Revision as of 16:15, 14 October 2008 by Zcurosh (Talk)

Most Difficult Problem on First Test

The problem that I found most difficult was problem number 4.

4. Compute the coefficients $ a_{k} \! $ of the Fourier series signal $ x(t) \! $ periodic with period $ T = 4 \! $ defined by

$ x(t)= \left\{ \begin{array}{ll}0&, -2<t<-1\\ 1&, -1\leq t\leq 1\\ 0&, 1<t\leq 2\end{array}\right. $


Solution

We know that

$ a_{k} = \frac{1}{T} \int_{0}^{T}x(t)e^{-jk\omega _{o}t}dt $, and since T = 4,

$ a_{k}= \frac{1}{4} \int_{0}^{4}x(t)e^{-jk\frac{\pi}{2}t}dt $


By the definition of $ a_{0} \! $, we know that it is the average of the signal over the period. In this case,

$ a_{0} = \frac{1}{2} \! $

Now using the formula for $ a_{k} \! $ given above, we can find the remaining $ a_{k} \! $s.


$ a_{k}= \frac{1}{4} \int_{-2}^{2}x(t)e^{-jk\frac{\pi}{2}t}dt $

$ = \frac{1}{4} \int_{-1}^{1}e^{-jk\frac{\pi}{2}t}dt $

$ = \frac{1}{4} \frac{e^{-jk\frac{\pi}{2}t}} {-jk\frac{\pi}{2}}|_{-1}^{1} $

$ = -\frac{1}{2jk\pi} (e^{-jk\frac{\pi}{2}} - e^{jk\frac{\pi}{2}}) $

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