---
title: "Energy or power: x(t) = e^(−a|t|)"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:energy-power-two-sided-exponential"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/energy-power-two-sided-exponential
site: BCT Engineering
---

# Energy or power: x(t) = e^(−a|t|)

Signals and Systems · Chapter 1: Signals

- Asked in: 2082 Chaitra, 4 marks
- Video: [Energy and Power Signals | Solved Problems / Examples](https://www.youtube.com/watch?v=bZgQD_7d5VM) — Shrenik Jain
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/energy-power-two-sided-exponential
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Determine whether the given signal is energy signal or power signal. x(t) = e^(−a|t|), a > 0

## Short answer

Compute energy: E = ∫ e^(-2a|t|) dt = 2∫0∞ e^(-2at) dt = 1/a, finite and non-zero. Average power P = lim_{T→∞} (1/(2T))∫_(-T)^T e^(-2a|t|) dt = 0. Therefore x(t)=e^(-a|t|) is an energy signal with E=1/a.

## Answer

$x(t) = e^{-a|t|}$, $a > 0$, is a two-sided decaying exponential (even signal).

$|x(t)|^2 = e^{-2a|t|}$.

**Energy:** using even symmetry,

$$
\begin{aligned}
E &= \int_{-\infty}^{\infty} e^{-2a|t|}\,dt = 2\int_{0}^{\infty} e^{-2at}\,dt \\
&= 2\left[\frac{e^{-2at}}{-2a}\right]_0^{\infty} = 2\cdot\frac{1}{2a} = \frac{1}{a}
\end{aligned}
$$

This is finite and non-zero since $a > 0$.

**Power:**

$$
P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}e^{-2a|t|}\,dt = \lim_{T\to\infty}\frac{1}{2T}\cdot\frac{1-e^{-2aT}}{a} = 0
$$

**Answer:** $x(t) = e^{-a|t|}$ is an **energy signal** with $E = 1/a$ and $P = 0$.

## Check yourself

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

1. For x(t) = e^(-a|t|), E = integral of e^(-2a|t|) dt is computed as:
   A. Integral of e^(-2at) from 0 to infinity only
   B. 2 times integral of e^(-2at) from 0 to infinity
   C. Integral of e^(2at) from -infinity to 0 minus the other half
   D. Zero by oddness
   Answer: B. The squared signal is even, so the integral is twice the one-sided integral. (Key point: Using even symmetry)

2. The energy is:
   A. 1/(2a)
   B. 1/a
   C. 2a
   D. Infinite
   Answer: B. 2 x 1/(2a) = 1/a, finite and non-zero for a > 0. (Key point: Energy value)

3. The average power is:
   A. 1/a
   B. 1/2
   C. 0
   D. Infinite
   Answer: C. Finite energy divided by 2T tends to zero as T grows. (Key point: Power value)

4. The signal is classified as:
   A. Power signal
   B. Energy signal
   C. Neither
   D. Both
   Answer: B. 0 < E < infinity and P = 0 define an energy signal. (Key point: Classification)

## Close variants set in other exams

- [Energy or power: x(t) = e^(−at)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aenergy-power-one-sided-exponential) — Signals and Systems, chapter 1, asked 6×

## Related in Signals and Systems

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- [Energy or power: x(t) = e^(−3t)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aenergy-power-decaying-exponential) — Signals and Systems, chapter 1, asked 5×
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