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<math> x(t) = a*x1(t) + b*x2(t) = a*y1(t) + b*y2(t) </math>
 
<math> x(t) = a*x1(t) + b*x2(t) = a*y1(t) + b*y2(t) </math>
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Take for a simple example:
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Ex) What is the output of:
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<math> x[n] = e^{j*\pi*n}  ->  n*e^{-j*\pi*n} </math>
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<math> x[n] \to Sys 1 \to n*x[-n] </math>
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from:
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<math> e^{j*n*y} = cos(n*y) + j*sin(n*y)  </math>
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we determine:
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<math> x[n] = cos(\pi*n) = \frac{e^{j*\pi*n} + e^{-j*\pi*n}}{2} </math>
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yields to <math> x[t*n] = \frac{te^{j*\pi*n} + t*e^{-j*\pi*n}}{2} </math>
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<math> == cos(\pi*n) \to -1 </math> on every even integer interval.

Latest revision as of 15:35, 18 September 2008

The Basics of Linearity

A system is linear if its inputs are sequentially equal to the outputs for a certain function:

$ x(t) = a*x1(t) + b*x2(t) = a*y1(t) + b*y2(t) $


Take for a simple example:

Ex) What is the output of:

$ x[n] = e^{j*\pi*n} -> n*e^{-j*\pi*n} $

$ x[n] \to Sys 1 \to n*x[-n] $


from: $ e^{j*n*y} = cos(n*y) + j*sin(n*y) $

we determine: $ x[n] = cos(\pi*n) = \frac{e^{j*\pi*n} + e^{-j*\pi*n}}{2} $


yields to $ x[t*n] = \frac{te^{j*\pi*n} + t*e^{-j*\pi*n}}{2} $

$ == cos(\pi*n) \to -1 $ on every even integer interval.

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn