# Exercise 2.6Z: Synchronous Demodulator

Amplitude modulator (top),
synchronous demodulator

The depicted block diagram shows a transmission system

Let the source signal consist of two harmonic oscillations with frequencies  $f_2 = 2 \ \rm kHz$  and  $f_5 = 5 \ \rm kHz$:

$$q(t) = {2 \, \rm V} \cdot {\rm cos}(\omega_2 t )+ {1 \, \rm V} \cdot {\rm sin}(\omega_5 t ) .$$
• This signal is multiplied by the dimensionless carrier signal  $z(t) = \cos(\omega_{\rm T} \cdot T)$  of carrier frequency  $f_{\rm T} = 50 \ \rm kHz$.
For DSB–AM, the dashed block is irrelevant so that the following holds for the transmission signal:
$$s(t) = q(t) \cdot {\rm cos}(\omega_{\rm T} t ) .$$
• In the synchronous demodulator, the received signal  $r(t)$  – in an ideal channel identical to the signal  $s(t)$  – is multiplied by the receive-site carrier signal  $z_{\rm E}(t)$  where the following applies:
$$z_{\rm E}(t) = K \cdot {\rm cos}(\omega_{\rm T} t - \Delta \varphi ) .$$
• This signal should not only be frequency-synchronous with  $z(t)$  but also phase-synchronous   ⇒   hence the name "synchronous demodulator".
• The above approach takes into account a phase shift between  $z(t)$  and  $z_{\rm E}(t)$, which should ideally be  $\Delta \varphi = 0$  but often cannot be avoided in real systems.
• The output signal  $b(t)$  of the second multiplier includes components around twice the carrier frequency in addition to the desired low-frequency component.
• Using an ideal low-pass filter  $\rm (LP)$  (e.g. with cut-off frequency  $f_{\rm T}$)  the sink signal  $v(t)$,  which ideally should be equal to the source signal  $q(t)$,  can be obtained.

For the transmitter, multiplication by the carrier signal  $z(t)$  generally results in two sidebands. In  Single-Sideband Modulation  (ESB–AM), only one of the two bands is transmitted, for example the lower sideband (USB). Thus, the following is obtained with an ideal channel:

$$r(t) = s(t)= {1 \, \rm V} \cdot {\rm cos}\big [(\omega_{\rm T} - \omega_2 )\cdot t \big ] - {0.5 \, \rm V} \cdot {\rm sin}\big [(\omega_{\rm T} - \omega_5 )\cdot t \big ] .$$
• Here, synchronous demodulation results in the following distorted sink signal considering a phase shift  $\Delta \varphi$, the constant  $K = 4$  and the downstream low-pass filter:
$$v(t)= {1 \, \rm V} \cdot {1}/{2}\cdot 4 \cdot{\rm cos}( \omega_2 t - \Delta \varphi)+ {0.5 \, \rm V} \cdot {1}/{2}\cdot 4 \cdot{\rm sin}( \omega_5 t - \Delta \varphi)$$
$$\Rightarrow \hspace{0.5cm}v(t)= {2 \, \rm V} \cdot{\rm cos}( \omega_2 t - \Delta \varphi)+ {1 \, \rm V} \cdot{\rm sin}( \omega_5 t - \Delta \varphi)$$
• In the ideal case of phase-synchronous demodulation  $(\Delta \varphi = 0)$,    $v(t) = q(t)$ holds again.

• The exercise belongs to the chapter  Linear Distortions.
• The topic  "amplitude modulation/synchronous demodulator"  is discussed in detail in the book  Modulation Methods.
• The following trigonometric relationships are given:
$$\cos^2(\alpha) = {1}/{2} \cdot \big [ 1 + \cos(2\alpha) \big ] \hspace{0.05cm},$$
$$\cos(\alpha) \cdot \cos(\beta) = {1}/{2} \cdot \big[ \cos(\alpha - \beta)+ \cos(\alpha + \beta) \big],$$
$$\sin(\alpha) \cdot \cos(\beta) = {1}/{2} \cdot \big[ \sin(\alpha - \beta)+ \sin(\alpha + \beta) \big] \hspace{0.05cm}.$$
• The signal designations result from the German original of this exercise.  Here again as a listing   $q(t)$   ⇒   source signal,   $v(t)$   ⇒   sink signal,   $z(t)$   ⇒   transmit-site carrier signal,   $s(t)$   ⇒   transmission signal (BP),  $r(t)$   ⇒   received signal (BP),   $z_{\rm E}(t)$   ⇒   receive-site carrier signal,   $b(t)$   ⇒   BP signal before low-pass.

### Questions

1

What is the sink signal  $v(t)$  for DSB-AM and phase-synchronous modulation  ⇒   $\Delta \varphi = 0$?
How should  $K$  be chosen such that  $v(t) = q(t)$  holds?

 $K \ = \$

2

The following holds:  $K = 2$.  Specify the sink signal  $v(t)$  considering a phase shift $\Delta \varphi$.  Which of the following statements are true?

 $v(t) = q(t)$  holds independently of  $\Delta \varphi$ . $\Delta \varphi \ne 0$  results in frequency-independent attenuation. A phase shift  $\Delta \varphi \ne 0$  results in attenuation distortions. A phase shift  $\Delta \varphi \ne 0$  results in phase distortions. $v(t) = q(t)/2$  holds with  $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$.

3

Which statements hold for synchronous demodulation of the SSB signal if a phase shift of $\Delta \varphi$  is considered?

 Regardless of  $\Delta \varphi$,  $v(t) = q(t)$  holds. $\Delta \varphi \ne 0$  results in frequency-independent attenuation. A phase shift  $\Delta \varphi \ne 0$  results in attenuation distortions. A phase shift  $\Delta \varphi \ne 0$  results in phase distortions. $v(t) = q(t)/2$  holds with  $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$ .

### Solution

#### Solution

(1)  The following holds for the band-pass signal after the second multiplier:

$$b(t) = r(t) \cdot z_{\rm E}(t)= q(t) \cdot z(t) \cdot z_{\rm E}(t)= K \cdot q(t)\cdot \cos^2(\omega_{\rm T} t).$$
• Using the trigonometric relation  $\cos^2(\omega_{\rm T} t) = {1}/{2} \cdot\big[ 1 + \cos(2\omega_{\rm T} t)\big]$,  one obtains:
$$b(t) = {K}/{2} \cdot q(t) + {K}/{2} \cdot q(t)\cdot \cos(2\omega_{\rm T} t).$$
• The second component is located at around twice the carrier frequency  ⇒   $2 f_{\rm T}$.
• This is removed by the low-pass filter  $($with the cut-off frequency  $f_{\rm G} = f_{\rm T})$ .
• Hence, the following is obtained:   $v(t) = {K}/{2} \cdot q(t) .$
• An ideal demodulation  ⇒   $v(t) = q(t)$  is obtained with  $\underline {K = 2}$ .

(2)  Considering the relation

$$\cos(\omega_{\rm T} t) \cdot \cos(\omega_{\rm T} t - \Delta \varphi) = {1}/{2} \cdot \big[ \cos(\Delta \varphi)+ \cos(2\omega_{\rm T} t - \Delta \varphi) \big]$$

and the downstream low-pass filter,  which removes the component at around twice the carrier frequency, the following is obtained here with  ${K = 2}$:

$$v(t) = q(t) \cdot \cos(\Delta \varphi).$$

Proposed solutions 2 and 5  are correct:

• A phase shift $\Delta \varphi$  only results in frequency-independent attenuation and not in attenuation distortions or phase distortions.
• A phase shift by $\varphi =\pm 60^\circ$  results in halving of the signal amplitude.

(3)  Here, proposed solution 4  is correct.

• Exactly the same phase shift  $\Delta \varphi$  occurs for both summands,  and phase distortions occur here:
$$v(t)= {2 \, \rm V} \cdot{\rm cos}\big[ \omega_2 \cdot (t - \tau_2) \big]+ {1 \, \rm V} \cdot{\rm sin}\big[ \omega_5 t \cdot (t - \tau_5)\big],$$
$${\rm where}\hspace{0.5cm}\tau_2 = \frac{\Delta \varphi}{\omega_2} \hspace{0.5cm}\ne \hspace{0.5cm} \tau_5 = \frac{\Delta \varphi}{\omega_5}.$$
• A phase shift of  $\varphi =60^\circ$ corresponding to  $\pi/3$  leads to the following delay times here:
$$\tau_2 = \frac{\pi/3}{2 \pi \cdot 2\,\,{\rm kHz }} \approx 83.3\,{\rm µ s }, \hspace{0.5cm} \tau_5 = \frac{\pi/3}{2 \pi \cdot 5\,\,{\rm kHz }} \approx 33.3\,{\rm µ s }.$$
• The lower-frequency signal is thus delayed more.