Lecture Opnet

8/10/2019 Lecture Opnet

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Dynamical Systems Analysis II:

Evaluating Stability, Eigenvalues

By Peter Woolf ([email protected])

University of Michigan

Michigan Chemical Process

Dynamics and Controls

Open Textbook

version 1.0

Creative commons



Problem: Given a large and complex system

of ODEs describing the dynamics and

control of your process, you want to know:

(1)Where will it go?

(2)What will it do?Is there anything fundamental you can say

about it?

E.g. With my control architecture, this processwill always ________.

Solution: Stability Analysis

Steady state from last lecture.

Topic for today!



Exponential increase Increase w/ oscillation

Stable oscillation

Periodic solution Non-periodic solution

(chaotic)

Only possible for

nonlinear systems

Decay w/ oscillationExponential decay

What will your system do?



How can we know where the

system will go?

Possible approaches:

1. Simulate system and observe

Disadvantages:• Can’t provide guaranteed behavior, just samples of

possible trajectories.

• Requires simulations starting from many points

• Assumes we have all variables defined, thus hard to

use to design controllers.

Advantages:

• Works for any system you can simulate

• Intuitive--you see the results



How can we know where the

system will go?

Possible approaches:

1. Simulate system and observe

2. Stability Analysis (this class)

Disadvantages:

• Only works for linear models

• Linear approximations of nonlinear models

break down away from the point of

linearization

Advantages:

• Provides strong guarantees for linear systems

• General



dA

dt = 3 A " A

2

" AB

dB

dt = 2 B " AB " 2 B

2

Nonlinear model Linear approximation at A=0, B=0

JacobianOr in a

different

format

dA

dt = 3 A

dB

dt

= 2 B

From last class…

"

A" B

#

$ % &

' ( = 3 0

0 2#

$ % &

' ( A B#

$ % &

' ( + 0

0#

$ % &

' (

Intuitively, what will the linear system do if A is perturbed slightly from 0?

dA

dt = 3(0+ ")

Increase in A above 0

yields a positive

derivative

Increase

in A

Increase in

slope of A

Exponential increase



But what if our model is more complex?

" A

" B

#

$ %

&

' ( =

3 )22 )2#

$ %

&

' ( A

B

#

$ %

&

' ( +

3

4

#

$ %

&

' (

E.g. (note: example below is made up)

Or in adifferent

format

dA

dt = 3 A " 2 B + 3

dB

dt = 2 A " 2 B + 4

What will happen if A or B are increased slightly

from the steady state value of A=1, B=3?

Result : increase A, A and B

increase!

Result : increase B, A and B

decrease!

dA

dt = 3(1+ ") # 2(3)+ 3 = +3"

dB

dt = 2(1+ ") # 2(3)+ 4 = +2"

Increase A by !:

dA

dt = 3(1) " 2(3+ #) + 3 = "2#

dB

dt = 2(1)" 2(3+ #) + 4 = "2#

Increase B by !:



Observations:

1. It is easy to predict where a linear system will go if the

variables are decoupled

2. Coupling between variables makes it harder to predict

what will happen

3. Coupling is determined by the Jacobian

dA

dt = 3 A

dB

dt

= 2 B

A only influences A, B only influences B.

-> Variables are decoupled

dA

dt = 3 A " 2 B + 3

dB

dt = 2 A " 2 B + 4

Changes in A influence changes in A

and B. Changes in B influence

changes in A and B.

--> Variables are coupled



Is it possible to change a coupled system to a decoupled one?

" A

" B

#

$ %

&

' ( =

k 11

k 12

k 21

k 22

#

$ %

&

' (

A

B

#

$ %

&

' ( +

k 13

k 23

#

$ %

&

' (

k 11

k 12

k 21 k 22

"

#

$ %

&

' A

B

"

#

$ %

&

' " 1 0

0 1

#

$

%

&

'

(

A

B

#

$

%

&

'

(

??

Can we find a ! value that satisfies this relationship?

k 11

k 12

k 21

k 22

"

# $

%

& '

()

1 0

0 1

"

# $

%

& '

*

+ ,

-

. /

A

B

"

# $

%

& ' = 0

k 11 " # k

12

k 21

k 22 " #

$

% &

'

( )

*

+

, -

.

/ A

B

$

% &

'

( ) = 0

k 11

" # ( ) A + k 12 B = 0

k 21 A + k 22 " # ( ) B = 0

Written

differently..

This is aneigenvalue



k 11

" # ( ) A + k 12 B = 0

k 21

A + k 22

" #

( ) B = 0

k 11 A " # A + k

12 B = 0

k 21 A + Bk

22 " # B = 0

expand

B =

"k 11 A + # A

k 12

k 21 A + "k 11 A + # A

k 12

$

% & '

( ) k 22 " # "k 11 A + # A

k 12

$

% & '

( ) = 0

Solve for B

k 21 A "

k 11 Ak

22

k 12

+

# Ak 22

k 12

+

# k 11 A

k 12

+

# 2 A

k 12

= 0

A k 21 "

k 11

k 22

k 12

+

# k 22

k 12

+

# k 11

k 12

+

# 2

k 12

$

% &

'

( ) = 0



A k 21 "

k 11

k 22

k 12

+

# k 22

k 12

+

# k 11

k 12

+

# 2

k 12

$

%

& '

(

) = 0

Solve for !

" =1

2k 11+ k

22± k

11

2

+ 4k 12

k 21# 2k

11k 22+ k

22

2[ ]

Observations:

1) Yes! There is always a way decouple a coupled linear

system

2) Direct approach involves lots of algebra

There is an easier way..



A bit of linear algebra background

Goal: solve this system for !

k 11 " # k

12

k 21

k 22 "

#

$

%

& '

(

) *

+

, -

.

/ A

B

$

%

& '

(

) = 0

Determinant : a property of any square matrix that

describes the degree of coupling between the equations.

Determinant equals zero when the system is not linearly

independent, meaning one of the equations can be cast as

a linear combination of the others.

Det

a b c

d e f

g h i

"

#

$ $

$

%

&

' '

'

= a * Det e f

h i

"

# $

%

& ' ( b* Det

d f

g i

"

# $

%

& ' + c * Det

d e

g h

"

# $

%

& '

Det a b

c d "

# $ %

& ' = a* d ( b*c



A bit of linear algebra background

Goal: solve this system for !

k 11 " # k

12

k 21

k 22 "

#

$

%

& '

(

) *

+

, -

.

/ A

B

$

%

& '

(

) = 0

Determinant : a property of any square matrix that

describes the degree of coupling between the equations.

Determinant equals zero when the system is not linearly

independent, meaning one of the equations can be cast as

a linear combination of the others.

Det k 11 " # k

12

k 21

k 22 " #

$

% &

'

( ) = 0

Revised Goal: find!

that satisfies

k 11

" # ( ) k 22 " # ( ) " k

12k 21

= 0

" =1

2

k 11+ k

22± k

11

2

+ 4k 12

k 21# 2k

11k 22+ k

22

2

[ ]



Similar Analysis can be done in Mathematica:

Det[{a,b},{c,d}] :Find the determinant of a matrix

Solve [{eqn1, eqn2,..},{var1, var2,..} ] : Solve algebraically

Eigenvalues[{a,b},{c,d}] : Automatically find the eigenvalues



What do eigenvalues tell us

about stability?Eigenvalues tell us the exponential part of the

solution of the differential equation system

Three possible values for an eigenvalue

1) Positive value: system will increaseexponentially

2) Negative value: system will decayexponentially

3) Imaginary value: system will oscillate

(note combinations of the above are possible)



What do eigenvalues tell us

about stability?

Effect: If any eigenvalue has a positive real part, the system

will tend to move away from the fixed point



Marble Analogy

Small perturbations

left or right will

cause the marble to

decay back to the

steady state position

Negative real

eigenvalue

Small perturbations

left or right will

cause the marble to

decay away from

the steady stateposition (xss)

Positive real

eigenvalue

Small perturbations

in y are stable, while

perturbations in x

are unstable (saddle

point), thus overallpoint is unstable!

Positive and

negative real

eigenvalues

xssx

Case I: stablexss

x

Case II: unstable

xss,,yss

x

y

Case III: Saddle point



" A

" B

#

$ %

&

' ( =

3 )2

2 )2

#

$ %

&

' ( A

B

#

$ %

&

' ( +

3

4

#

$ %

&

' (

Revisit our example: What will happen here?

1) Calculate eigenvalues

Eigenvalues: ! 1=2, ! 2 = -1

2) Classify stability:

At least one eigenvalue is positive,

so the point is unstable and a

saddle point.

Exponential increase



" A

" B

" C

#

$

%

% %

&

'

(

( (

=

3 )2 1

2 2 )2

)1 2 0

#

$

%

% %

&

'

(

( (

A

B

C

#

$

%

% %

&

'

(

( (

+

8

)2

4

#

$

%

% %

&

'

(

( (

A more complex example: What will happen here?

1) Calculate eigenvalues

Force Mathematica to find a numerical value using N[ ]

Using the Eigenvalue[ ] function in Mathematica

Given these eigenvalues what will it do?



" A

" B

" C

#

$

%

% %

&

'

(

( (

=

3 )2 1

2 2 )2

)1 2 0

#

$

%

% %

&

'

(

( (

A

B

C

#

$

%

% %

&

'

(

( (

+

8

)2

4

#

$

%

% %

&

'

(

( (

2) Classify stability:

• The real component of at least

one eigenvalue is positive, so

the system is unstable.

• There are imaginary eigenvalue

components, so the response

will oscillate.Increase w/ oscillation

A more complex example: What will happen here?



Exponential increase Increase w/ oscillation

Stable oscillation

Decay w/ oscillationExponential decay

What will your system do?

(according to eigenvalues)

All ! s are real and

negative

All ! s are real and

at least one positive

All ! s have negative

real parts, some

imaginary parts

At least one ! has

positive real parts,

some imaginary parts

All ! s have zero

real parts and

nonzero

imaginary parts



Take Home Messages

• Stability of linear dynamical systemscan be determined from eigenvalues

• Complicated sounding terms like

eigenvalues and determinant can bederived from algebra alone--fear not!

• Stability of nonlinear dynamical systemscan be locally evaluated using

eigenvalues

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