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Finally,
 
Finally,
  
<math>x[n] = 1e^{j\pi n}</math>
+
'''<math>x[n] = 1e^{j\pi n}</math>'''

Revision as of 17:31, 23 September 2008

For periodic DT signal, x[n] with fundamental period N:

$ x[n]=\displaystyle\sum_{k=0}^{n-1}a_ke^{jk\frac{2\pi}{N}n} $

The Fourier series coefficients can be calculated with:

$ a_k = \frac{1}{N}\displaystyle\sum_{n=0}^{N-1}x[n]e^{-jk\frac{2\pi}{N}n} $

Let us look for the Fourier series coefficients for the DT signal $ x[n] = cos(3\pi n) $

$ x[n] = cos(5\pi n) = \frac{e^{j5\pi n}+e^{-j5\pi n}}{2} = \frac{1}{2}e^{j4\pi n}e^{j\pi n} + \frac{1}{2}e^{-j4\pi n}e^{-j\pi n} = \frac{1}{2}e^{j\pi n} + \frac{1}{2}e^{-j\pi n} = \frac{1}{2}e^{j\pi n} + \frac{1}{2}e^{j\pi n} $

Finally,

$ x[n] = 1e^{j\pi n} $

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009