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Settling time and overshoot

  1. Dec 15, 2012 #1

    MathematicalPhysicist

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    1. The problem statement, all variables and given/known data + the attempt at solution
    I have the next transfer function:

    [tex]H(s)=\frac{P(s)}{1+C(s)P(s)}[/tex]

    where [tex]P(s)=\frac{20}{(s^2-2s+9)(s+100)}[/tex]

    [tex] C(s)=m(s+40)[/tex]

    Now I want to find the settling time and overshoot of [tex]H(s)[/tex] as [tex]m\rightarrow \infty[/tex] to whithin 5 percent accuracy.

    Now what I thought when [tex]m\rightarrow \infty[/tex] [tex]H(s)\approx \frac{1}{C(s)}[/tex].

    So basically in this case the quadratic equation (in the notation of next webpage:
    http://wikis.lib.ncsu.edu/index.php/Second_Order_Dynamics) is:

    [tex]s^2+(-2s+20ms)+(9+800m)[/tex]

    Is this right or am I just rambling nonsense here, I am lost here.



    2. Relevant equations
    I know that when we have the Quadratic ploynomial in the denominator of [tex]G(s)=P(s)C(s)[/tex] s.t:
    [tex] s^2+2\zeta \omega_n s +\omega_n ^2[/tex]

    then the settling time is given by:

    [tex]t_s \approx \frac{3}{\zeta \omega_n}[/tex]

    and the overshoot is given by:

    [tex] \sigma = exp(\frac{-\zeta \pi}{\sqrt{1-\zeta^2}})[/tex]

    Any hints or advice are welcomed and much appreciated.

    Thanks.
     
  2. jcsd
  3. Dec 16, 2012 #2
    Are you familiar with the root locus method? It describes the path the roots take when a single parameter, like m, is varied. I would need to do a little review before giving a better answer but maybe this will get you started.
     
  4. Dec 17, 2012 #3

    MathematicalPhysicist

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    Yes I know about the RL.

    I am not sure how to calculate the overshoot and settling time here, though I know that I need to look at the Bode graph of [tex]\frac{1}{m(s+40)}[/tex]

    And I know that the frequency that the asymptote to the the magnitude of the above expression intersects the [tex]\omega[/tex] axis is given by: [tex]\omega= 40[/tex].

    Other than that, I am not sure how to procceed, any advice?
     
  5. Dec 19, 2012 #4

    rude man

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    My reaction is to agree with you that as m → ∞, H(s) = 1/C(S) = 1/m(s+40).

    The impulse response to this H(s) is simply (1/m)exp(-40t). The step response is simply (1/m)[1 - exp(-40t)]. In neither case is there any overshoot. So I am as confused as you ... :smile:
     
  6. Dec 20, 2012 #5

    MathematicalPhysicist

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    And what of the settling time of 1/C(s)?

    Thank you for your help.
     
  7. Dec 20, 2012 #6

    rude man

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    Since the output is an exponential decay you can compute settling time depending on your definition of "settling time". For example, if the input is a step function, and "settling time" ts is defined as 95% of "final" value, then you solve for ts from 1 - exp(-40ts) = 0.95.
     
  8. Dec 20, 2012 #7

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    OK, thanks.

    Then the overshoot is zero.
     
  9. Dec 20, 2012 #8
    The problem with m→∞ is that the response goes to zero. So you can't let m→∞. Maybe you can say m becomes large enough that it dominates the response. Then I do agree with what you have been saying. You can call the (s+100) pole insignificant as I think you did or you could do this, which is another way to deal with these sorts of boundary case problems:

    [itex]0=1+C(s)P(s)=s^3+98s^2+(20m-191)s+(900+800m)[/itex]

    Making m large can simplify this by making most terms too small to be counted. Dividing by m yields this:

    [itex]0=\frac{1}{m}(s^3+98s^2-191s+900)+20s+800\approx s+40[/itex]

    This means you have a single pole response as you found.




    Your root equation can be written in the standard form with the parameter m in the right place for root locus:

    [itex]1+C(s)P(s)=1+m\frac{20(s+40)}{(s^2-2s+9)(s+100)}=1+mQ(s)[/itex]

    The root locus method says as m begins at 0 and approaches ∞, the roots of this equation begin at the poles of Q(s) and end at the zeroes of Q(s). Q(s) has two zeroes at ∞ and one zero at -40. It has three poles, one at -100 and two complex conjugate ones. The conclusion from root locus is that with m→∞, there will be two roots at ∞ and one at -40. The finite root is (s+40) as found above.
     
    Last edited: Dec 21, 2012
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