How to calculate maximum overshoot? 2

[hmmy]

Hello all. I have the open loop transfer function G(s) = (5s+2) / [s(s-2)]. How to calculate the maximum overshoot of the closed loop system when I have a unit step input? How many overshoots exist? Only the maximum one? Thank you.

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[IRstuff]

For school?

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[hmmy]

For my mind. We cannot use the formula exp[(-ζ/(1-ζ^2)^0.5)π] ?

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[IRstuff]

It'll depend on the damping. Have you looked here:

http://lpsa.swarthmore.edu/Transient/TransInputs/TransStep.html

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[LiteYear]

The numbers involved are so simple I can't help but feel I'm doing someone's homework, but it's such an easy problem that I haven't wasted much time on it.

Why bother with formula short-cuts? There might be some esoteric formula that gives the answer in this specific case, but it's very easy just to find the exact solution and see for yourself the result. Given they're asking about multiple overshoots and G(s) has a zero, I wouldn't bother with zeta and damping factor and specific formulae. Here:

G(s) = (5s+2) / [s(s-2)]
T(s) = (5s+2) / [s(s-2) + 5s+2] (closed loop)
= (5s+2) / [(s+2)(s+1)] (factor the quadratic on the denominator)
Y(s) = (5s+2) / [s(s+2)(s+1)] (step response)
= 1/s + 3/(s+1) - 4/(s+2) (partial fraction decomposition)
y(t) = 1 + 3e^(-t) - 4e^(-2t) (inverse laplace)

Plot it. One overshoot. Find maximum by differentiating w.r.t. t and setting =0:

y'(t) = -3e^(-t) + 8e^(-2t)
y'(t) = 0 => 8e^(-2t) = 3e^(-t)
=> e^(-t) = 3/8
=> t = ln(8/3)

So max value is y(ln(8/3)) = 1 + 3*(3/8) - 4*(3/8)^2 = 25/16


Any errors are deliberately placed to prevent casual copying.

 

[hmmy]

Thank you for answers. And if I want to reduce the maximum overshoot 20% with a
compesator λ * (s+α) / (s+β), then what I can say for λ,α,β ?

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[LiteYear]

Sorry mate, you've exhausted the work I'll do for you. Perhaps have a go yourself and let us know where you get stuck. I think you'll have a hard time convincing people it's "for my mind" though.