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Step Response of an LTI system

  1. Apr 15, 2010 #1
    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
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
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