Mathematical Models and Block Diagrams of Systems Regulation And Control Engineering

Mathematical Models and Block Diagrams of Systems

Regulation And Control Engineering

CONTENTS Introduction Differential Equations of Physical

Systems The Laplace Transform Transfer Function of Linear Systems Block Diagram

STEP AND PROCEDURE Define the system and its components Formulate the mathematical model and list

the necessary assumptions Write the differential equations describing

the model Solve the equations for the desired output

variables Examine the solutions and the

assumptions If necessary, reanalyze or redesign the

system

INTRODUCTIONS A mathematical model is a set of equations (usually

differential equations) that represents the dynamics of systems.

In practice, the complexity of the system requires some assumptions in the determination model.

The equations of the mathematical model may be solved using mathematical tools such as the Laplace Transform.

Before solving the equations, we usually need to linearize them.

DIFFERENTIAL EQUATIONS

Physical law of the process Differential Equation

Mechanical system (Newton’s laws)Electrical system (Kirchhoff’s laws)

How do we obtain the equations?

Examples:i. ii.

DIFFERENTIAL EQUATIONS Example: Springer-mass-damper system

Assumption: Wall friction is a viscous force.

The time function of r(t) sometimes called forcing function

Linearly proportional to the velocity

)()( tbvtf

DIFFERENTIAL EQUATIONS Example: Springer-mass-damper system

Newton’s 2nd Law:

)()()()( tMatrtkytbv

)()()()(2

2

trtkydt

tdybdt

tydM

DIFFERENTIAL EQUATIONS Example: RLC Circuit

t

tvdiC

tRidt

tdiL0

)()(1)()(

0)( cLR VVVtv

THE LAPLACE TRANSFORM The differential equations are transformed into

algebraic equations, which are easier to solve. The Laplace transformation for a function of time,

f(t) is:

If, , then,

Similarly,

Thus,

0

)}({)()( tfLdtetfsF st

dtdytf )( )0()}({)}({ ytysL

dtdyLtfL

dtdy

dttdysL

dttydL )0()()(2

2

dtdysytyLs

dttydL )0()0()}({)( 22

2

THE LAPLACE TRANSFORM Example: Spring-mass-damper dynamic equation

)()()()(2

2

trtkydt

tdybdt

tydM

)()()]0()([)]0()0()([ 2 sRskYyssYbysysYsM

0)()()( 002 skYbysbsYMsysYMs

Laplace Transform for the equation above:

When r(t)=0, y(0)= y0 and (0)=0:

y

)()()()( 2

0

sqsp

kbsMsybMssY

THE LAPLACE TRANSFORM Example: Spring-mass-damper dynamic equation

Some Definitionsi. q(s) = 0 is called characteristic

equation (C.E.) because the roots of this equation determine the character of the time response.

ii. The roots of C.E are also called the poles of the system.

iii. The roots of numerator polynomial p(s) are called the zeros of the system.

)()()()( 2

0

sqsp

kbsMsybMssY

THE LAPLACE TRANSFORM Transform table:

f(t) F(s)

1. δ(t) 1

2. u(t)

3. t u(t)

4. tn u(t)

5. e-at u(t)

6. sin t u(t)

7. cos t u(t)

s1

2

1s

1

!ns

n

as 1

22 s

22 ss

Impulse functionStep functionRamp function

THE LAPLACE TRANSFORM Transform

Properties

THE LAPLACE TRANSFORM Example: Find the Laplace Transform for the

following.

i. Unit function:

ii. Ramp function:

iii. Step function:

1)( tf

ttf )(

atAetf )(

THE LAPLACE TRANSFORM Transform Theorem

i. Differentiation Theorem

ii. Integration Theorem:

iii. Initial Value Theorem:

iv. Final Value Theorem:

ssFdfL

t )()(0

)(lim)(lim0

ssFtfst

)(lim)0( ssFft

)0()(})({ fssFdt

tdfL

)0()0()(})({ 22

2

fsfsFsdt

tfdL

THE LAPLACE TRANSFORM The inverse Laplace Transform can be obtained using:

Partial fraction method can be used to find the inverse Laplace Transform of a complicated function.

We can convert the function to a sum of simpler terms for which we know the inverse Laplace Transform.

j

j

stdsesFj

tf )(21)(

)()()()( 21 sFsFsFsF n

)()()()( 12

11

1 sFLsFLsFLtf n

)()()( 21 tftftf n

THE LAPLACE TRANSFORM We will consider three cases and show that F(s) can

be expanded into partial fraction:i. Case 1:

Roots of denominator A(s) are real and distinct.ii. Case 2:

Roots of denominator A(s) are real and repeated.iii. Case 3:

Roots of denominator A(s) are complex conjugate.

THE LAPLACE TRANSFORM Case 1: Roots of denominator A(s) are real and

distinct.Example:

Solution:

)2)(1(2)(

sssF

21)(

sB

sAsF It is found that:

A = 2 and B = -2

22

12

sstt eetf 222)(

THE LAPLACE TRANSFORM Case 1: Roots of denominator A(s) are real and

distinct.

Problem: Find the Inverse Laplace Transform for the following.

)2)(1(3)(

ssssF

THE LAPLACE TRANSFORM Case 2: Roots of denominator A(s) are real and

repeated.Example:

Solution:

2)2)(1(2)(

sssF

2)2(21)(

sC

sB

sAsF It is found that:

A = 2, B = -2 and C = -2

2)2(2

22

12

sss

ttt teeetf 22 222)(

THE LAPLACE TRANSFORM Case 3: Roots of denominator A(s) are complex

conjugate.Example:

Solution:

)52(3)( 2

sss

sF

52)( 2

ssCBs

sAsF

It is found that:A = 3/5, B = -3/5and C = -6/5

52

25353

2 sss

s

22 2)1(

)2)(21()1(5353

ss

s

THE LAPLACE TRANSFORM Case 3: Roots of denominator A(s) are complex

conjugate.Example:

Solution:

)52(3)( 2

sss

sF

)2sin212(cos

53

53)( ttetf t

THE LAPLACE TRANSFORM Problem: Find the solution x(t) for the following

differential equations.

i.

ii.

,023 xxx

,352 xxx

bxax )0(,)0(

bxax )0(,)0(

THE TRANSFER FUNCTION The transfer function of a linear system is the ratio of the

Laplace Transform of the output to the Laplace Transform of the input variable.

Consider a spring-mass-damper dynamic equation with initial zero condition.

)()()(

sInputsOutputsG

)()()()(2 sRskYsbsYsYMs

THE TRANSFER FUNCTION

The transfer function is given by the following.

kbsMssRsYsG

2

1)()()(

Y(s)R(s)kbsMs 2

1

THE TRANSFER FUNCTION Electrical Network Transfer Function

Component V-I I-V V-Q Impedance

Admittance

THE TRANSFER FUNCTION Problem: Obtain the transfer function for the following RC

network.

THE TRANSFER FUNCTION Problem: Obtain the transfer function for the following

RLC network.

Answer:

THE TRANSFER FUNCTION Mechanical System Transfer Function

Problem: Find the transfer function for the mechanical system below.

)(tukyybym

The external force u(t) is the input to the system, and the displacement y(t) of the mass is the output.

The displacement y(t) is measured from the equilibrium position.

The transfer function of the system.

BLOCK DIAGRAM A block diagram of a system is a practical representation

of the functions performed by each component and of the flow of signals.

Cascaded sub-systems:

Transfer Function G(s) Outpu

tInpu

t

BLOCK DIAGRAM Feedback Control System

BLOCK DIAGRAM Feedback Control System

Therefore,

The negative feedback of the control system is given by:Ea(s) = R(s) – H(s)Y(s)Y(s) = G(s)Ea(s)

)]()()()[()( sYsHsRsGsY

)()(1)(

)()(

sHsGsG

sRsY

BLOCK DIAGRAM Reduction Rules

BLOCK DIAGRAM Reduction Rules

BLOCK DIAGRAM Problem:

BLOCK DIAGRAM Problem:

FURTHER READING… Chapter 2

i. Dorf R.C., Bishop R.H. (2001). Modern Control Systems (9th Ed), Prentice Hall.

ii. Nise N.S. (2004). Control System Engineering (4th Ed), John Wiley & Sons.

THE END…

“The whole of science is nothing more than a refinement of everyday thinking…”