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Remember: Time delay only occurs on function, not variable on equations. | Remember: Time delay only occurs on function, not variable on equations. | ||
− | Concluded, it is time invariant if we say n represents time, | + | Concluded, it is time invariant if we say n represents time, and it is time variant if we say k represents time. |
− | + |
Revision as of 18:45, 10 September 2008
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] ,\ $
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.