---
title: "Convolution integral; e^(−t)u(t) * e^(−2t)u(t−3)"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:convolution-integral-exponentials"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-integral-exponentials
site: BCT Engineering
---

# Convolution integral; e^(−t)u(t) * e^(−2t)u(t−3)

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2076 Bhadra, 5+6 marks
- Video: [Convolution integral example - graphical method](https://www.youtube.com/watch?v=ShPguecXaaU) — bioMechatronics Lab
- 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-integral-exponentials
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

Derive formula to calculate convolution integral for continuous time LTI system. Find convolution between following continuous time signals x(t) = e^(−t)u(t) and y(t) = e^(−2t)u(t−3)

## Short answer

Convolution integral is y(t)=∫x(τ)h(t−τ)dτ. For x(t)=e^{-t}u(t) and y(t)=e^{-2t}u(t−3), overlap is 0≤τ≤t−3 for t≥3, giving z(t)=(e^{-(t+3)}−e^{-2t})u(t−3).

## Answer

#### Derivation of the convolution integral

The **convolution integral** gives the output $y(t)$ of a continuous-time LTI system for any input $x(t)$, once the impulse response $h(t)$ is known.

**Step 1 – Represent the input by impulses (sifting property):**

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

So $x(t)$ is a continuous sum of shifted impulses $\delta(t-\tau)$, each weighted by $x(\tau)\,d\tau$.

**Step 2 – Response to one impulse:** by definition, $\delta(t) \rightarrow h(t)$.

**Step 3 – Time invariance:** $\delta(t-\tau) \rightarrow h(t-\tau)$.

**Step 4 – Linearity (homogeneity):** $x(\tau)\,\delta(t-\tau)\,d\tau \rightarrow x(\tau)\,h(t-\tau)\,d\tau$.

**Step 5 – Linearity (additivity):** adding (integrating) the responses to all the weighted impulses,

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

This is the convolution integral. By changing the variable $\lambda = t-\tau$ it can also be written $y(t)=\int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau$.

**Graphical steps:** (1) change $t$ to $\tau$; (2) fold $h(\tau)$ to get $h(-\tau)$; (3) shift by $t$ to get $h(t-\tau)$; (4) multiply by $x(\tau)$; (5) integrate over the overlap; (6) repeat for all $t$.

#### Convolution of $x(t)=e^{-t}u(t)$ and $y(t)=e^{-2t}u(t-3)$

Let $z(t) = x(t) * y(t)$:

$$
z(t) = \int_{-\infty}^{\infty} e^{-\tau}u(\tau)\; e^{-2(t-\tau)}u(t-\tau-3)\,d\tau
$$

**Limits:**
- $u(\tau) = 1$ only for $\tau \ge 0$.
- $u(t-\tau-3) = 1$ only for $\tau \le t-3$.

So the product is non-zero only for $0 \le \tau \le t-3$, which needs $t \ge 3$.

**Case 1: $t < 3$** – no overlap, so $z(t) = 0$.

**Case 2: $t \ge 3$:**

$$
\begin{aligned}
z(t) &= \int_{0}^{t-3} e^{-\tau}\,e^{-2t}\,e^{2\tau}\,d\tau \\
&= e^{-2t}\int_{0}^{t-3} e^{\tau}\,d\tau \\
&= e^{-2t}\left[e^{\tau}\right]_0^{t-3} \\
&= e^{-2t}\left(e^{t-3} - 1\right) \\
&= e^{-(t+3)} - e^{-2t}
\end{aligned}
$$

**Check:** at $t = 3$, $z(3) = e^{-6} - e^{-6} = 0$, so the output starts smoothly from zero. At $t = 4$, $z(4) = e^{-7} - e^{-8} \approx 0.000576$.

**Answer:**

$$
z(t) = x(t)*y(t) = \left(e^{-(t+3)} - e^{-2t}\right)u(t-3)
$$

**Shortcut check:** $e^{-2t}u(t-3) = e^{-6}\,e^{-2(t-3)}u(t-3)$, so it is $e^{-6}$ times $e^{-2t}u(t)$ delayed by 3. Since $e^{-t}u(t) * e^{-2t}u(t) = (e^{-t}-e^{-2t})u(t)$, the result is $e^{-6}\left(e^{-(t-3)} - e^{-2(t-3)}\right)u(t-3) = \left(e^{-(t+3)} - e^{-2t}\right)u(t-3)$, the same answer.

## Check yourself

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

1. The convolution integral follows from:
   A. Parseval's theorem
   B. Causality and stability
   C. Sifting property, time invariance and linearity
   D. Fourier series only
   Answer: C. Impulse decomposition, then shift, scale, and sum. (Key point: Derivation)

2. For e^(-τ)u(τ) and e^(-2(t-τ))u(t-τ-3), the integrand is non-zero for:
   A. 0 ≤ τ ≤ t
   B. 3 ≤ τ ≤ t
   C. t-3 ≤ τ ≤ t
   D. 0 ≤ τ ≤ t-3
   Answer: D. u(τ) needs τ≥0, and u(t-τ-3) needs τ≤t-3. (Key point: Limits)

3. The convolution is:
   A. (e^(-(t+3)) - e^(-2t))u(t-3)
   B. (e^(-t) - e^(-2t))u(t-3)
   C. e^(-(t+3))u(t-3)
   D. (e^(-(t-3)) - e^(-2(t-3)))u(t)
   Answer: A. e^(-2t)(e^(t-3)-1). (Key point: Result)

4. Value of z(3)?
   A. 1
   B. 0
   C. e^(-3)
   D. e^(-6)
   Answer: B. e^(-6)-e^(-6)=0 so the output starts from zero. (Key point: Check)

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