5 Fourier transform of the unit step function

We have already pointed out that although

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

we cannot simply replace $s$ by $\text{i}\omega$ to obtain the Fourier transform of the unit step.

We proceed via the Fourier transform of the signum function $\text{sgn}\left(t\right)$ which is defined as

$\text{sgn}t=\left\{\begin{array}{ccc}\hfill 1& \hfill & \hfill t>0\hfill \\ \hfill -1& \hfill & \hfill t<0\hfill \end{array}\right.$

Figure 10

We obtain $\mathcal{F}\left\{\text{sgn}\left(t\right)\right\}$ as follows. Consider the odd two-sided exponential function $f_{\alpha }\left(t\right)$ defined as

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

where $\alpha >0$ :

Figure 11

By slightly adapting our earlier calculation for the even two-sided exponential function we find

The parameter $\alpha$ controls how rapidly the exponential function varies:

Figure 12

As we let $\alpha \to 0$ the exponential function resembles more and more closely the signum function.

This suggests that

Task!

Write the unit step function in terms of the signum function and hence obtain $\mathcal{F}\left\{u\left(t\right)\right\}.$

First express $u\left(t\right)$ in terms of $\text{sgn}\left(t\right)$ :

Now, using the linearity property of Fourier transforms and previously obtained Fourier transforms, find $\mathcal{F}\left\{u\left(t\right)\right\}:$

Answer

Now, using the linearity property and previously obtained Fourier transforms, find $\mathcal{F}\left\{u\left(t\right)\right\}:$

Answer

Exercises

1. Use Parserval’s theorem and the Fourier transform of a ‘two-sided’ exponential function to show that

   ${\int \; <!--nolimits\; -->}_{-\infty }^{\infty }\frac{d\omega }{{\left(a^2+{\omega }^2\right)}^2}=\frac{\pi }{2{\left|a\right|}^3}$

2. Using $\mathcal{F}\left\{\text{sgn}\left(t\right)\right\}=\frac{2}{i\omega }$ find the Fourier transforms of
   1. $f_1\left(t\right)=\frac{1}{t}$
   2. $f_2\left(t\right)=\left|t\right|$

      Hence obtain the transforms of  

   3. $f_3\left(t\right)=-\frac{1}{t^2}$
   4. $f_4\left(t\right)=\frac{2}{t^3}$
3. Show that

   $\mathcal{F}\left\{sin{\omega }_{0}t\right\}=i\pi \left[\delta \left(\omega +{\omega }_0\right)-\delta \left(\omega -{\omega }_0\right)\right]$

   Verify your result using inverse Fourier transform properties.

Answer