Methods for analysis and control of. Lecture 4: The root locus design method


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1 Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Lead Lag 17th March 2010
2 Outline Lead Lag Lead Lag
3 References Some interesting books: G. Franklin, J. Powell, A. EmamiNaeini, Feedback Systems, Prentice Hall, 2005 R.C. Dorf and R.H. Bishop, Modern Control Systems, Prentice Hall, USA, P.Borne, G.DauphinTanguy, J.P.Richard, F.Rotella, I.Zambettakis, Analyse et Régulation des processus industriels. Tome 1: Régulation continue, Méthodes et pratiques de l ingénieur, Editions Technip, Lead Lag
4 Definition Consider a simple one degreeoffreedom control structure, with G(s) the plant model, C(s) the controller, and K a constant gain, which is the parameter of the controller to be analyzed. Let us write L(s) = C(s)G(s) the looptransfer function. Then the closedloop transfer function is : T (s) = KL(s) 1 + KL(s) and the solution of the characteristic equation 1 + K.L(s) = 0 is: Then the solution of this equation is given by: Lead Lag {s Cs.t.L(s) = 1 K } which implies KL(s) = 1 and KL(s) = π + 2.k.π
5 Definition (2) Rewrite L as L(s) = B(s) A(s) where B(s) and A(s) are the numerator and denominator of L. Then the characteristic equation becomes: In all what follows: A(s) + K.B(s) = 0 Lead Lag B(s) = A(s) = m i=1 n j=1 (s z i ) (s p j )
6 Definition (3) The root locus of L(s) is the set of points in the splane where the phase of L(s) is π. If we define the angle to the test point from a zero as ψ i and the angle to the test point from a pole as φ i, then: i ψ i φ j = π + 2π(k 1) j Lead Lag
7 First example Consider the looptransfer function: L(s) = 1 s(s + 1) Write the characteristic equation and calculate the set of solutions {s C as a function of K. Illustration in Matlab using rlocus Lead Lag
8 Guidelines for the root locus plot There are 6 major rules to sketch a rootlocus. The beginning equation is : n j=1 (s p j ) + K. m i=1 Two different cases have to be considered: K is positive and then L(s) = π + 2.k.π K is negative and then L(s) = k.π (s z i ) = 0 (1) The number of branches (loci) is equal to the number of poles (i.e. the number of solutions of the characteristic equation). Lead Lag
9 Rule 1: departure and arrival Rule 1 The n branches of the locus start at the poles of L(s) and m of these branches end on the zeros of L(s) Indeed: 1. When K = 0, the solution of equation (1) are the poles p j. 2. from L(s) = 1 K, as K, the equation reduces to L(s) = 0, i.e. B(s) = 0 which corresponds to the zeros of L Example : L(s) = 1 s(s 2 + 8s + 32) Lead Lag
10 Rule 2: branches on the real axis Rule 2 The loci are on the real axis to the left of an odd number of poles and zeros. On the examples: L(s) = 1 s(s 2 + 8s + 32) Lead Lag L(s) = (s + 1) s(s + 2)(s + 4) 2
11 Rule 3: asymptotes We consider here large s and K, i.e when they reaches infinity. Rule 3 Let N = n m 0 of the loci are asymptotic to lines at angles φ A centered at a point on the real axis given by: σ A = p j z i n m Lead Lag with φ A = (2k + 1)π n m, k = 0,1,2,...,n m
12 Rule 4: angle of departures of the branch Rule 4 The angle of locus departure from a pole is the difference between the net angle due to all other poles and zeros and the criterion angle of ±π. The angles of departure of a branch of the locus from a pole of multiplicity q is given by: qφ l,dep = ψ i φ j π 2π(k 1) j l Lead Lag and the angles of arrival of a branch a a zero of multiplicity q is given by: qψ l,arr = ψ i φ j + π + 2π(k 1) i l
13 Rule 5 Rule 5 The root locus crosses the jω axis at points where the Routh criterion shows a transition from roots in the leftplane to roots in the right half plane. 1 Example: L(s) = s(s 2 +8s+32). Routh s s 2 8 K s 1 s K 8 0 For K = 0, there is a root at s = 0. No RHP roots for 0 < K < 8 32 = 256. When K = 256 roots on the jω axis (ω c = 5.66). K Lead Lag
14 Rule 6: breakaway point Rule 6 The equation to be solved can be rewritten as: A(s) B(s) = K which means that the set of solutions changes of direction (tangents) whenever d ds ( A(s) B(s) ) = 0 The locus will have multiple roots at points on the locus where the derivative is zero, or: Lead Lag B da ds AdB ds = 0
15 Root locus for satellite attitude control with PD control The characteristic equation is : 1 + [k p + k d s] 1 s 2 = 0 Assume k p /k d = α is known, and note K = k d. Then it leads 1 + K s + α s 2 = 0 Lead Lag R1: There are 2 branches that start at s = 0 and 1 which ends at s = α R2: The real axis to the left of s = α is on the locus R3: There is one asymptote (nm=1) at σ A = α of angle φ A = π
16 Example (cont.) R4: the angles of departure from the double pole at s = 0 are ±π/2 R5: Routh criterion: s 2 1 K.α s 1 K 0 s 0 K α. Lead Lag which does not cross the imaginary axis. R6: using B = s + α and A = s 2 it leads that the breakaway points stand at s = 0 and s = 2.α.
17 The required closedloop performances should be chosen in the following zone Lead Lag which ensures a damping greater than ξ = sinφ. γ implies that the real part of the CL poles are sufficiently negatives.
18 (2) Some useful rules for selection the desired pole/zero locations (for a second order system): Rise time : t r 1.8 ω n Seetling time : t s 4.6 ξ ω n Overshoot M p = exp( πξ /sqrt(1 ξ 2 )): ξ = 0.3 M p = 35%, ξ = 0.5 M p = 16%, ξ = 0.7 M p = 5%. Lead Lag
19 PID controller A PID controller is given by: C(s) = K p (1 + 1 T i s + T d s) = K p + K D s + K I s For convenience it will be rewritten as : C(s) = K D (s + z 1 )(s + z 2 ) s which has one pole at the origine and two stable zeros. Lead Lag
20 Lead with z < p. C(s) = K ( s + z s + p ) Derivativetype controller : if p is placed well outside the frequency range of the design, the controller looks like a PD controller. The effect of the zero is to move the locus to the left (towards more stable zones). To be chosen directly below the desired root location. p should be located left far on the real axis. It should ensure that the total angle at the desired root location is π Lead Lag
21 Lag with z > p. C(s) = K ( s + z s + p ) Integrationtype controller: p should be closed to the origine. z sufficiently far. z/p is chosen to be between 3 and 10 (according to the need of boosting the steadystate gain). Lead Lag
22 Matlab example a small airplane (pitch attitude): where G(s) = θ = G(s)(δ + M d ) 160(s + 2.5)(s + 0.7) (s 2 + 5s + 40)(s s ) θ is the pitch attitude, δ the elevator angle and M d the disturbance moment. : rise time 1sec and overshoot less than 10% Design an autopilot so that the steady state value of δ is zero for an arbitrary constant moment. Lead Lag
23 Matlab example t r 1sec implies ω n 1.8. M p 0.1 implies ξ 0.6. Steps Polynomial controller lead Compensation Leadlag Lead Lag
Methods for analysis and control of dynamical systems Lecture 4: The root locus design method
Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.inpg.fr www.gipsalab.fr/ o.sename 5th February 2015 Outline
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