Mathematical Madeling and Block Diagram

8/13/2019 Mathematical Madeling and Block Diagram

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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 useLaplace transform methods.

We will then proceed to obtain the inputoutput 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

Tat( ) Ts t( ) 0

Tat( ) Ts t( )

t( ) s t( ) at( )

Tat( ) = through - variable

angular rate difference = across-variable


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v21 Lti

d

d E

1

2L i

2

v211

k tFd

d E 1

2

F2

k

211

k tT

d

d E

1

2

T2

k

P21 ItQ

d

d E

1

2I Q

2

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 C tv21d

d E 1

2 M v212

F Mt

v2d

d E

1

2M v2

2

T Jt2

d

d E

1

2J 2

2

Q CftP21d

d E 1

2Cf P212

q CttT2

d

d E CtT2


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Differential Equation of Physical SystemsElectrical Resistance

Translational Damper

Rotational Damper

Fluid Resistance

Thermal Resistance

F b v21 P b v212

i 1

Rv21 P

1

Rv21

2

T b21 P b212

Q

1

Rf P21 P 1

Rf P21

2

q 1

RtT21 P

1

RtT21


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M 2t

y t( )dd

2

b ty t( )d

d k y t( ) r t( )


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v t( )

RC

tv t( )

d

d

1

L0

t

tv t( )

d r t( )

y t( ) K1e 1 t

sin 1t 1


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K2 1 2 .5 2 10 2 2

y t( ) K2 e 2 t

sin 2 t 2

y1 t( ) K2 e

2 t

y2 t( ) K2 e

2 t

0 1 2 3 4 5 6 71

0

1

y t( )

y1 t( )

y2 t( )

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
http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node46.htmlhttp://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node46.html


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Linear ApproximationsExample 2.1

M 200gm g 9.8m

s2 L 100cm 0 0rad

15

16

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

5

0

5

10

T 1 ( )

T 2 ( )

Students are encouraged to inves tigate linear approximation accuracy for dif ferent values of 0


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The Laplace Transform

Historical Perspective - Heavisides Operators

Origin of Operational Calculus (1887)


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pt

d

d

1

p0

t

u1

d

i v

Z p( )Z p( ) R L p

i 1

R L p H t( )

1

L p 1 R

L p

H t( ) 1

R

R

L

1

p

R

L

21

p2

R

L

31

p3

.....

H t( )

1

pn

H t( ) t

n

n

i 1

R

R

Lt

R

L

2t2

2

R

L

3t3

3 ..

i

1

R1 e

R

L

t

Expanded in a power series

v = H(t)

Historical Perspective - Heavisides Operators

Origin 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 f requency is s j w

The Laplace Transform exists w hen

0

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=

1

s 1 e

s a( ) t[ ] F2 s( )

1

s a


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

L

t

f t( )d

d

s

0

tf t( ) e s t( )

d-f(0+) +=

0

tf t( ) s e s t( )

df t( ) e s t( )=0

vu

d

w e obtain

v f t( )anddu s e

s t( ) dt

and, from w hich

dv df t( )u e

s t( )w here

u v uv

d=vu

dby the use of

Ltf t( )

d

d

0

t

tf t( ) e

s t( )d

d

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( ) 2t

i t( )d

d 0 assume i(0+) = 5 A

Applying the Laplace Transf orm, w e have

0

t4 i t( ) 2t

i t( )d

d

e s t( )

d 0 4

0

ti t( ) e s t( )

d 2

0

tt

i t( ) e s t( )

dd

d 0

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, w ith our present know ledge of

Laplace transf orm, w e may say thatI s( )

5

s 2

0 1 20

2

4

6

i t( )

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( )

or

F s( )K1

s p1

K2

s p2

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( )( )Limf t( )( )7. Final-value Theorem

s -> inf initet -> 0

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

0

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 dif ferentiation

f at( )2. Time scaling

1

aF

s

a

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 b

M

yo

s2 b

M

s k

M

s 2 n

s

22 n n

2

s1 n n 2

1n

k

M

b

2 k M

s2 n n 2

1

Roots

Real

Real repeated

Imaginary (conjugates)

Complex (conjugates)

s1 n jn 1 2

s2 n jn 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( ) R 1

Cs

I s( ) Z1 s( ) R

Z2 s( ) 1

CsV2 s( ) 1

Cs

I s( )

V2 s( )

V1 s( )

1

Cs

R 1

Cs

Z2 s( )

Z1 s( ) Z2 s( )


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Example 2.2

2t

y t( )d

d

2

4ty 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, w e obtain:

Y s( ) s 4( )

s2

4s 3 2

s s2

4s 3

The partial fraction expansion yields:

Y s( )

3

2

s 1( )

1

2

s 3( )

1

s 1( )

1

3

s 3( )

2

3

s

Therefore the transient response is:

y t( ) 3

2e

t 1

2e

3 t

1 e t 1

3e

3 t

2

3

The steady-state response is:

ty t( )lim

2

3


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Kf if

Tm K1Kf if t( ) ia t( )

f ield controled motor - Lapalce Transform

Tm s( ) K1Kf Ia If s( )

Vf s( ) Rf Lfs 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( ) KmIf s( )

If s( ) Vf s( )

Rf Lfs


Td s( ) 0

s( )

Vfs( )

Km

s J s b( ) Lfs Rf


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V 2s( )

V 1s( )

1

RCs

V 2s( )

V 1s( )

RCs


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V 2s( )

V 1s( )

R2R1C s 1 R1

V 2s( )

V 1s( )

R1C 1 s 1 R2C 2 s 1 R1C 2 s


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s( )

V fs( )

Km

s J s b( ) L fs Rf

s( )

V as( )

Km

s Ra L as J s b( ) KbKm


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Vos( )

Vcs( )

K

RcRq

sc 1 sq 1

cLc

Rcq

Lq

Rq

For the unloaded case:id 0 c q

0.05s c 0.5s

V12 Vq V34 Vd

s( )

Vcs( )

Km

s s 1

J

b m( )

m = slope of linearized

torque-speed curve

(normally negative)


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Y s( )

X s( )

K

s Ms B( )

K A kxkp

B b A2

kp

kxx

gd

dkp

Pg

d

d

g g x P( ) flow

A = area of piston

Gear Ratio = n = N1/N2

N2 L N1 m

L n m

L nm


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V2s( )

V1s( )

R2

R

R2

R1 R2

R2

R

max

V2s( ) ks 1s( ) 2s( ) V2s( ) ks errors( )

ksVbattery

max


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V2s( ) Kt s( ) Kts s( )Kt constant

V2 s( )

V1 s( )

ka

s 1

Ro = output resistance

Co = output capacitance

Ro Co 1s

and is often negligible

for controller amplifier


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T s( )

q s( )

1

Cts Q S 1

R

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

QS

Rtq s( )

xot( ) y t( ) xin t( )

Xos( )

Xin s( )

s2

s2 b

M

s k

M

For low frequency oscillations, w here n

Xo j Xin j

2

k

M


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x r

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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Y1s( ) G11s( ) R1s( ) G12s( ) R2s( )

Y2s( ) G21s( ) R1s( ) G22s( ) R2s( )


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a11x1 a12x2 r1 x1

a21x1 a22x2 r2 x2


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Example 2.8

Y s( )

R s( )

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

Si G


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Example 2.10

Y s( )R s( )

G1G

2 G

3 G

4

1 G 2G 3 H 2 G 3G 4 H 1 G 1G 2 G 3 G4 H 3

Si l Fl G h M d l


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Y s( )

R s( )

P1 P22 P3

P1 G1G2 G3 G4 G5 G6 P2 G1G2 G7 G6 P3 G1G2 G3 G4 G8

1 L1 L2 L3 L4 L5 L6 L7 L8 L5L7 L5L4 L3L4

1 3 1 2 1 L5 1 G4H4

D i E l


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Design Examples


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D i E l


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Design Examples

D i E l


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Design Examples

D i E l


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Design Examples

D i E l


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Design Examples

The Simulation of Systems Using MATLAB


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error

y g

Sys1 = sysh2 / sysg4



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y g



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error

y g

Num4=[0.1];



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y g



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y g

Sequential Design Example: Disk Drive Read System


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q g p y



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q g p y



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=

q g p y

P2.11


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P2.11


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1

L cs Rc

Vc

Ic

K1

1

L qs Rq

Vq

K2

K3-Vb

+Vd

Km

Id

1

L d L a s Rd Ra

Tm

1

J s b

1

s


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


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Mathematical Madeling and Block Diagram