GATE 2020 EC Q.41: solution and answer

Official answer: DSource: Official answer key, GATE 2020 EC (IIT Delhi)

1. Assumption: ideal op-amp (infinite open-loop gain, infinite input impedance) with negative feedback through $R\parallel C$. Since $V_{+}=0$, the virtual short holds node X (the inverting input) at $V_{-}=0\text{V}$, a virtual ground.

2. The shunt elements $2C$ and $2R$ are connected between node X ($0\text{V}$) and ground ($0\text{V}$). With no voltage across them, neither carries any current, so they have no effect on the transfer function.

3. KCL at node X (the ideal op-amp input draws no current): the input current $I=\frac{V_i-0}{R}=\frac{V_i}{R}$ flows entirely through the feedback impedance $$Z_f=R\parallel \frac{1}{j\omega C}=\frac{R}{1+j\omega RC}$$

4. Inverting-amplifier gain: $$\frac{V_o}{V_i}=-\frac{Z_f}{R}=-\frac{1}{1+j\omega RC}$$ This is a first-order low-pass response with DC gain magnitude $1$.

5. The $-3$ dB point is where the magnitude falls to $\frac{1}{\sqrt{2}}$ of its DC value, i.e. $|1+j{\omega }_{c}RC|=\sqrt{2}$, so ${\omega }_{c}RC=1$: $$f_c=\frac{1}{2\pi RC}=\frac{1}{2\pi (2\times 10^3)(1\times 10^{-6})}=\frac{1}{0.012566}\approx 79.58\text{Hz}$$

6. This matches option (D).