GATE 2025 IN Q.39: solution and answer

Official answer: CSource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Initial rest + causality: the input $x[n]=\delta [n-p]$ is zero for every $n<p$, and the system starts at rest, so $y[n]=0$ for all $n<p$. In particular $y[p-1]=0$.

2. Iterate the recursion at $n=p$: $y[p]=\frac{1}{\alpha }y[p-1]+\delta [p-p]=\frac{1}{\alpha }\cdot 0+\delta [0]=1$.

3. Iterate at $n=p+1$: $y[p+1]=\frac{1}{\alpha }y[p]+\delta [1]=\frac{1}{\alpha }\cdot 1+0=\frac{1}{\alpha }$.

4. Check with the impulse response: with $x[n]=\delta [n]$ the same recursion gives $h[n]={\alpha }^{-n}u[n]$. Equivalently, $H(z)=\frac{1}{1-{\alpha }^{-1}{z}^{-1}}$, and causality fixes the ROC as $|z|>1/\alpha$.

5. Time invariance: a shifted input gives a shifted output, so $x[n]=\delta [n-p]\Rightarrow y[n]=h[n-p]={\alpha }^{-(n-p)}u[n-p]$. Therefore $$y[p+1]=h[1]={\alpha }^{-1}=\frac{1}{\alpha }.$$

6. The value of $p$ does not matter. Because the system is time invariant, $y[p+1]$ is always $h[1]$, so the condition $p>10$ changes nothing. The condition $\alpha >1$ only makes the system stable (pole at $z=1/\alpha$, inside the unit circle) and does not affect $h[1]$ either. The answer is option (C).