Aufgaben:Exercise 2.10: SSB-AM with Channel Distortions

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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  $f_{\rm M} = 48 \ \rm kHz$.  This is different from the carrier frequency  $f_{\rm T} = 50 \ \rm kHz$ used in modulation.  The frequencies  $f_2$  and  $f_4$  stand for  $f = 2 \ \rm kHz$  und  $f = 4 \ \rm kHz$, respectively.

We will now investigate the following modulation methods with respect to the spectrum  $S_+(f)$  of the analytical signal as shown in the upper graph:

• DSB–AM  $($all four spectral lines at  $46 \ \rm kHz$,  $48 \ \rm kHz$,  $52 \ \rm kHz$  and  $54 \ \rm kHz)$,
• USB–AM  $($only blue spectral lines at  $52 \ \rm kHz$  and  $54 \ \rm kHz)$,
• LSB–AM  $($only green spectral lines at  $46 \ \rm kHz$  and  $48 \ \rm kHz)$.

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  $H_{\rm K}(f) = 1$ ,  $v(t) = q(t)$  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  $(Δϕ_{\rm T} = 0)$  as well as for a phase offset of  $Δϕ_{\rm T} = 30^\circ$ .  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   Single-Sideband Modulation.
• Reference will also be made to the chapter nbsp; Synchronous Demodulation.

Questions

1

Calculate the amplitudes for   DSB–AM  and  perfect synchronization  $(Δϕ_{\rm T} = 0)$.

$A_2 \ = \ $ $\ \rm V$

$A_4 \ = \ $ $\ \rm V$

2

What are the values for  $A_2$  and  $τ_2$  for  DSB–AM  and a  phase offset of $(Δϕ_{\rm T} = 30^\circ)$?

$A_2 \ = \ $ $\ \rm V$

$τ_2 \hspace{0.25cm} = \ $ $\ \rm µ s$

3

Calculate the amplitudes $A_2$  and  $A_4$  for  USB–AM  and  perfect synchronization  with  $(Δϕ_{\rm T} = 0)$.

$A_2 \ = \ $ $\ \rm V$

$A_4 \ = \ $ $\ \rm V$

4

Give the signal amplitudes for   LSB–AM  and  perfect synchronization with  $(Δϕ_{\rm T} = 0)$.

$A_2 \ = \ $ $\ \rm V$

$A_4 \ = \ $ $\ \rm V$

5

In contrast, what are the signal parameters for  LSB–AM  and a  phase offset  of  $(Δϕ_{\rm T} = 30^\circ)$?

$A_2 \ = \ $ $\ \rm V$

$τ_2 \hspace{0.25cm} = \ $ $\ \rm µ s$

$A_4 \ = \ $ $\ \rm V$

$τ_4 \hspace{0.25cm} = \ $ $\ \rm µ s$

6

Which of these statements are true given your results?  Here, "channel distortions" should always be understood as attenuation distortions.

	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.

Solution

Solution

(1)  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   $A_2\hspace{0.15cm}\underline{ = 1.882 \ \rm V}$  and  $A_4\hspace{0.15cm}\underline{ = 1.722 \ \rm V}$.

(2)  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   $τ_2\hspace{0.15cm}\underline {= 0}$  and  $τ_4 = 0$.

(3)  For USB–AM, the attenuation factor   $α_2$  is only determined by   $H_{\rm K}(f = 52\ \rm kHz)$ .

• Since the principal USB amplitude loss by a factor of r  $2$  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)  Analogous to the solution in subtask  (3) , 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)  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   $z_{\rm E}(t) = 4 \cdot \cos( \omega_{\rm 50} \cdot t - \Delta \phi_{\rm T})$ , 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  (4) .  For the transmit times when   $Δϕ_{\rm T} = π/6$, we get:

$$ \tau_2 = \frac {\Delta \phi_{\rm T}}{2 \pi \cdot f_2} = \frac {\pi /6}{2 \pi \cdot 2\,{\rm kHz}} \hspace{0.15cm}\underline {\approx 41.6\,{\rm µ s}},\hspace{0.5cm} \tau_4 = \frac {\Delta \phi_{\rm T}}{2 \pi \cdot f_4}= \frac {\tau_2}{2}\hspace{0.15cm}\underline {\approx 20.8\,{\rm µ s}} \hspace{0.05cm}.$$

(6)  The first and last answers are correct:

• Attenuation distortion on the channel exclusively lead to attenuation distortions with respect to   $v(t)$, 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   $v(t)$  can also arise in DSB–AM and SSB–AM, if these already occur on the channel.