View source for HW6.3 Zachary Curosh ECE301Fall2008mboutin - Rhea

== Most Difficult Problem on First Test ==
The problem that I found most difficult was problem number 4.

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

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


== Solution ==
We know that 

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

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


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

<math>a_{0} = \frac{1}{2} \!</math>

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


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

<math>     = \frac{1}{4} \int_{-1}^{1}e^{-jk\frac{\pi}{2}t}dt</math>

<math>     = \frac{1}{4} \frac{e^{-jk\frac{\pi}{2}t}} {-jk\frac{\pi}{2}}|_{-1}^{1}</math>

<math>     = -\frac{1}{2jk\pi} (e^{-jk\frac{\pi}{2}} - e^{jk\frac{\pi}{2}})</math>

<math>     = \frac{sin(k\frac{\pi}{2})}{k\pi}</math>