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<math>x(t)\to system\to tx(-t)</math> | <math>x(t)\to system\to tx(-t)</math> | ||
| + | |||
| + | From Euler's formula | ||
| + | <math>e^{iy}=cos{y}+i sin{y}</math> | ||
| + | |||
| + | We know that: | ||
| + | x(t) = cos(2t)= <math>\frac{e^{2jt}+e^{-2jt}}{2}</math> | ||
| + | Now taking the system into consideration and applying the above equation to it | ||
| + | <math>\frac{te^{-2jt}+te^{2jt}}{2}=t\frac{e^{-2jt}+e^{2jt}}{2}=tcos(2t)</math> | ||
| + | |||
| + | So the system's respone to '''cos(2t)''' is '''tcos(2t)''' | ||
Latest revision as of 11:51, 18 September 2008
As discussed in class,a system is called linear if for any constants a,b belongs to phi and for anputs x1(t), x2(t) (x1[n],x2[n]) yielding output y1(t) , y2(t) respectively the response to
ax1(t) + bx2(t) is ay1(t)+by2(t).
Consider the following system: $ e^{2jt}\to system\to te^{-2jt} $
$ e^{-2jt}\to system\to te^{2jt} $
From the given system:
$ x(t)\to system\to tx(-t) $
From Euler's formula $ e^{iy}=cos{y}+i sin{y} $
We know that: x(t) = cos(2t)= $ \frac{e^{2jt}+e^{-2jt}}{2} $ Now taking the system into consideration and applying the above equation to it $ \frac{te^{-2jt}+te^{2jt}}{2}=t\frac{e^{-2jt}+e^{2jt}}{2}=tcos(2t) $
So the system's respone to cos(2t) is tcos(2t)
