Step response of an LTI system

Discussion in 'Homework Help' started by qwertydump, Apr 15, 2010.

  1. qwertydump

    Thread Starter New Member

    Apr 15, 2010
    Hi lads, need help with this question from a signals and systems course.

    LTI system is defined as follows

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

    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:

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

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

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

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

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

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

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

    this one:

    \frac{A(s+a) + Bw}{(s+a)^{2}+w^{2}}

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

    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.

    Last edited: Apr 15, 2010
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Strange problem with that -ve sign in H(s) denominator. It means there is a complex root in the right half of the complex plane - an unstable system.
  3. qwertydump

    Thread Starter New Member

    Apr 15, 2010
    Thanks t_n_k for your reply ... now you make me wonder if it was a typo from my lecturer the negative part in the denominator .... maybe it should of been positive. But I can't really make sence of it anyway so yes, I'd agree that it's a strange one.

    I tried to solve it using partial fraction and got the below result.

    Whether this result is correct or not I don't know, don't even know if thats what I was supposed to do. But I need to get a valid result to be able to do the next part which consists of feeding in a pulse signal to the input of H(s) and finding an expresion for the output.

    Can someone please tell me if my approach and solution is correct and if not what other way can I approach this problem


    note (th) = theta

     H(s) = \frac{1}{s^{2}-2rscos(th)+r^{2}} + \frac{1}{s}

     = \frac{1}{re^{jth}(s^{2}-2rscos(th)+r^{2})} + \frac{1}{r^{2}s}

    From laplace


    = [​IMG]

    so ...

     h(t) = \frac{1}{r^{2}}u(t) + \frac{1}{re^{jth}}e^{2rcos(th)t}(\frac{1}{r}sin(rt))

     h(t) = \frac{1}{r^{2}}(u(t) + e^{2rcos(th)t-j(th)}sin(rt))

    but is that what I'm looking for?
    Last edited: Apr 18, 2010
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009