# Exercise 3.2: Spectrum with Angle Modulation

Table of Bessel functions

The following equations are assumed here:

• Source signal:
$$q(t) = 2\,{\rm V} \cdot \sin(2 \pi \cdot 3\,{\rm kHz} \cdot t)\hspace{0.05cm},$$
• Transmitted signal:
$$s(t) = 1\,{\rm V} \cdot \cos\hspace{-0.1cm}\big[2 \pi \cdot 100\,{\rm kHz} \cdot t + K_{\rm M} \cdot q(t)\big ]\hspace{0.05cm},$$
$$r(t) = s(t) = 1\,{\rm V} \cdot \cos\hspace{-0.1cm}\big[2 \pi \cdot 100\,{\rm kHz} \cdot t + \phi(t)\big ]\hspace{0.05cm},$$
• ideal demodulator:
$$v(t) = \frac{1}{ K_{\rm M}} \cdot \phi(t)\hspace{0.05cm}.$$

The graphs shows the   $n$–th order Bessel functions of the first kind   ${\rm J}_n (\eta)$  in table form.

Hints:

### Questions

1

Which modulation method is used here?

 Amplitude modulation. Phase modulation. Frequency modulation.

2

Which modulation method would you choose if the channel bandwidth was only  $B_{\rm K} = 10 \ \rm kHz$ ?

 Amplitude modulation. Phase modulation. Frequency modulation.

3

How should one choose the modulator constant $K_{\rm M}$  for a phase deviation of  $η = 1$ ?

 $K_{\rm M} \ = \$ $\ \rm 1/V$

4

Calculate the spectrum  $S_{\rm TP}(f)$  of the equivalent low-pass signal  $s_{\rm TP}(t)$.  What are the weights of the spectral lines at  $f = 0$  and  $f = -3 \ \rm kHz$?

 $S_{\rm TP}(f = 0)\ = \$ $\ \rm V$ $S_{\rm TP}(f = -3\ \rm kHz) \ = \$ $\ \rm V$

5

Calculate the spectra of the analytical signal $s_{\rm +}(t)$  and the physical signal  $s(t)$.  What are the weights of the spectral lines at  $f = 97 \ \rm kHz$?

 $S_+(f = 97 \ \rm kHz)\ = \$ $\ \rm V$ $S(f = 97 \ \rm kHz)\hspace{0.32cm} = \$ $\ \rm V$

6

What is the required channel bandwidth  $B_{\rm K}$  for  $η = 1$, if one ignores pulse weights smaller (in magnitude) than $0.01$ ?

 $η = 1\text{:} \ \ \ B_{\rm K}\ = \$ $\ \rm kHz$

7

What would be the channel bandwidths for  $η = 2$  and  $η = 3$ ?

 $η = 2\text{:} \ \ \ B_{\rm K}\ = \$ $\ \rm kHz$ $η = 3\text{:} \ \ \ B_{\rm K}\ = \$ $\ \rm kHz$

### Solution

#### Solution

(1)  The phase  $ϕ(t)$  is proportional to the source signal  $q(t)$   ⇒   this is a phase modulation   ⇒   Answer 2.

(2)  An angle modulation  (PM, FM)  always results in nonlinear distortion when the channel is bandlimited.

• In contrast, double-sideband amplitude modulation  (DSB-AM)  here enables distortion-free transmission with  $B_{\rm K} = 6 \ \rm kHz$ ; ⇒   Answer 1.

(3)  The modulation index (or phase deviation) is equal to  $η = K_{\rm M} · A_{\rm N}$ for phase modulation.

• Thus, the modulator constant must be set to  $K_{\rm M} = 1/A_{\rm N}\hspace{0.15cm}\underline { = 0.5 \rm \cdot {1}/{V}}$  to give   $η = 1$ .

(4)  A so-called Bessel spectrum is present:

$$S_{\rm TP}(f) = A_{\rm T} \cdot \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot \delta (f - n \cdot f_{\rm N})\hspace{0.05cm}.$$
• This is a discrete spectrum with components at   $f = n · f_{\rm N}$, where  $n$  is an integer.
• The weights of the Dirac functions are given by the Bessel functions.  When  $A_{\rm T} = 1\ \rm V$ , one obtains:
PM spectrum in the equivalent low-pass range
$$S_{\rm TP}(f = 0) = A_{\rm T} \cdot {\rm J}_0 (\eta = 1) \hspace{0.15cm}\underline {= 0.765\,{\rm V}},$$
$$S_{\rm TP}(f = f_{\rm N}) = A_{\rm T} \cdot {\rm J}_1 (\eta = 1)\hspace{0.15cm} = 0.440\,{\rm V},$$
$$S_{\rm TP}(f = 2 \cdot f_{\rm N}) = A_{\rm T} \cdot {\rm J}_2 (\eta = 1) = 0.115\,{\rm V} \hspace{0.05cm}.$$
• Due to the symmetry   ${\rm J}_{-n} (\eta) = (-1)^n \cdot {\rm J}_{n} (\eta)$ , the spectral line at   $f = -3 \ \rm kHz$ is obtained as:
$$S_{\rm TP}(f = -f_{\rm N}) = -S_{\rm TP}(f = +f_{\rm N}) =\hspace{-0.01cm}\underline { -0.440\,{\rm V} \hspace{0.05cm}}.$$

Note:  For the spectral value at  $f = 0$  we should actually write:

$$S_{\rm TP}(f = 0) = 0.765\,{\rm V} \cdot \delta (f) \hspace{0.05cm}.$$
• This is therefore infinite due to the Dirac function, and only the weight of the Dirac function is finite.
• The same applies for all discrete spectral lines.

(5)  $S_+(f)$  is obtained from   $S_{\rm TP}(f)$  by shifting   $f_{\rm T}$ to the right.  Therefore

$$S_{\rm +}(f = 97\,{\rm kHz}) = S_{\rm TP}(f = -3\,{\rm kHz}) \hspace{0.15cm}\underline {=-0.440\,{\rm V}} \hspace{0.05cm}.$$
• The actual spectrum differs from  $S_+(f)$  by a factor of   $1/2$ at positive frequencies:
$$S(f = 97\,{\rm kHz}) = {1}/{2} \cdot S_{\rm +}(f = 97\,{\rm kHz}) \hspace{0.15cm}\underline {=-0.220\,{\rm V}} \hspace{0.05cm}.$$
• In general, we can write:
$$S(f) = \frac{A_{\rm T}}{2} \cdot \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot \delta (f \pm (f_{\rm T}+ n \cdot f_{\rm N}))\hspace{0.05cm}.$$

(6)  Under the suggested conditions, all the Bessel lines  ${\rm J}_{|n|>3}$  can be disregarded.

• This gives  $B_{\rm K} = 2 · 3 · f_{\rm N}\hspace{0.15cm}\underline { = 18 \ \rm kHz}$.

(7)  The numerical values in the table given on the exercise page show that the following channel bandwidths would now be required:

• für $η = 2$:     $B_{\rm K} \hspace{0.15cm}\underline { = 24 \ \rm kHz}$,
• für $η = 3$:     $B_{\rm K} \hspace{0.15cm}\underline { = 36 \ \rm kHz}$.