GATE 2022 IN Q.16: solution and answer

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

1. The modulated wave is $s(t)=A_c[1+m(t)]\cos (2\pi f_{c}t)$, so its envelope is $A_c|1+m(t)|$.

2. An envelope detector outputs this envelope. The output equals $A_c[1+m(t)]$, which is linear in $m(t)$ and so returns the message, only when $1+m(t)\ge 0$ for all $t$.

3. This requires $|m(t){|}_{\max }\le 1$, that is, a modulation index $\mu \le 1$. If $\mu >1$ (overmodulation), $1+m(t)$ becomes negative, the carrier phase reverses, and the envelope $|1+m(t)|$ no longer follows $1+m(t)$.

4. Since $f_m\ll f_c$, the envelope varies slowly compared with the carrier, so the detector can follow it. Only the amplitude condition in the previous step matters.

5. (A) $m(t)=0.5\cos (2\pi f_{m}t)$ has peak $0.5\le 1$. The envelope is $A_c(1+0.5\cos 2\pi f_{m}t)$, which stays between $0.5A_c$ and $1.5A_c$ and is always positive. The message is recovered completely.

6. (B) $m(t)=1.5\sin (2\pi f_{m}t)$ has peak $1.5>1$. At times, $1+m(t)$ falls to $-0.5$, so the output is the distorted $|1+m(t)|$.

7. (C) $m(t)=2\sin (4\pi f_{m}t)$ and (D) $m(t)=2\cos (4\pi f_{m}t)$ both have peak $2>1$, so $1+m(t)$ reaches $-1$ and they are overmodulated. The message frequency $2f_m$ does not change this condition.

8. Only option (A) satisfies $|m(t)|\le 1$, so the answer is (A).