GATE 2026 EC Q.47: solution and answer

Official answer: BSource: Official answer key, GATE 2026 EC (IIT Guwahati)

1. Name the internal signals. Let $w[n]=(h_1\ast x)[n]$ be the output of $h_1$.

2. Lower summing junction: its '+' input is $x[n]$ and its '−' input is $w[n]$ passed through $z^{-1}$. So $v[n]=x[n]-w[n-1]$. This drives $h_2$, so $u[n]=(h_2\ast v)[n]$.

3. Output summing junction: its '+' input is $w[n]$ and its '−' input is $u[n]$ passed through $z^{-1}$. So $y[n]=w[n]-u[n-1]$.

4. Every block ($h_1$, $h_2$, $z^{-1}$) is LTI, and summing junctions and pick-offs are LTI too. There is no feedback loop: signals only flow forward from $x$ to $y$. Series, parallel and summing connections of LTI systems give an LTI system. To see it directly, set $x[n]=\delta [n]$: then $w[n]=h_1[n]$, $v[n]=\delta [n]-h_1[n-1]$ and $u[n]=h_2[n]-(h_1\ast h_2)[n-1]$. The overall impulse response is $$h[n]=h_1[n]-h_2[n-1]+(h_1\ast h_2)[n-2]$$

5. In the $z$-domain (where the transforms exist), the same result is $$\begin{array}{rl}Y(z)& =H_{1}X-z^{-1}{H}_2\left(X-z^{-1}{H}_{1}X\right)\\ H(z)& =H_1(z)-z^{-1}{H}_2(z)+z^{-2}{H}_1(z)H_2(z)\end{array}$$ So $y[n]=(h\ast x)[n]$ is a convolution with one fixed impulse response. That makes $S$ linear and time invariant, which is option (B), and it rules out (C) and (D).

6. Causality: nothing says $h_1[n]$ or $h_2[n]$ is zero for $n<0$. For example, take $h_1[n]=\delta [n+1]$, so $H_1(z)=z$. Then $$H(z)=z-z^{-1}{H}_2(z)+z^{-2}\cdot z{H}_2(z)=z$$ This holds for every $h_2$, so $y[n]=x[n+1]$, which is non-causal. $S$ is causal when both $h_1$ and $h_2$ are causal, but the question does not say they are. So (A) is not necessarily true.

7. Only (B) holds for every LTI $h_1$ and $h_2$.