Revision as of 15:05, 12 September 2008 by Nbrowdue (Talk)

Question 6a

The system is time variant because of the following example:

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 = 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_1[n] \longrightarrow X_1[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_1[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]\, $.

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