GATE 2024 IN Q.43: solution and answer

Official answer: ASource: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. Assumption: ideal op-amp. The non-inverting input is grounded, so the virtual short holds the inverting input at $0\text{V}$, and no current flows into the op-amp's input terminals.

2. Feedback impedance. The feedback network is $R_f$ in parallel with $C$: $$Z_f=R_f\parallel \frac{1}{j\omega C}=\frac{R_f}{1+j\omega R_{f}C}.$$

3. Transfer function. For an inverting amplifier, $\frac{V_{out}}{V_{in}}=-\frac{Z_f}{R}$, so $$\frac{V_{out}}{V_{in}}=-\frac{R_f}{R}\cdot \frac{1}{1+j\omega R_{f}C}.$$ This is a first-order low-pass filter.

4. DC gain fixes $R_f$. At $\omega =0$ the capacitor is an open circuit, so the gain is $-\frac{R_f}{R}=-10$. That gives $R_f=10R=10\text{k}\Omega$.

5. 3 dB frequency. The gain magnitude falls to $\frac{1}{\sqrt{2}}$ of its dc value when $\omega R_{f}C=1$, so $f_{3\text{dB}}=\frac{1}{2\pi R_{f}C}$. The pole is set by $R_f$ and $C$ only. The input resistor $R$ changes the gain, not the cut-off.

6. $$f_{3\text{dB}}=\frac{1}{2\pi (10^4\Omega )(0.1\times 10^{-6}\text{F})}=\frac{1}{2\pi \times 10^{-3}\text{s}}=159.15\text{Hz}$$

7. Matching the options. Strict rounding of $159.155\text{Hz}$ to one decimal place gives $159.2\text{Hz}$, but no option prints that. Option (A), $159.1\text{Hz}$, is the same value cut off after one decimal place. All four options were produced the same way. (B) $1591.5\text{Hz}$ is $\frac{1}{2\pi RC}$, which uses the wrong resistor $R=1\text{k}\Omega$ (exact value $1591.55$). (C) $1750.7\text{Hz}$ is $\frac{1}{2\pi (R\parallel R_f)C}$ with $R\parallel R_f=909.1\Omega$. (D) $175.0\text{Hz}$ is one-tenth of (C) (exact value $175.07$). Only (A) comes from the correct expression $\frac{1}{2\pi R_{f}C}$, so the answer is (A).