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<math>Y_k[n] = (k + 1)^2X_k[n-1]</math> | <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> | + | <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> |
<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> | <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> | ||
Revision as of 18:16, 12 September 2008
E-a
Yes,the system can be time-invariant.
the system is $ Y_k[n] = (k + 1)^2X_k[n-1] $
$ 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] $
$ 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] $
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
