GATE EC 2017 Set 1 — Question 6

Let's analyze the causality and stability of the given discrete-time system.Causality:
A system is causal if its output $y[n]$ at any time $n$ depends only on the present input $x[n]$ and past inputs $x[n-k]$ (where $k > 0$).The given system is defined as:
$y[n] = x[n] - x[n-1]$ for $0 \leq n \leq 10$
$y[n] = 0$ otherwiseFor $0 \leq n \leq 10$, the output $y[n]$ depends on $x[n]$ (present input) and $x[n-1]$ (past input). For $n < 0$ or $n > 10$, the output is $0$, which does not depend on future inputs. Since the output never depends on future inputs, the system is causal.Stability (BIBO Stability):
A discrete-time LTI system is BIBO stable if every bounded input produces a bounded output. This condition is met if and only if the impulse response $h[n]$ is absolutely summable, i.e., $\sum_{n=-\infty}^{\infty} |h[n]| < \infty$.To find the impulse response $h[n]$, we set $x[n] = \delta[n]$ (the unit impulse).
For $0 \leq n \leq 10$:
$h[n] = \delta[n] - \delta[n-1]$Let's evaluate $h[n]$ for different values of $n$ within the range $0 \leq n \leq 10$:
$h[0] = \delta[0] - \delta[-1] = 1 - 0 = 1$
$h[1] = \delta[1] - \delta[0] = 0 - 1 = -1$
$h[2] = \delta[2] - \delta[1] = 0 - 0 = 0$
...
$h[10] = \delta[10] - \delta[9] = 0 - 0 = 0$And $h[n] = 0$ for $n < 0$ or $n > 10$.So, the impulse response is:
$h[n] = \begin{cases} 1, & n=0 \\ -1, & n=1 \\ 0, & \text{otherwise} \end{cases}$Now, let's check for absolute summability:
$\sum_{n=-\infty}^{\infty} |h[n]| = |h[0]| + |h[1]| + \sum_{n \neq 0,1} |h[n]|$
$= |1| + |-1| + 0$
$= 1 + 1 = 2$Since $\sum_{n=-\infty}^{\infty} |h[n]| = 2 < \infty$, the system is BIBO stable.Therefore, the system is both causal and stable.The final answer is $\boxed{\text{A}}$