(New page: = Linearity and Time invariant system = == Time invariance == The system shown is not time invariance system. Although there is no <math>t\,</math> variable as the coefficient of the outp...) |
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Both signal doesn't match, as the coeefeicent change according to the time when the signal is inputted. Therefore, the system is not time invariant | Both signal doesn't match, as the coeefeicent change according to the time when the signal is inputted. Therefore, the system is not time invariant | ||
| − | == Finding the input == | + | == Finding the input == |
| − | + | ||
| − | + | <math>u[n-1] = \sum^\inf_{k=1} \delta[n-k]</math> | |
| − | <math> | + | <math>= \sum^\inf_{k=0} \delta[n-(k+1)]</math> |
| + | |||
| + | <math>= \sum^\inf_{k=0} \frac{(k+1)^2}{(k+1)^2}\delta[n-(k+1)]</math> | ||
| + | |||
| + | <math>= \frac{1}{(k+1)^2}\sum^\inf_{k=0} (k+1)^2\delta[n-(k+1)]</math> | ||
| + | |||
| + | <math>= \frac{1}{(k+1)^2}\sum^\inf_{k=0} Y_k[n]</math> | ||
| + | |||
| + | As the system is linear, thus the input would be | ||
| + | |||
| + | <math>\frac{1}{(k+1)^2}\sum^\inf_{k=0} X_k[n]</math> | ||
| + | |||
| + | <math>=\frac{1}{(k+1)^2}\sum^\inf_{k=0} \delta[n-k]</math> | ||
Latest revision as of 17:25, 12 September 2008
Linearity and Time invariant system
Time invariance
The system shown is not time invariance system. Although there is no $ t\, $ variable as the coefficient of the output or the system, the system depends on the delay of the input.
The input is shifted before undergoing the transformation
$ X[n] = X_{k + \alpha} = \delta[n - k - \alpha] = \delta[n - (k + \alpha)] \, $
$ Y[n] = Y_{k + \alpha} = (k + \alpha + 1)^2\delta[n - (k + \alpha + 1)] \, $
If the signal is shifted at the output, $ Y[n] = Y_k[n - \alpha] = (k+1)^2\delta[n-(k+1) - \alpha] = (k+1)^2\delta[n-(k + 1 + \alpha)]\, $
Both signal doesn't match, as the coeefeicent change according to the time when the signal is inputted. Therefore, the system is not time invariant
Finding the input
$ u[n-1] = \sum^\inf_{k=1} \delta[n-k] $
$ = \sum^\inf_{k=0} \delta[n-(k+1)] $
$ = \sum^\inf_{k=0} \frac{(k+1)^2}{(k+1)^2}\delta[n-(k+1)] $
$ = \frac{1}{(k+1)^2}\sum^\inf_{k=0} (k+1)^2\delta[n-(k+1)] $
$ = \frac{1}{(k+1)^2}\sum^\inf_{k=0} Y_k[n] $
As the system is linear, thus the input would be
$ \frac{1}{(k+1)^2}\sum^\inf_{k=0} X_k[n] $
$ =\frac{1}{(k+1)^2}\sum^\inf_{k=0} \delta[n-k] $
