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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 : | 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 : | ||
| − | <math>\,\!X[n]=\delta [n | + | <math>\,\!X[n]=\delta [n]</math> yields <math>\,\!Y_0[n]=\delta [n-1]</math> , then |
| − | <math>\,\!X[n]=u[n | + | <math>\,\!X[n]=u[n]</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] $ yields $ \,\!Y_0[n]=\delta [n-1] $ , then
$ \,\!X[n]=u[n] $ will yield $ \,\!Y_0[n]=u[n-1] $
