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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>X_1[n] \longrightarrow X_1[n-n_0-1] \longrightarrow Y_1[n]=4\delta[n-n_0-2]\,</math> | |
If I run it through the system, then time shift the output by <math> n_0\,</math> I obtain: | If I run it through the system, then time shift the output by <math> n_0\,</math> I obtain: | ||
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| + | <math>X_1[n] \longrightarrow X_1[n-n_0-1] \longrightarrow Y_1[n]=4\delta[n-n_0-2]\,</math> | ||
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| + | Consider the system: <math>y(t)=x(t^2-3) \,</math> | ||
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| + | If <math>x(t) \,</math> is first time shifted, then put into the system: | ||
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| + | <math>x(t) \longrightarrow x(t-t_0) \longrightarrow y(t)=x(t^2-3-t_0)\,</math> | ||
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| + | If <math>x(t) \,</math> is first entered into the system, then time shifted: | ||
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| + | <math>x(t) \longrightarrow y(t)=x(t^2-3) \longrightarrow y(t-t_0)=x((t-t_0)^2-3)\,</math> | ||
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| + | Thus this system isn't T.I. | ||
Revision as of 08:39, 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]=4\delta[n-2] \, $
If I time shift the input by $ n_0\, $ , then run it through the system I obtain:
$ X_1[n] \longrightarrow X_1[n-n_0-1] \longrightarrow Y_1[n]=4\delta[n-n_0-2]\, $
If I run it through the system, then time shift the output by $ n_0\, $ I obtain:
$ X_1[n] \longrightarrow X_1[n-n_0-1] \longrightarrow Y_1[n]=4\delta[n-n_0-2]\, $
Consider the system: $ y(t)=x(t^2-3) \, $
If $ x(t) \, $ is first time shifted, then put into the system:
$ x(t) \longrightarrow x(t-t_0) \longrightarrow y(t)=x(t^2-3-t_0)\, $
If $ x(t) \, $ is first entered into the system, then time shifted:
$ x(t) \longrightarrow y(t)=x(t^2-3) \longrightarrow y(t-t_0)=x((t-t_0)^2-3)\, $
Thus this system isn't T.I.
