(New page: Given the following system: Input <math>\to</math> Output <math>X0[n] = \delta[n] \to Y0[n] = \delta[n-1]</math> <math>X1[n] = \delta[n-1] \to Y1[n] = 4\delta[n-2]</math> <math>X2[n] = ...) |
(→Part B) |
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Assuming that this system is linear, what input <math>X[n]</math> would yield the output <math>Y[n]=u[n-1]</math>? | Assuming that this system is linear, what input <math>X[n]</math> would yield the output <math>Y[n]=u[n-1]</math>? | ||
| − | Since the system is linear, the input would be <math>X[n] = u[n]</math> | + | Since the system is linear, the input would be <math>X[n] = u[n]</math> based on the system we were given. |
Latest revision as of 16:18, 11 September 2008
Given the following system: Input $ \to $ Output
$ X0[n] = \delta[n] \to Y0[n] = \delta[n-1] $
$ X1[n] = \delta[n-1] \to Y1[n] = 4\delta[n-2] $
$ X2[n] = \delta[n-2] \to Y2[n] = 9\delta[n-3] $
$ X3[n] = \delta[n-3] \to Y3[n] = 16\delta[n-4] $
...
$ Xk[n] = \delta[n-k] \to Yk[n] = (k+1)^{2}\delta[n-(k+1)] $
Part A
Is this system Time Invariant?
Using $ X1[n] = \delta[n-1] $
Delay by 1: $ Y1[n] = \delta[n-2] $
Feed into system: $ Z1[n] = 9\delta[n-3] $
This time, feed into the system first: $ Y1[n] = 4\delta[n-2] $
Delay by 1: $ Z1[n] = 4\delta[n-3] $
Since $ 9\delta[n-3] \ne 4\delta[n-3] $ the system is NOT Time Invariant.
Part B
Assuming that this system is linear, what input $ X[n] $ would yield the output $ Y[n]=u[n-1] $?
Since the system is linear, the input would be $ X[n] = u[n] $ based on the system we were given.
