Correct Answer - Option 4 : Causal and not stable
Concept:
Condition for Causality:
A system with impulse response h[n] is stable if it satisfies:
h[n] = 0 ; n < 0
Condition for stability:
A system is said to be stable if the impulse response is absolutely integrable, i.e.
\(\mathop \sum \limits_{n = - \infty }^\infty \left| {h\left[ n \right]} \right| < \infty \)
Application:
Given h[n] = cos (n) u[n]
Multiplication of u[n] ensures that h[n] = 0 for n < 0
Hence given system is causal.
\(\mathop \sum \limits_{n = - \infty }^\infty \left| {h\left[ n \right]} \right| = \mathop \sum \limits_{h = - \infty }^\infty \left| {\cos \left[ n \right]u\left[ n \right]} \right|\)
\( = \mathop \sum \limits_{h = 0}^\infty \left| {\cos \left[ n \right]} \right|\)
Since -1 ≤ cos [n] ≤ 1
|cos [n]| < 1
\(\mathop \sum \limits_{n = - \infty }^\infty \left| {h\left[ n \right]} \right| = \mathop \sum \limits_{h = 0}^\infty \left| {\cos \left[ n \right]} \right|\)
For every n, the value of cos [n] is finite but the summation is going for n → ∞, which causes the
\(\mathop \sum \limits_{h = - \infty }^\infty \left| {h\left[ n \right]} \right| \to \infty \)
So, the given system is not stable.
