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Q. The input to the differentiator circuit is a sinusoidal voltage of peak value 5mv and frequency 1kHZ. find out the output if R=100K and C=10^-6F
Solution.
The equation of the input voltage is
V1 = 5*sin2*3.14*1000*t = 5 sin 200 0*3.14*t mv
= CR = (10^-6)*10^5 = 0.1
V0 = .1*(d(5sin 2000*3.14*t)
= (.5*2000*3.14)cos 2000*3.14*t
= 1000*3.14cos2000*3.14t mv
Hence output is a cosinusoidal voltage of ferquency 1 kHZ and peak value 1000 *3.14nV.
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