Exercise 3.7: Angular Modulation of a Harmonic Oscillation

From LNTwww

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.




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  $f_{\rm N} = 10 \ \rm kHz$.


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) \ = \ $

$\ \rm V$


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

$ϕ_{\rm N} \ = \ $

$\ \rm Grad$


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

$K\ = \ $

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


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.


(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:

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