Latest revision as of 18:56, 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.

Since it is linear, we can say that

$Y[n] = u[n-1] \,$

with an input $X[n] = u[n] - u[1] \,$

@@ Line 17: / Line 17: @@
 ===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>
+<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>
+<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.
@@ Line 30: / Line 35: @@
 <math>Y[n] = u[n-1] \,</math>
-with an input <math>X[n] = u[n] + u[1] \,</math>
+with an input <math>X[n] = u[n] - u[1] \,</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.