Aufgaben:Exercise 2.2: Modulation Depth - Revision history

Hwang at 14:17, 18 January 2023

2023-01-18T14:17:44Z

Reed at 15:32, 24 March 2022

2022-03-24T15:32:52Z

Reed at 15:27, 24 March 2022

2022-03-24T15:27:57Z

Guenter at 15:13, 29 November 2021

2021-11-29T15:13:27Z

Guenter at 14:58, 29 November 2021

2021-11-29T14:58:49Z

Guenter at 14:46, 29 November 2021

2021-11-29T14:46:50Z

Guenter at 14:46, 29 November 2021

2021-11-29T14:46:26Z

Reed at 12:13, 24 November 2021

2021-11-24T12:13:01Z

Reed: /* Solution */

2021-11-24T12:07:36Z

Solution

Reed at 11:48, 24 November 2021

2021-11-24T11:48:19Z

@@ Line 89: / Line 89: @@
 [[File:P_ID990__Mod_A_2_2_c.png|right|frame|Analytical signal's spectrum ]]
-'''(3)'''&nbsp; The spectrum&nbsp; $S_+(f)$&nbsp; is composed of three Dirac lines for each modulation depth &nbsp; $m$&nbsp; with the following weights:
+'''(3)'''&nbsp; The spectrum&nbsp; $S_+(f)$&nbsp; is composed of three Dirac delta lines for each modulation depth &nbsp; $m$&nbsp; with the following weights:
 *$A_{\rm T}$&nbsp; $($at&nbsp; $f = f_{\rm T})$,
 * $m/2 · A_{\rm T}$&nbsp; $($at&nbsp; $f = f_{\rm T} ± f_{\rm N})$.

← Older revision		Revision as of 15:32, 24 March 2022
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−	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Zweiseitenband~~-~~Amplitudenmodulation~~	+	{{quiz-Header\|Buchseite=Modulation_Methods/Double-Sideband_Amplitude_Modulation
	}}		}}

@@ Line 100: / Line 100: @@
-'''(4)'''&nbsp; The power (root mean square) of a harmonic oscillation with amplitude &nbsp; $A_{\rm T} = 1 \ \rm V$&nbsp; referenced to the &nbsp; $1 \ Ω$&nbsp; resistor is:
+'''(4)'''&nbsp; The power (mean square) of a harmonic oscillation with amplitude &nbsp; $A_{\rm T} = 1 \ \rm V$&nbsp; referenced to the &nbsp; $1 \ Ω$&nbsp; resistor is:
 :$$P_{\rm T} ={A_{\rm T}^2}/{2} = 0.5\,{\rm V}^2 \hspace{0.05cm}.$$
 *In the same way,&nbsp; for the powers of the lower and the upper sideband we obtain:

@@ Line 77: / Line 77: @@
 :$$m = \frac{A_{\rm max} - A_{\rm min}}{A_{\rm max} + A_{\rm min}}\hspace{0.05cm}.$$
 *With the given numerical values,&nbsp; one obtains:
-$$ m_1  =  \frac{2\,{\rm V} - 0.667\,{\rm V}}{2\,{\rm V} + 0.667\,{\rm V}} \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}, \hspace{0.5cm} m_2 = \frac{2\,{\rm V} - 0\,{\rm V}}{2\,{\rm V} + 0\,{\rm V}} \hspace{0.15cm}\underline {= 1.0}\hspace{0.05cm}, \hspace{0.5cm}
+:$$ m_1  =  \frac{2\,{\rm V} - 0.667\,{\rm V}}{2\,{\rm V} + 0.667\,{\rm V}} \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}, \hspace{0.5cm} m_2 = \frac{2\,{\rm V} - 0\,{\rm V}}{2\,{\rm V} + 0\,{\rm V}} \hspace{0.15cm}\underline {= 1.0}\hspace{0.05cm}, \hspace{0.5cm}
 m_3 = \frac{2\,{\rm V} -(-1\,{\rm V})}{2\,{\rm V} + (-1\,{\rm V})} \hspace{0.15cm}\underline{=3.0}\hspace{0.05cm}.$$

@@ Line 74: / Line 74: @@
 :$$A_{\rm N}  =  (A_{\rm max} - A_{\rm min})/2,\hspace{0.3cm}
 A_{\rm T}  = (A_{\rm max} + A_{\rm min})/2\hspace{0.05cm}.$$
-*Thus, the modulation depth is
+*Thus,&nbsp; the modulation depth is
 :$$m = \frac{A_{\rm max} - A_{\rm min}}{A_{\rm max} + A_{\rm min}}\hspace{0.05cm}.$$
-*With the given numerical values, one obtains:
+*With the given numerical values,&nbsp; one obtains:
 $$ m_1  =  \frac{2\,{\rm V} - 0.667\,{\rm V}}{2\,{\rm V} + 0.667\,{\rm V}} \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}, \hspace{0.5cm} m_2 = \frac{2\,{\rm V} - 0\,{\rm V}}{2\,{\rm V} + 0\,{\rm V}} \hspace{0.15cm}\underline {= 1.0}\hspace{0.05cm}, \hspace{0.5cm}
 m_3 = \frac{2\,{\rm V} -(-1\,{\rm V})}{2\,{\rm V} + (-1\,{\rm V})} \hspace{0.15cm}\underline{=3.0}\hspace{0.05cm}.$$
@@ Line 82: / Line 82: @@
-'''(2)'''&nbsp; <u>Answers 1 and 3</u> are correct:
+'''(2)'''&nbsp; <u>Answers 1 and 3</u>&nbsp; are correct:
-*In this case,&nbsp; $A_{\rm T} = 0$, which means it is indeed "DSB-AM without a carrier".
+*In this case,&nbsp; $A_{\rm T} = 0$,&nbsp; which means it is indeed&nbsp; "DSB-AM without carrier".
 *The modulation depth&nbsp; $m = A_{\rm N}/A_{\rm T}$&nbsp; is infinitely large.
+[[File:P_ID990__Mod_A_2_2_c.png|right|frame|Analytical signal's spectrum ]]
 '''(3)'''&nbsp; The spectrum&nbsp; $S_+(f)$&nbsp; is composed of three Dirac lines for each modulation depth &nbsp; $m$&nbsp; with the following weights:
 *$A_{\rm T}$&nbsp; $($at&nbsp; $f = f_{\rm T})$,
@@ Line 95: / Line 94: @@
-For $m = 1$, the weights are obtained according to the graph:
+For&nbsp; $m = 1$,&nbsp;  the weights are obtained according to the graph:
 *$S_+(f_{\rm T}) = 1\ \rm V$,
 *$S_+(f_{\rm T} ± f_{\rm T}) = 0.5\ \rm V$.
@@ Line 104: / Line 102: @@
 '''(4)'''&nbsp; The power (root mean square) of a harmonic oscillation with amplitude &nbsp; $A_{\rm T} = 1 \ \rm V$&nbsp; referenced to the &nbsp; $1 \ Ω$&nbsp; resistor is:
 :$$P_{\rm T} ={A_{\rm T}^2}/{2} = 0.5\,{\rm V}^2 \hspace{0.05cm}.$$
-*In the same way, for the powers of the lower and the upper sideband we obtain:
+*In the same way,&nbsp; for the powers of the lower and the upper sideband we obtain:
-:$$P_{\rm USB} = P_{\rm OSB} =({A_{\rm N}}/{2})^2/2 = 0.125\,{\rm V}^2 \hspace{0.05cm}.$$
+:$$P_{\rm LSB} = P_{\rm USB} =({A_{\rm N}}/{2})^2/2 = 0.125\,{\rm V}^2 \hspace{0.05cm}.$$
-*Thus, for&nbsp; $m=1$, the ratio we are looking for is:
+*Thus,&nbsp; for&nbsp; $m=1$,&nbsp; the ratio we are looking for is:
 :$${P_{\rm T}}/{P_{\rm S}}= \frac{P_{\rm T}}{P_{\rm USB} + P_{\rm T}+ P_{\rm OSB}}= \frac{0.5\,{\rm V}^2}{0.125\,{\rm V}^2 + 0.5\,{\rm V}^2+ 0.125\,{\rm V}^2}= 2/3\hspace{0.15cm}\underline { = 0.667}\hspace{0.05cm}.$$
-'''(5)'''&nbsp; Using the Dirac weights&nbsp; $m/2 · A_{\rm T}$&nbsp; of the two sidebands corresponding to subtask&nbsp; '''(3)'''&nbsp;, we get:
+'''(5)'''&nbsp; Using the Dirac weights&nbsp; $m/2 · A_{\rm T}$&nbsp; of the two sidebands corresponding to subtask&nbsp; '''(3)''',&nbsp; we get:
 :$${P_{\rm T}}/{P_{\rm S}}= \frac{A_{\rm T}^2/2}{A_{\rm T}^2/2 + 2 \cdot (m/2)^2 \cdot A_{\rm T}^2/2}= \frac{2}{2 + m^2}\hspace{0.05cm}.$$
-*This leads to the numerical values &nbsp; $8/9 = 0.889$&nbsp; $($for&nbsp; $m = 0.5)$, &nbsp; &nbsp; $2/11 = 0.182$&nbsp; $($for&nbsp; $m = 3)$&nbsp; und  &nbsp; &nbsp; $0$&nbsp;  $($for&nbsp; $m \to ∞$).
+*This leads to the numerical values &nbsp; $8/9 = 0.889$&nbsp; $($for&nbsp; $m = 0.5)$, &nbsp; &nbsp; $2/11 = 0.182$&nbsp; $($for&nbsp; $m = 3)$,  &nbsp; &nbsp; $0$&nbsp;  $($for&nbsp; $m \to ∞$).
-'''(6)'''&nbsp; <u>Both statements</u> are true:
+'''(6)'''&nbsp; <u>Both statements</u>&nbsp; are true:
-*The addition of the carrier only makes sense in order to use the simpler envelope demodulator.  This is only possible for&nbsp; $m \le 1$.
+*The addition of the carrier only makes sense in order to use the simpler envelope demodulator.&nbsp;  This is only possible for&nbsp; $m \le 1$.
-*However, should the modulation depth be&nbsp; $m > 1$&nbsp; and the use of a synchronous demodulator therefore be required, the carrier should be (almost) completely omitted for energy reasons.
+*However,&nbsp; should the modulation depth be&nbsp; $m > 1$&nbsp; and the use of a synchronous demodulator therefore be required,&nbsp; the carrier should be (almost) completely omitted for energy reasons.
-*Similarly due to energy concerns, if an envelope demodulator is used, the largest possible modulation depth&nbsp; $m < 1$&nbsp;  &nbsp; &rArr;  &nbsp; $m \to 1$&nbsp; should be aimed for.
+*Similarly due to energy concerns,&nbsp; if an envelope demodulator is used,&nbsp; the largest possible modulation depth&nbsp; $m < 1$&nbsp;  &nbsp; &rArr;  &nbsp; $m \to 1$&nbsp; should be aimed for.
-*However, a small residual carrier can facilitate carrier recovery, which is needed in a synchronous demodulator for frequency and phase synchronization.  Thus, the second statement is only conditionally correct.
+*However,&nbsp; a small residual carrier can facilitate carrier recovery,&nbsp; which is needed in a synchronous demodulator for frequency and phase synchronization.&nbsp;  Thus,&nbsp; the second statement is only conditionally correct.
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:P_ID989__Mod_A_2_2.png|right|frame|Definition of modulation depth for DSB–AM]]
-The graph shows&nbsp; 2DSB amplitude-modulated signals" &nbsp; $s_1(t)$&nbsp; to &nbsp;$s_4(t)$&nbsp; with differing modulation depth &nbsp;$m$.&nbsp; Let the message signal &nbsp;$q(t)$&nbsp; and the carrier signal &nbsp;$z(t)$&nbsp; each be cosine:
+The graph shows&nbsp; "DSB amplitude-modulated signals" &nbsp; $s_1(t)$&nbsp; to &nbsp;$s_4(t)$&nbsp; with differing modulation depth &nbsp;$m$.&nbsp; Let the message signal &nbsp;$q(t)$&nbsp; and the carrier signal &nbsp;$z(t)$&nbsp; each be cosine:
 :$$q(t) = A_{\rm N} \cdot \cos (2 \pi f_{\rm N} t),\hspace{0.2cm} f_{\rm N} = 4\,{\rm kHz}\hspace{0.05cm},$$
 :$$ z(t) = \hspace{0.2cm}1 \hspace{0.15cm} \cdot \cos (2 \pi f_{\rm T} t),\hspace{0.2cm} f_{\rm T} = 50\,{\rm kHz}\hspace{0.05cm}.$$

@@ Line 4: / Line 4: @@
 [[File:P_ID989__Mod_A_2_2.png|right|frame|Definition of modulation depth for DSB–AM]]
-The graph shows DSB amplitude-modulated signals &nbsp;$s_1(t)$&nbsp; to &nbsp;$s_4(t)$&nbsp; with differing modulation depth &nbsp;$m$. Let message signal &nbsp;$q(t)$&nbsp; and carrier signal &nbsp;$z(t)$&nbsp; each be cosine:
+The graph shows&nbsp; 2DSB amplitude-modulated signals" &nbsp; $s_1(t)$&nbsp; to &nbsp;$s_4(t)$&nbsp; with differing modulation depth &nbsp;$m$.&nbsp; Let the message signal &nbsp;$q(t)$&nbsp; and the carrier signal &nbsp;$z(t)$&nbsp; each be cosine:
 :$$q(t) = A_{\rm N} \cdot \cos (2 \pi f_{\rm N} t),\hspace{0.2cm} f_{\rm N} = 4\,{\rm kHz}\hspace{0.05cm},$$
 :$$ z(t) = \hspace{0.2cm}1 \hspace{0.15cm} \cdot \cos (2 \pi f_{\rm T} t),\hspace{0.2cm} f_{\rm T} = 50\,{\rm kHz}\hspace{0.05cm}.$$
-The modulated signal (transmitted signal) with the DC component added in the modulator is &nbsp;$A_{\rm T}$:
+The modulated signal&nbsp; (transmitted signal)&nbsp; with the DC component added in the modulator is &nbsp;$A_{\rm T}$:
 :$$s(t ) = A(t) \cdot z(t), \hspace{0.2cm} A(t) = q(t) + A_{\rm T}\hspace{0.05cm}.$$
-In the graphs, the chosen normalization was:
+In the graphs,&nbsp; the chosen normalization was:
 :$$A_{\rm T}+ A_{\rm N} = 2\,{\rm V}\hspace{0.05cm}.$$
-*If the modulation depth is &nbsp;$m ≤ 1$, then &nbsp;$A(t)= q(t) + A_{\rm T}$&nbsp;  is equal to the envelope &nbsp;$a(t)$.
+*If the modulation depth is &nbsp;$m ≤ 1$,&nbsp; then &nbsp;$A(t)= q(t) + A_{\rm T}$&nbsp;  is equal to the envelope &nbsp;$a(t)$.
-*In contrast, for a modulation depth &nbsp;$m > 1$:
+*In contrast,&nbsp; for a modulation depth &nbsp;$m > 1$:
 :$$a(t ) = |A(t)|\hspace{0.05cm}.$$
+*The cosine curve &nbsp;$A(t)$&nbsp; varies between &nbsp;$A_{\rm max}$&nbsp; and &nbsp;$A_{\rm min}$; because of normalization, &nbsp; &rArr; &nbsp; $A_{\rm max} = 2 \ \rm  V$.
-*The cosine curve &nbsp;$A(t)$&nbsp; varies between &nbsp;$A_{\rm max}$&nbsp; and &nbsp;$A_{\rm min}$; because of normalization, &nbsp;$A_{\rm max} = 2 \ \rm  V$ is always the case.
+*The minimum values of &nbsp;$A(t)$&nbsp; occur at half the period of the source signal&nbsp; $($i.e., for&nbsp; $t = 125 \ \rm &micro; s)$:
 :$$A_{\rm min} = q(T_0/2)+ A_{\rm T} = A_{\rm T}-A_{\rm N}.$$
 *The numerical values are given in the graph.
@@ Line 24: / Line 23: @@
 Hints:
 *This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Double-Sideband_Amplitude_Modulation|Double-Sideband Amplitude Modulation]].
 *Particular reference is made to the page&nbsp;  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Double-Sideband_Amplitude_Modulation_with_carrier|DSB-AM with carrier]].
@@ Line 39: / Line 33: @@
 <quiz display=simple>
-{Determine the modulation depth for each of the signals &nbsp;$s_1(t)$, &nbsp;$s_2(t)$, &nbsp;$s_3(t)$&nbsp;.
+{Determine the modulation depth for each of the signals &nbsp;$s_1(t)$, &nbsp;$s_2(t)$, &nbsp;$s_3(t)$.
 |type="{}"}
 $m_1 \ = \ $ { 0.5 3% }
@@ Line 45: / Line 39: @@
 $m_3 \ = \ $ { 3 3% }
-{Which statements are true for the signal &nbsp;$s_4(t)$&nbsp;?
+{Which statements are true for the signal &nbsp;$s_4(t)$?
 |type="[]"}
-+ This is a case of "DSB–AM without a carrier".
++ This is a case of&nbsp; "DSB–AM without carrier".
 - The modulation depth is &nbsp;$m = 0$.
 + The modulation depth &nbsp;$m$&nbsp; is infinite.
-{Let &nbsp;$A_{\rm T} = A_{\rm N} = 1\ \rm V$, so &nbsp;$m = 1$.&nbsp; What is the spectrum &nbsp;$S_+(f)$&nbsp; of the analytical signal? <br>Which Dirac weights occur at &nbsp;$f_{\rm T}$&nbsp; as well as at &nbsp;$f_{\rm T}± f_{\rm N}$&nbsp;?
+{Let &nbsp;$A_{\rm T} = A_{\rm N} = 1\ \rm V$ &nbsp; &rArr; &nbsp; $m = 1$.&nbsp; What is the spectrum &nbsp;$S_+(f)$&nbsp; of the analytical signal?&nbsp; Which Dirac weights occur at &nbsp;$f_{\rm T}$&nbsp; as well as at &nbsp;$f_{\rm T}± f_{\rm N}$?
 |type="{}"}
 $S_+(f_{\rm T})  \ = \ $ { 1 3% } $\ \text{V}$
 $S_+(f_{\rm T} ± f_{\rm N})  \ = \ $ { 0.5 3% }  $\ \text{V}$
-{Now let &nbsp;$m = 1$.&nbsp; Which fraction &nbsp;$P_{\rm T}/P_{\rm S}$&nbsp; of the total transmission power &nbsp;$P_{\rm S}$&nbsp; is due to the carrier alone, and thus cannot be used for demodulation??
+{Now let &nbsp;$m = 1$.&nbsp; Which fraction &nbsp;$P_{\rm T}/P_{\rm S}$&nbsp; of the total transmission power &nbsp;$P_{\rm S}$&nbsp; is due to the carrier alone,&nbsp; and thus cannot be used for demodulation??
 |type="{}"}
 $P_{\rm T}/P_{\rm S}  \ = \ $ { 0.667 3% }
-{Generalize the result from &nbsp; '''(4)'''&nbsp; for an arbitrary modulation depth&nbsp;$m$. <br>What are the power ratios for  &nbsp;$m = 0.5$, &nbsp;$m = 3$&nbsp; and &nbsp;$m → ∞$ ?
+{Generalize the result from &nbsp; '''(4)'''&nbsp; for an arbitrary modulation depth&nbsp; $m$.&nbsp; What are the power ratios for  &nbsp;$m = 0.5$, &nbsp;$m = 3$&nbsp; and &nbsp;$m → ∞$?
 |type="{}"}
 $m = 0.5\text{:}\hspace{0.3cm} P_{\rm T}/P_{\rm S}  \ = \ $ { 0.889 3% }
@@ Line 67: / Line 61: @@
 $m → ∞ \text{:}\hspace{0.3cm} P_{\rm T}/P_{\rm S}  \ = \  $ { 0. }
-{Based on the calculations so far, which of the following assessments seem reasonable to you?
+{Based on the calculations so far,&nbsp; which of the following statements seem reasonable to you?
 |type="[]"}
-+ $m ≈ 1$&nbsp; s more favorable than a small&nbsp; $m$ for energy reasons.
++ $m ≈ 1$&nbsp; is more favorable than a small&nbsp; $m$&nbsp; for energy reasons.
 + The carrier is only useful for envelope demodulation.
 </quiz>