(New page: == Linearity and Time Variance == From the given equation: <math>\,\!Y_k[n]=(k+1)^2\delta [n-(k+1)]</math> , It is clear that this is a non-time-invariant, or more simply, a time varia...)
 
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It is clear that this is a non-time-invariant, or more simply, a time variant function because the amplitude changing <math>(k+1)^2</math> term.
 
It is clear that this is a non-time-invariant, or more simply, a time variant function because the amplitude changing <math>(k+1)^2</math> term.
  
The inputs to this system are all shifted <math>\delta</math> functions, this means that if the <math>X[n]</math> inputs were changed to unit step functions <math>u[n]</math> , then the output will be a time shifted step function.
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The inputs to this system are all shifted <math>\delta</math> functions, this means that if the <math>X[n]</math> inputs were changed to unit step functions <math>u[n]</math> , then the output will be a time shifted step function. For this specific example we want an output of <math>Y[n]=u[n-1]</math> , so we can notice that when :
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<math>\,\!X[n]=\delta [n-1]</math> yields <math>\,\!Y_0[n]=\delta [n-1]</math> , then
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<math>\,\!X[n]=u[n-1]</math> will yield <math>\,\!Y_0[n]=u[n-1]</math>

Revision as of 15:22, 11 September 2008

Linearity and Time Variance

From the given equation:

$ \,\!Y_k[n]=(k+1)^2\delta [n-(k+1)] $ ,

It is clear that this is a non-time-invariant, or more simply, a time variant function because the amplitude changing $ (k+1)^2 $ term.

The inputs to this system are all shifted $ \delta $ functions, this means that if the $ X[n] $ inputs were changed to unit step functions $ u[n] $ , then the output will be a time shifted step function. For this specific example we want an output of $ Y[n]=u[n-1] $ , so we can notice that when :

$ \,\!X[n]=\delta [n-1] $ yields $ \,\!Y_0[n]=\delta [n-1] $ , then


$ \,\!X[n]=u[n-1] $ will yield $ \,\!Y_0[n]=u[n-1] $

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