B_lecture2 the Laplace Transform Automatic control System

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


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


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


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


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


)()](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


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


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


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


| 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.


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.

B_lecture2 the Laplace Transform Automatic control System

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