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<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> | + | <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}} $