LINEAR SYSTEM THEORY
LECTURE NOTES
Prof. Dr. Mario Edgardo Magaña
School of Electrical Engineering
and
Computer Science
Oregon State University
The course deals with the theoretical and practical aspects of linear dynamic systems as
they apply to engineering modeling, analysis and design. The mathematical concepts of
time and complex frequency domain representation of linear dynamic systems are covered
in detail. Furthermore, the theoretical foundations and application of dynamic system
stability are discussed thoroughly. Finally, the properties of controllability and
observability are studied in order to apply them to both feedback controller and state
estimator design.
MATLAB and Simulink are heavily used in the homework and term project to emphasize
the practical aspects of the course material.
Fundamental knowledge of linear algebra (matrices, determinants, vectors, eigenvalues,
eigenvectors, similarity transformations, etc.), differential equations, and signal and
system analysis is required. Also, knowledge of computer simulation will be helpful, as
you will often be performing dynamic system simulation in order to verify theoretical
results.
2
The behavior of physical systems can be characterized using heuristic or empirical methods
by applying a variety of input signals and observing their outputs. Furthermore, if such
behavior is not satisfactory, a compensating mechanism based on heuristic knowledge can
be introduced to modify the system behavior to meet design criteria. Such an approach is
based on experience and requires the use of trial and error. However, this methodology
may take an inordinate amount of time to achieve the desired behavior of the system. This
is further complicated when the complexity of the system increases to a point where trial
and error is no longer feasible.
This shortcoming may be addressed by using mathematical models of the system
components that take into account the limitations of the physical systems. When doing so,
the process of modifying the behavior of complex physical systems becomes manageable
because a large number of formal analysis and design approaches have been developed
over the years and the need for trial and error is practically removed from the design
process.
3
4
Mathematical System Description:
A system N is a device that maps a set of admissible inputs U to a set of
admissible output responses Y.
Mathematically, N: U Y or y() = N [u()].
Alternatively, a system can be described either by differential or difference
equations or by the impulse or unit-sample response in the time domain, or by
algebraic equations in the complex frequency domain.
Example: Let us describe the relationship between the input voltage and the
output voltage of the following active filter system:
where vi(t) is the input voltage and v0(t) is the output voltage.
5
Using nodal analysis and assuming ideal amplifiers, we get
Adding the currents at the nodes of the second amplifier yields
Assuming infinite input impedance (ideal amplifiers) , we get . Hence,
Furthermore, the two capacitor currents are described by
021
6
01
2
1
1
1
cc
i iiR
vv
R
v
R
vv
03
21
1
R
veic
0
5
02
4
22
R
ve
R
ve
1 2 0e e
1
52 3 0 2
4
= and c
Rv R i v v
R
6
Thus,
1
1
1
1 1 1 1 1
c
c
dv dvdi c c v e c
dt dt dt
2
2
1 2
2 2 1 2 2 2
c
c
dv dv dvdi c c v v c c
dt dt dt dt
12 3 1
dvv R c
dt dv
cRv 2
13
1
1
5 3 51 10 3 1 1
4 4
R R Rdv dvv R c c
R dt R dt
02 4
5
dvdv R
dt R dt
1 4 40 1 0
3 5 1 3 5 1
dv R Rv v v d
dt R R c R R c
7
In terms of the input and output voltages,
0 1 1 21 1 2
1 2 6 1 6
1 1 10iv v dv dv dv
v c cR R R R R dt dt dt
0 04 4 4 2 40 0 0 2
3 5 1 1 2 6 1 6 3 5 3 5 1 5
1 1 10iv v dvR R R c R
v d v v cR R c R R R R R R R R R c R dt
2
0 0 54 2 4 4 20 2
3 5 1 1 2 6 1 1 3 5 6 5 4
1 1 1 1 11 0idv dv d v RR c R R c
vR R c R R R R dt c R R R dt R dt R
2
5 5 0 020 2 2
3 1 1 2 6 1 4 1 3 4 6 2
1 1 1 1 1 11 0iR dv R dv d vc
v cR c R R R R R dt c R R R dt dt c
or
2
0 5 0 502
2 1 3 4 6 2 3 1 2 1 2 6 1 4 2
1 1 1 1 1 1 1 id v R dv R dvv
dt c c R R R c dt R c c R R R R R c dt
8
The last differential equation represents a time domain model of the active filter.
In the complex frequency domain, assuming zero initial conditions, the algebraic
relationship between the input and the output voltage is
Since both models assume linear behavior of the active amplifier circuits, we
could also obtain an input-output model in terms of the convolution relationship in
the time domain, namely,
where
5
1 4 20
2 5
2 1 3 4 6 2 3 1 2 1 2 6
( ) ( )1 1 1 1 1 1 1
( ) ( )
i
i
Rs
R R cV s V s
Rs s
c c R R R c R c c R R R
H s V s
0 ( ) ( ) ( )iv t h t v t
1( ) ( )h t H s L
9
Input-Output Linearity:
A system N is said to be linear if whenever the input u1 yields the output N[u1],
the input u2 yields the output N[u2], and
for arbitrary real or complex constants c1 and c2
Example:
Let the spring force be described by fk(x) = Kx, then
is an external force, and are the first and second derivatives of x
with respect to time.
22112211 uNcuNcucucN
)(tfKxxBxM a
)(tf a
M
B
kf x
x
af t
x x
10
Let x1(t) be the solution when fa(t) is replaced by fa1(t) and x2(t) be the solution
when fa(t) is replaced be fa2(t) .
Then, if , the total response is given by
x(t) =c1x1(t) + c2x2(t) .
Let the spring force be now described by fk(x) = Kx2, then
This time, however, x(t) ≠ c1x1(t) + c2x2(t) , i.e., the linearity property does not
hold because the system is now nonlinear.
Time Invariance and Causality
Let N represent a system and y() be the response of such system to the input
stimulus u(), i.e., y() = N[u()]. If for any real T, y( - T) = N[u( - T)], then the
system is said to be time invariant . In other words, a system is time invariant if
delaying the input by T seconds merely delays the response by T seconds.
)(2 tfKxxBxM a
1 1 2 2( ) ( ) ( )a a af t c f t c f t
11
Let the system be linear and time invariant with impulse response h(t), then
If the same system is also causal, then for t ≥ ≥ 0, (h(t) = 0, t < 0)
Example: Let a system be described by the ordinary, constant coefficients
differential equation
then the system is said to be a lumped-parameter system.
Systems that are described by either partial differential equations or linear
differential equations that contain time delays are said to be distributed-
parameter systems.
( ) ( ) ( ) ( ) ( )
t t
y t u h t d h u t d
0 0
( ) ( ) ( ) ( ) ( )
t t
y t h u t d u h t d
)()()('...)()( 1
)1(
1
)( tutyatyatyaty nn
nn
12
Example: Consider the dynamic system described by
According to the previous definition, this equation describes a distributed-
parameter system because of the presence of the time delays.
Definition: The state of a system at time t0 is the minimum (set of internal
variables) information needed to uniquely specify the system response given
the input variables over the time interval [t0, ).
Example:
vi(t): Input voltage (external stimulus or excitation)
i(t): Current flowing through circuit
y(t): Output (measured) variable (current flowing through inductor)
( ) 2 ( 1) 5 ( ) 4 ( 2)y t y t u t u t
13
Let the state be i(t) and the output be y(t) = i(t), then for t ≥ t0, i(t0) = i0,
and the solution is given by
Hence, regardless of what vi(t) is for t < t0, all we need to know to predict the
future of the output y(t) = i(t) is the initial state i(t0) and the input voltage vi(t),
for t ≥ t0.
State Models
They are elegant, though practical mathematical representations of the
behavior of dynamical systems. Moreover,
• A rich theory has already been developed
• Real physical systems can be cast into such a representation
( ) ( ) 1( ) ( ) 0 ( ) ( )i i
di t di t Rv t Ri t L i t v t
dt dt L L
0
0
( ) ( )
0 0
1( ) ( ) ( ) ,
t R Rt t t
L Li
t
i t e v d e i t t tL
14
Example:
By KVL, for t t0,
After taking the time derivative of the last equation and dividing by L, we get
0R L C
0
0
( ) ( ) 1( ) ( ) ( ) ( ) ( ) 0
t
C C
t
di t di tRi t L v t Ri t L i d t
dt dt C
2
2
( ) ( ) 1( ) 0
d i t R di ti t
dt L dt LC
15
To solve this second-order homogeneous differential equation, we may proceed
as follows:
Let 1 and 2 (1 ≠ 2) be the roots of the auxiliary equation
then, for t ≥ t0,
C1 and C2 can be uniquely obtained as follows:
From the knowledge of i(t0) and vc(t0) we can compute
and therefore C1 and C2.
012
LCL
R
)(
2
)(
10201)(
tttteCeCti
210 )( CCti
22110
)( CC
dt
tditt
0
)(tt
dt
tdi
16
Using a state variable approach, let x1(t) = vc(t) and x2(t) = i(t), then for t ≥ t0
or
or
This is a first-order linear, constant coefficient vector differential equation! In
principle, its solution should be easy to find.
)(1
)(1)()(
21 tx
Cti
Cdt
tdv
dt
tdx c
)()(1
)()(11
)()()(
21
)(
02
0
txL
Rtx
Ltvdi
CLti
L
R
dt
tdi
dt
tdx
tv
C
t
t
C
,)(
)(
1
10
)(
)(
2
1
2
1
tx
tx
L
R
L
Ctx
tx
dt
d
)(
)(
)(
)(
0
0
02
01
ti
tv
tx
tx c
0( ) ( ) , ( ).t A t tx x x
17
Specifically, for t ≥ t0 ,
The solution to the vector state equation is more elegant, easier to obtain
(provided there is an algorithm to compute eAt) and it specially makes the role
of the initial conditions (initial state) clear.
Linear State Models for Lumped-Parameter Systems
Consider the system described by the following block diagram
Mathematically, for t t0,
0( )0( ) ( )
A t tt e t
x x
0( ) ( ) ( ) ( ) ( ) , ( )t A t t B t t t x x u x
( ) ( ) ( ) ( ) ( )t C t t D t t y x u
B(t) C(t)
A(t)
D(t)
u(t) y(t)
x(t)x(t)
18
where x(t) Rn is the state vector, u(t) Rm is the input vector, y(t) Rr is the
output vector, A(t) Rnxn is the feedback (system) matrix, B(t) Rnxm is the
input distribution matrix, C(t) Rrxn is the output matrix and D(t) Rrxm is the
feed-forward matrix. Also, A(), B(), C() and D() are matrices whose entries
are piecewise continuous functions of time.
Definitions:
The zero-input state response is the response x() given x(t0) 0 and u() 0.
The zero-input system response is the response y() given x(t0) 0 and u() 0.
The zero-state state response is the response x() to an input u() 0 and
x(t0)=0.
The zero-state system response is the response y() to an input u() 0 and
x(t0)=0.
Let yzi() be the zero-input system response and yzs() be the zero-state system
response, then the total system response is given by y() = yzi() + yzs().
19
Example:
Now,
or
or
Let and , then for t ≥ t0, the state model is
3 ( ) 0.5 ( ) ( )v t v t u t
)(2)(6)( tutvtv
.)(,)(,)(2)(6)( 00 tytytutyty
)()(1 tytx )()()(2 tvtytx
1 01 1
0
2 02 2
( )( ) ( )0 1 0( ) ( ) ( ) ( ) , ( )
( )( ) ( )0 6 2
x tx t x tt u t A t B t t
x tx t x t
x x u x
1
2
( )( ) 1 0 ( )
( )
x ty t C t
x t
x
0.5m Kg
input force u t
3friction force v t
output position y t
( )velocity v t y t
20
Now, assuming we know eAt, the solution of the state equation is given by
where for t ≥ t0 = 0, the matrix exponential eAt is described by the 2x2 matrix
(how this was obtained will be explained later)
1. Zero-input state response: u(t) = 0, t ≥ 0 and x(0) ≠0.
0
0
( ) ( )
0( ) ( ) ( )
t
A t t A t
t
t e t e B d x x u
t
tAt
e
ee
6
6
06
1
6
11
6610 2010
6 20 6
20
1 11 11
( ) (0) 6 66 6
0
tt
At
t t
x e xxet e
xe e x
x x
21
2. Zero-input system response: u(t) = 0, t ≥ 0 and x(0) ≠0.
3. Zero-state state response: x(0) = 0. Let the input be the unit
step function, then.
6
10 20 610 20
620
1 11 1
6 6( ) ( ) (0) 1 06 6
t
At t
t
x e xt C t Ce x e x
e x
y x x
6( )
( ) ( )
6( )0 0 0
1 101
( ) ( ) 6 62
0
tt t t
A t A t
s
t
et e Bu d e Bd d
e
x
t
t
t
t
t
t
t
t
t
e
et
de
de
d
e
e
6
6
0
)(6
0
)(6
0)(6
)(6
13
1
118
1
3
1
2
3
1
3
1
23
1
3
1
( ) ( ),su t u t
22
4. Zero-state system response: x(0) = 0.
For t ≥ 0, the complete state response is then given by
and the complete system response by
6
6
6
1 11
1 13 18( ) ( ) 1 0 1
1 3 181
3
t
t
t
t e
y t C t t e
e
x
66
10 20( )
66020
1 11 1 1
3 18( ) (0) ( ) 6 6
11
3
ttt
At A t
tt
t ex e x
t e e Bu d
ee x
x x
1 6 6
1 10 20
2
( ) 1 1 1 1( ) ( ) 1 0 ( ) 1
( ) 6 6 3 18
t tx t
y t C t x t x e x t ex t
x
23
State Models from Ordinary Differential Equations
Let a dynamic system be described by the nth order scalar differential equation
with constant coefficients, i.e.,
where m ≤ n (because of physical realizability).
Let the initial energy of the system be zero, then with n = 3 and m = 2,
Let us implicitly solve this equation, namely,
( ) ( 1) ( )
1 1 1 1( ) ( ) ... ( ) ( ) ( ) ( ) ... ( )n n m
n n m my t a y t a y t a y t b u t b u t b u t
2
2
123322
2
13
3 )()()()(
)()()(
dt
tudb
dt
tdubtubtya
dt
tdya
dt
tyda
dt
tyd
2
2
123322
2
13
3 )()()()(
)()()(
dt
tudb
dt
tdubtubtya
dt
tdya
dt
tyda
dt
tyd
)()()()()()( 3322112
2
tyatubtyatubdt
dtyatub
dt
d
24
Integrating both sides of the last equation one step at a time, we get
This is the implicit solution of the original differential equation. This solution is
obtained via nested integration.
To obtain a state variable representation, we need to represent this implicit
solution in block diagram form (traditional analog simulation diagram).
t
dyaubtyatubtyatubdt
d
dt
tyd )()()()()()(
)(3322112
2
t
ddyaubyaubtyatubdt
tdy
)()()()()()()(
332211
dddyaubyaubyaubtyt
)()()()()()()( 332211
25
Block diagram representation:
If we select the output of the integrators as the state variables. Then
3
13123
23212
3331
xy
ubxaxx
ubxaxx
ubxax
b3
a1
b2 b1
a3 a2
Σ Σ Σx1 x3x21x 3x2x
u
y
+ ++++
_ __
26
In matrix form,
This is the so-called observable canonical form representation.
Alternative state variable representation:
Let us apply the Laplace transform to the original scalar ordinary differential
equation with constant coefficients, assuming zero initial conditions, i.e.,
or
3 3
2 2 0 0
1 1
0 0
1 0
0 1
a b
a b u A B u
a b
x x x
00 0 1y C x x
3 2 2
1 2 3 3 2 13 2 2
( ) ( ) ( ) ( ) ( )( ) ( )
d y t d y t dy t du t d u ta a a y t b u t b b
dt dt dt dt dt
L
)()( 2
12332
2
1
3 sUsbsbbsYasasas
11
0
0
( ) ( )( )
n knn n k
tn kk
d f t d f ts F s s
dt dt
L
27
In transfer function form,
Let us rewrite the last equation as follows:
where
and
Observation:
The overall transfer function is a cascade of two transfer functions.
2 2 1 2 33
3 2 1 3 2 1 1 2 3
3 2 3 2 3 1 2 3
1 2 3 1 2 3 1 2 3
( )
( ) 1
b b s b s b b s b s b s b s b sY s s
U s s a s a s a s a s a s a s a s a s a s
1 2 3
1 2 31 2 3
1 2 3
ˆ( ) ( ) ( ) 1,
ˆ( ) ( ) 1( )
Y s Y s Y sb s b s b s
U s U s a s a s a sY s
3
3
2
2
1
11
1
)(
)(ˆ
sasasasU
sY
3
3
2
2
1
1)(ˆ
)( sbsbsbsY
sY
28
Each of the transfer functions can be expressed in block diagram form, i.e.,
Observation: the term s-1 in the complex frequency domain corresponds to an
ideal integrator in the time domain.
( )U s ˆ( )Y s1 ˆ( )s Y s 2 ˆ( )s Y s 3 ˆ( )s Y s
1
s
1
s
1
s
1a
2a
3a
_ _ _
ˆ( )Y s
1 ˆ( )s Y s 2 ˆ( )s Y s 3 ˆ( )s Y s
1
s
1
s
1
s
1b 2b 3b
( )Y s
29
Putting the two diagrams together, yields
Again, choosing the outputs of the integrators as the state variables, we get
312213
3122133
32
21
xbxbxby
uxaxaxax
xx
xx
( )u t 3x 3x 2x 1x1
s
1
s
1
s
1a
2a
3a
_ _ _
1b 2b 3b
( )y t
2x 1x
30
In matrix form,
This form of the state equation is the so-called controllable canonical form.
Observation:
Both canonical forms are the dual of each other.
Consider the controllable canonical form of some linear time invariant dynamic
system, i.e.,
3 2 1
0 1 0 0
0 0 1 0
1
c cu A B u
a a a
x x x
3 2 1 cy b b b C x x
c c c c
c c
A B u
C
x x
y x
31
Then the observable canonical form is given by
Controllability and Observability (a conceptual introduction):
Suppose now that the initial conditions of an nth order scalar ordinary differential
equation are not equal to zero. How do we build state models such that their
responses will be the same as that of the original scalar model?
Method 1: Given the nth order scalar differential equation
with state model
0 0 0 0 0
0 0 0
0
,
T T
T T Tc c
c c cT
c
A B A CA A B C and C B
B
x x u x u
y x
( ) ( 1) ( 1)
1 1 1 1( ) ( ) ... ( ) ( ) ( ) ( ) ... ( )n n n
n n n ny t a y t a y t a y t b u t b u t b u t
( ) ( ) ( )
( ) ( ) ( )
t A t Bu t
y t C t Du t
x x
x
32
where x(t) Rn, u(t), y(t) R, A,B,C and D are constant matrices of appropriate
dimensions.
Observability Concept: Determine the initial state vector x(0)=[x1(0) … xn(0)]T
from the initial conditions and the initial input values
.
Remark: In the derivation of both observable and controllable canonical forms
from an ordinary linear differential equation with scalar constant coefficients and
m<n, we found out that D = 0, hence, for m = n-1,
)0(),...,0(),0( )1( nyyy
)0(),...,0(),0( )1( nuuu
2
( 1) 1 2 3 ( 3) ( 2)
(0) (0)
(0) (0) (0) (0)
(0) (0) (0) (0) (0) (0) (0)
(0) (0) (0) (0) (0) (0)n n n n n n
y C
y C CA CBu
y C CA CBu CA CABu CBu
y CA CA Bu CA Bu CABu CBu
x
x x
x x x
x
33
In matrix form,
where Rnxn, T Rnxn.
To get a unique solution x(0) for the last algebraic equation, it will be necessary
that the matrix be non-singular ( has full rank), i.e., its inverse exists and
x(0) = -1[Y(0) – TU(0)].
The existence of -1 is directly related to the property of observability of a system.
Hence, to uniquely reconstruct the initial state vector x(0) from input and
output measurements, the system must be observable, i.e., -1 must exist.
In fact, is called the observability matrix.
2
( 2)
( 1) 1 2 3 ( 1)
(0) 0 0 0 (0)
(0) 0 0 (0)
(0) (0)(0)
(0)
(0) 0 (0)
(0) (0)
n
n n n n n
y C u
y CA CB u
y CA CAB CB
u
y CA CA B CA B CB u
T
Y x
x U
34
Controllability Concept: Suppose now that instead of using the input-output
measurements to reconstruct the state at time t = 0 we use impulsive inputs to
change the value of the state instantaneously,
Let
with x(0-) = x0, A Rnxn, B Rn, describe an nth order scalar differential
equation and let the input be described by
Clearly, the input u(t) is described by a linear combination of impulsive inputs
where the scalars are unknown. Now, we know that for t 0-
( ) ( ) ( ), 0t A t Bu t t x x
( 1)
0 1 1( ) ( ) ( ) ( ), , 0, , 1n
n iu t t t t i n
( )
0
( ) (1) ( 1)
0 1 1
0
( ) ( ) ( ) ( ) ( 1)
0 1
0 0 0
( ) (0 ) ( )
(0 ) ( ) ( ) ( )
(0 ) ( ) ( ) ( )
t
At A t
t
At A t n
n
t t t
At A t A t i A t n
i n
t e e Bu d
e e B d
e e B d e B d e B d
x x
x
x
i
35
Integrating the ith term by parts recursively, we get
Therefore, x(t) is given by
where is an unknown vector.
( ) ( ) ( ) ( 1) ( 1)
00 0
( ) ( 2) 2 ( 2)
00
( ) 1
0
( ) ( ) ( )
( ) ( )
( )
t tt
A t i A t i At A ii i
tt
A t i At A ii
tA t i At A
e B d e B e e AB d
e AB e e A B d
e A B e e
0
0
00
( )
t
i At ii iA B d e A B
1
0 1 1
1
( ) (0 )
(0 )
At At At At n
n
At n
t e e B e AB e A B
e B AB A B
x x
x
0 1 1
T
n
36
At time t = 0+, we get
where Q is the so-called controllability matrix.
Clearly, an impulsive input that will take the state from x(0-) to x(0+) will exist if
and only if the inverse of Q exists, namely, we can uniquely compute the vector
Digital Simulation of State Models
Dynamic systems are nonlinear in general, therefore, let us begin with the
following nonlinear time-varying dynamic system which is described by
1(0 ) (0 ) (0 )n
Q
B AB A B Q
x x x
1 (0 ) (0 ) 0Q x x
0 0 0( ) ( , ( ), ( )), ( ) ,
( ) ( , ( ), ( ))
t t t t t t t
t t t t
x f x u x x
y g x u
1 2 , namely,T
n
37
Objective: We would like to know the behavior of the system over the time
interval t [t0, tn] for a given initial state x(t0) and input u(t), t [t0, tn].
In principle, for t [t0, tn],
However, to compute the integral analytically is very difficult in most cases.
Let’s examine the following numerical approximations to the integral. Let n = 10.
Case 1 (forward Euler formula, scalar case):
0
0( ) ( ) ( , ( ), ( ))
t
t
t t d x x f x u
38
Over the kth time interval,
Hence, the integral under the curve over the whole interval is given by
Now, at time tk, starting at tk-1, we get
Case 2 (backward Euler formula, scalar case):
9910112001 )()()())(),(,(10
0
fttfttfttduxf
t
t
11)())(),(,(
1
kkk
t
t
fttduxfk
k
))(),(,()()()( 11111 kkkkkkk tutxtftttxtx
39
For the kth time interval,
Therefore,
and the approximate solution at tk, starting at tk-1, is given by
Case 3 (trapezoidal rule, scalar case):
10910212101 )()()())(),(,(10
0
fttfttfttduxf
t
t
kkk
t
t
fttduxfk
k
)())(),(,( 1
1
))(),(,()()()( 11 kkkkkkk tutxtftttxtx
40
In this case, the integral over the entire interval is approximately equal to
Therefore, the solution at time tk , starting at tk-1 is approximately equal to
Example: Obtain an approximate solution over one time interval of the following
linearized pendulum state model at equally spaced time instants, tk – tk-1 = 0.5.
The linear model with l=g/4 is . Let
2
)()(
2
)()(
2
)()())(),(,( 109
91021
12
10
01
10
0
fftt
fftt
ffttduxf
t
t
))(),(,())(),(,()(2
1)()( 11111 kkkkkkkkkk tutxtftutxtftttxtx
1 2
2 1
( ) ( )
( ) 4 ( )
x t x t
x t x t
1
2
(0)40
(0)0
x
x
sin( ) sin( )g
mg ms mll
1 2
1 2 2 1
Let and , then
and sin( ) sin( )
x x
g gx x x x
l l
41
Forward Euler Method:
Backward Euler Method:
Trapezoidal Rule Method:
1 1 1 2 1 1 1 2 1
2 2 1 1 1 2 1 1 1
( ) ( ) ( ) ( ) 0.5 ( )0.5
( ) ( ) 4 ( ) ( ) 2 ( )
k k k k k
k k k k k
x t x t x t x t x t
x t x t x t x t x t
1 1 1 2 1 1 2
2 2 1 1 2 1 1
( ) ( ) ( ) ( ) 0.5 ( )0.5
( ) ( ) 4 ( ) ( ) 2 ( )
k k k k k
k k k k k
x t x t x t x t x t
x t x t x t x t x t
1 1 1 2 2 1
1
2 2 1 1 1 1
1 1 2 2 1
2 1 1 1 1
( ) ( ) ( ) ( )0.5( )
( ) ( ) 4 ( ) 4 ( )
( ) 0.25 ( ) ( )
( ) ( ) ( )
k k k k
k k
k k k k
k k k
k k k
x t x t x t x tt t
x t x t x t x t
x t x t x t
x t x t x t
42
Linear Discrete-Time Systems
Implementation of dynamic systems is actually done using digital devices like
computers and/or DSPs. Moreover, there are some naturally occurring processes
which are discrete-time. Hence, it is convenient to model such systems as
discrete-time systems.
In most cases, a continuous-time linear time-invariant dynamic system is
discretized at time t = tk, where tk+1 - tk may be fixed or variable. When such
system is described by its impulse response, the discretization process is
illustrated in the figure below (a sampled-data system).
t=tk t=tk
u(t) u(tk) y(t) y(tk)
h(t)
43
The equivalent single-input, single-output, discrete-time system can be depicted
by
Since the system is linear, its behavior is described by the convolution relation.
Let tk = kT, then
Let u(kT) = (kT), the unit sample, i.e.,
then,
is called the unit sample response.
n
nTunTkThkTy )()()(
otherwise
kkT
,0
0,1)(
)()()()( kThnTnTkThkTyn
u(tk) y(tk)h(tk)
44
Let the system be causal, i.e., h(kT) = 0, k < 0 then
In addition, if u(kT) = 0 for k < 0, then
k
n
nTunTkThkTy )()()(
k
n
nTunTkThkTy0
)()()(
Consider now a continuous-time linear time-invariant multiple-input, multiple-
output dynamic system in state-space form, i.e.
0 0 0( ) ( ) ( ), ( ) ,
( ) ( ) ( )
t A t B t x t x t t
t C t D t
x x u
y x u
We know that the solution of the state equation is
0
0
( ) ( )
0( ) ( ) ( )
t
A t t A t
t
t e t e B d x x u .
45
Let 0t kT and t kT T , then the last equation becomes
( )( ) ( ) ( )
kT T
AT A kT T
kT
kT T e kT e B d
x x u .
Suppose ( ) ( ),u t u kT kT t kT T , namely, the input does not change in the interval
[ , )kT kT T , (equivalent to the presence of a sample-and-hold device) then
( )( ) ( ) ( )
kT T
AT A kT T
kT
kT T e kT e d B kT
x x u .
Let kT T , then d d and
0
( )
0
( )
kT T T
A kT T A A
kT T
e d e d e d
.
The discretized system now becomes
0
0
( ) ( ) ( ), (0) , 0,1,2,
T
AT AkT T e kT e d B kT k x x u x x .
46
State representation of discrete-time dynamic systems:
Consider a linear discrete-time dynamic system described by the difference equation
where the sampling interval has been normalized, i.e., T = 1 sec.
)()1()2()1()( 121 kyakyankyankyanky nn
)()1()( 11 mkubkubkub mm
Define the discrete-time equivalent system and input distribution matrices as AT
dA e
and 0
T
A
dB e d B , then the discrete-time equivalent system is described by
0( ) ( ) ( ), (0) , 0,1,2,
( ) ( ) ( )
d dkT T A kT B kT k
kT C kT D kT
x x u x x
y x u
Note that this discretization of the original continuous-time system is exact when a
sample-and-hold device processes the input u(t) and the input and output samplers (D/A
and A/D) clocks are in synchronism.
47
If we now replace differentiations with forward shift operators and integrators with
backward shift operators then we can construct the same type of canonical
realizations that we built for continuous-time systems.
Example: Let n = 3, m = 2 and y(0) = y(1) = y(2) = u(0) = u(1) = u(2) = 0, then
Let us apply the backwards shift operator to this equation recursively, i.e.
The solution y(k) can now be computed implicitly using a simulation block
diagram.
)()()1()1()2()2()3( 332211 kyakubkyakubkyakubky
)()()()()1()1()2()3( 33
1
2211
1 kyakubqkyakubkyakubkykyq
)()()()()()()1()2( 33
1
22
1
11
1 kyakubqkyakubqkyakubkykyq
)()()()()()()()1( 33
1
22
1
11
11 kyakubqkyakubqkyakubqkykyq
48
Simulation diagram implementation:
Using the outputs of the shift operators as the state variables, we get
)()()()1(
)()()()1(
)()()1(
13123
23212
3331
kubkxakxkx
kubkxakxkx
kubkxakx
)()( 3 kxky
b3
a1
b2 b1
a3 a2
Σ Σ Σx1 x3x2
u
y
+ ++++
_ __q-1 q-1 q-1
49
In matrix form, we obtain the observable canonical form, i.e.
In general, a multiple-input, multiple-output linear shift-invariant discrete-time
dynamic system can be represented by (assuming T = 1)
where x(k) Rn, u(k) Rm, y(k) Rr, A, B, C and D are constants matrices of
appropriate dimensions.
As in the continuous-time case, we may reconstruct the state from input-output
measurements when the system is observable.
3 3
2 2
1 1
0 0
( 1) 1 0 ( ) ( )
0 1
a b
k a k b u k
a b
x x
( ) 0 0 1 ( )y k k x
( 1) ( ) ( )
( ) ( ) ( )
k A k B k
k C k D k
x x u
y x u
50
Iteratively,
In matrix form,
2
1 2 3
( ) ( ) ( )
( 1) ( 1) ( 1) ( ) ( ) ( 1)
( 2) ( 2) ( 2) ( ) ( ) ( 1) ( 2)
( 1) ( ) ( ) ( 1) ( 2) ( 1)n n n
k C k D k
k C k D k CA k CB k D k
k C k D k CA k CAB k CB k D k
k n CA k CA B k CA B k CB k n D k n
y x u
y x u x u u
y x u x u u u
y x u u u u
2
1 2 3
( ) 0 0 0 ( )
( 1) 0 0 ( 1)
( ) ( )( 2) ( 2)
0
( 1) ( 1)
( ) ( )
n n n
k C D k
k CA CB D k
k kk CA CAB CB D k
k n CA CA B CA B CB D u k n
k T k
y u
y u
Y xy u
y
x U
51
If has full rank, then x(k) = -# [Y(k)-TU(k)], i.e., if the system is observable then
we can reconstruct the state at time k using input, output measurements up to time
k+n-1, where -# is the pseudoinverse of .
Solution of the discrete-time state equation
Iteratively,
and
2
3 2
4 3 2
11
0
(1) (0) (0)
(2) (1) (1) (0) (0) (1)
(3) (2) (2) (0) (0) (1) (2)
(4) (3) (3) (0) (0) (1) (2) (3)
( ) (0) ( ), 0k
k k l
l
A B
A B A AB B
A B A A B AB B
A B A A B A B AB B
k A A B l k
x x u
x x u x u u
x x u x u u u
x x u x u u u u
x x u
11
0
( ) ( ) ( ) (0) ( ) ( )k
k k l
l
k C k D k CA C A B l D k
y x u x u u
52
From the solution of the discrete-time state equation, starting at time index k, the
value of the state at time index k+n is
where the controllability matrix Q and the vector are given by
To assure the existence of an input such that the state of the system can reach a
desired state at time k+n given the value of the state at time k, the matrix Q must
have rank n. Thus, the following relationship is satisfied
, where is the pseudo inverse of Q
In other words, the system must be controllable.
( ) ( ) ( ), hence, ( ) ( ) ( )n n
r rk n A k Q k Q k k n A k x x U U x x
1
( 1)
( 2)[ ] and ( )
( )
n
r
k n
k nQ B AB A B k
k
u
uU
u
#( ) ( ) ( )n
r k Q k n A k U x x#Q
( )r kU
53
Linearization of Nonlinear Systems
Consider the following scalar (single-input, single-output) nonlinear
memoryless system
where g(·):.
Let the nominal operating point be and let N be a neighborhood of it,
i.e., N = {x : a < x < b} and a < x0 < b. If the function g(·) is analytic
on N, i.e., it is infinitely differentiable on N, then for x such that
x0+xN, we get the following Taylor series expansion
For small x,
0x
0 0
22
0 0 2
1( ) ( )
2!x x x x
dg d gg x x g x x x
dx dx
0 00 0 0( ) ( ) x x x x
dg dgy g x x g x x y x
dx dx
)x(t) y(t)g(·)
54
Therefore, the linear approximation of a nonlinear system y = g(x) near
the operating point x0 has the form
or
where
Example: Consider a semiconductor diode described by
where vo=vt.
In this case,
00 x x
dgy y x
dx
0y y y m x
0x x
dgm
dx
0 ln 1 ( )s
iv v g i
i
0 0 0
0
0
( ) 1 1
1i i i i o i d
s s
s
vdg i dvm v r
idi di i i i
i
55
The linearized model is then given by
Using the parameters = 35.4, vt = 0.026 V, is = 1 nA, i0 = 0.05 A, v0 = 0.92 V,
rd = 18.44 , we get the following linear approximation:
0
0d
s
vv m i i r i
i i
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
i in Amps
v in V
olts
56
If, on the other hand, y=g(x1,x2,,xn) , then if g(·) is analytic on the
set N={xn:a<||x||<b}, x0N and x0+xN, x0=[x10,,xn0]T,
where and
Systems with memory:
I. Scalar Case. Consider a dynamic system described by
Suppose xn(t) is the system response resulting from a nominal
operating input un(t), assuming a nominal initial state xn(t0) = xn0
0 0
0
10 0 11
0
( ) ( )( , , ) higher order terms
( ) ( ) higher order terms
n nn
g gy g x x x x
x x
g g
x x x x
x x
x x
x x x
1 2
( ) ( ) ( )( )
n
g g gg
x x x
x x xx
0 0 0( ) ( ( ), ( ), ), ( ) , ( ), ( ) forx t f x t u t t x t x x t u t t t
1 2
T
nx x x x
57
In other words,
Assume that we know xn(t). Perturb the state and the input by taking
x(t) = xn(t) + x(t) and u(t) = un(t) + u(t).
We want to find the solution to the differential equation
For fixed values of t, and f(·) an analytic function on some neighborhood
of xn(t) and un(t), using a Taylor series expansion, we get
where
0 0 0( ) ( ( ), ( ), ) , ( ) , forn n n n nx t f x t u t t x t x t t
0 0 0 0( ) ( ( ) ( ), ( ) ( ), ) , ( ) ( ) , forn n nx t f x t x t u t u t t x t x t x t t
( ( ) ( ), ( ) ( ), ) ( ( ), ( ), )
( )( , , ) higher order terms
( )n
n
n n n n
x xu u
f x t x t u t u t t f x t u t t
x tf x u t
u t
( , , )f f
f x u tx u
58
Let and
Then, if we neglect the higher order terms in the expansion, we get
Therefore,
Example: Let
Clearly,
Now, if the nominal operating state and input are xn(t) and un(t) = 0, t 1,
( ( ), ( ), )( )
n
n
x xu u
f x t u t ta t
x
( ( ), ( ), )( )
n
n
x xu u
f x t u t tb t
u
( )( ) ( )( ( ), ( ), ) ( ) ( ) ( ) ( )n
n n
dx tdx t d x tf x t u t t a t x t b t u t
dt dt dt
0 0
( )( ) ( ) ( ) ( ), ( )
d x ta t x t b t u t x t x
dt
20 0( ) ( ) ( ) , ( )x t x t u t x t x
2( ( ), ( ), ) ( ) ( )f x t u t t x t u t
0 0
( ( ), ( ), )( ) 2 ( ) 2 ( )
n n
n n
x x x x nu u u u
f x t u t ta t x t x t
x
59
and
So,
With
Finally,
0 0
( )2 ( ) ( ) ( ), ( )n
d x tx t x t u t x t x
dt
0 0( ) 0, 1 and ( ) (1) 1, thenn n nu t t t x t x
2
1 2 1 22
( )( ) , (1) 1
( ) 1 1and ( ) , 1
( )( )
nn n
nn
nn
dx tx t x
dt
dx tdt c t c c c x t t
x t tx t
0
( ) 2( ) ( ) , (1) , 1
d x tx t u t x x t
dt t
0
( ( ), ( ), )( ) 1
n
n
x xu u
f x t u t tb t
u
60
II. Vector Case. Consider now the case of a system described by the following
nonlinear vector differential equation:
The ith element of the vector differential equation is described by
Moreover,
where
0 0 0( ) ( ( ), ( ), ) , ( ) , ( ) , ( ) forn mt t t t t t t t t x f x u x x x u
0 0 0( ) ( ( ), ( ), ) , ( ) , ( ) , ( ) forn mi ix t f t t t t t t t t x u x x x u
( ( ) ( ), ( ) ( ), ) ( ( ), ( ), )
( )( , , ) higher order terms
( )n
n
i n n i n n
i
f t t t t t f t t t
tf t
t
x xu u
x x u u x u
xx u
u
( , , ) ( , , ) ( , , )i i if t f t f t x ux u x u x u
61
and
or
1 2
( , , ) i i ii
n
f f ff t
x x x
x x u
1 2
( , , ) i i ii
m
f f ff t
u u u
u x u
1 1
1 1
2 2 2 2
( , , ) ( , , )
( , , ) ( , , )
( , , ) ( , , )
n n
n n
n n
n n
n n
n n
n n
n n
f t f t
x x
x f t x f td
dt
x x
f t f t
x x x u x xu u u u
x x x u x xu u u u
x x x u x xu u u u
x u x u
x u x u
x u x u
1
2
m
u
u
u
62
or
where
are real nxn and nxm matrices, respectively.
0 0
( )( ) ( ) , ( ),
d tA t B t x t t t
dt
xx u
1 1
2 2
( , , ) ( , , )
( , , ) ( , , ) and
( , , ) ( , , )
n n
n n
n n
n n
n n
n n
n n
f t f t
f t f tA B
f t f t
x x x u x xu u u u
x x x u x xu u u u
x x x u x xu u u u
x u x u
x u x u
x u x u
63
Finally, if the outputs of the nonlinear system are of the form
Then
( ) ( ( ), ( ), ) , ( ) pt t t t t y g x u y
1 1
1 1
2 2 2 2
( , , ) ( , , )
( , , ) ( , , )( )
( , , ) ( , , )
n n
n n
n n
n n
n n
n n
p n
p p
g t g t
y x
y g t x g tt
y x
g t g t
x x x u x xu u u u
x x x u x xu u u u
x x x u x xu u u u
x u x u
x u x uy
x u x u
1
2
m
u
u
u
64
or
where
are real pxn and pxm matrices, respectively.
( ) ( ) ( ) , ( ) pt C t D t t y x u y
1 1
2 2
( , , ) ( , , )
( , , ) ( , , ) and
( , , ) ( , , )
n n
n n
n n
n n
n n
n n
p p
g t g t
g t g tC D
g t g t
x x x u x xu u u u
x x x u x xu u u u
x x x u x xu u u u
x u x u
x u x u
x u x u
65
Example: Suppose we have an orbiting micro satellite modeled as a point mass
in an inverse square law force field, e.g., a gravity field as shown below
where r(t) is the radius of the orbit at time t
(t) is the angle relative to the horizontal axis at time t
u1(t) is the thrust in the radial direction at time t
u2(t) is the thrust in the tangential direction at time t
m is the mass of the orbiting body
m
u1(t)u2(t)
(t)r(t)
orbit
66
From the laws of mechanics and assuming m = 1kg, the total force in the radial
direction is described by
and the total force in the tangential direction is
Select the states as follows:
Then,
22
12 2
( ) ( )( ) ( )
( )
d r t d t Kr t u t
dtdt r t
2
22
( ) 2 ( ) ( ) 1( )
( ) ( )
d t d t dr tu t
r t dt dt r tdt
1 2 3 4
( ) ( )( ) ( ), ( ) , ( ) ( ), ( )
dr t d tx t r t x t x t t and x t
dt dt
1 2
22 1 4 12
1
( ) ( )
( ) ( ) ( ) ( )( )
x t x t
Kx t x t x t u t
x t
67
which implies that
For a circular orbit and u1n(t) = u2n(t) = 0 and t0 = 0, we have
3 4
2 44 2
1 1
( ) ( )
( ) ( ) 1( ) 2 ( )
( ) ( )
x t x t
x t x tx t u t
x t x t
2
21 4 12
1
4
2 4 2
1 1
( )( ( ), ( ), ) .
2
x
Kx x u
xd tt t t
dt x
x x u
x x
x= f x u
0 0( ) 0 , 0Tn t R t t x
68
where rn(t) = R and
Linearizing about xn(t) and un(t), yields
0 3
( ).nd t K
dt R
1( , , ) 0 1 0 0n
n
f t
x x xu u
x u
2 22 4 1 4 0 03
1
2( , , ) 0 0 2 3 0 0 2
n
n
n
n
Kf t x x x R
x
x x xu u
x xx u
u u
3( , , ) 0 0 0 1n
n
f t
x x xu u
x u
02 4 2 4 24 2
1 11
2( , , ) 2 0 2 0 2 0 0
n
n
n
n
x x u x xf t
x x Rx
x x xu u
x xx u
u u
69
Likewise,
In state form,
1( , , ) 0 0n
n
f t
u x xu u
x u
2 ( , , ) 1 0n
n
f t
u x xu u
x u
3( , , ) 0 0n
n
f t
u x xu u
x u
41
1 1( , , ) 0 0
n n
n n
f tx R
u x x x x
u u u u
x u
20 0
0
0 1 0 0 0 0
1 03 0 0 2
0 00 0 0 1
100 2 0 0
Rd
dt
RR
x x u
70
Theorem: Let A(t) Rnxn be piecewise continuous. Then the set of all solutions of
forms an n-dimensional vector space over the field of the real numbers.
Proof: Let { 1, 2, , n} be a set of linearly independent vectors in Rn, i.e.,
, if and only if i = 0, i = 1, 2, , n; and the i()’s be the
solutions of with initial conditions i(t0) = i, i = 1, 2, , n.
Suppose that t R+ the i’s, i = 1, 2, , n are linearly dependent, then i R,
i = 1, 2, , n, different than zero such that
But, at t = t0 R+,
the i’s are linearly dependent, which is an outright contradiction of the
hypothesis that the i’s are linearly independent. Therefore, the i’s are linearly
independent for all t R+.
1 1 2 2 0n n
1 1 2 2( ) ( ) ( ) 0n nt t t
0( ) ( ) ( ),t A t t t t x x
1 1 0 2 2 0 0( ) ( ) ( )n nt t t 1 1 2 2 0n n
71
Let (t) be any solution of and (t0) = . Since the i’s are linearly independent
vectors in Rn, can be uniquely represented by
But, is a solution of with initial condition .
This is because
In other words, the linear combination
satisfies the differential equation and the initial condition .
Therefore, () = implies that every solution of is a linear combination
of the basis of solutions i(), i = 1, 2, , n, i.e., the set of all solutions of forms
an n-dimensional vector space.
1
n
i i
i
0
1 1
( )n n
i i i i
i i
t
1 1 1 1
( ) ( ) ( ) ( ) ( ) ( )n n n n
i i i i i i i i
i i i i
dt t A t t A t t
dt
1
( )n
i i
i
1
( )n
i i
i
1
( )n
i i
i
1
n
i i
i
72
Example: Consider the dynamical system described by
Let the vectors 1 and 2 be described by and .
Then, and are two independent solutions to the system
with initial conditions 1(0) = 1 and 2(0) = 2.
Therefore, any solution (t) will be given by
i R, i = 1, 2.
0 0( ) ( )
0t t
t
x x
1
1
0
2
0
1
1 2
1
( ) 1
2
tt
2
0( )
1t
1
0
2
11
)()()( 2212211 t
ttt
73
Def. The state transition matrix of the differential eq. is given by
where the i’s, i = 1, 2, , n are the basis solutions and i = [0 … 0 1 0 … 0]T.
Properties of the state transition matrix:
1. (t0, t0) = I
Proof: Recall that
Thus (t0, t0) = I.
2. (t, t0) satisfies the differential eq. , M(t0) = I, M(t) Rnxn.
Proof: The time derivative of the state transition matrix is given by
However,
),;(),;(),;(),( 02021010 nn tttttttt
0 0( ; , ) 0 0 1 0 0T
i i it t
)()()( tMtAtM
0 1 0 1 2 0 2 0( , ) ( ; , ) ( ; , ) ( ; , )n n
dt t t t t t t t
dt
0 0( ; , ) ( ) ( ; , )i i i it t A t t t
( ) ( ) ( )t A t tx x
ith location
74
Therefore,
Also, from part (1), (t0, t0) = I.
3. (t, t0) is uniquely defined.
Proof: Since each i(t) is uniquely determined by A(t) for each initial condition i
then (t, t0) is also uniquely determined by A(t).
Proposition: The solution to , x(t0) = x0 is x(t) = (t, t0)x0 t.
Proof: At t = t0, (t0, t0)x0 = Ix0 = x0, it satisfies the initial condition.
We already know that .
Therefore, .
In other words, , satisfies the differential equation.
0 1 0 1 2 0 2 0( , ) ( ) ( ; , ) ( ) ( ; , ) ( ) ( ; , )n nt t A t t t A t t t A t t t
1 0 1 2 0 2 0 0( ) ( ; , ) ( ; , ) ( ; , ) ( ) ( , )n nA t t t t t t t A t t t
),()(),( 00 tttAtt
0 0 0 0( ) ( , ) ( ) ( , ) ( ) ( )t t t A t t t A t t x x x x
0 0( , )t t x
( ) ( ) ( )t A t tx x
75
If t and t0, and A(t) has the following commutative property:
Then, the state transition matrix is given by
Example: Compute the state transition matrix (t, t0) for the differential equation,
We can show that and are commutative, i.e.
)()()()(
00
tAdAdAtA
t
t
t
t
0
( )
0( , )
t
t
A d
t t e
21( ) ( )
0 1
tet t
x x
0
0 0
22 2
0 0
0
1( )
( ) ( ) ( ) ( ) 2
0
tt tt t
t t
t t e e t t eA t A d A d A t
t t
0
( ) ( )
t
t
A t A d 0
( ) ( )
t
t
A d A t
76
Hence,
and
0
0 0 0
2 3( )
0
1 1( , ) ( ) ( ) ( )
2! 3!
t
t
A d t t t
t t t
t t e I A d A d A d
!20
!2
1
!2
02
1
10
012
0
22
0
2
0
0
22
0
0
0
tt
eetttt
tt
eetttt
tt
!30
!32
3
!33
0
22
2
0
3
0 0
tt
eetttt tt
)(3
0
2
000110)(
!3
1)(
!2
1)(1),(
ttetttttttt
77
Finally,
Hence, the state transition matrix is given by
and .
We can show that the norm of the state transition matrix blows up as time t goes to
infinity, therefore, the system is unstable.
2
0
22
0
2222
012 )(!3
1
2
3)(
!2
1
2
1),( 000 tteetteeeett
tttttt
00000 3)(223
0
2
00
22
2
1
2
1)(
!3
1)(
!2
1)(1
2
1 tttttttttt eeeeettttttee
0),( 021 tt
0),(),( 011022
ttetttt
0 0 0
0
3
0
1
2( , )
0
t t t t t t
t t
e e et t
e
0 0
1 0( , )
0 1t t
78
Theorem: A(t) and commute if
1. A() is constant
2. A(t) = (t)M, () : R R and M is a constant square matrix
3. , i() : R R and the Mi’s are constant square matrices
such that MiMj = MjMi i, j.
Proof:
(1) If A() is a constant matrix, i.e., A() = A, then
(2) If A(t) = (t)M, then
t
t
dA
0
)(
k
i
ii MttA1
)()(
2
00
22 )()(
00
AttttAIdAAdA
t
t
t
t
2
0
2 )(
00
AttAIdAAd
t
t
t
t
2
0000
)()()()()()()()( MdtMIdMtMdMtdAtA
t
t
t
t
t
t
t
t
79
But,
(3) If then
But,
2
0000
)()()()()()()()( MdtMtMdIMtMdtAdA
t
t
t
t
t
t
t
t
k
i
ii MttA1
)()(
k
j
t
t
jj
k
i
ii
t
t
k
j
jj
k
i
ii
t
t
MdMtdMMtdAtA1111
000
)()()()()()(
k
i
t
t
jij
k
j
i
k
i
t
t
jj
k
j
ii MMdtMdMt1 11 1
00
)()()()(
k
i
ii
k
j
j
t
t
j
k
i
ii
t
t
k
j
jj
t
t
MMdMdMtAdA1111
)()()()()()(
000
0 01 1 1 1
( ) ( ) ( ) ( )
t tk k k k
j j i i i j j i
j i j it t
d M t M t d M M
80
and, if MiMj = MjMi i, j (i ≠ j), then
Corollary: If A(t) satisfies condition (3) of the previous theorem, then
Proof: Since when A(t) satisfies condition (3), then
But,
)()()()(
00
tAdAdAtA
t
t
t
t
0
( )
0
1
( , )
t
i i
t
M dk
i
t t e
)()()()(
00
tAdAdAtA
t
t
t
t
t
t
dA
ett 0
)(
0),(
k
i
t
t
ii
t
t
k
i
ii MddMk
i
ii eettMttA1
001
)()(
0
1
),()()(
81
or,
Def. Any nxn matrix M(t) satisfying the matrix differential equation
M(t0) = M0, where det(M0) ≠0 is a fundamental matrix of solutions.
Theorem: If det(M0) ≠0 then det(M(t)) ≠ 0 t R+
Proof: (By contradiction) Suppose there exists t1 R+ such that det(M(t1)) = 0.
Let v = [v1 v2 … vn]T ≠ 0 such that M(t1)v = 0 and x(t) = M(t)v be the solution to the
vector differential equation , x(t1) = 0. Notice also that z() 0 is a
solution to , z(t1) = 0. By the uniqueness theorem we conclude that
x(t) = z(t) everywhere A(t) is piecewise continuous.
But, z(t0) = x(t0) = M(t0)v = 0 det(M0) = 0, which is a contradiction. Hence,
det(M(t)) ≠ 0 t, i.e., M(t) is nonsingular t.
k
i
dMMdMdMd
t
t
ii
t
t
kk
t
t
t
t eeeett1
)()()()(
0000
22
0
11
),(
( ) ( ) ( )t A t tz z
,)()()( tMtAtM
( ) ( ) ( )t A t tx x
82
Def. Let M(t) be any fundamental matrix of .. Then, t R+, the
state transition matrix (t,t0) of is given by, (t,t0) = M(t)M-1(t0).
Theorem (Semigroup Property): For all t1, t0 and t, we have (t,t0) = (t,t1) (t1,t0).
Proof: It is known from the existence and uniqueness theorem that
x(t) = (t,t0)x(t0) for any t,t0 (a)
x(t1) = (t1,t0)x(t0) for any t1,t0 (b)
and x(t) = (t,t1)x(t1) for any t,t1 (c)
are solutions to the differential equation with initial conditions x(t0)
and x(t1). But, from (c) and (b)
x(t) = (t,t1)x(t1) = (t,t1) (t1,t0)x(t0) (d)
Comparing (a) and (d) leads us to conclude that
(t,t0) = (t,t1) (t1,t0) for any t, t1 and t0.
Theorem (The Inverse Property): (t,t0) is nonsingular t, t0 R+ and -1(t,t0) =
(t0,t).
( ) ( ) ( )t A t tx x
( ) ( ) ( )t A t tx x
( ) ( ) ( )t A t tx x
83
Proof: Since (t,t0) is a fundamental matrix of , then it is nonsingular
for all t, t0 R+.
Now, from the semigroup property we know that for arbitrary t0, t1, t R+ ,
(t,t0) = (t,t1) (t1,t0).
For t0 = t, we get,
(t,t) = I = (t,t1) (t1,t) -1(t,t1) = (t1,t) and since t1 is arbitrary we have that
-1(t,t0) = (t0,t).
Theorem (Liouville formula):
Consider now the linear, time-varying dynamic system modeled by
with x(t0) = x0 .
Theorem: The solution to the state equation is given by
t
t
dAtr
ett 0
)(
0 )],(det[
( ) ( ) ( ) ( ) ( )t A t t B t t x x u
( ) ( ) ( ) ( ) ( )t C t t D t t y x u
0
0 0( ) ( , ) ( , ) ( ) ( )
t
t
t t t t B d x x u
( ) ( ) ( )t A t tx x
84
where
1. (t,t0)x0 is the zero-input state response, and
2. is the zero-state state response.
Proof: At t = t0, the solution to the differential equation is given by
Now,
since
0
( , ) ( ) ( )
t
t
t B d u
0
0
1
0 0 0 0 0 0
0
( ) ( , ) ( , ) ( ) ( )
n n
n
t
tI
t t t t B d
x x u x
0 0
0 0 0 0( , ) ( , ) ( ) ( ) ( , ) ( , ) ( ) ( )
t t
t t
dt t t B d t t t B d
dt t
x u x u
0
0 0( ) ( , ) ( , ) ( ) ( ) ( , ) ( ) ( )
t
t
A t t t t t B t t t B dt
x u u
t
t
t
t
t
dtft
tfdtft
00
),(),(),(
85
Hence,
The complete response should be given by
Let us now consider the time-invariant case, i.e., A(t) = A, B(t) = B, C(t) = C and
D(t) = D, where A, B, C and D are constant matrices.
Theorem: The state transition matrix of the time-invariant state model is
0
0 0( ) ( , ) ( ) ( ) ( ) ( , ) ( ) ( )
t
t
dA t t t B t t A t t B d
dt x u u
0
0 0( ) ( , ) ( , ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
t
t
A t t t t B d B t t A t t B t t
x u u x u
0
0 0( ) ( ) ( , ) ( ) ( , ) ( ) ( ) ( ) ( )
t
t
t C t t t C t t B d D t t y x u u
)(
000)0,(),(
ttAetttt
86
Proof: Since A and commute, we have that
The complete state and system responses are now given by
This follows from the fact that (t,t0) = M(t)M-1(t0) since we can always let
M(t) = eAt, i.e.,
t
t
Ad
0
)()0,(exp),( 00000 ttttttAett
t
t
Ad
0 0
0 0
( ) ( )( )
0 0( ) ( ) ( )
t t
A t t A t tA t At A
t t
t e e B d e e e B d x x u x u
0
0
( )
0( ) ( ) ( )
t
A t t At A
t
t Ce Ce e B d D t y x u u
)(1
0
1 000 )()()(ttAAtAtAtAt eeeeetMtM
87
Def. If A is an nxn constant matrix, C, e Cn and the equation
Ae = e, e ≠0
is satisfied, then is called an eigenvalue of A and e is called an eigenvector of A
associated with . Also, the eigenvalues of A are the roots of its characteristics
polynomial, i.e.,
The set (A) = {1, 2,…, n} is called the spectrum of A. The spectral radius of A is
the non negative real number
The right eigenvector ei Cn of A associated with the eigenvalue i satisfies the
equation Aei = iei, whereas the left eigenvector wi Cn of A associated with i
satisfies the equation wi*A = iwi*, where ()* designates the complex conjugate
transpose of a vector. If (A) and is complex then * (A). The eigenvectors
associated with and * will be e and e*, respectively.
)())(()det()( 211
1
1 nnn
nn
A aaaAI
( ) max :i ii
A A
88
Example: Find the right eigenvectors of the matrix
The characteristic polynomial of A is given by
Therefore, its spectrum is described by (A) = {-1, -1 - j2, -1 + j2}
Now,
or
For 1 = -1, we get
010
011
552
A
)52)(1(573)det()( 223 AIA
( ) 0i i i i iAe e I A e
0~~
10
011
552
i
i
i
i
e
89
or
let e13 = 1, then e12 = -1, then
For 2 = -1 – j2,
1 1 1
1 5 5
( ) 1 0 0 0
0 1 1
I A e e
13121312
1111
131211
0
00
055
eeee
ee
eee
Te 110~1
2 2 2
1 2 5 5
( ) 1 2 0 0
0 1 1 2
j
I A e j e
j
90
or
let e23 = 1, then e22 = -1 – j2, e21 = -j2(-1 – j2) = -4 + j2
and
Finally, , because .
Theorem: Let A be an nxn constant matrix. Then A is diagonalizable if and only if
there is a set of n linearly independent vectors, each of which is an eigenvector of
A.
Proof: If A has n linearly independent eigenvectors e1, e2, … , en, form the
nonsingular matrix T = [e1 e2 … en].
23222322
22212221
232221
)21(0)21(
202
055)21(
ejeeje
ejeeje
eeej
Tjje 12124~2
3 2( ) 4 2 1 2 1T
e e j j 3 2
91
Now, T-1AT = T-1A [e1 e2 … en] = T-1[Ae1 Ae2 … Aen] = T-1[1e1 2e2 … nen]
= T-1[e1 e2 … en]D = T-1TD = D
where D = diag [1, 2, … , n], and i, i = 1, 2, … , n are the eigenvalues of A.
Conversely, suppose there exists a matrix T such that T-1AT = D is diagonal. Then
AT = TD.
Let T = [t1 t2 … tn], then
AT = [At1 At2 … Atn] = [t1d11 t2d22 … tndnn] = TD Ati = diiti, which implies
that the ith column of T is an eigenvector of A associated with the eigenvalue dii.
Since T is nonsingular, there are n linearly independent eigenvectors.
Now, if A is diagonalizable, then eAt = TeDtT-1 because
1 2 3
1 1 1 2 1 31 1
2! 3!
At TDT te e TT TDT t TDT t TDT t
1 1 1 2 1 1 1 31 1
2! 3!I TDT t TDT TDT t TDT TDT TDT t
3132121
!3
1
!2
1tTTDtTTDtTDTI
2 2 3 3 1 11 1
2! 3!
DtT I Dt D t D t T Te T
92
Example: For the given matrix, compute eAt.
We already know that (A) = {-1, -1 - j2, -1 + j2}. Now,
The inverse of T is
010
011
552
A
111
21211
24240
jj
jj
T
1
2 2 101
1 1 2 1 28
1 1 2 1 2
T j j
j j
93
Therefore,
which implies that
Finally,
2100
0210
001
j
jD
tj
tj
t
Dt
e
e
e
e)21(
)21(
00
00
00
1
4cos(2 ) 2sin(2 ) 10sin(2 ) 10sin(2 )
1 cos(2 ) 2sin(2 ) 1 5cos(2 ) 5 5cos(2 ) ( )4
1 cos(2 ) 1 cos(2 ) 2sin(2 ) 5 cos(2 ) 2sin( )
2
At Dtt
e Te T
t t t te
t t t t t
t t t t t
94
Proposition: Suppose D is a block diagonal matrix with square blocks Di , i = 1, 2, …, n, i.e.,
Then,
Example: For the same A matrix of the previous example, compute eAt.
Let the similarity transformation T be given by
nD
D
D
D
00
0
00
2
1
tD
tD
tD
Dt
ne
e
e
e
00
0
00
2
1
1 2 2
0 4 2
Re{ } Im{ } 1 1 2
1 1 0
T
e e e
95
then,
and
which implies that
But,
Finally,
220
111
511
4
11T
11
1 2
2
1 0 00 1 2
0 1 2 , 1,0 2 1
0 2 1
DD T AT D D
D
tD
tD
Dt
e
ee
2
1
0
0
tt
tteeee ttDttD
2cos2sin
2sin2cosand 21
1 1
1 0 0
0 cos 2 sin 2
0 sin 2 cos 2
At Dt te T e T e T t t T
t t
Explicitly,
4cos(2 ) 2sin(2 ) 10sin(2 ) 10sin(2 )
1 cos(2 ) 2sin(2 ) 1 5cos(2 ) 5 5cos(2 ) ( )4
1 cos(2 ) 1 cos(2 ) 2sin(2 ) 5 cos(2 ) 2sin(2 )
tAt
t t t te
e t t t t t
t t t t t
.
At 0t , 3 3(0) I , hence, it is a valid state transition matrix.
96