GATE 2022 EC Q.23: solution and answer

Official answer: CSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

1. What the figure gives. Only the magnitude is specified: $|H(f)|=1$ for $|f|\le \alpha$ and $0$ outside. The phase $\theta (f)=\angle H(f)$ is not given. So inside the band the response can be $H(f)=e^{j\theta (f)}$ with any phase function $\theta (f)$.

2. What a pure delay needs. A pure delay $y(t)=x(t-t_0)$ means $Y(f)=X(f)e^{-j2\pi f{t}_0}$. For every input bandlimited to $|f|\le \alpha$ this requires $H(f)=e^{-j2\pi f{t}_0}$ across the band: unit magnitude and a phase that is linear in $f$ and passes through the origin, $\theta (f)=-2\pi f{t}_0$. In particular $H(0)=1$.

3. Counter-example for Statement I. Take $H(f)=-1$ for $|f|\le \alpha$ and $0$ outside, i.e. $h(t)=-2\alpha \text{sinc}(2\alpha t)$, a real impulse response. Its magnitude is exactly the figure. For any input bandlimited to $|f|\le \alpha$ the output is $y(t)=-x(t)$. A pure delay would need $H(0)=e^0=1$, but here $H(0)=-1$. Concretely, the constant input $x(t)=1$ (spectrum $\delta (f)$, inside the band) gives $y(t)=-1$, while $x(t-t_0)=1$ for every $t_0$. Any other phase that is not linear through the origin fails in the same way (a Hilbert-type phase $-\frac{\pi }{2}\text{sgn}(f)$, for example). The magnitude alone cannot force a pure delay, so Statement I is incorrect.

4. PSD through an LTI system. For a WSS input, $R_Y(\tau )=h(\tau )\ast h^{\ast }(-\tau )\ast R_X(\tau )$. Taking Fourier transforms (the PSD is the Fourier transform of the autocorrelation) gives $$S_Y(f)=H(f)H^{\ast }(f)S_X(f)=|H(f){|}^2{S}_X(f).$$ Only $|H(f)|$ appears, so the phase has no effect.

5. Apply it to the band. For $-\alpha \le f\le \alpha$ we have $|H(f){|}^2=1$, so $S_Y(f)=S_X(f)$ for every WSS input, whatever the phase is. (In the counter-example, $y=-x$ gives $R_Y(\tau )=R_X(\tau )$, which agrees.) Statement II is correct.

6. Statement I is incorrect and Statement II is correct, so the answer is option (C).