GATE Papers >> EEE >> 2015 >> Question No 121

Question No. 121 EEE | GATE 2015

The operational amplifier shown in the figure is ideal. The input voltage (in Volt) is V1= 2sin2π×2000t. The amplitude of the output voltage V0 (in Volt) is ________.


Answer : 1.1 to 1.4


Solution of Question No 121 of GATE 2015 EEE Paper

$ {\mathrm V}_\mathrm i=2\sin\left(2\mathrm\pi\times200\mathrm t\right) $

The transfer function of the system is

$ \begin{array}{l}\mathrm H(\mathrm s)=\frac{1000}{(1000\times0.1\times10^{-6}\mathrm s+1)1000}\\\mathrm H(\mathrm s)=\frac1{(10^{-4}\mathrm s+1)}\end{array} $

The input is $ 2\sin\left(2\mathrm\pi\times2000\;\mathrm t\right) $

$ \mathrm\omega=4000\mathrm\pi $

Output will be $ 2\;\mathrm H\left(\mathrm{jω}\right)\vert_{\mathrm\omega=4000\mathrm\pi}.\sin\left(4000\mathrm{πt}\right) $

Output $ =1.245\;\sin\left(4000\mathrm{πt}-51.46^\circ\right) $

So, amplitude of output is 1.245.

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