Presentation is loading. Please wait.

Presentation is loading. Please wait.

Lecture 15: State Feedback Control: Part I Pole Placement for SISO Systems Illustrative Examples.

Similar presentations


Presentation on theme: "Lecture 15: State Feedback Control: Part I Pole Placement for SISO Systems Illustrative Examples."— Presentation transcript:

1 Lecture 15: State Feedback Control: Part I Pole Placement for SISO Systems Illustrative Examples

2 Feedback Control Objective In most applications the objective of a control system is to regulate or track the system output by adjusting its input subject to physical limitations. There are two distinct ways by which this objective can be achieved: open- loop and closed-loop.

3 Open-Loop vs. Closed-Loop Control System Output Controller Desired Output Command Input Open-Loop Control Closed-Loop Control - System Controller Output Desired Output Sensor

4 Closed-Loop Control Advantages –automatically adjusts input –less sensitive to system variation and disturbances –can stabilize unstable systems Disadvantages –Complexity –Instability

5 State Feedback Control Block Diagram Z -1 H G x C y -K r u

6 State-Feedback Control Objectives Regulation: Force state x to equilibrium state (usually 0) with a desirable dynamic response. Tracking: Force the output of the system y to tracks a given desired output y d with a desirable dynamic response.

7 Closed-loop System Plant: Control: Closed-loop System:

8 Pole Placement Problem Choose the state feedback gain to place the poles of the closed-loop system, i.e., At specified locations

9 State Feedback Control of a System in CCF Consider a SISO system in CCF: State Feedback Control

10 Closed-Loop CCF System Closed loop A matrix:

11 Choosing the Gain-CCF Closed-loop Characteristic Equation Desired Characteristic Equation: Control Gains:

12 Transformation to CCF Transform system To CCF First, find how new state z 1 is related to x: Where x + (k)=x(k+1) (for simplicity)

13 Transformed State Equations Necessary Conditions for p: Vector p can be found if the system is controllable:

14 State Transformation Invertibility State transformation: Matrix T is invertible since By the Cayley-Hamilton theorem.

15 Toeplitz Matrix Matrix on the right is called Toeplitz matrix The Cayley-Hamilton theorem can further be used to show that

16 State Transformation Formulas Formula 1: Formula 2:

17 State Feedback Control Gain Selection By Cayley Hamilton: or

18 Bass-Gura Formula

19 Double Integrator-Matlab Solution T=0.5; lam=[0;0]; G=[1 T;0 1]; H=[T^2/2;T]; C=[1 0]; K=acker(G,H,lam); Gcl=G-H*K; clsys=ss(Gcl,H,C,0,T); step(clsys);

20 Flexible System Example Consider the linear mass-spring system shown below: m1m1 u m2m2 x2x2 x1x1 k Parameters: m 1 =m 2 =1Kg. K=50 N/m Analyze PD controller based on a)x 1, b)x 2 Design state feedback controller, place poles at

21 Collocated Control Transfer Function: PD Control: Root- Locus

22 Non-Collocated Control Transfer Function: PD Control: Root- Locus Unstable

23 State Model Discretized Model: x(k+1)=Gx(k)+Hu(k)

24 Open-Loop System Information Controllability matrix: Characteristic equation: |zI-G|=(z-1) 2 (z z+1)=z z z z+6

25 State Feedback Controller Characteristic Equations: |zI-G|=(z-1) 2 (z z+1)=z z z z+6

26 Matlab Solution %System Matrices m1=1; m2=1; k=50; T=0.01; syst=ss(A,B,C,D); A=[ ; ; ; ]; B=[0; 0; 1; 0]; C=[ ; ]; D=zeros(2,1); cplant=ss(A,B,C,D); %Discrete-Time Plant plant=c2d(cplant,T); [G,H,C,D]=ssdata(plant);

27 Matlab Solution %Desired Close-Loop Poles pc=[-20;-20;-5*sqrt(2)*(1+j);- 5*sqrt(2)*(1-j)]; pd=exp(T*pc); % State Feedback Controller K=acker(G,H,pd); %Closed-Loop System clsys=ss(G-H*K,H,C,0,T); grid step(clsys,1)

28 Time Respone

29 Steady-State Gain Closed-loop system: x(k+1)=G cl x(k)+Hr(k), Y=Cx(k) Y(z)=C(zI-G cl ) -1 H R(z) If r(k)=r.1(k) then y ss =C(I-G cl ) -1 H Thus if the desired output is constant r=y d /gain, gain= C(I-G cl ) -1 H

30 Time Response

31 Integral Control Control law: Z -1 H G x C y -K s ydyd u KiKi Integral controller plant Automatically generates reference input r! 1/g

32 Closed-Loop Integral Control System Plant: Control: Integral state: Closed-loop system

33 Double Integrator-Matlab Solution T=0.5; lam=[0;0;0]; G=[1 T;0 1]; H=[T^2/2;T]; C=[1 0]; Gbar=[G zeros(2,1);C 1]; Hbar=[H;0]; K=acker(Gbar,Hbar,lam); Gcl=Gbar-Hbar*K; yd=1; r=0; %unknown gain clsys=ss(Gcl,[H*r;-yd],[C 0;K],0,T); step(clsys);

34 Closed-Loop Step Response


Download ppt "Lecture 15: State Feedback Control: Part I Pole Placement for SISO Systems Illustrative Examples."

Similar presentations


Ads by Google