Revision as of 10:55, 14 February 2011

Practice Question on Computing the Fourier Series continuous-time signal

Obtain the Fourier series the CT signal

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

x(t) periodic with period 20.

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Answer 1

$ k=0\, $

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

$ k\ne0 $

$ 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) $

$ 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} $

--Cmcmican 21:35, 7 February 2011 (UTC)

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

Another thing is that you may also further simplify $ x(t) $ and write it in terms of sin waves only. You will actually notice some pattern in the frequencies of these sin waves.

Answer 2

Write it here.

Answer 3

Write it here.

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