Control Systems Design of State Feedback Control chibum@seoultech.ac.kr
Outline Design of State feedback control Dominant pole design Symmetric root locus (linear quadratic regulation) 2
Selection of closed-loop poles Now, in principle, we can place the closed-loop poles (CLPs) arbitrarily Is it practical to make a truck behave like a sports car? The further we try to shift the open-loop poles (OLPs), the greater the control effort required expense? actuator saturation? component strength? If an OLP is nearly cancelled by an open-loop zero (OLZ), that mode will be almost uncontrollable will require large control gain (and effort) to shift 3
Ex. How zero location can affect the control law A specific thermal system is in OCF with a zero at s = z 0. x 1 x 2 = 7 1 12 0 x 1 x 2 + 1 z 0 u y = 1 0 x (a) Find a state feedback gain necessary for placing the poles of the system at the roots of s 2 + 2ζω n s + ω n 2 (b) Repeat the computation with MATLAB with z 0 = 2, ζ = 0.5 and ω n = 2rad/sec Sol) The closed-loop characteristic equation: det(si A BK 2 )=0 s 2 + 7 + K 1 z 0 K 2 s + 12 K 2 7z 0 + 12 K 1 z 0 =0 Equating the desired characteristic equation 7 + K 1 z 0 K 2 =2ζω n 2 12 K 2 7z 0 + 12 K 1 z 0 = ω n 4
Ex. The solutions are K 1 = z 0 14ζω n 37ω n 2 + 12(2ζω n 7) (z 0 + 3)(z 0 + 4) K 2 = z 0 7 2ζω n +12 ω n 2 (z 0 +3)(z 0 +4) (b) MATLAB Ao=[-7 1; -12 0]; zo=2; Bo=[1;-zo]; pc=roots([1 2 4]) K=place(Ao,Bo,pc) Open-loop Poles @ -3, -4 K=[-3.80 0.60] If zo=-2.99 (closed to open-loop pole) then K=[2052.5-688.1] 5
Ex. 2 Observation from the example 1 st. The gains grow as the zero z 0 approaches open loop poles (either - 3 or -4), the values where this system loses controllability. As controllability is almost lost, the control gains become very large. The system has to work harder and harder to achieve control as controllability slips away. 2 nd. Both K 1 and K 2 grow as the desired closed-loop bandwidth given by ω n is increased. From this we conclude that To move the poles a long way requires large gains 6
Basic approaches to design state feedback control Dominant pole design try to make CL system like a simple 2nd-order system Symmetric root locus LQR select CLPs with an explicit trade-off between control effort and performance 7
Dominant pole design-selection of CL poles Seek to make CL system behave like a simple secondorder system: Then choose dominant pair of CLPs based on wellknown relationships between ζ and ω n and performance measures such as: rise time overshoot, settling time, etc. ω n 2 G(s)= s 2 +2ζω n s+ω2 n Choose other CLPs to have well-damped responses 8
Ex. Tape Drive Dominant 2 nd order poles A simplified sketch of a computer tape drive system A = 0 2 0 0 0 0.1 0.35 0.1 0.1 0.75 0 0 0 2 0 0.4 0.4 0.4 1.4 0 0 0.03 0 0 1, B = 0 0 0 0 1 C = 0.5 0 0.5 0 0, D = 0 Design the tape servomotor by the dominant second-order poles method to have no more than 5% overshoot and a rise time of no more than 4 sec. Keep the peak tension as low as possible. 9
Ex. Tape Drive Dominant 2 nd order poles Solution) 2 dominant poles (complex conjugate poles): M p e 1 2 Damping ratio ζ = 0.7 overshoot 5% Natural frequency ω n = 1/1.5 rise time 4sec t r 1 tan d 1 1 2 1 ( ) d 10
Ex. Tape Drive Dominant 2 nd order poles The other three poles: Need to be placed far to the left of the dominant pair. far means the transients due to the fast poles should be over well before the transients due to the dominant poles. (Assume a factor of 4 in the respective undamped natural frequencies to be adequate.) pc = [ 0.707 + 0.707 j; 0.707 0.707 j; 4; 4; 4] / 1.5 Use function acker with A and B to find the control gains K = [8.5123 20.3457 1.4911 7.8821 6.1927]. 11
Ex. Tape Drive Dominant 2 nd order poles 12
Symmetric root locus For a SISO regulator, define a cost function J which penalizes (generalized) output excursions from the set-point, y control effort, u J = 0 ρz 2 t + u 2 t With x = Ax + Bu z = C 1 x dt Linear quadratic regulator Plant transfer function: Z(s) U(s) = C 1 si A 1 B = N(s) D(s) It can be shown that: J will be minimized by the control law u = Kx the eigenvalues of the closed-loop system are the left half-plane roots of the 2 nd degree polynomial equation α c s α c s = D s D s + ρn s N( s)=0 13
Optimal pole placement for SISO systems α c s α c s = D s D s + ρn s N( s)=0 Hence we can see the effects of the output-weighting factor ρ on the closed-loop poles by plotting a root locus for 1 + ρ N(s)N( s) D(s)D( s) = 0 If branches of the 180º SRL lie on the imaginary axis, plot the 0º locus instead Note that s and s affect in an identical manner. Any root s 0 of the above Eq., there will also be a root at s 0. The SRL provides a basis for specifying CLPs in a SISO pole placement design Increasing ρ places more emphasis on minimizing output excursions, at the expense of control effort and make us get a fast response. Decreasing ρ places more emphasis on minimizing input power with a low control effort and make us get a slow response It s a trade-off between performance and control effort!! 14
Ex. Servo speed control Plot the SRL for the servo control system with z = y. y = ay + u Sol) The Transfer function G 0 s = 1 s+a The SRL equation SRL 1 + ρ N(s)N( s) D(s)D( s) = 1 + ρ 1 ( s + a)(s + a) = 0 s = a 2 + ρ The closed loop poles s = a 2 + ρ minimizes J = 0 ρy 2 t + u 2 t dt 15
Ex. Satellite system Plot the SRL for the satellite system with z = y. x 1 x 2 = 0 1 0 0 x 1 x 2 + 0 1 u y = 1 0 x Sol) The transfer function G 0 s = 1 s 2 SRL equation 1 + ρ N(s)N( s) D(s)D( s) = 1 + ρ 1 s 2 s 2 = 1 + ρ 1 s 4 = 0 Note that the (stable) closed-loop poles has damping of ζ = 0.707 16
Ex. Satellite system Nyquist plot shows excellent stability PM=65, GM= General feature of LQR design but may not possible if not full statefeedback! 17
LQR design A general form for optimal design is linear quadratic regulator (LQR) J 0 T T ( x Qx u Ru) dt If we take Q = ρc T C and R = 1, the general LQR design becomes SRL Bryson s rule: An appropriate choice to obtain acceptable values x and u is to choose diagonal matrices Q and R such that Q ii = 1/maximum acceptable value of [x 2 i ] R ii = 1/maximum acceptable value of [u 2 i ] Feedback control law that minimizes the value of the cost u = K x where K = R 1 B T P P from solving the continuous time algebraic Riccati equation(are) A T P + PA PBR 1 B T P + Q = 0 18
LQR design The direct solution for the optimal control gain can be get from MATLAB function K=lqr(A,B,Q,R) More generalized form can be studied in optimal control class 19
Ex. Tape Drive system LQR design (a) Find the optimal control for the tape drive using the position x 3 as the output for the performance index. Let ρ = 1. Compare the results with that of dominant second order obtained before. (b) Compare the LQR designs for ρ = 0.1, 1, 10. Sol) (a) The performance index matrix R = 1 The state-cost matrix Q = C T C = where C = 0.5 0 0.5 0 0 0 2 0 0 0 0.1 0.35 0.1 0.1 0.75 0 0 0 2 0 0.4 0.4 0.4 1.4 0 0 0.03 0 0 1 The gain can be obtained by MATLAB K=lqr(A,B,Q,R) K = 0.6526 2.1667 0.3474 0.5976 1.0616 20
Ex. Tape Drive system LQR design 21