# 3.5 Fourier transform of common signals

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Fourier transform of common signals are derived.

## Fourier transform of common signals

Next, we'll derive the FT of some basic continuous-time signals. [link] summarizes these transform pairs.

## Rectangular pulse

Let's begin with the rectangular pulse

$\text{rect}\left(t,\tau \right)\equiv \left\{\begin{array}{cc}1,\hfill & t\le \tau /2\hfill \\ 0,\hfill & t>\tau /2\hfill \end{array}\right)$

The pulse function, $\text{rect}\left(t,\tau \right)$ is shown in [link] . Evaluating the Fourier transform integral with $x\left(t\right)=\text{rect}\left(t,\tau \right)$ gives

$\begin{array}{ccc}\hfill X\left(j\Omega \right)& =& {\int }_{-\tau /2}^{\tau /2}{e}^{-j\Omega t}dt\hfill \\ & =& {\left(\frac{-1}{j\Omega },{e}^{-j\Omega t}|}_{-\tau /2}^{\tau /2}\hfill \\ & =& \frac{1}{j\Omega }\left[{e}^{j\Omega \tau /2},-,{e}^{-j\Omega \tau /2}\right]\hfill \\ & =& \tau \frac{sin\left(\Omega \tau /2\right)}{\Omega \tau /2}\hfill \\ & =& \tau \text{sinc}\left(\Omega \tau /2\right)\hfill \end{array}$

A plot of $X\left(j\Omega \right)$ is shown in [link] .

Note that when $\Omega =0$ , $X\left(j\Omega \right)=\tau$ . We now have the following transform pair:

$\text{rect}\left(t,\tau \right)↔\tau \frac{sin\left(\Omega \tau /2\right)}{\Omega \tau /2}$

## Impulse

The unit impulse function was described in a previous section. From the sifting property of the impulse function we find that

$\begin{array}{ccc}\hfill X\left(j\Omega \right)& =& {\int }_{-\infty }^{\infty }\delta \left(t\right){e}^{-j\Omega t}dt\hfill \\ & =& 1\hfill \end{array}$

or

$\delta \left(t\right)↔1$

## Complex exponential

The complex exponential function, $x\left(t\right)={e}^{j{\Omega }_{0}t}$ , has a Fourier transform which is difficult to evaluate directly. It is easier to start with the Fourier transform itself and work backwards using the inverse Fourier transform. Suppose we want to find the time-domain signal which has Fourier transform $X\left(j\Omega \right)=\delta \left(\Omega -{\Omega }_{0}\right)$ . We can begin by using the inverse Fourier transform [link]

$\begin{array}{ccc}\hfill x\left(t\right)& =& \frac{1}{2\pi }{\int }_{-\infty }^{\infty }\delta \left(\Omega -{\Omega }_{0}\right){e}^{j\Omega t}d\Omega \hfill \\ & =& \frac{1}{2\pi }{e}^{\Omega t}\hfill \end{array}$

This result follows from the sifting property of the impulse function. By linearity, we can then write

${e}^{j\Omega t}↔2\pi \delta \left(\Omega -{\Omega }_{0}\right)$

## Cosine

The cosine signal can be expressed in terms of complex exponentials using Euler's Identity

$cos\left({\Omega }_{0}t\right)=\frac{1}{2}\left({e}^{j{\Omega }_{0}t},+,{e}^{-j{\Omega }_{0}t}\right)$

Applying linearity and the Fourier transform of complex exponentials to the right side of [link] , we quickly get:

$cos\left({\Omega }_{0}t\right)↔\pi \delta \left(\Omega -{\Omega }_{0}\right)+\pi \delta \left(\Omega +{\Omega }_{0}\right)$

## Real exponential

The real exponential function is given by $x\left(t\right)={e}^{-\alpha t}u\left(t\right)$ , where $\alpha >0$ . To find its FT, we start with the definition

$\begin{array}{ccc}\hfill X\left(j\Omega \right)& =& {\int }_{0}^{\infty }{e}^{-\alpha t}{e}^{-j\Omega t}dt\hfill \\ & =& {\int }_{0}^{\infty }{e}^{-\left(\alpha +j\Omega \right)t}dt\hfill \\ & =& \frac{-1}{\alpha +j\Omega }{\left({e}^{-\left(\alpha +j\Omega \right)t}|}_{0}^{\infty }\hfill \\ & =& \frac{-1}{\alpha +j\Omega }\left(0-1\right)\hfill \\ & =& \frac{1}{\alpha +j\Omega }\hfill \end{array}$

therefore,

${e}^{-\alpha t}u\left(t\right)↔\frac{1}{\alpha +j\Omega }$

## The unit step function

In a previous section, we looked at the unit step function,

$u\left(t\right)=\left\{\begin{array}{cc}1,\hfill & t\ge 0\hfill \\ 0,\hfill & t<0\hfill \end{array}\right)$

A closely related signal is the signum function, defined by

$sgn\left(t\right)=\left\{\begin{array}{cc}1,\hfill & t\ge 0\hfill \\ -1,\hfill & t<0\hfill \end{array}\right)$

from which it follows that

$u\left(t\right)=\left(sgn\left(t\right)+1\right)/2$

The signum function can be described as follows:

$sgn\left(t\right)=\underset{a\to 0}{lim}\left[{e}^{-at},u,\left(t\right),-,{e}^{at},u,\left(-t\right)\right]$

Since we already have the Fourier transform of the exponential signal,

$\begin{array}{cc}\hfill SGN\left(j\Omega \right)& =\underset{a\to 0}{lim}\left[\frac{1}{a+j\Omega },-,\frac{1}{a-j\Omega }\right]\hfill \\ & =\underset{a\to 0}{lim}\frac{-2j\Omega }{{a}^{2}+{\Omega }^{2}}\hfill \\ & =\frac{2}{j\Omega }\hfill \end{array}$

Using [link] and linearity then leads to

$u\left(t\right)↔\pi \delta \left(\Omega \right)+\frac{1}{j\Omega }$
 $x\left(t\right)$ $X\left(j\Omega \right)$ $\text{rect}\left(t,\tau \right)$ $\tau \frac{sin\left(\Omega \tau /2\right)}{\Omega \tau /2}$ $\delta \left(t\right)$ 1 ${e}^{j{\Omega }_{0}t}$ $2\pi \delta \left(\Omega -{\Omega }_{0}\right)$ $cos\left({\Omega }_{0}t\right)$ $\pi \delta \left(\Omega -{\Omega }_{0}\right)+\pi \delta \left(\Omega +{\Omega }_{0}\right)$ ${e}^{-\alpha t}u\left(t\right)$ $\frac{1}{\alpha +j\Omega }$ $u\left(t\right)$ $\pi \delta \left(\Omega \right)+\frac{1}{j\Omega }$

When working problems involving finding the Fourier transform, it is often preferable to use a table of transform pairs rather than to recalculate the Fourier transform from scratch. Often, transform pairs in can be combined with known Fourier transform properties to find new Fourier transforms.

Example 3.1 Find the Fourier transform of: $y\left(t\right)=2{e}^{5t}u\left(-t\right)$ . Clearly, we can write $y\left(t\right)=x\left(-t\right)$ where $x\left(t\right)=2{e}^{-5t}u\left(t\right)$ . Therefore, we can combine the known transform of $x\left(t\right)$ from [link] , namely,

$X\left(j\Omega \right)=\frac{2}{5+j\Omega }$

with the time reversal property:

$x\left(-t\right)↔X{\left(j\Omega \right)}^{*}$

to get the answer:

$Y\left(j\Omega \right)=\frac{2}{5-j\Omega }$

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