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

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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
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.

open loop vs closed loop control
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

closed loop control
Closed-Loop Control
  • Advantages
    • automatically adjusts input
    • less sensitive to system variation and disturbances
    • can stabilize unstable systems
  • Disadvantages
    • Complexity
    • Instability
state feedback control objectives
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.
closed loop system
Closed-loop System

Plant:

Control:

Closed-loop System:

pole placement problem
Pole Placement Problem

Choose the state feedback gain to place the poles of the closed-loop system, i.e.,

At specified locations

state feedback control of a system in ccf
State Feedback Control of a System in CCF

Consider a SISO system in CCF:

State Feedback Control

closed loop ccf system
Closed-Loop CCF System

Closed loop A matrix:

choosing the gain ccf
Choosing the Gain-CCF

Closed-loop Characteristic Equation

Desired Characteristic Equation:

Control Gains:

transformation to ccf
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:

transformed state equations
Transformed State Equations

Necessary Conditions for p:

0

0

0

1

Vector p can be found if the system is controllable:

state transformation invertibility
State Transformation Invertibility

State transformation:

Matrix T is invertible since

By the Cayley-Hamilton theorem.

toeplitz matrix
Toeplitz Matrix

The Cayley-Hamilton theorem can further be used to show that

Matrix on the right is called Toeplitz matrix

state transformation formulas
State Transformation Formulas

Formula 1:

Formula 2:

double integrator matlab solution
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);

flexible system example
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
collocated control
Collocated Control

Transfer Function:

PD Control:

Root-Locus

non collocated control
Non-Collocated Control

Transfer Function:

PD Control:

Root-Locus

Unstable

state model
State Model

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

open loop system information
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

state feedback controller
State Feedback Controller

Characteristic Equations:

|zI-G|=(z-1)2(z2-1.99z+1)=z4-3.99z3+5.98z2-3.99z+6

matlab solution
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);

matlab solution1
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)

steady state gain
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

integral control
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!

closed loop integral control system
Closed-Loop Integral Control System

Plant:

Control:

Integral state:

Closed-loop system

double integrator matlab solution1
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);