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<math> X_k[n]=Y_k[n] \,</math> | <math> X_k[n]=Y_k[n] \,</math> | ||
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where | where | ||
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| − | Consider the | + | Consider the input and output of the system when k = 1 |
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| + | <math> X_1[n]=\delta[n-1]\,</math> | ||
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| + | and | ||
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| + | <math> Y_1[n]=\delta[n-2] \,</math> | ||
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| + | If I time shift the input by <math> n_0\,</math> , then run it through the system I obtain: | ||
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| − | + | If I run it through the system, then time shift the output by <math> n_0\,</math> I obtain: | |
Revision as of 08:33, 11 September 2008
Question 6a
I'm assuming $ n\, $ is the variable I will be applying the time shift to. I looked at some other peoples work and although they all thought $ k\, $ was the time variable, I think $ k\, $ is just the time step moving the function forward relative to some time position $ n\, $. In other words , $ k=2\, $ doesn't mean time = 2 sec, it just means 2 time steps ahead of time $ n\, $. Another reason I chose $ n\, $ to be the time variable is because when you discussed the sifting property in class you sifted by $ n_0\, $, not $ k\, $.
$ 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 = 1
$ X_1[n]=\delta[n-1]\, $
and
$ Y_1[n]=\delta[n-2] \, $
If I time shift the input by $ n_0\, $ , then run it through the system I obtain:
If I run it through the system, then time shift the output by $ n_0\, $ I obtain:
