Root Locus. 1 Review of related mathematics. Ang Man Shun. October 30, Complex Algebra in Polar Form. 1.2 Roots of a equation

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1 Root Locus Ang Man Shun October 3, Review of relate mathematics 1.1 Complex Algebra in Polar Form For a complex number z, it can be expresse in polar form as z = re jθ 1 Im z Where r = z, θ = tan. Denote r = z an θ = z Re z For multiple complex number Similarly, for a function z = 1.2 Roots of a euation z 1 z 2 = z 1 z 1 z 2 z 2 = z 1 z 2 z 1 + z 2 ) z 1 z 2 = z 1 z 1 z 2 z 2 = z 1 z 2 z 1 z 2 ) zk wk = z k ) z k ) w k ) w k ) = ) zk zk ) w k wk Funamental Theorem of Algebra : a orer n polynomial euation has exactly n roots, incluing complex roots, eual roots, real roots. Complex roots occur in pairs If α is a ouble root of fx) =, then it is also a root of fx) x = 1

2 1.3 Review of transfer function The open loop gain is s)gs) Consier a control system with plant Gs), controller s) an feeback Hs), input X I s) an output X O s) The close loop gain is s)gs) 1 + s)gs)hs) Then the characteristic euation of the close loop gain is 1 + s)gs)hs) For simplicity, assume s) is a number The termgs)hs) can be represente by a rational function Ns) Ds) be n, an egree of Ns) be m, an let the egree of Ds) Let Gs)Hs) has overall egree of p, so p = n m. For proper fraction, n m, i.e. p m k=1 By Funamental Theorem of Algebra, an apply factorization, Gs)Hs) = b ks k k=n k=1 a ks = b m k=1 s z k) k a n k=1 s p k) So the characteristic euation is 1 + Gs)Hs) = 1 + b m k=1 s z k) a n k=1 s p k) Assume, b, a > 2 The Root Locus s)gs)hs) The close loop transfer function contains the characteristic euation 1+s)Gs)Hs) 1 + s)gs)hs) For s) = 1 + Gs)Hs) = 1 + b m k=1 s z k) a n k=1 s p k) = 1 + b s z 1 ) s z 2 )... s z m ) a s p 1 ) s p 2 )... s p n ) = i.e. b s z 1 ) s z 2 )... s z m ) a s p 1 ) s p 2 )... s p n ) = 1 It can be seen that, as varies, poles will change, the locus of poles when changes is the root locus. As this is a complex variable function, so we can consier the euality of both magnitue an phase of this euation Gs)Hs) = Ns) Ns) Ds) = 1 Ds) = 1 Ns) Ds) = 1 R + = o Ns) Ds) = 1 1 = 18 o + l 36 o Ns) Ds) = 18o + l 36 o 2

3 2.1 The Rules Root locus is symmetryic about real axis The characteristic euation, Ds) = has real coefficients, so it s rootust occur in complex conjugate paris that symmetric about real axis. An the root locus is a iagram of root of characteristic euation or euivalently poles of Gs) ), so it is also symmetric about real axis Number of branches of root locus = orer of characteristic euation number of poles of Gs) ) The root locus is locus of poles as varies, so number of branches iumber of poles orer of characteristic euation ), which i Root locus start at poles of Gs), an en at zeros of Gs). Ns) Ds) = 1 Ns) = Ds), at the beginning, when =, then Ds) =, i.e. the root locus start at poles of the characteristic euation. Ns) Ds) = 1 Ns) Ds) = 1, at the en, when, then Ns) =, i.e. the root locus ens at zeros of the characteristic euation Segments on real axis that o number of poles an zeros on its left are part of root locus Start with Ns) Ds) = 18o + l 36 o Ns) Ds) = 18 o + l 36 o = k 18 o k is o s z k ) s p k ) = k 18 o From the iagram, a z that is on the left of the s, its resultant s z has angle o From the iagram, a z that is on the right of the s, its resultant s z has angle 18 o Using same logic to poles, then The term s z k ) s p k ) means the angle iffernece between all the zero an poles, since any zero or pole on the left of the s result in o, so only consier the zero an pole of the right sie of. s Since one zero on the right sie of s contribute to 18 o an same as the pole oes. But since the angle s p) iegative, so it means subtract 18 o. By the s z k ) s p k ) = k 18 o, k is o number, then Segments on real axis that o number of poles an zeros on its right are part of root locus For poles an zeroot on the real axis, from the iagram, the angle cancel out by the conjugate pairs, so those are not counte. 3

4 pi z i The asymptotes as s is in α = Consier the characterisitic eaution, ϕ = 18o + l 36 o 1 + Gs)Hs) = 1 + b b m a + a a n Take sm out b 1+Gs)Hs) = 1+ sm + s b m a + a 1 s a = 1+ n Thus Then consier the phase lim 1 + Gs)Hs) = 1 + b = 1 + b s s s a a ) ) lim 1 + Gs)Hs) = lim 1 + Gs)Hs) s s = lim 1 + Ns) ) s Ds) = 1 + b ) s a = 1 + b s a } {{} R + By R + can be ignore an Ns) Ds) = 18o + l 36 o an 1 b + s b m m a + a 1 s a b + = 1+ s b m n s a + a 1 s a n e jϕ e jϕ) = ϕ = 18 o + l 36 o ϕ = 18o + l 36 o 4

5 2.1.6 The breakaway / break in points When more than one brancheeet in the break-away / break-in point, that point is a multiple root of 1 + Gs)Hs) = As the break-away / break-in point is a multiple root of 1 + Gs)Hs) =, so it is also a root of s Gs)Hs) = Let the break away / break in point be x Eliminate the 1 + Gx)Hx) = 1 + Nx) Dx) = { Dx) + Nx) = D x) + N x) = = Dx) Nx) = D x) N x) Dx)N x) = D x)nx) N x)dx) D x)nx) = Consier s Gs)Hs) = Ns) s Ds) = N s)ds) N s)ds) D 2 s) Plug in the euation N x)dx) D x)nx) =, so to fin the break-in / break-away point is to solve s Gs)Hs) = If the root locus corsses the stability bounary the imaginary axis), the corssing point can be foun by solving 1 + Gjω)Hjω) = Recall that, the location of the poles of the system T.F. etermine the amping property of the system. For stable system, the poles is in LHP Left Half Plane) When the poles lies on the imaginary axis, i.e. real part is zero), the system output io-ampe oscillation, it is the bounary case for stable system. When the poles has postive real part e pt = e Rep)t e jimp)t as t, so the system is unstable. So the bounary stable case can be foun by solving i.e. 1 + Gs)Hs) = s lies on imaginary axis 1 + Gjω)Hjω) = END 5