state space design - University of...

Amme 3500 : System Dynamics & Control

State Space Design

Dr. Stefan Williams

Slide 2 Dr. Stefan B. Williams Amme 3500 : Introduction

Course Outline Week Date Content Assignment Notes 1 1 Mar Introduction 2 8 Mar Frequency Domain Modelling 3 15 Mar Transient Performance and the s-plane 4 22 Mar Block Diagrams Assign 1 Due 5 29 Mar Feedback System Characteristics 6 5 Apr Root Locus Assign 2 Due 7 12 Apr Root Locus 2 8 19 Apr Bode Plots No Tutorials 26 Apr BREAK 9 3 May Bode Plots 2 10 10 May State Space Modeling Assign 3 Due 11 17 May State Space Design Techniques 12 24 May Advanced Control Topics 13 31 May Review Assign 4 Due 14 Spare

Slide 3

Higher Order Poles

•  Classical controller designs are based on selecting the location of the dominant second order pole locations

•  Higher order poles are generally not able to be set explicitly

Dr. Stefan Williams Amme 3500 : State Space Slide 4

Higher Order Poles: Example

•  For the following plant

•  Design a PD controller to yield an overshoot of 5% and a settling time of 1s

Dr. Stefan Williams Amme 3500 : State Space

G(s) =1

s(s + 2)(s +10)


•  Consider the design of a PD controller to achieve the required specifications

•  Desired root locations are at


s = !4 ± 4.08 j

x x x

x

x

o

!z "!1 "!2 "!3 = (2k +1)180°!z = 134.4 +116.1+ 34.2 "180!z = 104.7°# zc = 2.93 With K=44.7

Slide 6



Slide 7

Higher Order Poles

•  As can be seen, the position of the higher order poles can have a significant impact on the actual system behaviour

•  Classical techniques do not easily afford us with a means of choosing the position of higher order poles

Dr. Stefan Williams Amme 3500 : State Space Slide 8 Dr. Stefan Williams Amme 3500 : State Space

State Space Modelling

•  Recall that for the state space approach, we represent a system by a set of n first-order differential equations:

•  The output of the system is expressed as: !x = Ax +Bu

[ ]= +y Cx Dux - state vector y - output vector u - input vector

A - state matrix B - input matrix C - output matrix D - feedthrough or feedforward matrix (often zero)


State Space Control •  We can represent a

general state space system as a Block Diagram.

•  If we feedback the state variables, we end up with n controllable parameters.

•  State feedback with the control input

u=-Kx +r.

* N.S. Nise (2004) “Control Systems Engineering” Wiley & Sons

Slide 10 Dr. Stefan Williams Amme 3500 : State Space

State Space Control

•  We can then control the pole locations by finding appropriate values for K

•  This allows us to select the position of all the closed loop system roots during our design.

•  There are a number of methods for selecting and designing controllers in state space, including pole placement and optimal control methods via the Linear Quadratic Regulator algorithm.


State Space Control

•  Setting u=-Kx+r yields

•  Rearranging the state equation and taking LT yields

•  Essentially we select values of K so that the

eigenvalues (root locations) of (A-BK) are at a particular location

˙ x (t) = Ax(t) + B(!Kx(t) + r)y(t) = Cx(t)

sX(s) = A ! BK( )X(s) + BR(s)sI! A ! BK( )( )X(s) = BR(s)

X(s)R(s)

= sI! A ! BK( )( )!1B

Slide 12

Pole Placement

•  Given a system of the form

•  We can find the characteristic equation by solving for det(sI-(A-BK))


A =

0 1 0 !0 0 1 !" " " #

!a0 !a1 !a2 !

000

!an!1

"

#

$ $ $ $

%

&

' ' ' '

B =

00"1

"

#

$ $ $ $

%

&

' ' ' '

C = c1 c2 ! cn[ ]

Pole Placement

•  Using u=-Kx, we can find the system matrix A-BK

•  With characteristic equation of the form


A ! BK =

0 1 0 !0 0 1 !" " " #

!(a0 + k1) !(a1 + k2) !(a2 + k3) !

000

!(an!1 + kn )

"

#

$ $ $ $

%

&

' ' ' '

det(sI! (A ! BK)) = sn + (an!1 + kn )sn!1

+! + (a1 + k2)s + (a0 + k1) = 0Slide 14

Pole Placement

•  Given a desired pole placement of

•  and equating terms, we find

•  or

•  This will allow us to place all of the closed loop poles where we want them


sn + dn!1sn!1 +! + d1s + d0 = 0

di = ai + ki+1 i = 0,1, 2,…, n !1

ki+1 = di ! ai

Slide 15

Pole Placement •  Represent the plant in state space via the

phase-variable form •  Feed back each phase variable to the input

of the plant through a gain, ki •  Find the characteristic equation of the closed-

loop system •  Decide upon closed-loop pole locations and

determine an equivalent characteristic equation

•  Equate like coefficients of the characteristic equations and solve for ki


Pole Placement Example

•  For the following plant

•  Design a full state feedback controller to yield an overshoot of 5% and a settling time of 1s


G(s) =1

s(s + 2)(s +10)


•  The OL transfer function for the system is

•  Convert the transfer function to SS


˙ x 1˙ x 2˙ x 3

!

"

# # #

$

%

& & &

=0 1 00 0 10 '20 '12

!

"

# # #

$

%

& & &

x1

x2

x3

!

"

# # #

$

%

& & &

+001

!

"

# # #

$

%

& & & u

y = 1 0 0[ ]x

G(s) =1

s3 +12s2 + 20s( )

Slide 18

Pole Placement Example •  Decide on desired root locations •  From specifications, we require

– 5% OS and – settling time of 1s

•  Select dominant second order roots to be at

•  Select third root 5 times to left at •  Desired characteristic equation:


s = !4 ± 4 j

s = !20

s3 + 28s2 +192s + 640

Slide 19


•  Feeding back states we find


˙ x = Ax + Bu= Ax + B(!Kx + r)= A ! BK( )x + Br

˙ x 1˙ x 2˙ x 3

"

#

$ $ $

%

&

' ' '

=0 1 00 0 10 !20 !12

"

#

$ $ $

%

&

' ' ' !

0 0 00 0 0K1 K2 K3

"

#

$ $ $

%

&

' ' '

(

)

* *

+

,

- -

x1

x2

x3

"

#

$ $ $

%

&

' ' '

+001

"

#

$ $ $

%

&

' ' ' r

Slide 20


•  So

•  Solve for the characteristic equation


sX =0 1 00 0 1

!K1 !(20 + K2) !(12 + K3)

"

#

$ $ $

%

&

' ' '

(

)

* *

+

,

- - X +

001

"

#

$ $ $

%

&

' ' ' r

s !1 00 s !1K1 (20 + K2) s + (12 + K3)

= 0

s3 + (12 + K3)s2 + (20 + K2)s + K1 = 0


•  Equating like terms we find

•  So


s3 + 28s2 +192s + 640s3 + (12 + K3)s

2 + (20 + K2)s + K1

K1 = 640K2 = 172K3 = 16

Slide 22


•  Looks like we have met the overshoot and settling time requirements

•  However, steady state error is not addressed here


Slide 23

Steady State Error •  Full state feedback does not directly

address the issue of steady state error •  A common strategy is to feedback the

output and add additional state variables


* N.S. Nise (2004) “Control Systems

Engineering” Wiley & Sons

Slide 24

Steady State Error

•  Looking at the previous diagram, we find

•  We can augment the state equations yielding


˙ x N = e = r !Cx

˙ x ˙ x N

!

" #

$

% & =

A 0'C 0!

" #

$

% &

xxN

!

" #

$

% & +

B0!

" #

$

% & u +

01!

" # $

% & r

y = C 0[ ] xxN

!

" #

$

% &

Steady State Error

•  But

•  So

•  We can now choose gains to yield desired pole locations and zero steady state error


u = !Kx + KE xN = ! K KE[ ] xxN

"

# $

%

& '

˙ x ˙ x N

!

" #

$

% & =

A ' BK BKe

'C 0!

" #

$

% &

xxN

!

" #

$

% & +

01!

" # $

% & r

y = C 0[ ] xxN

!

" #

$

% &

Slide 26

Steady State Error Example

•  Let’s consider the previous example where we wanted the roots at

•  We now have an additional pole which we will place at s=100 so the new desired characteristic equation is


s3 + 28s2 +192s + 640

s4 +128s3 + 2992s2 +19840s + 64000

Slide 27


•  The system is now characterised by

•  Finding the eigenvectors of this equation yields


sXXN

!

" #

$

% & =

0 1 0 00 0 1 0

'K1 '(20 + K2) '(12 + K3) Ke

'1 0 0 0

!

"

# # # #

$

%

& & & &

(

)

* * * *

+

,

- - - -

XXN

!

" #

$

% & +

001

!

"

# # #

$

%

& & & r

s4 + (12 + K3)s3 + (20 + K2)s

2 + K1s + Ke

-C

BKe

Slide 28


•  Equating like terms we find

•  So


s4 + (12 + K3)s3 + (20 + K2)s

2 + K1s + Ke

s4 +128s3 + 2992s2 +19840s + 64000

K1 = 19840K2 = 2972K3 = 116Ke = 64000


•  We now have a system that meets all of our requirements

•  We haven’t considered the required control effort


Observer Design

•  The previous development assumes that we have access to all state variables for state feedback

•  In some instances, the state variables may not be available – They may be difficult to observe – Sensors may be expensive


Slide 31

Observer Design •  We may need to

design an Observer to estimate the state variables

•  We can compare the estimated output against the measured output to close the loop on our estimate

•  The estimated state is then used as the feedback in the controller


* N.S. Nise (2004) “Control Systems Engineering” Wiley & Sons


Linear Quadratic Regulator Control

•  Optimal control methods attempt to find optimal gains subject to certain conditions

•  One such algorithm is called the Linear Quadratic Regulator (LQR) algorithm.

•  An objective/cost function can be defined and the appropriate state feedback gain K that minimizes the cost function can be found.

•  Weighting factors/matrices are used to penalize the states and control input. This generally requires a good engineering judgment depending on the control performance requirements.


LQR Control •  Consider a dynamic system in a state space

form:

•  We can set a particular objective/cost function which governs the performance of the closed-loop system:

•  Here, Q and R are the weighting matrices that respectively penalize the states x and control input u (Q!0, R>0).

J = xTQx + uTRu( )0

!

" dt

˙ x = Ax + Bu


LQR Control •  Then, the state feedback gain that minimizes this

cost function is:

•  P can be obtained by solving the Riccati equation:

•  In Matlab, the feedback gain can be found via [K]=lqr(system,Q,R)

ATP + PA ! PBR!1BTP +Q = 0

u = !(R!1BTP)x = !Kx


An Example System


Inverted Pendulum

Inverted pendulum – video courtesy of Colorado State University

Inverted Pendulum

•  We will look at the modelling and control of the inverted pendulum solutions provided by the University of Michigan

•  http://www.engin.umich.edu/class/ctms/examples/pend/invpen.htm


State Space Systems

•  There are many other aspects of state space systems that we have not touched on here

•  These include accounting for non-linear or digital systems

•  We have also only touched briefly on optimal control methods



Conclusions

•  We have looked at techniques for modelling systems using State Space techniques.

•  We have provided a brief introduction to designing controllers using the State Space techniques using pole placement and the LQR algorithm.


Further Reading

•  Nise – Sections 3.1-3.7 and 12.1-12.8

•  Franklin & Powell – Section 2.2 and 7.1-7.6

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