(Brian Thomas Rhea HW4.3/4 (4 actually) - work in progress) |
(Brian Thomas Rhea HW4.4 done!) |
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'''Solution:''' | '''Solution:''' | ||
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| + | '''a)''' <math>h[n] = f(\delta(n)) = 2\delta[n+1] + \delta[n-1]</math> | ||
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| + | <math>H(z) = \sum_{k=-\infty}^{\infty} h[k] z^{-k}</math> | ||
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| + | <math> = \sum_{k=-\infty}^{\infty} (2\delta[k+1] + \delta[k-1]) z^{-k} </math> | ||
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| + | <math> = \sum_{k=-\infty}^{\infty} 2\delta[k+1] z^{-k} + \sum_{k=-\infty}^{\infty} \delta[k-1] z^{-k} </math> | ||
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| + | <math> = 2 z^{1} + z^{-1} </math> (By the sifting property) | ||
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| + | '''b)''' Given the Fourier coefficients, x[n] can be expressed as follows: | ||
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| + | <math>x[n] = \sum_{k=0}^3 a_k e^{jk \frac{\pi}{2} n} = \frac{1}{2} e^{0} + \frac{1-j}{4} e^{j \frac{\pi}{2} n} + 0 e^{j \pi n} + \frac{1+j}{4} e^{j \frac{3\pi}{2} n} = \frac{1}{2} + \frac{1-j}{4} e^{j \frac{\pi}{2} n} + \frac{1+j}{4} e^{j \frac{3\pi}{2} n}</math> | ||
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| + | Our output for our periodic DT system is <math>y[n] = f(x[n]) = \sum_{k} a_k H(z_k) {z_k}^n</math> | ||
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| + | Consider, then, <math>z_k = e^{jk \frac{\pi}{2}}</math>. Then, <math>y[n] = f(x[n]) = \sum_{k} a_k H(e^{jk \frac{\pi}{2}}) e^{jk \frac{\pi}{2}n}</math>. | ||
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| + | <math>f(x[n]) = \frac{1}{2} H(e^{0}) + \frac{1-j}{4} H(e^{j \frac{\pi}{2}}) e^{j \frac{\pi}{2} n} + \frac{1+j}{4} H(e^{j \frac{3\pi}{2}}) e^{j \frac{3\pi}{2} n}</math> | ||
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| + | <math> = \frac{1}{2} H(1) + \frac{1-j}{4} H(j) e^{j \frac{\pi}{2} n} + \frac{1+j}{4} H(-j) e^{j \frac{3\pi}{2} n}</math> | ||
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| + | <math> = \frac{1}{2} (3) + \frac{1-j}{4} (j) e^{j \frac{\pi}{2} n} + \frac{1+j}{4} (-j) e^{j \frac{3\pi}{2} n}</math> | ||
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| + | <math> = \frac{3}{2} + \frac{1+j}{4} e^{j \frac{\pi}{2} n} + \frac{1-j}{4} e^{j \frac{3\pi}{2} n}</math> | ||
Latest revision as of 15:48, 25 September 2008
Problem: Say you are given a DT LTI system f, where input x[n] yields output $ y[n] = f(x[n]) = 2x[n+1] + x[n-1] $.
a) Obtain the unit impulse response h[n] and the system function H(z) of f.
b) Compute the response of f to the signal x[n] found here by using H(z) and the Fourier series coefficients of x[n].
Solution:
a) $ h[n] = f(\delta(n)) = 2\delta[n+1] + \delta[n-1] $
$ H(z) = \sum_{k=-\infty}^{\infty} h[k] z^{-k} $
$ = \sum_{k=-\infty}^{\infty} (2\delta[k+1] + \delta[k-1]) z^{-k} $
$ = \sum_{k=-\infty}^{\infty} 2\delta[k+1] z^{-k} + \sum_{k=-\infty}^{\infty} \delta[k-1] z^{-k} $
$ = 2 z^{1} + z^{-1} $ (By the sifting property)
b) Given the Fourier coefficients, x[n] can be expressed as follows:
$ x[n] = \sum_{k=0}^3 a_k e^{jk \frac{\pi}{2} n} = \frac{1}{2} e^{0} + \frac{1-j}{4} e^{j \frac{\pi}{2} n} + 0 e^{j \pi n} + \frac{1+j}{4} e^{j \frac{3\pi}{2} n} = \frac{1}{2} + \frac{1-j}{4} e^{j \frac{\pi}{2} n} + \frac{1+j}{4} e^{j \frac{3\pi}{2} n} $
Our output for our periodic DT system is $ y[n] = f(x[n]) = \sum_{k} a_k H(z_k) {z_k}^n $
Consider, then, $ z_k = e^{jk \frac{\pi}{2}} $. Then, $ y[n] = f(x[n]) = \sum_{k} a_k H(e^{jk \frac{\pi}{2}}) e^{jk \frac{\pi}{2}n} $.
$ f(x[n]) = \frac{1}{2} H(e^{0}) + \frac{1-j}{4} H(e^{j \frac{\pi}{2}}) e^{j \frac{\pi}{2} n} + \frac{1+j}{4} H(e^{j \frac{3\pi}{2}}) e^{j \frac{3\pi}{2} n} $
$ = \frac{1}{2} H(1) + \frac{1-j}{4} H(j) e^{j \frac{\pi}{2} n} + \frac{1+j}{4} H(-j) e^{j \frac{3\pi}{2} n} $
$ = \frac{1}{2} (3) + \frac{1-j}{4} (j) e^{j \frac{\pi}{2} n} + \frac{1+j}{4} (-j) e^{j \frac{3\pi}{2} n} $
$ = \frac{3}{2} + \frac{1+j}{4} e^{j \frac{\pi}{2} n} + \frac{1-j}{4} e^{j \frac{3\pi}{2} n} $
