(New page: a) <pre> Yk[n]=(k+1)^2 d[n-(k+1)] </pre> Yes, the system is time invariant because the system does not have a scaling affect on the time part of the function. For example, if the system we...) |
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a) | a) | ||
| − | < | + | <math>Xk[n]=d[n-k]</math> -> syst. -> <math>Yk[n]=(k+1)^2 d[n-(k+1)]</math> |
| − | Yk[n]=(k+1)^2 d[n-(k+1)] | + | |
| − | </ | + | To see if the system is time invariant we must run it through time delay. |
| − | + | ||
| − | + | <math>Xk[n]=d[n-k]</math> -> syst. -> <math>Yk[n]=(k+1)^2 d[n-(k+1)]</math> -> delay -> <math>Zk[n] = Yk[n+n0] = (k+1)^2 d[n-(k+1)+n0]</math> | |
| − | < | + | |
| − | Yk[n]=(k+1)^2 d[ | + | Alternately: |
| − | </ | + | <math>Xk[n]=d[n-k]</math> -> delay -> <math>Yk[n]=d[n-k+no]</math> -> syst. -> <math>Zk[n]=(k+n0+1)^2 d[n-(k+no+1)+no]</math> |
| − | + | ||
| + | The outputs are not the same, therefore the system is not time invariant. | ||
b) | b) | ||
k would have to be 0 because only then is d[2n-(k+1)] not affected. | k would have to be 0 because only then is d[2n-(k+1)] not affected. | ||
Latest revision as of 09:42, 12 September 2008
a) $ Xk[n]=d[n-k] $ -> syst. -> $ Yk[n]=(k+1)^2 d[n-(k+1)] $
To see if the system is time invariant we must run it through time delay.
$ Xk[n]=d[n-k] $ -> syst. -> $ Yk[n]=(k+1)^2 d[n-(k+1)] $ -> delay -> $ Zk[n] = Yk[n+n0] = (k+1)^2 d[n-(k+1)+n0] $
Alternately: $ Xk[n]=d[n-k] $ -> delay -> $ Yk[n]=d[n-k+no] $ -> syst. -> $ Zk[n]=(k+n0+1)^2 d[n-(k+no+1)+no] $
The outputs are not the same, therefore the system is not time invariant.
b) k would have to be 0 because only then is d[2n-(k+1)] not affected.
