GATE 2022 IN Q.33: solution and answer

Official answer: 6.22 to 6.42Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. Identify the topology. $L_0$, $C_0$ and $R_0$ are all in series with the input source, and the output is taken across $R_0$. By voltage division, $$H(j\omega )=\frac{V_o}{V_i}=\frac{R_0}{R_0+j\left(\omega L_0-\frac{1}{\omega C_0}\right)}$$ This is a second-order bandpass response: $|H|\to 0$ as $\omega \to 0$ (capacitor open) and as $\omega \to \infty$ (inductor open), with its peak $|H|=1$ at series resonance.

2. Centre (resonant) frequency. At resonance the reactances cancel, ${\omega }_0{L}_0=\frac{1}{{\omega }_0{C}_0}$: $${\omega }_0=\frac{1}{\sqrt{L_0{C}_0}}=\frac{1}{\sqrt{(10^{-3})(10\times 10^{-9})}}=\frac{1}{\sqrt{10^{-11}}}\approx 3.162\times 10^5\text{rad/s}$$

3. Bandwidth (half-power points). $|H|=1/\sqrt{2}$ when $\left|\omega L_0-\frac{1}{\omega C_0}\right|=R_0$, i.e. $L_0{\omega }^2\mp R_0\omega -\frac{1}{C_0}=0$. Taking the positive root of each quadratic, $${\omega }_{2,1}=\frac{\pm R_0+\sqrt{R_0^2+4L_0/C_0}}{2L_0}$$ so $$B={\omega }_2-{\omega }_1=\frac{R_0}{L_0}=\frac{50}{10^{-3}}=5\times 10^4\text{rad/s}$$

4. Q of a series RLC. $$Q=\frac{{\omega }_0}{B}=\frac{1/\sqrt{L_0{C}_0}}{R_0/L_0}=\frac{1}{R_0}\sqrt{\frac{L_0}{C_0}}$$

5. Substitute. $$\sqrt{\frac{L_0}{C_0}}=\sqrt{\frac{10^{-3}}{10^{-8}}}=\sqrt{10^5}=316.23\Omega ,Q=\frac{316.23}{50}=6.3246$$

6. Round to two decimal places: $Q\approx 6.32$. Check: ${\omega }_0/B=(3.1623\times 10^5)/(5\times 10^4)=6.32$.