Principles of Fourier Analysis, 2nd Edition

24

Convolution and Transforms of Products

Earlier, we obtained

$y\left(t\right)={{\mathcal{F}}^{-1}\left[\frac{2\text{sinc}\left(2\pi \omega \right)}{3+i2\pi \omega }\right]|}_t$

as a solution to a differential equation (see example 22.2 on page 343). At the time we did not attempt to further evaluate it, because, well, it just looked too darned hard.

Let us reconsider this formula. It is the product of two relatively simple functions,

$F\left(\omega \right)=2\text{sinc}\left(2\pi \omega \right)\text{and}G\left(\omega \right)=\frac{1}{3+i2\pi \omega },$

whose inverse transforms,

$f\left(t\right)={{\mathcal{F}}^{-1}\left[F\right]|}_t={\text{pulse}}_1\left(t\right)\text{and}\text{g}\left(t\right)={{\mathcal{F}}^{-1}\left[G\right]|}_t=e^{-3t}\text{step}\left(t\right),$

can be found by such elementary means as looking them up in table 21.1 on page 320. An obvious question now arises: Is there a relatively simple formula of f (t) and g(t) that can be relied on to give the inverse transform of the product $F\left(\omega \right)G\left(\omega \right)$ ?

The answer ...