Explanation: Log amplifier requires the input and reference voltage to be of the same polarity. This restricts log-antilog multiplier to one quadrant operation. 4. An input of Vsinωt is applied to an ideal frequency doubler. Compute its output voltage?
A. Vo = [(Vx×Vy) /Vref2] × [1-cos2ωt/2]
B. Vo = [(Vx2×Vy2) /Vref] × [1-cos2ωt/2]
C. Vo = [(Vx×Vy)2 /Vref] × [1-cos2ωt/2]
D. Vo = [(Vx×Vy) /( Vref] × [1-cos2ωt/2]
Answer:A
Explanation: Log amplifier requires the input and reference voltage to be of the same polarity. This restricts log-antilog multiplier to one quadrant operation. 4. An input of Vsinωt is applied to an ideal frequency doubler. Compute its output voltage?
A. Vo = [(Vx×Vy) /Vref2] × [1-cos2ωt/2] B. Vo = [(Vx2×Vy2) /Vref] × [1-cos2ωt/2] C. Vo = [(Vx×Vy)2 /Vref] × [1-cos2ωt/2] D. Vo = [(Vx×Vy) /( Vref] × [1-cos2ωt/2]
Explanation: In an ideal frequency doubler, same frequency is applied to both inputs.