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...
The Wrong Way
$ 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 $
$ = \frac{1}{-2jj(\omega -\pi)}e^{-jt(\omega - \pi)}|_{-\infty}^{\infty}-\frac{1}{-2jj(\omega + \pi)}e^{-jt(\omega + \pi)}|_{-\infty}^{\infty} $
$ = \frac{1}{2(\omega -\pi)}e^{-jt(\omega - \pi)}|_{-\infty}^{\infty}-\frac{1}{2(\omega + \pi)}e^{-jt(\omega + \pi)}|_{-\infty}^{\infty} $
And now you realize that you're in trouble because you have infinite terms as a consequence of the limits of integration, and you can't solve the problem. The ten minutes is up and you receive 0 points for all your work, maybe 1 if you're lucky. Don't do this! You have to look ahead to see what to do.
The Right Way
Just because you can't simply look at the F.T. table and write down the answer doesn't mean you can't use the table to your advantage. Also, by this time you should know that the CT F.T. of $ sin(\omega _{o} t) $ is $ \frac{\pi}{j}[\delta(\omega - \omega _{o}) - \delta(\omega + \omega _{0})] $. If not, though, chances are it would be provided in the F.T. table at the back of the test.
A quick look ahead would reveal that the F.T. of x(t) has deltas in it. Do you know how to get deltas out of an integral? Spending 10 seconds thinking about the problem can save you 10 minutes down the road.
The key is to start with the F.T. of x(t) found in the table or from memory and use the inverse F.T. to work backward. In this manner, you can prove the F.T. of x(t).
$ X(\omega) = \frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)] $
$ x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t} d\omega $
$ = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)]e^{j\omega t} d\omega $
$ = \frac{1}{2j}\int_{-\infty}^{\infty}[\delta(\omega - \pi) - \delta(\omega + \pi)]e^{j\omega t} d\omega $
$ = \frac{1}{2j}\int_{-\infty}^{\infty}\delta(\omega - \pi)e^{j\omega t} d\omega -\frac{1}{2j}\int_{-\infty}^{\infty} \delta(\omega + \pi)e^{j\omega t} d\omega $
$ = \frac{1}{2j}e^{j\pi t} -\frac{1}{2j}e^{-j\pi t} $
$ = \frac{e^{j\pi t} - e^{-j\pi t}}{2j} $
$ = sin(\pi t) $
$ = x(t) $
Therefore, the F.T. of x(t) is $ \frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)] $.
As it turned out, the solution to this problem was almost trivial. Remember to take time to plan before jumping into the problem. It could end up being the difference of a letter grade on an exam.
