---
title: "Output for x = e^(−at)u(t), h = u(t)"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:convolution-exponential-step"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-exponential-step
site: BCT Engineering
---

# Output for x = e^(−at)u(t), h = u(t)

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2075 Baisakh, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-exponential-step
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

Find output y(t) of LTI system if input x(t) = exp(−at) u(t) and impulse response h(t) = u(t).

## Short answer

Output is convolution x*h. For x=e^{-at}u(t) and h=u(t), overlap is 0≤τ≤t for t≥0, giving y(t)=(1/a)(1−e^{-at})u(t).

## Answer

For an LTI system the output is the convolution of input and impulse response:

$$
y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau
$$

**Given:** $x(t) = e^{-at}u(t)$ (take $a > 0$), $h(t) = u(t)$.

$$
y(t) = \int_{-\infty}^{\infty} e^{-a\tau}u(\tau)\,u(t-\tau)\,d\tau
$$

**Limits:** $u(\tau) = 1$ for $\tau \ge 0$ and $u(t-\tau) = 1$ for $\tau \le t$. The overlap is $0 \le \tau \le t$, which exists only for $t \ge 0$.

**For $t < 0$:** no overlap, $y(t) = 0$.

**For $t \ge 0$:**

$$
\begin{aligned}
y(t) &= \int_{0}^{t} e^{-a\tau}\,d\tau \\
&= \left[\frac{e^{-a\tau}}{-a}\right]_0^{t} \\
&= \frac{1}{a}\left(1 - e^{-at}\right)
\end{aligned}
$$

**Answer:**

$$
y(t) = \frac{1}{a}\left(1 - e^{-at}\right)u(t)
$$

```text
 y(t)
 1/a |        .-------------
     |     .-'
     |   .'
     |  /
     | /
   0 +/----------------------> t
     0
```

The output starts at 0 at $t = 0$ and rises to the final value $1/a$ as $t \to \infty$. This is expected, because $h(t) = u(t)$ is an ideal integrator, so $y(t) = \int_{-\infty}^{t} x(\tau)\,d\tau$, the running area under $e^{-at}$, whose total area is $1/a$.

## Check yourself

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

1. For e^(-aτ)u(τ) and u(t-τ), the overlap is:
   A. τ ≥ t
   B. -∞<τ<t
   C. 0 ≤ τ ≤ t
   D. τ ≤ 0
   Answer: C. u(τ) needs τ≥0, u(t-τ) needs τ≤t. (Key point: Overlap)

2. y(t) =
   A. e^(-at)u(t)
   B. (1/a)e^(-at)u(t)
   C. (1/a)(1+e^(-at))u(t)
   D. (1/a)(1-e^(-at))u(t)
   Answer: D. Integral of e^(-aτ) from 0 to t. (Key point: Result)

3. As t→∞, y(t) →
   A. 1/a
   B. 0
   C. 1
   D. ∞
   Answer: A. Total area under e^(-at) is 1/a. (Key point: Final value)

4. h(t)=u(t) is:
   A. A differentiator
   B. An ideal integrator
   C. A pure delay
   D. An ideal LPF
   Answer: B. y is the running integral of x. (Key point: Meaning of u(t))

## Close variants set in other exams

- [y(t) for h = e^(−t)u(t−3), x = u(t−2)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-delayed-exponential-step) — Signals and Systems, chapter 5, asked 5×
- [Output of e^(−2t)u(t) to a rectangular input](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-exponential-rect-input) — Signals and Systems, chapter 5, asked 4×

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