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Chapter 2
Mathematical Models of Systems
2.1 Introduction
To understand and control complex systems, one must obtain quantitative mathematical models of these systems. It is necessary therefore to analyze the relationship between the system variables and to obtain a mathematical model.
Example Electrical resistance
R
U
Mathematical model
i=U/R
2.1 Introduction
In practice, the complexity of systems and our ignorance of all the relevant factors necessitate the introduction of assumption concerning the system operation.
2.1 Introduction
2.2 Differential Equations of Physical Systems
ky(t)dt
dy(t)br(t)
dt
y(t)dM
2
dt
dy(t)v(t)
Example 1 : Spring-mass-damper system
Wall
friction, b
r(t) Force
y
k
r(t)v(t)dtkbv(t)dt
dv(t)M
t
0
r(t)
Current source v(t)
r(t)v(t)dtL
1
dt
dv(t)C
R
v(t) t
0
2.2 Differential Equations of Physical Systems
Example 2 : RLC Circuit
Analogous systems
)()()()(
0trdttvktbv
dt
tdvM
t
r(t)v(t)dtLdt
dv(t)C
R
v(t) t 0
1
Spring-mass-damper system
RLC circuit
2.2 Differential Equations of Physical Systems
Principle of superposition
system x(t)
excitation
y(t)
response
x(t) = x1(t) y(t) = y1(t)
x(t) = x2(t) y(t) = y2(t)
x(t) = x1(t)+x2(t) y(t) = y1(t)+y2(t)
system x(t)
excitation
y(t)
response
x(t) = x1(t) y(t) = y1(t)
x(t) = ax1(t) y(t) = ay1(t)
Property of homogeneity
Linear system
A linear system satisfies the properties of superposition and homogeneity.
Is it a linear relationship ?
2xy
bmxy
Operating point: x0,y0
bmxy
xxx 0 yyy 0
bxmmxyy 00
xmy
2.3 Linear Approximation
))((( txgty
2
02
2
00
)(|
)(|)()(
0
0
xxdx
gd
xxdx
dgxgxgy
xx
xx
Taylor series
2
02
2
00
)(|
)(|)()(
0
0
xxdx
gd
xxdx
dgxgxgy
xx
xx
)(y | 000 xgmdx
dgxx
)( 00 xxmyy
Linear Approximation
Example: nonlinear spring
0| yy
dy
df
y
f0
y0
Equilibrium
(operating point) yyf 02
Pendulum oscillator model
0 0 00 TEquilibrium point
The torque on the Mass is
)(|sin
00 0
MgLTT
)(cos 000 MgLTT
0 0 00 T MgLT
sinMgLT
2.4 Laplace Transform
0
)()( dtetfsF st
0|)(| dtetf t
for some finite, real , the Laplace transform of is
defined as
)(tf
or
)]([)( of transform)( tftfLaplacesF L
condition thesatisfies that )(function real Given the tf
Inverse Laplace Transform
Given the Laplace transform F(s), the operation
of obtaining f(t) is termed the inverse Laplace
transformation, and is denoted by
)]([)( 1 sFtf L
The inverse Laplace transform integral is given as
jc
jc
stdsesFj
tf )(2
1)(
where c is a real constant that is greater than the
real parts of all the singularities of F(s).
Theorems of the Laplace Transform
Theorem 1. Multiplication by a constant
Let k be a constant, and F(s) be the Laplace
transform of f(t). Then
)()]([ skFtkf L
Theorem 2. Sum and Difference
Let F1(s) and F2(s) be the Laplace transform of
f1(t) and f2(t), respectively. Then
)()()]()([ 2121 sFsFtftf L
Theorems of the Laplace Transform
Theorem 3. Differentiation
Let F(s) be the Laplace transform of f(t), and f(0) is
the limit of f(t) as t approaches 0. The Laplace
transform of the time derivative of f(t) is
)0()()(lim)()(
0fssFtfssF
dt
tdf
t
L
In general, for higher-order derivatives of f(t),
1
121
0
)()()(lim)(
)(n
nnn
t
n
n
n
dt
tfd
dt
tdfstfssFs
dt
tfdL
)0()0()0()( )1()1(21 nnnn ffsfssFs
Theorems of the Laplace Transform
Theorem 4. Integration
The Laplace transform of the first integral of f(t) with
respect to t is the Laplace transform of f(t) divided
by s; that is,
s
sFdf
o
)()(
t
L
For nth-order integration,
n
n
t tn
s
sFduuf
)())((
0 0
Theorems of the Laplace Transform
)()](1)([ sFeTtTtf TsL
where 1(t-T) denotes the unit-step function that is
shifted in time to the right by T.
Theorem 5. Shift in Time
The Laplace transform of f(t) delayed by time T
is equal to the Laplace transform f(t) multiplied
by ; that is Tse
Theorems of the Laplace Transform
Theorem 6. Initial-Value Theorem
If the Laplace transform of f(t) is F(s), then
)(lim)(lim0
ssFtfst
if the limit exists.
Theorem 7. Final-Value Theorem
If the Laplace transform of f(t) is F(s), and if sF(s)
is analytic on the imaginary axis and the right half of
the s-plane, then
)(lim)(lim0
ssFtfst
Theorems of the Laplace Transform
Theorem 8. Complex Shifting
The Laplace transform of f(t) multiplied by , where
is a constant, is equal to the Laplace transform F(s),
with s replaced by ; that is,
te
s
)()]([ sFtfe tL
Theorem 9. Real Convolution
Let F1(s) and F2(s) be the Laplace transform of f1(t)
and f2(t), respectively, and f1(t)=0, f2(t)=0, for t
Inverse Laplace Transform
Consider the function
)(
)()(
sP
sQsG
where P(s) and Q(s) are polynomial of s. It is
assumed that the order of P(s) in s is greater than
that of Q(s). The polynomial P(s) may be written
Partial-Fraction Expansion
01
1
1)( asasassPn
n
n
where are real coefficients. 110 ,, naaa
Partial-Fraction Expansion
P(s) Has Simple Roots If all the roots of P(s) are simple and real, the function G(s) can be written as
)(
)()(
sP
sQsG 01
1
1)( asasassPn
n
n
n
n
n ss
K
ss
K
ss
K
ssssss
sQsG
2
2
1
1
21 )())((
)()(
where
issii
sGssK )()(
P(s) Has Multiple-Order Roots If r of the n roots of P(s) are identical, G(s) is written
r
irn ssssssss
sQsG
))(())((
)()(
21
| roots simple of r terms-n|
)(2
2
1
1
rn
rn
ss
K
ss
K
ss
KsG
| roots repeated of r terms |
)()( 221
r
i
r
ii ss
A
ss
A
ss
A
),,2,1( rni , then G(s) can be expanded as
Partial-Fraction Expansion
| roots simple of r terms-n|
)(2
2
1
1
rn
rn
ss
K
ss
K
ss
KsG
| roots repeated of r terms |
)()( 221
r
i
r
ii ss
A
ss
A
ss
A
),,2,1( )()( rnjsGssKjssjj
where
iss
r
ir sGssA )()(
)1,2,1( )()(!
1 rksGss
ds
d
kA
iss
r
ik
k
kr
The determination of the coefficients that correspond
to the multiple-order roots is described as follows.
Partial-Fraction Expansion
Example
)2()1(
1)(
3
ssssG
G(s) can be written as
3
3
2
2121
)1()1(12)(
s
A
s
A
s
A
s
K
s
KsG
The coefficients corresponding to the simple roots and
those of the third-order root are
2
1 )( 01 sssGK
2
1 )()2( 22 ssGsK
1 )()1( 133 ssGsA 0 )()1( 13
2 ssGsds
dA
1 )()1(2
11
3
2
2
1 ssGsds
dA
Application of the Laplace
transform to the solution of linear
ordinary differential equations
Transform the differential equation to the s-domain by Laplace transform using the Laplace transform table.
Manipulate the transformed algebraic equation and solve for the output variable.
Perform partial-fraction expansion to the transformed algebraic equation.
Obtain the inverse Laplace transform from the Laplace transform table.
Example
)(523 tuyyy
Taking the Laplace transform on both sides:
ssYyssYysysYs /5)(2)0(3)(3)0()0()(2
Substituting the values of the initial conditions into the
last equation. Then solving for Y(s) and expanding by
partial-fraction, we get
)2(2
3
1
5
2
5
)2)(1(
5)(
2
ssssss
sssY
2)0(,1)0(),(1)( yyttu where
Taking the inverse Laplace transform, we get
0 5.155.2)( 2 teety tt
Laplace Transform
In contrast with the classical methods of solving linear differential equations, the Laplace transform method has following two features:
1. The homogeneous solution and the particular integral of the differential equation are obtained in one operation.
2. The Laplace transform converts the differential equation into an algebraic equation in s. It is then possible to manipulate the algebraic equation by simple algebraic rules to obtain the solution in the s-domain. The final solution is obtained by taking the inverse Laplace transform.