| Line 5: | Line 5: | ||
| − | <math>X_{k}[n] \to timedelay \to sys \to Z_{k}[n]=</math> | + | <math>X_{k}[n] \to timedelay \to sys \to Z_{k}[n]=(k+1)^2 \delta [n-N-(k+1)]</math> |
| − | <math>X_{k}[n] \to sys \to timedelay \to Z_{k}[n]=</math> | + | <math>X_{k}[n] \to sys \to timedelay \to Z_{k}[n]=(k+1)^2 \delta [n-N-(k+1)]</math> |
| − | Since <math></math> is equal to <math></math>, the system is time-invariant.</font> | + | Since <math>(k+1)^2 \delta [n-N-(k+1)]</math> is equal to <math>(k+1)^2 \delta [n-N-(k+1)]</math>, the system is time-invariant.</font> |
| + | |||
| + | == Part b == | ||
Revision as of 14:23, 10 September 2008
Part a
System: $ X_{k}[n-k] \to Y_{k}[n] = (k+1)^2 \delta [n-(k+1)] $
$ X_{k}[n] \to timedelay \to sys \to Z_{k}[n]=(k+1)^2 \delta [n-N-(k+1)] $
$ X_{k}[n] \to sys \to timedelay \to Z_{k}[n]=(k+1)^2 \delta [n-N-(k+1)] $
Since $ (k+1)^2 \delta [n-N-(k+1)] $ is equal to $ (k+1)^2 \delta [n-N-(k+1)] $, the system is time-invariant.
