Question

If the Impulse response is $h[n] = (0.5)^n u(n)$ then, the system is

• A. unstable
• B. stable
• C. marginally stable
• D. time-varying

Hint:

For any causal first-order LTI system with an impulse response of the form $h[n] = a^n u(n)$, the system is BIBO stable if and only if the pole lies inside the unit circle, meaning $|a| < 1$.
Here, $|a| = 0.5 < 1$, which immediately guarantees system stability.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question asks us to analyze the stability of a discrete-time linear time-invariant (LTI) system from its given impulse response $h[n]$.
A discrete-time LTI system is bounded-input bounded-output (BIBO) stable if and only if its impulse response is absolutely summable.

Step 2: Key Formula or Approach:
The mathematical condition for absolute summability of a discrete-time impulse response $h[n]$ is:
\[ S = \sum_{n=-\infty}^{\infty} |h[n]| < \infty \]

Step 3: Detailed Explanation:

• We are given the impulse response of the system as:
\[ h[n] = (0.5)^n u(n) \]

• The unit step function $u[n]$ is defined as equal to $1$ for $n \geq 0$ and $0$ for $n < 0$.

• Thus, we can rewrite the absolute summation limits from $0$ to $\infty$:
\[ S = \sum_{n=0}^{\infty} |(0.5)^n| \]

• Since the base $0.5$ is positive, we can omit the absolute value signs:
\[ S = \sum_{n=0}^{\infty} (0.5)^n \]

• This expression represents an infinite geometric series with the first term $a = 1$ and the common ratio $r = 0.5$.

• The sum of an infinite geometric series converges to a finite value if and only if the absolute value of the common ratio is strictly less than 1 ($|r| < 1$).

• Since $|0.5| < 1$, the series converges, and its sum is given by:
\[ S = \frac{a}{1 - r} = \frac{1}{1 - 0.5} = \frac{1}{0.5} = 2 \]

• Since the sum $S = 2$ is finite ($S < \infty$), the system satisfies the absolute summability condition.

• Therefore, the LTI system is BIBO stable.

Step 4: Final Answer
Thus, the system is stable, which corresponds to option (B).