# Exercise 2.4Z: Low-pass Influence with Synchronous Demodulation

Let us consider the same communication system as in Exercise 2.4. But this time, we will assume perfect frequency and phase synchronization for the synchronous demodulator $\rm (SD)$ .

The source signal $q(t)$, the transmitted signal $s(t)$ and the signal $b(t)$ in the synchronous demodulator before the low-pass filter are given as follows:

- $$q(t) = q_1(t) + q_2(t)\hspace{0.2cm}{\rm with }$$
- $$q_1(t) = 2\,{\rm V} \cdot \cos(2 \pi \cdot 2\,{\rm kHz} \cdot t)\hspace{0.05cm},$$
- $$q_2(t) = 1\,{\rm V} \cdot \sin(2 \pi \cdot 5\,{\rm kHz} \cdot t)\hspace{0.05cm},$$

- $$s(t) = q(t) \cdot \sin(2 \pi \cdot 50\,{\rm kHz} \cdot t)\hspace{0.05cm},$$
- $$b(t) = s(t) \cdot 2 \cdot \sin(2 \pi \cdot 50\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$

The graph shows the source signal $q(t)$ at the top and the transmission signal $s(t)$ in the middle.

The sink signal $v(t)$ is shown at the bottom (purple waveform).

- This obviously does not match the source signal (blue dashed curve).
- The reason for this undesired result $v(t) ≠ q(t)$ could be a missing or wrongly dimensioned low-pass filter.

In the subtasks **(3)** and **(4)** , a "trapezoidal low-pass filter" is used, whose frequency response is as follows:

- $$H_{\rm E}(f) = \left\{ \begin{array}{l} \hspace{0.25cm}1 \\ \frac{f_2 -|f|}{f_2 -f_1} \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{10}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\hspace{0.94cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_1,} \\ {f_1 \le \left| \hspace{0.005cm}f\hspace{0.05cm} \right| \le f_2,} \\ {\hspace{0.94cm}\left|\hspace{0.005cm} f \hspace{0.05cm} \right| > f_2.} \\ \end{array}$$

Hints:

- This exercise belongs to the chapter Synchronous Demodulation.
- Particular reference is made to the page Block diagram and time domain representation.
- In contrast to Exercise 2.4 , $f_1$ and $f_2$ do not describe signal frequencies, but instead relate to the low-pass filter.

### Questions

### Solution

**(1)**

__The first statement__is correct:

- The sink signal $v(t)$ shown in the bottom graphic exactly matches the signal $b(t)$ given in the equation and thus also contains components around twice the carrier frequency.
- The filter $H_{\rm E}(f)$ is either missing completely or its upper cutoff frequency $f_2$ is too high.
- Regarding the lower cutoff frequency $f_1$, the only statement possible is that it is smaller than the smallest frequency $\text{(2 kHz)}$ occurring in the signal $b(t)$.
- Whether or not a DC component is removed by the filter is unclear, since such a component is not present in the signal $b(t)$.

**(2)** __Answers 1 and 3__ are correct:

- A prerequisite for distortion-free demodulation is that all spectral components up to a certain frequency $f_1$ are transmitted equally and as unattenuated as possible, and all components at frequencies $f > f_2$ are completely suppressed.
- The rectangular and trapezoidal low-pass filters satisfy this condition.

**(3)** It must be ensured that the $\text{5 kHz}$ component still lies in the passband:

- $$f_{\text{1, min}}\hspace{0.15cm}\underline{ =5 \ \rm kHz}.$$

**(4)** All spectral components in the vicinity of twice the carrier frequency – more precisely between $\text{95 kHz}$ and $\text{ 105 kHz}$ – must be completely suppressed:

- $$f_{\text{2, max}}\hspace{0.15cm}\underline{ =95 \ \rm kHz}.$$

- Otherwise nonlinear distortion would arise.

**(5)** __Answer 2__ is correct:

- The cutoff frequency} $f_{\rm G} = \text{ 4 kHz}$ would result in (linear) distortions, since the $\text{5 kHz}$ component would be cut off.
- The low-pass with cutoff frequency $f_{\rm G} = \text{6 kHz}$ is preferable, since with $f_{\rm G} = \text{10 kHz}$ more noise components would be superimposed on the signal $v(t)$.