GATE 2026 IN – Question 46
In the following active-filter circuit with ideal op-amps, $R=1\ \mathrm{k}\Omega$ and $C=1\ \mu\mathrm F$. The input $V_{in}$ has an amplitude of 1 V at 1000 rad/s. The amplitude of $V_{out}$ is ________ V.

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Correct answer: (C) 1.0
Explanation
Let $p=sRC$ and denote the first two stage outputs by $v_1$ and $v_2$. The second stage is an integrator, so $v_2=-v_1/p$. The last stage is a unity-gain inverter, so $v_o=-v_2=v_1/p$.
KCL at the first summing node gives $v_i+(1+p)v_1+v_o=0$. Substituting $v_1=pv_o$ yields $H(s)=v_o/v_i=-1/(p^2+p+1)$.
Here $RC=10^{-3}\ \mathrm s$ and $\omega RC=1$, so $p=j$ and $|H|=|-1/j|=1$. The output amplitude is **1.0 V (C)**.