Fourier transform of heaviside functions.

• Apr 23rd 2008, 08:43 AM
Fourier transform of heaviside functions.
Well i found this question in my tutorial:
f(t) = 5[H(t-3) - H(t-11)], find the fourier transform.
Ans: (-10/w)(sin w)exp(-2iw)

i am totally blur on how to even start. I tried the general method of fourier transform but got no closer to the answer.

integrate from 3 to 5 for 5.exp(-iwt)

Couldn't find any table with Fourier transform for H(t-a).

All help are appreciated.
• Apr 23rd 2008, 09:42 AM
Peritus
$\displaystyle \begin{gathered} F\left\{ {5\left[ {u(t - 5) - u(t - 11)} \right]} \right\} = \hfill \\ \hfill \\ = 5\int\limits_3^{11} {e^{ - jwt} dt = \frac{5} {{jw}}\left( {e^{ - 3jw} - e^{ - 11jw} } \right) = } \hfill \\ \end{gathered}$

$\displaystyle \begin{gathered} \frac{{10}} {w}e^{ - 7jw} \frac{{e^{4jw} - e^{ - 4jw} }} {{2j}} = \frac{{10}} {w}e^{ - 7jw} \sin (4w) = \hfill \\ \hfill \\ = 40e^{ - 7jw} \sin c(4w) \hfill \\ \end{gathered}$

what we get is a pretty classic result: a square signal in the time domain transforms into sinc in the frequency domain.
The Fourier transform of a step function includes something called Dirac's delta function you can read more about this in the following article:

Fourier Transform--Heaviside Step Function -- from Wolfram MathWorld
• Apr 23rd 2008, 09:48 AM
$\displaystyle \frac{{10}} {w}e^{ - 7jw} \sin (4w)$