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.**