(New page: == E-a == Yes,the system can be time-invariant. the system is <math>Y_k[n] = (k + 1)^2 *X_k[n-1]</math> <math>X_k[n]\rightarrow system \rightarrow Y_k[n] = (k + 1)^2*X_k[n-1]\rightarrow...) |
(→E-a) |
||
| Line 3: | Line 3: | ||
the system is | the system is | ||
| − | <math>Y_k[n] = (k + 1)^ | + | <math>Y_k[n] = (k + 1)^2X_k[n-1]</math> |
| − | <math>X_k[n]\rightarrow system \rightarrow Y_k[n] = (k + 1)^ | + | <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)^ | + | <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. | Since the outputs match, the system is time-invariant. | ||
Revision as of 17:16, 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
