3 Inversion of the Fourier transform

Formal inversion of the Fourier transform, i.e. finding $f\left(t\right)$ for a given $F\left(\omega \right)$ , is sometimes possible using the inversion integral (4). However, in elementary cases, we can use a Table of standard Fourier transforms together, if necessary, with the appropriate properties of the Fourier transform.

The following Examples and Tasks involve such inversion.

Example 3

Find the inverse Fourier transform of   $F\left(\omega \right)=20\frac{sin5\omega }{5\omega }.$

Solution

The appearance of the sine function implies that $f\left(t\right)$ is a symmetric rectangular pulse.

We know the standard form   $\mathcal{F}\left\{p_a\left(t\right)\right\}=2a\frac{sin\omega a}{\omega a}$  or   ${\mathcal{F}}^{-1}\left\{2a\frac{sin\omega a}{\omega a}\right\}=p_a\left(t\right).$

Putting $a=5$   ${\mathcal{F}}^{-1}\left\{10\frac{sin5\omega }{5\omega }\right\}=p_5\left(t\right).$  Thus, by the linearity property

$f\left(t\right)={\mathcal{F}}^{-1}\left\{20\frac{sin5\omega }{5\omega }\right\}=2p_5\left(t\right)$

Figure 4

Example 4

Find the inverse Fourier transform of  $G\left(\omega \right)=20\frac{sin5\omega }{5\omega }\text{exp}\left(-3\text{i}\omega \right).$

Solution

The occurrence of the complex exponential factor in the Fourier transform suggests the time-shift property with the time shift $t_0=+3$ (i.e. a right shift).

From Example 3

${\mathcal{F}}^{-1}\left\{20\frac{sin5\omega }{5\omega }\right\}=2p_5\left(t\right)$ so $g\left(t\right)={\mathcal{F}}^{-1}\left\{20\frac{sin5\omega }{5\omega }{e}^{-3\text{i}\omega }\right\}=2p_5\left(t-3\right)$

Figure 5

Task!

Find the inverse Fourier transform of

$H\left(\omega \right)=6\frac{sin2\omega }{\omega }{e}^{-4\text{i}\omega }.$

Firstly ignore the exponential factor and find the inverse Fourier transform of the remaining terms:

Answer

Now take account of the exponential factor:

Answer

Example 5

Find the inverse Fourier transform of

$K\left(\omega \right)=\frac{2}{1+2\left(\omega -1\right)\text{i}}$

Solution

The presence of the term $\left(\omega -1\right)$ instead of $\omega$ suggests the frequency shift property.

Hence, we consider first

$\stackrel{̂}{K}\left(\omega \right)=\frac{2}{1+2\text{i}\omega }.$

The relevant standard form is

$\mathcal{F}\left\{e^{-\alpha t}u\left(t\right)\right\}=\frac{1}{\alpha +\text{i}\omega }\text{or}{\mathcal{F}}^{-1}\left\{\frac{1}{\alpha +\text{i}\omega }\right\}=e^{-\alpha t}u\left(t\right).$

Hence, writing $\stackrel{̂}{K}\left(\omega \right)=\frac{1}{\frac{1}{2}+\text{i}\omega }$ $\stackrel{̂}{k}\left(t\right)=e^{-\frac{1}{2}t}u\left(t\right).$

Then, by the frequency shift property with ${\omega }_0=1k\left(t\right)={\mathcal{F}}^{-1}\left\{\frac{2}{1+2\left(\omega -1\right)\text{i}}\right\}=e^{-\frac{1}{2}t}{e}^{\text{i}t}u\left(t\right).$

Here $k\left(t\right)$ is a complex time-domain signal.

Task!

Find the inverse Fourier transforms of

1. $L\left(\omega \right)=2\frac{sin\left\{3\left(\omega -2\pi \right)\right\}}{\left(\omega -2\pi \right)}$
2. $M\left(\omega \right)=\frac{e^{\text{i}\omega }}{1+\text{i}\omega }$

Answer