3 Fourier transform and Laplace transforms

Suppose $f\left(t\right)=0$ for $t<0$ . Then the Fourier transform of $f\left(t\right)$ becomes

$\mathcal{F}\left\{f\left(t\right)\right\}={\int \; <!--nolimits\; -->}_0^{\infty }f\left(t\right)e^{-\text{i}\omega t}dt.$ (1)

As you may recall from earlier units, the Laplace transform of $f\left(t\right)$ is

$\mathcal{L}\left\{f\left(t\right)\right\}={\int \; <!--nolimits\; -->}_0^{\infty }f\left(t\right)e^{-st}dt.$ (2)

Comparison of (1) and (2) suggests that for such one-sided functions, the Fourier transform of $f\left(t\right)$ can be obtained by simply replacing $s$ by $\text{i}\omega$ in the Laplace transform.

An obvious example where this can be done is the function

$f\left(t\right)=e^{-\alpha t}u\left(t\right).$

In this case $\mathcal{L}\left\{f\left(t\right)\right\}=\frac{1}{\alpha +s}=F\left(s\right)$ and, as we have seen earlier,

$\mathcal{F}\left\{f\left(t\right)\right\}=\frac{1}{\alpha +\text{i}\omega }=F\left(\text{i}\omega \right).$

However, care must be taken with such substitutions. We must be sure that the conditions for the existence of the Fourier transform are met. Thus, for the unit step function,

$\mathcal{L}\left\{u\left(t\right)\right\}=\frac{1}{s},$

whereas, $\mathcal{F}\left\{u\left(t\right)\right\}\ne \frac{1}{\text{i}\omega }.$ (We shall see that $\mathcal{F}\left\{u\left(t\right)\right\}$ does actually exist but is not equal to $\frac{1}{\text{i}\omega }.$ )

We should also point out that some of the properties we have discussed for Fourier transforms are similar to those of the Laplace transforms e.g. the time-shift properties:

Fourier: $\mathcal{F}\left\{f\left(t-t_0\right)\right\}=e^{-\text{i}\omega t_0}F\left(\omega \right)$ Laplace: $\mathcal{L}\left\{f\left(t-t_0\right)\right\}=e^{-s{t}_0}F\left(s\right).$