3 Existence of the Fourier transform

We will discuss this question in a little detail at a later stage when we will also consider briefly the relation between the Fourier transform and the Laplace Transform ( HELM booklet  20). For now we will use (5) to obtain the Fourier transforms of some important functions.

Example 1

Find the Fourier transform of the one-sided exponential function

$f\left(t\right)=\left\{\begin{array}{ccc}\hfill 0\hfill & \hfill & t<0\hfill \\ \hfill e^{-\alpha t}\hfill & \hfill & t>0\hfill \end{array}\right.$

where $\alpha$ is a positive constant, shown below:

Figure 1

Solution

Using (5) then by straightforward integration

since $e^{-\alpha t}\to 0$ as $t\to \infty$ for $\alpha >0$ .

This important Fourier transform is written in the following Key Point:

Note that if $u\left(t\right)$ is used to denote the Heaviside unit step function :

$u\left(t\right)=\left\{\begin{array}{ccc}0\hfill & \hfill & t<0\hfill \\ 1\hfill & \hfill & t>0\hfill \end{array}\right.$

then we can write the function in Example 1 as:   $f\left(t\right)=e^{-\alpha t}u\left(t\right).$  We shall frequently use this concise notation for one-sided functions.

Task!

Write down the Fourier transforms of

1. $e^{-t}u\left(t\right)$
2. $e^{-3t}u\left(t\right)$
3. $e^{-\frac{t}{2}}u\left(t\right)$

Use Key Point 1:

Answer

Task!

Obtain, using the integral definition (5), the Fourier transform of the rectangular pulse

$p\left(t\right)=\left\{\begin{array}{ccc}1\hfill & \hfill & -a<t<a\hfill \\ 0\hfill & \hfill & \text{otherwise}\hfill \end{array}\right..$

Note that the pulse width is $2a$ as indicated in the diagram below.

First use (5) to write down the integral from which the transform will be calculated:

Answer

Now evaluate this integral and write down the final Fourier transform in trigonometric, rather than complex exponential form:

Answer

$P\left(\omega \right)=2a\frac{sin\omega a}{\omega a},$

we can say

$P\left(\omega \right)=2a\text{sinc}\left(\omega a\right).$

Using the result (6) in (4) we have the Fourier integral representation of the rectangular pulse.

$p\left(t\right)=\frac{1}{2\pi }{\int \; <!--nolimits\; -->}_{-\infty }^{\infty }2\frac{sin\omega a}{\omega }{e}^{\text{i}\omega t}d\omega .$

As we have already mentioned, this corresponds to a Fourier series representation for a periodic function.

Clearly, if the rectangular pulse has width 2, corresponding to $a=1$ we have:

$P_1\left(\omega \right)\equiv \mathcal{F}\left\{p_1\left(t\right)\right\}=2\frac{sin\omega }{\omega }.$

As $\omega \to 0$ , then $2\frac{sin\omega }{\omega }\to 2$ . Also, the function $2\frac{sin\omega }{\omega }$ is an even function being the product of two odd functions $2sin\omega$ and $\frac{1}{\omega }$ . The graph of $P_1\left(\omega \right)$ is as follows:

Figure 2

Task!

Obtain the Fourier transform of the two sided exponential function

$f\left(t\right)=\left\{\begin{array}{ccc}{e}^{\alpha t}\hfill & \hfill & t<0\hfill \\ e^{-\alpha t}\hfill & \hfill & t>0\hfill \end{array}\right.$

where $\alpha$ is a positive constant.

Answer

Note that, as in the case of the rectangular pulse, we have here a real even function of $t$ giving a Fourier transform which is wholly real. Also, in both cases, the Fourier transform is an even (as well as real) function of $\omega$ .

Note also that it follows from the above calculation that

$\mathcal{F}\left\{e^{-\alpha t}u\left(t\right)\right\}=\frac{1}{\alpha +\text{i}\omega }$ (as we have already found)

and

$\mathcal{F}\left\{e^{\alpha t}u\left(-t\right)\right\}=\frac{1}{\alpha -\text{i}\omega }$ where $e^{\alpha t}u\left(-t\right)=\left\{\begin{array}{ccc}\hfill e^{\alpha t}\hfill & \hfill \hfill & \hfill t<0\hfill \\ \hfill 0\hfill & \hfill \hfill & \hfill t>0\hfill \end{array}\right..$