Line 3: | Line 3: | ||
Input: <math>X_k[n] = d[n-k] \,</math> | Input: <math>X_k[n] = d[n-k] \,</math> | ||
+ | |||
+ | Prob: Which variable represent time ? | ||
+ | |||
+ | ===1st Assumption: n represents time=== | ||
+ | |||
+ | <math> d[n-k] --> [system] --> (k+1)^2 d[n - (k+1)] --> [timedelay -1] --> (k+1)^2 d[(n-1) -( k+1)] ,\</math> | ||
+ | |||
+ | yields the same result as: | ||
+ | |||
+ | <math> d[n-k] --> [timedelay -1] --> d[(n-1) - k] --> [system] --> (k+1)^2 d[(n-1) - (k+1)] ,\</math> | ||
+ | |||
+ | |||
+ | ===2st Assumption: k represents time=== | ||
+ | |||
+ | <math> d[n-k] --> [system] --> (k+1)^2 d[n - (k+1)] --> [timedelay -1] --> (k+1)^2 d[n-k] ,\</math> | ||
+ | |||
+ | yields not the same result as: | ||
+ | |||
+ | <math> d[n-k] --> [timedelay -1] --> d[n - (k-1)] --> [system] --> k^2 d[n-k] ,\</math> | ||
+ | |||
+ | |||
+ | |||
+ | 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. |
Revision as of 18:44, 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.