(New page: <math>x[n]=\displaystyle\sum_{k=0}^{n-1}a_ke^{jk\frac{2\pi}{N}n}</math> <math>a_k=\frac{1}{N}\sum_{n=0}^{N-1} x[n] e^{-jk\frac{2\pi}{N} n} </math> <math>\frac{2\pi}{N} =\omega_0</math> ...) |
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<math>x[n]=3+ \frac{1}{2i} e^{i\frac{\pi}{2}n}-\frac{1}{2i}e^{-i\frac{\pi}{2}n}</math> | <math>x[n]=3+ \frac{1}{2i} e^{i\frac{\pi}{2}n}-\frac{1}{2i}e^{-i\frac{\pi}{2}n}</math> | ||
− | + | <math>x[0]=3,x[1]=4,x[2]=3,x[3]=2,x[4]=3, x[5]=4\!</math> | |
− | + | ||
So, the fundamental period is 4. | So, the fundamental period is 4. | ||
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== Fourier Series Coefficients == | == Fourier Series Coefficients == | ||
− | + | <math>a_0=3\!</math> | |
<math>a_1=\frac{1}{2i}</math> | <math>a_1=\frac{1}{2i}</math> | ||
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<math>a_{-1}=-\frac{1}{2i}</math> | <math>a_{-1}=-\frac{1}{2i}</math> | ||
− | + | <math>a_k,even=3\!</math> | |
− | <math>a_k=\frac{1}{2i}</math> for all <math>k=1,5,9,13,..., 4n+1</math> | + | <math>a_k=\frac{1}{2i}</math> for all <math>k=1,5,9,13,..., 4n+1\!</math> |
− | <math>a_k=-\frac{1}{2i}</math> for all <math>k=3,7,11,15,..., 4n+3</math> | + | <math>a_k=-\frac{1}{2i}</math> for all <math>k=3,7,11,15,..., 4n+3\!</math> |
Latest revision as of 13:35, 26 September 2008
$ x[n]=\displaystyle\sum_{k=0}^{n-1}a_ke^{jk\frac{2\pi}{N}n} $
$ a_k=\frac{1}{N}\sum_{n=0}^{N-1} x[n] e^{-jk\frac{2\pi}{N} n} $
$ \frac{2\pi}{N} =\omega_0 $
Signal
$ x[n]=3+sin\bigg(\frac{\pi}{2}n\bigg) $
$ \frac{w_0}{2\pi}=\frac{1}{4} $, which is rational, so it is periodic.
Fourier Series
$ x[n]=3+sin\bigg(\frac{\pi}{2}n\bigg) $
$ x[n]=3+\frac{e^{i\frac{\pi}{2}n}-e^{-i\frac{\pi}{2}n}}{2i} $
$ x[n]=3+ \frac{1}{2i} e^{i\frac{\pi}{2}n}-\frac{1}{2i}e^{-i\frac{\pi}{2}n} $
$ x[0]=3,x[1]=4,x[2]=3,x[3]=2,x[4]=3, x[5]=4\! $
So, the fundamental period is 4.
Fourier Series Coefficients
$ a_0=3\! $
$ a_1=\frac{1}{2i} $
$ a_{-1}=-\frac{1}{2i} $
$ a_k,even=3\! $
$ a_k=\frac{1}{2i} $ for all $ k=1,5,9,13,..., 4n+1\! $
$ a_k=-\frac{1}{2i} $ for all $ k=3,7,11,15,..., 4n+3\! $