(New page: == Part E == <font size ="4">Input_______________________________Output <math>X_{0}[n] = \delta[n]</math>__________________________<math>Y_{0}[n] = \delta[n-1]</math> <math>X_{1}[n] = ...) |
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<font size ="4"><math>X_{k}[n] = \delta[n-k]</math>_______________________<math>Y_{k}[n] = (k+1)^2\delta[n-(k+1)]</math></font> | <font size ="4"><math>X_{k}[n] = \delta[n-k]</math>_______________________<math>Y_{k}[n] = (k+1)^2\delta[n-(k+1)]</math></font> | ||
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| + | == First Part == | ||
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| + | The system is time-invariant because any kind of response to the shifted input <math>X_{k}[n] = \delta[n-N-k]</math> is of the shifted output <math>Y_{k}[n] = (k+1)^2\delta[n-N-(k+1)]</math>. | ||
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| + | == Second Part == | ||
Revision as of 14:28, 10 September 2008
Part E
Input_______________________________Output
$ X_{0}[n] = \delta[n] $__________________________$ Y_{0}[n] = \delta[n-1] $
$ X_{1}[n] = \delta[n-1] $_______________________$ Y_{1}[n] = 4\delta[n-2] $
$ X_{2}[n] = \delta[n-2] $_______________________$ Y_{2}[n] = 9\delta[n-3] $
$ X_{3}[n] = \delta[n-3] $_______________________$ Y_{3}[n] = 16\delta[n-4] $
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$ X_{k}[n] = \delta[n-k] $_______________________$ Y_{k}[n] = (k+1)^2\delta[n-(k+1)] $
First Part
The system is time-invariant because any kind of response to the shifted input $ X_{k}[n] = \delta[n-N-k] $ is of the shifted output $ Y_{k}[n] = (k+1)^2\delta[n-N-(k+1)] $.
