(→Question 6a) |
(→Question 6a) |
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The system is time variant because of the following example: | The system is time variant because of the following example: | ||
| − | + | Dealing with the following system | |
| − | + | ||
<math> X_k[n]=Y_k[n] \,</math> | <math> X_k[n]=Y_k[n] \,</math> | ||
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If I time shift the input by <math> n_0\,</math> , then run it through the system I obtain: | If I time shift the input by <math> n_0\,</math> , then run it through the system I obtain: | ||
| − | <math> | + | <math>X_0[n] \longrightarrow X_0[n-n_0] \longrightarrow Y_1[n]=\delta[n-n_0-1],</math> |
but this isn't equal to running the input through the system, then time shifting the output by <math> n_0\,</math> | but this isn't equal to running the input through the system, then time shifting the output by <math> n_0\,</math> | ||
| − | <math> | + | <math>X_0[n] \longrightarrow Y_1[n]=4\delta[n-1] \longrightarrow Y_1[n-n_0]=4\delta[n-n_0-1]\,</math> |
== Question 6b == | == Question 6b == | ||
Latest revision as of 16:07, 12 September 2008
Question 6a
The system is time variant because of the following example:
Dealing with the following system
$ X_k[n]=Y_k[n] \, $
where
$ X_k[n]=\delta[n-k]\, $
and
$ Y_k[n]=(k+1)^2 \delta[n-(k+1)] \, $
Consider the input and output of the system when k = 0
$ X_0[n]=\delta[n]\, $
and
$ Y_0[n]=\delta[n-1] \, $
If I time shift the input by $ n_0\, $ , then run it through the system I obtain:
$ X_0[n] \longrightarrow X_0[n-n_0] \longrightarrow Y_1[n]=\delta[n-n_0-1], $
but this isn't equal to running the input through the system, then time shifting the output by $ n_0\, $
$ X_0[n] \longrightarrow Y_1[n]=4\delta[n-1] \longrightarrow Y_1[n-n_0]=4\delta[n-n_0-1]\, $
Question 6b
Assuming this system is linear, an input $ X_0[n]=u[n]\, $ would result in an output $ Y_0[n]=u[n-1]\, $.
