Aufgaben:Exercise 2.10: SSB-AM with Channel Distortions - Revision history

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Reed: /* Solution */

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Solution

← Older revision		Revision as of 15:55, 31 March 2022
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−	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Einseitenbandmodulation~~	+	{{quiz-Header\|Buchseite=Modulation_Methods/Single-Sideband_Modulation
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 '''(6)'''&nbsp; The&nbsp; <u>first and last answers</u>&nbsp; are correct:
-*Also for&nbsp; "single-sideband AM"&nbsp; holds:&nbsp; Attenuation distortion on the channel lead only to attenuation distortions with respect to&nbsp; $v(t)$.
+*Also for&nbsp; "single-sideband AM"&nbsp;:&nbsp; Attenuation distortions on the channel lead only to attenuation distortions with respect to&nbsp; $v(t)$.
 *Phase distortions are only present for a demodulator with a phase offset in the case of&nbsp; "single-sideband AM".
 *For&nbsp; "double-sideband AM",&nbsp; such a phase offset would not result in any distortions,&nbsp; but only in frequency-independent attenuation.

@@ Line 70: / Line 70: @@
 $τ_4 \hspace{0.25cm} = \ $ { 20.8 3% } $\ \rm &micro;  s$
-{Which of these statements are true given your results?&nbsp; Here,&nbsp; "channel distortions"&nbsp; should always be understood as a kind of attenuation distortions.
+{Which of these statements are true given your results?&nbsp; Here,&nbsp; "channel distortions"&nbsp; should always be understood as a kind of attenuation distortion.
 |type="[]"}
 + In&nbsp; "double-sideband AM",&nbsp; each channel distortion leads to attenuation distortions.

@@ Line 19: / Line 19: @@
 In each case,&nbsp;  a synchronous demodulator is used to first convert the receiver-side carrier signal
 :$$ z_{\rm E} (t) = \left\{ \begin{array}{c} 2 \cdot z(t) \\ 4 \cdot z(t) \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\rm DSB} \hspace{0.05cm}, \\ {\rm USB, LSB} \hspace{0.05cm} \\ \end{array}$$
-by multiplication and then completely suppresses the components by twice the carrier frequency.&nbsp;  With an ideal channel &nbsp;$H_{\rm K}(f) = 1$&nbsp;, &nbsp;$v(t) = q(t)$&nbsp; would hold in all cases.
+by multiplication and then completely suppresses the components at twice the carrier frequency.&nbsp;  With an ideal channel &nbsp;$H_{\rm K}(f) = 1$&nbsp;, &nbsp;$v(t) = q(t)$&nbsp; would hold in all cases.
 The Gaussian channel considered here is given by the following auxiliary values:

@@ Line 83: / Line 83: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; For DSB–AM, the following attenuation factors are to be taken into account:
+'''(1)'''&nbsp; For DSB–AM,&nbsp; the following attenuation factors are to be taken into account:
 :$$\alpha_2  =  {1}/{2} \cdot \left[ H_{\rm K}(f = 48\,{\rm kHz}) + H_{\rm K}(f = 52\,{\rm kHz})\right] = 0.981,$$
 :$$\alpha_4  =  {1}{2} \cdot \left[ H_{\rm K}(f = 46\,{\rm kHz}) + H_{\rm K}(f = 54\,{\rm kHz})\right] = 0.861\hspace{0.05cm}.$$
-*Thus, we get the amplitudes &nbsp; $A_2\hspace{0.15cm}\underline{ = 1.882 \ \rm V}$&nbsp; and&nbsp; $A_4\hspace{0.15cm}\underline{ = 1.722 \ \rm V}$.
+*Thus,&nbsp; we get the amplitudes &nbsp; $A_2\hspace{0.15cm}\underline{ = 1.882 \ \rm V}$&nbsp; and&nbsp; $A_4\hspace{0.15cm}\underline{ = 1.722 \ \rm V}$.
-'''(2)'''&nbsp; For DSB, a phase offset between the carrier frequencies at the transmitter and the receiver, respectively, leads to one and the same attenuation for all frequencies:
+'''(2)'''&nbsp; For DSB-AM,&nbsp; a phase offset between the carrier frequencies at transmitter and receiver, resp.,&nbsp; leads to one and the same attenuation for all frequencies:
 :$$A_2  =  \cos (30^\circ) \cdot 1.882\,{\rm V} \hspace{0.15cm}\underline {= 1.630\,{\rm V}},$$
 :$$A_4  =  \cos (30^\circ) \cdot 1.722\,{\rm V} = 1.491\,{\rm V}\hspace{0.05cm}.$$
-*The transmit times are &nbsp; $τ_2\hspace{0.15cm}\underline {= 0}$&nbsp; and&nbsp; $τ_4 = 0$.
+*The delay times are &nbsp; $τ_2\hspace{0.15cm}\underline {= 0}$&nbsp; and&nbsp; $τ_4 = 0$.
-'''(3)'''&nbsp; For USB–AM, the attenuation factor &nbsp; $α_2$&nbsp; is only determined by &nbsp; $H_{\rm K}(f = 52\ \rm  kHz)$&nbsp;.
+'''(3)'''&nbsp; For USB–AM,&nbsp; the attenuation factor &nbsp; $α_2$&nbsp; is only determined by &nbsp; $H_{\rm K}(f = 52\ \rm  kHz)$.
-*Since the principal USB amplitude loss by a factor of &nbsp; $2$&nbsp; is compensated for by a larger carrier amplitude, the following holds:
+*Since the principal USB amplitude loss by a factor of &nbsp; $2$&nbsp; is compensated for by a larger carrier amplitude,&nbsp; the following holds:
 :$$A_2  =  0.882 \cdot 2\,{\rm V}\hspace{0.15cm}\underline {= 1.764\,{\rm V}},$$
 :$$A_4  =  0.754 \cdot 2\,{\rm V}\hspace{0.15cm}\underline {= 1.508\,{\rm V}} \hspace{0.05cm}.$$
@@ Line 104: / Line 104: @@
-'''(4)'''&nbsp; Analogous to the solution in subtask&nbsp; '''(3)'''&nbsp;, here we obtain:
+'''(4)'''&nbsp; Analogous to the solution in subtask&nbsp; '''(3)''',&nbsp; we obtain here:
 :$$ A_2  =  H_{\rm K}(f = 48\,{\rm kHz}) \cdot 2\,{\rm V}\hspace{0.15cm}\underline {= 2\,{\rm V}},$$
 :$$A_4  =  H_{\rm K}(f = 46\,{\rm kHz}) \cdot 2\,{\rm V}\hspace{0.15cm}\underline {= 1.936\,{\rm V}} \hspace{0.05cm}.$$
@@ Line 110: / Line 110: @@
-'''(5)'''&nbsp; For LSB–AM, the received signal is:
+'''(5)'''&nbsp; For LSB–AM,&nbsp; the received signal is:
 :$$r(t) = 1\,{\rm V} \cdot \cos( \omega_{\rm 48} \cdot t) + 0.968\,{\rm V} \cdot \cos( \omega_{\rm 46} \cdot t)\hspace{0.05cm}.$$
-*By multiplication with the receiver-side carrier signal &nbsp; $z_{\rm E}(t) = 4 \cdot \cos( \omega_{\rm 50} \cdot t - \Delta \phi_{\rm T})$&nbsp;, applying the trigonometric addition theorem gives:
+*By multiplication with the receiver-side carrier signal &nbsp; $z_{\rm E}(t) = 4 \cdot \cos( \omega_{\rm 50} \cdot t - \Delta \phi_{\rm T})$,&nbsp; applying the trigonometric addition theorem gives:
 :$$v(t) = r(t) \cdot z_{\rm E}(t) =  \hspace{0.15cm}\underline { 2.000\,{\rm V}} \cdot \cos( \omega_{\rm 2} \cdot t - \Delta \phi_{\rm T})+\hspace{0.15cm}\underline { 1.936\,{\rm V}} \cdot \cos( \omega_{\rm 4} \cdot t - \Delta \phi_{\rm T})
   +  {\rm components \hspace{0.15cm}around\hspace{0.15cm}} 2f_{\rm T}\hspace{0.05cm}$$
 :$$ \Rightarrow \hspace{0.3cm} A_2 \hspace{0.15cm}\underline {= 2\,{\rm V}},\hspace{0.5cm} A_4 \hspace{0.15cm}\underline {= 1.936\,{\rm V}}.$$
-*Considering the downstream lowpass filter, this can also be written as:
+*Considering the downstream lowpass filter,&nbsp; this can also be written as:
 :$$ v(t) = A_2 \cdot \cos( \omega_{\rm 2} \cdot (t - \tau_2))+ A_4 \cdot \cos( \omega_{\rm 4} \cdot (t - \tau_4))\hspace{0.05cm}.$$
-*The amplitudes are unchanged compared to subtask&nbsp; '''(4)'''&nbsp;.&nbsp; For the transmit times when &nbsp; $Δϕ_{\rm T} = π/6$, we get:
+*The amplitudes are unchanged compared to subtask&nbsp; '''(4)'''.&nbsp; For the delay times when &nbsp; $Δϕ_{\rm T} = π/6$,&nbsp; we get:
 :$$ \tau_2  =  \frac {\Delta \phi_{\rm T}}{2 \pi \cdot f_2} = \frac {\pi /6}{2 \pi \cdot 2\,{\rm kHz}} \hspace{0.15cm}\underline {\approx 41.6\,{\rm &micro; s}},\hspace{0.5cm} \tau_4  =  \frac {\Delta \phi_{\rm T}}{2 \pi \cdot f_4}= \frac {\tau_2}{2}\hspace{0.15cm}\underline {\approx 20.8\,{\rm &micro; s}} \hspace{0.05cm}.$$
-'''(6)'''&nbsp; The <u>first and last answers</u> are correct:
+'''(6)'''&nbsp; The&nbsp; <u>first and last answers</u>&nbsp; are correct:
-*Attenuation distortion on the channel exclusively lead to attenuation distortions with respect to &nbsp; $v(t)$, also for SSB.
+*Also for&nbsp; "single-sideband AM"&nbsp; holds:&nbsp; Attenuation distortion on the channel lead only to attenuation distortions with respect to&nbsp; $v(t)$.
-*Phase distortions are only present for a demodulator with a phase offset in the case of single-sideband modulation.
+*Phase distortions are only present for a demodulator with a phase offset in the case of&nbsp; "single-sideband AM".
-*For DSB–AM, such a phase offset would not result in any distortions, but only in frequency-independent damping.
+*For&nbsp; "double-sideband AM",&nbsp; such a phase offset would not result in any distortions,&nbsp; but only in frequency-independent attenuation.
-*Phase distortions with respect to &nbsp; $v(t)$&nbsp; can also arise in DSB–AM and SSB–AM, if these already occur on the channel.
+*Phase distortions with respect to&nbsp; $v(t)$&nbsp; can also arise in&nbsp; "DSB–AM"&nbsp; and&nbsp; "SSB–AM",&nbsp; if these already occur on the channel.

@@ Line 24: / Line 24: @@
 :$$ H_{\rm K}(f = 46\ {\rm kHz}) = 0.968,\hspace{0.3cm}H_{\rm K}(f = 48\ {\rm kHz}) = 1.000,\hspace{0.3cm}
   H_{\rm K}(f = 52\ {\rm kHz}) = 0.882,\hspace{0.3cm}H_{\rm K}(f = 54\ {\rm kHz}) = 0.754\hspace{0.05cm}.$$
-In each case, write the sink signal in the form
+In each case,&nbsp; write the sink signal in the form
 :$$v(t) = A_2 \cdot \cos(2 \pi f_2 \cdot (t - \tau_2)) + A_4 \cdot \cos(2 \pi f_4 \cdot (t - \tau_4))\hspace{0.05cm}.$$
 All calculations are to be carried out for both a perfect phase synchronization &nbsp;$(Δϕ_{\rm T} = 0)$&nbsp; as well as for a phase offset of &nbsp;$Δϕ_{\rm T} = 30^\circ$.&nbsp; This is present,&nbsp; for example,&nbsp; if the transmit-side carrier signal is cosine-shaped and the receiver-side carrier is:
@@ Line 32: / Line 32: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Single-Sideband_Modulation|Single-Sideband Modulation]].
+*Reference will also be made to the chapter&nbsp;   [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]].
@@ Line 47: / Line 43: @@
 <quiz display=simple>
-{Calculate the amplitudes for &nbsp; <u>DSB–AM</u>&nbsp; and&nbsp; <u>perfect synchronization</u> &nbsp;$(Δϕ_{\rm T} = 0)$.
+{Calculate the amplitudes for &nbsp; <u>double-sideband AM</u>&nbsp; and&nbsp; <u>perfect synchronization</u> &nbsp;$(Δϕ_{\rm T} = 0)$.
 |type="{}"}
 $A_2 \ = \ $ { 1.882 3% } $\ \rm V$
 $A_4 \ = \ $ { 1.722 3% } $\ \rm V$
-{What are the values for &nbsp;$A_2$&nbsp; and &nbsp;$τ_2$&nbsp; for&nbsp; <u>DSB–AM</u>&nbsp; and a&nbsp; <u>phase offset</u> of&nbsp;$(Δϕ_{\rm T} = 30^\circ)$?
+{What are the values for &nbsp;$A_2$&nbsp; and &nbsp;$τ_2$&nbsp; for&nbsp; <u>double-sideband AM</u>&nbsp; and a&nbsp; <u>phase offset</u> &nbsp;$(Δϕ_{\rm T} = 30^\circ)$?
 |type="{}"}
 $A_2 \ = \ $ { 1.63 3% } $\ \rm V$
 $τ_2 \hspace{0.25cm} = \ $ { 0. } $\ \rm &micro; s$
-{Calculate the amplitudes&nbsp;$A_2$&nbsp; and &nbsp;$A_4$&nbsp; for&nbsp; <u>USB–AM</u>&nbsp; and&nbsp; <u>perfect synchronization</u>&nbsp; with &nbsp;$(Δϕ_{\rm T} = 0)$.
+{Calculate the amplitudes&nbsp;$A_2$&nbsp; and &nbsp;$A_4$&nbsp; for&nbsp; <u>upper-sideband AM</u>&nbsp; and&nbsp; <u>perfect synchronization</u>&nbsp; $(Δϕ_{\rm T} = 0)$.
 |type="{}"}
 $A_2 \ = \ $ { 1.764 3% } $\ \rm V$
 $A_4 \ = \ $ { 1.508 3% } $\ \rm V$
-{Give the signal amplitudes for &nbsp; <u>LSB–AM</u>&nbsp; and&nbsp; <u>perfect synchronization</u> with &nbsp;$(Δϕ_{\rm T} = 0)$.
+{Give the signal amplitudes for &nbsp; <u>lower-sideband AM</u>&nbsp; and&nbsp; <u>perfect synchronization</u> &nbsp;$(Δϕ_{\rm T} = 0)$.
 |type="{}"}
 $A_2 \ = \ $ { 2 3% } $\ \rm V$
 $A_4 \ = \ $ { 1.936 3% } $\ \rm V$
-{In contrast, what are the signal parameters for&nbsp; <u>LSB–AM</u>&nbsp; and a&nbsp; <u>phase offset</u>&nbsp; of &nbsp;$(Δϕ_{\rm T} = 30^\circ)$?
+{In contrast,&nbsp; what are the signal parameters for&nbsp; <u>lower-sideband AM</u>&nbsp; and a&nbsp; <u>phase offset</u> &nbsp;$(Δϕ_{\rm T} = 30^\circ)$?
 |type="{}"}
 $A_2 \ = \ $  { 2 3% } $\ \rm V$
@@ Line 74: / Line 70: @@
 $τ_4 \hspace{0.25cm} = \ $ { 20.8 3% } $\ \rm &micro;  s$
-{Which of these statements are true given your results?&nbsp; Here, "channel distortions" should always be understood as attenuation distortions.
+{Which of these statements are true given your results?&nbsp; Here,&nbsp; "channel distortions"&nbsp; should always be understood as a kind of attenuation distortions.
 |type="[]"}
-+ In DSB-AM, each channel distortion leads to attenuation distortions.
++ In&nbsp; "double-sideband AM",&nbsp; each channel distortion leads to attenuation distortions.
-- In SSB-AM, each channel distortion leads to phase distortions.
+- In&nbsp; "single-sideband AM",&nbsp; each channel distortion leads to phase distortions.
-- In DSB-AM, a phase offset leads to attenuation distortions.
+- In&nbsp; "double-sideband AM",&nbsp; a phase offset leads to attenuation distortions.
-+ In SSB-AM, a phase offset leads to phase distortions.
++ In&nbsp; "single-sideband AM",&nbsp; a phase offset leads to phase distortions.

@@ Line 6: / Line 6: @@
 Let us consider the transmission of the source signal
 :$$q(t) = 2\,{\rm V} \cdot \cos(2 \pi f_2 t) + 2\,{\rm V} \cdot \cos(2 \pi f_4 t)$$
-over a Gaussian bandpass channel with center frequency  &nbsp;$f_{\rm M} = 48 \ \rm  kHz$.&nbsp; This is different from the carrier frequency &nbsp;$f_{\rm T} = 50 \ \rm  kHz$ used in modulation.&nbsp; The frequencies &nbsp;$f_2$&nbsp; and &nbsp;$f_4$&nbsp; stand for  &nbsp;$f = 2 \ \rm   kHz$&nbsp; und &nbsp;$f = 4 \ \rm   kHz$, respectively.
+over a Gaussian bandpass channel with center frequency  &nbsp;$f_{\rm M} = 48 \ \rm  kHz$.&nbsp;
 We will now investigate the following modulation methods with respect to the spectrum &nbsp;$S_+(f)$&nbsp; of the analytical signal as shown in the upper graph:
-* DSB–AM&nbsp; $($all four spectral lines at &nbsp;$46 \ \rm   kHz$,&nbsp; $48 \ \rm   kHz$,&nbsp; $52 \ \rm   kHz$&nbsp; and&nbsp; $54 \ \rm   kHz)$,
+* DSB–AM&nbsp; $($all four spectral lines at &nbsp;$46 \ \rm   kHz$,&nbsp; $48 \ \rm   kHz$,&nbsp; $52 \ \rm   kHz$&nbsp; and&nbsp; $54 \ \rm   kHz)$ &nbsp; &rArr; &nbsp; "double-sideband" ,
-*USB–AM&nbsp; $($only blue spectral lines at &nbsp;$52 \ \rm   kHz$&nbsp; and&nbsp; $54 \ \rm   kHz)$,
+*USB–AM&nbsp; $($only blue spectral lines at &nbsp;$52 \ \rm   kHz$&nbsp; and&nbsp; $54 \ \rm   kHz)$&nbsp; &rArr; &nbsp; "upper-sideband",
-*LSB–AM&nbsp; $($only green spectral lines at &nbsp;$46 \ \rm   kHz$&nbsp; and&nbsp; $48 \ \rm   kHz)$.
+*LSB–AM&nbsp; $($only green spectral lines at &nbsp;$46 \ \rm   kHz$&nbsp; and&nbsp; $48 \ \rm   kHz)$&nbsp; &rArr; &nbsp; "lower-sideband".
-In each case, a synchronous demodulator is used to first convert the receiver-side carrier signal
+In each case,&nbsp;  a synchronous demodulator is used to first convert the receiver-side carrier signal
 :$$ z_{\rm E} (t) = \left\{ \begin{array}{c} 2 \cdot z(t) \\ 4 \cdot z(t) \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\rm DSB} \hspace{0.05cm}, \\ {\rm USB, LSB} \hspace{0.05cm} \\ \end{array}$$
-by multiplication and then completely suppresses the components by twice the carrier frequency.  Thus, with an ideal channel &nbsp;$H_{\rm K}(f) = 1$&nbsp;, &nbsp;$v(t) = q(t)$&nbsp; would hold in all cases.
+by multiplication and then completely suppresses the components by twice the carrier frequency.&nbsp;  With an ideal channel &nbsp;$H_{\rm K}(f) = 1$&nbsp;, &nbsp;$v(t) = q(t)$&nbsp; would hold in all cases.
 The Gaussian channel considered here is given by the following auxiliary values:
@@ Line 23: / Line 26: @@
 In each case, write the sink signal in the form
 :$$v(t) = A_2 \cdot \cos(2 \pi f_2 \cdot (t - \tau_2)) + A_4 \cdot \cos(2 \pi f_4 \cdot (t - \tau_4))\hspace{0.05cm}.$$
-All calculations are to be carried out for both a perfect phase synchronization &nbsp;$(Δϕ_{\rm T} = 0)$&nbsp; as well as for a phase offset of &nbsp;$Δϕ_{\rm T} = 30^\circ$&nbsp;.&nbsp; This is present, for example, if the transmit-side carrier signal is cosine-shaped and the receiver-side carrier is:
+All calculations are to be carried out for both a perfect phase synchronization &nbsp;$(Δϕ_{\rm T} = 0)$&nbsp; as well as for a phase offset of &nbsp;$Δϕ_{\rm T} = 30^\circ$.&nbsp; This is present,&nbsp; for example,&nbsp; if the transmit-side carrier signal is cosine-shaped and the receiver-side carrier is:
 :$$ z_{\rm E} (t) = A_{\rm E} \cdot \cos(\omega_{\rm T} \cdot t - 30^\circ) . $$