---
title: "Derive the convolution integral"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:convolution-integral-derivation"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-integral-derivation
site: BCT Engineering
---

# Derive the convolution integral

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2082 Chaitra, 6 marks; 2081 Chaitra, 5 marks; 2077 Chaitra, 4 marks; 2075 Bhadra, 6 marks; 2074 Bhadra, 6 marks; 2073 Bhadra, 6 marks; 2071 Bhadra, 8 marks; 2082 Bhadra (new course), 4 marks (asked 8 times)
- Video: [Signals & Systems - Derivation of Convolution Integral - UNIT III](https://www.youtube.com/watch?v=XJg0oR9uDnA) — 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-integral-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

Derive the convolution integral for continuous time LTI system.

## Short answer

The convolution integral is y(t) = ∫ x(τ)h(t-τ)dτ. Derivation: approximate input as sum of scaled narrow pulses, use linearity and time invariance to sum shifted impulse responses, then let pulse width → 0, turning sum into integral.

## Answer

The **convolution integral** gives the output $y(t)$ of a continuous-time LTI system for any input $x(t)$ in terms of its impulse response $h(t)$:
$$
y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau = x(t) * h(t)
$$

#### Step 1: Represent the input with narrow pulses
Define a unit-area pulse of width $\Delta$:
$$
\delta_\Delta(t) = \begin{cases} 1/\Delta, & 0 \le t < \Delta \\ 0, & \text{otherwise}\end{cases}
$$
The staircase approximation of $x(t)$ is a sum of shifted, scaled pulses:
$$
\hat{x}(t) = \sum_{k=-\infty}^{\infty} x(k\Delta)\,\delta_\Delta(t - k\Delta)\,\Delta
$$
```text
x(t)     __
      __|  |__
   __|        |__     staircase of width-Delta
  |              |    pulses, height x(k.Delta)
--+--+--+--+--+--+--> t
     kD (k+1)D
```

#### Step 2: Response to one pulse
Let $\hat{h}_k(t)$ be the response to $\delta_\Delta(t - k\Delta)$. By **linearity** (superposition and scaling),
$$
\hat{y}(t) = \sum_{k=-\infty}^{\infty} x(k\Delta)\,\hat{h}_k(t)\,\Delta
$$
By **time invariance**, the response to a shifted pulse is the shifted response: $\hat{h}_k(t) = \hat{h}_0(t - k\Delta)$.

#### Step 3: Take the limit $\Delta \to 0$
As $\Delta \to 0$:
- $\hat{x}(t) \to x(t)$,
- $\delta_\Delta(t) \to \delta(t)$, so $\hat{h}_0(t) \to h(t)$, the impulse response,
- $k\Delta \to \tau$ (continuous variable), $\Delta \to d\tau$, and the sum becomes an integral.

$$
\begin{aligned}
y(t) &= \lim_{\Delta\to0}\sum_{k=-\infty}^{\infty} x(k\Delta)\,\hat{h}_0(t-k\Delta)\,\Delta \\
&= \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau
\end{aligned}
$$
The same result follows directly from the sifting property: $x(t) = \int x(\tau)\delta(t-\tau)d\tau$; the system maps $\delta(t-\tau) \to h(t-\tau)$ (time invariance) and the integral passes through (linearity).

#### Meaning and evaluation
- The output is a weighted sum of shifted impulse responses; $x(\tau)$ is the weight of the impulse at $\tau$.
- Steps to evaluate: (1) change variable 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; repeat for each range of $t$.
- Using $\tau \to t - \tau$, $y(t) = \int h(\tau)x(t-\tau)d\tau$ also holds (commutative property).

**Example:** $x(t) = u(t)$, $h(t) = e^{-at}u(t)$, $a>0$: $y(t) = \int_0^t e^{-a(t-\tau)}d\tau = \frac{1}{a}(1 - e^{-at})u(t)$.

## Check yourself

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

1. How is the input approximated first?
   A. As a sum of sinusoids
   B. As a staircase of narrow pulses x(kΔ) δΔ(t - kΔ) Δ
   C. As a single impulse
   D. As a polynomial in t
   Answer: B. The input is a sum of shifted, scaled unit-area pulses. (Key point: Input representation)

2. Which properties let the response to the pulse sum be built from one pulse response?
   A. Causality and stability
   B. Memorylessness and invertibility
   C. Linearity (superposition) and time invariance (shifts)
   D. Only linearity
   Answer: C. Linearity sums scaled responses and time invariance shifts h0 to h0(t - kΔ). (Key point: Role of properties)

3. What happens when Δ -> 0?
   A. The sum becomes the integral ∫x(τ)h(t-τ)dτ
   B. The output becomes zero
   C. The sum becomes ∫x(τ)h(τ)dτ
   D. The output becomes x(t)h(t)
   Answer: A. kΔ -> τ, Δ -> dτ and the pulse response -> h(t). (Key point: Limit)

4. What is the correct order to evaluate convolution graphically?
   A. Shift x, add h, differentiate
   B. Multiply x and h, then fold
   C. Integrate x, then shift h
   D. Fold h, shift by t, multiply by x, integrate over overlap
   Answer: D. h(t-τ) is h(τ) folded then shifted by t; multiply and integrate over overlap. (Key point: Evaluation steps)

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