GATE EC 2026 Set 1 — Question 47

Let's analyze the system in the Z-domain. The input is $X(z)$ and the output is $Y(z)$.The upper path: $X(z)H_1(z)$.The lower path:
Input to the first summer is $X(z)$.
Output of the first summer is $X(z) + Y(z)z^{-1}$. (Feedback from output $y[n]$ delayed by one sample)
This is then multiplied by $H_2(z)$, so $(X(z) + Y(z)z^{-1})H_2(z)$.
This signal is then delayed by $z^{-1}$, so $(X(z) + Y(z)z^{-1})H_2(z)z^{-1}$.The final output $Y(z)$ is the sum of the upper path and the lower path:
$Y(z) = X(z)H_1(z) + (X(z) + Y(z)z^{-1})H_2(z)z^{-1}$
$Y(z) = X(z)H_1(z) + X(z)H_2(z)z^{-1} + Y(z)H_2(z)z^{-2}$Rearrange to find the transfer function $H(z) = Y(z)/X(z)$:
$Y(z) - Y(z)H_2(z)z^{-2} = X(z)H_1(z) + X(z)H_2(z)z^{-1}$
$Y(z)(1 - H_2(z)z^{-2}) = X(z)(H_1(z) + H_2(z)z^{-1})$
$H(z) = \frac{Y(z)}{X(z)} = \frac{H_1(z) + H_2(z)z^{-1}}{1 - H_2(z)z^{-2}}$Since $H_1[n]$ and $H_2[n]$ are impulse responses of linear and time-invariant (LTI) systems, their Z-transforms $H_1(z)$ and $H_2(z)$ represent LTI systems. The operations of addition, multiplication by $z^{-k}$ (delay), and feedback (which results in a rational transfer function) all preserve linearity and time-invariance. Therefore, the overall system S is also linear and time-invariant.Regarding causality:
For the system to be causal, the impulse response $h[n]$ must be zero for $n < 0$. This means all poles of $H(z)$ must be inside the unit circle for a stable and causal system, or the region of convergence (ROC) must be outside the outermost pole for a causal system.The transfer function is $H(z) = \frac{H_1(z) + H_2(z)z^{-1}}{1 - H_2(z)z^{-2}}$.
If $H_1(z)$ and $H_2(z)$ are causal, then $H_1(z)$ and $H_2(z)z^{-1}$ are causal. The numerator is causal.
The denominator $1 - H_2(z)z^{-2}$ also corresponds to a causal system if $H_2(z)$ is causal.
However, the presence of feedback can make a system non-causal if the feedback loop contains an advance element or if the system is unstable and the ROC for causality does not exist. In this case, the feedback is $Y(z)z^{-1}$ which is a delay, and the overall system has $z^{-2}$ in the denominator. This structure is typical of recursive systems, which are generally causal if the individual components are causal and the system is stable.However, the question asks what is necessarily TRUE. While the system can be causal, it's not necessarily causal without more information about $H_1(z)$ and $H_2(z)$ (e.g., if they are FIR or IIR, or if the system is stable). For example, if $H_2(z)$ has poles outside the unit circle, the system might be unstable, and a causal ROC might not exist or might not be the one chosen.However, the linearity and time-invariance are preserved by the interconnection of LTI systems. The operations (summation, scaling, delay) are all linear and time-invariant operations. Therefore, the overall system S is necessarily linear and time-invariant.The final answer is $\boxed{B}$.