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<math>a_1=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-j\pi n} </math> | <math>a_1=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-j\pi n} </math> | ||
| − | <math>a_1=\frac{1}{2}(x[0] e^{0} + x[1]e^{-j\pi}) </math> | + | <math>a_1=\frac{1}{2}(x[0] e^{0} + x[1]e^{-j\pi})=1 </math> |
| − | <math> | + | |
| + | Finally, | ||
| + | |||
| + | <math>a_{even k}=0</math> | ||
| + | |||
| + | <math>a_{odd k}=1</math> | ||
Revision as of 04:34, 25 September 2008
Fourier Series for DT signals
Let $ x[n]\, $ be a periodic DT signal with fundamental period N.
Then $ x[n]=\sum_{k=0}^{N-1} a_k e^{jk\frac{2\pi}{N} n} $
where $ a_k=\frac{1}{N}\sum_{n=0}^{N-1} x[n] e^{-jk\frac{2\pi}{N} n} $
note that $ \frac{2\pi}{N} =\omega_0 $
Now consider the signal $ x[n]=sin(3 \pi)\, $
It's periodic because $ \frac{\omega_0}{2\pi} = \frac{3\pi}{2\pi} =1.5 $ is a rational number.
Notice that $ x[0]= 1, x[1]=-1, x[2]=1, x[3]=-1 \, $ etc
Thus the fundamental period is 2.
Then...
$ a_0=\frac{1+(-1)}{2}=0 $
$ a_1=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-j1\frac{2\pi}{2} n} $
$ a_1=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-j\pi n} $
$ a_1=\frac{1}{2}(x[0] e^{0} + x[1]e^{-j\pi})=1 $
Finally,
$ a_{even k}=0 $
$ a_{odd k}=1 $
