(New page: ==Problems== ##If a discrete time singal x[n] is periodic with period N, then the Fourier series coefficients <math>a_k</math> of the signal x[n] is also periodic with period N. For the pe...) |
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==Problems== | ==Problems== | ||
| − | + | 1)a)If a discrete time singal x[n] is periodic with period N, then the Fourier series coefficients <math>a_k</math> of the signal x[n] is also periodic with period N. For the periodic signal x[n], find the values of <math>a_0,a_1,...,a_{N-1}.</math> Express your answer in a + jb form. | |
| − | <math>x[n]=1-2je^{5+j( | + | <math>x[n]=1 - 2je^{5+j(3\pi/2*n+\pi/4)}</math> |
| − | + | ||
| + | 1)b) | ||
==Solutions== | ==Solutions== | ||
| − | + | 1)a) | |
| − | + | Change the equation to look like a forier series. <math>x[n]=1-2je^{5+j(3\pi/2*n+\pi/4)}=</math> | |
| + | <math>x[n]=1 - 2je^{\pi/4}e^{5}e^{j(3\pi/2)*n}=</math> | ||
| + | <math>x[n]=1 + (\sqrt{2}e^{5} - j\sqrt{2}e^5)e^{j(3\pi/2*n)}</math> | ||
| + | |||
| + | find the period N. <math>\omega_0=3\pi/2</math>. <math>P=2\pi/\omega_0=4/3</math>. The LCM of 1 and 4/3 is 12. Therefore N=12 | ||
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| + | Choose values of <math>a_k</math> that match the forier series repersentation of x[n] | ||
| + | |||
| + | <math>x[n]=\sum_{k=<n>}^{}a_ke^{jk(2\pi/12)n}</math> | ||
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| + | <math>a_0=1</math> | ||
| + | |||
| + | <math>a_9=\sqrt{2}e^{5} - j\sqrt{2}e^5</math> | ||
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--[[User:Krtownse|Krtownse]] 15:53, 18 July 2008 (EDT) | --[[User:Krtownse|Krtownse]] 15:53, 18 July 2008 (EDT) | ||
| + | [[Category:ECE 301 San Summer 2008]] | ||
Revision as of 16:51, 18 July 2008
Problems
1)a)If a discrete time singal x[n] is periodic with period N, then the Fourier series coefficients $ a_k $ of the signal x[n] is also periodic with period N. For the periodic signal x[n], find the values of $ a_0,a_1,...,a_{N-1}. $ Express your answer in a + jb form.
$ x[n]=1 - 2je^{5+j(3\pi/2*n+\pi/4)} $
1)b)
Solutions
1)a) Change the equation to look like a forier series. $ x[n]=1-2je^{5+j(3\pi/2*n+\pi/4)}= $ $ x[n]=1 - 2je^{\pi/4}e^{5}e^{j(3\pi/2)*n}= $ $ x[n]=1 + (\sqrt{2}e^{5} - j\sqrt{2}e^5)e^{j(3\pi/2*n)} $
find the period N. $ \omega_0=3\pi/2 $. $ P=2\pi/\omega_0=4/3 $. The LCM of 1 and 4/3 is 12. Therefore N=12
Choose values of $ a_k $ that match the forier series repersentation of x[n]
$ x[n]=\sum_{k=<n>}^{}a_ke^{jk(2\pi/12)n} $
$ a_0=1 $
$ a_9=\sqrt{2}e^{5} - j\sqrt{2}e^5 $
--Krtownse 15:53, 18 July 2008 (EDT)
