Revision as of 19:44, 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 ?

$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)] ,\$

$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.

@@ Line 3: / Line 3: @@
 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.