(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] = ...) |
(→Second Part) |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 21: | Line 21: | ||
<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> | ||
| + | |||
| + | == First Part == | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | == Second Part == | ||
| + | |||
| + | Assuming that this system is linear, the X[n] input would be u[n] to yield the output Y[n] = u[n-1]. | ||
| + | |||
| + | <math>X[n] = u[n] = \delta[n] - \delta[n-N]</math> | ||
| + | |||
| + | So k = 0 and N = 1, which will give the output <math>Y[n] = \delta[n-1] - \delta[n-2] = u[n-1]</math> after the input goes through the system. | ||
Latest revision as of 11:51, 11 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] $
.
.
.
.
$ 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)] $.
Second Part
Assuming that this system is linear, the X[n] input would be u[n] to yield the output Y[n] = u[n-1].
$ X[n] = u[n] = \delta[n] - \delta[n-N] $
So k = 0 and N = 1, which will give the output $ Y[n] = \delta[n-1] - \delta[n-2] = u[n-1] $ after the input goes through the system.
