Explanation: The error voltage is zero when the phase shift between the two inputs is 90o. So, for the perfect lock, the VCO output should be 90o out of phase with respect to the input signal. 3. Find the error voltage of phase comparator whose input signal is Vs= Vssin(2πfst) and the output signal Vo= Vosin(2πfot+φ).
A. Ve=[k×(Vs/2)]×[cos(-φ)-cos(2πfot+φ)]
B. Ve=[k×Vs×(Vo/2)]×[cos(-φ)-cos(2πfot+φ)]
C. Ve=[k×Vs×(Vo/2)]×[cos(-φ)+cos(2πfot+φ)]
D. Ve=[k×Vs×Vo]×[cos(-φ)-cos(2πfot+φ)]
Answer:A
Explanation: The error voltage is zero when the phase shift between the two inputs is 90o. So, for the perfect lock, the VCO output should be 90o out of phase with respect to the input signal. 3. Find the error voltage of phase comparator whose input signal is Vs= Vssin(2πfst) and the output signal Vo= Vosin(2πfot+φ).
A. Ve=[k×(Vs/2)]×[cos(-φ)-cos(2πfot+φ)] B. Ve=[k×Vs×(Vo/2)]×[cos(-φ)-cos(2πfot+φ)] C. Ve=[k×Vs×(Vo/2)]×[cos(-φ)+cos(2πfot+φ)] D. Ve=[k×Vs×Vo]×[cos(-φ)-cos(2πfot+φ)]
Explanation: A phase comparator is basically a multiplier which multiplies the input signal by the VCO signal. Thus, the phase comparator output = Vs×Vo = Vs×Vosin(2πfst)× sin(2πfot+φ) =k×Vs×Vo ×sin(2πfst)×sin(2πfot+φ)