GATE 2025 EC Q.28: solution and answer

Official answer: A, DSource: Official answer key, GATE 2025 EC (IIT Roorkee)

1. An envelope detector (with $f_c\gg W$) outputs the magnitude of the slowly varying amplitude factor: for $x(t)=a(t)\cos (2\pi f_{c}t)$ the output is $|a(t)|$.

2. DSB-SC: $a(t)=A_{c}m(t)$, so the output is $A_c|m(t)|$. If $m(t)>0$ for all $t$, then $|m(t)|=m(t)$, so the output is $A_{c}m(t)$ and the message is recovered. Statement (i) is TRUE, so (A) is correct and (B) is wrong.

3. AM: $a(t)=A_c(1+\mu m(t))$, so the output is $A_c|1+\mu m(t)|$. This equals $A_c(1+\mu m(t))$, which is the message plus a DC term, exactly when $1+\mu m(t)\ge 0$ for all $t$. That is the no-overmodulation condition $\mu |m(t){|}_{\max }\le 1$ for a message with negative values.

4. This condition does not need $m(t)>0$. Take $m(t)=\cos (2\pi Wt)$ and $\mu \le 1$. Then $1+\mu m(t)\ge 1-\mu \ge 0$, so the envelope detector works, although $m(t)$ is negative for part of each cycle.

5. So the claim that an envelope detector works in AM only if $m(t)>0$ for all $t$ is not necessary. Statement (ii) is FALSE, so (D) is correct and (C) is wrong.

6. Correct options: (A) and (D).