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<math> d[n-k] --> [timedelay -1] --> d[n-k-1)] --> [system] --> k^2*d[n-k] \,</math> | <math> d[n-k] --> [timedelay -1] --> d[n-k-1)] --> [system] --> k^2*d[n-k] \,</math> | ||
| + | |||
| + | =Input X[n]= | ||
| + | |||
| + | Since it is linear, we can say that | ||
| + | |||
| + | <math>Y[n] = u[n-1] \,</math> | ||
| + | |||
| + | with an input <math>X[n] = u[n] + u[1]</math> | ||
Revision as of 19:54, 10 September 2008
Contents
Time Invariance? or Time Variance?
System: $ Y_k[n] = (k+1)^2 d[n - (k+1)] \, $
Input: $ X_k[n] = d[n-k] \, $
Prob: Which variable represent time ?
1st Assumption: n represents time
$ d[n-k] --> [system] --> (k+1)^2*d[n-(k+1)] --> [timedelay -1] --> (k+1)^2*d[(n-1)-(k+1)] \, $
yields the same result as:
$ d[n-k] --> [timedelay -1] --> d[(n-1)-k] --> [system] --> (k+1)^2*d[(n-1)-(k+1)] \, $
2st Assumption: k represents time
$ d[n-k] --> [system] --> (k+1)^2*d[n-k+1)] --> [timedelay -1] --> (k+1)^2*d[n-k] \, $
yields not the same result as:
$ d[n-k] --> [timedelay -1] --> d[n-k-1)] --> [system] --> k^2*d[n-k] \, $
Input X[n]
Since it is linear, we can say that
$ Y[n] = u[n-1] \, $
with an input $ X[n] = u[n] + u[1] $
Remember: Time delay only occurs on function, not variable on equations.
Concluded, it is time invariant if we say n represents time, and it is time variant if we say k represents time.
