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

Transcription

1 Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsa-lab, CNRS-INPG, FRANCE Olivier.Sename@gipsa-lab.inpg.fr Lead Lag 17th March 2010

2 Outline Lead Lag Lead Lag

3 References Some interesting books: G. Franklin, J. Powell, A. Emami-Naeini, Feedback Systems, Prentice Hall, 2005 R.C. Dorf and R.H. Bishop, Modern Control Systems, Prentice Hall, USA, P.Borne, G.Dauphin-Tanguy, 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 degree-of-freedom 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 loop-transfer function. Then the closed-loop 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 s-plane 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 loop-transfer 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 root-locus. 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 left-plane 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 (n-m=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 closed-loop 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 ) Derivative-type 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 ) Integration-type 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 steady-state 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 Lead-lag Lead Lag