Fa2004 16.33385 Root locus basics U Gc(s o Assume that the plant transfer function is of the form S-2 G=K K S-p and the controller transfer function is Ge(s)=k=kIl(s-zci Signals are u control commands y output/measurements r reference input e response error This is the unity feedback form. We could add the controller gc in the feedback path without changing the pole locations Will add disturbances later
� � Fall 2004 16.333 8–5 Root Locus Basics r e u y /OO / Gc(s) / Gp(s) / − • Assume that the plant transfer function is of the form Gp = Kp Np = Kp� i (s − zpi) Dp i (s − ppi) and the controller transfer function is Nc Gc(s) = Kc = Kc� i (s − zci) Dc i (s − pci) • Signals are: u control commands y output/measurements r reference input e response error • This is the unity feedback form. We could add the controller Gc in the feedback path without changing the pole locations. • Will add disturbances later
Fa2004 16.33386 Basic questions Analysis: Given N and De, where do the closed loop poles go as a function of k Synthesis: Given Kp, Np and Dp, how should we chose Kc, to put the closed loop poles in the desired locations? Block diagram analysis: Since y=G Gce and e=r-g, then easy to show that S r 1+gg vhere c(s KcKpNNp DDp+ KcKpNCNp is the closed loop transfer function The denominator is called the characteristic equation oc(s)and the roots of s)=0 are called the closed-loop poles(CLp) The CLp are clearly functions of Kc for a given Kp, Np, Dp, Nc, Dc a" locus of roots"[Evans, 1948
Fall 2004 16.333 8–6 • Basic questions: – Analysis: Given Nc and Dc, where do the closed loop poles go as a function of Kc? – Synthesis: Given Kp, Np and Dp, how should we chose Kc, Nc, Dc to put the closed loop poles in the desired locations? • Block diagram analysis: Since y = GpGce and e = r − y, then easy to show that y GcGp =r 1 + GcGp ≡ Gcl(s) where KcKpNcNp Gcl(s) = DcDp + KcKpNcNp is the closed loop transfer function • The denominator is called the characteristic equation φc(s) and the roots of φc(s) = 0 are called the closedloop poles (CLP) . • The CLP are clearly functions of Kc for a given Kp, Np, Dp, Nc, Dc ⇒ a “locus of roots” [Evans, 1948]
Fa2004 16.3338-7 Root Locus Analysis General root locus is hard to determine by hand and requires matlab tools such as locus (num, den) to obtain full result but we can get some important insights by developing a short set of plotting rules Full rules in FPE, page 260 e basic questions 1. What points are on the root locus? 2. Where does the root locus start? 3. Where does the root locus end? 4. When/ where is the locus on the real line? 5. Given that so is found to be on the locus, what gain is need for that to become the closed-loop pole location? Question #1: is point so on the root locus? Assume that Ne and De are known let DD and K=Kkn then (s)=1+KLd(s)=0 or equivalently for values of s for which La(s)=-1/K, with K real For K positive, So is on the root locus if ∠L(50)=180°±l·360°,l=0.,1 If K negative, so is on the root locus if [0 locus] ∠L(s0)=0°±·360°,7=0,1 These are known as the phase conditions
Fall 2004 16.333 8–7 Root Locus Analysis • General root locus is hard to determine by hand and requires Matlab tools such as rlocus(num,den) to obtain full result, but we can get some important insights by developing a short set of plotting rules. – Full rules in FPE, page 260. • Basic questions: 1. What points are on the root locus? 2. Where does the root locus start? 3. Where does the root locus end? 4. When/where is the locus on the real line? 5. Given that s0 is found to be on the locus, what gain is need for that to become the closedloop pole location? • Question #1: is point s0 on the root locus? Assume that Nc and Dc are known, let Ld = Nc Dc Np Dp and K = KcKp then φc(s) = 1 + KLd(s) = 0 or equivalently for values of s for which Ld(s) = −1/K, with K real. – For K positive, s0 is on the root locus if ∠Ld(s0) = 180◦ ± l · 360◦ , l = 0, 1, . . . – If K negative, s0 is on the root locus if [0◦ locus] ∠Ld(s0) = 0◦ ± l · 360◦ , l = 0, 1, . . . These are known as the phase conditions
Fa2004 16.33388 Question#2: Where does the root locus start? Q=1+K D. →DD+ KN.N=0 So if K-0. then locus starts at solutions of dd,=0 which are the poles of the plant and compensator Question #3: Where does the root locus end? Already shown that for so to be on the locus, must have La(so) K So if K→∞, the poles must satisfy NN Ld DD There are several possibilities 1. Poles are located at values of s for which NN=0, which are the zeros of the plant and the compensator 2. If Loop la( s) has more poles than zeros As|s→∞,|La(s)→0, but we must ensure that the phase condition is still satisfied
Fall 2004 16.333 8–8 • Question #2: Where does the root locus start? NcNp φc = 1 + K = 0 DcDp ⇒ DcDp + KNcNp = 0 So if K → 0, then locus starts at solutions of DcDp = 0 which are the poles of the plant and compensator. • Question #3: Where does the root locus end? Already shown that for s0 to be on the locus, must have 1 Ld(s0) = −K So if K → ∞, the poles must satisfy: NcNp Ld = = 0 DcDp • There are several possibilities: 1. Poles are located at values of s for which NcNp = 0, which are the zeros of the plant and the compensator 2. If Loop Ld(s) has more poles than zeros – As | | s → ∞ | , Ld(s)| → 0, but we must ensure that the phase condition is still satisfied
Fa2004 16.3338-9 More details as k→: Assume there are n zeros and p poles of la(s) Then for large s d(s So the root locus degenerates to 1+ (s-a)-n0 So n poles head to the zeros of la(s) Remaining p-n poles head to s= oo along asymptotes defined by the radial lines 1809+3600·(-1) P-7 so that the number of asymptotes is governed by the number of poles compared to the number of zeros(relative degree) If zi are the zeros if ld and p; are the poles, then the centroid of the asymptotes is given by ∑n-∑ Exampl pIe: L(S)
� � Fall 2004 16.333 8–9 • More details as K → ∞: – Assume there are n zeros and p poles of Ld(s) – Then for large |s|, 1 Ld(s) ≈ (s − α)p−n – So the root locus degenerates to: 1 1 + = 0 n (s − α)p− – So n poles head to the zeros of Ld(s) – Remaining p − n poles head to | | s = ∞ along asymptotes defined by the radial lines 180◦ + 360◦ · (l − 1) φl = l = 1, 2, . . . p − n so that the number of asymptotes is governed by the number of poles compared to the number of zeros (relative degree). – If zi are the zeros if Ld and pj are the poles, then the centroid of the asymptotes is given by: p n pj − zi α = p − n • Example: L(s) = s−4 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 Root Locus Real Axis Imaginary Axis