| Line 12: | Line 12: | ||
---- | ---- | ||
Ans: | Ans: | ||
| − | <math> a_0= </math>average of x(t) = <math> | + | |
| + | <math>a_0 = </math>average of x(t) = <math>\frac{1x2}{4}=\frac{1}{2}; </math> | ||
T=4, <math>\omega _o = \frac{2 \pi} {4} = \frac{\pi}{2}</math> | T=4, <math>\omega _o = \frac{2 \pi} {4} = \frac{\pi}{2}</math> | ||
Latest revision as of 08:12, 15 October 2008
Question 4 Compute the coefficients $ a_k $ of the Fourier series of the signal $ x(t) $ periodic with period $ T=4 $ defined by $ \,x(t)=\left\{\begin{array}{cc} 0, & -2<t<-1 \\ 1, & -1\leq t\leq 1 \\ 0, & 1<t\leq 2 \end{array} \right. \, $
(Simplify your answer as much as possible.)
Ans:
$ a_0 = $average of x(t) = $ \frac{1x2}{4}=\frac{1}{2}; $
T=4, $ \omega _o = \frac{2 \pi} {4} = \frac{\pi}{2} $
for k<>0,
$ a_k=\frac{1}{T}\int_{0}^{T}x(t)e^{-jk\frac{2\pi}{T}t}dt $
$ a_k=\frac{1}{4}\int_{-2}^{2}x(t)e^{-jk\frac{\pi}{2}t}dt $
$ a_k=\frac{1}{4}\int_{-1}^{1}x(t)e^{-jk\frac{\pi}{2}t}dt $
$ a_k=\frac{1}{4} \left.\frac{e^{-jk\frac{\pi}{2}t}}{-jk\frac{\pi}{2}}\right|_{-1}^{1} $
$ a_k=\frac{1}{4} (\frac{e^{-jk\frac{\pi}{2}t}}{-jk\frac{\pi}{2}}-\frac{e^{jk\frac{\pi}{2}t}}{jk\frac{\pi}{2}}) $
$ a_k=\frac{1}{2jk\pi}(e^{jk\frac{\pi}{2}}-e^{-jk\frac{\pi}{2}}) $
$ a_k=\frac{sin(k\frac{\pi}{2})}{k\pi} $
