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==DT LTI System==
 
==DT LTI System==
  
<math>y[n] = \sum_{n=-\infty}^{\infty}\frac{1}{2}x[n] \; \;</math> &nbsp; &nbsp; (scaled DT integral)
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<math>y[n] = x[n] + 2x[n-1]</math>
  
 
==h[n]==
 
==h[n]==
  
<math>h[n] = \sum_{n=-\infty}^{\infty}\frac{1}{2}\delta [n] = \frac{1}{2}u[n]</math>
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<math>h[n] = \delta [n] + 2 \delta [n-1]</math>
  
 
==H(z)==
 
==H(z)==
  
<math>H(z) = \sum_{m=-\infty}^{\infty}h[m] e^{-j \omega m} = \sum_{m=-\infty}^{\infty} \frac{1}{2}u[m] e^{-j \omega m} = \sum_{m=0}^{\infty} \frac{1}{2}e^{-j \omega m} = \sum_{m=0}^{\infty} (\frac{1}{2 e^{j \omega}})^m = \frac{1}{1-\frac{1}{2 e^{j \omega}}}</math> &nbsp; &nbsp; (geometric series <math>r^n</math> where <math>|r| < 1</math>)
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<math>H(z) = \sum_{m=-\infty}^{\infty}h[m] e^{-j \omega m} = \sum_{m=-\infty}^{\infty} (\delta [m] + 2 \delta [m-1]) e^{-j \omega m} = \sum_{m=-\infty}^{\infty} \delta [m] e^{-j \omega m} + \sum_{m=-\infty}^{\infty} 2 \delta [m-1] e^{-j \omega m} = 1 + 2 e^{-j \omega}</math>
  
 
==Response to x[n]==
 
==Response to x[n]==
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<math>a_{3} = -\frac{1}{2}</math>
 
<math>a_{3} = -\frac{1}{2}</math>
  
<math>y[n] = \sum_{k=0}^{3} a_{k} F(e^{jk \omega_{o}}) e^{jk \omega_{o}n}</math>
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<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} (1 + 2 e^{-jk \frac{\pi}{2}}) e^{jk \frac{\pi}{2}n} = (1)(3)(1) + (-\frac{1}{2})(1-2j)e^{j \frac{\pi}{2} n} + 0 + (-\frac{1}{2})(1+2j)e^{j \frac{3 \pi}{2}n}</math>
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<math>y[n] = 3 + (-\frac{1}{2}+j)e^{j \frac{\pi}{2} n} + (-\frac{1}{2}-j)e^{j \frac{3 \pi}{2}n}</math>
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 +
The system response y[n] is shown below:
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[[Image:RespSawDTJP_ECE301Fall2008mboutin.jpg]]

Latest revision as of 13:55, 24 September 2008

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 [m] + 2 \delta [m-1]) e^{-j \omega m} = \sum_{m=-\infty}^{\infty} \delta [m] e^{-j \omega m} + \sum_{m=-\infty}^{\infty} 2 \delta [m-1] e^{-j \omega m} = 1 + 2 e^{-j \omega} $

Response to x[n]

Input $ x[n] $ is the following signal:

SawDTJP ECE301Fall2008mboutin.jpg

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} (1 + 2 e^{-jk \frac{\pi}{2}}) e^{jk \frac{\pi}{2}n} = (1)(3)(1) + (-\frac{1}{2})(1-2j)e^{j \frac{\pi}{2} n} + 0 + (-\frac{1}{2})(1+2j)e^{j \frac{3 \pi}{2}n} $

$ y[n] = 3 + (-\frac{1}{2}+j)e^{j \frac{\pi}{2} n} + (-\frac{1}{2}-j)e^{j \frac{3 \pi}{2}n} $

The system response y[n] is shown below:

RespSawDTJP ECE301Fall2008mboutin.jpg

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