(DT signal & its Fourier Coefficients)
(DT signal & its Fourier Coefficients)
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<math>\ x[n] = 5sin(3 \pi n + \frac{4}{\pi})</math>
 
<math>\ x[n] = 5sin(3 \pi n + \frac{4}{\pi})</math>
  
<math>\ x[n] = \frac{5}{2j} (e^{j 3\pi n}-e^{-j 3\pi n})
+
<math>\ x[n] = \frac{5}{2j} (e^{j 3\pi n - \frac{4}{\pi}}-e^{-j 3\pi n})

Revision as of 16:06, 26 September 2008

DT signal & its Fourier Coefficients

$ \ x[n] = 5sin(3 \pi n + \frac{4}{\pi}) $

$ \ x[n] = \frac{5}{2j} (e^{j 3\pi n - \frac{4}{\pi}}-e^{-j 3\pi n}) $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood