Line 3: Line 3:
 
   <math>x[n] = e^{jw_{o}n}</math>         
 
   <math>x[n] = e^{jw_{o}n}</math>         
 
   <math>x[n] = x[n+N] = e^{jw_{o}(n+N)} = e^{jw_{o}n}e^{jw_{o}N}</math>
 
   <math>x[n] = x[n+N] = e^{jw_{o}(n+N)} = e^{jw_{o}n}e^{jw_{o}N}</math>
   <math>Insert formula here</math>
+
   <math>\therefore e^{jw_{o}N = 1 = e^{j2\pi k}}</math>

Revision as of 15:07, 30 June 2008

(a) Derive the condition for which the discrete time complex exponetial signal x[n] is periodic.

 $ x[n] = e^{jw_{o}n} $         
 $ x[n] = x[n+N] = e^{jw_{o}(n+N)} = e^{jw_{o}n}e^{jw_{o}N} $
 $ \therefore e^{jw_{o}N = 1 = e^{j2\pi k}} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood