GATE 2022 EC Q.51: solution and answer

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

1. A de-emphasis filter must undo the pre-emphasis, so $H_{de}(\omega )=\frac{1}{H_{pe}(\omega )}=\frac{1}{1+j\omega /{\omega }_0}$.

2. The network is a series $R$ with the output taken across the shunt $C$, so by the voltage divider $H(\omega )=\frac{1/(j\omega C)}{R+1/(j\omega C)}=\frac{1}{1+j\omega RC}$.

3. Comparing the two forms, $RC=\frac{1}{{\omega }_0}=\frac{1}{10^4}=10^{-4}\text{s}$.

4. (A) $1\text{k}\Omega \times 0.1\mu \text{F}=10^3\times 10^{-7}=10^{-4}\text{s}$, which matches.

5. (B) $2\times 10^3\times 10^{-6}=2\times 10^{-3}\text{s}$, which does not match.

6. (C) $10^3\times 2\times 10^{-6}=2\times 10^{-3}\text{s}$, which does not match.

7. (D) $2\times 10^3\times 0.5\times 10^{-6}=10^{-3}\text{s}$, which does not match.

8. Only (A) gives $RC=10^{-4}\text{s}$, so (A) is the only correct pair.