The inttrans (integral transformation) package contains a number of functions, but engineers are generally only interested in two pairs of these: the Fourier transform and its inverse, and the Fourier transform and its inverse.
Before we begin, however, we need to set up an environment familiar to engineers:
[> interface( 'imaginaryunit' = 'j' ):
alias( 'u' = 'Heaviside' ):
alias( 'delta' = 'Dirac' ):
_EnvUseHeavisideAsUnitStep := true:
Now we proceed:
[> inttrans[fourier]( exp(-t^2), t, omega );;
$e^{-\frac{\omega^2}{4}} \sqrt{\pi}$
The second argument is the time variable, and the second is the complex frequency variable, generally $t$ and $\omega$, respectively. Recall that there are tables of Fourier transforms in text books, and you can replicate many of these in Maple:
[> inttrans[fourier]( u(t)*exp(-t), t, omega );
$\frac{1}{1 + j\omega}$
[> inttrans[fourier]( u(t - 3)*t*exp(-t), t, omega );
$\frac {{e^{-3-3\,j\omega}} \left( 3\,j\omega+4 \right) }{ \left( 1+j\omega \right) ^2}$
We can also use symbols:
[> inttrans[fourier]( u(t - a)*t*exp(-t), t, omega );
${\frac { \left( ia\omega+a+1 \right) {{\rm e}^{-a}}{{\rm e}^{-ia\omega }}}{ \left( 1+i\omega \right) ^{2}}}$
[> inttrans[fourier]( u(t)*exp(-a*t), t, omega );
$\mathcal{F}(u(t)e^{-at}, t, \omega)$
In this case, Maple cannot return a solution, because the solution depends on whether $a$ is positive or not, so we can give Maple an assumption:
[> inttrans[fourier]( u(t)*exp(-a*t), t, omega ) assuming a > 0;
$\frac{1}{j\omega + a}$
To be continued...