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== Question 6a == | == Question 6a == | ||
| + | The system is time variant because of the following example: | ||
I'm assuming <math>n\,</math> is the variable I will be applying the time shift to. I looked at some other peoples work and although they all thought <math>k\,</math> was the time variable, I think <math>k\,</math> is just an arbitrary step moving the function forward relative to some time position <math>n\,</math>. In other words , <math>k=2\,</math> doesn't mean time = 2 sec, it just means 2 steps ahead of time <math>n\,</math>. Another reason I chose <math>n\,</math> to be the time variable is because when you discussed the sifting property in class you sifted by <math>n_0\,</math>, not <math>k\,</math>. | I'm assuming <math>n\,</math> is the variable I will be applying the time shift to. I looked at some other peoples work and although they all thought <math>k\,</math> was the time variable, I think <math>k\,</math> is just an arbitrary step moving the function forward relative to some time position <math>n\,</math>. In other words , <math>k=2\,</math> doesn't mean time = 2 sec, it just means 2 steps ahead of time <math>n\,</math>. Another reason I chose <math>n\,</math> to be the time variable is because when you discussed the sifting property in class you sifted by <math>n_0\,</math>, not <math>k\,</math>. | ||
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| − | Consider the input and output of the system when k = | + | Consider the input and output of the system when k = 0 |
| − | <math> | + | <math> X_0[n]=\delta[n]\,</math> |
and | and | ||
| − | <math> | + | <math> Y_0[n]=\delta[n-1] \,</math> |
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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 | + | <math>X_1[n] \longrightarrow X_1[n-n_0] \longrightarrow Y_1[n]=\delta[n-n_0-1],</math> |
| − | |||
| + | but this isn't equal to running the input through the system, then time shifting the output by <math> n_0\,</math> | ||
| − | + | <math>X_1[n] \longrightarrow Y_1[n]=4\delta[n-1] \longrightarrow Y_1[n-n_0]=4\delta[n-n_0-1]\,</math> | |
| − | + | ||
| − | + | ||
| − | <math>X_1[n] \longrightarrow Y_1[n]=4\delta[n- | + | |
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== Question 6b == | == Question 6b == | ||
Revision as of 15:05, 12 September 2008
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]\, $.
