(Define a periodic DT signal and compute its Fourier series coefficients.)
Line 2: Line 2:
 
For DT,  
 
For DT,  
  
<math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math>
+
<math>x[n]=\sum_{k=0}^{N-1} a_k e^{jk\frac{2\pi}{N} n}</math>
  
 
where
 
where
  
<math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math>.
+
<math>a_k=\frac{1}{N}\sum_{n=0}^{N-1} x[n] e^{-jk\frac{2\pi}{N} n} </math>.
  
 
Let the signal be
 
Let the signal be

Revision as of 10:17, 25 September 2008

Define a periodic DT signal and compute its Fourier series coefficients.

For DT,

$ 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} $.

Let the signal be

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