---
title: "CTFT convolution property"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ctft-convolution-property"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctft-convolution-property
site: BCT Engineering
---

# CTFT convolution property

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2081 Chaitra, 5 marks; 2081 Asoj, 5 marks; 2080 Asoj, 6 marks; 2079 Asoj, 5 marks; 2078 Baisakh, 4 marks; 2080 Chaitra, 4 marks; 2079 Chaitra, 5 marks; 2076 Bhadra, 5 marks; 2071 Bhadra, 6 marks (asked 9 times)
- Video: [Fourier Transform Properties CONVOLUTION](https://www.youtube.com/watch?v=7ApZ8cjNfSI) — Francesco Fedele
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctft-convolution-property
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

State and prove the convolution property of the continuous time Fourier transform (show that convolution in time domain results in multiplication in frequency domain).

## Short answer

Convolution in time corresponds to multiplication in frequency: x(t)*h(t) ↔ X(ω)H(ω). The proof substitutes the convolution integral into the Fourier definition, interchanges integration order, uses the time-shift property to obtain e^{-jωτ}H(ω), and integrates over τ to yield X(ω)H(ω).

## Answer

#### Statement
If $x(t) \xleftrightarrow{F} X(\omega)$ and $h(t) \xleftrightarrow{F} H(\omega)$, then
$$
y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)\,d\tau \;\xleftrightarrow{F}\; Y(\omega) = X(\omega)H(\omega)
$$
Convolution in the time domain becomes multiplication in the frequency domain.

#### Proof
By the definition $X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt$:
$$
Y(\omega) = \int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau\right]e^{-j\omega t}dt
$$
Change the order of integration:
$$
Y(\omega) = \int_{-\infty}^{\infty} x(\tau)\left[\int_{-\infty}^{\infty} h(t-\tau)e^{-j\omega t}dt\right]d\tau
$$
In the inner integral put $\sigma = t - \tau$ ($dt = d\sigma$, limits unchanged):
$$
\int_{-\infty}^{\infty} h(\sigma)e^{-j\omega(\sigma + \tau)}d\sigma = e^{-j\omega\tau}H(\omega)
$$
This is the time-shift property. Substituting back:
$$
Y(\omega) = H(\omega)\int_{-\infty}^{\infty} x(\tau)e^{-j\omega\tau}d\tau = H(\omega)X(\omega)
$$
$$
\boxed{x(t)*h(t) \xleftrightarrow{F} X(\omega)H(\omega)}
$$

#### Significance
- **LTI analysis:** the output of an LTI system with impulse response $h(t)$ is $Y(\omega) = H(\omega)X(\omega)$. $H(\omega)$ is the **frequency response**: each frequency component of the input is scaled by $|H(\omega)|$ and phase-shifted by $\angle H(\omega)$.
- **Filtering:** an ideal low-pass filter has $H(\omega) = 1$ for $|\omega| < \omega_c$ and 0 elsewhere. It removes the high-frequency parts of $X(\omega)$ simply by multiplication.
- **Cascades:** systems in series give $H(\omega) = H_1(\omega)H_2(\omega)$.
- **Computation:** a convolution integral can be replaced by: transform, multiply, inverse transform.

#### Example
$x(t) = e^{-at}u(t)$ is applied to $h(t) = e^{-bt}u(t)$, with $a \ne b$ and both positive:
$$
Y(\omega) = \frac{1}{(a+j\omega)(b+j\omega)} = \frac{1}{b-a}\left[\frac{1}{a+j\omega} - \frac{1}{b+j\omega}\right]
$$
So $y(t) = \frac{1}{b-a}\left(e^{-at} - e^{-bt}\right)u(t)$, found without evaluating the convolution integral.

**Dual (multiplication/modulation) property:** $x(t)p(t) \xleftrightarrow{F} \frac{1}{2\pi}X(\omega) * P(\omega)$.

## Check yourself

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

1. If y(t) = x(t) * h(t), what is Y(ω)?
   A. X(ω) * H(ω)
   B. X(ω)H(ω)
   C. X(ω)/H(ω)
   D. X(ω) + H(ω)
   Answer: B. Convolution in time becomes multiplication in frequency. (Key point: Statement)

2. In the proof, after interchanging the integrals, the inner integral over t of h(t - τ)e^{-jωt} is evaluated as:
   A. H(ω)
   B. e^{-jωτ}H(ω) by the time-shift property
   C. e^{jωτ}H(ω)
   D. h(τ)
   Answer: B. Putting σ = t - τ gives e^{-jωτ} times the integral of h(σ)e^{-jωσ}, i.e. e^{-jωτ}H(ω). (Key point: Proof step)

3. In an LTI system, what does H(ω) do to each input frequency component?
   A. Scales it by |H(ω)| and shifts its phase by angle H(ω)
   B. Only scales its amplitude by H(0)
   C. Delays all components by the same time
   D. Adds new frequencies
   Answer: A. Y(ω) = H(ω)X(ω), so each component is multiplied by |H| and phase-shifted by angle H. (Key point: Frequency response)

4. For x(t) = e^{-at}u(t) applied to h(t) = e^{-bt}u(t), Y(ω) is:
   A. 1/(a + b + jω)
   B. 1/((a+jω)(b+jω))
   C. 1/(a+jω) + 1/(b+jω)
   D. (a+jω)(b+jω)
   Answer: B. X = 1/(a+jω), H = 1/(b+jω), and Y = XH. (Key point: Worked example)

## Close variants set in other exams

- [DTFT convolution property](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-convolution-property) — Signals and Systems, chapter 3, asked 12×

## Related in Signals and Systems

- [Frequency shifting property of CTFT](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-frequency-shifting) — Signals and Systems, chapter 3, asked 3×
- [CTFT convolution; DTFT of 2ⁿu[n]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-convolution-dtft-2n) — Signals and Systems, chapter 3
- [Time shifting property of CTFT](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-time-shifting) — Signals and Systems, chapter 3, asked 5×
- [Duality property of CTFT](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-duality) — Signals and Systems, chapter 3, asked 3×
- [Convolution property of CTFS](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actfs-convolution-property) — Signals and Systems, chapter 2
- [DTFT properties; convolution property](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-properties-convolution) — Signals and Systems, chapter 3, asked 2×

## Related in other subjects

- [Fourier transform property: Convolution](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aft-property-convolution) — Communication Systems, chapter 1
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