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==DT Perodic function== | ==DT Perodic function== | ||
| − | <math>x[n]= | + | |
| + | Find the Fourier Series coefficients of x[n] | ||
| + | :<math>x[n]=5cos(5/2\pi n +\pi) \, </math> | ||
| + | :<math>N=\dfrac{2\pi}{5/2\pi}K </math>, where K is the smallest integer, that makes N an integer. | ||
| + | :<math> K = 5,N = 4\,</math> | ||
| + | :<math>x[n]=5cos(5/2 j \pi n +\pi) = \dfrac{5(e^{5/2 j \pi n+\pi}-e^{-5/2\pi n-\pi})}{2} \,</math> | ||
| + | :<math>x[n]= \dfrac{5}{2}(e^{5/2\pi n}e^{\pi}-e^{-5/2\pi n}e^{-\pi}) \,</math> | ||
| + | :<math>e^{\pi}=-1,e^{-\pi}=1 \,</math> | ||
| + | :<math>x[n]= \dfrac{5}{2}(e^{-5/2\pi n}-e^{5/2\pi n}) \,</math> | ||
Revision as of 10:19, 23 September 2008
DT Perodic function
Find the Fourier Series coefficients of x[n]
- $ x[n]=5cos(5/2\pi n +\pi) \, $
- $ N=\dfrac{2\pi}{5/2\pi}K $, where K is the smallest integer, that makes N an integer.
- $ K = 5,N = 4\, $
- $ x[n]=5cos(5/2 j \pi n +\pi) = \dfrac{5(e^{5/2 j \pi n+\pi}-e^{-5/2\pi n-\pi})}{2} \, $
- $ x[n]= \dfrac{5}{2}(e^{5/2\pi n}e^{\pi}-e^{-5/2\pi n}e^{-\pi}) \, $
- $ e^{\pi}=-1,e^{-\pi}=1 \, $
- $ x[n]= \dfrac{5}{2}(e^{-5/2\pi n}-e^{5/2\pi n}) \, $
