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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 | ||
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| + | <math>\,y[n]=\sum_{k=0}^{N-1}a_kF(e^{jk\frac{2\pi}{N}})e^{jk\frac{2\pi}{N}n}\,</math> | ||
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| + | <math>\,y[n]=\sum_{k=0}^{3}a_kF(e^{jk\frac{2\pi}{4}})e^{jk\frac{2\pi}{4}n}\,</math> | ||
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| + | <math>\,y[n]=\sum_{k=0}^{3}a_k(1+e^{-jk\frac{\pi}{2}})e^{jk\frac{\pi}{2}n}\,</math> | ||
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| + | <math>\,y[n]=a_0(1+e^{0})e^{0} + a_1(1+e^{-j\frac{\pi}{2}})e^{j\frac{\pi}{2}n} + a_2(1+e^{-j\pi})e^{j\pi n} + a_3(1+e^{-j\frac{3\pi}{2}})e^{j\frac{3\pi}{2}n}\,</math> | ||
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| + | <math>\,y[n]=2a_0 + a_1(1-j)e^{j\frac{\pi}{2}n} + 0a_2e^{j\pi n} + a_3(1+j)e^{j\frac{3\pi}{2}n}\,</math> | ||
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| + | <math>\,y[n]=2a_0 + a_1(1-j)e^{j\frac{\pi}{2}n} + a_3(1+j)e^{j\frac{3\pi}{2}n}\,</math> | ||
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| + | Therefore, the system's response is | ||
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| + | <math>\,y[n]=\frac{1}{2}(-2 + \pi + \sqrt[e]{\frac{\pi^3}{\ln(5)}}) + \frac{1}{4}(4 - j\pi + j\sqrt[e]{\frac{\pi^3}{\ln(5)}})(1-j)e^{j\frac{\pi}{2}n} + \frac{1}{4}(4 + j\pi - j\sqrt[e]{\frac{\pi^3}{\ln(5)}})(1+j)e^{j\frac{3\pi}{2}n}\,</math> | ||
Latest revision as of 17:54, 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
$ \,y[n]=\sum_{k=0}^{N-1}a_kF(e^{jk\frac{2\pi}{N}})e^{jk\frac{2\pi}{N}n}\, $
$ \,y[n]=\sum_{k=0}^{3}a_kF(e^{jk\frac{2\pi}{4}})e^{jk\frac{2\pi}{4}n}\, $
$ \,y[n]=\sum_{k=0}^{3}a_k(1+e^{-jk\frac{\pi}{2}})e^{jk\frac{\pi}{2}n}\, $
$ \,y[n]=a_0(1+e^{0})e^{0} + a_1(1+e^{-j\frac{\pi}{2}})e^{j\frac{\pi}{2}n} + a_2(1+e^{-j\pi})e^{j\pi n} + a_3(1+e^{-j\frac{3\pi}{2}})e^{j\frac{3\pi}{2}n}\, $
$ \,y[n]=2a_0 + a_1(1-j)e^{j\frac{\pi}{2}n} + 0a_2e^{j\pi n} + a_3(1+j)e^{j\frac{3\pi}{2}n}\, $
$ \,y[n]=2a_0 + a_1(1-j)e^{j\frac{\pi}{2}n} + a_3(1+j)e^{j\frac{3\pi}{2}n}\, $
Therefore, the system's response is
$ \,y[n]=\frac{1}{2}(-2 + \pi + \sqrt[e]{\frac{\pi^3}{\ln(5)}}) + \frac{1}{4}(4 - j\pi + j\sqrt[e]{\frac{\pi^3}{\ln(5)}})(1-j)e^{j\frac{\pi}{2}n} + \frac{1}{4}(4 + j\pi - j\sqrt[e]{\frac{\pi^3}{\ln(5)}})(1+j)e^{j\frac{3\pi}{2}n}\, $
