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<math>= \int_{-\infty}^{\infty}\frac{e^{j\pi t}-e^{-j\pi t}}{2j}e^{-j\omega t} dt</math> | <math>= \int_{-\infty}^{\infty}\frac{e^{j\pi t}-e^{-j\pi t}}{2j}e^{-j\omega t} dt</math> | ||
| − | <math>= \frac{1}{2j}\int_{-\infty}^{\infty}e^{-jt(\omega | + | <math>= \frac{1}{2j}\int_{-\infty}^{\infty}e^{-jt(\omega - \pi)}dt-\frac{1}{2j}\int_{-\infty}^{\infty}e^{-jt(\omega + \pi)}dt</math> |
Revision as of 18:43, 23 October 2008
Introduction
There are 10 minutes left in the exam. You have done pretty well so far and get to the last problem. You hope it's doable in ten minutes. The question is something like this:
Find the F.T. of $ x(t) = sin(\pi t) $. Simply using the formula in the table will result in no points. You must prove the F.T. of x(t).
Without thinking, you rush into the problem using the definition of the Fourier Transform.
$ X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t} dt $
$ = \int_{-\infty}^{\infty}sin(\pi t)e^{-j\omega t} dt $
You don't know how to integrate that, but you remember that you can rewrite $ sin(\pi t) $.
$ = \int_{-\infty}^{\infty}\frac{e^{j\pi t}-e^{-j\pi t}}{2j}e^{-j\omega t} dt $
$ = \frac{1}{2j}\int_{-\infty}^{\infty}e^{-jt(\omega - \pi)}dt-\frac{1}{2j}\int_{-\infty}^{\infty}e^{-jt(\omega + \pi)}dt $
