Step Response of an LTI system

  • Thread starter qwertydump
  • Start date
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Hi lads, need help with this question from a signals and systems course.





LTI system is defined as follows



[tex] H(s) = \frac{1}{ s^2 - 2rs cos(theta) + r^2}[/tex]



where r = 20pi, theta = 1.47.



Find an expression for the step response of the system





I think what I should do is multiply H(s) by 1/s cos it's a step response

and then mix and match it with the laplace tables to get the inverse transform.



So my attempt was:



[tex]\frac{1}{s^{2}-2rs cos(th) + r^{2}}[/tex]



= [tex]\frac{1}{s^{2}-2rscos(th)+r^{2}cos^{2}(th) + r^{2}- r^{2}cos^{2}(th)}[/tex]



= [tex]\frac{1}{(s-rcos(th))^{2}+r^{2}-r^{2}cos^{2}(th)}[/tex]



= [tex]\frac{1}{(s-rcos(th))^{2}+r^{2}(1-cos^{2}(th))}[/tex]



= [tex]\frac{1}{(s-rcos(th))^{2}+r^{2}sin^{2}(th) }[/tex]



= [tex]\frac{1}{(s-rcos(th))^{2}+(rsin(th))^{2}}[/tex]



This was great cos then I could match it to a transform on my laplace transform table



this one:



[tex]\frac{A(s+a) + Bw}{(s+a)^{2}+w^{2}}[/tex]



= [tex] e^{-at}[Acos(wt)+Bsin(wt)] [/tex]



but the problem was that I realized that A & B would be zero and i'd get zero as a result and also I hadn't multiplied my equation by 1/s and that screws up everything. I haven't a clue what else to do then.



so ... am I approching this thing all wrong?



any guidance or help very much appreciated

please bear in mind that I don't really know if I should be trying to do the laplace transform or what.



Berty

Homework Statement





Homework Equations





The Attempt at a Solution

 

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