GATE 2024 EC Q.28: solution and answer

Official answer: 2.49 to 2.51Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Write the carrier term as $A\cos (400\pi t)$, so the carrier frequency is $f_c=400\pi /2\pi =200\text{Hz}$. Its power across $1\Omega$ is $P_c=\frac{A^2}{2}=50\text{W}$, so $A=10\text{V}$.

2. The other two terms have frequencies $360\pi /2\pi =180\text{Hz}$ and $440\pi /2\pi =220\text{Hz}$. These are the lower and upper sidebands at $f_c\mp 20\text{Hz}$, each of amplitude $B$.

3. A cosine of amplitude $B$ across $1\Omega$ has power $\frac{B^2}{2}$. The total sideband power is $P_s=2\cdot \frac{B^2}{2}=B^2$.

4. The total power is $P_t=P_c+P_s=50+B^2$.

5. The given ratio is $\frac{P_s}{P_t}=\frac{B^2}{50+B^2}=\frac{1}{9}$.

6. Cross-multiplying gives $9B^2=50+B^2$, so $8B^2=50$ and $B^2=6.25$.

7. Hence $B=2.50\text{V}$, taking the positive root as an amplitude. Check: $\mu =2B/A=0.5$ and $P_s/P_c={\mu }^2/2=0.125=1/8$, so $P_s/P_t=\frac{1/8}{1+1/8}=\frac{1}{9}$.