Course Outline
Week
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8
Amme 3500 :
System Dynamics & Control
State Space Design
9
10
11
12
13
14
Dr. Stefan Williams
Date
1 Mar
8 Mar
15 Mar
22 Mar
29 Mar
5 Apr
12 Apr
19 Apr
26 Apr
3 May
10 May
17 May
24 May
31 May
Dr. Stefan B. Williams
Higher Order Poles
Amme 3500 : State Space
Amme 3500 : Introduction
Assignment Notes
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Slide 2
Higher Order Poles: Example
•  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
Content
Introduction
Frequency Domain Modelling
Transient Performance and the s-plane
Block Diagrams
Feedback System Characteristics
Root Locus
Root Locus 2
Bode Plots
BREAK
Bode Plots 2
State Space Modeling
State Space Design Techniques
Advanced Control Topics
Review
Spare
Slide 3
•  For the following plant
1
G(s) =
s(s + 2)(s + 10)
•  Design a PD controller to yield an
overshoot of 5% and a settling time of 1s
Dr. Stefan Williams
Amme 3500 : State Space
Slide 4
Higher Order Poles: Example
•  Consider the design
of a PD controller to
achieve the required
specifications
•  Desired root locations
are at
Higher Order Poles: Example
x
o
x
x
x
s = !4 ± 4.08 j
! z " !1 " ! 2 " ! 3 = (2k + 1)180° x
! z = 134.4 + 116.1 + 34.2 " 180
! z = 104.7°
# zc = 2.93
Dr. Stefan Williams
With K=44.7
Amme 3500 : State Space
Slide 5
Dr. Stefan Williams
Amme 3500 : State Space
Slide 6
Higher Order Poles
State Space Modelling
•  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
•  Recall that for the state space approach,
we represent a system by a set of n firstorder differential equations:
x! = Ax + Bu
•  The output of the system is expressed as:
y = Cx + [Du]
x - state vector
y - output vector
u - input vector
Dr. Stefan Williams
Amme 3500 : State Space
Slide 7
Dr. Stefan Williams
A - state matrix
B - input matrix
C - output matrix
D - feedthrough or feedforward matrix (often zero)
Amme 3500 : State Space
Slide 8
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.
Dr. Stefan Williams
State Space Control
* N.S. Nise (2004) Control Systems Engineering Wiley & Sons
Amme 3500 : State Space
Slide 9
•  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.
Dr. Stefan Williams
•  Setting u=-Kx+r yields
x˙ (t) = Ax(t) + B(!Kx(t) + r)
•  Given a system of the form
y(t) = Cx(t)
•  Rearranging the state equation and taking LT yields
sX(s) = (A ! BK ) X(s) + BR(s)
(sI ! (A ! BK ))X(s) = BR(s)
1
0
" 0
$ 0
0
1
A=$
"
"
$ "
$!a !a !a
# 0
1
2
C = [c1 c 2 !
X(s)
!1
= ( sI ! (A ! BK )) B
R(s)
•  Essentially we select values of K so that the
eigenvalues (root locations) of (A-BK) are at a
particular location
Amme 3500 : State Space
Slide 10
Pole Placement
State Space Control
Dr. Stefan Williams
Amme 3500 : State Space
Slide 11
cn ]
! 0 %
"0 %
'
$0 '
! 0
'B= $ '
# 0 '
$" '
'
$1 '
! !an !1 &
# &
•  We can find the characteristic equation by
solving for det(sI-(A-BK))
Dr. Stefan Williams
Amme 3500 : State Space
Slide 12
Pole Placement
Pole Placement
•  Using u=-Kx, we can find the system
matrix A-BK
0
1
0
"
$
0
0
1
A ! BK = $
"
"
"
$
$!(a + k ) !(a + k ) !(a + k )
# 0 1
1
2
2
3
•  Given a desired pole placement of
!
0
%
'
!
0
'
#
0
'
! !(an !1 + k n ) '&
n !1
det(sI ! (A ! BK)) = s + (an !1 + kn )s
+ ! + (a1 + k 2 )s + (a0 + k1 ) = 0
Dr. Stefan Williams
Amme 3500 : State Space
Slide 13
•  or
Dr. Stefan Williams
Amme 3500 : State Space
Slide 14
Pole Placement Example
•  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 closedloop system
•  Decide upon closed-loop pole locations and
determine an equivalent characteristic
equation
•  Equate like coefficients of the characteristic
equations and solve for ki
Amme 3500 : State Space
di = ai + k i+1 i = 0,1, 2,…, n ! 1
•  This will allow us to place all of the closed
loop poles where we want them
Pole Placement
Dr. Stefan Williams
•  and equating terms, we find
ki+1 = di ! ai
•  With characteristic equation of the form
n
sn + dn !1sn !1 + ! + d1s + d0 = 0
Slide 15
•  For the following plant
1
G(s) =
s(s + 2)(s + 10)
•  Design a full state feedback controller to
yield an overshoot of 5% and a settling
time of 1s
Dr. Stefan Williams
Amme 3500 : State Space
Slide 16
Pole Placement Example
Pole Placement Example
•  The OL transfer function for the system is
G(s) =
1
(s + 12s2 + 20s)
3
–  5% OS and
–  settling time of 1s
•  Convert the transfer function to SS
0 $! x1 $ !0 $
! x˙1 $ !0 1
# x˙ & = #0 0
1 &# x 2 & + #0 & u
2
# & #
&# & # &
#" x˙ 3 &% #"0 '20 '12 &%#" x 3 &% #"1 &%
y = [1 0 0]x
Dr. Stefan Williams
Amme 3500 : State Space
•  Decide on desired root locations
•  From specifications, we require
Slide 17
•  Select dominant second order roots to be
at s = !4 ± 4 j
•  Select third root 5 times to left at s = !20
•  Desired characteristic equation:
s3 + 28s2 + 192s + 640
Dr. Stefan Williams
Pole Placement Example
Amme 3500 : State Space
Slide 18
Pole Placement Example
•  Feeding back states we find
•  So
x˙ = Ax + Bu
= Ax + B(!Kx + r)
= (A ! BK ) x + Br
(" 0
1
0
%+
"0 %
$
'
*
sX =
0
0
1
X + $0 'r
'$ '
*$
!K
!(20
+
K
)
!(12
+
K
)
$
'
$#1 '&
)# 1
2
3 &,
•  Solve for the characteristic equation
0 % "0 0
0 %+ " x1 % "0 %
" x˙1 % ( "0 1
$ x˙ ' = * $0 0
1 '! $ 0 0
0 '- $ x 2 ' + $0 ' r
$ 2' *$
' $
'- $ ' $ '
$# x˙ 3 '& ) $#0 !20 !12 '& $#K1 K 2 K 3 '&, $# x 3 '& $#1 '&
s
!1
0
0
s
!1
=0
K1 (20 + K 2 ) s + (12 + K 3 )
s3 + (12 + K 3 )s2 + (20 + K 2 )s + K1 = 0
Dr. Stefan Williams
Amme 3500 : State Space
Slide 19
Dr. Stefan Williams
Amme 3500 : State Space
Slide 20
Pole Placement Example
Pole Placement Example
•  Looks like we
have met the
overshoot and
settling time
requirements
•  However, steady
state error is not
addressed here
•  Equating like terms we find
s3 + 28s2 + 192s + 640
s3 + (12 + K 3 )s2 + (20 + K 2 )s + K1
•  So
K1 = 640
K 2 = 172
K 3 = 16
Dr. Stefan Williams
Amme 3500 : State Space
Slide 21
Dr. Stefan Williams
Steady State Error
* N.S. Nise (2004)
Control Systems
Engineering Wiley &
Sons
Amme 3500 : State Space
Slide 22
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
Dr. Stefan Williams
Amme 3500 : State Space
Slide 23
•  Looking at the previous diagram, we find
x˙ N = e = r ! Cx
•  We can augment the state equations
yielding
! x˙ $ ! A 0 $ ! x $ !B $
!0 $
# x˙ & = #'C 0 & # x & + # 0 &u + #1 & r
" N% "
%" N % " %
" %
!x$
y = [C 0]# &
" xN %
Dr. Stefan Williams
Amme 3500 : State Space
Slide 24
Steady State Error
•  But
•  So
Steady State Error Example
"x%
u = !Kx + K E x N = ![K K E ]$ '
#xN &
•  Let s consider the previous example
where we wanted the roots at
s3 + 28s2 + 192s + 640
•  We now have an additional pole which we
will place at s=100 so the new desired
characteristic equation is
! x˙ $ !A ' BK BK e $! x $ !0 $
+
r
# x˙ & = # 'C
0 &%#" x N &% #"1 &%
" N% "
!x$
y = [C 0]# &
" xN %
•  We can now choose gains to yield desired
pole locations and zero steady state error
Dr. Stefan Williams
Amme 3500 : State Space
Slide 25
Steady State Error Example
•  The system is now characterised by
1
0
0 $+
(! 0
!0 $
#
*
0
1
0 &- ! X $ # &
!X$ # 0
&
-# & + 0 r
s# & = *
" X N % * #'K1 '(20 + K 2 ) '(12 + K 3 ) K e &- " X N % # &
#"1 &%
#
-C *
0
0
0 &%-,
) " '1
•  Finding the eigenvectors of this equation
yields
s4 + (12 + K 3 )s3 + (20 + K 2 )s2 + K1s + K e
Amme 3500 : State Space
Dr. Stefan Williams
Amme 3500 : State Space
Slide 26
Steady State Error Example
•  Equating like terms we find
BKe
Dr. Stefan Williams
s4 + 128s3 + 2992s2 + 19840s + 64000
Slide 27
s4 + 128s3 + 2992s2 + 19840s + 64000
s4 + (12 + K 3 )s3 + (20 + K 2 )s2 + K1s + K e
•  So
K1 = 19840
K 2 = 2972
K 3 = 116
K e = 64000
Dr. Stefan Williams
Amme 3500 : State Space
Slide 28
Steady State Error Example
•  We now
have a
system that
meets all of
our
requirements
•  We haven t
considered
the required
control effort
Dr. Stefan Williams
•  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
Amme 3500 : State Space
Slide 29
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
Dr. Stefan Williams
Observer Design
Dr. Stefan Williams
Amme 3500 : State Space
Slide 30
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.
* N.S. Nise (2004) Control Systems Engineering Wiley & Sons
Amme 3500 : State Space
Slide 31
Dr. Stefan Williams
Amme 3500 : State Space
Slide 32
LQR Control
LQR Control
•  Consider a dynamic system in a state space
form:
•  Then, the state feedback gain that minimizes this
cost function is:
x˙ = Ax + Bu
•  We can set a particular objective/cost function
which governs the performance of the closedloop system: ! T
J = " ( x Qx + uT Ru) dt
u = !(R!1 B T P)x = !Kx
•  P can be obtained by solving the Riccati
equation:
A T P + PA ! PBR !1 B T P + Q = 0
0
•  Here, Q and R are the weighting matrices that
respectively penalize the states x and control
input u (Q!0, R>0).
Dr. Stefan Williams
Amme 3500 : State Space
Slide 33
An Example System
•  In Matlab, the feedback gain can be found via
[K]=lqr(system,Q,R)
Dr. Stefan Williams
Amme 3500 : State Space
Slide 34
Inverted Pendulum
Inverted pendulum – video courtesy of Colorado State University
Dr. Stefan B. Williams
Amme 3500 : Introduction
Slide 35
Dr. Stefan B. Williams
Amme 3500 : Introduction
Slide 36
Inverted Pendulum
State Space Systems
•  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
Dr. Stefan Williams
Amme 3500 : State Space
•  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
Slide 37
Dr. Stefan Williams
Conclusions
Amme 3500 : State Space
Slide 38
Further Reading
•  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.
Dr. Stefan Williams
Amme 3500 : State Space
Slide 39
•  Nise
–  Sections 3.1-3.7 and 12.1-12.8
•  Franklin & Powell
–  Section 2.2 and 7.1-7.6
Dr. Stefan Williams
Amme 3500 : State Space
Slide 40