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== Part B == | == Part B == | ||
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| + | Compute the system's response to (from problem 2 [[HW4.2 Jeff Kubascik_ECE301Fall2008mboutin]]) | ||
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| + | <math>\,x[n]=\sum_{k=-\infty}^{\infty}\delta[n-4k] + \pi\delta[n-1-4k] - 3\delta[n-2-4k] + \sqrt[e]{\frac{\pi^3}{\ln(5)}}\delta[n-3-4k]\,</math> | ||
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| + | Using the Fourier coefficients calculated in problem 2, we can express the system's response to <math>\,x[n]\,</math> as | ||
Revision as of 17:24, 25 September 2008
Given the following LTI DT system
$ \,s[n]=x[n]+x[n-1]\, $
Part A
Find the system's unit impulse response $ \,h[n]\, $ and system function $ \,H(z)\, $.
The unit impulse response is simply (plug a $ \,\delta[n]\, $ into the system)
$ \,h[n]=\delta[n]+\delta[n-1]\, $
The system function can be found using the following formula (for LTI systems)
$ \,H(z)=\sum_{m=-\infty}^{\infty}h[m]z^{-m}\, $
$ \,H(z)=\sum_{m=-\infty}^{\infty}(\delta[m]+\delta[m-1])z^{-m}\, $
using the sifting property
$ \,H(z)=z^{0}+z^{-1}\, $
$ \,H(z)=1+z^{-1}\, $
Part B
Compute the system's response to (from problem 2 HW4.2 Jeff Kubascik_ECE301Fall2008mboutin)
$ \,x[n]=\sum_{k=-\infty}^{\infty}\delta[n-4k] + \pi\delta[n-1-4k] - 3\delta[n-2-4k] + \sqrt[e]{\frac{\pi^3}{\ln(5)}}\delta[n-3-4k]\, $
Using the Fourier coefficients calculated in problem 2, we can express the system's response to $ \,x[n]\, $ as
