Methods for analysis and control of dynamical systems 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 Gipsa-lab, CNRS-INPG, FRANCE o.sename 5th February 2015

2 Outline

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, 1993.

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: {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: A(s) + K.B(s) = 0 In all what follows: m B(s) = (s z i ) i=1 n A(s) = (s p j ) j=1

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 ({s Cs.t.L(s) = 1 K }) Then, arg(b(s)) arg(a(s)) = π + 2π(k 1) 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

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 Cs.t.L(s) = K 1 } as a function of K. Illustration in Matlab using rlocus(l) or rltool(l)

8 Guidelines for the root locus plot There are 6 major rules to sketch a root-locus. The beginning equation is : n m (s p j ) + K. (s z i ) = 0 (1) j=1 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 The number of branches (loci) is equal to the number of poles (i.e. the number of solutions of the characteristic equation).

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) = K 1, 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)

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) = L(s) = 1 s(s 2 + 8s + 32) (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: with φ A = σ A = p j z i n m (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 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 K 8 0 s 0 K 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).

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 ds d ( A(s) B(s) ) = 0 The locus will have multiple roots at points on the locus where the derivative is zero, or: B da ds A db 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 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. 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 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%.

19 (3) Some rules do exist to shape the transient response. The ITAE (Integral of Time multiplying the Absolute value of the Error), defined as: ITAE = t e(t) dt 0 can be used to specify a dynamic response with relatively small overshoot and relatively little oscillation (there exist other methods to do so). The optimum coefficients for the ITAE criteria are given below (see Dorf & Bishop 2005). Order Characteristic polynomials d k (s) 1 d 1 = [s + ω n ] 2 d 2 = [s ω n s + ω 2 n ] 3 d 3 = [s ω n s ω 2 n s + ω 3 n ] 4 d 4 = [s ω n s ω 2 n s ω 3 n s + ω 4 n ] 5 d 5 = [s ω n s 4 + 5ω 2 n s ω 3 n s ω 4 n s + ω 5 n ] 6 d 6 = [s ω n s ω 2 n s ω 3 n s ω 4 n s ω 5 n s + ω 6 n ] and the corresponding transfer function is of the form: H k (s) = ωk n, k = 1,...,6 d k (s)

20 (4) STEP RESPONSE OF TRANSFER FUNCTIONS WITH ITAE CHARACTERISTIC POLYNOMIALS H1 H2 H3 H4 H5 H NORMALIZED TIME ω n t

21 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.

22 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

23 with z > p. Integration-type controller: C(s) = K ( s + z s + p ) 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).

24 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.

25 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

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