laplace ( exponentials and sinusoids)

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tronicsworld

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i learned that the differnece between FT and LT is that FT uses division of a time doamain signal into its sinusoids frequency components we have chosen sinusoids because when passed through LTI its shape not changes so we can easily detect change in amplitude and phase . Now i do not understand why we break up into exponentials and and sinusoids both in laplace i read that it is necessary because differential equation solution consists of both sinusods and exponentials . but i do not understand physical significance of this ???
 


The physical significance comes from the solution of the differential equation. If you are solving a first-order differential equation, your general solution comes from the characteristic equation A+Bs=0, where you solve for s=-A/B and you end up with \[{Ke }^{-\frac{A }{B } t }\] for your general solution. For second (and higher) -order differential equations, you have characteristic equations that are second (and higher) order polynomials, from which you solve for the roots of the equation, r1, r2, r3,... These are then integrated and you find the solution \[{K_1e}^{-r_1 t}+{K_2e}^{-r_2 t}+{K_3e}^{r_3 t}+...\]. Now the really interesting part is when your polynomial has complex roots. If r1 and r2 are something like \[\alpha+j\omega\] and \[\alpha-j\omega\], you can find decaying complex sinusoidals in your general solution using Euler's identities: \[{K_1e^{-\alpha t}e^{-j\omega t}}+{K_2e^{-\alpha t}e^{j\omega t}} \equiv K_3e^{-\alpha t}(cos( \omega t + \phi))[\tex]. This is what leads into all those neat things like natural resonances.

Hope this helps!
 
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