GATE 2022 EC Q.26: solution and answer

Official answer: B, CSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

1. Assume an ideal op-amp working in its linear region (the input amplitude is small enough that the output stays inside the $\pm 15\text{V}$ supplies). The inputs are then virtually shorted, and because the $+$ input is grounded, the $-$ input sits at $0\text{V}$ (virtual ground).

2. Input impedance: the capacitor and the $1\text{k}\Omega$ resistor are in series, so $Z_1=R_1+\frac{1}{j\omega C}$ with $R_1=1\text{k}\Omega$ and $C=1\mu \text{F}$. The feedback impedance is $Z_f=R_f=2\text{k}\Omega$.

3. Inverting-amplifier gain: $$\frac{V_{out}}{V_{in}}=-\frac{Z_f}{Z_1}=-\frac{R_f}{R_1+\frac{1}{j\omega C}}=-\frac{R_f}{R_1}\cdot \frac{j\omega R_{1}C}{1+j\omega R_{1}C}$$

4. Limiting cases: as $\omega \to 0$ the capacitor acts as an open circuit and $|A|\to 0$. As $\omega \to \infty$ it acts as a short and $|A|\to R_f/R_1=2$. The gain rises from zero to a constant maximum, so the circuit is a first-order high-pass filter. (B) is correct and (A) is wrong.

5. Magnitude: $|A(j\omega )|=2\frac{\omega R_{1}C}{\sqrt{1+(\omega R_{1}C{)}^2}}$. The maximum is $2$. The gain is 3 dB below that maximum when $|A|=2/\sqrt{2}$, which happens when $\omega R_{1}C=1$.

6. 3 dB frequency: $${\omega }_{3\text{dB}}=\frac{1}{R_{1}C}=\frac{1}{(10^3)(10^{-6})}=1000\text{rad/s}$$ So (C) is correct. $R_f$ sets only the passband gain and has no effect on the corner, so the $1000/3\text{rad/s}$ in (D) is wrong.