The fourier(M) function computes the element-wise Fourier transform of M.  The result, R, is formed as R[i,j] = fourier(M[i,j], u, v).

fourier(f) is the Fourier transform of the scalar f with default independent variable x.  If f is not a function of x, then f is assumed to be a function of the independent variable returned by findsym(f,1). The default return is a function of w.

If f = f(w), then fourier returns a function of t.

By definition, $F\left(w\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right){\ⅇ}^{\mathrm{-I}wx}\ⅆx$, where the integration above proceeds with respect to x.

fourier(f,v) makes F a function of the variable v instead of the default w.

fourier(f,u,v) makes f a function of u instead of the default. The integration is then with respect to u.

$\mathrm{with}\left(\mathrm{MTM}\right)\:$

$\mathrm{fourier}\left(t\exp \left(-3t\right)\mathrm{Heaviside}\left(t\right)\right)$

$\frac{1}{{\left(3+\mathrm{I}w\right)}^2}$

$\mathrm{fourier}\left(w\exp \left(-3w\right)\mathrm{Heaviside}\left(w\right)\right)$

$\frac{1}{{\left(3+\mathrm{I}t\right)}^2}$

$\mathrm{fourier}\left(t\exp \left(-3t\right)\mathrm{Heaviside}\left(t\right)\,s\right)$

$\frac{1}{{\left(3+\mathrm{I}s\right)}^2}$

$\mathrm{fourier}\left(zt\exp \left(-3t\right)\mathrm{Heaviside}\left(t\right)\,z\,s\right)$

$2\mathrm{I}t{\ⅇ}^{-3t}\mathrm{Heaviside}\left(t\right)\mathrm{\pi }\mathrm{Dirac}\left(1\,s\right)$

$M≔\mathrm{Matrix}\left(\left[x\exp \left(-3x\right)\mathrm{Heaviside}\left(x\right)\,zx\exp \left(-3x\right)\mathrm{Heaviside}\left(x\right)\right]\right)\:$

$\mathrm{fourier}\left(M\right)$

$\left[\begin{array}{cc}\frac{1}{{\left(3+\mathrm{I}w\right)}^2}& \frac{z}{{\left(3+\mathrm{I}w\right)}^2}\end{array}\right]$