Chapter 2: Mathematical Models of Systems O bjectives

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We use quantitative mathematical models of physical systems to design and analyze control systems. The dynamic behavior is generally described by ordinary differential equations. We will consider a wide range of systems, including mechanical, hydraulic, and electrical. Since most physical systems are nonlinear, we will discuss linearization approximations, which allow us to use Laplace transform methods.

We will then proceed to obtain the input–output relationship for components and subsystems in the form of transfer functions. The transfer function blocks can be organized into block diagrams or signal-flow graphs to graphically depict the interconnections. Block diagrams (and signal-flow graphs) are very convenient and natural tools for designing and analyzing complicated control systems

Chapter 2: Mathematical Models of Systems Objectives

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Introduction Six Step Approach to Dynamic System Problems

• 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

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Differential Equation of Physical Systems

Ta t( ) Ts t( ) 0

Ta t( ) Ts t( )

t( ) s t( ) a t( )

Ta t( ) = through - variable

angular rate difference = across-variable

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v21 Ltid

v211k t

211k t

P21 ItQd

Electrical Inductance

Translational Spring

Rotational Spring

Fluid Inertia

Describing Equation Energy or Power

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Differential Equation of Physical SystemsElectrical Capacitance

Translational Mass

Rotational Mass

Fluid Capacitance

Thermal Capacitance

i Ctv21

M v212

F Mtv2

Q CftP21

Cf P212

q CttT2

dd E Ct T2

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Differential Equation of Physical SystemsElectrical Resistance

Translational Damper

Rotational Damper

Fluid Resistance

Thermal Resistance

F b v21 P b v212

v21 P1R

T b 21 P b 212

P21 P1Rf

T21 P1Rt

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M 2ty t( )d

ty t( )d

d k y t( ) r t( )

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v t( )R

Ctv t( )d

ttv t( )

d r t( )

y t( ) K 1 e 1 t

sin 1 t 1

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K2 1 2 .5 2 10 2 2

y t( ) K2 e2 t

sin 2 t 2

y1 t( ) K2 e2 t

y2 t( ) K2 e2 t

0 1 2 3 4 5 6 71

y t( )

y1 t( )

y2 t( )

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Linear Approximations

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Linear Approximations

Linear Systems - Necessary condition

Principle of Superposition

Property of Homogeneity

Taylor Serieshttp://www.maths.abdn.ac.uk/%7Eigc/tch/ma2001/notes/node46.html

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Linear Approximations – Example 2.1

M 200gm g 9.8m

s2 L 100cm 0 0rad

T0 M g L sin 0

T1 M g L sin

T2 M g L cos 0 0 T0

4 3 2 1 0 1 2 3 410

T1 ( )

T2 ( )

Students are encouraged to investigate linear approximation accuracy for different values of0

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The Laplace Transform

Historical Perspective - Heaviside’s Operators

Origin of Operational Calculus (1887)

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Z p( )Z p( ) R L p

R L pH t( )

L p 1R

H t( )

p3 .....

H t( )

pnH t( )

Expanded in a power series

v = H(t)

Historical Perspective - Heaviside’s OperatorsOrigin of Operational Calculus (1887)

(*) Oliver Heaviside: Sage in Solitude, Paul J. Nahin, IEEE Press 1987.

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Definition

L f t( )( )0

tf t( ) e s t

d = F(s)

Here the complex frequency is s j w

The Laplace Transform exists when

tf t( ) e s t

d this means that the integral converges

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Determine the Laplace transform for the functions

a) f1 t( ) 1 for t 0

F1 s( )0

te s t

d = 1s

e s t( )1s

b) f2 t( ) e a t( )

F2 s( )0

te a t( ) e s t( )

d = 1s 1

e s a( ) t[ ] F2 s( )

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note that the initial condition is included in the transformationsF(s) - f(0+)=Ltf t( )d

tf t( ) e s t( )

d-f(0+) +=

tf t( ) s e s t( )

df t( ) e s t( )=0

we obtain

v f t( )anddu s e s t( ) dt

and, from which

dv df t( )u e s t( )where

u v uv

dby the use of

Ltf t( )d

ttf t( ) e s t( )d

Evaluate the laplace transform of the derivative of a function

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The Laplace TransformPractical Example - Consider the circuit.

The KVL equation is

4 i t( ) 2ti t( )d

d 0 assume i(0+) = 5 A

Applying the Laplace Transform, we have

t4 i t( ) 2ti t( )d

e s t( )

d 0 40

ti t( ) e s t( )

tti t( ) e s t( )d

4 I s( ) 2 s I s( ) i 0( )( ) 0 4 I s( ) 2 s I s( ) 10 0

transforming back to the time domain, with our present knowledge of Laplace transform, we may say thatI s( )

0 1 20

i t( )

t 0 0.01 2( )

i t( ) 5 e 2 t( )

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The Partial-Fraction Expansion (or Heaviside expansion theorem)

Suppose that

The partial fraction expansion indicates that F(s) consists of a sum of terms, each of which is a factor of the denominator. The values of K1 and K2 are determined by combining the individual fractions by means of the lowest common denominator and comparing the resultant numerator coefficients with those of the coefficients of the numerator before separation in different terms.

F s( )s z1

s p1( ) s p2( )

F s( )K1

Evaluation of Ki in the manner just described requires the simultaneous solution of n equations. An alternative method is to multiply both sides of the equation by (s + pi) then setting s= - pi, the right-hand side is zero except for Ki so that

Kis pi( ) s z1( )

s p1( ) s p2( )s = - pi

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s -> 0t -> infinite

Lim s F s( )( )Lim f t( )( )7. Final-value Theorem

s -> infinitet -> 0

Lim s F s( )( )Lim f t( )( ) f 0( )6. Initial-value Theorem

sF s( )

df t( )

t5. Frequency Integration

F s a( )f t( ) e a t( )4. Frequency shifting

sF s( )d

dt f t( )3. Frequency differentiation

f at( )2. Time scaling

f t T( ) u t T( )1. Time delaye s T( ) F s( )

Property Time Domain Frequency Domain

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Useful Transform Pairs

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y s( )s

s 2 n s2 2 n n

s1 n n 2

2 k M s2 n n

RealReal repeatedImaginary (conjugates)Complex (conjugates)

s1 n j n 1 2

s2 n j n 1 2

Consider the mass-spring-damper system

Y s( )Ms b( ) yo

Ms2 bs k

equation 2.21

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The Transfer Function of Linear Systems

V1 s( ) R1

I s( ) Z1 s( ) R

Z2 s( )1

CsV2 s( )1

I s( )

V2 s( )

V1 s( )

Z2 s( )

Z1 s( ) Z2 s( )

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Example 2.2

2ty t( )d

ty t( )d

d 3 y t( ) 2 r t( )

Initial Conditions: Y 0( ) 1ty 0( )d

d0 r t( ) 1

The Laplace transform yields:

s2 Y s( ) s y 0( ) 4 s Y s( ) y 0( )( ) 3Y s( ) 2 R s( )

Since R(s)=1/s and y(0)=1, we obtain:

Y s( )s 4( )

s2 4s 3 2

s s2 4s 3

The partial fraction expansion yields:

Y s( )

s 1( )

s 3( )

1s 1( )

s 3( )

Therefore the transient response is:

y t( )32

1 e t 13

The steady-state response is:

ty t( )lim

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Tm K1 Kf if t( ) ia t( )

field controled motor - Lapalce Transform

Tm s( ) K1 Kf Ia If s( )

Vf s( ) Rf Lf s If s( )

Tm s( ) TL s( ) Td s( )

TL s( ) J s2 s( ) b s s( )

rearranging equations

TL s( ) Tm s( ) Td s( )

Tm s( ) Km If s( )

If s( )Vf s( )

Rf Lf s

Td s( ) 0

s( )Vf s( )

Kms J s b( ) Lf s Rf

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V 2 s( )

V 1 s( )1

V 2 s( )

V 1 s( )RCs

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V 2 s( )

V 1 s( )

R 2 R 1 C s 1 R 1

V 2 s( )

V 1 s( )

R 1 C 1 s 1 R 2 C 2 s 1 R 1 C 2 s

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s( )V f s( )

s J s b( ) L f s R f

s( )V a s( )

s R a L a s J s b( ) K b K m

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Vo s( )

Vc s( )

KRc Rq

s c 1 s q 1

RqFor the unloaded case:id 0 c q

0.05s c 0.5s

V12 Vq V34 Vd

s( )Vc s( )

b m( )

m = slope of linearized torque-speed curve (normally negative)

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The Transfer Function of Linear SystemsY s( )X s( )

Ks Ms B( )

dg g x P( ) flow

A = area of piston

Gear Ratio = n = N1/N2

N2 L N1 m

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V2 s( )

V1 s( )

V2 s( ) ks 1 s( ) 2 s( ) V2 s( ) ks error s( )

ksVbattery

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V2 s( ) Kt s( ) Kt s s( )

Kt constant

V2 s( )

V1 s( )

Ro = output resistanceCo = output capacitance

Ro Co 1s

and is often negligible for controller amplifier

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T s( )q s( )

Ct s Q S1R

T To Te = temperature difference due to thermal process

Ct = thermal capacitance= fluid flow rate = constant= specific heat of water= thermal resistance of insulation= rate of heat flow of heating element

q s( )

xo t( ) y t( ) xin t( )

Xo s( )

Xin s( )s2

For low frequency oscillations, where n

Xo j Xin j

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converts radial motion to linear motion

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Block Diagram Models

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Original Diagram Equivalent Diagram

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Example 2.7

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Block Diagram Models Example 2.7

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Signal-Flow Graph Models

For complex systems, the block diagram method can become difficult to complete. By using the signal-flow graph model, the reduction procedure (used in the block diagram method) is not necessary to determine the relationship between system variables.

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Y1 s( ) G11 s( ) R1 s( ) G12 s( ) R2 s( )

Y2 s( ) G21 s( ) R1 s( ) G22 s( ) R2 s( )

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a11 x1 a12 x2 r1 x1

a21 x1 a22 x2 r2 x2

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Example 2.8

Y s( )R s( )

G 1 G 2 G 3 G 4 1 L 3 L 4 G 5 G 6 G 7 G 8 1 L 1 L 2

1 L 1 L 2 L 3 L 4 L 1 L 3 L 1 L 4 L 2 L 3 L 2 L 4

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Example 2.10

Y s( )R s( )

G 1 G 2 G 3 G 4

1 G 2 G 3 H 2 G 3 G 4 H 1 G 1 G 2 G 3 G 4 H 3

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Y s( )R s( )

P1 P2 2 P3

P1 G1 G2 G3 G4 G5 G6 P2 G1 G2 G7 G6 P3 G1 G2 G3 G4 G8

1 L1 L2 L3 L4 L5 L6 L7 L8 L5 L7 L5 L4 L3 L4

1 3 1 2 1 L5 1 G4 H4

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Design Examples

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Speed control of an electric traction motor.

Design Examples

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Design Examples

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Design Examples

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Design Examples

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Design Examples

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The Simulation of Systems Using MATLAB

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Sys1 = sysh2 / sysg4

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Num4=[0.1];

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Sequential Design Example: Disk Drive Read System

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1L c s R c

1L q s R q

1L d L a s R d R a

1J s b

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http://www.jhu.edu/%7Esignals/sensitivity/index.htm

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http://www.jhu.edu/%7Esignals/

Chapter 2: Mathematical Models of Systems O bjectives

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