View source for Aufgaben:Exercise 2.10: SSB-AM with Channel Distortions


{{quiz-Header|Buchseite=Modulationsverfahren/Einseitenbandmodulation
}}

[[File:P_ID1045__Mod_A_2_9.png|right|frame|Transmission spectrum of the analytical signal and channel frequency response]]
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.

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; $54 \ \rm   kHz)$,
*USB–AM&nbsp; $($only blue spectral lines at &nbsp;$52 \ \rm   kHz$&nbsp; and&nbsp; $54 \ \rm   kHz)$,
*LSB–AM&nbsp; $($only green spectral lines at &nbsp;$46 \ \rm   kHz$&nbsp; and&nbsp; $48 \ \rm   kHz)$.


In each case, 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. 

The Gaussian channel considered here is given by the following auxiliary values:
:$$ H_{\rm K}(f = 46\ {\rm kHz}) = 0.968,\hspace{0.3cm}H_{\rm K}(f = 48\ {\rm kHz}) = 1.000,\hspace{0.3cm}
 H_{\rm K}(f = 52\ {\rm kHz}) = 0.882,\hspace{0.3cm}H_{\rm K}(f = 54\ {\rm kHz}) = 0.754\hspace{0.05cm}.$$
In each case, write the sink signal in the form
:$$v(t) = A_2 \cdot \cos(2 \pi f_2 \cdot (t - \tau_2)) + A_4 \cdot \cos(2 \pi f_4 \cdot (t - \tau_4))\hspace{0.05cm}.$$
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:
:$$ z_{\rm E} (t) = A_{\rm E} \cdot \cos(\omega_{\rm T} \cdot t - 30^\circ) . $$








''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]].
 



===Questions===

<quiz display=simple>

{Calculate the amplitudes for &nbsp; <u>DSB–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;$(Δϕ_{\rm T} = 30^\circ)$?
|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;$(Δϕ_{\rm T} = 0)$.
|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;$(Δϕ_{\rm T} = 0)$.
|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;$(Δϕ_{\rm T} = 30^\circ)$?
|type="{}"}
$A_2 \ = \ $  { 2 3% } $\ \rm V$
$τ_2 \hspace{0.25cm} = \ $  { 41.6 3% } $\ \rm &micro; s$
$A_4 \ = \ $ { 1.936 3% } $\ \rm V$
$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. 
|type="[]"}
+ In DSB-AM, each channel distortion leads to attenuation distortions.
- In SSB-AM, each channel distortion leads to phase distortions.
- In DSB-AM, a phase offset leads to attenuation distortions.
+ In SSB-AM, a phase offset leads to phase distortions.



</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; For DSB–AM, 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; $A_4\hspace{0.15cm}\underline{ = 1.722 \ \rm V}$.



'''(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:
:$$A_2  =  \cos (30^\circ) \cdot 1.882\,{\rm V} \hspace{0.15cm}\underline {= 1.630\,{\rm V}},$$ 
:$$A_4  =  \cos (30^\circ) \cdot 1.722\,{\rm V} = 1.491\,{\rm V}\hspace{0.05cm}.$$
*The transmit times are &nbsp; $τ_2\hspace{0.15cm}\underline {= 0}$&nbsp; and&nbsp; $τ_4 = 0$.



'''(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;. 
*Since the principal USB amplitude loss by a factor of r&nbsp; $2$&nbsp; is compensated for by a larger carrier amplitude, the following holds:
:$$A_2  =  0.882 \cdot 2\,{\rm V}\hspace{0.15cm}\underline {= 1.764\,{\rm V}},$$ 
:$$A_4  =  0.754 \cdot 2\,{\rm V}\hspace{0.15cm}\underline {= 1.508\,{\rm V}} \hspace{0.05cm}.$$



'''(4)'''&nbsp; Analogous to the solution in subtask&nbsp; '''(3)'''&nbsp;, here we obtain:
:$$ A_2  =  H_{\rm K}(f = 48\,{\rm kHz}) \cdot 2\,{\rm V}\hspace{0.15cm}\underline {= 2\,{\rm V}},$$ 
:$$A_4  =  H_{\rm K}(f = 46\,{\rm kHz}) \cdot 2\,{\rm V}\hspace{0.15cm}\underline {= 1.936\,{\rm V}} \hspace{0.05cm}.$$



'''(5)'''&nbsp; For LSB–AM, 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:
:$$v(t) = r(t) \cdot z_{\rm E}(t) =  \hspace{0.15cm}\underline { 2.000\,{\rm V}} \cdot \cos( \omega_{\rm 2} \cdot t - \Delta \phi_{\rm T})+\hspace{0.15cm}\underline { 1.936\,{\rm V}} \cdot \cos( \omega_{\rm 4} \cdot t - \Delta \phi_{\rm T})
 +  {\rm components \hspace{0.15cm}um \hspace{0.15cm}} 2f_{\rm T}\hspace{0.05cm}$$
:$$ \Rightarrow \hspace{0.3cm} A_2 \hspace{0.15cm}\underline {= 2\,{\rm V}},\hspace{0.5cm} A_4 \hspace{0.15cm}\underline {= 1.936\,{\rm V}}.$$ 
*Considering the downstream lowpass filter, this can also be written as:
:$$ v(t) = A_2 \cdot \cos( \omega_{\rm 2} \cdot (t - \tau_2))+ A_4 \cdot \cos( \omega_{\rm 4} \cdot (t - \tau_4))\hspace{0.05cm}.$$
*The amplitudes are unchanged compared to subtask&nbsp; '''(4)'''&nbsp;.&nbsp; For the transmit times when &nbsp; $Δϕ_{\rm T} = π/6$, we get:
:$$ \tau_2  =  \frac {\Delta \phi_{\rm T}}{2 \pi \cdot f_2} = \frac {\pi /6}{2 \pi \cdot 2\,{\rm kHz}} \hspace{0.15cm}\underline {\approx 41.6\,{\rm &micro; s}},\hspace{0.5cm} \tau_4  =  \frac {\Delta \phi_{\rm T}}{2 \pi \cdot f_4}= \frac {\tau_2}{2}\hspace{0.15cm}\underline {\approx 20.8\,{\rm &micro; s}} \hspace{0.05cm}.$$



'''(6)'''&nbsp; The <u>first and last answers</u> are correct: 
*Attenuation distortion on the channel exclusively lead to attenuation distortions with respect to &nbsp; $v(t)$, also for SSB.
*Phase distortions are only present for a demodulator with a phase offset in the case of single-sideband modulation. 
*For DSB–AM, such a phase offset would not result in any distortions, but only in frequency-independent damping.
*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.


{{ML-Fuß}}



[[Category:Modulation Methods: Exercises|^2.4  Single Sideband Amplitude Modulation^]]