GATE 2025 EC Q.17: solution and answer

Official answer: DSource: Official answer key, GATE 2025 EC (IIT Roorkee)

1. Input–output relation. The output of Subsystem 1 is $x[n]\ast h_1[n]$. After the first adder the signal is $x[n]\ast h_1[n]+b\delta [n]$. Subsystem 2 is LTI, so its response to a sum is the sum of its responses, and $\delta [n]\ast h_2[n]=h_2[n]$:
   $$y[n]=x[n]\ast h_1[n]\ast h_2[n]+b{h}_2[n]-b\delta [n]$$

2. Write $h[n]=h_1[n]\ast h_2[n]$ and $g[n]=b\left(h_2[n]-\delta [n]\right)$, so
   $$y[n]=x[n]\ast h[n]+g[n]$$
   $g[n]$ does not depend on the input. Since $b>0$ and $h_2[n]\ne \delta [n]$, $g[n]$ is not the zero signal. The $+b\delta [n]$ and $-b\delta [n]$ do not cancel, because the first one passes through Subsystem 2 and comes out as $b{h}_2[n]$.

3. Linearity (zero in, zero out). Put $x[n]=0$. Then $y[n]=g[n]\ne 0$. A linear system must give zero output for zero input (homogeneity with scale factor $0$), so the system is nonlinear. Superposition gives the same result: input $x_1+x_2$ produces $x_1\ast h+x_2\ast h+g$, while $y_1+y_2=x_1\ast h+x_2\ast h+2g$, and these differ because $g\ne 0$.

4. Time invariance (shift test). Apply $x[n-n_0]$. The LTI part delays its output by the same amount, so the response is $x[n-n_0]\ast h[n]+g[n]$. The original output delayed by $n_0$ is $y[n-n_0]=x[n-n_0]\ast h[n]+g[n-n_0]$.

5. The two match for every input and every $n_0$ only if $g[n-n_0]=g[n]$ for all $n$ and $n_0$. That means $g[n]$ has to be a constant sequence. A constant added offset would not break time invariance, so the question to settle is whether $g[n]$ is constant.

6. $g[n]$ is not constant. Suppose $g[n]=K$ for every $n$. Since $g\ne 0$, $K\ne 0$, and then $h_2[n]=\delta [n]+\frac{K}{b}$. That impulse response keeps the same nonzero value at every $n$, forever in both directions: it never dies out, it is not absolutely summable, and it is nonzero for $n<0$. This degenerate case is the only way out, and it is not what the question means by an LTI subsystem. For any $h_2[n]$ whose samples die out as $|n|\to \infty$, $g[n]\to 0$ while $g$ is nonzero somewhere, so $g$ cannot be constant. In other words, $g[n]$ is tied to the time origin through $\delta [n]$ and $h_2[n]$ and does not move when the input is delayed. So for some $n_0$, $y[n-n_0]$ differs from the response to $x[n-n_0]$, and the system is time variant.

7. The system is nonlinear and time variant, so the answer is option (D).