Thursday, 3 March 2016

frequency - 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)\$




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