(New page: ==Periodic Signal== Let <math>x[n] = e^{jn}</math> For x[n] to be periodic, the following must hold true: <math>e^{jn} = e^{j(n+N)}</math> ==Non Periodic Signal==)
 
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A discrete time signal is periodic if there exists T > 0 such that x(t + T) = x(t)
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A continuous time signal is periodic if there exists some integer N > 0 such that x[n + N] = x[n]
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==Periodic Signal==  
 
==Periodic Signal==  
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Let <math>Insert formula here</math>
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==Non Periodic Signal==
  
 
Let <math>x[n] = e^{jn}</math>
 
Let <math>x[n] = e^{jn}</math>
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For x[n] to be periodic, the following must hold true:
 
For x[n] to be periodic, the following must hold true:
  
<math>e^{jn} = e^{j(n+N)}</math>
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<math>e^{jn} = e^{j(n+N)}</math> for some integer N
  
==Non Periodic Signal==
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<math>e^{jn} = e^{jn}  e^{jN}</math>
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<math>1 = e^{jN}</math>
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1 = cos(N) + jsin(N)
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This equation only holds true if <math>N = 2\pi</math> or some multiple of <math>2\pi</math>
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Therefore<math>x[n] = e^{jn}</math> is not periodic because <math>2\pi</math> is not an integer.

Revision as of 08:23, 4 September 2008

A discrete time signal is periodic if there exists T > 0 such that x(t + T) = x(t)

A continuous time signal is periodic if there exists some integer N > 0 such that x[n + N] = x[n]

Periodic Signal

Let $ Insert formula here $

Non Periodic Signal

Let $ x[n] = e^{jn} $

For x[n] to be periodic, the following must hold true:

$ e^{jn} = e^{j(n+N)} $ for some integer N

$ e^{jn} = e^{jn} e^{jN} $

$ 1 = e^{jN} $

1 = cos(N) + jsin(N)

This equation only holds true if $ N = 2\pi $ or some multiple of $ 2\pi $

Therefore$ x[n] = e^{jn} $ is not periodic because $ 2\pi $ is not an integer.

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Seraj Dosenbach