4 Basic properties of the Fourier transform

4.1 Real and imaginary parts of a Fourier transform

Using the definition (5) we have,

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

If we write $e^{-\text{i}\omega t}=cos\omega t-\text{i}sin\omega t$ , then

$F\left(\omega \right)={\int \; <!--nolimits\; -->}_{-\infty }^{\infty }f\left(t\right)cos\omega tdt-\text{i}{\int \; <!--nolimits\; -->}_{-\infty }^{\infty }f\left(t\right)sin\omega tdt$

where both integrals are real, assuming that $f\left(t\right)$ is real. Hence the real and imaginary parts of the Fourier transform are:

Task!

Recalling that if $h\left(t\right)$ is even and $g\left(t\right)$ is odd then ${\int \; <!--nolimits\; -->}_{-a}^{a}h\left(t\right)dt=2{\int \; <!--nolimits\; -->}_0^{a}h\left(t\right)dt$ and ${\int \; <!--nolimits\; -->}_{-a}^{a}g\left(t\right)dt=0$ , deduce Re $\left(F\left(\omega \right)\right)$ and Im $\left(F\left(\omega \right)\right)$ if

1. $f\left(t\right)$ is a real even function
2. $f\left(t\right)$ is a real odd function.

Answer

Answer

These results are summarised in the following Key Point:

4.2 Polar form of a Fourier transform

Task!

The one-sided exponential function   $f\left(t\right)=e^{-\alpha t}u\left(t\right)$  has Fourier transform   $F\left(\omega \right)=\frac{1}{\alpha +\text{i}\omega }.$ Find the real and imaginary parts of $F\left(\omega \right)$ .

Answer

We can rewrite $F\left(\omega \right)$ , like any other complex quantity, in polar form by calculating the magnitude and the argument (or phase). For the Fourier transform in the last Task

$\left|F\left(\omega \right)\right|=\sqrt{R^2\left(\omega \right)+I^2\left(\omega \right)}=\sqrt{\frac{{\alpha }^2+{\omega }^2}{{\left({\alpha }^2+{\omega }^2\right)}^2}}=\frac{1}{\sqrt{{\alpha }^2+{\omega }^2}}$

$\text{and}\text{arg}F\left(\omega \right)={tan}^{-1}\frac{I\left(\omega \right)}{R\left(\omega \right)}={tan}^{-1}\left(\frac{-\omega }{\alpha }\right).$

Figure 3

In general, a Fourier transform whose Cartesian form is $F\left(\omega \right)=R\left(\omega \right)+\text{i}I\left(\omega \right)$  has a polar form $F\left(\omega \right)=\left|F\left(\omega \right)\right|e^{\text{i}\varphi \left(\omega \right)}$  where $\varphi \left(\omega \right)\equiv \text{arg}F\left(\omega \right).$

Graphs, such as those shown in Figure 3, of $\left|F\left(\omega \right)\right|$ and $\text{arg}F\left(\omega \right)$ plotted against $\omega$ , are often referred to as magnitude spectra and phase spectra, respectively.

Exercises

1. Obtain the Fourier transform of the rectangular pulses
   1. $f\left(t\right)=\left\{\begin{array}{cc}\hfill 1\hfill & \hfill \left|t\right|\le 1\hfill \\ \hfill 0\hfill & \hfill \left|t\right|>1\hfill \end{array}\right.$
   2. $f\left(t\right)=\left\{\begin{array}{cc}\hfill \frac{1}{4}\hfill & \hfill \left|t\right|\le 3\hfill \\ \hfill \hfill \\ \hfill 0\hfill & \hfill \left|t\right|>3\hfill \end{array}\right.$
2. Find the Fourier transform of

   $f\left(t\right)=\left\{\begin{array}{cc}\hfill 1-\frac{t}{2}\hfill & \hfill 0\le t\le 2\hfill \\ \hfill \hfill \\ \hfill 1+\frac{t}{2}\hfill & \hfill -2\le t\le 0\hfill \\ \hfill \hfill \\ \hfill 0\hfill & \hfill \left|t\right|>2\hfill \end{array}\right.$

Answer