GATE 2018 EC Q.15: solution and answer

Official answer: 0.12 to 0.13Source: Official answer key, GATE 2018 EC (IIT Guwahati)

1. Convert the angular frequencies to Hz: $2000\pi ,2400\pi ,2800\pi$ rad/s are $1000,1200,1400$ Hz. The middle component is the carrier, with $f_c=1200$ Hz and amplitude $A_c=4$. The outer two are the lower and upper sidebands at $f_c\mp 200$ Hz.

2. Combine the sidebands: $\cos (2000\pi t)+\cos (2800\pi t)=2\cos (2400\pi t)\cos (400\pi t)$.

3. Then $s(t)=4\cos (2400\pi t)\left[1+\frac{1}{2}\cos (400\pi t)\right]=A_c[1+m(t)]\cos (2\pi f_{c}t)$, with $A_c=4$ and $m(t)=0.5\cos (400\pi t)$. This is single-tone AM with $f_m=200$ Hz and modulation index $\mu =0.5$.

4. Carrier power: $P_c=\frac{A_c^2}{2}=\frac{16}{2}=8$.

5. Power in the message-dependent part (the two sidebands): $P_{SB}=\frac{1^2}{2}+\frac{1^2}{2}=1$. Check: $P_c{\mu }^2/2=8\times 0.25/2=1$.

6. Ratio: $\frac{P_{SB}}{P_c}=\frac{1}{8}=0.125$. The same value comes from the message $m(t)$ itself, since its power is $\frac{0.5^2}{2}=0.125$ relative to a unit-amplitude carrier term.