Presentation is loading. Please wait.

Presentation is loading. Please wait.

Lecture 15: State Feedback Control: Part I

Similar presentations

Presentation on theme: "Lecture 15: State Feedback Control: Part I"— 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
Open-Loop Control Output Desired Output Command Input Controller System Closed-Loop Control Output Desired Output Controller System - Sensor

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

5 State Feedback Control Block Diagram
u x y H Z-1 C 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: 1 Vector p can be found if the system is controllable:

14 State Transformation Invertibility
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

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 m1 m2 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=[ ; ; ; ]; 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)=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 yd u x y  Ki H C G plant
Z-1 C 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

Download ppt "Lecture 15: State Feedback Control: Part I"

Similar presentations

Ads by Google