GATE 2025 EE Q.20: solution and answer

Official answer: ASource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Impulse response. The response of the system to $\delta [n]$ is, by definition, its impulse response: $h[n]=1$ for $n=0,1,2$ and $0$ otherwise. So $$h[n]=\delta [n]+\delta [n-1]+\delta [n-2].$$

2. Convolution sum. The output of an LTI system is the convolution of the input with the impulse response: $$y[n]=\sum _{k=-\infty }^{\infty }h[k]x[n-k].$$

3. Evaluate. Only $k=0,1,2$ contribute, and each has weight $h[k]=1$: $$y[n]=h[0]x[n]+h[1]x[n-1]+h[2]x[n-2]=x[n]+x[n-1]+x[n-2].$$

4. Check with time invariance. The input $\delta [n-1]$ must produce $h[n-1]$, which is nonzero at $n=1,2,3$. Putting $x[n]=\delta [n-1]$ into $y[n]=x[n]+x[n-1]+x[n-2]$ gives $\delta [n-1]+\delta [n-2]+\delta [n-3]$, which is nonzero at exactly $n=1,2,3$. It matches, so option (A) is correct.

5. Why the other options fail. Put $x[n]=\delta [n]$ into each one to get its impulse response. (B) gives $\delta [n-1]+\delta [n]+\delta [n+1]$, which is nonzero at $n\in \{-1,0,1\}$. (C) gives support $\{0,-1,-2\}$ and (D) gives $\{-1,-2,-3\}$. All three use a future sample such as $x[n+1]$, so their impulse responses reach into negative $n$. None of them has the given support $\{0,1,2\}$.