Convolution Theorem

The convolution theorem for Fourier transforms states that convolution in the time domain equals multiplication in the frequency domain. The continuous-time convolution of two signals $x(t)$ and $y(t)$ is defined by

$$(x\ast y)(t) \isdef \ensuremath{\int_{-\infty}^{\infty}}x(\tau)y(t-\tau)d\tau.$$ (B.15)

The Fourier transform is then

$$\begin{eqnarray*} \hbox{\sc FT}_\omega(x\ast y) &\isdef & \int_{-\infty}^\infty \left[\ensuremath{\int_{-\infty}^{\infty}}x(\tau)y(t-\tau)d\tau\right] e^{-j\omega t}dt\\ &=& \int_{-\infty}^\infty d\tau\, x(\tau) \ensuremath{\int_{-\infty}^{\infty}}dt\, y(t-\tau)e^{-j\omega t}\\ &=& \int_{-\infty}^\infty d\tau\, x(\tau) e^{-j\omega\tau}Y(\omega) \quad\mbox{(by the \emph{shift theorem})}\\ &=& X(\omega)Y(\omega), \end{eqnarray*}$$

or,

$$\zbox {x\ast y \;\longleftrightarrow\;X\cdot Y.}$$ (B.16)

Exercise: Show that

$$\zbox {x\cdot y \;\longleftrightarrow\;\frac{1}{2\pi}X\ast Y}$$ (B.17)

when frequency-domain convolution is defined by

$$(X\ast Y)(\omega) \isdef \ensuremath{\int_{-\infty}^{\infty}}X(\nu) Y(\omega-\nu) d\nu,$$ (B.18)

where $\nu$ is in radians per second, and that

$$\zbox {x\cdot y \;\longleftrightarrow\;X\ast Y}$$ (B.19)

when frequency-domain convolution is defined by

$$(X\ast Y)(f) \isdef \ensuremath{\int_{-\infty}^{\infty}}X(\nu) Y(f-\nu) d\nu,$$ (B.20)

with $\nu$ in Hertz.