ISRO EC 2025 Q.20: solution and answer

Official answer: ASource: ISRO's final answer key of 15 May 2026

1. The output of a discrete-time LTI system is the convolution $y[n]=x[n]\ast h[n]$.

2. Convolving a signal with a shifted impulse shifts the signal: $x[n]\ast \delta [n-k]=x[n-k]$.

3. Here $h[n]=\delta [n]+\delta [n-1]$. Convolution is linear, so $y[n]=x[n]\ast \delta [n]+x[n]\ast \delta [n-1]=x[n]+x[n-1]$.

4. Put $x[n]=u[n]$: $y[n]=u[n]+u[n-1]$.

5. Check the values: $y[n]=0$ for $n<0$, $y[0]=1+0=1$, and $y[n]=1+1=2$ for $n\ge 1$. This agrees with $u[n]+u[n-1]$.

6. The other options fail. $2u[n]$ would give $y[0]=2$, but the true value is $1$. $\delta [n]+\delta [n-1]$ is only $h[n]$, the response to an impulse, not to a step. $u[n]$ alone ignores the delayed branch, so it gives $y[n]=1$ for $n\ge 1$ instead of $2$.