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

Revision as of 16:13, 26 September 2008

DT signal & its Fourier Coefficients

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

Knowing its Fourier series is

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

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood