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Lecture 15: State Feedback Control: Part I

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

JasminFlorian
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Lecture 15: State Feedback Control: Part I

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  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 Open-Loop Control Output Desired Output Command Input System Controller Closed-Loop Control Output Desired Output Controller System - 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 y x u r C H Z-1 G -K

  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 yd 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 To CCF Transform system Where x+(k)=x(k+1) (for simplicity) First, find how new state z1 is related to x:

  13. Transformed State Equations Necessary Conditions for p: 0 0 0 1 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 The Cayley-Hamilton theorem can further be used to show that Matrix on the right is called Toeplitz matrix

  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: x1 x2 Parameters: m1=m2=1Kg. K=50 N/m k u m2 m1 • Analyze PD controller based on a)x1, b)x2 • 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(z2-1.99z+1)=z4-3.99z3+5.98z2-3.99z+6

  25. State Feedback Controller Characteristic Equations: |zI-G|=(z-1)2(z2-1.99z+1)=z4-3.99z3+5.98z2-3.99z+6

  26. Matlab Solution %System Matrices m1=1; m2=1; k=50; T=0.01; syst=ss(A,B,C,D); A=[0 0 1 0;0 0 0 1;-50 50 0 0;50 -50 0 0]; B=[0; 0; 1; 0]; C=[1 0 0 0;0 1 0 0]; 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)=Gclx(k)+Hr(k), Y=Cx(k) Y(z)=C(zI-Gcl)-1H R(z) If r(k)=r.1(k) then yss=C(I-Gcl)-1H Thus if the desired output is constant r=yd/gain, gain= C(I-Gcl)-1H

  30. Time Response

  31. Integral Control Control law: 1/g y x u yd Ki Z-1 C  H G plant Integral controller -Ks Automatically generates reference input r!

  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

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