Phase Locked Loop (PLL)

Demodulation of an FM signal using PLL:-

Let the input to PLL is an FM signal S(t) = A_{c} \sin (2 \pi f_{c}t+2\pi k_{f} \int_{0}^{t}m(t)dt)

let  \Phi _{1} (t) = 2\pi k_{f} \int_{0}^{t}m(t)dt ------Equation (I)

 Now the signal at the output of VCO is FM signal (another FM signal, which is different from input FM signal) Since Voltage Controlled Oscillator is an FM generator.

\therefore b(t) = A_{v} \cos (2 \pi f_{c}t+2\pi k_{v} \int_{0}^{t}v(t)dt)

the corresponding phase    \Phi _{2} (t) = 2\pi k_{v} \int_{0}^{t}v(t)dt ------Equation (II)

It is observed that S(t) and b(t) are out of phase by 90^{o}. Now these signals are applied to a phase detector , which is basically a multiplier

\therefore the error signal e(t) =S(t) .b(t)

e(t) =A_{c} \sin (2 \pi f_{c}t+2\pi k_{f} \int_{0}^{t}m(t)dt). A_{v} \cos (2 \pi f_{c}t+2\pi k_{v} \int_{0}^{t}v(t)dt)

e(t) =A_{c}A_{v} \sin (2 \pi f_{c}t+\phi _{1}(t)). \cos (2 \pi f_{c}t+\phi _{2}(t))

on further simplification , the product yields a higher frequency term (Sum) and a lower frequency term (difference)

e(t) =A_{c}A_{v}k_{m} \sin (4 \pi f_{c}t+\phi _{1}(t)+\phi _{2}(t))- A_{c}A_{v}k_{m}\sin (\phi _{1}(t)-\phi _{2}(t))

e(t) =A_{c}A_{v}k_{m} \sin (2 \omega _{c}t+\phi _{1}(t)+\phi _{2}(t))- A_{c}A_{v}k_{m}\sin (\phi _{1}(t)-\phi _{2}(t))

This product e(t) is given to a loop filter , Since the loop filter is a LPF it allows the difference and term and rejects the higher frequency term.

the over all output of a loop filter is 

 

Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.