(New page: == Fourier Series for DT signals == Let <math>x[n]\,</math> be a periodic DT signal with fundamental period N. Then <math>x[n]=\sum_{k=0}^{N-1} a_k e^{jk\frac{2\pi}{N} n}</math> where ...) |
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| − | where <math>a_k=\frac{1}{ | + | where <math>a_k=\frac{1}{N}\sum_{n=0}^{N-1} x_n e^{-jk\frac{2\pi}{N} n} </math> |
note that <math>\frac{2\pi}{N} =\omega_0</math> | note that <math>\frac{2\pi}{N} =\omega_0</math> | ||
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| + | Now consider the signal <math>x[n]=sin(3 \pi)\,</math> | ||
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| + | It's periodic because <math>\frac{\omega_0}{2\pi} = \frac{\3\pi}{2\pi} =1.5</math>is a rational number. | ||
Revision as of 04:15, 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.
