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Fourier Transform Convolution Property Derivation

November 17, 2021

Introduction

The convolution property of the Fourier transform states that the convolution of x(t) and h(t) is the product of the frequency-responses X(f) and H(f),

(1) $\begin{equation*}\mathcal{F} \{ x(t) \ast h(t) \} = X(f) \cdot H(f).\end{equation*}$

Proof

The end result needs to be two Fourier transform integrals of x(t) and h(t) in order to arrive at (1). Start by writing the convolution using the integral

(2) $\begin{equation*}x(t) \ast h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau.\end{equation*}$

Take the Fourier transform of (2)

(3) $\begin{equation*}\mathcal{F} \left\{ x(t) \ast h(t) \right\} = \mathcal{F} \left\{ \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau \right\}.\end{equation*}$

Expand the Fourier transform of (3) by writing out the integral,

(4) $\begin{equation*}\mathcal{F} \left\{ x(t) \ast h(t) \right\} = \int_{\tau=-\infty}^{\infty} \int_{t=-\infty}^{\infty} x(\tau)h(t-\tau) e^{-j2\pi f t} dt ~ d\tau.\end{equation*}$

There are now two integrals over time, both t and $\tau$ which will form the basis for the two Fourier transforms. The variables dependent on t and $\tau$ need to be rearranged so their integrals are independent of one another. Start by separating the terms only dependent on $\tau$ by rearranging the terms in the right hand of (4),

(5) $\begin{equation*}\mathcal{F} \left\{ x(t) \ast h(t) \right\} = \int_{\tau=-\infty}^{\infty} x(\tau) \left( \int_{t=-\infty}^{\infty} h(t-\tau) e^{-j2\pi f t} dt \right) d\tau.\end{equation*}$

The variable $h(t-\tau)$ is a problem because it has dependencies on both t and $\tau$ . Use a variable substitution

(6) $\begin{equation*}v = t - \tau\end{equation*}$

such that

(7) $\begin{equation*}h(t-\tau) = h(v).\end{equation*}$

As a result of (6),

(8) $\begin{equation*}t = v + \tau\end{equation*}$

and

(9) $\begin{equation*}dt = dv.\end{equation*}$

The derivative $d\tau = 0$ because it is a constant within the integral over t in (5). Substituting (8) and (9) into (5),

(10) $\begin{equation*}\mathcal{F} \left\{ x(t) \ast h(t) \right\} = \int_{\tau=-\infty}^{\infty} x(\tau) \left( \int_{v=-\infty}^{\infty} h(v) e^{-j2\pi f (v + \tau)} dv \right) d\tau.\end{equation*}$

The exponential can be expanded as the product of two exponentials of $\tau$ and v,

(11) $\begin{equation*}e^{-j2\pi f (v + \tau)} = e^{-j2\pi f v} e^{-j 2\pi f \tau}.\end{equation*}$

Substituting (11) into (10),

(12) $\begin{equation*}\mathcal{F} \left\{ x(t) \ast h(t) \right\} = \int_{\tau=-\infty}^{\infty} x(\tau) \left( \int_{v=-\infty}^{\infty} h(v) e^{-j2\pi f v} e^{-j 2\pi f \tau} dv \right) d\tau.\end{equation*}$

Rearranging (12) into integrals of v and $\tau$ ,

(13) $\begin{equation*}\mathcal{F} \left\{ x(t) \ast h(t) \right\} = \left( \int_{\tau=-\infty}^{\infty} x(\tau) e^{-j2\pi f \tau} d\tau \right) \left( \int_{v=-\infty}^{\infty} h(v) e^{-j 2\pi f v} dv \right).\end{equation*}$

The two integrals are now Fourier transforms of x(t) and h(t),

(14) $\begin{equation*}X(f) = \left( \int_{\tau=-\infty}^{\infty} x(\tau) e^{-j2\pi f \tau} d\tau \right) = \mathcal{F} \{ x(t) \},\end{equation*}$

(15) $\begin{equation*}H(f) = \left( \int_{v=-\infty}^{\infty} h(v) e^{-j 2\pi f v} dv \right) = \mathcal{F} \{ h(t) \}.\end{equation*}$

Therefore $\mathcal{F} \{ x(t) \ast h(t) \} = X(f) \cdot H(f)$ as in (1).

Conclusion

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Table of Contents

Introduction

Proof

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