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== Second Part == | == Second Part == | ||
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| + | Assuming that this system is linear, the X[n] input would be u[n] to yield the output Y[n] = u[n-1]. | ||
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| + | <math>X[n] = u[n] = \delta[n] - \delta[n-N]</math> | ||
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| + | 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 10:51, 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] $
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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)] $.
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.
