Prof.christenson July17 APSS2010

8/20/2019 Prof.christenson July17 APSS2010

1/11

Advanced Hazards Mitigation LabAdvanced Hazards Mitigation Lab

Department of Civil & Environmental EngineeringDepartment of Civil & Environmental Engineering

Theory of Control: ITheory of Control: I

Richard ChristensonRichard Christenson

University of ConnecticutUniversity of Connecticut

Asia Asia‐‐Pacific Summer SchoolPacific Summer School

on Smart Structures Technology on Smart Structures Technology



OverviewOverview

Introduction to structural control

Control theory

Basic feedback control

Optimal control – state feedback control

Observers and LQG controllers



Passive Control SystemsPassive Control Systems

StructureExcitation Response

Passive Device

M

Passive Damper

M

Base Isolation

M

m

Tuned Mass Damper

3

Types of Structural ControlTypes of Structural Control



Active Control Systems Active Control Systems


4


ActuatorsM

Active Bracing

M

m

Active Mass Damper

Sensor Sensor



Active Control Systems Active Control Systems


Actuators

5


Controller SensorsSensors

feedforward feedback



Hybrid Control SystemsHybrid Control Systems


Actuators

6




Passive Device

M

Active Base Isolation

Sensor

Actuator


2/11



Semiactive Control SystemsSemiactive Control Systems


Actuators

7



Passive Device




Functionally Upgraded Passive SystemsFunctionally Upgraded Passive Systems


Actuators

8



Passive Devicefeedforward feedback




Actuators

9




Our focus today Our focus today ……

Controller



Introduction to Structural ControlIntroduction to Structural Control

Introduction to theory theory behind automatic

control systems – closed‐loop control

Control is used primarily for:

1. Reduce sensitivity to variations

2. Reduce sensitivity to output disturbance

3. Ability to control system bandwidth

4. Stabilization of an unstable system5. Control system transient response

Segway




Controlling the temperature of fluid in a tank

Open‐loop control

In open‐loop control the command signal alone is

selected to achieve the desired response

Controller

G(s)

Plant

H(s)

r(t)

reference

input

u(t)

control

input

y(t)

output

Example: Filling a bathtub with water*Example: Filling a bathtub with water*

*taken from Linear Control Systems, (*taken from Linear Control Systems, (RohrsRohrs, et al.), et al.)Advanced Hazards Mitigation LabAdvanced Hazards Mitigation Lab



Open‐loop control

Open hot water tap specified amount

Open cold water tap specified amount

If you have done this many time before, you might

know rather well the necessary settings

However, a number of factors might affect the

control of the output

Example: Filling a bathtub with waterExample: Filling a bathtub with water


3/11




Closed‐loop control

In closed‐loop control, feedback measurements

are included to achieve the desired response

Controller

G(s)

Plant

H(s)

r(t)

reference

input

u(t)

control

input

y(t)

output






In closed‐loop control feedback measurements are

included to achieve the desired response

Feel the water at several intervals while the tub is

filling

If water is not at right temperature, adjust hot or

cold water faucets

In this manner, the system output affects the control

of the system






Use of the state of the output is termed feedback feedback

More measurements (temp. of each faucet, rate

of change of temp.) can achieve better results

Closed‐loop may be more complex than open‐loop,

but can provide better performance

Compromise between stability stability and performance performance





Compromise between stability stability and performance performance

Controlling only hot water (cold predetermined

level) by turning fully on or fully off ;

Our slow response time with the dramatic

response may cause oscillations in temperature

Common causes of instability in automatic

control systems: (1) delay; and (2) high gain




Take the human out of the closed‐loop control

Automatic closed‐loop control

Sensor to measure the required variables

Actuator to adjust control valves

Controller Controller to interpret sensors and send control

signal (which would then be amplified) to actuator




Modeling the system is a crucial step in thedesign of a controller

The quality of the controller is linked to the

quality of the model used in the control design

Since no system can be perfectly modeled,

care must be taken in designing the controller

Parameter inaccuracies

Unmodeled dynamics

Nonlinearities


4/11



OverviewOverview


Control theory






Closed Loop ControlClosed Loop Control

General form of the closed‐loop control system

Controller

G(s)

Plant

H(s)

r(t)

reference

input

u(t)

control

input

y(t)

output

Feedback control can take many forms…




Let’s begin with an example examining the

effect of control gains in the forward path:

Controller

G(s)

Plant

H(s)

r(t)

reference

input

u(t)

control

input

y(t)

output

When G(s) = K , this is called a proportional

controller with unity gain feedback

e(t)

error +-




Let’s begin with an example examining the

effect of control gains in the forward path:

K H(s)r(t)

reference

input

u(t)

control

input

y(t)

output

When G(s) = K , this is called a proportional

controller with unity gain feedback

e(t)

error +-

Close loop system

H cl (s)




The goal is to choose the control gain (K) to

stabilize the system and improve response time

Using the block diagram, we can write

K H(s)

r(t) u(t) y(t)e(t)+-

)()()( sU sH sY = )()()( sKE sH sY =

[ ])()()()( sY sR K sH sY −=[ ]

)()(1

)()( sR

sKH

sKH sY

+=

[ ])(1)(

)(

)()(

sKH

sKH

sR

sY sH cl +

==Advanced Hazards Mitigation LabAdvanced Hazards Mitigation Lab



Consider a simple example of a dynamic system In the Laplace domain, the transfer function of the

plant is

Simple model of an electric motor or hydraulic actuator with theSimple model of an electric motor or hydraulic actuator with the

command/voltage as the input and the position/command/voltage as the input and the position/dispdisp. as the output. as the output

Note that this system is marginally stable because one pole is at the origin

)(

1)(

τ +=

sssH


5/11




The close loop system is:

[ ] K ssK

sKH

sKH sH cl ++

=+

=τ 2)(1

)()(

tao = 1;

K = 1;

num = K;

den = [1 tao K];

sys = tf(num,den);

[y,t]=step(sys,t);

plot(t,y)

K=1

K=0.1

K=10

uncontrolled system H(s)

improved response time

overshoot




Let’s look at the closed loop poles

has poles at

Note:

The system is stable when

The system is underdamped when

2

42

2,1

K p

−±−=

τ τ

K ss

K sH cl

++

=

τ

2)(

τ τ K




We can use the pole placement approach to

assign K values to achieve the specific behavior

K=1

K=0.1

K=10

K=1

K=0.1

K=10




This system can equivalently be considered in

state space

)(

1

)(

)()(

τ +==

sssU

sY sH

( ) ( )sU sssY =+ )( 2 τ

)()()( t ut y t y =+ &&& τ

)()()( t ut y t y +−= &&& τ




This system can equivalently be considered instate space

)()()(

)()()(

t Dut Cz t y

t But Az t z

+=

+=

[ ] [ ] )(0)(

)(01)(

)(1

0

)(

)(

0

10

)(

)(

t ut y

t y t y

t ut y

t y

t y

t y

+

=

+

−

=

&

&&&

&

τ

)()()( t ut y t y +−= &&& τ

−=

τ 0

10 A




This system can equivalently be considered instate space

Poles of the transfer function are equal to the

eigenvalues of the state space A matrix

−

=τ 0

10 A

( )

+

−=

−−

=−

τ λ

λ

τ λ

λ λ

0

1

0

10

0

0det AI

λτ λ τ λ λ +=−−+= 2)0)(1()(


6/11




Let’s look at the poles and the step response of

a second order differential equation

The Laplace Transform (zero IC) is

Consider the poles of the system

)()()(2)( 22

t r t y t y t y nnn ω ω ζω =++ &&&

22

2

2)(

)(

nn

n

sssR

sY

ω ζω

ω

++=

( ) 22 1 nns ω ζ ζω −±−=Advanced Hazards Mitigation LabAdvanced Hazards Mitigation Lab



Assuming the system is underdamped21 ζ ω ζω −±−= nn j s

real imaginary

21 ζ ω −n

nζω

( ) ( )222 1 ζ ω ζω −+ nnMagnitude:

( ) ( )222 1 ζ ω ζω −+ nn

nω =

22222

nnn ω ζ ω ω ζ −+

nω




Assuming the system is underdamped21 ζ ω ζω −±−= nn j s

real imaginary

21 ζ ω −n

nζω

( ) ζ ω

ζω θ ==

n

nsin

Angle:

nω ζ=sinθ




Consider the response of the system

To a step input

The step response can be determined as

)(2

)(22

2

sR ss

sY nn

n

ω ζω

ω

++=

ssR

1)( =

( ) ( )Ψ+−−−= −

t et y nt n 2

2 1sin1

1

1 ζ ω ζ

ζω

−=Ψ

ζ

ζ 21arctan




The peak response occurs at

The peak value of y is then

And the overshoot is

Note: overshoot is only a function of dampingNote: overshoot is only a function of damping

21 ζ ω

π

−= nt

−−

+= 21

max 1)( ζ

ζπ

et y

−−

21 ζ

ζπ

e




The settling time (defined as time required forresponse to remain within 5% of final value) is

Note: increasingNote: increasing wnwn decreases the rise timedecreases the rise time

05.0=− t ne ζω

3=t nζω

n

st ζω

3=


7/11




Let’s look at the poles and the step response

m a g

= ω

θ = ζ

Optimal poles move away from origin at desired damping

overshoot =

settling time = nst

ζω

3=

−−

21 ζ

ζπ

e



Theory of Control: IITheory of Control: II

Richard ChristensonRichard Christenson

University of ConnecticutUniversity of Connecticut

Asia Asia‐‐Pacific Summer SchoolPacific Summer School

on Smart Structures Technology on Smart Structures Technology



OverviewOverview


Control theory






Optimal ControlOptimal Control

Modern control theory uses the approach that

an optimal controller can be obtained for a plant

taking the form

Controller

G(s)

Plant

w(t)

excitationu(t)

control

y(t)

output)()()()(

)()()()(

t Fv t Dut Cz t y

t Ew t But Az t z

++=

++=&

v(t)



State Feedback ControlState Feedback Control

Assume that all states are measured and a fullstate feedback control law takes the form

The closed loop dynamics are given by

The poles of this system may be placedarbitrarily if the system is controllable

However, optimal placement is possible with aproperly chosen cost function

)()( t Kz t u −=

( ) )()()( t z At z BK At z cl =−=&




Consider the Linear Quadratic Regulator (LQR) We seek a state feedback controller (K) that

minimizes the cost function

Where Q is positive semidefinate, R is positive

definite, and subject to

dt RuuQz z J T t

T f

)(0

∫ +=

0)0( z z Bu Az z =+=&


8/11




The solution to the LQR problem is given by

Where the control gain matrix K is given by

Where P is the Riccati matrix which is goverened

by the Riccati equation

)()( t Kz t u −=

0)(,)()()()( 1 =−++=− − f T T t P P BBR t P R At P t P At P &

P BR K T 1−=




As t f goes to infinity, we see that P becomes

constant and can be determined by solving the

algebraic Riccati equation (ARE)

We can use MATLAB to readily obtain this

solution

The matrices Q and R provide the mechanisms

to design an effective controller

P BBR t P R At P t P A T T 1)()()(0 −−++=




Let’s consider an example of a sdof building with

active bracing

M

Active Bracing

Sensor

wn = 1*2*pi; % rad/sec

xsi = 5/100; % damping 5%

M = 100;

K = M*wn^2;

C = 2*xsi*wn*M;

% State Space System

% dx = Ac*x + Bc*u + Ec*w

% y = Cc*x + Dc*u + Fc*w

Ac = [0 1;-inv(M)*K -inv(M)*C];

Bc = [0;1];

Ec = [0;inv(M)*1];

Cc = [eye(2);-inv(M)*K -inv(M)*C];

Dc = [1];

Fc = [0];




Plot the uncontrolled system’s response due to

Kobe earthquake [w(t)]

sys = ss(Ac,Ec,Cc,Fc);

t = linspace(0,30,1000);

load kobe

w = interp1(k(1,:),k(2,:),t);

y = lsim(sys,w,t);

figure(3);

subplot(311);plot(t,y(:,1),'g');






Plot the uncontrolled system’s response due toKobe earthquake [w(t)]

sys = ss(Ac,Ec,Cc,Fc);

t = linspace(0,30,1000);

load kobe

w = interp1(k(1,:),k(2,:),t);

y = lsim(sys,w,t);

figure(3);







Design an LQR controller to weight displacementand velocity equally

Q = diag([1 1]);

R = 1e-2;

Klqr = lqr(Ac,Bc,Q,R,[]);

sys = ss(Ac-Bc*Klqr,Ec,Cc-Dc*Klqr,Fc);

y = lsim(sys,w,t);

figure(3);

subplot(311);hold on;plot(t,y(:,1),'b');




9/11




Design an LQR controller to weight displacement

and velocity equally

Q = diag([1 1]);R = 1e-2;

Klqr = lqr(Ac,Bc,Q,R,[]);

sys = ss(Ac-Bc*Klqr,Ec,Cc-Dc*Klqr,Fc);

y = lsim(sys,w,t);

figure(3);




dt RuuQz z J T t

T f

)(0

∫ +=




Let’s look at the poles

R decreases

R decreases

Q – displacement weighting Q – velocity weighting

dt RuuQz z J T t

T f

)(0

∫ +=



State ObserversState Observers

In practice, it is not feasible or practical to

measure all of the states of the system

Thus, feedback control design often requires

that one estimate the state variables

Controller

G(s)

Plant

w(t)

excitationu(t)

control

y(t)

output)()()()(

)()()()(

t Fv t Dut Cz t y

t Ew t But Az t z

++=

++=&

v(t)




An observer is a dynamic system with inputs u

(control input) and y (measured responses), and

output that estimates the state vector (called

xhat)

Observer (linear, continuous time) category:

Open loop observer

Full order observer

Kalman (Bucy) filter



Open Loop ObserversOpen Loop Observers

The objective is: Observer (linear, continuous time) category:

Linear invariant system:

Auxiliary dynamical system:

Estimation error:

If A is stable, then e(t) approaches zero

Drawbacks:

Unbounded error for unstable state matrix

Fails in the presence of modeling errors and disturbances

0)(ˆ)(lim =−∞→ t x t x t

00 )()()()()()( x t x t Cx t y t But Ax t x ==+=&

)()(ˆ)(ˆ t But x At x +=&

)(ˆ)()( t x t x t e −≡

( ) 0)()()(ˆ)()(ˆ)( t t t Aet But x At But x At e ≥=+−+=&




Full order observer – “Luenberger observer”



Estimation error:

If (A‐LC) is stable, then e(t) approaches zero

Drawbacks:

Still fails in the presence of modeling errors and disturbances

00 )()()()()()( x t x t Cx t y t But Ax t x ==+=&

( ))(ˆ)()()(ˆ)(ˆ t x C t y Lt But x At x −++=&

)(ˆ)()( t x t x t e −≡

( ) 0)()( t t t eLC At e ≥−=&Observer feedback


10/11




Stochastic state observer, “Kalman‐Bucy filter”



)()()()()()()( t v t Cx t y t w t But Ax t x +=++=&

( ))(ˆ)()()(ˆ)(ˆ t x C t y K t But x At x −++=&

Process noise with

covariance Q(t)

Disturbance with

covariance R(t)

Both noise terms are assumed white, Gaussian and mutually independant

)()()()()()()( t K t R t K t Q At P t AP t P T T −++=&

)()()()( 1 t R t C t P t K T −=

Riccati equation:

Kalman Gain:




The Kalman filter provides the best estimate of

the states based on current available noisy

information

The solution P(t) to the associated differentialRiccati equation (DRE) is also the covariance of

estimation error

If only the steady‐state behavior is of interest,

the time derivative is eliminated in the DRE and

and results in an algebraic Riccati equation (ARE)



LQG ControlLQG Control

Kalman filter is often known as linear quadratic

estimation (LQE)

When we combine optimal state feedback with

estimator design, we realize a linear quadratic

Gaussian (LQG) controller

( ))(ˆ)(

ˆ)()(ˆ)(ˆ

t z K t u

z C y Lt But z At z

−=

−++=&




The closed loop system is thus

( ) ( )

)(ˆ)(ˆ)(

)()(ˆ)(ˆ

)()(ˆˆ)()(ˆ)(ˆ

t z C t z K t u

t y Bt z At z

t Ly t z BK LC Az C y Lt But z At z

e

ee

=−=+=

+−−=−++=

&

&

Controller

Plant

w(t)

excitationu(t)

control

y(t)

output)()()()(

)()()()(

t Fv t Dut Cz t y

t Ew t But Az t z

++=

++=&

v(t)

( ))(ˆ)(

ˆ)()(ˆ)(ˆ

t z K t u

z C y Lt But z At z

−=

−++=&




The closed loop system is thus

[ ]

=

+

+=

)(ˆ

)()(

)(0)(ˆ

)(

)(ˆ

)(

t z

t z DC C t y

t w E

t z

t z

DC B AC B

BC A

t z

t z

e

eeee

e

&

&

)(ˆ)()(

)()(ˆ)()(

t z DC t Cz t y

t Ew t z BC t Az t z

e

e

+=

++=&

)(ˆ)(ˆ)(

)()(ˆ)(ˆ

t z C t z K t u

t y Bt z At z

e

ee

=−=

+=&




Design an LQG controllerEww = 0.1;

Evv = 4e-5;

Lgain = lqe(Ac,Ec,Cc(3,:),Eww,Evv);

Ak = Ac-Bc*Klqr-Lgain*Cc(3,:);

Bk = Lgain;

Ck = -Klqr;

Dk = 0;

Acl = [Ac Bc*Ck;Bk*Cc(3,:)

Ak+Bk*Dc(3,:)*Ck];

Bcl = [Ec;zeros(2,1)];

Ccl = [Cc Dc*Ck];

Dcl = zeros(3,1);

sys = ss(Acl,Bcl,Ccl,Dcl);

y = lsim(sys,w,t);


11/11




Design an LQG controller

States (actual – blue; estimated – green) Response (lqr – blue; lqg– red)

Prof.christenson July17 APSS2010

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