(New page: Given the definition of Linear systems <math>\alpha x1(t) + \beta x2(t) <\math> is <math> \alpha y1(t)+ \beta y2(t).</math> Consider the following system: <math...) |
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| − | Given the definition of [[3.A David Hartmann | + | Given the definition of [[3.A David Hartmann - Linear System_ECE301Fall2008mboutin| Linear systems]] we know the response to <math>\alpha x_1(t) + \beta x_2(t) </math> is <math> \alpha y_1(t)+ \beta y_2(t).</math> |
| − | + | Consider the system: | |
| − | |||
| − | |||
| + | <math>e^{-2jt}\to F ( e^{-2jt} ) \to te^{2jt}</math> | ||
| − | + | From the given system we know | |
| − | + | <math>x(t)\to F( x(t) )\to tx(-t)</math> | |
| − | <math> | + | Euler's formula allows to rewrite <math>e^{iy} </math> as |
| + | <math>e^{iy}=cos{y}+i*sin{y}</math> | ||
| − | + | Using this we deduct that | |
| − | <math>e^{ | + | x(t) = cos(2t)= <math>\frac{e^{2jt}+e^{-2jt}}{2}</math> |
| + | |||
| + | and when we plug it into our system we see | ||
| + | <math>\frac{te^{-2jt}+te^{2jt}}{2}=t\frac{e^{-2jt}+e^{2jt}}{2}=tcos(2t)</math> | ||
| + | |||
| + | Therefore we can confidently say that our system yields the output y(t) = t*cos(2t) when the input | ||
| + | |||
| + | x(t) = cos(2t). | ||
Latest revision as of 14:32, 19 September 2008
Given the definition of Linear systems we know the response to $ \alpha x_1(t) + \beta x_2(t) $ is $ \alpha y_1(t)+ \beta y_2(t). $
Consider the system:
$ e^{-2jt}\to F ( e^{-2jt} ) \to te^{2jt} $
From the given system we know
$ x(t)\to F( x(t) )\to tx(-t) $
Euler's formula allows to rewrite $ e^{iy} $ as $ e^{iy}=cos{y}+i*sin{y} $
Using this we deduct that x(t) = cos(2t)= $ \frac{e^{2jt}+e^{-2jt}}{2} $
and when we plug it into our system we see $ \frac{te^{-2jt}+te^{2jt}}{2}=t\frac{e^{-2jt}+e^{2jt}}{2}=tcos(2t) $
Therefore we can confidently say that our system yields the output y(t) = t*cos(2t) when the input
x(t) = cos(2t).
