Peak Overshoot of a 3rd Order System PDF

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Peak Overshoot of a Generalized 3rd order system...

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For a standard 3rd order system determine ζ, ωn , ωd and tp . Let us consider a standard 3rd order system of characteristic equation b3 m3 + b2 m2 + b1 m + b0 = 0 We can also think of it as a 2nd order dominant 3rd system. That means, b3 m3 + b2 m2 + b1 m + b0 = (m + p)(m2 + 2ζωn m + ωn2 ) = m3 + m2 (2ζωn + p) + m(ωn2 + 2ζωn p) + ωn2 p Now, equating both side we get, b3 = 1 2ζωn + p = b2 Hence, 2ζωn

=

b2 − p

ωn2 + 2ζωn p = b1 ωn2 p = b0 From (1), (2) and (3), b0 + (b2 − p)p = b1 p =⇒ b0 + (b2 − p)p2 = pb1 =⇒ b0 + b2 p − p3 − pb1 = 0 =⇒ p3 − b2 p2 + pb1 − b0 = 0 ...(5) Now solving (5) we get the values of p. Then, ωn =

s

b0 p

b2 − p 2ωn p ωd = ωn 1 − ζ 2 ζ=

To determine

C(s) ωn2 = 2 (s + p)(s + 2ζωn s + ωn2 ) R(s) (s + p)(s2 + 2ζωn s + ω 2n ) − s(s2 + 2ζωn s + ωn2) − [p(s2 + 2ζωn s + ωn2 ) − ω n2 ] = s(s + p)(s2 + 2ζωn s + ωn2 ) 1 1 1 ω 2n 1 − + + 2 = − s s+p (s + 2ζωn s + ωn2 ) s s+p 2 ωn = (s + ζωn )2 + ωn2 (1 − ζ ) ωn2 ωd = ωd [(s + ζωn )2 + ωd2] ω2 c(t) = n .e−ζωn t .sinωd t ωd Now for tp , dc(t) =0 dt p p p ωn ωn =⇒ p (−ζωn )e−ζωn t sinωn 1 − ζ 2 t + p e−ζωn t ωn 1 − ζ 2 cosωn 1 − ζ 2 t = 0 2 2 1−ζ 1−ζ 1

(1) (2) (3)

Hence the necessary condition, ωn Peak overshoot occurs at first value after 0

p

1 − ζ 2 = 0, π, 2π

p

1 − ζ 2 tp = π π p ∴ tp = ωn 1 − ζ 2 ωn

2...