Difference between revisions of "HW2-E Nathan Najdek ECE301Fall2008mboutin" - Rhea

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.