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Assuming this system is linear an input | Assuming this system is linear an input | ||
| − | <math> X_0[n]= | + | <math> X_0[n]=u[n]\,</math> |
would result in an output | would result in an output | ||
| − | <math> Y_0[n]= | + | <math> Y_0[n]=u[n-1]\,</math> |
Revision as of 07:52, 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 an arbitrary 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 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]\, $
Which shifts the system's value when k = 1 a time length of $ n_0\, $ forward.
If I run the input through the system, then time shift the output by $ n_0\, $ I obtain:
$ X_1[n] \longrightarrow Y_1[n]=4\delta[n-2] \longrightarrow Y_1[n-n_0]=4\delta[n-n_0-2]\, $
This also shifts the system's value when k = 1 a time length of $ n_0\, $ forward. Thus the system is T.I.
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]\, $
