View source for Fourier series coefficients pulse train CT ECE301S11 - Rhea

[[Category:ECE301Spring2011Boutin]]
[[Category:ECE301]]
[[Category:signals and systems]]
[[Category:Problem_solving]] 
[[Category:Fourier series]]

= [[:Category:Problem_solving|Practice Question]] on Computing the Fourier Series continuous-time signal=
Obtain the Fourier series the CT signal

<math>
x(t) = \left\{ 
\begin{array}{ll}
1, & \text{ for } -5\leq t \leq 5,\\
0, & \text{ for } 5< |t| \leq 10,
\end{array}
\right.  \ </math>

x(t) periodic with period 20.

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===Answer 1===
<math>k=0\,</math>

<math>a_0=\frac{1}{20}\int_{-10}^{10}x(t)e^{-0}dt=\frac{1}{20}\int_{-5}^{5}1dt=\frac{1}{2}</math>

<math>k\ne0</math> 

<math>a_k=\frac{1}{20}\int_{-10}^{10}x(t)e^{-jkw_0t}dt=\frac{1}{20}\int_{-5}^{5}e^{-jk\frac{\pi}{10}t}dt=\frac{1}{20}\Bigg[\frac{e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}}{-jk\frac{\pi}{10}}\Bigg]=\frac{1}{-jk2\pi}\Bigg(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}\Bigg)</math>

<math>x(t)=\frac{1}{2}e^{-jk\frac{\pi}{10}t}+\sum_{k=-\infty}^-1\frac{1}{-jk2\pi}\Bigg(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}\Bigg)e^{-jk\frac{\pi}{10}t}+\sum_{k=1}^\infty\frac{1}{-jk2\pi}\Bigg(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}\Bigg)e^{-jk\frac{\pi}{10}t}</math>

--[[User:Cmcmican|Cmcmican]] 21:35, 7 February 2011 (UTC)

:TA's comment: That looks fine. The expression for <math>a_k</math> is better written in terms of a sin function, though. Regarding the synthesis of <math>x(t)</math>, you got it wrong actually. The complex exponentials should not have a minus sign in their exponents and for <math>k=0</math> the complex exponential has a frequency of zero (DC). 

:Another thing is that you may also further simplify <math>x(t)</math> and write it in terms of sin waves only. You will actually notice some pattern in the frequencies of these sin waves.
===Answer 2===
First I plotted it out, thanks to WolframAlpha [[Image:EthanHall_ECE301_S11_Wolframalpha-20110215211454118.gif‎]]

Then I used the equation <math class="inline"> a_{k} = \frac{1}{T}\int_{0}^{20}x(t)e^{-jk(\frac{2\pi}{T})t}dt </math> where T = 20. You can change the limits of the integral to -10 and 10 since the function is periodic. We just need it over one period.

Solving for <math class="inline">a_{0}</math> first we get <math class="inline"> a_{0} = \frac{1}{20} \int_{-10}^{10}x(t)e^{-j(0)(\frac{2\pi}{20})t}dt </math>. We know that x(t) = 0 except for -5 < t < 5 where it is 1.

<math> a_{0} = \frac{1}{20} \int_{-5}^{5}(1)e^{(0)}dt = \frac{1}{20}x \bigg|_{-5}^{5} = \frac{1}{2} </math>

<math > a_{k} = \frac{1}{20}\int_{-5}^{5}(1)e^{-jk(\frac{2\pi}{20})t}dt; \text{ where } -5 < k <5 \text{ and }k \neq 0 </math>

<math > a_{k} = \frac{1}{20} \left[ \frac{e^{-jk(\frac{\pi}{10})t}}{-jk(\frac{\pi}{10})}\right]_{-5}^{5} </math>

<math >a_{k} = \frac{1}{20}\left(\frac{1}{-jk\frac{\pi}{10}}\right) \left[ e^{-jk(\frac{\pi}{10})5} - e^{-jk(\frac{\pi}{10})(-5)}  \right] </math> 

<math>a_{k}= \frac{-1}{2jk}\left[ e^{-\frac{jk\pi}{2}} - e^{\frac{jk\pi}{2}}  \right]\text{ for }-5 < k < 5\text{ and }k \neq 0\text{, all others }0</math>

<math>x(t) = \sum_{k=-\infty}^{\infty}a_{k}e^{jk\frac{2\pi}{T}t}</math>

<math>x(t) = \sum_{k=-5}^{-5} a_{k}e^{jk\frac{ \pi}{10}t}</math> I changed the limits of the sum because <math class="inline">a_{k}</math> is 0 for anything outside that range.

===Answer 3===
Write it here.
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