Advanced Control Theory

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1 State Feedback Control Design

2 Outline State feedback control design Benefits of CCF 2

3 Conceptual steps in controller design We begin by considering the regulation problem the task of the controller is to maintain the plant output at a fixed set point (r = 0), in the face of disturbance inputs, w. For simplicity, we consider SISO systems x = Ax + Bu + Bw y = Cx 3

4 Conceptual steps in controller design Now assume that all of the states are available for feedback full state feedback The control law u = K 1 K 2 K n x involves n gains enough degrees of freedom to place the n closed-loop poles at any desired locations in the s-plane 4

5 Conceptual steps in controller design If all states are not accessible, we can design an estimator (or observer) an estimate of the entire state vector from measurements of the control input u, the plant output y, and a model of the plant xˆ full state estimation We then use the estimated states in the original control law: u Kxˆ 5

6 Control law design The plant: x = Ax + Bu + Bw y = Cx The characteristic polynomial (=characteristic eq.) of the plant is a s = det si A = s n + a n 1 s n a 0 = s λ 1 s λ 2 (s λ n ) By use of full state feedback u = Kx we aim to place the closedloop poles (eigenvalues) at specified locations in the s-plane That is, we can specify a desired closed-loop characteristic polynomial. α c s = s s 1 s s 2 s s n i.e. α c s = s n + α n 1 s n α 0 These s 1 s 2,.. s n are desired closed loop poles (eigenvalue) 6

7 Control law design Desired characteristic polynomial: α c s = s n + α n 1 s n α 0 For the closed-loop system x = Ax + Bu + Bw = Ax BKx + Bw = (A BK)x + Bw Hence, the closed-loop characteristic polynomial is a s = det si A + BK = s n + a kn 1 s n a k0 The coefficients a kn 1,, a k0 involve the n feedback gains to be determined Our task is to choose the elements of K such that a k (s) = α c (s) The simplest approach is to match the coefficients of s in a k (s) and α c (s) 7

8 Ex1: Position regulation of DC servo motor Task of this example is to hold motor position at = 0 despite fluctuating disturbance torque Here, just accept the equation for DC servo motor and don t care about the procedures inducing it. 8

9 Ex1: Position regulation of DC servo motor Suppose That is, Choose state variables: Then Specify closed-loop dynamics: i.e., Desired characteristic polynomial: 9

10 Ex1: Position regulation of DC servo motor Full state feedback: Compare with desired characteristic polynomial: Then, match the coefficients Hence: 10

11 Ex1: Implementation Closed-loop dynamics: Closed-loop transfer function: dc gain = 0.25; i.e., a unit step disturbance will cause a steadystate position error of

12 Ex1: Impulse disturbance 12

13 Ex1: Step disturbance 13

14 Response calculations with MATLAB A = [0 1;0-2]; B = [0 0;4 4]; C = [1 0]; D = [0 0]; G = ss(a, B, C, D); set(g, 'InputName', {'w', 'u'}, 'OutputName', 'y') Ga = augstate(g); % augment output with states K = [4 0.5]; feedin = [2]; % feedback to input 'u' feedout = [2:3]; % feedback from state outputs of Ga Gca = feedback(ga, -K, feedin, feedout, +1); % +ve f.b. 14

15 Response calculations with MATLAB Gc = Gca(1,:); % select only output 'y' set(gc, 'InputName', {'w', 'r'}) tf(gc) Transfer function from input "w" to output "y": s^2 + 4 s + 16 Transfer function from input "r" to output "y": s^2 + 4 s + 16 % apply impulse to 'w' % input only impulse(gc(:,1), G(:,1)) Amplitude Impulse Response From: w To: y Time (sec) 15

16 Ex2: Control law for a pendulum Suppose you have a pendulum with frequency ω 0 and a state-space description given by x 1 = x 2 ω 0 0 x 1 x u Find the control law that places the closed-loop poles of the system so that they are both at 2ω 0. In the other words, you wish to double the natural frequency and increase the damping ratio ζ from 0 to 1 Sol) Desired characteristic polynomial : α c s = s + 2ω 2 0 = s ω 0 s + 4ω 0 Full state feedback det si A BK = det s 0 0 s ω K 1 K 2 =0 or s 2 + K 2 s + ω K 1 = 0 16

17 Ex2: Control law for a pendulum Equating the coefficient: K 1 = 3ω 0 2, K 2 = 4ω 0 The control law : K = [K 1 K 2 ]= [3ω 0 2 4ω 0 ] The following figure show the response of the closed-loop system to the IC x 1 = 0. 1, x 2 = 0. 0 and ω 0 = 1. (Well damped with roots at s = 2.) 17

18 Benefit of controller canonical form Calculation of control-law gains by coefficient matching is tedious for n > 3 The process is simplified if the state equations are in controller canonical form Consider general 3rd-order plant from before: y + a 2 y + a 1 y + a 0 y = b 2 u + b 1 u + b 0 u The controller canonical realization is: x 1 x 2 x 3 = a 2 a 1 a x 1 x 2 x u y = b 2 b 1 b 0 x 2 x 3 x 1 18

19 Benefit of controller canonical form The closed-loop system matrix is A c B c K = A c K 1 K 2 K 3 = a 2 a 1 a K 1 K 2 K = a 2 K 1 a 1 K 2 a 0 K Note that this is still in upper companion form 19

20 Control gains by inspection Hence, we can write the closed-loop characteristic polynomial by inspection: a k s = s 3 + (a 2 +K 1 )s 2 + (a 1 +K 2 )s 1 + (a 0 +K 3 ) If the desired characteristic polynomial is α c s = s 3 + α 2 s 2 + α 1 s + α 0 then equating coefficients gives: K 1 = a 2 + α 2 K 2 = a 1 + α 1 K 3 = a 0 + α 0 20

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