GATE 2022 EC Q.46: solution and answer

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

1. Eigenfunction property of LTI systems: write $\sin (t)=\frac{e^{jt}-e^{-jt}}{2j}$. An LTI system maps $e^{j\omega t}$ to $H(j\omega )e^{j\omega t}$, so its response to $\sin (t)$ is $\frac{H(j1)e^{jt}-H(-j1)e^{-jt}}{2j}$. For a real system this is $|H(j1)|\sin \left(t+\angle H(j1)\right)$. Either way the output contains only the same frequency $\omega =1\text{rad/s}$, scaled and phase-shifted. An LTI system cannot create a new frequency or a DC term.

2. $S_1$: $\sin (-t)=-\sin (t)=\sin (t+\pi )$. This is the same frequency with gain $|H(j1)|=1$ and phase $\pi$. The LTI system $y(t)=-x(t)$ produces exactly this output. (So could the time reversal $y(t)=x(-t)$, which is not time-invariant, but one input cannot tell the two apart.) So $S_1$ is not definitely non-LTI.

3. $S_2$: $\sin (t+1)$ is the input phase-shifted by $1\text{rad}$ at the same frequency. The LTI system $y(t)=x(t+1)$ (a time advance with $H(j\omega )=e^{j\omega }$) produces it. That system is non-causal, but causality is not part of being LTI. So $S_2$ is not definitely non-LTI.

4. $S_3$: $\sin (2t)$ has frequency $2\text{rad/s}$, and that frequency is not in the input. No LTI system can do this, so $S_3$ is definitely NOT LTI.

5. $S_4$: $${\sin }^2(t)=\frac{1}{2}-\frac{1}{2}\cos (2t)$$ This has a DC term and a $2\text{rad/s}$ component, and neither is in the input. So $S_4$ is definitely NOT LTI.

6. Cross-check without frequency response. Since $\sin (t-\pi )=-\sin (t)$, the input delayed by $\pi$ is the same signal as the input negated. If a system is LTI and $\sin (t)\to y(t)$, time-invariance says the response to $\sin (t-\pi )$ is $y(t-\pi )$, and linearity says the response to $-\sin (t)$ is $-y(t)$. So we need $y(t-\pi )=-y(t)$ for all $t$. $S_1$: $\sin (\pi -t)=\sin (t)=-\sin (-t)$, which holds. $S_2$: $\sin (t+1-\pi )=-\sin (t+1)$, which holds. $S_3$: $\sin (2t-2\pi )=\sin (2t)\ne -\sin (2t)$, which fails. $S_4$: ${\sin }^2(t-\pi )={\sin }^2(t)\ne -{\sin }^2(t)$, which fails. This agrees with steps 2 to 5.

7. One input-output pair can prove that a system is not LTI, but it can never prove that a system is LTI. That is why the question says "definitely NOT": $S_1$ and $S_2$ are consistent with an LTI system, while $S_3$ and $S_4$ rule one out. Answer: (C) and (D).