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I
Contents 1 Lecture 1 .......................................................................................................................... 1
Introduction to Design of Physical Systems ............................................................ 1 1.1
System Classification ............................................................................................... 2 1.2
Applications ............................................................................................................. 3 1.3
I/O Description of Dynamical Systems.................................................................... 4 1.4
Definition: Linearity ................................................................................................. 5 1.5
2 Lecture 2 .......................................................................................................................... 6
Definition: Causality ................................................................................................ 7 2.1
Definition: Time-Invariance ..................................................................................... 8 2.2
3 Lecture 3 ........................................................................................................................ 11
Impulse Response................................................................................................... 11 3.1
Computation of the Convolution Integral for Linear Systems ............................... 13 3.2
3.2.1 For a Causal, Linear, Relaxed, and Time-Varying System ............................ 13
3.2.2 For a Linear, Relaxed, Causal, Time-Invariant System .................................. 14
4 Lecture 4 ........................................................................................................................ 15
Computation of Impulse Response Using Differential Equation ........................... 15 4.1
Computation of Impulse Response Using Laplace Transform .............................. 18 4.2
5 Lecture 5 ........................................................................................................................ 21
State Diagram ......................................................................................................... 21 5.1
5.1.1 Concept of State .............................................................................................. 24
5.1.2 State Vector ..................................................................................................... 25
5.1.3 State Space ...................................................................................................... 25
5.1.4 State Trajectories ............................................................................................ 25
General State Space Representation ....................................................................... 26 5.2
II
6 Lecture 6 ....................................................................................................................... 27
State Models .......................................................................................................... 27 6.1
Canonical Representations .................................................................................... 31 6.2
6.2.1 Observable Canonical Form ........................................................................... 31
7 Lecture 7 ....................................................................................................................... 33
Controllable Canonical Form ................................................................................ 33 7.1
Jordan Canonical Form .......................................................................................... 34 7.2
7.2.1 Characteristic Polynomial D(s) = 0 Contains N Distinct Poles ..................... 35
7.2.2 Characteristic Polynomial D(s) = 0 Contains Multiply Repeated Poles ........ 36
8 Lecture 8 ....................................................................................................................... 39
Transformation From State Equations to Transfer Function ................................. 39 8.1
Examples for C.C.F., O.C.F., J.C.F. ...................................................................... 42 8.2
8.2.1 Observable C.F. ............................................................................................. 42
8.2.2 Controllable C.F. ........................................................................................... 42
8.2.3 Jordan Canonical Form .................................................................................. 43
9 Lecture 9 ....................................................................................................................... 46
Solution of State Equation ..................................................................................... 46 9.1
Fundamental Matrix .............................................................................................. 48 9.2
9.2.1 Definition: ...................................................................................................... 48
9.2.2 Examples: ....................................................................................................... 48
State Transition Matrix .......................................................................................... 50 9.3
9.3.1 Definition ....................................................................................................... 50
9.3.2 Proposition ..................................................................................................... 50
9.3.3 Properties of ),( 0tt .................................................................................. 50
10 Lecture 10 ..................................................................................................................... 53
State Transition Matrix for LTI Systems ............................................................... 53 10.1
III
10.1.1 Definition ........................................................................................................ 53
10.1.2 Approaches to Compute )(t ....................................................................... 54
Cayley-Hamilton Technique .................................................................................. 56 10.2
10.2.1 Cayley-Hamilton Theorem ............................................................................. 56
10.2.2 Cayley-Hamilton Technique ........................................................................... 57
10.2.3 Algorithm to Apply Cayley-Hamilton Technique .......................................... 57
11 Lecture 11 ...................................................................................................................... 61
Ф(t, t0) for LTI System ........................................................................................... 61 11.1
Ф(t, t0) for LTV Systems ........................................................................................ 65 11.2
Solution of the State Equations and Output Equation ............................................ 69 11.3
12 Lecture 12 ...................................................................................................................... 70
Definitions .............................................................................................................. 70 12.1
12.1.1 Controllability ................................................................................................. 70
12.1.2 Observability ................................................................................................... 70
12.1.3 Examples for Controllability and Observability ............................................. 71
Canonical Transformations .................................................................................... 72 12.2
12.2.1 A Canonical Decomposition of Dynamic System .......................................... 72
12.2.2 Controllable Canonical Form (C.C.F.) ........................................................... 73
12.2.3 Observable Canonical Form (O.C.F.) ............................................................. 74
13 Lecture 13 ...................................................................................................................... 78
Diagonalization of a Matrix ................................................................................. 78 13.1
13.1.1 Jordan Canonical Form (J.C.F.) – Distinct Eigenvalues Case ........................ 78
13.1.2 Jordan Canonical Form (J.C.F.) – Repeated Eigenvalues Case ...................... 81
13.1.3 Matrix Reduction to Jordan Form ................................................................... 83
Controllability of JCF and Observability of JCF ................................................... 87 13.2
13.2.1 Theorem .......................................................................................................... 87
IV
13.2.2 Examples ........................................................................................................ 88
14 Lecture 14 ..................................................................................................................... 92
Stable Feedback: General Case ............................................................................. 92 14.1
State Estimation: Full-Order Observer .................................................................. 95 14.2
14.2.1 Theorem ......................................................................................................... 95
14.2.2 Design of the State Estimator ......................................................................... 97
14.2.3 Closed-Loop System Characteristics ........................................................... 100
Reduced-Order Estimators .................................................................................. 101 14.3
Reduced-Order State Estimator: General Case ................................................... 104 14.4
15 Lecture 15 ................................................................................................................... 106
Tracking Problem ................................................................................................ 106 15.1
Closed-Loop Pole-Zero Assignment ................................................................... 108 15.2
General Case Using Controllable Canonical Form ............................................. 110 15.3
V
Table of Figures
Figure 1-1. Vehicle suspension system. ............................................................................... 2
Figure 1-2. System classification. ........................................................................................ 2
Figure 1-3. Applications #1. ................................................................................................ 3
Figure 1-4. Applications #2. ................................................................................................ 3
Figure 1-5. Applications #3. ................................................................................................ 3
Figure 1-6. Applications #4. ................................................................................................ 4
Figure 1-7. Applications #5. ................................................................................................ 4
Figure 1-8. Input/output system representation. .................................................................. 4
Figure 1-9. RC network. ...................................................................................................... 4
Figure 2-1. Time varying systems ....................................................................................... 8
Figure 3-1. Delta function. ................................................................................................. 11
Figure 3-2. Impulse at t = τ. ............................................................................................... 11
Figure 3-3. Property of step function. ................................................................................ 12
Figure 3-4. Behavior of rectangular pulse. ........................................................................ 12
Figure 3-5. Multiplication of a function with a delta function. ......................................... 12
Figure 3-6. Causal versus non-causal system. ................................................................... 13
Figure 3-7. Delta function and its shifted value. ................................................................ 14
Figure 4-1. Unit impulse at t = 0 , 0, 0+. .......................................................................... 15
Figure 4-2. Mass-spring-dashpot system. .......................................................................... 17
Figure 5-1. Basic state diagram operations. ....................................................................... 21
Figure 5-2. Non-minimal internal state representation. ..................................................... 21
Figure 5-3. External representation. ................................................................................... 22
Figure 5-4. Minimal internal state representation. ............................................................. 22
Figure 5-5. System state diagram for example 2. .............................................................. 22
Figure 5-6. System state diagram for example 3. .............................................................. 23
Figure 5-7. System state diagram for example 4. .............................................................. 23
Figure 5-8. RLC network. ................................................................................................. 24
Figure 5-9. System state trajectory for example 1. ............................................................ 25
Figure 6-1. State diagram for a linear time invariant system. ............................................ 27
VI
Figure 6-2. RC network. ................................................................................................... 28
Figure 6-3. Mass-Spring system. ....................................................................................... 29
Figure 6-4. State diagram. ................................................................................................. 31
Figure 7-1. Decomposition of the transfer function. ......................................................... 33
Figure 7-2. State diagram. ................................................................................................. 35
Figure 7-3. State diagram. ................................................................................................. 35
Figure 7-4. Parallel configuration. .................................................................................... 35
Figure 7-5. State diagram of the term in bracket ............................................................... 37
Figure 8-1. State diagram for example 2. .......................................................................... 42
Figure 8-2. Transfer function decomposition for example 3. ............................................ 42
Figure 8-3. System state diagram. ..................................................................................... 43
Figure 8-4. State diagram. ................................................................................................. 44
Figure 8-5. Armature controlled DC motor. ...................................................................... 45
Figure 9-1. Transition property. ........................................................................................ 51
Figure 10-1. System plot. .................................................................................................. 60
Figure 12-1. State trajectory as a result of the control u. .................................................. 70
Figure 12-2. System diagram. ........................................................................................... 70
Figure 12-3. System classification. ................................................................................... 72
Figure 13-1. Block diagram. .............................................................................................. 89
Figure 14-1. System state diagram. ................................................................................... 92
Figure 14-2. System plot for example 2. ........................................................................... 95
Figure 14-3. Block diagram of combined plant & observer. ............................................. 95
Figure 14-4. Plant and observer diagram. ......................................................................... 96
Figure 14-5. Block diagram for example 4. ...................................................................... 99
Figure 14-6. Signal flow graph. ......................................................................................... 99
Figure 14-7. Estimator response. ....................................................................................... 99
Figure 14-8. Block diagram of the combined plant and reduced order observer. ........... 102
Figure 14-9. System signal flow graph. .......................................................................... 103
Figure 14-10. Closed-loop combined plant and controller estimator. ............................. 104
Figure 14-11. Block diagram for system with reduced order estimator. ......................... 105
Figure 15-1. State feedback diagram. .............................................................................. 106
Figure 15-2. State diagram for example 1. ...................................................................... 107
VII
Figure 15-3. ADC motor system schematic. .................................................................... 108
Figure 15-4. System close-loop block diagram. .............................................................. 108
Figure 15-5. System block diagram. ................................................................................ 108
Figure 15-6. Root Locus. ................................................................................................. 109
Figure 15-7. State diagram for example 1. ...................................................................... 111
Figure 15-8. Block diagram of the closed-loop system for example 1. ........................... 111
Figure 15-9. Closed-loop diagram 3 for a unity feedback of example 1. ........................ 111
Figure 15-10. Root Locus for the standard feedback control system of example 2. ....... 112
Figure 15-11. Root Locus for the state feedback control system of example 2. .............. 112
Figure 15-12. System sate diagram for example 3. ......................................................... 113
Figure 15-13. Closed-loop block diagram for example 3. ............................................... 113
Figure 15-14. Root Locus for example 3. ........................................................................ 113
Figure 15-15. Root Locus for example 3. ........................................................................ 114
Figure 15-16. System state diagram for example 4. ........................................................ 115
Figure 15-17. Closed-loop block diagram for example 4. ............................................... 115
Figure 15-18. Root Locus for example 4. ........................................................................ 115
1
Objectives:
1) Introduction to design of physical systems
2) System classification
3) Applications
4) I/O description of systems
5) Linearity
1 Lecture 1
Introduction to Design of Physical Systems 1.1
a) Performance Specification
b) Modeling
A tool for study
A representation of the physical system
ODE or difference equations
c) Simulation
A tool for studying behavior of ODE’s or difference equations
d) Analysis
Qualitative: stability, controllability, observability
Quantitative: simulation
e) Optimization/control
Optimize parameters of the system
f) Realization
Page 2 of 115
2 Linear system: Lecture 1
Example 1: Vehicle suspension system
System Classification 1.2
Figure 1-2. System classification.
Seat Mass
Mass of Body
Wheel Mass
Road Profile
Seat springness
Seat damping
Spring Shock absorber
x1
x2
x3
Tires springness
f(t)
System classes
Distributed parameter Lumped parameter
Stochastic Deterministic
Continuous time Discrete time
Nonlinear Linear
Time-varying Time invariant
Figure 1-1. Vehicle suspension system.
Page 3 of 115
1.3 Applications 3
Linear spring: Nonlinear spring:
21
212,121
222
111
)(
ff
xxkfxx
kxfx
kxfx
21
3
212,121
3
222
3
111
)(
ff
xxkfxx
kxfx
kxfx
Applications 1.3
a) Deterministic Control Problem
Figure 1-3. Application #1.
b) Estimation Problems
Figure 1-4. Application #2.
Internal representation: State characterization
c) Stochastic Control Problem
Figure 1-5. Application #3.
Known plant x
Unknown
input
y
Output Sensor
Parameters known I/O description known
z
Measured
output
Known plant x
Known
input
y
States
(unknown)
Sensor z
Measured
output
Known plant x
Unknown
input
y
Output Sensor
z
Measured
output Noise input Noise input
Page 4 of 115
4 Linear system: Lecture 1
d) System Identification
Figure 1-6. Application #4.
e) Adaptive Control Problem
Figure 1-7. Application #5.
I/O Description of Dynamical Systems 1.4
Figure 1-8. Input/output system representation.
Figure 1-9. RC network.
oo
i edt
deRCe
Unknown
plant x
Unknown
input
y
Output Sensor
z
Measured
output
Noise Noise
Parameters unknown, however I/O description known
Unknown
plant
x
Unknown
input
y
Output Sensor
z
Measured
output
Noise Noise
Plant
(system)
L{.}
input u(t)
output y(t)
y(t) = L{u(t)} (*) ,
Under what conditions (*) holds.
R
C
+
-
+
-
i
dt
deci
c
i
dt
de oo
idtc
e
idtc
Rie
o
i
1
1
Page 5 of 115
1.5 Definition: Linearity 5
1) 0ie and 0)0( oe zero input response
2) 0ie and 0)0( oe zero state response
3) 0ie and 0)0( oe general response
Case 1: 00 oo
i edt
deRCe
RCt
oo eete /)0()(
(zero input response)
Case 2: 0)0( oe and Aei (constant) 0)0(; ooo eAe
dt
deRC
RCt
o eAte /1)(
(zero state response)
Case 3: RCtRCt
oo eAeete // 1)0()(
(general response)
Relaxed (at rest):
A system is relaxed (or at rest), if the initial conditions (i.c.) are set to zero.
Definition: Linearity 1.5
A relaxed system is said to be linear, if and only if
)}({)}({)}()({ 22112211 tuLatuLatuatuaL
(Superposition principle)
Condition for defining the I/O
map
6
Objectives:
1) Examples
2) Introduction to causality and time-invariance system
2 Lecture 2
Examples of linear systems :
Example 1 : RC net.
ioo ee
dt
deRC io
o eRC
eRCdt
de 11 (*)
Since Integrating factor o
RCtoRCt
o
RCt eeRCdt
deeee
dt
d /// 1 (**)
(*)(**) i
RCt
o
RCt eeRC
eedt
d // 1
0
/
0
/ 1dtee
RCdtee
dt
di
RCt
o
RCt
0
/0/ 1)0()( dtee
RCeeee i
RCt
oo
RC
Since the system is relaxed 0)0( oe
0
// 1)( dtee
RCee i
RCt
o
RC
0
//1)( dteee
RCe i
RCtRC
o
0
/)(1)( dtee
RCe i
RCt
o
(I/O representation)
Page 7 of 115
2.1 Definition: Causality 7
Q: Show if the above I/O representation is linear?
To show this, we use superposition principal
2
02
1
01 ; eeee LL
0
1
/)(1
0
1)( dtee
RCe RCt
;
0
2
/)(2
0
1)( dtee
RCe RCt
02211 eeaea L
0
2211
/)(
0 )(1
)( dteaeaeRC
e RCt
0
11
/)(1dteae
RC
RCt
0
22
/)(1dteae
RC
RCt
2
02
1
010 eaeae The system is linear.
Example 2: System t
duty0
)()( , is this system linear?
2211 yuyu ; t
duy0
11 ; t
duy0
22
yuauau 2211
2211
0
22
0
11
0
2211 )( yayaduaduaduauay
ttt
Therefore, the system is linear.
Example 3: t
duty0
2 )()( is nonlinear
Definition: Causality 2.1
The system is said to be causal if the output at time t does not depend on the input at times
after t.
A system which is not causal is said to be noncausal or anticipatory.
Page 8 of 115
8 Linear system: Lecture 2
Definition: Time-Invariance 2.2
Example 4: RC net.
t
i
RCt deeRC
te0
/)(
0 )(1
)(
Show if this system is time varying or not.
Change )()( Tee ii
t
i
RCt dTeeRC
te0
/)(
0 )(1
)(
Let TT , dd
Tt
T
i
RCTt deeRC
te )(1
)( /)(
0
Since 0)( ie for all 0 !
Tt
i
RCTt deeRC
te0
/)(
0 )(1
)(
)()(0 Ttete o
The system is time-invariant
L u y
u(t)
t
y(t)
t 0 0
t
t 0 0
Time-invariant
t
t 0 0
Time-varying
Figure 2-1. Time varying systems.
Page 9 of 115
2.2 Definition: Time-Invariance 9
Example 5: )()(1
)( tutyt
ty
Is this system time-invariant or time-varying?
Is this system linear or nonlinear?
In general, for ( )y a t y , by using integrating factor da
e)(
,
since
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
a t dt a t dt a t dtdy t e y t a t e y t e
dt
y t ay t u t y t u t ay t
dttadtta
ueetydt
d )()(
)(
For this example t
ta1
)( teee tt
dtdtta
ln)(
)( )(1)( tutyttytytdt
d
so,
bb
utdttydt
ddt
b
utbbyy dt )()(
For a relaxed system, y(b) = 0
bb
ut
uty dt dt 1
)(
L u y
y(t)=L{u(t)}
b is the initial time
Page 10 of 115
10 Linear system: Lecture 2
Q: Linear or nonlinear?
t
b
dut
ty
)()(
11 yu t
b
dut
y
11 ; 22 yu t
b
dut
y
22
yuaua 2211
t
b
t
b
t
b
yayaduat
duat
duauat
y 221122112211 )(
Therefore, the system is linear.
Q: Time-invariant or time-varying?
t
b
dut
ty
)()( , y(t) = L{u(t)}
)()( Tuu Due to the shifted input, the resulting output y1 is
t
b
dTut
ty
)()(1 Question: Is ?)()(1 Ttyty
Tt
b
duTt
Tty
)()(
Chang of variable ddTT , (T is constant)
at Tbb ; at Ttt
Tt
Tb
dut
Tty
)()(1
)()(1 Ttyty
The system is time-varying
11
Objectives
1) Impulse Response of LTI and LTV systems
2) Computation of the Convolution Integral for Linear Systems
3 Lecture 3
ubdt
udb
dt
udbya
dt
yda
dt
yda
m
m
mm
m
mn
n
nn
n
n 01
1
101
1
1
This is a linear system since y… )(ny and u…
)(mu enter linearly.
If the coefficients 0aan and 0bbm are constant, then the system is time-invariant.
If they are time dependent, then the system is time-varying.
A nonlinear example is yy ,2y ,
2y , uy, yu , ....
Impulse Response 3.1
Let ),( th be the response of the system at time t due to an impulse applied at time .
),( th is called the impulse response of the system.
0
t
Unit impulse
0 t
Figure 3-1. Delta function.
Figure 3-2. Impulse at t = τ.
Page 12 of 115
12 Linear system: Lecture 3
Figure 3-3. Property of step function.
Figure 3-4. Behavior of rectangular pulse.
In the limit as )]()([1
lim)(00
tutut
For a given u , we can always write
dtutu )()()( , since
)()()()( 000 xxxfxxxf
)(
)()(
)()()(
1
tu
dttu
dttutu
Figure 3-5. Multiplication of a function with a delta function.
0 t
0 t
0 t
f(x)
x
Page 13 of 115
3.2 Computation of the Convolution Integral for Linear Systems 13
Computation of the Convolution Integral for Linear 3.2
Systems
The output y(t) due to u(t) is )}({)( tuLty
dtuLty )()()( L is an operator in time t
dutLdtuLty
th
)()()()()(
),(
dthuty ),()()(
(General Convolution Integral for a Linear, Relaxed LTI or LTV systems).
3.2.1 For a Causal, Linear, Relaxed, and Time-Varying System
For a causal, linear, relaxed, and time-varying system we now get
0
),()(),()()(
t
t
dthudthuty
t
dthuty ),()()(
For a causal system 0),( th for all t
Figure 3-6. Causal versus non-causal system.
0 t
0 t
noncausal
t
causal
Page 14 of 115
14 Linear system: Lecture 3
If the input is applied at time 0tt and is zero before that, then
t
t
dthuty
0
),()()(
Let h(t) be the impulse response due to )(t , ),()( tht
Figure 3-7. Delta function and its shifted value.
For a time-invariant system )()0,()( ththt
Example 1:
are time-varying
While 2)( t
, are time-invariant
3.2.2 For a Linear, Relaxed, Causal, Time-Invariant System
For a linear, relaxed, causal, time-invariant system, we now get
t
dthuty0
)()()(
We want to show that alternatively this is the same as
t
dtuhty0
)()()(
Let tddt ,
t
tt
dtuh
dtuhdhtuty
0
00
)()(
)()())(()()(
0
)()0,()( ththt
15
Objectives
1) Impulse response computation using differential equation
2) Impulse response computation using Laplace transform
4 Lecture 4
Computation of Impulse Response Using Differential 4.1
Equation
For a given general system given by,
ubdt
udb
dt
udbya
dt
yda
dt
yda
m
m
mm
m
mn
n
nn
n
n 01
1
101
1
1
to find the impulse response, let u(t) be an impulse.
Assuming a’s and b’s are constant this is a linear time-invariant system
System is assumed to be relaxed, that is initial conditions (I.C.’s) are zero at .
Figure 4-1. Unit impulse at t = 0 , 0, 0+.
0
u(t)
1
t
Page 16 of 115
16 Linear system: Lecture 4
Example 1: For RC network we have: ioo ee
dt
deRC . Let us find the impulse response:
For finding the impulse response )()( 0 thetei
(**) 0dt
dhRC 0 t
(*) )(dt
dhRC 0 t
hfor
thfor
From (**) the system is relaxed 0)0(
oe
)0(0)(0)0( tthh
from (*), integrate it from 0t to
0t . We do this to get the initial
condition at 0t , in other words, we need to find )0( h .
10)0()0()(
0
0
0
0
0
0
0
hhRCdtthdtdtdt
dhRC
RChRChRCh
1)0(1)0(0)0(
For t > 0, the system becomes 0 hdt
dhRC with
RCh
1)0(
Therefore, 01
)( / teRC
th RCt
Since h(t) does not contain an impulse, therefore 0
0
0-
dth
So, {
Page 17 of 115
4.1 Computation of Impulse Response Using Differential Equation 17
Example 2: For the system FKxxBxM ,
find impulse response.
Since this system is relaxed
0)0()0( xx
We need )0( x and )0( x for )(tF
00
)(0
KhhBhMtfor
tKhhBhMtfor
for t < 0 0)0()0( hh h(t) = 0, t < 0
for t ≥ 0 integrate from 0t to
0t
∫
∫
∫ ∫
10)0()0()0()0( hhBhhM
Since 1)0()0( BhhM
By double integrating the equation from 0t to
0t
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
dtdtKhdtdtdtdthBdtdthM
000)0()0( hhM
0)0( h , now since 1)0()0( BhhM M
h1
)0(
For t > 0 0 KhhBhM with 0)0( h M
h1
)0(
0 0
0
)(
t
t
th
The explicit solution depends on
particular value of M, B and K.
M
K
B
x
F
Figure 4-2. Mass-spring-dashpot system.
Page 18 of 115
18 Linear system: Lecture 4
Computation of Impulse Response Using Laplace 4.2
Transform
This method is only applicable to LTI systems.
Example 3: ioo ee
dt
deRC Find impulse response using Laplace transform (L.T.)
method.
Let )(tei )(theo thdt
dhRC
Take L.T. of both sides
hdt
dhRCL
0
)()()( dtetxsXtx stL 1)()0()( sHhssHRC
Since the system is relaxed 0)0( h 1HRCsH RCs
sH
1
1)(
01
)()( /1 teRC
sHLth RCt
For system
t
duthty0
)()()( , take L.T.
t
duthLtyL0
)()()(
0 0
)()()( dtduthesY
t
st
Since the system is causal
)(
00
0
)(
00 0
)()()()(
)()()()()(
sU
ss
esH
stst
duesHduesH
dudtethdtduthesY
s
Y(s) = H(s)∙U(s) )(
)()(
sU
sYsH
Page 19 of 115
4.2 Computation of Impulse Response Using Laplace Transform 19
Example 4: For system uuyyy 434
1) Find T.F. and impulse response.
Taking the L.T of the system,
)(5)(4)(3)(4)(2 sUssUsYssYsYs
54)(34)( 2 ssUsssY
34
54
)(
)()(..
2
ss
s
sU
sYsHFT
Impulse response )()( 1 sHLth
34
54)(
2
1
ss
sLth
Using partial fraction expansion
1
1
3
2/7)( 1
ssLth
02
1
2
7)( 3 teeth tt
2) Find the output for a step input using the convolution integral for a
relaxed system.
t
duthty0
)()()()(
00
01)()(
02
1
2
7)( 3
t
ttutu
teeth
s
tt
t
tt deety0
)()(3 12
1
2
7)(
Page 20 of 115
20 Linear system: Lecture 4
tt eety 2
1
6
7
3
5)( 3
Step Response
3) Find the output for a step input using T.F. for a relaxed system.
Y(s) = H(s)U(s)
34
54)(
2
ss
ssH ;
ssU
1)(
34
54)(
2
sss
ssY
tt eety 2
1
6
7
3
5)( 3
4) If y(0) = 1 and 0)0( y , find y(t) for a step input.
uuyyyL 5434
)0()0()(}{ 2 ysysYsyL
)0()(}{ yssYyL
)(5)(4)(31)(401)(2 sUssUsYssYssYs
54)(434)( 2 ssUssssY
2 2
. . 0
4 5 4( ) ( )
4 3 4 3due to i c
s sY s U s
s s s s
22
4 5 4( )
4 34 3
s sY s
s ss s s
tt eety 3
3
5
3
5)(
21
Objectives
1) State Diagram
2) State Space Representation
ʃ x Integrator
Differentiator
a ax Scalar
∑
Summary
×
Multiplier
+ +
x
x
5 Lecture 5
State Diagram 5.1
State diagram basic mathematical operations are
shown in figure 5-1.
Example1: Draw state diagram for system
cudt
dubay
dt
dy .
Solve for dt
dy in terms of the rest.
Minimal Representation:
aycudt
dub
dt
dy
Integrate both sides of the equation
to get
dtadtcudtdt
dubdt
dt
dyy y
dtaycubuy )( state diagram is shown as below.
Figure 5-2. Non-minimal internal state representation.
Figure 5-1. Basic state diagram operations.
b ʃ
c a
u y +
+ -
cu ay
Page 22 of 115
22 Linear system: Lecture 5
Figure 5-3. External representation.
Figure 5-4. Minimal internal state representation.
Example 2: Single Input Single Output (SISO) uuyyy 4853
uuyyy 4853
uuyyy 4853
)38(54 yuyuy
State diagram is drawn as below:
Figure 5-5. System state diagram for example 2.
Example 3: For a multi - input - multi - output(MIMO) system
69
368532
211222
112111
uyyyyy
uuyyyyy y
draw state diagram for it.
H U Y
y 4
u
ʃ
8
+
5
-
∫
+
3
(8u-3y)
+
-
- ʃ
y + c
u
ʃ cu - ay
b bu
∫ +
a
-
ay
cu +
y(0)
Page 23 of 115
5.1 State Diagram 23
69
368532
211222
1122111
uyyyyy
uuyyyyy
211222
1122111
69
368532
uyyyyy
uuyyyyy
96
652383
122212
1211211
yyyuyy
uyyuyyy
Therefore, state diagram of this system is drawn below:
Figure 5-6. System state diagram for example 3.
Example 4: )sin()sin( yuyuyy (this is a nonlinear system)
Figure 5-7. System state diagram for example 4.
3
ʃ
6
+
3
- +
2
+
-
- ʃ
ʃ +
+
-
ʃ
6
8
9
5
-
-
+
+
u +
sin
-
ʃ y
sin(y)
Page 24 of 115
24 Linear system: Lecture 5
5.1.1 Concept of State
Internal Representation (Time) State Equations
External Representation I/O Representation
I/O is partitioned into (a) Convolution Integral (time) and (b) Transfer Function (frequency)
State: The minimal number of variables x(t) that their knowledge at time 0tt , and the
input u(t) for 0tt will completely characterize the response of the system.
Example 5: )(; 0teeedt
deRC oio
o oe is the state
response state zero
/)(
responseinput
/
0
0
)(1
)()(
t
t
i
RCt
zero
RCt
oo deeRC
etete
Example 6:
Figure 5-8. RLC network.
dt
de
dt
diRi
Cdt
idL i
12
2
)(tei ( 0tt ), )( 0ti , )( 0tdt
di should be specified !
States are i(t) and dt
di (*)
dt
dee
RCdt
ed
R
L
dt
de oo
i 1
2
0
2
States are oe and dt
deo (**)
Case (*) and (**) reveal that states are NOT UNIQUE , there is another
choice for the states !
Let 1
dtdv i
v iC dt C
dt
dvRCv
dt
vdLCei
2
2
R
+
-
+
-
i
C L
dt 1
Rie
RiiCdt
diLe
o
v
i
Page 25 of 115
5.1 State Diagram 25
In order to solve for v(t) , we need )( 0tv , )( 0tdt
dv and ie , since
C
tit
dt
dv )()( 0
0
so, to solve the system we need )( 0tv and )( 0ti .
Therefore, for an RLC network the voltage across the capacitor and the current
through an inductor are the states.
5.1.2 State Vector
Given n states we define the state vector as an (n×1)-dimensional column vector
1
2
1
)(
)(
)(
)(
nn tx
tx
tx
tx
, nRtx )(
5.1.3 State Space
The space where the states are evolving in.
5.1.4 State Trajectories
0 0( ) ; , ( ), ( )x t f t t x t u t
Example 7: 023 yyy
Let
ytx
ytx
)(
)(
2
1,
2
1
x
xx is the state vector, 2Rx is the state space.
)()( 12
2
1 txececty tt with c1 and c2 depending on )( 0ty and )( 0ty
)(2)( 22
2
1 txececty tt
tt
tt
ecectx
ecectx
2
2
12
2
2
11
2)(
)(
let 00 t , y(0) =0 and 1)0( y
tt
tt
eetx
eetx2
2
2
1
2)(
)(
Figure 5-9. System state trajectory for example 1.
(
(t = 0)
Page 26 of 115
26 Linear system: Lecture 5
General State Space Representation 5.2
( ) ( ( ), ( ), ) Equations
( ) ( ( ), ( ), ) Equations
x t f x t u t t State
y t g x t u t t Output
n dim state
m dim output
diminput
n
m
q
x R
y R
u R q
This is a nonlinear, time varying, causal system.
Nonlinear, time invariant, causal:
))(),(()(
))(),(()(
tutxgty
tutxftx
Linear, time varying system:
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
x t A t x t B t u t
y t C t x t D t u t
MIMO system
nn x
x
x
xn
x
x
x
x
2
1
2
1
vector 1
( )
11 12 11 11 12 1 1
21 222 21 22 2 2
1 2
1
1
( ) ( ) ( )( ) ( ) ( )
( ) (( ) ( ) ( )
( ) ( ) ( )
A t
qn
n
n n n nn n
nn n
n
b t b t b tx a t a t a t x
b t b tx a t a t a t x
x a t a t a t x
( )
1
2 2
1 2
1
1
) ( )
( ) ( ) ( )
B t
q
n n nq q
qn q
n
u
b t u
b t b t b t u
1
1
2
1
)(
21
22221
11211
1
1
2
1
)(
21
22221
11211
2
1
)()()(
)()()(
)()()(
)()()(
)()()(
)()()(
m
q
q
qm
tD
mqmm
q
q
m
n
n
nm
tC
mnmm
n
n
mu
u
u
tdtdtd
tdtdtd
tdtdtd
x
x
x
tctctc
tctctc
tctctc
y
y
y
27
Objectives
1) State Models
2) Canonical Representations:
Observable Canonical Form (OCF)
6 Lecture 6
State Models 6.1
Linear Time Invariant System
Equations )()()(
Equations )()()(
OutputtDutCxty
StatetButAxtx
A, B, C and D are now constant matrices.
State Diagram
Figure 6-1. State diagram for a linear time invariant system.
Example 1: uytyty 2)1( , find the state space representation,
in another word, state equation and output equation
Let x1= y, 212 xxyx , now we need ?2 x
ytytuyx )1(2
2 21
2
2 )1( xtxtux
21
2
2
21
)1(
xtxtux
xx
State Equation
1xy Output Equation
y + B
u
ʃ x
D
+
A
+
+
C
Page 28 of 115
28 Linear system: Lecture 6
R
C
+
- i
R
C
+
-
y u
i
In order to get A, B, C and D, define
2
1
x
xx
11
12122
1
22
2
122
1
1
0
)1(
10
u
x
x
ttx
x
BA
⏟
[
]
⏟
Example 2: For system shown as below, find the state and output equations.
1) dtiiC
Riu )(1
211
2) 0)(11
1222 dtiiC
dtiC
Ri
Let dtiC
x 11
1, dti
Cx 22
1 11 ixc , 22 ixc
1222
211
0 xxxxRC
xxxRCu
21
111
212
121
xRC
xRC
x
uRC
xRC
xRC
x
State Equation
22
1xdti
Cy 2xy Output Equation
Find A, B, C and D
22
21
11
RCRC
RCRCA ;
120
1
RC
B
21
10
C ; 110 D
This system is Time invariant since A, B, C and D are constant.
Now with the voltages across the capacitors as states, we get
Figure 6-2. RC network.
Page 29 of 115
6.1 State Models 29
M K
M
K K
F
x x
1 1 1 2 1 2 1
1 1 2
2 2 2 2 2
1( )
1
x v i i dt i i CxC i Cx Cx
x v i dt i CxC
1 2 1
2 2 1
( ) 1) (*)
0 2) (**)
u RC x x x from
RCx x x from
212
11x
RCx
RCx
121 2(**)(*) xxxRCu 121
211x
RCx
RCu
RCx
11
112
212
211
xRC
xRC
x
uRC
xRC
xRC
x
State Equation
222 xyxvy Output Equation
RCRC
RCRCA
11
12
;
0
1
RCB ; 10C ; 0D
Example 3: System as below:
0)(
)(
1222
2111
yykkyym
Fyykkyym
Choose the states as 24231211 , , , yxyxyxyx
2
2
1324
43
3112
21
xm
kx
m
kyx
xx
xm
kx
m
k
m
Fyx
xx
State Equations, let
4
3
2
1
x
x
x
x
x
Figure 6-3. Mass-Spring system.
Page 30 of 115
30 Linear system: Lecture 6
(i)
3
1
2
1
x
x
y
y Output Equations or
(ii) 11 xyy Output Equation (valid) or
(iii) 32 xyy Output Equation (valid) or
(iv) 3121 xxyyy Output Equation (valid) etc.
44
02
0
1000
002
0010
m
k
m
k
m
k
m
k
A ;
140
0
1
0
mB
1114144414 FBxAx 1112144212 FDxCy
(i)
420100
0001
C ,
120
0
D , 12 42 12 DCy
(ii) 41
0001
C , 110 D , 11 41 11 DCy
(iii) 41
0100
C , 110 D
(iv) 41
0101
C , 110 D
Now choose the states as follows: 2142131211 , , , yyxyyxyxyx
3
3214
43
3112
21
xm
k
m
Fyyx
xx
xm
kx
m
k
m
Fyx
xx
State Equations
44
03
00
1000
00
0010
m
k
m
k
m
k
A
14
1
0
1
0
m
mB
Let as an example,
(Output Equation)
Page 31 of 115
6.2 Canonical Representations 31
Canonical Representations 6.2
a) Observable Canonical Form (O.C.F)
b) Controllable Canonical Form (C.C.F) (also known as Phase Variable CF)
c) Jordan Canonical Form (J.C.F)
6.2.1 Observable Canonical Form
For system ubdt
udb
dt
udbya
dt
yda
dt
ydn
n
nn
n
nn
n
nn
n
01
1
101
1
1
we define dt
dD , then ubuDbuDbyayDayD n
n
n
n
n
n
n
0
1
10
1
1
ubuDbuDbyayDayD n
n
n
n
n
n
n
0
1
10
1
1
yaubD
yaubD
ubynnnn 0011
11
State Diagram:
Figure 6-4. State diagram.
Define the states as the output of each integrator.
00
111
2232
1121
yaubx
yaubxx
yaubxx
yaubxx
n
nn
nn
nn
and 1xuby n
)(
)(
)(
0010
223122
112111
ubabxax
ubabxxax
ubabxxax
nn
nnnn
nnnn
State Equations
ʃ
ʃ
ʃ
+
-
+
-
+
- + + +
+ y
u
Page 32 of 115
32 Linear system: Lecture 6
100
11
22
11
nn
n
nnn
nnn
bab
bab
bab
bab
B
Outputubxy
bD
C
n
n
n
1
11
1001
The pair (A, C) is known as one of the Observable Canonical forms.
Alternatively, from the state diagram, reverse the assignment for the states, i.e. the output
of the left integral is denoted by 1x and consecutively increased to x then we get,
The pair (A, C) is now considered as the selected Observable Canonical Form (OCF) in
this course.
The coefficients of
the characteristic
polynomial
0 0 0 1 .C
33
Objectives
1) Controllable Canonical Forms (CCF)
2) Jordan Canonical Form (JCF)
7 Lecture 7
Controllable Canonical Form 7.1
0
1
1
1
1
0
1
1
1
1
)(
)(
asasas
bsbsbsb
sU
sYn
n
n
n
n
n
n
Figure 7-1. Decomposition of the transfer function.
0
1
1
1
)(
)(
asassU
sZn
n
n
uzazaz n
n
n
0
)1(
1
)(
Let
)1(
)2(
1
2
1
n
n
n
n
zx
zx
zx
zx
)(
1
32
21
n
n
nn
zx
xx
xx
xx
nn
n
n
n
xaxaxau
zazazauz
12110
)1(
110
)(
U Y
U Z Y
all pole system all zeros system
Page 34 of 115
34 Linear system: Lecture 7
Therefore,
nnn
nn
xaxaxaux
xx
xx
xx
12110
1
32
21
State Equation
Output Equation
0
1
1 bsbsbZ
Y n
n
n
n zbzbzby n
n
n
n 0
)1(
1
)(
10112110 xbxbxaxaxauby nnnnn
0 0 1 1 1 2 1 1( ) ( ) ( )n n n n n n ny b b a x b b a x b b a x b u Output equation
)()()( 111100 nnnnn abbabbabbC nbD
The pair (A,B) is in the Controllable Canonical Form (CCF) also known as the Phase
Variable Canonical Form
Jordan Canonical Form 7.2
1 1
1 1 0
1 1
1 1 0
( )( ) ( ) ( ) ( )
( )
n n
n nnn n
n
b s b s b s b N sY s U s b U s U s
s a s a s a D s
Where )()()()( 00
2
22
1
11 n
n
nnn
n
nnn babsbabsbabsN
and
0
1
1)( asassD n
n
n
1
0
0
0
B
Page 35 of 115
7.2 Jordan Canonical Form 35
γ ʃ
λ
+
+
u ẏ/λ y/λ y
y/λ
7.2.1 Characteristic Polynomial D(s) = 0 Contains N Distinct Poles
)())(()( 21 nssssD
then )()()(2
2
1
1 sUsss
sUbsYn
nn
Construct the state diagram.
Example 1: Consider Us
Y
Figure 7-3. State diagram.
yuyy
uy
Figure 7-4. Parallel configuration.
γ ʃ
λ
+
+
u y ẏ
ʃ
+
+
u
y
ʃ
+
+
ʃ
+
+
+
+ +
+
Figure 7-2. Alternative state diagram.
Page 36 of 115
36 Linear system: Lecture 7
1 1 1
2 2 2
1 1 1n n n
n n n
x x u
x x u
x x u
x x u
Sate Equation
nnn xxxuby 2211 Output Equation
n
A
000
000
000
000
3
2
1
;
1
1
1
1
B ; nC 321 ; nnD
Jordan Canonical Form (JCF)
7.2.2 Characteristic Polynomial D(s) = 0 Contains Multiple Repeated
Poles
Suppose that D(s) can be expressed as
)())(()()()( 4321
nssssssD
UbY n Usss
)()()( 1
1
1
2
1
1
1 2
1
2 2 2
1 2
3 4
( ) ( ) ( )
n
n
Us s s
Us s s
This bracket is
represented in the next
state diagram
Page 37 of 115
7.2 Jordan Canonical Form 37
First, construct sate diagram for )( 1s terms in
)( 1
1
sU
Yas shown below..
Then, similarly construct state diagram for
)( 2s and )()(),( 43 nsss
parallel with the above state diagram (detail omitted).
Finally, the combined state diagram leads to the following A and B matrices.
n
J
J
A
000
000
000
000
3
2
1
,
1
1
2
1
B
B
B
where
i
i
i
i
i
iJ
0000
1000
0000
0010
0001
,
1
0
0
0
0
iB for i = 1 and i = 2
ʃ
+
+
ʃ
+
+
u ʃ
+
+
+
terms
+
+
+ +
Jordan Block
Figure 7-5. State diagram of the term in bracket.
Page 38 of 115
38 Linear system: Lecture 7
In other words,
1
1
1
1
1
1
0000
1000
0000
0010
0001
J ,
1
1
1
0
0
0
0
B
and
2
2
2
2
2
2
0000
1000
0000
0010
0001
J ,
1
2
1
0
0
0
0
B
39
Objectives
1) Transformation from state equations to Transfer Function
2) Examples for C.C.F., O.C.F., J.C.F
8 Lecture 8
Transformation From State Equations to Transfer Function 8.1
Given
DuCxy
BuAxx, find the transfer function (TF) ?
)(
)(
sU
sY
Take Laplace transform (L.T.) ( x(0) = 0 )
(**))()()(
(*))()()(
sDUsCXsY
sBUsAXssX
)()()((*) sBUsAXssX )()()( sBUsXAsI
nn
Assuming )( AsI is nonsingular, then 1)( AsI exists.
)()()( 1 sBUAsIsX
)()()()( 1 sDUsBUAsICsY
11
1
11
1
11
1
)()()(
qqpp
sUDBAsICsY
.).()()(
)( 1 FTDBAsICsU
sY
GUY
u
u
u
y
y
y
p
MatrixTransfer
p
***
***
***
2
1
2
1
T.F. for SISO system Transfer Matrix for MIMO system
Page 40 of 115
40 Linear system: Lecture 8
Example 1: Find T.F. for system u
x
x
yu
x
x
x
x
2
1
2
1
2
1
11;
0
1
12
01
First approach: Brute Force
Write down equations for system:
uxxy
xxx
uxx
21
212
11
2
Since 1 2 1 2y x x u y x x u
uuxxuxxuxy 21211 2
uuxxuxuuxxy 21121 2
uuuxxy 213
Since uuxyy 22 2)(
2
1
2
12 yyuux
Since uxxy 21 uxyx 21
yyuyx 5.05.05.01
uuuyyuuyyuyy )5.05.05.0()5.05.05.0(3
uuuyyy 22
uuuyyy 22
Therefore, Transfer Function (T.F.):
usuusysyys 22 22
12
2
)(
)(2
2
ss
ss
sU
sY
Page 41 of 115
8.1 Transformation From State Equations to Transfer Function 41
Second Approach: Using formula DBAsICsU
sY 1)(
)(
)(
With
12
01A ,
0
1B , 11 C , 1D to find ?)( AsI
s
ssI
0
0,
12
01
12
01
0
0
s
s
s
sAsI
12
01
0)1)(1(
1)( 1
s
s
ssAsI
10
01
)1)(1(
1
)1)(1(
2
0)1)(1(
1
12
01
ss
s
ss
ss
s
s
s
2
1
)1(
1
0
1
12
01
)1(
1)(
22
1s
ss
s
sBAsI
222
2
2
1
)1(
3
)1(
2
)1(
1
)1(
2
)1(
)1(
11)(
s
s
ss
s
s
s
s
BAsIC
2
2
2 )1(
)1(31
)1(
3
)(
)(
s
ss
s
s
sU
sY
12
2
12
123
)(
)(2
2
2
2
ss
ss
ss
sss
sU
sY
Page 42 of 115
42 Linear system: Lecture 8
U X Y
Examples for C.C.F., O.C.F., J.C.F. 8.2
8.2.1 Observable C.F.
Example 2: uuyyy 265 , for observable C.F. draw the state diagram
uuyyyuuyyy 265265
Figure 8-1. State diagram for example 2.
212
21
52
6
xuxx
xux
State Equations 2xy Output Equation
2
1
x
xx ;
uDxCy
uBxAx
oo
oo
51
60oA ,
2
1oB , 10oC , 0oD Observable C.F.
8.2.2 Controllable C.F.
Example 3:
uxx
ccBA
1
0
56
10
uxy
ccDC
021
T
oc AA , T
oc CB , T
oc BC , oc DD
To get C.C.F, directly from the differential equation, we find the T.F.
65
12
)(
)(2
ss
s
sU
sY
Figure 8-2. Transfer function decomposition for example 3.
ʃ -
-
2
ʃ +
+
5
6
-
y u
Page 43 of 115
8.2 Examples for C.C.F., O.C.F., J.C.F. 43
uxxxssU
X
65
65
12
Let
xx
xx
2
1, then
xx
xx
2
21 122 6565 xxuxxuxx
1 2
2 1 26 5
x x
x x x u
State Equations
Since (2 1)
2 (2 1) 21
Y sy x x Y s X sX X
X
122 xxy Output Equation
56
10
cA ,
1
0
cB , 21cC , 0cD
8.2.3 Jordan Canonical Form
For system 65
122
ss
s
U
Y , in order to find the J.C.F., follow the steps listed below:
1) Using partial fraction expansion, 3
7
2
5
ssU
Y
2) Draw the state diagram
Figure 8-3. System state diagram.
Therefore, for Jordan C.F.
30
02
A ,
1
1
B , 75C , 0D
ʃ +
-
2
-5
ʃ +
-
3
7
+
+
u y
21
22
11
75
3
2
xxy
uxx
uxx
Page 44 of 115
44 Linear system: Lecture 8
Example 4: uuyyy 22 , show
21
10
cA ,
1
0
cB , 12cC , 0cD
21
10
oA ,
1
2
oB , 10oC , 0oD
For Jordan Canonical Form
2 2 2 2 2
( ) 2 2 3 3
( ) 2 1 ( 1) ( 1) ( 1) ( 1) ( 1) ( 1)
Y s s s a b a as a
U s s s s s s s s s
2
1 31
1 ( 1)
Ya
U s s
Figure 8-4. State diagram.
Jordan Block:
10
11
A ,
1
0
B , 13C , 0D
Example 5: Given poles: -1, -1, 3,what is A and B?
ʃ +
-
1
3 ʃ +
-
1
1
- +
u y
T.F.=
T.F.=
T.F.=
A=
, B=
21
22
121
3
xxy
xux
xxx
Page 45 of 115
8.2 Examples for C.C.F., O.C.F., J.C.F. 45
Example 6: Armature Controlled DC Motor
Figure 8-5. Armature controlled DC motor.
From the above figure, the governing equations are
This is a 3rd
order system,
let 321 ,, xxix a
eKiRdt
diL
iKBJ
baaa
a
aT
eLL
Ki
L
R
dt
di
aa
ba
a
aa 1
1
313
32
311
xJ
Bx
J
Kx
xx
eL
xL
Kx
L
Rx
T
aa
b
a
a
Defining the output as the angular position of the motor
2xy , Jordan Canonical Form is:
J
B
J
K
L
K
L
R
A
T
a
b
a
a
0
100
0
,
0
0
1
aL
B , 010C , 0D
R
+
-
i L
+
R L
i
J
B T
bb
ba
aaa
aT
Ke
edt
diLiRe
iKT
TBJ
46
Objectives
1) Solution of State Equation
2) Fundamental Matrix
3) State Transition Matrix
9 Lecture 9
Solution of State Equation 9.1
Given
utDxtCy
utBxtAx
)()(
)()(
,
u , ,
q
mn
Ry
RRx
Example 1: RyRRx , u , ; utbxtax )()( , )( 0tx is given
[
]
∫
[
]
∫
∫
∫
⏟
∫
⏟
Integrating factor
Page 47 of 115
9.1 Solution of State Equation 47
Example 2: Let ata )( ( constant ) utbaxx )(
∫
Example 3: Let
t
taat udbexetxt0
)(
0 )()0()(0
Show this by using Laplace Transform method !
We have , 0 t, )( utbaxx
b(t)u(t)LaX(s)x(0)-sX(s) )( utbaxxL
)()(1
)0(1
)( tutbLas
xsa
sX
taking 1L
)()(
1)0()( 1 tutbL
asLxetx at
)()()()()()( 21
0
2121
1 tgtgdgtgsGsGL
t
t
taat dubexetx0
)( )()()0()(
Example 4: tuxt
x
1
with u = t , x(1) =1
We know
0
1
1 1
1 1( ) ( ) ( ) ln ln
t t tt
t
da t b t t a d d t
t
,
0
1( ) ln
ln 1t
ta d
t te e et
, ln ln
1
1( ) 1 ( )
t
tx t e dt
for 1t
Page 48 of 115
48 Linear system: Lecture 9
Fundamental Matrix 9.2
9.2.1 Definition:
GivennRxxtxxtAx ,(,)( 00) , an n×n matrix M(t) is said to be a fundamental
matrix of xtAx )( , if and only if the n columns of M(t) , denoted as
)(M , ),(M ),( n21 tttM are any n linearly independent solutions of xtAx )( .
9.2.2 Examples:
Example 5: Given x
t
x
20
01
, find M(t).
Let
12
2
12
1 )()()( tMtMtM ,
12
11
1
x
x
M , since 1M is a solution, it has to
satisfy the dynamic equation, namely 11 )( MtAM .
0,2
0x, x
12121212
11111111
2
xextxx
exx
t
t
let
01
te
M ;
2
0
2te
M
Why not
2
1t
t
e
e
M ,
0
0
2M ? Linear combination of M1 and M2 is
0
0
0
0
212
t
t
e
e
, since 02 these vectors are linearly dependent
NOT ACCEPTABLE FOR M(t) !
Now
0
00
02
21t
t
e
e
since 021 linearly independent ⇒M(t) √
2
0
0
)(t
t
e
e
tM is the fundamental matrix.
Page 49 of 115
9.2 Fundamental Matrix 49
Fundamental matrix is not unique! Why?
Choose
2
40
02
)(t
t
e
e
tM , now it is still an acceptable fundamental
matrix as it satisfies the differential equations.
Example 6: Given x
tt
tx
22
01
2
, find M(t).
Let )()()( 21 tMtMtM ,
12
11
1
x
x
M
with 122
12121211 xt
xt
xtx
with 2
12121211
20 txx
txx
11
t
M ,
2
2
0
t
M
21
0
)(
t
t
tM
Check to see if M(t) satisfies the dynamic equation: MtAtM )()(
tt
t
tt
t
t 20
01
1
0
22
01
20
01
2
2
?
√
Page 50 of 115
50 Linear system: Lecture 9
State Transition Matrix 9.3
9.3.1 Definition
Given xtAx )( , an n×n matrix ),( 0tt is a state transition matrix if and only if
)(),()( 00 txtttx is a solution of the state equation with )( 0tx given.
9.3.2 Proposition
Let M(t) be the fundamental matrix of xtAx )( , then
ttMtMtt
uniquenotunique
)()(),( 0
1
0
Example 7:
2
0
0
)(
20
01
t
t
e
e
tMx
t
x , find ),( 0tt .
We need
20
0
0
0
)( 0
t
t
e
e
tM , then
20
0
0
0
)( 0
1
t
t
e
e
tM
20
2
0
20
0
2
0
0
0
0
0
0
),( 0
tt
tt
t
t
t
t
e
e
e
e
e
e
tt
Moreover, )(),()( 00 txtttx
20
10
20
2
0
0
0
)(
x
x
e
e
txtt
tt
202
101
20
2
0
)(
)(
xetx
xetx
tt
tt
9.3.3 Properties of ),( 0tt
1) ),()(),(
00 tttA
t
tt
2) Itt ),( 00
Page 51 of 115
9.3 State Transition Matrix 51
3) )(),(
00
0
tAt
tt
tt
4) Transition property: ),(),(),( 011202 tttttt
5) Inversion property: ),(),( 1001
1 tttt
1)
)(tA
t
Since )(),()( 00 txtttx and x(t) is a solution, therefore it should satisfy the differential
equation (d.e.), i.e.
)(),()()(),( 0000 txtttAtxttx
),()(),( 00 tttAtt √
2) Itt ),( 00
Since )()(),( 0
1
0 tMtMtt
ItMtMttttat )()(),(, 0
1
0000 √
3) Follows from (1) and (2), since ),()(),( 00 tttAtt
)(),(),()(),(, 000000000
tAtttttAttttattt
I
√
4) ),(),(),( 011202 tttttt
000 )(),()( tttxtttx (*) )(),()( 0011 txtttx for 01 tt
111 )(),()( tttxtttx (**) )(),()( 1122 txtttx for 12 tt
)(),(),()( 001122 txtttttx
From )(),()((*) 0022 txtttx
)(),()(),(),( 00200112 txtttxtttt √
)
Figure 9-1. Transition property.
Page 52 of 115
52 Linear system: Lecture 9
5) ),(),( 1001
1 tttt
Since ),(),(),(),(),( 0110000101
1 ttttttItttt
),(),(),(),( 01100101
1 tttttttt
),(),( 1001
1 tttt √
Example 8:
x
t
x
20
01
20
2
0
0
0
),( 0
tt
tt
e
e
tt , check the above properties.
),()(),( 00 tttAtt
20
2
0
20
2
0
20
0
20
01
20
0
tt
tt
tt
tt
te
e
tte
e
√
Itt ),( 00
10
01
0
0
),(20
20
00
00
tt
tt
e
e
tt √
)(),( 000
tAtttt
)(
20
01
20
0
20
0
0
00
0
0
0
20
2
0
tA
tet
e
te
e
tt
tt
tt
√
53
Objectives
1) State Transition Matrix for LTI Systems
2) Cayley-Hamilton technique
10 Lecture 10
State Transition Matrix for LTI Systems 10.1
10.1.1 Definition
Assuming system Axx , )( 0tx is given, for a scalar system axx , Rx , )( 0tx
)()( 0
)( 0 txetxtta
, generalization to an n-th order system Axx , )( 0tx , nRx ,
suggests that the solution can be expressed as )()( 0
)( 0 txetxttA
. Check to see if x
satisfies the differential equation )()( 0
)( 0 txAtxeAxttA
.
Since )(),()( 00 txtttx
)(
00),(
ttAett
is a state transition matrix for a LTI system
)(),( 00 tttt
Now at 00 t , ( ,0) ( ) Att t e )0( 0 t , from property (5), ?)(1 t
1 1
0 0 0 0( , ) ( , ), ( ) ( )t t t t t t t t for 1
0 0 ( ) ( )t t t
)()(1 tt
Page 54 of 115
54 Linear system: Lecture 10
10.1.2 Approaches to Compute )(t
1) 11 )()( AsILet At
2) Cayley-Hamilton technique
3) The Jordan Form technique
4)
0 !k
kkAt
k
tAe
For approach 1), given Axx with )( 0tx , take L-transform
)()()( 0 sAXtxssX )()()( 0txsXAsI
1
0
1
1
)()()(
nnnn
txAsIsX
)()()( 0
11 txAsILtx
Since )(),()( 00 txtttx 11
0 )(),( AsILtt
11 )( AsILeAt
Computation of 1)( AsI , using Leverrier’s algorithm:
01
1
1
01
2
2
1
11
)det(
)()(
asasas
PsPsPsP
AsI
AsIadjAsI
n
n
n
n
n
n
n
scalarsaresamatrixnnaresPwhere ii ','
Algorithm:
1) nnn IP 1
2) )(1 Atran
3) IaAPP nnn 112
Page 55 of 115
10.1 State Transition Matrix for LTI Systems 55
4)
)(1
)(2
1
11
22
APtrkn
a
IaAPP
APtra
kk
kkk
nn
Final verification test is 00 IaAP o .
Example 1:
1
2
2 2
2
3 0 1
1 3 0
1 15
4
P P A a IP I
a
1 1 0 1 1
2 1 2
1 1 1
3
1( ) 7
22 4
a tr P A P P A a I
2
1 2 1 00 0 3 2
2 1 0
11
3
P s Ps Pa tr P A sI A
s a s a s a
2
2
3 2
2
3 2 1 21
1 3 1 15 7 1
3 2 4 4
s s s
s s ss s s
s s s s
Example 2: System xx
44
10 , xy 51 , i.c. y(0) = 0 , 1)0( y , find y(t).
1 2
1 2
( ) ( ) 5 ( )
( ) ( ) 5 ( )
y t x t x tt
y t x t x t
1 2
2 1 2
2 1 2 1 2
4 4
5( 4 4 ) 20 19
x x
x x x t
y x x x x x
11 2
1 22
5(0)
(0) (0) 5 (0) 0 81
(0) 20 (0) 19 (0) 1 1(0)
81
xy x x
y x xx
,
Page 56 of 115
56 Linear system: Lecture 10
11 )( AsILeAt
44
1
44
10
0
0
)(
s
s
s
s
AsI
22
221
)2()2(
4
)2(
1
)2(
4
)(
s
s
s
ss
s
AsI
ttt
ttt
At
teete
teete
e222
222
24
2
)0()( xetx At
81/1
81/5
24
2
)(
)(
222
222
2
1
ttt
ttt
teete
teete
tx
tx
2
1 2( ) ( ) 5 ( ) ( ) ty t x t x t y t te
Cayley-Hamilton Technique 10.2
10.2.1 Cayley-Hamilton Theorem
For a given n×n matrix A with characteristic polynomial 0)det()( AIA , the
matrix A satisfies its characteristic polynomial (C.P.), i.e. 0)( AA .
In other words, if 0)( 1
1
1
nn
nn
A then
01
1
1
IAAA nn
nn .
Example 3:
21
10
A , 012
21
1
det)det()( 2
AIA
We want to show that (from C.H. Theorem): 0122 AA
00
00
10
01
42
20
21
10?
2
00
00
32
21
32
21
√
Page 57 of 115
10.2 Cayley-Hamilton Technique 57
10.2.2 Cayley-Hamilton Technique
Given an n×n matrix A with characteristic polynomial (C.P.) 0)( A , the matrix
polynomial P(A) can be computed by considering the scalar polynomial )(P .
We compute ( ) ( )
( )( ) ( )A A
P RQ
at i ( i is the eigenvalue of A or the roots of the C.P.)
)()()()(
0
iiiAi RQP
1)()( ii RP i = 1,…, n (distinct si ' )
Now, from C.H. Theorem )()()()(
0
ARAQAAP A
)()( ARAP
10.2.3 Algorithm to Apply Cayley-Hamilton Technique
1) Find the eigenvalues of A.
2) If the eigenvalues are distinct, use equation (1) for the coefficients of )(R by
solving n simultaneous equations. If the eigenvalues are repeated, let’s say for an
eigenvalue with multiplicity m, then instead of (1) use:
3) Use the constants found in step (2) to get )()( ARAP .
[Examples of tAAt eAAeAP
2
;;; )( 2101 ]
Polynomial of degree n-1
with unknown coefficients
)()()()( RQP A
ii
ii
ii
m
m
m
m
ii
d
Rd
d
Pd
d
Rd
d
Pd
d
dR
d
dP
RP
1
1
1
1
2
2
2
2
)()(
Page 58 of 115
58 Linear system: Lecture 10
Example 4: Find 101A for
43
10
A
1) 101)( AAP ; 101)( P
2) 2 ( )n R
3) ( ) det( ) 0 1, 3A I A
4) Apply ( ) ( ) 1,2, , . .i iP R for i i e
101
101
1 ( 1)
3 ( 3) 3
2
)3()1( 101101 ;
2
)3()1(3 101101
5) Apply )()( ARAP
IAARAAP )()( 101
10
01
2
)3()1(3
43
10
2
)3()1( 101101101101101A
2
)3(3)1(1
2
)3(3)1(3
2
)3()1(
2
)3()1(3
101101101101
101101101101
101A √
Example 5: Find Ate for
21
10
A
2, ( ) , ( ) , ( )At tn P A e P e R
1,10)1()( 2 A (Repeated)
Page 59 of 115
10.2 Cayley-Hamilton Technique 59
( ) ( )
@ 1 1
t
t t
P R e
applydP dR
te at ted d
,t t tte e te
Apply P(A) = R(A) , hence
IeteteIAe tttAt
)(
21
10
ttt
ttt
At
teete
tetee
e √
Example 6:
22
10
A , find Ate .
21
( ) det( ) det 2 2 0 12 2
A I A j
jA
1
22
10
(distinct)
( ) ( ) ( )At tP A e P e R , ,
Apply ( ) ( ) 1P R at j
)1(
)1(
)1(
)1(
je
je
tj
tj
Page 60 of 115
60 Linear system: Lecture 10
4/
4/
2)sin(cos
2)sin(cos
jt
jt
etjte
etjte
t
j
t
jt
t
j
t
jt
ee
ee
ee
ee
4/
4/
2
2
)2(2sin22 4/4/ je
tjeee
eet
jj
t
jtjt
tet
sin2
10
01
22
10sin
2t
ee
tAt
Figure 10-1. System plot.
.8
2
.6 .4 .2
5
-2
4
0
2
1
6
8
-4
-6
-8
1.2 1.4 1.6 1.8 2
xjxe jx sincos
4/21 jej
Use
√ to find β,
it will be a real number
61
11 Lecture 11
Ф(t, t0) for LTI System 11.1
From the scalar case: 22
!21 t
te t, we get
22
0
*2! !
k kAt
k
t t Ae I tA A
k
( )
Some of the properties of Ate are :
1) Ie 0
2) 2121 )( AtAtttA
eee
21,tt scalars
3) BtAttBA eee )(
if and only if (iff.) AB = BA (A & B commute)
4) AeAeedt
d AtAtAt
Proof of (1): from (*) with either t= 0 or A= 0!
Objectives
1) Using Cayley Hamilton technique to calculate 𝜱 𝒕 𝒕𝟎 for LTI Systems
2) Using Cayley Hamilton technique to calculate 𝜱 𝒕 𝒕𝟎 for LTV Systems
3) Solution of the state equations and output equation
Page 62 of 115
62 Linear system: Lecture 11
Proof of (2): 2
21
2
21
)()(
!2)(21 tt
AttAIe
ttA
2 22 2 2
1 2 1 2 1 2 *2! 2!
A AI At At t t A t t ( )
!2!2
2
2
2
2
2
1
2
121
tAAtI
tAAtIee
AtAt
2 2 2 222 1
2 1 1 2 **2! 2!
A t A tI At At A t t ( )
Since (*) = (**) 1 2 1 2( )A t t At At
e e e
Proof of (3):
!2)()(
22)( t
BAtBAIe tBA
!2!2
2222 tBBtI
tAAtIee BtAt
2 2 2 3 2 2 2 32
2 22
2! 2! 2! 2!
( )2! 2!
B t AB t A t A BtI Bt At ABt
A BI A B t t AB
ABBABA
222
)( 22?2
)2(2
1)(
2
1 22?
22 ABBABAABBA
If and only if AB = BA AB+BA=2AB that is, A & B should commute!
Page 63 of 115
11.1 Ф(t, t0) for LTI System 63
Proof of (4):
1 1
11
0 )!1(!! k k
kkkk
k
kkAt
k
At
k
Akt
k
At
dt
de
dt
d
Ak
At
k
AAt
k
AAt
k
At
k
kk
k
kk
k
kk
k
kk
0000
1
!!!!
AtAt
k
kk
AeAek
AtA
0 ! A &
Ate commute!
Jordan Canonical Representation Ate
1) A has distinct eigenvalues
2) A has repeated eigenvalues
Case 1: (Distinct λ) For LTI System
, find
{
[
]
[
]
{
[
]
Page 64 of 115
64 Linear system: Lecture 11
Case 2: ( Repeated λ)
nn
J
1
1
1
1
1
0000
1000
0000
0010
0001
, find Jte
Using Cayley-Hamilton technique:
1) n
J JI )()det()( 1
2) JteJP )( ,
teP )(
1
11
2
12110 )()()()(
n
nR
3) )()( 11 RP 01
te √
11
d
dR
d
dP 1
1
tte √
11
2
2
2
2
d
Rd
d
Pd 2
2
1
2
t
et
√
11
1
1
1
1
n
n
n
n
d
Rd
d
Pd 1
1
1
)!1(
n
tn
en
t
√
Page 65 of 115
11.2 Ф(t, t0) for LTV Systems 65
4) 1
11110 )()(
n
n
Jt IJIJIe
t
tt
tn
tn
t
tn
tn
tt
tn
tn
ttt
e
tee
en
te
n
te
en
te
n
ttee
en
te
n
te
ttee
1
11
111
1111
11111
0000
000
)!3()!4(00
)!2()!3(0
)!1()!2(!2
34
23
122
Ф(t, t0) for LTV Systems 11.2
1) )()(),( 1
0 tMtMtt
2) Direct solution of xtAx )( to get )(),()( 00 txtttx . Brute Force (Hyperlink)
reference to lecture 8 Example 1.
3) If A(t) and )( 0tA commute, then
t
t
dA
ett 0
)(
0 ),(
4) If ∑ are constant matrices where
are scalars and
, then
k
i
dM
t
t
ii
ett1
)(
00),(
Page 66 of 115
66 Linear system: Lecture 11
Example 3: [
]
⏟
[
]
⏟
⏟
[
]
⏟
Since
[
]
[
]
[
] [
]
(This can be shown by applying Cayley-Hamilton technique -- details not shown)
Therefore, [
]
Example 4:
tt
ttA
22
01
)(
2
Check to see that A(t) and )( 0tA do not commute, therefore #3 does not apply.
Use method #2 11
1x
tx ; 2122
22x
tx
tx
)()()( 01
0
01
1
10 tx
t
ttxetx
t
t
d
)()(
01
0
2
)(
2 )(2
2)(
tutbta
txtt
xt
tx
Page 67 of 115
11.2 Ф(t, t0) for LTV Systems 67
t
t
dd
dtxt
etxetx
tt
t
0
0 )(2
)()( 01
0
2
02
2
2
dtxt
etxe
t
t
ttt
)(
2)( 01
0
lnln
02
lnln
0
2220
2
t
t
dt
txttx
t
t
00
01
2
2
022
0
2 )(2)(
t
t
dtxt
ttx
t
t
0
301
0
2
022
0
2 2)()(
2
0
201
0
2
022
0
2
2
11)()()(
tttx
t
ttx
t
ttx
)(
)(
1
0
)(
)(
02
01
),(
2
0
2
3
0
2
0
0
2
1
0
tx
tx
t
t
t
t
t
t
t
tx
tx
tt
Example 5:
0
020
01)(
ttA )()()()( 00 tAtAtAtA , therefore, use method #3
[
]
∫[
]
[
]
[
]
[
] √
This follows from Cayley-Hamilton technique.
1 0( )
0 2A t
t
Page 68 of 115
68 Linear system: Lecture 11
Example 6:
t
ttA
1
1)(
0
0
0
1
1)(
t
ttA
( )0 0 0
0 0
0 0 0
1 11( ) ( ) ( ) ( )
1 11
showst tt t ttA t A t A t A t
t t t ttt
√
Method #3 applies
dA
t
tett
0
)(
0 ),(
d
t
t01
1
B
tttt
tttt
2
220
2
0
0
20
2
, ?2
20
2
0
02
20
2
tttt
tttt
e
?Be
10)(,)(,)( RePeBP B
Find eigenvalues of B
)(2
0
2
2det)det( 0
2
0
2
2,12
0
2
0
0
2
0
2
tttt
tttt
tttt
BI
)sinh(
12
)sinh(
)()(
)()(
0
021
0
02
0
22
11
20
2
0
20
2
tt
tte
ttttee
RP
RP
tt
tt
tt
BIeB
10
2
00
00
0
20
2
)cosh()sinh(
)sinh()cosh(),(
tt
B etttt
ttttett
√
Page 69 of 115
11.3 Solution of the State Equations and Output Equation 69
Solution of the State Equations and Output Equation 11.3
For system
DuCxy
BuAxx,
the complete solution is given by
respond state zero
t
trespondinput zero
00
0
dττ)B(τ)u(τ)Φ(t,+) tΦ(t,= )x(tx(t)
Proof:
∫
∫
Left=
{ ⏟
Since
{
∫
Leibnitz’s Rule
t
utuf
t
vtvfdx
t
txfdxtxf
t
tv
tu
tv
tu
),(),(
),(),(
)(
)(
)(
)(
∫
Given , constantx Ax Bu A and B are
t
t
tAttAdBuetxex(t)
0
0 )()( )(
0
)(
?
70
Objectives
1) Controllability; Observability
2) Canonical Transformations: C.C.F., O.C.F
12 Lecture 12
Definitions 12.1
12.1.1 Controllability
Given
DuCxy
BuAxx, RuyRx n ,, (SISO), the system is controllable if and only if
the n×n matrix (known as the controllability matrix)
is nonsingular; i.e. nCRank x .
Figure 12-1. State trajectory as a result of the control u.
12.1.2 Observability
Given the above system, we say it is observable if and only if the n×n matrix Ox (known
as the observability matrix) is nonsingular, in another word, nORank x , where
nn
n
x
CA
CA
C
O
1
Figure 12-2. System diagram.
System
Observer
y u
x
Controller
Control loop
Page 71 of 115
12.1 Definitions 71
12.1.3 Examples for Controllability and Observability
Example 1:
01
10A , 1 2
1 1
1 1B B B
, is this system controllable ?
[
]
2)( xCRank full rank controllable !
Q: check to see if the system is controllable from individual inputs.
[
] 21)( 1 xCRank not controllable from u1
[
] 21)( 2 xCRank not controllable from u2
Example 2:
40
21A ,
2
1
10
01
C
CC , is this system observable?
CA
COx
40
21
10
01
xO 2)( xORank Observable
Q: is the system observable from each output alone?
21
01
1
11
AC
COx 2)( 1 xORank Observable from y1
40
10
2
22
AC
COx 21)( 2 xORank Unobservable from y2
Page 72 of 115
72 Linear system: Lecture 12
Canonical Transformations 12.2
12.2.1 A Canonical Decomposition of Dynamic System
Figure 12-3. System classification.
If Ox singular system is not observable
If Cx singular system is not controllable
Example 3: uxx
0
1
20
01 , xy 11 (n = 2)
For Controllability
21)(
00
11
xx CRankABBC Uncontrollable
{
⇒
For Observability
2)(
21
11
xx ORank
CA
C
O Observable
Let us set C as 01C , then
21)(
01
01
xx ORankO Unobservable
2
1
22
11
xy
xx
uxx
C/O
C/NO NC/O
NC/NO
u y
Uncontrollable mode is also unobservable
(in addition to being unstable)
Page 73 of 115
12.2 Canonical Transformations 73
12.2.2 Controllable Canonical Form (C.C.F.)
12.2.2.1 Theorem
DuCxy
BuAxx
RuyRx n ,, , assume that the system is controllable i.e.
is nonsingular. We want to find a nonsingular transformation
xTx cc
1 , so that in the new coordinate system we have
uDxCy
uBxAx
ccc
cccc
where
[
]
[ ]
cC does not have any particular form and DDc .
For xTx cc
1 , BuTAxTBuAxTxTx ccccc
1111 )( .
Since ccxTx
uBTxATTx
ccB
cc
A
ccc
11
, uDxCTy
cc
D
c
C
c
DD
CTC
TB
ATTA
c
cc
cc
ccc
B1
1
The controllability matrix for the new system
)()())(( 1111111 BTATTBTTATBTBABABC c
n
cc
I
ccccc
n
ccccxc
xc
C
n
c
n
ccc CTBAABBTBATABTBT
x
1111111 )
)1)(1()1)(1()1(1)1(
1
B
I
cT
cT
A
A
I
cT
I
cTA
I
cT
cTA
cTB
cTn
cAT
cT
n
Page 74 of 115
74 Linear system: Lecture 12
By assumption, original system is controllable, nCRank x )(
existCandCnCTRankCRankcc xxccx
111 )(
111 xxcxxc CCTorCCTcc
12.2.2.2 Procedure for Controllable Canonical Form
0) Check to see if the system is controllable.
If yes, proceed to the following steps; if not, stop as there is no CCF possible.
1) Determine the coefficients of the C.P. of matrix A.
2) Form cA and cB .
3) Calculate xC andcxC , using A, B and cA , cB .
4) Find the transformation1cxxc CCT .
5) Use cc CTC to get cC .
6) As a check, verify that ccc ATTA 1 and BTB cc
1 .
7) DDc .
12.2.3 Observable Canonical Form (O.C.F.)
12.2.3.1 Theorem
DuCxy
BuAxx
Given this system, find a transformation that takes the system into an
observable canonical form. The transformed system takes the form
uDxCy
uBxAx
ooo
oooo
Where
[
]
oB does not have any particular structure, DDo
Page 75 of 115
12.2 Canonical Transformations 75
Define xTx oo
1 , ooxTx BuTxATTBuAxTxTx ooooooo
1111 )(
DuxCTDuCxy oo
DuxCTy
BuTxATTx
oo
ooooo
11
DD
CTC
BTB
ATTA
o
oo
oo
ooo
1
1
Form xO and oxO as
1n
x
CA
CA
C
O
and
1n
oo
oo
o
x
AC
AC
C
Oo
, it follows that oxx TOO
o .
)()( oxx TORankORanko . Since oT is full rank, xO is full rank nORank
ox )(
oxO is nonsingular! oo xxoxxo OOTorOOT 111
12.2.3.2 Procedure for Observable Canonical Form
0) Check to see if the system is observable.
If yes, proceed to the following steps; if not, stop as there is no possible OCF.
1) Find the coefficients of the C.P. of A.
2) Form oA and oC .
3) Calculate xO and 0xO using A, C and oA , oC .
4) Find the transformation xxo OOTo
11 .
5) Use BTo
1 to get BTB oo
1 .
6) As a check, verify that ooo ATTA 1 and oo CTC .
7) DDo .
Page 76 of 115
76 Linear system: Lecture 12
Example 4: System given as uxx
1
1
2
020
113
021
, xy 100
transform it into C.C.F and O.C.F.
Controllable form
0) Check that xC is nonsingular? √
1) C.P. = 0290)det( 3 AI
2)
092
100
010
cA ;
1
0
0
cB then ?cC
1221
861
1642
2BAABBCx ;
901
010
100
2
cccccx BABABCc
123
161
242
1
cxxc CCT
123
123
161
242
100
cc CTC
123
1
0
0
092
100
010
c
cc
xy
uxx
Page 77 of 115
12.2 Canonical Transformations 77
Observable Canonical Form
0)
126
020
100
2CA
CA
C
Ox nonsingular Observable
T
co AA
010
901
200
; T
co BC 100
901
010
100
2
oo
oo
o
x
AC
AC
C
Oo
T
cooxxo CBTBOOTo
1
2
3
11
78
Objectives
1) Diagonalization of a matrix – Jordan canonical form
2) Controllability of J.C.F; Observability of J.C.F.
13 Lecture 13
Diagonalization of a Matrix 13.1
13.1.1 Jordan Canonical Form (J.C.F.) – Distinct Eigenvalues Case
Given an nxn matrix A with distinct eigenvalues, the diagonal form of A denoted by A
takes the form
n
A
00
00
00
ˆ2
1
where AMMA 1ˆ , with 1 2 nM x x x
( known as modal transformation).
Given
DuCxy
BuAxx, transformation into Jordan Canonical Form is achieved through
JJ MxxxMx ,1, BuMAMxMBuMAxMxMx JJ
11111
11
DuCMxy
BuMAMxMx
J
JJ
, if A has n distinct eigenvalues, then choose M as the modal
matrix, then
00
00
00
21
2
1
2
1
Duxcccy
u
b
b
b
xx
Jn
n
J
n
J
Eigenvectors corresponding to eigenvalues
Page 79 of 115
13.1 Diagonalization of a Matrix 79
The above system in Jordan Canonical Form is
Controllable if and only if niibi ,...,2,1,0 and
Observable if and only if niici ,...,2,1,0
Example 1: uxx
0
1
43
10
, √ √
Transform it into J.C.F. and also check its controllability and observability
3
10)det(
43
10
2
1
AIA
Given a matrix A and eigenvalue λ, the eigenvalue and eigenvector v
relations are given by: reigenvectoisvwherevAv ,
1
1
1
1
033
0
1
43
10
1
2
1
21
21
2
1
2
1
1
1
1
vx
x
xx
xx
x
x
x
x
vvA
Similarly,
Without loss of generality, normalize to a unit length, named
√
√
3
1
3
1
03
03
3
43
10
2
2
1
21
21
2
1
2
1
2
2
2
vx
x
xx
xx
x
x
x
x
vvA
Normalizing to unit length
√
√
1 1 2 2
1 1 1 31 3
2 2 10 10
TT
nv nv
,
Page 80 of 115
80 Linear system: Lecture 13
Modal matrix:
2
10
2
10
2
2
2
23
2
1
2
1
10
1
10
3
2
20
10
3
2
1
10
1
2
1
1MM
30
011 AMM
uxx JJ
2
10
2
23
30
01
Jxy
10
220
43
10A ,
0
1B , √ √
30
01
ABBCx rank = 2 Controllable
2323
22
CA
COx rank = 1 Unobservable
≠
𝑖 𝑜𝑛 𝑟𝑜𝑙𝑙 𝑙
≠
𝑖 𝑜𝑛 𝑟𝑜𝑙𝑙 𝑙
𝑖 𝑛𝑜 𝑟𝑣 𝑙
≠ 𝑖 𝑜 𝑟𝑣 𝑙
Page 81 of 115
13.1 Diagonalization of a Matrix 81
13.1.2 Jordan Canonical Form (J.C.F.) – Repeated Eigenvalues Case
13.1.2.1 Diagonalization of Matrices
An n×n matrix A may be diagonalized by a similarity transformation where AMMA 1ˆ
if and only if nullity of )( IA i (which is )( IArankn i ) is equal to the multiplicity
of the eigenvalues, and moreover the modal matrix M is the matrix of eigenvectors.
Therefore, if multiple eigenvalues, then same number of linearly independent
eigenvectors. If this is not the case, then a diagonal form is not possible, and Jordan form is
used as the standard form.
Example 2:
21
10A 1,10)1()det( 2 IA
111
11)(
rankIArank ,
therefore, nullity of one linearly independent eigenvector.
For 1 A cannot be diagnolized.
We will show later in example3 that the proper form is
10
11A
13.1.2.2 Generalized Eigenvectors
A nonzero vector x for which 0)( 1 xIA k but 0)( xIA k is called a generalized
eigenvector of rank k of A associated with the eigenvalue .
Generalized eigenvectors are defined as
xxk
kk xIAxIAx )()(1
1
2
2 )()( kk xIAxIAx
2
1
1 )()( xIAxIAx k
generalized eigenvectors kxxx ,,, 21
Page 82 of 115
82 Linear system: Lecture 13
Theorem:
The set of generalized eigenvectors kxxx ,...,, 21 for the same eigenvalue are
linearly independent.
The generalized eigenvectors of A associated with different eigenvalues are
linearly independent.
Example 3:
21
10A , 1,1
Here k = 2, 2xx is computed from 0)( 2 xIA and 0)( 2
2 xIA
0
00
000
11
11
22
2
xx Any vector that satisfies 0)( 2 xIA
can be used. Choose Tx 012
011)( 2 T
xIA
Calculate
1
1
0
1
11
11)( 2
12
1 xIAx
21
01
11xxM
10
11ˆˆ 1 AAMMA
Example 4:
201
010
001
A
1,1,20)2()1()det( 2 IA
201
010
001
IA
Page 83 of 115
13.1 Diagonalization of a Matrix 83
Q : Nullity ? IA i when 1i
101
000
000
IA , 1)( IArank ,
1 of 213)( tymultipliciIArankn
2 linearly independent eigenvectors for 1
0
1
0
,
1
0
1
0
101
000
000
)( 2111 xxxxIA
For
1
0
0
0
001
010
001
2 33 xx
200
010
001
ˆ
101
010
001
1
321 AMMAxxxM
13.1.3 Matrix Reduction to Jordan Form
13.1.3.1 Definition
The Jordan form J of an n×n matrix A is an upper triangular matrix whose diagonal entries
are of the matrix form
j
j
j
j
j
jijL
1
1
1
1
)(
0
0
Page 84 of 115
84 Linear system: Lecture 13
The general case of the Jordan Matrix J is
[
]
)( jijL matrices are called Jordon blocks.
13.1.3.2 General Rules for Constructing the Jordan Matrices
1) There is one and only one linearly independent eigenvector associated with each
Jordan block. In general, there are )( IArankn i linearly independent
eigenvectors for a general n×n matrix A with its associated eigenvalue i .
2) The number of 1’s directly above the diagonal terms equal the order of A minus the
number of linearly independent eigenvectors r, in another words, rnns
'1
3) The number of Jordan blocks associated with each eigenvalue is equal to the
number of linearly independent eigenvectors associated with each eigenvalue r.
Example 5:
[
]
[
]
For JCF.in one 1 134)# r(with 1 onlyrnblocksofJ
For JCF.in sone' 2 224)#(2 onlyrnblocksofrwithJ
1 generalized eigenvector; 1 linearly independent eigenvector for
1 l.i.e.v. for
1 l.i.e.v. for
2 g.e.v.;
1 l. i.e.v. for
1 l.i.e.v. for
Page 85 of 115
13.1 Diagonalization of a Matrix 85
Example 6: Transform A into Jordan form
5972
1000
0100
0010
A
2,1,1,102795)det( 234 IA
)(....# IAranknveilof i
3
0000
0100
0010
0001
rank
4-9-7-2-
1100
0110
0011
rank)( 1
IArank
-1 with associated l.i.e.v. 13-4only .
Thus
[ ]
( # of 1’s in J are n - 2 = 4-2 = 2,
where A has 2 l.i.e.v.) Check: Let 4321 xxxxM
Define 3x by
0)(
0)(
3
2
1
3
3
1
xIA
xIA
and subsequently
312 )( xIAx , 211 )( xIAx and 0)( 42 xIA .
For 3x , we have: 0)( 3
3 xIA 0
824248
412124
2662
1331
3
x
only 1 linearly independent vector
Txdcba 1001033 3
TxIAx 2101)( 32 ,
TxIAx 1111)( 21
4x is calculated from TxxIA 84210)2( 44
AMMAxxxxM 1
4321ˆ √
Page 86 of 115
86 Linear system: Lecture 13
13.1.3.3 A Procedure for Generating a JCF
1) Compute the eigenvalues of A by solving 0)det( IA . Let n ,, 21 be these
distinct eigenvalues with multiplicity nnnn ,, 21 , respectively.
2) Compute the number of linearly independent eigenvectors for1 . This is also the
number of Jordan blocks for1 .
3) For eigenvalue 1 determine its Jordan blocks from the information in (1) and (2)
and the general rules.
4) Determine the linearly independent eigenvector for 1 associated with each Jordan
block and the remaining generalized eigenvectors for each block. Denote the
linearly independent one with the subscript of the order of the Jordan block. Thus if
the Jordan block were of order 3, then denote the linearly independent vector as 3x
and the generalized eigenvectors as 2x and
1x ..
5) Repeat steps (2), (3), and (4) for eigenvectors n ,, 32 to generate a Modal
matrix by placing the eigenvectors as columns. The order of the column placement
is important and is directly related to how they are numbered in step (4).
As an example, suppose that A is 55 with eigenvalues 32111 ,,,, . Also, suppose
there are two linearly independent eigenvectors associated with1 , that is
2)-(A ofnullity 3)( 11 IIArank . Therefore, there are two Jordan blocks
associated with 1 as 1111 )( J &
1
1
1210
1)(
J
The Jordan form is
[
]
[
]
Page 87 of 115
13.2 Controllability of JCF and Observability of JCF 87
Assume we have found for block )( 111 J an eigenvector1x from
11 xAx .For block
)( 121 J , we need two generalized eigenvectors 3x from 0)( 3
2
1 xIA and
231 )( xxIA . Eigenvalues 2 and 3 will contribute the two eigenvectors
4x and 5x ,
respectively from 424 xAx and 535 xAx .
The Modal matrix is formed by 54321 xxxxxM
Controllability of JCF and Observability of JCF 13.2
13.2.1 Theorem
DuCxy
BuAxx qpn RyRuRx ,,
m
nn
A
A
A
A
2
1
,
m
pn
B
B
B
B2
1
, mnq CCCC 21
A has m distinct eigenvalues m ,, 21 ; iA denotes the Jordan blocks associated with i .
r(i) is the number of Jordan blocks in iA ; ijA is the j-th Jordan block in iA .
1 2 1 2 ( ), , , ; , , ,m i i i ir iA diagn A A A A diagn A A A
)(
2
1
)(
iir
i
i
nni
A
A
A
Aii
;
)(
2
1
)(
iir
i
i
pni
B
B
B
Bi
( ) 1 2 ( )ii q n i i ir iC C C C
m
i
ir
j
ij
m
i
i nnn1
)(
11
i
i
i
i
nnij ijijA
1
1
1
)( ,
lij
ij
ij
pnij
b
b
b
Bij
2
1
)(,
T
lij
ij
ij
pnij
c
c
c
Cij
2
1
)(
Page 88 of 115
88 Linear system: Lecture 13
The n-dimensional LTI Jordan form is controllable if and only if for each i =1, 2,…,m,
the rows of the pir )( matrix
)(
2
1
ilir
li
li
l
i
b
b
b
B
are linearly independent.
and the system is observable if and only if for each i = 1, 2,…,m the columns of the
)(irq matrix are linearly independent.
13.2.2 Examples
Example 7:
[
]
[ ]
lecontrollabBrankfullB ll mode
100
010
001
111
lecontrollabBl modetindependenlinealry 100 22
observable modeCrank full
321
211
021
1
1
1
1
1
C
𝑙
𝑙
𝑙
𝑙
Page 89 of 115
13.2 Controllability of JCF and Observability of JCF 89
leunobservab mode0 rank
0
0
0
2
1
2
C
Block diagram of this system:
Figure 13-1. Block diagram.
If 021lb all modes in the chain can be controlled;
If 0121c all modes in the chain can be observed.
Example 8:
[ ]
[ ]
What is the JCF for a system with complex conjugate eigenvalues?
𝑙
+
u
u
y
u y
u y
u
u u
y y
y
𝑙
𝑙
𝑙
+
+
𝑙
uncontrollable.
≠ controllable.
≠ observable. ≠ observable.
Page 90 of 115
90 Linear system: Lecture 13
Let u
b
b
x
x
A
A
x
x
2
1
2
1
1
1
2
1
0
0
,
2
1
11
x
x
ccy
where 1A is the Jordan block associated with and 1A is the complex
conjugate of 1A , in another word, 1A is the Jordan block associated with 1 .
Introduce the equivalence transformation Pxx
where
jIjI
II
P and
jII
jII
P2
11
It can be verified that
1 11 1 1
2 21 1 1
1
1 12
Re Im 2Re
Im Re 2Im
Re Im
A A bx xu
A A bx x
xy c c
x
This system can now be used in computer simulations.
Example 9: jjA 21 , 21 , 2
[
]
[
]
Let Pxx , where
[ ]
Page 91 of 115
13.2 Controllability of JCF and Observability of JCF 91
[
]
[ ]
All the coefficients are real!
Example 10: jjA 32 , 1
[
]
[
]
A
��
92
Objectives
State Space Control Method
1) State Feedback: General Case
2) State Estimation: Full Order Observer
3) State Estimation: Reduced Order Observer
14 Lecture 14
Stable Feedback: General Case 14.1
State: RuRxBuAxx n ,,
Control: Kxu with nnKKKK 121
Figure 14-1. Closed-loop system state diagram.
matrix loop-closed
A )( f
BKAA
xxBKAx
f
,
the characteristic polynomial for the closed-loop system is
0)det()det( BKAsIAsI f
Let the Design Specification require closed-loop eigenvalues at n ,,, 21 .
0)())(()( 01
1
121
sssssss n
n
n
nc
Pole-placement design procedure is achieved by setting
01
1
1)det(
sssBKAsI n
n
n
y + B
r ʃ
D
+
A
+
Ax
+
+
K
-
C u
Page 93 of 115
14.1 Stable Feedback: General Case 93
Solve for n unknown nKK ,...,1 . The equations are linear!
K can be computed directly from this formula if and only if (A,B) is controllable.
Consider the transfer function 01
1
1
01
1
1)(asasas
bsbsbsG
n
n
n
n
np
The controllable canonical form is
u
aaaa
x
n
1
0
0
0
1000
0100
0010
1210
xbbby n 110
The control law is Kxu
Closed - loop fA matrix is
nn
f
KaKaKaKa
BKAA
1322110
1000
0100
0010
Characteristic polynomial 0)det( BKAsI
0)()()( 1021
1
1
KasKasKas n
nn
n
Since desired characteristic polynomial is
0)( 01
1
1
ssss n
n
n
c
niaK iii ,...2,111
Alternatively, K can be computed according to the Ackermann’s Formula as follows:
Page 94 of 115
94 Linear system: Lecture 14
1
2 1
1
1 1 0
0 0 0 1 ( )
( )
x
n n
c
C
n n
c n
K B AB A B A B A
where A A A A I
Example 1: uxx
1
0
00
10
Let the desired characteristic polynomial be
0)()( 2121
2 sssc
Based on Ackermann’s formula, need to compute xC
01
10
01
101
xx CABBC
21
2121
2121
2
0)()(
IAAAc
212121
1)(10
KKKAABBK c
Example 2: Let the design specifications require a critically damped system with a
settling time of 1 sec., i.e, 14 .
4damped),y (criticall141
nnn
168)4)(4()( 2 sssssc
8,16 212211 KK
If we compute the closed-loop transfer function, we get the figure 14-2.
Also, 168
101)(
2
1
ssBBKAsIsT as desired!
As a different example, let2.707, .25 4 4 ( ) 8 32cs j s s s
1 1 2
2 1 2
3232 8
8
KK
K
Page 95 of 115
14.2 State Estimation: Full-Order Observer 95
Plant
x(t)
State
Estimator u(t)
y(t)
Figure 14-2. System plot for example 2.
State Estimation: Full-Order Observer 14.2
14.2.1 Theorem
RyCxy
RuRxBuAxx n ,
The estimator equation is expressed as
yGuHxFx nnnn 11ˆˆ
We need to choose matrices F, G, and H such that )(ˆ tx is accurate estimate of x(t). Then
in the control system, the estimated states are used to generate the feedback control, i.e.
xKu ˆ . To do this, the transfer function from u to ix must be equal to the transfer
function from u to ix , i =1,2,…,n, that is
nisU
sX
sU
sX
i
i
i
i ,...2,1)(
)(
)(
)(ˆ
The Laplace Transform of state equations are
( ) (0) ( ) ( ), 0
( ) ( ),
sX s x AX s BU s x( ) is unknown
Y s CX s A, B, C, and D known
By neglecting i.c. )()()( 1 sBUAsIsX
)(
)(
sU
sX i can be obtained.
The Laplace Transform of estimator is (neglecting i.c.)
)()()(ˆ)()()(ˆ)(ˆ sGCXsHUsXFsGYsHUsXFsXs
Since )()( sCXsY
)()()()()()()(ˆ 111 sUBAsIGCHFsIsGCXsHUFsIsX
t .5 1
1.5
Figure 14-3. Block diagram of combined plant & observer.
Page 96 of 115
96 Linear system: Lecture 14
We want )(
)(ˆ
)(
)(
sU
sX
sU
sX ii
BAsIGCHFsIBAsI 111 )()()(
Collecting BAsI 1)( terms
HFsIBAsIGCFsII 111 )()()(
Since IFsIFsI )()( 1 HFsIBAsIGCFsIFsI 111 )()()(
HGCFsIBAsIHBAsIGCFsI 111 )()()(
This equation is satisfied if we choose AGCFandBH
BHandGCAF
G is chosen such that an acceptable transient response or frequency response for the state
estimator is achieved.
xGCGyBuxAx ˆˆˆ
xCyGyBuxGCAx ˆˆ,ˆ)(ˆ
ˆˆ
)ˆ(ˆˆ
xCy
yyGBuxAx
Figure 14-4. Plant and observer diagram.
Let )(ˆ)()( txtxte GyBuxGCABuAxxxe ˆ)(
Since Cxy , hence )ˆ()(ˆ)( xxGCAxGCxGCAAxe
eGCAe )(
y + B
u
ʃ x
A
+
C
observer gain
+ B ʃ
A
+ +
C
+
G -K
Plant
Page 97 of 115
14.2 State Estimation: Full-Order Observer 97
The error in the estimation of the states is governed by the above dynamic equation with
the characteristic polynomial: 0)det( GCAsI
The gain vector G is chosen to make the dynamics of the estimator faster than the open-
loop system dynamics (~ 2 to 4 times faster).
14.2.2 Design of the State Estimator
1) GyBuxGCAx ˆ)( , the characteristic polynomial is 0)det( GCAsI
2) Let the desired char. poly. be )(se , that reflects the desired transient behavior:
0)( 01
1
1
ssss n
n
n
e
3) The gain matrix G is calculated to satisfy )()det( sGCAsI e
4) Using the transformation xTx oo
1 transforms the system into observable
Canonical form
{
[
]
[
]
Now GCA becomes
[
]
01
1
1)()det(
ssssGCAsI n
n
n
e
niag iii ,...2,111
Now using Ackermann’s formula we have [if and only if (A, C) is observable]
)(
1
0
0
)(
01
1
1
1
1
1
IAAAA
CA
CA
C
AG
n
n
n
e
n
en
Page 98 of 115
98 Linear system: Lecture 14
Example 3:
xy
uxx
01
1
0
00
10
A controller was designed for the characteristic polynomial 328)( 2 sssc ,
which yielded a time constant 707.sec,25.
We choose the estimator to be critically damped with a .sec1.
10020)10()( 22 sssse
1000
2010010020)( 2 IAAAe
Now
10
01,
10
011
xx OCA
CO
100
20
1
0)(
1
CA
CAG e
2
1
21
g ; 100g , 20
gGg
Example 4: Combining previous two examples:
100
20,832 GK
The estimator equations GyxBKGCAGyBuxGCAx ˆ)(ˆ)(
Since xKu ˆ , now
8132
120BKGCA
Observer/Controller
The T.F. from output u to input y is GGCBKAsIKsGec
1)()(
29228
32001440)(
2
ss
ssGec
The transfer function of the controller-estimator is
GGCBKAsIKsGec
1)()(
ˆ832
100
20ˆ
8132
120ˆ
xu
yxx
acting as input
acting as output
Page 99 of 115
14.2 State Estimation: Full-Order Observer 99
Figure 14-5. Block diagram for example 4.
The characteristic polynomial of the closed-loop system is
0)()(1 sGsG pec
Example 5: xyuxx 01,1
0
00
10
The controller-estimator is given by
Signal flow graph of the controller-estimator
Figure 14-6. Signal flow graph.
Defining
[
]
[
]
[
]
[
]
Figure 14-7. Estimator response.
Controller-Estimator Plant
+
-
R = 0 y
ˆ832
100
20ˆ
8132
120ˆ
xu
yxx
R 1 1 1 u 20
100
32
8
-20
+1
-132
-8 -1
t .5 1 1.5
y
≠
Page 100 of 115
100 Linear system: Lecture 14
14.2.3 Closed-Loop System Characteristics
The characteristic polynomial with full state feedback is
and the characteristic polynomial of a state estimator is
We now derive the characteristic polynomial of the closed-loop system.
First consider , the plant equations are {
Hence,
The error dynamic equation is , therefore the closed-loop system is
[
] [
] [
]
The characteristic polynomial of the closed-loop system is
[
]
The 2n roots of the closed-loop characteristic polynomial are then the n roots of the pole-
placement design plus the n roots of the estimator. This is fortunate, since otherwise the
roots from the pole-placement would have been shifted by the addition of the estimator.
This is called the separation principle.
Example 6: From the previous examples we have
We have
or equivalently from ( )
[
]
The above characteristic polynomial is obtained.
Page 101 of 115
14.3 Reduced-Order Estimators 101
Reduced-Order Estimators 14.3
Estimators developed so far are called full-order estimators since all states of the plant are
estimated. However, usually from output measurement some states are directly available.
Hence it is not logical to estimate states that are directly measured.
Assume without loss of generality that
Partition x into [
], where are the states to be estimated.
Partition also[
] [
] [
] [
]
The equation of the states to be estimated
And the equation of the state that is measured where is
unknown and to be estimated and and u are known.
To derive the equation of the estimator observer that for the full state estimation
{
For the reduced order observer we have
unknown
ee
knownknownknown
eeeee
unknown
e
xAubxax
uBxAxAx
1111111
11
)(
)(
Comparing we have
e
ee
ee
ee
AC
ubxaxy
uBxABu
AA
xwithxreplacetxtx
1
11111
1
)()(
Making these substitutions for the full-order observer equations yield
]ˆ)(ˆ[ GyBuxGCAx
)(ˆ)(ˆ11111 ubyayGuByAxAGAx eeeeeeeee
Page 102 of 115
102 Linear system: Lecture 14
Thus the error dynamic is ee xxe ˆ eAGAe eeee )( 1
Thus, the characteristic polynomial of the reduced-order estimator is
0)det()( 1 eeeee AGAsIs
We then choose to satisfy this equation where is chosen to give the estimator desired
dynamics. This can be achieved if and only if ),( ieee AA is observable.
Ackermann’s formula yields
1
0
0
)(
1
2
1
1
1
n
eee
eee
e
eeee
AA
AA
A
AG
The above estimator requires measurement of y . To relax this requirement, define a new
variable yGxxoryGxx eeeeee 11ˆˆˆˆ
Substituting in observer dynamic yields
)()ˆ)((ˆ1111111 ubyayGuByAyGxAGAyGx eeeeeeeeeee
ubGByGAGGAaGAxAGAx eeeeeeeeeeeeeeee )()(ˆ)(ˆ11111111
The control u is
)ˆ(ˆ111 yGxKyKxKyKu eeeee
Where , .
Figure 14-8. Block diagram of the combined plant and reduced order observer.
Plant
x
Reduced Order
Observer
u y
- -
+ +
+
Page 103 of 115
14.3 Reduced-Order Estimators 103
Example 7: { [
] [
]
the controller designed earlier yielded
As for the estimator of reduced order (1st order), let
From Ackermann’s formula:
The estimated value of is then
Example 8: If we combine plant equations with the estimator and the control we have
The signal flow graph becomes
Figure 14-9. System signal flow graph.
The characteristic polynomial is
⏟
⏟
Now opening the graph from u we calculate
1 -8 u 1
-32 -10 1
-100 10
Page 104 of 115
104 Linear system: Lecture 14
Figure 14-10. Closed-loop combined plant and controller estimator.
√
Reduced-Order State Estimator: General Case 14.4
Consider
q
pn
RyCxy
RuRxBuAxx
,
,,
Assumption: C has full rank,
Define: [
] , where arbitrary as long as P is nonsingular.
Define
[
] [
] [
]
Define the equivalence transformation
{
Partition above system as
{
[
] [
] [
] [
]
Therefore, only the last (n - q) elements of need to be estimated.
We need only an (n - q)-dimensional state estimator.
R = 0
Controller - estimator plant
+
-
Page 105 of 115
14.4 Reduced-Order State Estimator: General Case 105
Using {
Define
Lemma:
The pair (A, C) or equivalently is observable, iff. the pair is observable.
Therefore, a (n - q)-dimensional estimator of in the form
)()(ˆ)(ˆ221111212222 uByAuByAyGxAGAx
such that the eigenvalues of can be arbitrarily assigned by proper choice
of
To eliminate define
yGxz 2ˆ
Define
by proper choice of .
Now 1
2
ˆˆ
ˆ
yxx
Gy zx
, since
1
1 2 1 2
0
ˆˆ [ ] [ ]
q
n q
Iy yx Px x P x Qx x Qx Q Q Q Q
zGy z G I
which is an estimate of the original vector x.
Figure 14-11. Block diagram for system with reduced order estimator.
�� ��
ʃ
�� 𝐴
+ + + u
y
z
106
Objectives
1) Tracking Problem
2) Closed-Loop Pole-Zero Assignment
3) General Case Using Phase Variable Canonical Form
15 Lecture 15
Tracking Problem 15.1
1xy
rKKxu
BuAxx
r
Figure 15-1. State feedback diagram.
It is assumed that K is designed to meet a desired closed-loop characteristics. Problem is to
design to satisfy the tracking requirement.
⏟
, where a choice for is to defined as
Plant
r e u y
+ -
+
- -
Page 107 of 115
15.1 Tracking Problem 107
Example 1: System { [
] [
]
from lecture 14
Example 2. (Hyperlink)
Figure 15-2. State diagram for example 1.
For the general case, if ≠ , instead ,
then we express
since we want the system to be driven by the difference between the input
and output.
Comparing with
{
Hence one gain has to be determined from other design criteria.
32
8
r e y +
-
+
-
Page 108 of 115
108 Linear system: Lecture 15
Closed-Loop Pole-Zero Assignment 15.2
Figure 15-3. ADC motor system schematic.
{
Figure 15-5. System block diagram.
( ) ( )
( ) 1 ( ) ( )eq
Y s G s
R s G s H s
,
2
200( )
( 6 5)
AG s
s s s
3 2
3 2 3 1
( ) 200
( ) (6 100 ) (5 200 100 ) 200
Y s A
R s s k A s k A k A s k A
𝑖 + e
-
Amplifier
A
Current
Sensor
+
-
+
-
+ -
𝑖
Amp.
Fixed Field
B
Velocity Sensor
(Tachometer)
Position Sensor
(Potentiometer)
+
-
+
-
A
R U E
0.1
y
+
+ -
+
-
+ +
+
Figure 15-4. System close-loop block diagram.
Page 109 of 115
15.2 Closed-Loop Pole-Zero Assignment 109
2
3 2 3 1(2 ) 2( )
2eq
k s k k s kH s
must be designed such that | and other design specifications are
satisfied.
Observations:
1) To have zero steady state error due to
.
2) Poles of are poles of G.
3) Zeros of are zeros of
4) Use of state feedback produces additional zeros in loop transfer function without
adding poles! Additional zeros move the root locus to the left system is more
stable and time response is improved.
Figure 15-6. Root Locus.
Let the desired closed-loop poles be located at
Therefore, solve for
-3
j4
-j4
s - plane
-8
Page 110 of 115
110 Linear system: Lecture 15
General Case Using Controllable Canonical Form 15.3
Let
1
1 1 0
1
1 1 0
( )( )( ) , m n
( )
m m
m
n n
n
K s c s c s cY sG s
U s s a s a s a
0 1 2 1
0 1
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
1
0 0
n
m
x x u
a a a a
y c c c x Cx
1 1
0 1 1
( ) ( )( ) ( )( )
( ) ( ) ( ) ( )
T T
n neq
m
k X s k X sk X s k X sH s
Y s CX s c X s X s
Since
1 21 1 1 2 1
1 1
1 1 0
( ) ( ) ( ) ( )n n
j j n nj eq n m
m
k s k s k s kX s s X s s Y s H s
s c s c s c
1 2 1 2
1 2 1 2 1 0
1 1
1 1 0 1 1 0
( )
n n n n
n n n n
eq n n n n
n n
K k s k s k s k k K s s sGH s
s a s a s a s a s a s a
Now
1
1 0
1
1 0 1
( )( )
( ) ( ) ( )
m m
m
n n
n n
K s c s cY s
R s s a Kk s a Kk
Example 2: Given system as below,
design a state feedback such that due to a unit step is zero and
%OS = 4.3% and
A
U(s) Y(s)
Page 111 of 115
15.3 General Case Using Controllable Canonical Form 111
Figure 15-7. State diagram for example 1.
Figure 15-8. Block diagram of the closed-loop system for example 1.
To get 0 10, ( ) 1ss sunit step
e with R R u t k
Given 1,2% 4.3 0.708 0.7064
5.65, 0.708s n
OS s j
T
Now we place the 3rd
pole arbitrarily at 3 100s
Therefore, the desired closed–loop transfer function is
3 2
( ) 10 10
( ) ( 100)( 0.708 0.7064) 101.4 142.6 100
Y s A A
R s s s j s s s
3
1
2 3
2
3
106 2 101.4
15 ( )10 142.6
3.39310 100
4.77
AAk
kk k A
kA
k
2
100( ) ( 1 )
( 101.4 142.7)eqG s type systm
s s s
Figure 15-9. Closed-loop diagram 3 for a unity feedback of example 1.
A
R U
+
+ + -
+ + +
Y(s)
𝐴
R U Y
+
-
G
R E Y
Page 112 of 115
112 Linear system: Lecture 15
Figure 15-10. Root Locus for the standard feedback control system of example 2.
1 0
1 1
( )( 1)( 5)
2 , 0
(2 1) , 0
KGH
GH GHKK K G s
s s sGH k k
GH k k
Figure 15-11. Root Locus for the state feedback control system of example 2.
9.54 ( 0.721 0.726)( )
( 1)( 5)eq
A s jGH s
s s s
As a result of vicinity of desired pole locations with the zero of the controller,
a system is insensitive to variations in A.
s - plane
-2 -3 -4 -5 -6
-1
-2
2
0
3
×
×
×
1
Root Locus
-1
s - plane
-2 -5 -100
-1
-2
2
0
3
×
× 𝐴 𝐴
×
1
Root Locus
𝐴
𝐴
-1
𝐴
𝐴
𝐴
𝐴
Page 113 of 115
15.3 General Case Using Controllable Canonical Form 113
Example 3: Given system as below,
Figure 15-12. System sate diagram for example 3.
Figure 15-13. Closed-loop block diagram for example 3.
3 2
3 2 3 2 1 1
( ) ( 2)
( ) [6 ( ) ] [5 ( 2 ) ] 2
Y s A s
R s s k k A s k k k A s k A
A desired closed-loop T.F. is selected as
1.22 2
3 2
( ) 2( 2) 2; 1 , 2, 0.707
( ) ( 2 2)( 2) 2 2
2( 2)
4 6 4
n
Y s ss j
R s s s s s s
s
s s s
1 2 31, 0.5, 1.5k k k
Figure 15-14. Root Locus for example 3.
A
R U
+
+ + -
+ + +
Y(s)
𝐴
R U Y +
-
×
s - plane
-2 -5
-1
-2
2
0
3
× 𝐴 𝐴
×
1
Root Locus
𝐴
-1
𝐴 𝐴
-1.18
1.68
𝐴
𝐴
𝐴
No non-dominant pole.
Page 114 of 115
114 Linear system: Lecture 15
2( 0.5 2)( )
( 1)( 5)eq
A s sGH s
s s s
Unfortunately the system is sensitive to variations in A. In order to achieve
this we must have one non-dominant pole. Let the desired closed-loop T.F.
be 1 2 32
( ) 2( 2), 50, 0.5, 47.5
( ) ( 2 2)( 100)
Y s sk k k
R s s s s
2 248 98.5 100 48 98.5 100, ,
( 2) ( 1)( 5)eq eq
s s s sH Hence GH
s s s s
Figure 15-15. Root Locus for example 3.
Example 4: Alternatively, since in no dominant pole cases there are 3 dominant poles
at must have 3 zeros to be insensitive to A.
However, to have 3 zeros in requires having 4 poles in G. Therefore, one
pole is added to G by using a cascade compensator
Then
can have one non-dominant pole p and
2
( ) ( 2)
( ) ( 2 2)( 2)( )desired
Y s A s
R s s s s s p
Selection of A and p are independent since
The closer poles and zeros are the less sensitivity of changes in A.
Selecting is non-dominant.
4 3 2
( ) 100( 2)
( ) 54 206 304 200
Y s s
R s s s s s
s - plane
-1 -5
-1
1
0
×
× 𝐴 𝐴
×
Root Locus
𝐴
𝐴
𝐴
Sensitivity to variations in
A is reduced significantly
Page 115 of 115
15.3 General Case Using Controllable Canonical Form 115
Figure 15-16. System state diagram for example 4.
Figure 15-17. Closed-loop block diagram for example 4.
To guarantee a controllable and observable plant ≠ , let a = 1.
Therefore, with A = 100, a =1
4 3 2
4 4 3 2 4 3 2 1 1
( )
( )
100( 2)
(7 100 ) [11 100(6 )] [5 100(5 )] 200
Y s
R s
s
s k s k k k s k k k k s k
Due to a unit step reference input, with
3 2
4 3 2
0.47 ( 4.149 6.362 4.255)
7 11 5eq
A s s sGH
s s s s
Figure 15-18. Root Locus for example 4.
A
R U
+
+
-
+ + +
Y(s)
+
+
𝐴
R U Y +
-
×
s - plane
-1 -2
-1
1
0
× × 𝐴 𝐴
Root Locus
𝐴 𝐴
𝐴
-5 ×
-50
𝐴
𝐴
-2.07
𝐴
𝐴
𝐴
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