(Solved):   We use a phase locked loop (PLL) to demodulate a frequency modulated (FM) signal \( \mathr ...

 

We use a phase locked loop (PLL) to demodulate a frequency modulated (FM) signal \( \mathrm{v}_{\text {in }}(\mathrm{t}) \). The carrier is a sine function with \( \mathrm{f}_{\mathrm{c}}=500 \mathrm{~Hz} \) and \( \mathrm{A}_{\mathrm{c}}=1 \) (the phase \( =0 \) at \( \mathrm{t}=0 \) ). The message \( \mathrm{m}(\mathrm{t} \) ) is a cosine function with \( \mathrm{f}_{\mathrm{m}}=10 \mathrm{~Hz} \) and \( \mathrm{A}_{\mathrm{m}}=1 \mathrm{~V}( \) the phase \( =0 \) at \( \mathrm{t}=0 \) ) and the index of modulation \( \beta=50 \). When the loop is locked, we have exactly \( \pi / 2 \) difference between the output of the VCO \( v_{0}(t) \) and \( v_{\text {in }}(t) \). The phase of \( v_{0}(t) \) is given by: \( \theta_{0}(t)=K_{v} \int_{-\infty}^{t} v_{2}(\tau) d \tau \) PD gain \( K_{d}=0.5 \mathrm{~V} / \mathrm{rad} \), VCO gain constant \( K_{v}=1000 \pi \mathrm{rad} /(\mathrm{s} . \mathrm{V}) \), the low-pass loop filter is an ideal one with a transfer function \( F(f)=1 \) (in the passband) 1) Sketch the block diagram of the PLL and express \( v_{\text {in }}(t) \) 2) Calculate the open loop transfer function and deduce the total loop transfer function. 3) Find the output signal of the loop \( v_{2}(t) .\left(K_{v} K_{d} \gg>2 \pi f\right) \)