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Steady state solution

  1. Nov 6, 2007 #1
    [SOLVED] Steady state solution

    I was wondering if I did this question correctly, solving for y(t) and putting t = infinity to get a steady state solution. Or is this wrong or is there an alternative way that is much quicker (as solving for y(t) would take a page of working, where the working out on the exam paper was less then half a page)

    1. The problem statement, all variables and given/known data
    A continuous linear time invariant system with input x(t) and output y(t) related by:

    y''(t) + y'(t) - 2y(t) = x(t)

    Find the stead state output of the system for x(t) = 2cos(t) + sin(200t)

    3. The attempt at a solution

    Initially used laplace transform to get Y(s) and inverse transform and put t = infinity to see what part died out over time (transient reponse), however it did not seem correct (used mathematica to double check) as I got a e^t term.

    Using ODE solving methods with characteristic equation r^2 + r - 2 = 0, I solved initially the homogeneous solution which had a Ce^t term and got a similar total solution to when using laplace transforms.
  2. jcsd
  3. Nov 6, 2007 #2
    I haven't tried but you may want to assume y = a*cos(t)+b*sin(t)+c*sin(200t)+d*cos(200t) or something like that.
  4. Nov 6, 2007 #3


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    Yes, the solution has a e^t part, as you might expect since 1 is a root of the characteristic equation. You aren't making any mistakes. I would say that means the system is unstable to runaway solution, but I'm not sure what to do in your context. Do you just ignore it by setting the coefficient=0?
  5. Nov 6, 2007 #4


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    Yes, the characteristic equation r2+ r- 2= (r+ 2)(r-1) has solutions 1 and -2. In order to have a "steady state solution", you cannot have unbounded solutions. Take the "C" in "Cet" to be 0. Then the other solution, De-2t goes to 0. Your steady state solution is just the specific solution corresponding to y = a*cos(t)+b*sin(t)+c*sin(200t)+d*cos(200t).
  6. Nov 6, 2007 #5
    Thanks for the replies.

    I see how it works when using ODE methods now. Just a final question, how do you do this with laplace transforms.

    I got:
    Y(s)(s^2 + s - 2) = (2s/(1 + s^2) + 200/(40000 + s^2))

    This was taking to long to evaluate by hand so I used mathematica, and the e^x term had a coefficient of 40201/120003. How can laplace transforms used to make this coefficient zero ?
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