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<math>a * x1(t) + b * x2(t) ---> Sys ---> a * y1(t) + b * y2(t)</math> | <math>a * x1(t) + b * x2(t) ---> Sys ---> a * y1(t) + b * y2(t)</math> | ||
| + | '''Problem: ''' | ||
We are given a linear system that behaves as follows, | We are given a linear system that behaves as follows, | ||
<math> e^{2jt} --> Sys --> t*e^{-2jt} </math> | <math> e^{2jt} --> Sys --> t*e^{-2jt} </math> | ||
| + | |||
| + | and asked to find the response to find the response to cos(2t). | ||
| + | |||
| + | |||
| + | '''Solution:''' | ||
| + | Using the properties of cosine we can convert cos(2t) to an exponential function. | ||
| + | |||
| + | cos(2t) = <math>\frac{e^{2tj}+e^{-2tj}}{2}</math> | ||
| + | |||
| + | = <math>\frac{1}{2}*(x1(t) + x2(t))</math> | ||
| + | |||
| + | = <math>\frac{1}{2}*x1(t) + \frac{1}{2}*x2(t))</math> | ||
| + | |||
| + | Using the defition of linearity above, | ||
| + | |||
| + | a = <math>\frac{1}{2}</math> | ||
| + | |||
| + | b = <math>\frac{1}{2}</math> | ||
| + | |||
| + | x1(t) = <math>e^{2tj}</math> | ||
| + | |||
| + | x2(t) = <math>e^{-2tj}</math> | ||
| + | |||
| + | <math>a * x1(t) + b * x2(t) ---> Sys ---> a * y1(t) + b * y2(t)</math> | ||
| + | |||
| + | yields the following results, | ||
| + | |||
| + | = <math>\frac{1}{2}*t*x1(-t) + \frac{1}{2}*t*x2(-t))</math> | ||
| + | |||
| + | plugging in for x1(-t) and x2(-t) and factoring out the t, | ||
| + | |||
| + | <math>t*\frac{e^{2tj}+e^{-2tj}}{2}</math> | ||
| + | |||
| + | converting the exponential back to cosine yields the output of the sytem, | ||
| + | |||
| + | '''System output = t * cos(2t) ''' | ||
Latest revision as of 07:29, 18 September 2008
Basics of Linearity
Definition of Linearity: For any constants a and b (that are complext numbers), and inputs x1(t) and x2(t) which yield outputs y1(t) and y2(t),
$ a * x1(t) + b * x2(t) ---> Sys ---> a * y1(t) + b * y2(t) $
Problem: We are given a linear system that behaves as follows,
$ e^{2jt} --> Sys --> t*e^{-2jt} $
and asked to find the response to find the response to cos(2t).
Solution:
Using the properties of cosine we can convert cos(2t) to an exponential function.
cos(2t) = $ \frac{e^{2tj}+e^{-2tj}}{2} $
= $ \frac{1}{2}*(x1(t) + x2(t)) $
= $ \frac{1}{2}*x1(t) + \frac{1}{2}*x2(t)) $
Using the defition of linearity above,
a = $ \frac{1}{2} $
b = $ \frac{1}{2} $
x1(t) = $ e^{2tj} $
x2(t) = $ e^{-2tj} $
$ a * x1(t) + b * x2(t) ---> Sys ---> a * y1(t) + b * y2(t) $
yields the following results,
= $ \frac{1}{2}*t*x1(-t) + \frac{1}{2}*t*x2(-t)) $
plugging in for x1(-t) and x2(-t) and factoring out the t,
$ t*\frac{e^{2tj}+e^{-2tj}}{2} $
converting the exponential back to cosine yields the output of the sytem,
System output = t * cos(2t)
