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== Question 6a == | == Question 6a == | ||
| − | I'm assuming k is the variable | + | 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. |
<math> X_k[n]=Y_k[n] \,</math> | <math> X_k[n]=Y_k[n] \,</math> | ||
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| − | + | Consider the value of the system at when time = 0s | |
Under this assumption the following system cannot possibly be time invariant because of the <math>(k+1)^2</math> term. | Under this assumption the following system cannot possibly be time invariant because of the <math>(k+1)^2</math> term. | ||
Revision as of 08:26, 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 value of the system at when time = 0s
Under this assumption the following system cannot possibly be time invariant because of the $ (k+1)^2 $ term.
