Difference between revisions of "HW2.E Hetong Li ECE301Fall2008mboutin" - Rhea

Latest revision as of 17:35, 12 September 2008

No,the system can not be time-invariant.

Based on the data, if X has been delayed by 1, Y is not only delayed by 1,but also scale in the magnitude.

Since if we operate delay, the outputs won't match, the system is not time-invariant.

Since the system is linear, the input should be X[n] = u[n] in order to get Y[n]=u[n-1], where k=0

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 == E-a ==
-Yes,the system can be time-invariant.
+No,the system can not be time-invariant.
-the system is
+Based on the data, if X has been delayed by 1, Y is not only delayed by 1,but also scale in the magnitude.
-<math>Y_k[n] = (k + 1)^2X_k[n-1]</math>
-<math>X_k[n]\rightarrow system \rightarrow Y_k[n] = (k + 1)^2X_k[n-1]\rightarrow Time Delay by m \rightarrow Z_k[n]=(k+1)^2X_k[n-m-1]</math>
+Since if we operate delay, the outputs won't match, the system is not time-invariant.
-<math>X_k[n]\rightarrow Time Delay by m\rightarrow Y_k[n] = X_k[n-m]\rightarrow Time  System\rightarrow Z_k[n]=Y_k[n-m]=(k+1)^2X_k[n-m-1]</math>
-Since the outputs match, the system is time-invariant.
 == E-b ==
 Since the system is linear, the input should be X[n] = u[n] in order to get Y[n]=u[n-1],
 where k=0