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<math>a_{3} = -\frac{1}{2}</math> | <math>a_{3} = -\frac{1}{2}</math> | ||
| − | <math>y[n] = \sum a_{k} | + | <math>y[n] = \sum a_{k} H(e^{jk \omega_{o}}) e^{jk \omega_{o}n} = \sum_{k=0}^{3} a_{k} H(e^{jk \frac{\pi}{2}}) e^{jk \frac{\pi}{2}n} = \sum_{k=0}^{3} a_{k} \frac{1}{1-\frac{1}{2 e^{jk \frac{\pi}{2}}}} e^{jk \frac{\pi}{2}n} = </math> |
Revision as of 21:58, 23 September 2008
Contents
DT LTI System
$ y[n] = x[n] + 2x[n-1] $
h[n]
$ h[n] = \delta [n] + 2 \delta [n-1] $
H(z)
$ H(z) = \sum_{m=-\infty}^{\infty}h[m] e^{-j \omega m} = \sum_{m=-\infty}^{\infty} (\delta [n] + 2 \delta [n-1]) e^{-j \omega m} = \sum_{m=-\infty}^{\infty} \delta [n] e^{-j \omega m} + \sum_{m=-\infty}^{\infty} 2 \delta [n-1] e^{-j \omega m} = \delta [n] e^{-j \omega m} + 2 \delta [n-1] e^{-j \omega m} $
In other words:
$ H(z) = e^{-j \omega m} $ for n = 0
$ = 2 e^{-j \omega m} $ for n = 1
$ = 0 $ otherwise
Response to x[n]
Input $ x[n] $ is the following signal:
The Fourier series coefficients for $ x[n] $ are:
$ a_{0} = 1 $
$ a_{1} = -\frac{1}{2} $
$ a_{2} = 0 $
$ a_{3} = -\frac{1}{2} $
$ y[n] = \sum a_{k} H(e^{jk \omega_{o}}) e^{jk \omega_{o}n} = \sum_{k=0}^{3} a_{k} H(e^{jk \frac{\pi}{2}}) e^{jk \frac{\pi}{2}n} = \sum_{k=0}^{3} a_{k} \frac{1}{1-\frac{1}{2 e^{jk \frac{\pi}{2}}}} e^{jk \frac{\pi}{2}n} = $
