GATE 2026 IN Q.46: solution and answer

Official answer: CSource: Official answer key, GATE 2026 IN (IIT Guwahati)

1. Assumptions: the op-amps are ideal. Every non-inverting input is grounded, so each inverting input is a virtual ground, and no current flows into either input. Call the output of op-amp 1 $V_1$ and the output of op-amp 2 $V_2$. Work in the Laplace domain with $s=j\omega$.

2. Stage 3 is an inverting amplifier with gain $-R/R=-1$, so $V_{out}=-V_2$.

3. Stage 2 is an inverting integrator with input resistor $R$ and feedback capacitor $C$, so $V_2=-\frac{V_1}{sRC}$. Combining stages 2 and 3 gives $V_{out}=\frac{V_1}{sRC}$, that is, $V_1=sRC{V}_{out}$.

4. Stage 1 is a lossy summing integrator. Its inverting input receives current from $V_{in}$ through $R$ and from $V_{out}$ through the top resistor $R$, and its feedback path is $R$ in parallel with $C$. KCL at the virtual ground: $$\frac{V_{in}}{R}+\frac{V_{out}}{R}+V_1\left(\frac{1}{R}+sC\right)=0\Rightarrow V_1(1+sRC)=-(V_{in}+V_{out})$$

5. Substitute $V_1=sRC{V}_{out}$: $$sRC(1+sRC)V_{out}=-V_{in}-V_{out}\Rightarrow H(s)=\frac{V_{out}}{V_{in}}=\frac{-1}{(sRC{)}^2+sRC+1}$$ This is the second-order low-pass response of a three-op-amp (Tow-Thomas) biquad. Its DC gain is $-1$. Written as $\frac{-{\omega }_0^2}{s^2+({\omega }_0/Q)s+{\omega }_0^2}$, it has ${\omega }_0=\frac{1}{RC}$ and $\frac{{\omega }_0}{Q}=\frac{1}{RC}$, so $Q=1$.

6. Numbers: $RC=(10^3\Omega )(10^{-6}\text{F})=10^{-3}\text{s}$, so ${\omega }_0=1000\text{rad/s}$. The input is at exactly ${\omega }_0$, so $sRC=j\omega RC=j(1000)(10^{-3})=j$.

7. Then $$H(j{\omega }_0)=\frac{-1}{j^2+j+1}=\frac{-1}{-1+j+1}=\frac{-1}{j}=j$$ so $|H(j{\omega }_0)|=1$. This agrees with the general result for a low-pass biquad at ${\omega }_0$: $|H(j{\omega }_0)|=|\text{DC gain}|\times Q=1\times 1$.

8. The output amplitude is $|V_{out}|=|H(j{\omega }_0)|\times 1\text{V}=1.0\text{V}$. It leads the input by $90^{\circ }$, but only the amplitude is asked for. The answer is option (C).