GATE 2022 IN Q.56: solution and answer

Official answer: 15.80 to 16.00Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. Assume an ideal op-amp (infinite open-loop gain, infinite input impedance). The + input is grounded, so by the virtual short the − input sits at 0 V and no current enters the op-amp. The circuit is an inverting amplifier with input impedance $Z_1=R_1=10\text{k}\Omega$ and feedback impedance $Z_f=R_f\parallel \frac{1}{j\omega C}$, where $R_f=100\text{k}\Omega$ and $C=1\text{nF}$.

2. Feedback impedance: $$Z_f=\frac{R_f\cdot \frac{1}{j\omega C}}{R_f+\frac{1}{j\omega C}}=\frac{R_f}{1+j\omega R_{f}C}$$ so the inverting-amplifier gain is $$\frac{V_{out}}{V_{in}}=-\frac{Z_f}{R_1}=-\frac{R_f/R_1}{1+j\omega R_{f}C}$$ This is a first-order low-pass (lossy integrator) with DC gain magnitude $R_f/R_1=10$, i.e. $20\text{dB}$.

3. Time constant: $\tau =R_{f}C=(100\times 10^3)(1\times 10^{-9})=10^{-4}\text{s}$.

4. Magnitude of the gain: $$\left|\frac{V_{out}}{V_{in}}\right|=\frac{10}{\sqrt{1+(\omega \tau {)}^2}}$$

5. 0 dB means $|V_{out}/V_{in}|=1$: $$\sqrt{1+(\omega \tau {)}^2}=10\Rightarrow (\omega \tau {)}^2=99\Rightarrow \omega \tau =\sqrt{99}\approx 9.9499$$

6. $\omega =\frac{9.9499}{10^{-4}}\approx 9.9499\times 10^4\text{rad/s}$, so $$f=\frac{\omega }{2\pi }=\frac{99498.7}{2\pi }\approx 15835.7\text{Hz}$$

7. Rounded to two decimal places in kilohertz: $f\approx 15.84\text{kHz}$.