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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 | 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 | + | <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> |
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<math>a_{0} = \frac{1}{2} \!</math> | <math>a_{0} = \frac{1}{2} \!</math> | ||
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| + | Now using the formula for <math>a_{k} \!</math> given above, we can find the remaining <math>a_{k} \!</math>s. | ||
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| + | <math>a_{k}= \frac{1}{4} \int_{-2}^{2}x(t)e^{-jk\frac{\pi}{2}t}dt</math> | ||
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| + | <math> = \frac{1}{4} \int_{-1}^{1}e^{-jk\frac{\pi}{2}t}dt</math> | ||
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| + | <math> = \frac{1}{4} \frac{e^{-jk\frac{\pi}{2}t}} {-jk\frac{\pi}{2}}|_{-1}^{1}</math> | ||
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| + | <math> = -\frac{1}{2jk\pi} (e^{-jk\frac{\pi}{2}} - e^{jk\frac{\pi}{2}})</math> | ||
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| + | <math> = \frac{sin(k\frac{\pi}{2})}{k\pi}</math> | ||
Latest revision as of 16:19, 14 October 2008
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
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}}) $
$ = \frac{sin(k\frac{\pi}{2})}{k\pi} $
