GATE 2023 EC Q.59: solution and answer

Official answer: 0.24 to 0.26Source: Official answer key, GATE 2023 EC (IIT Kanpur)

1. The input is white noise with two-sided power spectral density $S_X(f)=\frac{N_0}{2}=\frac{1}{2}\text{W/Hz}$, flat for all $f$.

2. The output PSD is $S_Y(f)=S_X(f)|H(f){|}^2$.

3. The impulse response $h(t)=e^{-t}u(t)$ has frequency response $H(f)=\frac{1}{1+j2\pi f}$, so $|H(f){|}^2=\frac{1}{1+4{\pi }^2{f}^2}$.

4. The output power is $$P_Y={\int }_{-\infty }^{\infty }{S}_Y(f)df=\frac{1}{2}{\int }_{-\infty }^{\infty }\frac{df}{1+4{\pi }^2{f}^2}.$$

5. By Parseval's theorem, $${\int }_{-\infty }^{\infty }|H(f){|}^{2}df={\int }_{-\infty }^{\infty }{h}^2(t)dt={\int }_0^{\infty }{e}^{-2t}dt=\frac{1}{2}.$$

6. Therefore $P_Y=\frac{1}{2}\times \frac{1}{2}=0.25\text{W}$.

7. Check with the autocorrelation: $R_Y(\tau )=\frac{N_0}{2}(h(\tau )\ast h(-\tau ))$, so $R_Y(0)=\frac{N_0}{2}\int h^2(t)dt=0.25\text{W}$. Direct integration agrees: ${\int }_{-\infty }^{\infty }\frac{df}{1+4{\pi }^2{f}^2}=\frac{1}{2\pi }\cdot \pi =\frac{1}{2}$.