# Exercise 3.7: Angular Modulation of a Harmonic Oscillation

Demodulator
for FM

The signal arriving at a receiver is:

$$r(t) = 3\,{\rm V} \cdot \cos \hspace{-0.05cm} \big[2 \pi \cdot 1\,{\rm MHz} \cdot t + 3 \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\big]\hspace{0.05cm}.$$

$r(t)$  is an angle-modulated signal that was neither distorted nor influenced by noise during transmission.

The signals  $v_{\rm PM}(t)$  and  $v_{\rm FM}(t)$  result after ideal demodulation by means of

• a phase demodulator, given by the equation
$$v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) \hspace{0.05cm},\hspace{0.3cm} {K_{\rm PM}} = 2\,{\rm V}^{-1}\hspace{0.05cm},$$
• a frequency demodulator, consisting of a PM demodulator, a differentiator and a constant $K$.

In order for all signals to have equal units, this constant $K$ is dimensionally constrained.

Hints:

### Questions

1

Which statements are definitely true?

 There could be a PM modulation. There could be a FM modulation. The message phase is definitely  $ϕ_{\rm N} = 0$. The message phase is definitely  $f_{\rm N} = 10 \ \rm kHz$.

2

Calculate the signal $v_{\rm PM}(t)$  after the phase demodulator.  What is the signal value at time  $t = 0$?

 $v_{\rm PM}(t = 0) \ = \$ $\ \rm V$

3

Calculate the signal $v_{\rm FM}(t)$. What is the message phase  $ϕ_{\rm N}$?

 $ϕ_{\rm N} \ = \$ $\ \rm Grad$

4

How should  $K$  be chosen so that the amplitude of  $v_{\rm FM}(t)$  is equal to  $1.5 \ \rm V$ ?

 $K\ = \$ $\ \rm \cdot 10^4 \ 1/s$

5

Which of the following statements is true for the FM-modulated signal?

 The phase deviation is  $ϕ_{\rm max} = 3$. The frequency deviation is  $Δf_{\rm A} = 30 \ \rm kHz$. The instantaneous frequencies are between  $0.97\ \rm MHz$  and  $1.03 \ \rm MHz$ . If  $f_{\rm N} = 5 \ \rm kHz$ , the phase deviation would be unchanged. If  $f_{\rm N} = 5 \ \rm kHz$  the frequency deviation would be unchanged.

### Solution

#### Solution

(1)  Answers 1, 2 and 4 are correct:

• From the equation for  $r(t)$  it can only be ascertained that it is an angle modulation,
• but not whether it is a phase modulation (PM) or a frequency modulation (FM).
• Based on the equation, it is clear that the message frequency is  $f_{\rm N} = 10 \ \rm kHz$ .
• The phase  $ϕ_{\rm N} = 0$  of the source signal would then only apply, if phase modulation were present.

(2)  With the modulator constant  $K_{\rm PM} = 2 \ \rm V^{–1}$  this is given by:

$$v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) = \frac{3}{2\,{\rm V}^{-1}} \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
• At time   $t = 0$  it therefore holds that:
$$v_{\rm PM}(t = 0) = {A_{\rm N}} \hspace{0.15cm}\underline {= 1.5\,{\rm V}}\hspace{0.05cm}.$$

(3)  The output signal   $v_{\rm FM}(t)$  of the FM demodulator – consisting of a PM–demodulator and differentiator – can be written as:

$$v_{\rm FM}(t) = \frac{{\rm d}v_{\rm PM}(t)}{{\rm d}t} \cdot K = \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot (- \sin(2 \pi \cdot {f_{\rm N}} \cdot t))= \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot \cos(2 \pi \cdot {f_{\rm N}} \cdot t + 90^\circ)\hspace{0.05cm}.$$
• The message phase is thus   $ϕ_{\rm N} \hspace{0.15cm}\underline {= 90^\circ}$.

(4)  In this case, it must hold that:

$$K ={2 \pi \cdot f_{\rm N}} \hspace{0.15cm}\underline { = 6.28 \cdot 10^{4} \,\,{1}/{ s}} \hspace{0.05cm}.$$

(5) Answers 1, 2, 3 and 5 are correct:

• The phase deviation is identical to the modulation index, which can be discerned from the equation given:
$$\phi_{\rm max} = \eta = 3 = \frac{\Delta f_{\rm A}}{ f_{\rm N}} \hspace{0.05cm}.$$
• This leads to the frequency deviation  $Δf_{\rm A} = 3 · f_{\rm N} = 30 \ \rm kHz$.
• With a carrier frequency of   $f_{\rm T} = 1 \ \rm MHz$ , the instantaneous frequency  $f_{\rm T}(t)$  can only take values between  $1±0.03 \ \rm MHz$ .

Thus, the following statement is also valid::

At half the message frequency, the phase deviation  $η$ doubles, while the frequency deviation  $Δf_{\rm A}$ is unaffected:

$$\eta = \frac{K_{\rm PM} \cdot A_{\rm N}}{ f_{\rm N}} = 6 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\Delta f_{\rm A} = \eta \cdot f_{\rm N} = 6 \cdot 5\,{\rm kHz} = 30\,{\rm kHz}\hspace{0.05cm}.$$