fourier - Maple Help
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MTM

 fourier
 Fourier integral transform

 Calling Sequence fourier(M) fourier(M,v) fourier(M,u, v)

Parameters

 M - array v - variable u - variable

Description

 • 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, $F\left(w\right)={{\int }}_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right){ⅇ}^{-Iwx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}x$, 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.

Examples

 > $\mathrm{with}\left(\mathrm{MTM}\right):$
 > $\mathrm{fourier}\left(t{ⅇ}^{-3t}\mathrm{Heaviside}\left(t\right)\right)$
 $\frac{{1}}{{\left({3}{+}{I}{}{w}\right)}^{{2}}}$ (1)
 > $\mathrm{fourier}\left(w{ⅇ}^{-3w}\mathrm{Heaviside}\left(w\right)\right)$
 $\frac{{1}}{{\left({3}{+}{I}{}{t}\right)}^{{2}}}$ (2)
 > $\mathrm{fourier}\left(t{ⅇ}^{-3t}\mathrm{Heaviside}\left(t\right),s\right)$
 $\frac{{1}}{{\left({3}{+}{I}{}{s}\right)}^{{2}}}$ (3)
 > $\mathrm{fourier}\left(zt{ⅇ}^{-3t}\mathrm{Heaviside}\left(t\right),z,s\right)$
 ${2}{}{I}{}{t}{}{{ⅇ}}^{{-}{3}{}{t}}{}{\mathrm{Heaviside}}{}\left({t}\right){}{\mathrm{\pi }}{}{\mathrm{Dirac}}{}\left({1}{,}{s}\right)$ (4)
 > $M≔\mathrm{Matrix}\left(\left[x{ⅇ}^{-3x}\mathrm{Heaviside}\left(x\right),zx{ⅇ}^{-3x}\mathrm{Heaviside}\left(x\right)\right]\right):$
 > $\mathrm{fourier}\left(M\right)$
 $\left[\begin{array}{cc}\frac{{1}}{{\left({3}{+}{I}{}{w}\right)}^{{2}}}& \frac{{z}}{{\left({3}{+}{I}{}{w}\right)}^{{2}}}\end{array}\right]$ (5)