---
title: "Output of e^(−2t)u(t) to a rectangular input"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:convolution-exponential-rect-input"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-exponential-rect-input
site: BCT Engineering
---

# Output of e^(−2t)u(t) to a rectangular input

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2079 Jestha, 2+8 marks
- Video: [Signals & Systems - Convolution of two signals - working examples -1 - UNIT III](https://www.youtube.com/watch?v=oHDC-DXxXa4) — Dr.P.Prasanna Murali krishna
- 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-rect-input
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

What is an LTI system? Find the output of the LTI system with impulse response h(t) = e^(−2t)u(t) when the input to the system is x(t) = 1 for |t| < 5; 0 otherwise.

## Short answer

For t<-5, y(t)=0; for -5≤t<5, y(t)=0.5[1 - e^{-2(t+5)}]; for t≥5, y(t)=0.5[e^{-2(t-5)} - e^{-2(t+5)}]. Output rises to ~0.5 while input is on, then decays.

## Answer

#### LTI system
An **LTI system** is a system that is both linear (obeys superposition) and time-invariant (a shift in input gives the same shift in output). It is fully described by its impulse response $h(t)$, and the output for any input is the convolution $y(t) = x(t) * h(t)$.

#### Output for $h(t) = e^{-2t}u(t)$, $x(t) = 1$ for $|t| < 5$
$$
y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau
$$
- $x(\tau) = 1$ for $-5 < \tau < 5$.
- $h(t - \tau) = e^{-2(t-\tau)}$ for $t - \tau > 0$, i.e. $\tau < t$.

So the integrand is non-zero for $-5 < \tau < \min(t, 5)$.

```text
   x(tau)                  h(t-tau)
    ___________          .-'|
   |           |     .-'    |
 --+-----+-----+--.--------+---> tau
  -5     0     5           t
```

**Case 1: $t < -5$.** No overlap.
$$
y(t) = 0
$$

**Case 2: $-5 \le t < 5$.** Overlap from $-5$ to $t$.
$$
\begin{aligned}
y(t) &= \int_{-5}^{t} e^{-2(t-\tau)}\,d\tau = e^{-2t}\left[\frac{e^{2\tau}}{2}\right]_{-5}^{t} \\
&= \frac{e^{-2t}}{2}\left(e^{2t} - e^{-10}\right) = \frac{1}{2}\left(1 - e^{-2(t+5)}\right)
\end{aligned}
$$

**Case 3: $t \ge 5$.** Overlap from $-5$ to $5$.
$$
\begin{aligned}
y(t) &= \int_{-5}^{5} e^{-2(t-\tau)}\,d\tau = \frac{e^{-2t}}{2}\left(e^{10} - e^{-10}\right) \\
&= \frac{1}{2}\left(e^{-2(t-5)} - e^{-2(t+5)}\right)
\end{aligned}
$$

#### Result
$$
y(t) = \begin{cases} 0, & t < -5 \\[2pt] \dfrac12\left(1 - e^{-2(t+5)}\right), & -5 \le t < 5 \\[6pt] \dfrac12\left(e^{-2(t-5)} - e^{-2(t+5)}\right), & t \ge 5 \end{cases}
$$

```text
 y(t)
 0.5 |      .------------.
     |    /               \
     |   /                  `.
  ---+--/--------+-----------+-`---___--> t
       -5        0           5
```
Checks: $y(-5) = 0$; at $t = 5$ both pieces give $\tfrac12(1 - e^{-20}) \approx 0.5$ (continuous); $y \to 0$ as $t \to \infty$. Physically, the output rises toward 0.5 with time constant 0.5 s while the input is on, then decays after $t = 5$.

## Check yourself

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

1. What defines an LTI system and how is its output found?
   A. It is linear and time-invariant; output is x(t)*h(t)
   B. It is linear only; output is x(t)h(t)
   C. It is time-invariant only; output is the derivative of x(t)
   D. It is linear and causal; output is h(t)/x(t)
   Answer: A. An LTI system obeys superposition and time invariance, so it is fully described by h(t) and y = x*h. (Key point: LTI definition)

2. For x = 1 on (-5,5) and h(t-τ)=e^(-2(t-τ)) for τ<t, over what range is the integrand non-zero?
   A. 0 < τ < t
   B. -5 < τ < min(t, 5)
   C. -5 < τ < 5 for every t
   D. t < τ < 5
   Answer: B. h(t-τ) needs τ < t and x(τ) needs -5 < τ < 5, giving -5 < τ < min(t,5). (Key point: Overlap limits)

3. What is y(t) for -5 ≤ t < 5?
   A. (1/2)(e^(-2(t-5)) - e^(-2(t+5)))
   B. 1 - e^(-2t)
   C. (1/2)(1 - e^(-2(t+5)))
   D. (1/2)e^(-2(t+5))
   Answer: C. Integrating e^(-2(t-τ)) from -5 to t gives (1/2)(1 - e^(-2(t+5))). (Key point: Result for -5 ≤ t < 5)

4. For t ≥ 5 the output is (1/2)(e^(-2(t-5)) - e^(-2(t+5))). What does it do as t → ∞?
   A. Stays at 0.5
   B. Grows without bound
   C. Settles at 1
   D. Decays to 0 since the input has ended
   Answer: D. After t=5 the input is off, so the output decays exponentially with time constant 0.5 s. (Key point: Result for t ≥ 5)

## 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 for x = e^(−at)u(t), h = u(t)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-exponential-step) — Signals and Systems, chapter 5, asked 4×

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