Advanced Control Theory - Chibum Lee · 2015-04-07 · Chibum Lee -Seoultech Advanced Control...

Advanced Control Theory

State Feedback ControlDesign

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Advanced Control TheoryChibum Lee -Seoultech

Outline

State feedback control design

Benefits of CCF

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Conceptual steps in controller design

We begin by considering the regulation problem

• the task of the controller is to maintain the plant output at a fixed set point

(𝒓 = 𝟎), in the face of disturbance inputs, 𝒘.

• For simplicity, we consider SISO systems

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𝒙 = 𝑨𝒙 + 𝑩𝒖 + 𝑩𝒘𝒚 = 𝑪𝒙



Now assume that all of the states are available for feedback

The control law 𝒖 = − 𝑲𝟏 𝑲𝟐 …𝑲𝒏 𝒙 involves 𝑛 gains

enough degrees of freedom to place the 𝑛 closed-loop poles

at any desired locations in the s-plane

full state feedback

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x̂

xKu ˆ

full state

estimation

If all states are not accessible, we can design an estimator (or

observer) an estimate of the entire state vector from measurements

of the control input 𝑢, the plant output 𝑦, and a model of the plant

We then use the estimated states in the original control law:

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Control law design

The plant:

The characteristic polynomial (=characteristic eq.) of the plant is

By use of full state feedback 𝒖 = −𝑲𝒙 we aim to place the closed-

loop poles (eigenvalues) at specified locations in the s-plane

That is, we can specify a desired closed-loop characteristic polynomial.

These 𝑠1 𝑠2, . . 𝑠𝑛 are

desired closed loop

poles (eigenvalue)

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𝒙 = 𝑨𝒙 + 𝑩𝒖 + 𝑩𝒘𝒚 = 𝑪𝒙

𝒂 𝒔 = 𝒅𝒆𝒕 𝒔𝑰 − 𝑨 = 𝒔𝒏 + 𝒂𝒏−𝟏𝒔𝒏−𝟏 +⋯+ 𝒂𝟎

= 𝒔 − 𝝀𝟏 𝒔 − 𝝀𝟐 …(𝒔 − 𝝀𝒏)

𝜶𝒄 𝒔 = 𝒔 − 𝒔𝟏 𝒔 − 𝒔𝟐 … 𝒔 − 𝒔𝒏i.e. 𝜶𝒄 𝒔 = 𝒔𝒏 + 𝜶𝒏−𝟏𝒔

𝒏−𝟏 +⋯+ 𝜶𝟎


Control law design

Desired characteristic polynomial:

For the closed-loop system

Hence, the closed-loop characteristic polynomial is

The coefficients 𝒂𝒌𝒏−𝟏, … , 𝒂𝒌𝟎 involve the 𝑛 feedback gains to be

determined

Our task is to choose the elements of 𝐾 such that 𝑎𝑘(𝑠) = 𝛼𝑐(𝑠)

The simplest approach is to match the coefficients of s in 𝑎𝑘(𝑠) and

𝛼𝑐(𝑠)

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𝒙 = 𝑨𝒙 + 𝑩𝒖 + 𝑩𝒘= 𝑨𝒙 − 𝑩𝑲𝒙 + 𝑩𝒘= (𝑨 − 𝑩𝑲)𝒙 + 𝑩𝒘

𝜶𝒄 𝒔 = 𝒔𝒏 + 𝜶𝒏−𝟏𝒔𝒏−𝟏 +⋯+ 𝜶𝟎

𝒂 𝒔 = 𝒅𝒆𝒕 𝒔𝑰 − 𝑨 + 𝑩𝑲 = 𝒔𝒏 + 𝒂𝒌𝒏−𝟏𝒔𝒏−𝟏 +⋯+ 𝒂𝒌𝟎


Ex1: Position regulation of DC servo motor

• Task of this example is to hold motor position at = 0 despite

fluctuating disturbance torque

• Here, just accept the equation for DC servo motor and don’t care

about the procedures inducing it.

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Suppose That is,

Choose state variables:

Then

Specify closed-loop dynamics:

i.e.,

Desired characteristic polynomial:

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Full state feedback:

Compare with desired characteristic polynomial:

Then, match the coefficients

Hence:


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Ex1: Implementation

Closed-loop dynamics:

Closed-loop transfer function:

dc gain = 0.25; i.e., a unit step disturbance will cause a steady-

state position error of 0.25

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Ex1: Impulse disturbance

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Ex1: Step disturbance

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Response calculations with MATLAB

A = [0 1;0 -2]; B = [0 0;4 4]; C = [1 0]; D = [0 0];

G = ss(A, B, C, D);

set(G, 'InputName', {'w', 'u'}, 'OutputName', 'y')

Ga = augstate(G); % augment output with states

K = [4 0.5];

feedin = [2]; % feedback to input 'u'

feedout = [2:3]; % feedback from state outputs of Ga

Gca = feedback(Ga, -K, feedin, feedout, +1); % +ve f.b.

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Response calculations with MATLAB

Gc = Gca(1,:); % select only output 'y'

set(Gc, 'InputName', {'w', 'r'})

tf(Gc)

Transfer function from input "w" to output "y":

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--------------

s^2 + 4 s + 16

Transfer function from input "r" to output "y":

4

--------------

s^2 + 4 s + 16

% apply impulse to 'w'

% input only

impulse(Gc(:,1), G(:,1))0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-0.5

0

0.5

1

1.5

2

From: w To: y

Impulse Response

Time (sec)A

mplit

ude

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Ex2: Control law for a pendulum

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Suppose you have a pendulum with frequency 𝜔0 and a state-space

description given by

𝑥1 𝑥2=

0 1−𝜔0

2 0

𝑥1𝑥2

+01𝑢

Find the control law that places the closed-loop poles of the system

so that they are both at −2𝜔0. In the other words, you wish to

double the natural frequency and increase the damping ratio 𝜁 from

0 to 1

Sol) Desired characteristic polynomial :

𝜶𝒄 𝒔 = 𝒔 + 𝟐𝝎𝟎𝟐 = 𝒔𝟐 + 𝟒𝝎𝟎𝒔 + 𝟒𝝎𝟎

𝟐

Full state feedback

𝒅𝒆𝒕 𝒔𝑰 − 𝑨 − 𝑩𝑲 = 𝒅𝒆𝒕𝑠 00 𝑠

−0 1

−𝜔02 0

−01

𝑲𝟏 𝑲𝟐 =0

or 𝒔𝟐 +𝑲𝟐 𝒔 + 𝝎𝟎𝟐 +𝑲𝟏 = 𝟎


Ex2: Control law for a pendulum

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Equating the coefficient:

𝑲𝟏 = 𝟑𝝎𝟎𝟐 , 𝑲𝟐 = 𝟒𝝎𝟎

The control law :

𝑲 = [𝑲𝟏 𝑲𝟐]= [𝟑𝝎𝟎𝟐 𝟒𝝎𝟎]

The following figure show the response of the closed-loop system to the IC 𝒙𝟏 = 𝟎. 𝟏, 𝒙𝟐 = 𝟎. 𝟎 and 𝝎𝟎 = 𝟏.(Well damped with roots at 𝒔 = −𝟐.)


Benefit of controller canonical form

Calculation of control-law gains by coefficient matching is tedious for

𝑛 > 3

The process is simplified if the state equations are in controller

canonical form

Consider general 3rd-order plant from before:

𝑦 + 𝑎2 𝑦 + 𝑎1 𝑦 + 𝑎0𝑦 = 𝑏2 𝑢 + 𝑏1 𝑢 + 𝑏0𝑢

The controller canonical realization is:

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𝑥1 𝑥2 𝑥3

=−𝑎2 −𝑎1 −𝑎01 0 00 1 0

𝑥1𝑥2𝑥3

+100𝑢

y = 𝑏2 𝑏1 𝑏0

𝑥1𝑥2𝑥3


Benefit of controller canonical form

The closed-loop system matrix is

Note that this is still in upper companion form

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𝐴𝑐 − 𝐵𝑐𝐾 = 𝐴𝑐 −100

𝐾1 𝐾2 𝐾3

=−𝑎2 −𝑎1 −𝑎01 0 00 1 0

−𝐾1 𝐾2 𝐾31 0 00 1 0

=−𝑎2 − 𝐾1 −𝑎1 − 𝐾2 −𝑎0 − 𝐾3

1 0 00 1 0


Control gains by inspection

Hence, we can write the closed-loop characteristic polynomial by

inspection:

If the desired characteristic polynomial is

𝜶𝒄 𝒔 = 𝒔𝟑 + 𝜶𝟐𝒔𝟐 + 𝜶𝟏𝒔 + 𝜶𝟎

then equating coefficients gives:

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𝒂𝒌 𝒔 = 𝒔𝟑 + (𝒂𝟐+𝑲𝟏)𝒔𝟐 + (𝒂𝟏+𝑲𝟐)𝒔

𝟏 + (𝒂𝟎+𝑲𝟑)

𝑲𝟏 = −𝒂𝟐 + 𝜶𝟐

𝑲𝟐 = −𝒂𝟏 + 𝜶𝟏

𝑲𝟑 = −𝒂𝟎 + 𝜶𝟎

Advanced Control Theory - Chibum Lee · 2015-04-07 · Chibum Lee -Seoultech Advanced Control...

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Transcript of Advanced Control Theory - Chibum Lee · 2015-04-07 · Chibum Lee -Seoultech Advanced Control...