Aufgaben:Exercise 2.7Z: DSB-AM and Envelope Demodulator - Revision history

Hwang at 14:19, 18 January 2023

2023-01-18T14:19:35Z

Reed at 15:27, 31 March 2022

2022-03-31T15:27:39Z

Guenter at 15:36, 22 February 2022

2022-02-22T15:36:44Z

Guenter at 12:31, 15 February 2022

2022-02-15T12:31:34Z

Guenter at 11:42, 15 February 2022

2022-02-15T11:42:26Z

Guenter at 10:49, 15 February 2022

2022-02-15T10:49:10Z

Guenter at 10:47, 15 February 2022

2022-02-15T10:47:04Z

Reed at 19:07, 20 December 2021

2021-12-20T19:07:59Z

Reed at 19:07, 20 December 2021

2021-12-20T19:07:32Z

Reed at 19:07, 20 December 2021

2021-12-20T19:07:03Z

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-The graph shows the spectrum&nbsp; $R_{\rm TP}(f)$&nbsp; of the received signal in the equivalent low-pass region,&nbsp; which is composed of Dirac lines
+The graph shows the spectrum&nbsp; $R_{\rm TP}(f)$&nbsp; of the received signal in the equivalent low-pass region,&nbsp; which is composed of Dirac delta lines
 *at &nbsp; $f = 0$&nbsp; (originating from the carrier),&nbsp;
 *at &nbsp;$±2\ \rm  kHz$&nbsp; (originating from the cosine component)&nbsp; and
@@ Line 71: / Line 71: @@
-'''(2)'''&nbsp; In the graph on the information page,&nbsp; the weight of the Dirac line at&nbsp; $f = 0$&nbsp; indicates the amplitude of the added carrier.&nbsp; This is&nbsp; $A_{\rm T}\hspace{0.15cm}\underline{ = 4\ \rm V }$.
+'''(2)'''&nbsp; In the graph on the information page,&nbsp; the weight of the Dirac delta line at&nbsp; $f = 0$&nbsp; indicates the amplitude of the added carrier.&nbsp; This is&nbsp; $A_{\rm T}\hspace{0.15cm}\underline{ = 4\ \rm V }$.
 *From this,&nbsp; we get the modulation depth&nbsp; $m = q_{\rm max}/A_{\rm T} \hspace{0.15cm}\underline{ = 1}$.

← Older revision		Revision as of 15:27, 31 March 2022
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−	{{quiz-Header\|Buchseite=Modulationsverfahren/~~Hüllkurvendemodulation~~	+	{{quiz-Header\|Buchseite=Modulationsverfahren/Modulation_Methods/Envelope_Demodulation
	}}		}}

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 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID1035__Mod_Z_2_7_a.png|right|frame|Source signal in the region up to&nbsp; $1\text{ ms}$]]
+[[File:EN_Mod_Z_2_7_a.png|right|frame|Source signal in the region up to&nbsp; $1\text{ ms}$]]
 '''(1)'''&nbsp; The graph shows that the source signal can take on all values between &nbsp; $–4 \ \rm V$&nbsp; and&nbsp; $+3.667\ \rm  V$.&nbsp;
 *For example,&nbsp; the maximum magnitude occurs at time &nbsp; $t = t_0 =0.75\ \rm  ms$:

← Older revision		Revision as of 12:31, 15 February 2022
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−	The graph shows the spectrum  $R_{\rm TP}(f)$  of the received signal in the equivalent low-pass region,  which is composed of Dirac lines at   $f = 0$  (originating from the carrier),  at  $±2\ \rm kHz$  (originating from the cosine component)  and at  $±5\ \rm kHz$  (originating from the sine component).	+	The graph shows the spectrum  $R_{\rm TP}(f)$  of the received signal in the equivalent low-pass region,  which is composed of Dirac lines
		+	*at   $f = 0$  (originating from the carrier),
		+	*at  $±2\ \rm kHz$  (originating from the cosine component)  and
		+	*at  $±5\ \rm kHz$  (originating from the sine component).


−	*The locus curve is the plot of the equivalent low-pass signal  $r_{\rm TP}(t)$  in the complex plane,	+	The locus curve is the plot of the equivalent low-pass signal  $r_{\rm TP}(t)$  in the complex plane,  where  $r_{\rm TP}(t)$  is the Fourier retransform of  $R_{\ \rm TP}(f)$ .
−	* where  $r_{\rm TP}(t)$  is the Fourier retransform of  $R_{\ \rm TP}(f)$ .

@@ Line 61: / Line 61: @@
 {{ML-Kopf}}
 [[File:P_ID1035__Mod_Z_2_7_a.png|right|frame|Source signal in the region up to&nbsp; $1\text{ ms}$]]
-'''(1)'''&nbsp; The graph shows that the source signal can take on all values between &nbsp; $–4 \ \rm V$&nbsp; and&nbsp; $+3.667\ \rm  V$&nbsp; annehmen kann.
+'''(1)'''&nbsp; The graph shows that the source signal can take on all values between &nbsp; $–4 \ \rm V$&nbsp; and&nbsp; $+3.667\ \rm  V$.&nbsp;
-*For example, the maximum magnitude occurs at time &nbsp; $t = t_0 =0.75\ \rm  ms$&nbsp;:
+*For example,&nbsp; the maximum magnitude occurs at time &nbsp; $t = t_0 =0.75\ \rm  ms$:
 :$$q(t = t_0)  =  2 \,{\rm V} \cdot \cos(2 \pi \cdot 2\,{\rm kHz} \cdot t_0 ) + 2 \,{\rm V} \cdot \sin(2 \pi \cdot 5\,{\rm kHz} \cdot t_0 )$$
 :$$\Rightarrow \hspace{0.3cm}q(t = 0.75 \,{\rm ms})  =  2 \,{\rm V} \cdot \cos(3 \pi) + 2 \,{\rm V} \cdot \sin(7.5 \pi)= -4 \,{\rm V}\hspace{0.05cm}.$$
+'''(3)'''&nbsp;<u>Answers 2 and 3</u>&nbsp; are correct:
-'''(3)'''&nbsp;<u>Answers 2 and 3</u> are correct:
+*Since the modulation depth is not greater than&nbsp; $m = 1$,&nbsp; the envelope demodulator does not cause distortion either.
 *The main advantage of envelope demodulation is that no frequency and phase synchronization is necessary.
 *A disadvantage is that a significantly higher power must be applied at the transmitter relative to synchronous demodulation.
-*When&nbsp; $m = 1$&nbsp;, this results in three times the transmit power compared to DSB-AM without a carrier.
+*When&nbsp; $m = 1$,&nbsp; this results in three times the transmit power compared to DSB-AM without a carrier.
-[[File:P_ID1036__Mod_Z_2_7_d.png|right|frame|Equivalent low-pass Signal <br>in the complex plane]]
-'''(4)'''&nbsp; <u>Answers 1 and 3</u> are correct:
+[[File:P_ID1036__Mod_Z_2_7_d.png|right|frame|Equivalent low-pass signal <br>in the complex plane]]
-*When&nbsp; $ω_2 = 2 π · 2 \ \rm kHz$&nbsp; and&nbsp; $ω_5 = 2 π · \ \rm 5 kHz$&nbsp;:
+'''(4)'''&nbsp; <u>Answers 1 and 3</u>&nbsp; are correct:&nbsp; When&nbsp; $ω_2 = 2 π · 2 \ \rm kHz$&nbsp; and&nbsp; $ω_5 = 2 π · \ \rm 5 kHz$:
 :$$ r_{\rm TP}(t) = 4 \,{\rm V} \hspace{-0.05cm}+\hspace{-0.05cm} 1 \,{\rm V} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 2}\cdot \hspace{0.03cm}t} \hspace{-0.05cm}+\hspace{-0.05cm} 1 \,{\rm V} \cdot {\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 2}\cdot \hspace{0.03cm}t}
 \hspace{-0.05cm}-\hspace{-0.05cm} \hspace{0.15cm}{\rm j} \cdot1 \,{\rm V} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 5}\cdot \hspace{0.03cm}t} \hspace{-0.05cm}+\hspace{-0.05cm} {\rm j} \cdot1 \,{\rm V} \cdot {\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 5}\cdot \hspace{0.03cm}t} \hspace{0.05cm}. \hspace{0.1cm}$$
-*Thus, in constructing the locus &nbsp; $r_{TP}(t)$&nbsp;, there are exactly five pointers to consider &nbsp; &rArr; &nbsp; answer 1 is correct. The graph shows a snapshot at time &nbsp; $t = 0$.
+Thus,&nbsp; in constructing the locus &nbsp; $r_{TP}(t)$,&nbsp; there are exactly five pointers to consider &nbsp; &rArr; &nbsp; answer 1 is correct.&nbsp; The graph shows a snapshot at time &nbsp; $t = 0$.
-:*The (red) carrier is given by the real pointer of length&nbsp; $4 \ \rm V$ for all time points.  In contrast to the pointer diagram (showing the analytic signal), this does not rotate &nbsp; &rArr; &nbsp;  Answer 2 is false.
+*The (red) carrier is given by the real pointer of length&nbsp; $4 \ \rm V$ for all time points.&nbsp;  In contrast to the pointer diagram&nbsp; (showing the analytic signal),&nbsp; this does not rotate &nbsp; &rArr; &nbsp;  Answer 2 is false.
-:*The third statement is similarly correct: The rotating pointers at negative frequencies rotate in mathematically negative direction (clockwise) in contrast to the two pointers with $f > 0$.
+*The third statement is similarly correct:&nbsp; The rotating pointers at negative frequencies rotate in mathematically negative direction&nbsp; ("clockwise")&nbsp; in contrast to the two pointers with&nbsp; $f > 0$.
-:*The last statement is not true.  The larger the frequency &nbsp; $f$&nbsp;, the faster the associated pointer rotates.
+*The last statement is false.&nbsp;  The larger the frequency &nbsp; $f$,&nbsp; the faster the associated pointer rotates.
-[[File:P_ID1037__Mod_Z_2_7_e.png|right|frame|Locus curve for distortionless envelope demodulation]]
+[[File:P_ID1037__Mod_Z_2_7_e.png|right|frame|Locus curve for <br>distortionless envelope demodulation]]
-'''(5)'''&nbsp; <u>Statements 1 and 2</u> are correct:
+'''(5)'''&nbsp; <u>Statements 1 and 2</u>&nbsp; are correct:
-*In the example considered, the equivalent low-pass signal can be written as:
+*In the example considered,&nbsp; the equivalent low-pass signal can be written as:
 :$$r_{\rm TP}(t) = q(t) + A_{\rm T} \hspace{0.05cm}.$$
-*Thus, it is obvious that &nbsp; $r_{\rm TP}(t)$&nbsp; is always real.&nbsp; Moreover, it follows from subtasks&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)'''&nbsp; that &nbsp; $r_{\rm TP}(t) ≥ 0$.
+*Thus,&nbsp; it is obvious that &nbsp; $r_{\rm TP}(t)$&nbsp; is always real.&nbsp;
 This means:
-:*Here, the locus curve is a horizontal line on the real plane and always lies in the right half-plane.
+#Here, the locus curve is a horizontal line on the real plane and always lies in the right half-plane.
-:*These are the two necessary conditions for an envelope demodulator to recover the message signal without distortion.
+#These are the two necessary conditions for an envelope demodulator to recover the signal without distortion.
-:*If one of these conditions is not satisfied, &nbsp; <u>'''nonlinear'''</u>&nbsp; distortions arise, not linear ones &nbsp; &rArr; &nbsp; Answer 3 is wrong.
+#If one of these conditions is not satisfied, &nbsp; <u>'''nonlinear'''</u>&nbsp; distortions arise,&nbsp; not linear ones &nbsp; &rArr; &nbsp; Answer 3 is wrong.
 {{ML-Fuß}}

@@ Line 41: / Line 41: @@
 |type="[]"}
 - With the envelope demodulator,&nbsp; distortion-free demodulation is not possible in the example considered.
-+ One can do without frequency and phase synchronization.
++ One can do demodulation without frequency and phase synchronization.
 + A smaller transmission power would be needed using a synchronous demodulator.

@@ Line 6: / Line 6: @@
 Assume a source signal
 :$$ q(t)  =  2 \,{\rm V} \cdot \cos(2 \pi \cdot 2\,{\rm kHz} \cdot t )   +  2 \,{\rm V} \cdot \sin(2 \pi \cdot 5\,{\rm kHz} \cdot t )\hspace{0.05cm}.$$
-This is modulated according to the modulation method "DSB-AM with carrier" and transmitted through an ideal channel. The influence of noise can be disregarded.
+This is modulated according to the modulation method&nbsp; "DSB-AM with carrier"&nbsp; and transmitted through an ideal channel.&nbsp; The influence of noise can be disregarded.
-The graph shows the spectrum&nbsp;$R_{\rm TP}(f)$&nbsp; of the received signal in the equivalent low-pass region, which is composed of Dirac lines at &nbsp; $f = 0$&nbsp; (originating from the carrier),&nbsp; at &nbsp;$±2\ \rm  kHz$&nbsp; (originating from the cosine component)&nbsp; and at &nbsp;$±5\ \rm   kHz$&nbsp; (originating from the sine component)&nbsp;.
+The graph shows the spectrum&nbsp; $R_{\rm TP}(f)$&nbsp; of the received signal in the equivalent low-pass region,&nbsp; which is composed of Dirac lines at &nbsp; $f = 0$&nbsp; (originating from the carrier),&nbsp; at &nbsp;$±2\ \rm  kHz$&nbsp; (originating from the cosine component)&nbsp; and at &nbsp;$±5\ \rm   kHz$&nbsp; (originating from the sine component).&nbsp;
@@ Line 19: / Line 19: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Envelope_Demodulation|Envelope Demodulation]].
-*Particular reference is made to the page&nbsp;[[Modulation_Methods/Envelope_Demodulation#Description_using_the_equivalent_low-pass_signal|Description using the equivalent low-pass signal]].
+*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 33: / Line 29: @@
 <quiz display=simple>
-{Estimate the maximum magnitude &nbsp;$q_{\rm max} = {\rm Max} |q(t)|$&nbsp; of the source signal S.
+{Estimate the maximum magnitude &nbsp;$q_{\rm max} = {\rm Max} |q(t)|$&nbsp; of the source signal.
 |type="{}"}
 $q_{\rm max} \ = \ $ { 4 3%  } $\ \rm V$
@@ Line 42: / Line 38: @@
 $m \ = \ $ { 1 3% }
-{Which of these are arguments for or against using an envelope demodulator?  Assume the alternative would be a synchronous demodulator.
+{Which of these are arguments for or against using an envelope demodulator?&nbsp;  Assume the alternative would be a synchronous demodulator.
 |type="[]"}
-- With the envelope demodulator, distortion-free demodulation is not possible in the example considered.
+- With the envelope demodulator,&nbsp; distortion-free demodulation is not possible in the example considered.
 + One can do without frequency and phase synchronization.
-+ A smaller transmit power would be needed using a synchronous demodulator.
++ A smaller transmission power would be needed using a synchronous demodulator.
-{Calculate the equivalent low-pass signal &nbsp;$r_{\rm TP}(t)$  &nbsp; ⇒ &nbsp; "locus curve", &nbsp;using the Fourier retransform of &nbsp;$R_{\rm TP}(f)$&nbsp;. Which statements are true?
+{Calculate the equivalent low-pass signal &nbsp;$r_{\rm TP}(t)$  &nbsp; ⇒ &nbsp; "locus curve", &nbsp;using the Fourier retransform of &nbsp;$R_{\rm TP}(f)$.&nbsp; Which statements are true?
 |type="[]"}
 + The locus curve &nbsp;$r_{\rm TP}(t)$&nbsp; is composed of five pointers.
@@ Line 55: / Line 51: @@
 - The pointer for &nbsp;$2 \ \rm  kHz$&nbsp; rotates twice as fast as the one for &nbsp;$5 \ \rm  kHz$.
-{Which statements can be made based on the locus curve? Answer the following questions by considering the application of envelope demodulation.
+{Which statements can be made based on the locus curve?&nbsp; Answer the following questions by considering the application of envelope demodulation.
 |type="[]"}
 + A distortionless demodulation is only possible when &nbsp;$r_{\rm TP}(t)$&nbsp; is real at all times.
 + A distortionless demodulation is only possible when &nbsp;$r_{\rm TP}(t)$&nbsp; does not become negative at any point in time.
-- If the first two conditions mentioned are not met, linear distortions will occur.
+- If the first two conditions mentioned are not met,&nbsp; linear distortions will occur.
 </quiz>