| Line 1: | Line 1: | ||
==Test Problem 4== | ==Test Problem 4== | ||
| + | 4. Compute the coefficients <math>a_{k}</math> of the Fourier series of the signal x(t) with period T = 4 defined by | ||
| + | |||
| + | <math>x(t) = \left\{ \begin{array}{ll} 0 &, -2<t<-1\\ n+1 &, \text{ else}\end{array}\right. </math> | ||
<math>a_{k} = \frac{1}{T} \int_{0}^{T}x(t)e^{-jk\omega _{o}t}dt</math> | <math>a_{k} = \frac{1}{T} \int_{0}^{T}x(t)e^{-jk\omega _{o}t}dt</math> | ||
Revision as of 19:30, 8 October 2008
Test Problem 4
4. Compute the coefficients $ a_{k} $ of the Fourier series of the signal x(t) with period T = 4 defined by
$ x(t) = \left\{ \begin{array}{ll} 0 &, -2<t<-1\\ n+1 &, \text{ else}\end{array}\right. $
$ a_{k} = \frac{1}{T} \int_{0}^{T}x(t)e^{-jk\omega _{o}t}dt $
From the problem statement we know that T=4
$ = \frac{1}{4} \int_{0}^{4}x(t)e^{-jk\frac{2\pi}{4}t}dt $
Knowing that T=4 we can visualize the periodic signal in the range $ 0 \leq t \leq 4 $. x(t) = 1 for $ 0 \leq t \leq 1 $ and $ 3 \leq t \leq 4 $. Otherwise, x(t) = 0. Therefore:
$ = \frac{1}{4} \int_{0}^{1}e^{-jk\frac{\pi}{2}t}dt + \frac{1}{4} \int_{3}^{4}e^{-jk\frac{\pi}{2}t}dt $
$ = \frac{1}{4}\frac{j}{\frac{\pi}{2}k}e^{-jk\frac{\pi}{2}t}|_{0}^{1} + \frac{1}{4}\frac{j}{\frac{\pi}{2}k}e^{-jk\frac{\pi}{2}t}|_{3}^{4} $
$ =\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}} - 1 + e^{-jk2\pi} - e^{-jk\frac{3\pi}{2}}) $
$ =\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}} - 1 + 1 - e^{-jk\frac{4\pi}{2}}e^{jk\frac{\pi}{2}}) $
$ =\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}) $
$ =\frac{1}{\pi k}(\frac{e^{jk\frac{\pi}{2}}-e^{-jk\frac{\pi}{2}}}{2j}) $
$ =\frac{1}{\pi k}sin(\frac{\pi}{2}k) $
But we are not done yet!
A quick glance at the above result reveals that $ a_{0} $ is undefined.
But $ a_{0} $ is simply equal to the average value of the function over 1 period.
$ a_{0}=\frac{2}{4} = \frac{1}{2} $
So in conclusion:
$ a_{k}=\frac{1}{\pi k}sin(\frac{\pi}{2}k) $ for all $ k \neq 0, \; a_{0}=\frac{1}{2} $
