---
title: "Stability range for h[n] = e^(an)u[n] + e^(bn)u[−n]"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:stability-range-two-sided"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/stability-range-two-sided
site: BCT Engineering
---

# Stability range for h[n] = e^(an)u[n] + e^(bn)u[−n]

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2078 Poush, 4 marks
- Video: [Stable and Unstable Systems](https://www.youtube.com/watch?v=ngJ2QWvMPbI) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/stability-range-two-sided
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Determine the range of a and b for which the given LTI system becomes stable whose impulse response is: h[n] = e^(an) u[n] + e^(bn) u[−n], where u[n] is discrete-time unit step function.

## Short answer

For BIBO stability, the geometric series for the causal part requires a < 0, and for the anti-causal part requires b > 0.

## Answer

An LTI system is **BIBO stable** if and only if its impulse response is absolutely summable:

$$
S=\sum_{n=-\infty}^{\infty}|h[n]|<\infty
$$

Here $h[n]=e^{an}u[n]+e^{bn}u[-n]$ (take $a,b$ real). Split the sum:

$$
\begin{aligned}
S &= \sum_{n=0}^{\infty} e^{an} + \sum_{n=-\infty}^{0} e^{bn} \\
&= \sum_{n=0}^{\infty} (e^{a})^{n} + \sum_{m=0}^{\infty} (e^{-b})^{m} \qquad (m=-n)
\end{aligned}
$$

(At $n=0$ both terms are present; that adds only a finite value 2, so it does not affect convergence.)

#### Right-sided (causal) part
A geometric series $\sum r^n$ converges only if $|r|<1$:

$$
e^{a}<1 \;\Rightarrow\; a<0
$$

Its sum is then $\frac{1}{1-e^{a}}$.

#### Left-sided (anti-causal) part

$$
e^{-b}<1 \;\Rightarrow\; -b<0 \;\Rightarrow\; b>0
$$

Its sum is then $\frac{1}{1-e^{-b}}$.

#### Result
Both series must converge, so

$$
S=\frac{1}{1-e^{a}}+\frac{1}{1-e^{-b}}<\infty \quad \text{only when } a<0 \text{ and } b>0
$$

**Answer: the system is stable for $a<0$ and $b>0$** (if $a,b$ are complex: $\mathrm{Re}(a)<0$ and $\mathrm{Re}(b)>0$). Physically, the part for $n\ge0$ must decay as $n\to+\infty$ and the part for $n\le0$ must decay as $n\to-\infty$. Note that the system is non-causal because $h[n]\ne0$ for $n<0$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. BIBO stability of an LTI system requires:
   A. h[n] = 0 for n < 0
   B. sum of |h[n]| < infinity
   C. h[n] bounded
   D. h[0] = 1
   Answer: B. An absolutely summable impulse response is the necessary and sufficient condition. (Key point: Stability criterion)

2. For the causal part e^(an)u[n], the series sum (e^a)^n converges only if:
   A. a > 0
   B. a < 0
   C. a = 0
   D. any a
   Answer: B. A geometric series needs e^a < 1, which means a < 0. (Key point: Condition on a)

3. For the left-sided part e^(bn)u[-n], putting m = -n gives sum (e^(-b))^m, which converges only if:
   A. b < 0
   B. b > 0
   C. b = 0
   D. any b
   Answer: B. e^(-b) < 1 requires b > 0. (Key point: Condition on b)

4. Is the system causal?
   A. Yes, because a < 0
   B. No, because h[n] is nonzero for n < 0
   C. Yes, because it is stable
   D. Only when b = 0
   Answer: B. The term e^(bn)u[-n] is nonzero for negative n, so the system is non-causal. (Key point: Causality)

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