Aufgaben:Exercise 2.9: Symmetrical Distortions - Revision history

Hwang at 14:20, 18 January 2023

2023-01-18T14:20:00Z

Reed at 15:40, 31 March 2022

2022-03-31T15:40:33Z

Guenter at 13:54, 15 February 2022

2022-02-15T13:54:25Z

Guenter at 13:25, 15 February 2022

2022-02-15T13:25:49Z

Reed at 20:25, 20 December 2021

2021-12-20T20:25:03Z

Reed at 20:24, 20 December 2021

2021-12-20T20:24:37Z

Reed at 20:23, 20 December 2021

2021-12-20T20:23:47Z

Reed at 20:23, 20 December 2021

2021-12-20T20:23:10Z

Reed at 20:15, 20 December 2021

2021-12-20T20:15:45Z

Reed at 20:15, 20 December 2021

2021-12-20T20:15:27Z

← Older revision		Revision as of 14:20, 18 January 2023
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−	The upper graph shows the spectrum  $S_{\rm TP}(f)$  of the equivalent low-pass signal in schematic form.  This means that the lengths of the Dirac lines drawn do not correspond to the actual values of  $A_{\rm T}$,  $A_1/2$  and  $A_2/2$.	+	The upper graph shows the spectrum  $S_{\rm TP}(f)$  of the equivalent low-pass signal in schematic form.  This means that the lengths of the Dirac delta lines drawn do not correspond to the actual values of  $A_{\rm T}$,  $A_1/2$  and  $A_2/2$.

← Older revision		Revision as of 15:40, 31 March 2022
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−	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Hüllkurvendemodulation~~	+	{{quiz-Header\|Buchseite=Modulation_Methods/Envelope_Demodulation
	}}		}}

@@ Line 61: / Line 61: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; On the basis of the graphs on the exercise page, the following statements can be made:
+'''(1)'''&nbsp; On the basis of the graphs on the exercise page,&nbsp; the following statements can be made:
 :$${A_{\rm T}} \cdot 0.5 = 2 \,{\rm V}\hspace{0.3cm}  \Rightarrow  \hspace{0.3cm}A_{\rm T} \hspace{0.15cm}\underline {= 4 \,{\rm V}},$$
 :$${A_{\rm 1}}/{2} \cdot 0.4 = 0.6\,{\rm V}\hspace{0.3cm}  \Rightarrow  \hspace{0.3cm}A_{\rm 1} \hspace{0.15cm}\underline {= 3 \,{\rm V}},$$
@@ Line 68: / Line 68: @@
-'''(2)'''&nbsp; <u>Answer 3</u> is correct:
+'''(2)'''&nbsp; <u>Answer 3</u>&nbsp; is correct:
-*The resulting modulation depth is &nbsp; $m = (A_1 + A_2)/A_T = 1.75$.
+*The resulting modulation depth is &nbsp; $m = (A_1 + A_2)/A_T = 1.75 >1$.
 *This leads to strong nonlinear distortion when using an envelope demodulator.
 *A distortion factor cannot be specified because the source signal contains two frequency components.
@@ Line 80: / Line 80: @@
 :$$ r_{\rm TP}(t) = 2 \,{\rm V} + 1.2 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 0.8 \,{\rm V} \cdot \cos(2 \pi f_2 t )\hspace{0.05cm}.$$
 *This function is always real and non-negative.
-*Thus, &nbsp; $ϕ(t) = 0$.&nbsp; holds simultaneously, whereas&nbsp; $ϕ(t) = 180^\circ$&nbsp; is not possible.
+*Thus,&nbsp; $ϕ(t) = 0$&nbsp; holds simultaneously,&nbsp; whereas&nbsp; $ϕ(t) = 180^\circ$&nbsp; is not possible.
@@ Line 88: / Line 88: @@
 :$$q(t)  =  3 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 4 \,{\rm V} \cdot \cos(2 \pi f_2 t ),$$
 :$$ v(t)  =  0.4 \cdot 3 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 0.2 \cdot 4 \,{\rm V} \cdot \cos(2 \pi f_2 t )$$
-:shows, that linear distortions now arise – attenuation distortions to be precise  &nbsp; &rArr; &nbsp;  <u>Answer 2</u>.
+:shows,&nbsp; that linear (attenuation) distortions now arise   &nbsp; &rArr; &nbsp;  <u>Answer 2</u>.
-*Here, the channel&nbsp; $H_{\rm K}(f)$&nbsp; has the positive effect, that instead of irreversible nonlinear distortions, only linear distortions arise, and these can be eliminated by a downstream filter.
+*Here, the channel&nbsp; $H_{\rm K}(f)$&nbsp; has the positive effect,&nbsp; that instead of irreversible nonlinear distortions,&nbsp; only linear distortions arise,&nbsp; and these can be eliminated by a downstream filter.
-*This is due to the fact that the higher attenuation of the source signal&nbsp; $q(t)$&nbsp; compared to the carrier signal&nbsp; $z(t)$&nbsp; lowers the modulation depth from &nbsp; $m = 1.75$&nbsp; to&nbsp; $m = (0.4 · 3 \ \rm  V + 0.2 · 4 \ \rm  V)/(0.5 · 4 \ \rm  V) = 1$&nbsp;.
+*This is due to the fact that the higher attenuation of the source signal&nbsp; $q(t)$&nbsp; compared to the carrier signal&nbsp; $z(t)$&nbsp; lowers the modulation depth from &nbsp; $m = 1.75$&nbsp; to&nbsp;

@@ Line 5: / Line 5: @@
 The source signal made up of two components
 :$$q(t) = A_1 \cdot \cos(2 \pi f_1 t ) + A_2 \cdot \cos(2 \pi f_2 t )$$
-is amplitude modulated and transmitted through a linearly distorting transmission channel.&nbsp; The carrier frequency is  &nbsp;$f_{\rm T}$&nbsp; and the added DC component &nbsp;$A_{\rm T}$.&nbsp; Thus, a &nbsp;''Double-sideband amplitude moduluation''&nbsp; $\rm (DSB–AM)$ ''with carrier''&nbsp; is present.
+is amplitude modulated and transmitted through a linearly distorting transmission channel.&nbsp;
-The channel frequency response is characterised with sufficient accuracy with a few auxiliary values:
+The spectral function &nbsp;$R(f)$&nbsp; of the received signal was measured.&nbsp; In the lower graph we can observe the equivalent low-pass spectrum &nbsp;$R_{\rm TP}(f)$ calculated from this.
 :$$ H_{\rm K}(f = f_{\rm T}) = 0.5,$$
 :$$H_{\rm K}(f = f_{\rm T} \pm f_1) = 0.4,$$
@@ Line 20: / Line 23: @@
 Hints:
+*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Envelope_Demodulation|Envelope Demodulation]].
+*Particular reference is made to the section&nbsp; [[Modulation_Methods/Envelope_Demodulation#Description_using_the_equivalent_low-pass_signal|Description using the equivalent low-pass signal]].
@@ Line 46: / Line 45: @@
 + Nonlinear distortions.
-{Calculate the equivalent lowpass signal and answer the following questions. Is it true that...
+{Calculate the equivalent low-pass signal and answer the following questions. Is it true that...
 |type="[]"}
 + $r_{\rm TP}(t)$&nbsp; is always real,

← Older revision		Revision as of 20:25, 20 December 2021
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	*Here, the channel  $H_{\rm K}(f)$  has the positive effect, that instead of irreversible nonlinear distortions, only linear distortions arise, and these can be eliminated by a downstream filter.		*Here, the channel  $H_{\rm K}(f)$  has the positive effect, that instead of irreversible nonlinear distortions, only linear distortions arise, and these can be eliminated by a downstream filter.
−	*This is due to the fact that the higher attenuation of the source signal  $q(t)$  compared to the carrier signal  $z(t)$  lowers the modulation depth from n  $m = 1.75$  to  $m = (0.4 · 3 \ \rm V + 0.2 · 4 \ \rm V)/(0.5 · 4 \ \rm V) = 1$ .	+	*This is due to the fact that the higher attenuation of the source signal  $q(t)$  compared to the carrier signal  $z(t)$  lowers the modulation depth from   $m = 1.75$  to  $m = (0.4 · 3 \ \rm V + 0.2 · 4 \ \rm V)/(0.5 · 4 \ \rm V) = 1$ .

@@ Line 91: / Line 91: @@
 :shows, that linear distortions now arise – attenuation distortions to be precise  &nbsp; &rArr; &nbsp;  <u>Answer 2</u>.
-*Here, the channel&nbsp; $H_{\rm K}(f)$&nbsp; has the positive effect, that instead of irreversible nonlinear distortions, linear distortions occur that can be eliminated by a downstream filter.
+*Here, the channel&nbsp; $H_{\rm K}(f)$&nbsp; has the positive effect, that instead of irreversible nonlinear distortions, only linear distortions arise, and these can be eliminated by a downstream filter.
 *This is due to the fact that the higher attenuation of the source signal&nbsp; $q(t)$&nbsp; compared to the carrier signal&nbsp; $z(t)$&nbsp; lowers the modulation depth from n&nbsp; $m = 1.75$&nbsp; to&nbsp; $m = (0.4 · 3 \ \rm  V + 0.2 · 4 \ \rm  V)/(0.5 · 4 \ \rm  V) = 1$&nbsp;.

@@ Line 62: / Line 62: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Anhand der Grafiken auf der Angabenseite sind folgende Aussagen möglich:
+'''(1)'''&nbsp; On the basis of the graphs on the exercise page, the following statements can be made:
 :$${A_{\rm T}} \cdot 0.5 = 2 \,{\rm V}\hspace{0.3cm}  \Rightarrow  \hspace{0.3cm}A_{\rm T} \hspace{0.15cm}\underline {= 4 \,{\rm V}},$$
 :$${A_{\rm 1}}/{2} \cdot 0.4 = 0.6\,{\rm V}\hspace{0.3cm}  \Rightarrow  \hspace{0.3cm}A_{\rm 1} \hspace{0.15cm}\underline {= 3 \,{\rm V}},$$
@@ Line 69: / Line 69: @@
-'''(2)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 3</u>:
+'''(2)'''&nbsp; <u>Answer 3</u> is correct:
-*Der Modulationsgrad ergibt sich zu&nbsp; $m = (A_1 + A_2)/A_T = 1.75$.
+*The resulting modulation depth is &nbsp; $m = (A_1 + A_2)/A_T = 1.75$.
-*Damit ergeben sich bei Verwendung eines Hüllkurvendemodulators starke nichtlineare Verzerrungen.
+*This leads to strong nonlinear distortion when using an envelope demodulator.
-*Ein Klirrfaktor kann aber nicht angegeben werden, da das Quellensignal zwei Frequenzanteile beinhaltet.
+*A distortion factor cannot be specified because the source signal contains two frequency components.
-'''(3)'''&nbsp; Richtig sind <u>die Aussagen 1 und 2</u>:
+'''(3)'''&nbsp; <u>Answers 1 and 2</u> are correct:
-*Die Fourierrücktransformation von&nbsp; $R_{\rm TP}(f)$&nbsp; führt zum Ergebnis:
+*The Fourier retransform of &nbsp; $R_{\rm TP}(f)$&nbsp; gives us the result:
 :$$ r_{\rm TP}(t) = 2 \,{\rm V} + 1.2 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 0.8 \,{\rm V} \cdot \cos(2 \pi f_2 t )\hspace{0.05cm}.$$
-*Diese Funktion ist stets reell und nicht–negativ.
+*This function is always real and non-negativ.
-*Damit gilt gleichzeitig&nbsp; $ϕ(t) = 0$.&nbsp; Dagegen ist&nbsp; $ϕ(t) = 180^\circ$&nbsp; nicht möglich.
+*Thus, &nbsp; $ϕ(t) = 0$.&nbsp; holds simultaneously, whereas&nbsp; $ϕ(t) = 180^\circ$&nbsp; is not possible.
-'''(4)'''&nbsp; Ein Vergleich der beiden Signale
+'''(4)'''&nbsp; A comparison of the two signals
 :$$q(t)  =  3 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 4 \,{\rm V} \cdot \cos(2 \pi f_2 t ),$$
 :$$ v(t)  =  0.4 \cdot 3 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 0.2 \cdot 4 \,{\rm V} \cdot \cos(2 \pi f_2 t )$$
-:zeigt, dass nun lineare Verzerrungen – genauer gesagt Dämpfungsverzerrungen – auftreten &nbsp; &rArr; &nbsp;  <u>Lösungsvorschlag 2</u>.
+:shows, that linear distortions now arise – attenuation distortions to be precise  &nbsp; &rArr; &nbsp;  <u>Answer 2</u>.
-*Der Kanal&nbsp; $H_{\rm K}(f)$&nbsp; hat hier den positiven Effekt, dass anstelle von irreversiblen nichtlinearen Verzerrungen nun lineare Verzerrungen entstehen, die durch ein nachgeschaltetes Filter eliminiert werden können.
+*Here, the channel&nbsp; $H_{\rm K}(f)$&nbsp; has the positive effect, that instead of irreversible nonlinear distortions, linear distortions occur that can be eliminated by a downstream filter.
-*Dies ist darauf zurückzuführen, dass durch die stärkere Dämpfung des Quellensignals&nbsp; $q(t)$&nbsp; im Vergleich zum Trägersignal&nbsp; $z(t)$&nbsp; der Modulationsgrad von&nbsp; $m = 1.75$&nbsp; auf&nbsp; $m = (0.4 · 3 \ \rm  V + 0.2 · 4 \ \rm  V)/(0.5 · 4 \ \rm  V) = 1$&nbsp; herabgesetzt wird.
+*This is due to the fact that the higher attenuation of the source signal&nbsp; $q(t)$&nbsp; compared to the carrier signal&nbsp; $z(t)$&nbsp; lowers the modulation depth from n&nbsp; $m = 1.75$&nbsp; to&nbsp; $m = (0.4 · 3 \ \rm  V + 0.2 · 4 \ \rm  V)/(0.5 · 4 \ \rm  V) = 1$&nbsp;.

@@ Line 56: / Line 56: @@
 - No distortion.
 + Linear distortions.
-- Nonlinear distortions.
+- Nonlinear distortions.<br>
 </quiz>