# Whats the principle of a PLL used as demodulator of a FM signal?

I didnt quite understand te following:

A basic PLL consists of the following parts:

• phase detector
• low pass filter
• VCO

If you input a 1MHz sine the PLL will try to lock on it by controlling the VCO. According to what i've found it's possible to demodulate a FM modulated signal.

Assume:

Input signal for example: (Carrier: 1Mhz sine and signal of 50Khz). you get 2 side-band frequenties with the carrier frequention (0.95Mhz, 1.0Mhz and 1.05Mhz).

I want to demodulate the 50Khz signal from the input signal.

From what i've found a DC signal from the Phase Detector is fed to the VCO to keep the PLL locked to the input frequency. My assumption was (i might be wrong) that when you input a signal with multiple frequency components the PLL keeps "re-locking" and the DC voltage fed to the VCO is the same as the difference of the frequention components of my input signal (so 1.0Mhz - 0.95Mhz = 50Khz).

edit:

Ye, there are some misconceptions in my story. With AM modulation you get the frequency components i was talking about (Dual Side-Band Full Carrier).

With FM modulation you have the following formula:

$v_c$ = carrier, $v_m$ = modulator

$v_c = V_c \sin(2 \pi f_c t)$
$v_m = V_m \sin(2 \pi f_m t)$

$f_c$ depends on the modulator voltage so $f_c = f_c + k*v_m$, where k is a amplifier factor.

the complete formula becomes:

$v = V_c \sin(2 \pi (f_c + k v_m) t) \rightarrow v = V_c \sin(2 \pi (f_c + k V_m \sin(2 \pi f_m t)) t)$

• The spectrum of a FM signal can be radically different depending on the modulation index. The Bessel function is used for computing it's components. See the link. Commented Jun 11, 2013 at 18:07
• Barry are we done with this old question now or are you still requiring some help on this subject? Commented Apr 20, 2021 at 18:01