GATE 2024 IN – Question 43
In the circuit shown, R = 1 kΩ and C = 0.1 μF. For a DC gain of −10, the 3 dB cutoff frequency (one decimal place) is as follows. Assume the opamp is ideal.
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Correct answer: (A) 159.1 Hz
Explanation
The DC gain is $-R_f/R=-10$, so $R_f=10$ kΩ. The parallel feedback capacitor gives $H(s)=-(R_f/R)/(1+sR_fC)$.
Therefore $f_c=1/(2\pi R_fC)=1/(2\pi\times10000\times10^{-7})=159.15$ Hz, corresponding to **159.1 Hz, A**.