \rm kHz$ component is slightly more attenuated.
-*If one then chooses&nbsp; $f_{\rm T}  \approx 54.5 \ \rm kHz$ for example, both components are attenuated equally &nbsp; $($etwa um den Faktor $0.53)$&nbsp; and there is no or less distortion.
+*If one chooses&nbsp; $f_{\rm T}  \approx 54.5 \ \rm kHz$&nbsp; for example,&nbsp; both components are attenuated equally &nbsp; $($by about the factor&nbsp; $0.53)$&nbsp; and there is no or less distortion.
-*However, this result is only valid for the source signal considered.  &nbsp; Another &nbsp; $q(t)$&nbsp; also with two spectral components would require a different "optimal carrier frequency".
+*However,&nbsp; this result is only valid for the source signal considered.  &nbsp; Another&nbsp; $q(t)$&nbsp; with two spectral components would require a different&nbsp; "optimal carrier frequency".&nbsp; For a source signal with three or more spectral lines,&nbsp; linear distortions would always occur.
 {{ML-Fuß}}

@@ Line 33: / Line 33: @@
 <quiz display=simple>
-{Calculate the resulting frequency response &nbsp;$H_{\rm MKD}(f)$&nbsp; for &nbsp;$f_{\rm T} = 50 \ \rm  kHz$.&nbsp;
+{Calculate the resulting frequency response &nbsp;$H_{\rm MKD}(f)$&nbsp; for $f_{\rm T} = 50 \ \rm  kHz$.&nbsp;
-What are the values for &nbsp;$f = 1 \ \rm  kHz$&nbsp; and &nbsp;$f = 5 \ \rm  kHz$?
+What are the values for $f = 1 \ \rm  kHz$&nbsp; and &nbsp;$f = 5 \ \rm  kHz$?
 |type="{}"}
 $|H_{\rm MKD} (f = 1\ \rm  kHz)| \ = \ $ { 0.969 3% }
@@ Line 45: / Line 45: @@
-{Calculate the resulting frequency response &nbsp;$H_{\rm MKD}(f)$&nbsp; for &nbsp;$f_{\rm T} = 55 \ \rm  kHz$.&nbsp; Now, what are the values for &nbsp;$f = 1 \ \rm  kHz$&nbsp; and &nbsp;$f = 5 \ \rm  kHz$?
+{Calculate the resulting frequency response &nbsp;$H_{\rm MKD}(f)$&nbsp; for &nbsp;$f_{\rm T} = 55 \ \rm  kHz$. &nbsp; Now, what are the values for &nbsp;$f = 1 \ \rm  kHz$&nbsp; and &nbsp;$f = 5 \ \rm  kHz$?
 |type="{}"}
 $|H_{\rm MKD} (f = 1\ \rm  kHz)| \ = \ $ { 0.463 3% }
 $|H_{\rm MKD} (f = 5\ \rm  kHz)| \ = \ $ { 0.521 3% }
-{Calculate the sink signal &nbsp;$v(t)$.&nbsp;. Specify the amplitudes &nbsp;$A_1$&nbsp; and &nbsp;$A_5$&nbsp; of the &nbsp;$1\ \rm kHz$– and &nbsp;$5\ \rm kHz$ components, respectively.
+{Calculate the sink signal &nbsp;$v(t)$.&nbsp; Specify the amplitudes &nbsp;$A_1$&nbsp; and &nbsp;$A_5$&nbsp; of the two components.
 |type="{}"}
 $A_1 \ = \ $ { 0.926 3% } $\ \text{ V }$
 $A_5 \ = \ $ { 1.563 3% } $\ \text{ V }$
-{Is there a carrier frequency &nbsp;$f_{\rm T}$ that results in no distortion for the given source signal and channel?  Justify your answer.
+{Is there a carrier frequency $f_{\rm T}$ that results in no distortion for the given source signal and channel?&nbsp;  Justify your answer.
 |type="()"}
 + Yes,

@@ Line 4: / Line 4: @@
 [[File:EN_Mod_A_2_5.png|right|frame|DSB-AM over a distorting channel]]
 The communication system considered here is composed of the following blocks:
-*DSB-AM without carrier with  &nbsp;$f_{\rm T} = 50 \ \rm  kHz$&nbsp; and &nbsp;$f_{\rm T} = 55 \ \rm  kHz$:
+*DSB-AM without carrier  &nbsp;$(f_{\rm T} = 50 \ \rm  kHz$&nbsp; or &nbsp;$f_{\rm T} = 55 \ \rm  kHz)$:
 :$$ s(t) = q(t) \cdot \cos (2 \pi f_{\rm T} \hspace{0.05cm} t).$$
-* Gaussian bandpass channel; the magnitude &nbsp;$|f|$&nbsp;  in the exponent causes &nbsp;$H_K(–f) = H_K(f)$&nbsp; to hold:
+* Gaussian band-pass channel;&nbsp; the magnitude &nbsp;$|f|$&nbsp;  in the exponent causes &nbsp;$H_K(–f) = H_K(f)$&nbsp; to hold:
 :$$H_{\rm K}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (({|f| - f_{\rm M}})/{\Delta f_{\rm K}}\right)^2} ,\hspace{0.2cm} f_{\rm M} = 50\,{\rm kHz},\hspace{0.2cm} \Delta f_{\rm K} = 10\,{\rm kHz}\hspace{0.05cm}.$$
-*The synchronous demodulator has optimal parameters such that the sink signal &nbsp;$v(t)$&nbsp; completely coincides with the source signal &nbsp;$q(t)$&nbsp; when &nbsp;$H_{\rm K}(f) = 1$&nbsp; (ideal channel).
+*The synchronous demodulator has optimal parameters such that the sink signal &nbsp;$v(t)$&nbsp; completely coincides with the source signal &nbsp;$q(t)$&nbsp; when &nbsp;$H_{\rm K}(f) = 1$&nbsp; ("ideal channel").
 On the page &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Influence_of_linear_channel_distortions| Influence of linear channel distortions]]&nbsp; it was shown that the entire system is is sufficiently accurately characterized by the resulting frequency response
 :$$H_{\rm MKD}(f) = {1}/{2} \cdot \big[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\big]$$
-Here the index stands for '''M'''odulator–'''K''' (for german "Kanal" i.e. Channel) –'''D'''emodulator.
+Here the subscript stands for '''M'''odulator–'''K''' (for German "Kanal" i.e. channel) –'''D'''emodulator.
 The source signal &nbsp;$q(t)$&nbsp; is composed of two cosine oscillations:
@@ Line 23: / Line 23: @@
 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 42: / Line 39: @@
 $|H_{\rm MKD} (f = 5\ \rm  kHz)| \ = \ $ { 0.456 3% }
-{Calculate the sink signal &nbsp;$v(t)$.&nbsp;. Specify the amplitudes &nbsp;$A_1$&nbsp; and &nbsp;$A_5$&nbsp; of the &nbsp;$1\ \rm kHz$– and &nbsp;$5\ \rm kHz$ components, respectively.
+{Calculate the sink signal &nbsp;$v(t)$.&nbsp; Specify the amplitudes &nbsp;$A_1$&nbsp; and &nbsp;$A_5$&nbsp; of the &nbsp;$1\ \rm kHz$ component and the &nbsp;$5\ \rm kHz$&nbsp; component.
 |type="{}"}
 $A_1 \ = \ $ { 1.938 3% } $\ \text{ V }$

@@ Line 6: / Line 6: @@
 *DSB-AM without carrier with  &nbsp;$f_{\rm T} = 50 \ \rm  kHz$&nbsp; and &nbsp;$f_{\rm T} = 55 \ \rm  kHz$:
 :$$ s(t) = q(t) \cdot \cos (2 \pi f_{\rm T} \hspace{0.05cm} t).$$
-* GGaussian bandpass channel; the magnitude &nbsp;$|f|$&nbsp;  in the exponent causes &nbsp;$H_K(–f) = H_K(f)$&nbsp; to hold:
+* Gaussian bandpass channel; the magnitude &nbsp;$|f|$&nbsp;  in the exponent causes &nbsp;$H_K(–f) = H_K(f)$&nbsp; to hold:
 :$$H_{\rm K}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (({|f| - f_{\rm M}})/{\Delta f_{\rm K}}\right)^2} ,\hspace{0.2cm} f_{\rm M} = 50\,{\rm kHz},\hspace{0.2cm} \Delta f_{\rm K} = 10\,{\rm kHz}\hspace{0.05cm}.$$
 *The synchronous demodulator has optimal parameters such that the sink signal &nbsp;$v(t)$&nbsp; completely coincides with the source signal &nbsp;$q(t)$&nbsp; when &nbsp;$H_{\rm K}(f) = 1$&nbsp; (ideal channel).

@@ Line 60: / Line 60: @@
 {Is there a carrier frequency &nbsp;$f_{\rm T}$ that results in no distortion for the given source signal and channel?  Justify your answer.
 |type="()"}
-+ Ja,
++ Yes,
-- Nein.
+- No.
@@ Line 68: / Line 68: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID1011__Mod_A_2_5_a.png|right|frame|Resulierender Basisbandfrequenzgang für&nbsp; $f_{\rm T} = f_{\rm M}$]]
+[[File:P_ID1011__Mod_A_2_5_a.png|right|frame|Resulting baseband frequency response for&nbsp; $f_{\rm T} = f_{\rm M}$]]
-'''(1)'''&nbsp; Die angegebene Gleichung besagt, dass der BP–Frequenzgang&nbsp; $H_{\rm K}(f)$&nbsp; jeweils um die Trägerfrequenz&nbsp; $f_{\rm T}$&nbsp; nach links und rechts verschoben und die beiden Anteile aufaddiert werden müssen.&nbsp;
+'''(1)'''&nbsp; The equation given states that the BP frequency response &nbsp; $H_{\rm K}(f)$&nbsp; has to be shifted left and right by the carrier frequency &nbsp; $f_{\rm T}$&nbsp;, respectively, and the two components have to be added up.
-*Es ist noch der Faktor&nbsp; $1/2$&nbsp; zu berücksichtigen (siehe Skizze).
+*The factor &nbsp; $1/2$&nbsp; must still be taken into account (see plot).
-*Bei niedrigen Frequenzen ergibt sich dann eine Gaußfunktion um die Mittenfrequenz „0”:
+*At low frequencies, this results in a Gaussian function around the center frequency  "0":
 :$$H_{\rm MKD}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left ({f}/{\Delta f_{\rm K}}\right)^2} \hspace{0.05cm}.$$
-*Die beiden Anteile bei&nbsp; $±2f_{\rm T}$&nbsp; müssen nicht weiter betrachtet werden.&nbsp; Für die zwei gesuchten Frequenzen&nbsp; $f_1 = 1\ \rm  kHz$&nbsp; und&nbsp; $f_5 = 5 \ \rm kHz$&nbsp; erhält man:
+*The two components at &nbsp; $±2f_{\rm T}$&nbsp; need not be considered further.  For the two frequencies we are looking for &nbsp; $f_1 = 1\ \rm  kHz$&nbsp; and &nbsp; $f_5 = 5 \ \rm kHz$&nbsp;, we obtain:
 :$$ H_{\rm MKD}(f = f_1) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{1\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/100}\hspace{0.15cm}\underline {\approx 0.969} \hspace{0.05cm},$$
 :$$H_{\rm MKD}(f = f_5) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/4} \hspace{0.3cm}\hspace{0.15cm}\underline {\approx 0.456} \hspace{0.05cm}.$$
@@ Line 80: / Line 80: @@
-'''(2)'''&nbsp; Mit&nbsp; $ω_1 = 2π · 1\ \rm   kHz$&nbsp; und&nbsp; $ω_5 = 2π · 5 \ \rm  kHz$&nbsp; gilt:
+'''(2)'''&nbsp; With&nbsp; $ω_1 = 2π · 1\ \rm   kHz$&nbsp; and&nbsp; $ω_5 = 2π · 5 \ \rm  kHz$&nbsp;, it holds that:
 :$$ v(t)  =  0.969 \cdot 2\,{\rm V}\cdot \cos (\omega_1 \cdot t)+ 0.456 \cdot 3\,{\rm V}\cdot \cos (\omega_5 \cdot t) =  \underline { 1.938\,{\rm V}}\cdot \cos (\omega_1 \cdot t) + \hspace{0.15cm}\underline {1.368\,{\rm V}}\cdot \cos (\omega_5 \cdot t) \hspace{0.05cm}.$$
-*Man erkennt, dass nun –&nbsp; im Gegensatz zum Quellensignal&nbsp; $q(t)$&nbsp; – der Anteil bei&nbsp; $1 \ \rm kHz$ &nbsp; &rArr; &nbsp; $A_1 = 1.938 \ \rm V$&nbsp; größer ist als der&nbsp;  $5 \ \rm kHz$–Anteil &nbsp; &rArr; &nbsp; $A_5 = 1.368 \ \rm V$, da der Kanal die Frequenzen&nbsp; $49 \ \rm kHz$&nbsp; und&nbsp; $51 \ \rm kHz$&nbsp; weniger dämpft als die Spektralanteile bei&nbsp; $45 \ \rm kHz$&nbsp; und&nbsp; $55 \ \rm kHz$.
+*It can be seen that now, unlike the source signal &nbsp; $q(t)$&nbsp;, the component at &nbsp; $1 \ \rm kHz$ &nbsp; &rArr; &nbsp; $A_1 = 1.938 \ \rm V$&nbsp; is larger than the &nbsp;  $5 \ \rm kHz$ component &nbsp; &rArr; &nbsp; $A_5 = 1.368 \ \rm V$, because the channel attenuates the &nbsp; $49 \ \rm kHz$&nbsp; and&nbsp; $51 \ \rm kHz$&nbsp; frequencies less than the spectral components at &nbsp; $45 \ \rm kHz$&nbsp; und&nbsp; $55 \ \rm kHz$.
-'''(3)'''&nbsp; Die beiden um&nbsp; $±f_{\rm T}$&nbsp; verschobenen Spektralfunktionen kommen nun nicht mehr direkt übereinander zu liegen, sondern sind um&nbsp; $10 \ \rm kHz$&nbsp; gegeneinander versetzt.
+'''(3)'''&nbsp; The two spectral functions shifted by &nbsp; $±f_{\rm T}$&nbsp; now no longer come to lie directly on top of each other, but are offset from each other by &nbsp; $10 \ \rm kHz$&nbsp;.
-*Der resultierende Frequenzgang&nbsp; $H_{\rm MKD}(f)$&nbsp; ist somit nicht mehr gaußförmig, sondern es gilt entsprechend der unteren Skizze:
+*The resulting frequency response &nbsp; $H_{\rm MKD}(f)$&nbsp; is thus no longer Gaussian, but characterized according to the sketch below:
 :$$H_{\rm MKD}(f ) = {1}/{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f - 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f + 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right]\hspace{0.05cm}.$$
-*Für die Frequenzen&nbsp; $f_1$&nbsp; und&nbsp; $f_5$&nbsp; erhält man:
+*For the frequencies&nbsp; $f_1$&nbsp; and&nbsp; $f_5$&nbsp; we get:
 :$$H_{\rm MKD}(f = 1\,{\rm kHz}) = \frac{1}{2} \cdot \left[ H_{\rm K}(f = 56\,{\rm kHz}) + H_{\rm K}(f = -54\,{\rm kHz})\right]=$$
@@ Line 96: / Line 96: @@
 :$$H_{\rm MKD}(f = 5\,{\rm kHz}) = \frac{1}{2} \cdot \left[ H_{\rm K}(f = 60\,{\rm kHz}) + H_{\rm K}(f = -50\,{\rm kHz})\right]= \hspace{0.75cm}$$
 :$$\hspace{1.25cm}= \frac{1}{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{60\, {\rm kHz}- 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{-50\, {\rm kHz}+ 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right] = 0.022 + 0.500 \hspace{0.15cm}\underline {= 0.521}\hspace{0.05cm}.$$
-[[File:P_ID1012__Mod_A_2_5_c.png|right|frame|Resulierender Basisbandfrequenzgang für $f_{\rm T} \ne f_{\rm M}$]]
+[[File:P_ID1012__Mod_A_2_5_c.png|right|frame|Resulting baseband frequency response for $f_{\rm T} \ne f_{\rm M}$]]
-*Während bei&nbsp; $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$&nbsp; der Synchrondemodulator die Information über das Nachrichtensignal aus beiden Seitenbändern in gleicher Weise gewinnt, liefert mit&nbsp; $f_{\rm T} = 55\ \rm  kHz$&nbsp; das untere Seitenband (USB) den größeren Beitrag.
+*While the synchronous demodulator extracts information about the message signal from both sidebands in the same way at &nbsp; $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$&nbsp;, the lower sideband (LSB) provides the larger contribution at&nbsp; $f_{\rm T} = 55\ \rm  kHz$&nbsp;.
-*Zum Beispiel liegt das USB des&nbsp; $5 \ \rm kHz$–Anteils nun genau bei&nbsp; $f_{\rm M} = 50 \ \rm kHz$&nbsp; und wird ungedämpft übertragen, während das OSB bei&nbsp; $60 \ \rm kHz$&nbsp; starken Dämpfungen unterliegt.
+*For example, the LSB of the &nbsp; $5 \ \rm kHz$ component is now exactly at &nbsp; $f_{\rm M} = 50 \ \rm kHz$&nbsp; and is transmitted undamped, while the USB is subject to heavy attenuation at &nbsp; $60 \ \rm kHz$&nbsp;.
-'''(4)'''&nbsp; Mit dem Ergebnis der letzten Teilaufgabe erhält man:
+'''(4)'''&nbsp; With the result of the previous subtask, one obtains:
 :$$ A_1  = 0.463 \cdot 2\,{\rm V}\hspace{0.15cm}\underline { = 0.926\,{\rm V}}\hspace{0.05cm},$$
 :$$A_5  = 0.521 \cdot 3\,{\rm V} \hspace{0.15cm}\underline {= 1.563\,{\rm V}}\hspace{0.05cm}.$$
@@ Line 110: / Line 110: @@
-'''(5)'''&nbsp; Richtig ist JA:
+'''(5)'''&nbsp; YES is correct:
-*Mit der Trägerfrequenz&nbsp; $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$&nbsp; wird der&nbsp; $5 \ \rm kHz$–Anteil stärker gedämpft als der&nbsp; $1 \ \rm kHz$–Anteil, während mit&nbsp; $f_{\rm T}  = 55 \ {\rm kHz} \ne f_{\rm M}$&nbsp; der&nbsp; $1 \ \rm kHz$–Anteil etwas mehr gedämpft wird.
+*With the carrier frequency&nbsp; $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$&nbsp;, the &nbsp; $5 \ \rm kHz$component is more attenuated than the &nbsp; $1 \ \rm kHz$ component, while at&nbsp; $f_{\rm T}  = 55 \ {\rm kHz} \ne f_{\rm M}$&nbsp;, the &nbsp; $1 \ \rm kHz$ component is slightly more attenuated.
-*Wählt man nun zum Beispiel&nbsp; $f_{\rm T}  \approx 54.5 \ \rm kHz$, so werden beide Anteile gleich gedämpft&nbsp; $($etwa um den Faktor $0.53)$&nbsp; und es gibt keine / weniger Verzerrungen.
+*If one then chooses&nbsp; $f_{\rm T}  \approx 54.5 \ \rm kHz$ for example, both components are attenuated equally &nbsp; $($etwa um den Faktor $0.53)$&nbsp; and there is no or less distortion.
-*Dieses Ergebnis gilt allerdings nur für das betrachtete Quellensignal.&nbsp; Ein anderes&nbsp; $q(t)$&nbsp; mit ebenfalls zwei Spektralanteilen würde eine andere „optimale Trägerfrequenz” erfordern.
+*However, this result is only valid for the source signal considered.  &nbsp; Another &nbsp; $q(t)$&nbsp; also with two spectral components would require a different "optimal carrier frequency".
-*Bei einem Nachrichtensignal mit drei oder mehr Spektrallinien würde es stets zu linearen Verzerrungen kommen.
+*For a message signal with three or more spectral lines, linear distortions would always occur.
 {{ML-Fuß}}

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Modulationsverfahren/Synchrondemodulation}}
-[[File:EN_Mod_A_2_5.png|right|frame|ZSB-AM über einen verzerrenden Kanal]]
+[[File:EN_Mod_A_2_5.png|right|frame|DSB-AM over a distorting channel]]
-Das hier betrachtete Übertragungssystem setzt sich aus folgenden Blöcken zusammen:
+The communication system considered here is composed of the following blocks:
-*ZSB–AM ohne Träger mit &nbsp;$f_{\rm T} = 50 \ \rm  kHz$&nbsp; bzw. &nbsp;$f_{\rm T} = 55 \ \rm  kHz$:
+*DSB-AM without carrier with  &nbsp;$f_{\rm T} = 50 \ \rm  kHz$&nbsp; and &nbsp;$f_{\rm T} = 55 \ \rm  kHz$:
 :$$ s(t) = q(t) \cdot \cos (2 \pi f_{\rm T} \hspace{0.05cm} t).$$
-* Gaußförmiger Bandpass–Kanal; der Betrag &nbsp;$|f|$&nbsp; im Exponenten bewirkt, dass &nbsp;$H_K(–f) = H_K(f)$&nbsp; gilt:
+* GGaussian bandpass channel; the magnitude &nbsp;$|f|$&nbsp;  in the exponent causes &nbsp;$H_K(–f) = H_K(f)$&nbsp; to hold:
 :$$H_{\rm K}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (({|f| - f_{\rm M}})/{\Delta f_{\rm K}}\right)^2} ,\hspace{0.2cm} f_{\rm M} = 50\,{\rm kHz},\hspace{0.2cm} \Delta f_{\rm K} = 10\,{\rm kHz}\hspace{0.05cm}.$$
-* Synchrondemodulator mit optimalen Kenngrößen, so dass das Sinkensignal &nbsp;$v(t)$&nbsp; vollständig mit dem Quellensignal &nbsp;$q(t)$&nbsp; übereinstimmt, wenn &nbsp;$H_{\rm K}(f) = 1$&nbsp; ist (idealer Kanal).
+*The synchronous demodulator has optimal parameters such that the sink signal &nbsp;$v(t)$&nbsp; completely coincides with the source signal &nbsp;$q(t)$&nbsp; when &nbsp;$H_{\rm K}(f) = 1$&nbsp; (ideal channel).
-Auf der Seite &nbsp;[[Modulation_Methods/Synchrondemodulation#Einfluss_linearer_Kanalverzerrungen| Einfluss linearer Kanalverzerrungen]]&nbsp; wurde gezeigt, dass das gesamte System durch den resultierenden Frequenzgang
+On the page &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Influence_of_linear_channel_distortions| Influence of linear channel distortions]]&nbsp; it was shown that the entire system is is sufficiently accurately characterized by the resulting frequency response
-:$$H_{\rm MKD}(f) = {1}/{2} \cdot \big[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\big]$$
+:$$H_{\rm MKD}(f) = {1}/{2} \cdot \big[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\big]$$.
-ausreichend genau charakterisiert ist. Der Index steht hierbei für '''M'''odulator–'''K'''anal–'''D'''emodulator.
+Here the index stands for '''M'''odulator–'''K''' (for german "Kanal" i.e. Channel –'''D'''emodulator.
-Das Quellensignal &nbsp;$q(t)$&nbsp; setzt sich aus zwei Cosinus-Schwingungen zusammen:
+The source signal &nbsp;$q(t)$&nbsp; is composed of two cosine oscillations:
 :$$q(t) = 2\,{\rm V}\cdot \cos (2 \pi \cdot 1\,{\rm kHz} \cdot t)+ 3\,{\rm V}\cdot \cos (2 \pi \cdot 5\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
@@ Line 26: / Line 26: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Modulation_Methods/Synchrondemodulation|Synchrondemodulation]].
+*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]].
-*Bezug genommen wird insbesondere auf die Seite&nbsp;  [[Modulation_Methods/Synchrondemodulation#Einfluss_linearer_Kanalverzerrungen|Einfluss linearer Kanalverzerrungen]].
+*Particular reference is made to the page&nbsp;  [[Modulation_Methods/Synchronous_Demodulation#Influence_of_linear_channel_distortions|Influence of linear channel distortions]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>