5.updated draft review of modern control

7/28/2019 5.Updated Draft Review of Modern Control

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1

Control of Smart Structures 5. Review of Modern Control 1

Department of Mechanical Engineering

Dr. G. Song, Associate Professor

5. Review of Modern Control Theory

Draft Updated 09/28/03; 10:38pm


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2




System Representation

A linear system can be represented in Transfer function form as

Above system can also be represented in state space form as

A: System matrix B: Input Influence Matrix

C: Output/Measurement matrix D: Direct transmission matrix

Task is to determine, the state space matrices A,B,C, D

State space is representation and implementation helps us in MIMIcontrol problems.

2

( ) 12.5

( ) 0.24 101

Y s

U s s s= + +

x Ax Buy Cx Du

= += +


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3





[ ]

2

1 2 1

1 2

2 2 1

1 1

2 2

1

2

( ) 12.5

( ) 0.24 101

0.24 101 12.5Let, x ,

0.24 101 12.5

0 1 0

101 0.24 12.5

: 1 0

A B

C

Y s

U s s s

y y y uy x y x Then

x x

x x x u

x xu

x x

xOutput y

x

=+ +

+ + == = =

=

= +

= +

=

In Matlab: tf2ss


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4





Transformation of linear system in state space to transfer function form

-1

-1

Take Laplace Transform

( ) - (0) ( ) ( )

( ) ( ) ( )We assume (0) 0 for deriving transfer function matrix

( ) ( - ) ( )

Substitute ( ) in ( )

( ) [ ( - )

( )

(

] ( )

x Ax Bu

y Cx Du

sX s x AX s BU s

Y s CX s DU sx

X s sI A BU s

X s Y s

Y s C sI A B D

Y s

U

U

s

s

= +

= +

= +

= + =

=

= +

-1( ) [ ( - ) ])

G s C sI A B D= = +

In Matlab: ss2tf

( ) : Transfer function of the system or the

Transfer function matrix if MIMO systems

G s


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5




Solving The Time-Invariant State Equation

Solution of homogeneous state equations

1

2 2

0

(1)

( ) (0)

1 1... ...

2! ! !

( ) (0)

n n n

At

k kAt k k

k

X A X

X t e X

A te I At A t A t

k k

X t X

=

=

=

= + + + + + = =

=

Laplace transform approach can also be used to solve Equation 1.

1

2

2 3

2 2 3 31

( ) (0) ( )

( ) ( ) (0)

1

...

1...

2! 3!

( ) (0)

At

At

sX s X AX s

X s sI A X

I A A

sI A s s s

A t A tL I At e

sI A

X t e X

=

=

= + + +

= + + + + =

=


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6




[ ] [ ]

+=

=

+=

=

==

=

+=

t

At

ttAAt

tAAt

AtAtAt

rrnnnn

dBUtXttX

et

dBUeXetX

dBUeXtXe

tBUetXedt

dtAXtXe

tBUtAXtX

tUBXAX

0

0

)(

0

11

)()()0()()(

)(

)()0()(

)()0()(

)()()()(

)()()(

)(

Solution of non-homogeneous state equations



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7




stepunit)1()(

1

0

32

10

2

1

2

1

ttu

ux

x

x

x

=

+

=

Example


The state transition matrix is given by( )t

1 1

1

2 2

1 1

2 2

( ) [( ) ]

1( )

2 3

3 11( ) 2( 1)( 2)

2( ) [( ) ]

2 2 2

At

t t t t

At

t t t t

t e L sI A

ssI A

s

s

sI A ss s

e e e et e L sI A

e e e e

= =

= +

+

= + +

= = =

+ +


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8




[ ]( ) 2( ) ( ) 2( )

( ) 2( ) ( ) 2( )0

2 2

1 1

2 22 2

2 0( ) (0) 1

12 2 2

1( ) (0)22

( ) (0)2 2 2

t t t t t

At

t t t t

t t t t

t t t t

e e e et e X d

e e e e

ex t xe e e e

x t xe e e e

= +

+ +

= +

+ +

2

2

2

1

22

1

2

(0) 0

1 1( )

2 2( )

t t

t t

t t

t t

e

e e

If Initial Conditions X

e ex t

x te e

+

=

+ =

Advantage: we dont need to decouple the equation



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9




Consider an n x n matrixA, its Characteristics equation

The Cayley Hamilton theorem states that the matrix A satisfies its own characteristicequation. Proof can found in any standard book.

1

1 1... 0n n

n nI A a a a

= + + + + =

Cayley Hamilton Theorem

1

1 1... 0n n

n nA a A a A a I

+ + + + =


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10




Controllability

A system is said to be state controllable at t = t0 if it is possible by

means of an unconstrained control signal to transfer the system from

an initial state to any other state in a finite time interval

Completely state controllable: every state is controllable

Condition: The above system is completely state controllable if and

only if the vectorsB, AB, An-1B are linearly independent or the

n x n matrixMis of rankn

In Matlab, command is CTRB

UBXAXnnnn 11

+=

1| | ... | nn n

M B AB A B

=

0 1.t t t 0( )x t


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11




Example1 :

Is NOT Controllable, because

Q. Which of the following systems is NOT state controllable and Why?

Controllability

1 1

2 2

1 1 0

0 1 1

x xu

x x

= +

1 1[ ] ( ) 1 ( )

0 0B AB rank M rank A

= = =

1 1

2 2

1 1

2 2

1 0 2.

0 2 5

1 0 2.

0 2 0

x xu

x x

x xB u

x x

= +

= +

Remark: Uncontrollable system has a subsystem that is physically

disconnected from the input.

NOT state controllable


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Condition for complete state Controllability in s-plane

In terms of transfer function, necessary and sufficient condition for

controllability is that no cancellation occurs in the transfer function for

SISO systems or transfer function matrix for MIMO systems.

If cancellation occurs, the system cannot be controlled in the direction

of the cancelled mode.

Example: is Not Controllable.

Task: Represent the above system in State Space. (Remember, this system

has derivative terms in numerator or in other words the forcing function)

Controllability

( )2.5( )( )

sX sU s

+=( 2.5)s + ( 1)s

[ ]1 12 2

0 1 1 1 1

2.5 1.5 1 1 1

x xu M B AB Not Controllable

x x

= + = =

13


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13




In practical systems, we want to control the output of the system rather

than the states of the system. Complete state controllability is neither

necessary nor sufficient for controlling the output of the system. It is

desirable to have separate complete output controllability.

Condition: The matrix given below should have rankm.

Output Controllability

1 1 1

1 1 1

n n n n n r r

m m n n m r r

x A x B u

y C x D u

= +

= +

rnm

n DBCABCACABCB )1(12 ||...||| +

14


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14




Stabilizabilty

For a partially controllable system, if the uncontrollable modes are

stable and the unstable modes are controllable, the system is said to be

stabilizable.

Example:

The stable mode corresponding to eigen value of -1 is not controllable,

but the unstable mode corresponding to eigen value of 1 is

controllable. Thus this system is stabilizable.

1 1

2 2

1 0 1( )

0 1 0

x xu not controllable

x x

= +

C l f S S 5 R i f M d C l 15


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Observability

A system is said to be state observable if every statex(t) can be

determined from the observation ofy(t) over a finite interval of time

Completely state Observable: every state is Observable

Condition: The above system is completely state controllable if and

only if the vectors CT, ATCT, (AT)n-1CTare linearly independent

or the n x nm Observability matrix is of rankn

In Matlab command is OBSV

0 1.t t t

1 1 1

1 1 1

n n n n n r r

m m n n m r r

x A x B u

y C x D u

= +

= +

1Observability matrix N=[ ( ) ]T T T T n T C A C A C

C t l f S t St t 5 R i f M d C t l 16


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[ ]

[ ]

1 1 1

2 2 2

1 1 1

2 2 2

1 0. 1 3

0 2

1 0. 0 1

0 2

x x xA y

x x x

x x xB y

x x x

= =

= =

Example1 :

Is Observable, because

Q. Which of the following systems is NOT Observable and Why?

[ ]

1 1

2 2

1

2

1 1 0

2 1 1

1 0

x xu

x x

xy

x

= +

=

1 1[ ] ( ) 2 ( )

0 1

T T TN C A C rank N rank A

= = = =

Remark: Unobservable system has a zero element in C that

corresponds to distinct eigen value ofA

.

NOT Observable

Observability

C t l f S t St t 5 R i f M d C t l 17


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Condition for complete Observability in s-plane

In terms of transfer function, necessary and sufficient condition for

Observability is that no cancellation occurs in the transfer function for

SISO systems or transfer function matrix for MIMO systems.

If cancellation occurs, the system cannot be observed in the direction

of the cancelled mode.

For a partially observable system, if the unobservable modes are stable

and the observable modes are unstable, the system is said to be

detectable.

Observability

Detectability

Control of Smart Str ct res 5 Re ie of Modern Control 18


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18




Canonical Forms

For a system given by the differential equation

The Transfer function is given as

1

0 1 1

1

1 1

( )

( )

n n

n n

n n

n n

b

a a

s b s b s bY

U s s s s a

s

+ + + +=

+ + + +

( ) ( 1) ( ) ( 1)

1 1 0 1 1

n n n n

n n n ny a y a y a y b u b u b u b u

+ + + + = + + + +

Control of Smart Structures 5 Review of Modern Control 19


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Controllable canonical form

Canonical Forms

[ ]

1

2

0 1 1 0 1 1 0 0

1

.

.

.

n n n n

n

n

x

x

y b a b b a b b a b b u

x

x

= +

1 2

1

1

1

2 2

1 1

0 1 0 0 0

0 0 1 0 0

. . . . . . .

. . . . . . .

. . . . . . .

0 0 0 1 0

1

n n

nnn n na a

x x

x x

u

x x

x xa a

= +



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Observable canonical form

[ ]

1 1 0

2 2 1 1 0

1 1

1 1

1

1

1 0

0 0 0

1 0 0. . .. . . .

. . .. . . .

. . .. . . .

.. . . .0 0 1

0 0 0 1

n n

n n

n

n

n

n

n

n

x x b a b

x x b a b

u

x xx x b a b

x

x

a

a

a

y

= +

=

2

0

1

.

.

.

n

n

b u

xx

+

Canonical Forms



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21




Canonical Forms

[ ]

[ ]

2

1 1

2 2

1

2

1 1

2 2

1

2

( ) 3:

( ) 3 2

Controllable Canonical Form

0 1 0

2 3 1

3 1

Observable Canonical Form

0 2 31 3 1

0 1

Y s sExample

U s s s

x xu

x x

xyx

x x ux x

xy

x

+=

+ +

= +

=

= +

=



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22




Pole Placement offers the flexibility to the designer to place the poles

of the system so as to achieve desired performance of the system.

We assume all the states are measurable and available for feedback.

The system should be controllable.

If controllability condition is satisfied, we can find n independent

feedback gains in a system so that the arbitrarily assigned/desired poles

performance is achieved. For a SISO system, we designed a compensator so as to reassign the

dominant closed loop poles for desired performance specifications.

Pole placement controls can assign n poles and helps us in MIMO

control.

Pole Placement Control



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Control of Smart Structures 5. Review of Modern Control



A linear system can be represented in block diagram form as


x Ax Bu

y Cx Du= += +



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24




Review for poles in s-plane


Loci of Constant Damping Ratio

Loci of Constant Undamped Natural Frequency

Q. Which of the two systems, the one shown in red and the other in blue,

has good damping characteristics and faster response?

A. The one in blue.



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25




For a system given by

Pole placement feedback control law:

This is also called full state feedback pole placement control. (Why?)

Thus, Closed loop system and its solution is given as:

The eigen values of(A-BK) determine the stability and transient

response of the system. They are also called regulator poles.


Cxy

BuAxx

=

+=

u Kx=

x yx+

-

K

BuAxx += C

( )

( ) ( ) ( )

( ) (0)

A BK

x t A BK x t

x t e x

=

=

The underlying

assumption is all

the states x are

available for

feedback



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26




Now it can be represented in block diagram form as,

Or concisely,


yx+

-

K

BuAxx += C



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27Department of Mechanical Engineering


Step 1. Check Controllability condition. If the system is controllable,

then proceed further.

Step 2. Form the characteristic polynomial of A i.e.

Step 3. Find the Transformation matrix T, that transforms the system

to controllable canonical form.


1

1 1

n n

n nsI A s a s a s a

= + + + +

T MW= 1| | ... | nn n

Controllabilty Matrix M B AB A B

=

1 2 1

2 3

1

1

1 0. . . .

. . . .

. . . .

1 0 01 0 0 0

n n

n n

a a a

a a

W

a

=



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The new state vector defined by transforms the

system to controllable canonical form


1 1

1 1

1 2 1

controllable canonical form

0 1 0 0 0

0 0 1 0 0

. . . . .

. . . . .

. . . . .

0 0 0 1 0

1n n n

x T ATx T Bu

where T AT and T B

a a a a

= + = =

x Tx=



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Step 4: Using the desired eigen values (desired closed

loop poles) , write the desired characteristic polynomial.

Write,

The Characteristic Equation is


n ..., 21

0...))...()(( 11

121 =++++=

nn

nn

n ssssss

1 1

1 1

...

,

n nK KT

Then u KTx Kx

x T ATx T BKTx

= =

= =

=

1 1| | 0sI T AT T BKT + =



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1 1

1 1 1 1

| | 0

0 1 0 0 0 0

0

0 0 1 0 0

1 0 0

0

n n n n

sI T AT T BKT

sI

a a a

s

+ =

= +

=

1 1 2 2 1 1

1

1 1 1 1

0

0 0 1

( ) ... ( ) ( ) 0

n n n n

n n

n n n n

sa a a s a

s a s a s a

+ + + + +

= + + + + + + + =




31/72



are the coefficients of desired characteristic polynomial.

Step 5. The required state feedback gain matrixK can be determined

as below.


[ ] 11 1 2 2 1 1| | | |n n n na a a a T

=

We are able to place the poles anywhere

nia

nia

iii

iii

...,,2,1then

...,,2,1If

==

==+

'i s



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Ackermann's Method

The feedback gains can also be found by the following Ackermann

formula

are the coefficients of desired characteristic polynomial.

In MATLAB, the commands are ACKERand PLACE


[ ]1

10 0 0 1 ( )n

M

K B AB A B A =

1

1 1

where , by Cayley Hamilton Theorem

( ) n n n nA A A I

= + + + +

'i s



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Example:

2

0 1 0 0

0 0 1 0

1 5 6 1

0 0 1

[ | | ] 0 1 6 rank 3 n completely controllable

1 6 31

x Ax Bu A B

M B AB A B

= + = = = = = =


3 2

3 2

1 2 3 1 2 3

3 2

1 2 3

1 1

Method 1: | | 0 6 5 10 6, 5, 1

Desired Poles (given) 2 4, 10

Desired Characteristic Polynomial

( 2 4)( 2 4)( 10) 14 60 200

14, 60, 200

| | |n n n n

sI A s s ss a s a s a a a a

s j s

s j s j s s s s

K a a

= = + + + + + + = = = =

= =

+ + + + = + + + = = =

= [ ] 12 2 1 1|

[200 1, 60 5, 14 6] [199 55 8]

a a T T I

K

=

= =



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An experimental implementation of pole placement control is shown here

Base to Cantilever the 3.35 meter Composite I-Beam

Piezo Patches as Sensors and Actuators

PC with Real-time Control System

Power Amplifier for Peizo Actuators

The experimental setup




35/72



Experimental Setup Schematic

Strong Dir.

Weak Dir.

Power

Ampl-

-ifier

A/Dconverter

dSPACE

Interface

PC with

MATLABD/A

converter

PZT

sensor

PZT

Actuator




36/72



Allows the designer to position the closed-loop poles of a system to

prescribed values.

Cxy

BuAxx

=

+=State Space form of system

u Kx= Feedback Controller

State feedback gain matrix : G

( )x A BK x

y Cx

=

=

Closed Loop System

x : state variable vector; inthis case is the voltageoutput from the piezo sensorand its derivative.

y : output vector. This issame as the state variablevector.

u: Input vector, in this caseis the pre-amplified voltageto the piezo actuator.




37/72



+

-

Low Pass

Filter

G

Bux+= C

Pole Placement Scheme

=

=

= 10

01

,238

0

,03.032.2161

0.10

CBA

[ ]1.4230 0.1039K=

For I-beam reduced second order state space system

State Feedback Gain, by Ackerman Formula

Pole locations in s-plane Damping RatioExisting

j5.46139.0 0030.0

Desired

j4.485.12 0.25

Pole Locations




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Simulation with Pole Placement Controller

Uncontrolled

Controlled

control_signal

Y

To Workspace3

T

To Workspace2

Y1

To Workspace1Switch3

Switch2

Sum5

Sum1

x' = Ax+Bu

y = Cx+Du

State-Space2

x' = Ax+Bu

y = Cx+Du

State-Space1

Sine Wave1

Scope2

Scope1

K(1)

Gain5

K(2)

Gain4

-1

Gain2

-1

Gain1

m

m

0

Constant1

30Clock2

Band-Limited

White Noise1




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Closed and Open Loop Pole Locations Desired and Open loop Discrete Pole Locations

Pole-Zero Map

Real Axis

Imag

Ax

is

-10 -5 0 5-50

-40

-30

-20

-10

0

10

20

30

40

50

Desired Poles Existing Poles

Pole-Zero Map

Real Axis

ImagAxis

0.8 0.85 0.9 0.95 1

0

0.1

0.2

0.3

0.4

0.5

0.615 7

78.5

Existing Pole

Desired Pole




40/72




Simulated Comparative Time Response PlotWith and Without Pole Placement Control

Bode Plot of the I-beam With and WithoutPole Placement Control

Bode Diagram

Frequency (rad/sec)

Phase(deg)

Magnitude(dB)

-80

-60

-40

-200

20

40Without ControlWith PP Control

101

102

103

-180

-135

-90

-45

0

5 10 15 20 25 30-8

-6

-4

-2

0

2

4

6

8Simulation Results

Time (sec)

A

mp

litude

(Vo

lts

)

Without controlWith Pole Placement control




41/72



Pole Placement Experimental Results

Experimental Comparative Time ResponsePlot With and Without Pole Placement Control

PSD Comparison Plot for Pole PlacementControlled Strong Direction for 10-15second Period (Decibel drop of 29 dB)

0 20 40 60 80 100 120

-80

-60

-40

-20

0

20

40

Frequency (Hz)

Energy

leve

linDec

ibe

ls

with Pole Placement controlwithout control

5 10 15 20 25 3

-8

-6

-4

-2

0

2

4

6

8Time Response: Strong

Time (sec)

Amp

litude

(Vo

lts

)

without controlwith Pole Placement control




42/72



5 10 15 20 25 30-10

-8

-6

-4

-2

0

2

4

6

8

10Time Response: Strong

Time (seconds)

Amp

litude

(Vo

lts

)

with Pole Placement control

Pole Placement Experimental Results

Experimental Time Response Plot With Pole Placement Control when the beam isexcited manually for multi-modes at around 15 seconds. Initial 5-15 seconds depicts the

plot with the usual experimental excitation source.




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Observer Design

In pole placement control we assumed, all the states are available for

feedback. In practical systems not all the states are available for

feedback.

To get an estimate of unmeasured states we need an observer.

Consider the system :

The observer is a subsystem to reconstruct the state vector of the plant.

Cxy

BuAxx

=

+=



44/72



For observer mathematical model is same except we introduce a

term that accounts for the estimation error (difference between

observed output and measured output)

Thus, mathematical model of observer is

Observer Design

We use or to represent the estimation of

the state.

( )

( )

e

e e

x x

x Ax Bu K y Cx

A K C x Bu K y

= + +

= + +



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Observer Design

x Ax Bu

y Cx

= +

=

( )ex Ax Bu K y Cx= + +



46/72



Full order state Observer: This means all the states will be estimated

and the order is same as that of the plant.

Observer Design

( )( )( )

( )

e

e

e

x x Ax Ax K Cx CxA K C x x

e x x

e A K C e

= =

=

=

Determines convergence of error



47/72



Thus, if is stable, error will converge to zero

thereby implying that the estimated state converges to the actual

state.

To design an observer, the Observability condition should be

satisfied.

Then we can design an observer that has arbitrarily

assigned/desired poles.

Thus, this problem is a DUAL problem i.e. solve the pole

placement problem for the dual system.

We need to determine Ke such that the error dynamics areasymptotically stable with sufficient speed of response.

Observer Design

( )eK C



48/72



Dual Problem

The previous equation of error convergence, means we need to

determine Ke for convergence of .

This is tantamount to solving pole placement problem for a set of

desired eigen values of the observer.

Observer Design

( )eK C

Thus for a given system

We have to solve the Pole Placement problem (Dual Problem)

of the following system

where control signal is given by

-

TT

T

x Ax Bu

y Cx

z A z C

n B z

v Kz

= +

=

= +

=

=



49/72



How do we arrive at the dual problem

Observer Design

Given system

Pole Placement problem for above system is to find

such that is stableNow for observer problem we have to make stable

So, if we take

( -

the transpose of pole plac

( -

e

))e

K

KA

x x u

K C

y x

= +

=

( - ) ,We can solv

ment pr

e the observer p

ob

roble .

lem

m

T T T T

K K =



50/72



Observer Design

Now for observer problem, we have t

On comparing ( - ), we get follow

o solve the Pole Placement

probl

ing matrices for obse

em (Dual Problem) of t

rver probl

he following system (Chang

em

, ,

e th

e

T

e

T T

A K C

KA C K = = =

e variable names)

where control signal is given by

-

and

TT

T

Te

z A z C

n B z

Kz

K K

= +

=

=

=



51/72



Summary

Observer Design

Observer problem

The observer problem can be converted to Pole Placement

problem as (Dual Problem)

: Control Law

TT

T

T

e

x Ax Bu

y Cx

z A z C

n B z

KzK K

= +

=

= +

=

= =



52/72



Transformation approach to find observer gain Ke Step 1. Check Observability condition. If the system is observable,

then proceed further.

Step 2. Form the characteristic polynomial of A (same as AT) i.e.

Step 3. Find the Transformation matrix Q, that transforms the systemto Observable canonical form.

The new state vector defined by transforms thesystem to observable canonical form.

Observer Design

1( ) : ( )T T T T n T N C A C A C Condition rank N n = =

11 1n n n nsI A s a s a s a = + + + +

1

( )T

Q WN

=

Matlab: POLY gives ch. eqn

x Q=



53/72



Observer Design

1 2 1

2 3

1

... 1

... 1 0

where, ... ... ... ... ...

1 ... ... ...1 0 ... ... 0

n n

n n

a a a

a a

W

a

=

1 1

0

1

1 1 01 1

1 1 0

1

Observable Canonical form

,

0 ... ... 0

1 ... ... ...

0 ... ... ... ...

...... ... ... 0 ...

0 ... 0 1

[0 0 ...

n

n n

n

n n

x Q Q AQ Q Bu

y CQ

ab a b

ab a b

Q AQ Q B

b a ba

CQ

= = +

=

= =

=

0 1]



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Observer Design

1

( ) ( )

( )( )

,

( ) ( )( )

,

( )

e e e

e

e

e

x Ax Bu K y Cx A K C x Bu K Cx

x x A K C x x

Take x Q x Q

Q A K C

Take

Q A K C Q

= + + = + +

=

= =

= =

=



55/72



Observer Design

[ ]

1

1 11

1 1

1 11

11

( )

,... ...

0 ... 0

... ...0 0 ... 0 1

... ... ......0 ... 0

e

n n

n n

e e

n n

n n

e

Q A K C Q

Take Q K K Q

Q K CQ

=

= =

= =



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Observer Design

1

1 1

2 2

1 1

1 2

1 1 2 2

Chara. Eqn. ( ) 0

0 ... 01 ... ...

00 1 ... 0... ... ... ...

0 ... 0 1

( ) ( ) ...( ) 0

e

n n

n n

n n

n n n

n n

sI Q A K C Q

s as a

as

s a

s a s a s a

=

+ + = + + +

+ + + + + + =



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Step 4: Using the desired eigen values (desired closed

loop poles) , write the desired characteristic polynomial.

On comparing the desired characteristic polynomial, we get

From the above we can now solve for Ke

Observer Design

n ..., 21

0...))...()(( 11

121=++++=

nn

nn

n ssssss

1 1 1 1 1 1

2 2 2 2 2 2

... ...

n n n n n n

a a

a a

a a

+ = = + = =

+ = =



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Observer Design

1 1

1

1 1 1 1

1

1 1 1 1

( )...

( ). .. .

n

n T

e

n n n n

n n n n

T

e

K Q Q WN

a a

a a

K WN

a a

= =

= =



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Alternate approach to find observer gain Ke (We will arrive at the

same result)

If the desired eigen values of observer are known

(desired closed loop poles of observer) , we can write the desired

characteristic polynomial.

Observer Design

1 2( ) ( )( )...( )

T T

nsI A C K s s s =

e

Note that eigen values of ( - ) and ( - ) are same.

Thus, K and K are related by

T T T

Te

A K C A C K

K K=( ) ( )T T TsI A C K sI A K C

n ..., 21



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Following the approach as discussed in pole placement control, we

will derive the Observer Gain matrix.

For the system

The dual problem is given as

The state feedback law is given by

Recall, from the pole placement problem,Kis given as

Observer Design

Cxy

BuAxx

=

+=

TT

T

z A z C

n B z

= +

=

Kz =

( )T Tz A C K z =

[ ] 11 1 2 2 1 1| | | | wheren n n na a a a T T MW

= =

Desired Eigen values of

observer gain matrix aren ..., 21



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Thus,

For the original system, the Observability matrix is

Observer Design

1( )T T T T n T T MW C A C A C W = =

1

( )T T T T n T

N C A C A C

=

1 2 1

2 3

1

1

1 0

. . . .

where . . . .

. . . .

1 0 0

1 0 0 0

n n

n n

a a a

a a

T NW W

a

= =

1 1

Since,

( ) ( )

T T T T T

T T

W W T W N WN

T WN

= = =

=



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Observer Design

[ ] 11 1 2 2 1 1

1 1 1 1

1 1

1 1 1 1

1 1

| | | |

( ) ( ). .. .

For any matrix, ( ) ( )

n n n n

T

e

n n n n

n n n n

T T T

e

T T

K a a a a T

K Ka a

a a

K K T WN

a a

P P

=

=

= = =

=



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[ ]

1,2

12

2

Example:

0 20.6 00 1

1 0 1

desired pole for the observer

1.8 2.4

0 1| rank 2 completely observable

1 0

020.6| | 0 20.6

1 20.6

desired characteristic equation

(

T T T

A B C

j

N C A C

assI A s

s a

= = =

=

= = =

= = = + =

12

2

2 21 1

1 1

3.61.8 2.4)( 1.8 2.4) 3.6 9

9

9 20.6 29.6( ) ( ( ) )

3.6 0 3.6

T T

e

s j s j s s

aK WN Q WN I

a

=+ + + = + + =

+ = = = = =

Observer Design



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Observer in closed loop system

Observed states feedback

(all the states)

Output/Measured states

( not all the states)

Pole Placement Controller

Gain ( designed assuming

all the states)

Error between

observed states

and measured

state

Observer Gain

(All the states are

available in

observer)

u Kx= Controlled input( using the observed states)



65/72



The system :

The Observer:

Control Law:

Error Dynamics:

Write the closed loop equations in terms of errore(t) and statex(t)


Cxy

BuAxx

=

+=

( )

( )

e

e e

x Ax Bu K y Cx

x A K C x Bu K y

= + +

= + +

u Kx=

( ) ( )( )( )e e

e

x x Ax Ax K Cx Cx A K C x xe x x e A K C e

= =

= =

( ) ( )

( )

x Ax Bu Ax BKx A BK x BK x x

x A BK x BKe

= + = = + = +

Controller

System

Observer

y

x

u Kx=



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Thus the closed loop dynamics is

The eigen values of

Thus, we can design independently the controller gain and the observergain.

Practically, we choose the observer poles to be 2-5 times faster than

the desired closed loop poles of the system.


( )

0 ( )cl

e

A

A BK BKx x

A K Ce e

=

( ) ( )cl eeig A BK eig A K C =



67/72



Example:

For the above system, design an observer based pole placement

controller with closed loop poles at

and observer poles at . Simulate theclosed loop system for Initial conditions [0 0 2]T


[ ]

0 1 0 1

0 0 1 8

0 -24 -10 106

1 0 0

x x u

y x

= +

=

1 2, 1 2, 5s j s j s= + = =

5, 5, 5s s s= = =



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Step 1. Check Controllability and Observability

Step 2. Assuming all the states are available for feedback, find

feedback gain for the pole placement controller for the desired poles.


1 8 106

8 106 868 ( ) 3

106 868 6136

1 0 0

0 1 0 ( ) 3

0 0 1

M rank M

N rank N

= =

= =

[ ]0.5000 -0.0904 -0.0398K=



69/72



Step 3. Using the dual system, find the observer gain for the desired

observer poles.

For the given Initial conditions, form a closed loop system and

simulate it using SIMULINK/Matlab. The simulation block diagram is

shown next.


5

1-5

eK

=



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Blue: Observer

Black: Plant



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Remember the output of plant is 1y x=

Observer Output as compared to plant output



72/72


5.updated draft review of modern control

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