Final Exam Review

Final Exam Study Guide

Important Topics to Review

Final Exam Dates
1
Final Exam Guidelines




Review the lecture notes and text sections
corresponding to the final exam topics to
be discussed.
Review Homework problems and solutions
Review the posted sample final exam
The final exam will be closed-book but you
are allowed two(2) 8.5x11 sheets (both
sides) of notes
2
1
Final Exam Topics







State-Space Realizations (L3)
Time Response Analysis (L4)
Modal Decomposition (L9-10)
Stability (L11)
Controllability & Observability (L12-14)
State Feedback and Observers (L15-18)
Linear Quadratic Optimal Control (L19-20)
3
Review of Important Topics
Concepts & Examples
4
2
State & Transfer Function Models

State Space Representation of an LTI system:
x  Ax  Bu
y  Cx  Du

Transfer Function 
L(output)
L(input) zero initial conditions
G(s)  C sI  A B  D
1
5
General Linear State Transformation


State and output equations
State transformation xˆ  Tx
x  Ax  Bu

y  Cx  Du
x  Ax  Bu xˆ Tx Tx  TAx  TBu xˆ  TAT-1xˆ  TBu



y  Cx  Du
y  CT-1xˆ  Du
 y  Cx  Du

Transformation Formulas
A,B,C,DAˆ ,Bˆ , Cˆ ,D
T
ˆ  CT 1
ˆ  TAT1, B
ˆ  TB, C
A
6
3
Controllable Realization

General n-th order transfer function:
G(s) 

Y(s)
b s n 1  b2 s n  2    bn 1s  bn
 1 n
U(s)
s  a1s n 1    a n 1s  a n
Controllable Canonical form (CCF):
x  Ac x  Bcu

y  Cc x
1
0 
0 
 0
0

 0
0
0
1 
0 

 
Ac   

 
 , Bc    , Cc  bn bn1  b2 b1


 
0
1 
 0
 0
0
 an  an1   a2  a1
1
7
Observable Realization

General n-th order transfer function:
G(s) 

Y(s)
b s n 1  b2 s n  2    bn 1s  bn
 1 n
U(s)
s  a1s n 1    a n 1s  a n
Observable Canonical form (OCF):
  a1
 a
 2
Ao   

 an1
  an
x  Ao x  Bou

y  Co x
1 0  0
 b1 
b 

0 1  0
 2
    , Bo    , Co  1 0  0 0
 

0  0 1
 bn 
bn1
0  0 0
8
4
Observable Realization
General n-th order transfer function:
G(s) 
Y(s) b1 s n 1  b2 s n  2    bn 1 s  bn

U(s)
s n  a1 s n 1    a n 1 s  a n
Canonical observable form:
  a1
 a
 2
Ao   

 an1
  an
1
0

0
0
0
1






0
0
x  Aox  Bou

y  Cox
0
 b1 
b 

0
 2
  , Bo     , Co  1 0  0 0
 

1
 bn 
bn1 
0
9
Time Response of LTI Systems



x  Ax  B u
LTI System:
t
A t-t 
Time Response:
xt   e
xt 0    e A t- Bud
t
The matrix exponential (eAt)
0
0


is key in computing response
can be computed using Laplace Transform:
e At 
L  1 s I - A  1 
10
5
Jordan Decomposition Theorem (JDT)

Let A be a real nxn matrix with
eigenvalues i of multiplicity mi. Then
there exits an nxn invertible complex
matrix M such that
J1 0
0 J
2
1
M AM  
 

0 

0
i  1i 
0 



2
 
0 i  i
 

, Ji 
     mi 1 
 0
i



0 Jk 
i 
 0  0

where Jordan Blocks Ji are mi x mi
matrices with super diagonal elements
ij=1 or 0.
11
Comments on JDT




A defective matrix can only be block
diagonalized
The columns of modal matrix consists
of regular and generalized eigenvectors
Multiplicity of eigenvalues determines
size of each Jordan block
Number of super-diagonal 1’s of block i
is equal to number of generalized
eigenvectors for i
12
6
State Transition Matrix: eAt

State Transition Matrix:

Let M be the modal matrix of A. Then
(t , t 0 )  e A (t t0 )
A n  MJ n M  e A t  M e J t M  1
e At
 e J1t

 M





 M 1



e Jm t 
0
e
J2t
0
13
Exponential of Jordan Block

Given mxm Jordan block

Laplace Transform:

0

J 

0
0
1 0  0
 1  0
   

 0  1
 0 0  
e Jt  L1[( s I  J ) 1 ]
e t

0
Jt
e  

0
0

te t
t 2e t 2! 
e t
te t





0
e t

0
0
t m 1e t (m  1)!

t m 2e t (m  2)!



t
te

t

e

14
7
Exponential of Complex Jordan Block
Using Laplace Transformation

J
 

 e t cos  t e t sin  t 
Jt
 e   t

t
 
  e sin  t e cos  t 
15
Example
0
0
A
0

0
Find exp(At) of
Solution:
0
0
A
0

0
e At
1
0

0

0
0
0
1

0  1 0
1
0
0
0
0
0
  0 1 t
0
e  0 0 
0
At
e 
0 0
0 0 1


0  1 0
 0 0
1
0
0
0
0
t
1
0
0 cos t
0  sin t
0 0 

0 0 
 0 1 

t
e  1 0  

0 
0 
sin t 

cos t 
16
8
Stability of Continuous-Time LTI Systems

Theorem: The LTI system x  Ax is
stable iff



all the eigenvalues of A have non-positive
real parts
imaginary eigenvalues are distinct
It is asymptotically stable iff all
eigenvalues of A have negative real
parts
17
Illustration of Stability Criteria
Im
X
X
X
X
X
Re
X
X
X
X
X
X: Unstable, X: Stable,
X: Asy. Stable
18
9
Controllability & Observability


Controllability: Any initial state can be
transferred to any final state in finite
time by means of a control input
Observability: The initial state can be
uniquely determined from the output
and input measurements for any input
including 0.
19
Rank Test
Consider the LTI system
 (A,B) is controllable iff

rank B

AB

x  Ax  Bu

y  Cx  Du

A n 1B  n
(A,C) is observable iff
 C 


  CA  
rank
n
  


 CAn1 


20
10
PBH Rank Test
Consider the LTI system
 (A,B) is controllable iff
x  Ax  Bu

y  Cx  Du
ranksI  A  B  n, for all complex s

(A,C) is observable iff
 sI  A 


rank      n, for all complex s
 C 


21
Transfer Function Test
If the system
x  Ax  Bu

 y  Cx
is uncontrllable and/or unobservable then
its transfer G( s )  C( sI  A) B function has
pole-zero cancellations.
1
22
11
State Feedback Control Block Diagram


r
State Feedback Control:
Block Diagram
u  Kx  r
y
x
u

B
C
A
-K
23
Closed-loop System



Plant:
x  Ax  Bu
y  Cx
u  Kx  r
Control:
Closed-loop System:
x  A  BK x  Br
y  Cx

Open-loop numerator
Closed Loop TF: YR((ss ))  C(sI  A  BK )
1
B
N (s )
des (s )
24
12
Integral Control

t
Control law:
u  K s x   K I e(t ) dt
0
yd
Ki
y
x
u

B
Integral
controller

C
A
plant
-Ks
Automatically generates reference input r!
25
Closed-Loop Integral Control System

Plant:
x  Ax  Bu  w
y  Cx
t



u  K s x   K I e(t ) dt , e  y  y d
Control:
0
t
x I   e(  ) d 
Integral state:
0
Closed-loop system
d
dt
 x   A 0  x  B
x   C 0 x    0 K s
 I   
 I 
x  w 
K I    

 x I   y d 
26
13
Steady State Error

Observation: If the control gains are
chosen such that the closed-loop
system is asymptotically stable, then all
the state variable must reach a constant
value since the inputs w and yd are
constant. Thus
lim
t  
e (t )  0
27
Pole-Placement with Integral Control




Need to find K=[Ks KI] to place the
eigenvalues of A  B K
where
A 0
B 
A  
C
0 
,B   
0 
Pole placement technique can be used
to find K.
Matlab commands:


SISO Systems:
MIMO Systems:
K  acker
A , B , P 
K  place
A , B , P 
d
d
28
14
Pole Placement Problem

Choose the state feedback gain to place
the poles of the closed-loop system, i.e.,
Eigenvalue s of A : A  BK


At specified locations
Computation of K:


Direct method
Bass-Gura Formula:
des
, , des
1
n
 1
 
 0
K  ( d  a )T  P 

 
 0
 
a1

1

0



a n 1  

a n  2  
  

1  
1
29
Luenberger Observer
x  Ax  Bu




Plant
y  Cx
Observer:
xˆ  Axˆ  Bu  Ly  Cxˆ 
Closed-loop System: ~x  A  LC ~x, ~x  x  xˆ
If eigenvalues of A-LC are in LHP then
lim ~
x(t )  0  xˆ (t )  x(t ) as t  
t 

Observer pole placement problem is
equivalent to state feedback pole
placement of (AT,CT)
30
15
Observer Block Diagram
u
y
System
L
x

B
A
C
y
Observer
31
Observer Transfer Function

Observer State Equation:
xˆ  Axˆ  Bu  Ly  Cxˆ 
x
u
y


Observer
Sate-Space Realization: ss( A, [B L], I,0 )
Transfer Function:
Xˆ (s )  Gu (s )U(s )  G y (s )Y(s )
Gu (s )  sI  A  B,
1
A  A  LC
G y (s )  sI  A  L
1
32
16
State Equation Form for Reduced
Order Observer (ROO)

System State Equation Form:
 x 1   A 11 A 12   x 1 
x   A
  
 2   21 A 22   x 2 
x 
y  Iq q 0  1   x 1
x 2 
d
dt

ROO Summary:
z  A r z  Buu  B y y
xˆ 2  z  Ly

B 1 
B u
 2
A r  A 22  LA12
Bu  B2  LB1
B y  A r L  A 21  LA11
ROO State Realization: ss(Ar,[Bu By],I,[0 L])
33
Combined State Feedback Controller
and Observer

Combined State Feedback Controller and
Observer Formulation

Separation Principle

Transfer Function Representation


Combined State Feedback Controller and
Reduced Order Observer
Illustrative Examples
34
17
Motivation



In most control applications all state
variables are not measurable
A full or reduced order observer may be
used to estimate needed states
Separation principle allows independent
controller and observer design
35
Combined Controller-Observer Formulation

Plant:
x  Ax  Bu

Controller:
u  Kxˆ  r

Observer:
xˆ  Axˆ  Bu  Ly  Cxˆ 
y  Cx
Open-loop numerator

Closed-Loop TF:
Y (s )
N (s )

R ( s )  des ( s )
36
18
State Feedback Observer Block Diagram
u
r
y
System
L

B
A
x
C
y
Observer
K
37
Closed-Loop System



Closed-Loop Control
Subsystem:
Closed-Loop Observer
Subsystem:
Overall Closed-Loop
System:
x  Ax  B  Kxˆ  r 
~  Br
x  A  BK x  BKx
xˆ  Axˆ  Bu  L y  Cxˆ 
~  A  LC x
~
x
d x A  BK  BK  x B
~   0
~   0 r
A  LC x
dt x
  
 
38
19
Separation Principle

Eigenvalues of the closed loop systems
are the union of eigenvalues of



Closed-loop poles, i.e., eigenvalues of ABK and
Observer poles A-LC
This is known as the separation
principle
39
Block Digaram Representation

Block Diagram
Y
R
P
F1
F2

Feedforward and Feedback Blocks:


1
F1(s )  I  K(sI  A )1B , F2 (s )  K(sI  A )1L

Closed-Loop TF
Gclosed (s )  I  PF1F2  PF1  CsI  A  BK  B 
1
1
N (s )
 des (s )
40
20
State Realizations of F1 and F2
F2 (s )  K(sI  A )1L  ss( A,L,K,0 )
F2:
 What is the state space realization of F1
1
F1(s )  I  F1(s ) , F1(s )  K(sI  A )1B  ss( A,B,K,0 )


F1(s ) :
v
ss( A  BK,B,K, I)
z
F’1
u
41
Combined Controller Reduced Order
Observer (CCRO)

Plant State Equation:
 x 1   A 11 A 12   x 1   B 1 
x   A
     u
 2   21 A 22   x 2  B 2 
x 
y  I q q 0  1   x 1
x 2 
d
dt

Reduced Order Observer
z  A r z  B u u  B y y
xˆ 2  z  Ly
A r  A 22  LA 12
B u  B 2  LB 1
B y  A r L  A 21  LA 11
42
21
Transfer Function Block Digaram
Representation of CCRO

Block Diagram
Y
R
P
F1
F2

Block and Closed-Loop TF:

F1 ( s )  1  K 2 ( s I  A r ) 1B u

1
F2 ( s )  K 2 ( s I  A r ) 1B y  K 1  K 2L r
Gclosed (s )  I  PF1F2  PF1  CsI  A  BK  B
1

1
F1 and F2 are of order n-q (lower than FOO)
43
Alternative Output Feedback Control:
Polynomial Design

The block diagram representation of the
combined observer and state feedback
controller raises the possibility of a direct
s-plane design approach:
Y
R
P=N(s)/M(s)
F1
F2

F1(s ) 
( s )
s n 1     n 2s   n 1

(s )  0s n 1     n 2s   n 1
F2 (s ) 
(s ) 0s n 1     n 2s  n 1
 n 1
( s )
s     n 2s   n 1
Indeed for SISO systems, we can find the
transfer functions F1 and F2 directly.
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22
Determination of Compensation Coefficients



Choose the reduced observer characteristic
polynomial (z).
Solve the Diophantine equation
(s)A(s)+(s)B(s)=D(s)= (s)des(s)
for (s) and (s), where des(s) is the desired
closed-loop characteristic equation.
The Bezout equation may be rearranged as
shown next.
45
Linear Quadratic (LQ) Optimal Control

Given continuous-time state equation
x  Ax  Bu

Find the control function u(t) to
minimize
tf


J ( x, t0 )  x (tf )Sx(tf )   xT (t )Qx(t )  uT (t )Ru(t ) dt
T
0
S, Q  0, R  0 and symmetric
46
23
Summary of LQR Solution



x  Ax  Bu
Plant:
Control law: ut   K t x t ,
Ricatti Equation:
K t   R 1 BT P t 
P  PA  A T P  Q  PBR  1B T P  0, P ( t f )  S

J  x , t   x T P ( t ) x
Cost-to-go:
47
Infinite Horizon LQ

Given continuous-time state equation
x  Ax  Bu

Find the control function u(t) to minimize



J ( x, t0 )   xT (t )Qx(t )  uT (t )Ru(t ) dt
0
Q  0, R  0 and symmetric

Solution is obtained as the limiting case
of Ricatti Eq.
48
24
Summary of Steady-State LQR Solution



x  Ax  Bu
Plant:
u  Kx , K  R 1 BT P
Control law:
Algebraic Ricatti Equation (ARE):
PA  A T P  Q  PBR  1B T P  0

Cost-to-go:
J  x   x T Px
49
Existence Conditions


Theorem: If (A,B) is stabilizable and (A,Q1/2)
is detectable (note: Q=QT/2Q1/2) then the
ARE has a unique positive definite solution.
Reason:


(A,B) stabilizable implies J* (hence P) is finite
(A,C) detectable implies J* >0 for nonzero x(0)
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