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I. MATHEMATICAL MODELS

Desired

Performance ControllerPlant /

ProcessComparison

Sensor

Output

Feedback Loop

InputIII.

II.

I.Errorsignals

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I.A Model Building Revised September, 2004

The process of model building

Real LifeSystem

PhysicalModel

Set of DifferentialEquations relating

Input/output of

plant/process

x

Up/down

movementRotational

movement

Comprised ofbasic elements

relating relevantinputs and outputs

I.A Model Building

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Governing Equations

For BasicElements

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I.A.1 Mechanical SystemUse Newtons Law

maF

..

IExample: Spring-Mass-damper system

)()()()( tFtkytydt

dbty

dt

dM

2

2

)()( ty

dt

dtv Define velocity:

)()()()( tFdttvktbvtvdt

dM

t

0

Input: force F ; Output: positiony

(As given in Table 2.2)

Springconst.k

M y

F

Dampingcoeff. b

Equilibriumposition

Equation forDamper elementEquation for

Mass element

Equation forSpring element

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I.A.2 Electric Circuit

Use Kirchoffs Laws

KCL:Sums of currents leaving and entering a node are equal

KVL: Sum of voltage around a closed path is zero

Example:LRCcircuit

)()()(

)(1

0tItV

dt

dC

R

tVdttV

L

t

I(t)RL C

VoltageV(t)

Using KCL (and Table 2.2):

Input: CurrentI(t)Output: Voltage V(t)

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)()()()( tVdttiC

tRitidt

dL

t

0

1

Velocity v(t)Force F(t)MassM

Friction b

Spring constantk

Current i(t)Voltage V(t)L InductanceR Resistance

1/C Inverse of capacitance

Force-Voltage analogy

V(t)

RL

C

)()()()( tFdttvktbvtvdt

dMt

0

Consider anotherLRCcircuit Voltage driven

i(t)

Compare mechanical system example:

+

-

Using KVL:

Input: Voltage V(t)Output: Current i(t)

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One can build mechanical, electrical, fluid, or thermalanalog to simulate performance of a system One may thus use easier-to-build analog to test out

performance of a system first before actual construction

Early example: Moniac a hydraulic systemto study US money flow by Bill Philips atLondon School of Economics, 1949

Summary of Model Building:Plant/process + Objective of study

Differential equations of

basic elements from Table 2.2

Appropriate physicalmodel comprisingof ideal basic element

Set of differentialequation relatingrelevant inputs andoutputs

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A Linearprocess:

Given: input is u1, output is y1input is u2, output is y2

Then, if input is u1+u2, output is y1+y2

I.A.4 Linear, Time-Invariant (LTI) process

A Time-invariantprocess:

input-output relation is not

dependent on time

Note: most processes can reasonably

be assumed as Time-invariant

Linear differential equation

with Time-invariantcoefficients

Scope of MAE3050:

Process describable by

Linear differential equation

Process describable by differentialequation of constant coefficients

Linear differential equationwith Time-varying coefficients

Nonlinear differential equationwith Time-invariant coefficients

Note: most processes are actuallynot linear (i.e., nonlinear) by nature

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)()(cos)()(

)()()(

)()()(

)()()()(

2

tutyCtydt

dAt

tutCytydt

dA

tutCytydt

dA

tutCytydt

dAt

Describing different equation for S:

S is Linear, Time-varying system

S is Linear, Time-invariant system

S is Nonlinear, Time-invariant system

S is Nonlinear, Time-varying system

Sinput u outputy

Given a system S:

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I.B Linearization

Example 2.1: Pendulum oscillator model

Idea: Taylor expansion of function

Can have a linear approximation of (X) about a certain Describing differential equation (X) is nonlinear (is it?)

PendulumTorque Input Output

2ML

T

L

g sin

..

L

Mg

Inputtorque

(X)

Nonlinear system may be studied through linearization ofits nonlinear model

ooperating point

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Recall: Taylor Series expansion of functionf(x) about point

.....)(!

)(.....)(')()()( )(

o

nn

oooo xf

n

xxxfxxxfxf

ox

)(')()( ooo xfxxxf

Range of validity of linear approximation depends on function

f(x) and operating point ox

)(xf

)( oxf

ox xOperating point

Linear approximation of )(xf

about operating point ox

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- From 1st order terms of (XX),

2MLL

go

)cos(

..

This yields relation between o and operating point

- From 0

th

order terms of (XX),

gML

T

ML

T

L

goo

02

0 )sin()sin(

This gives the linear approximation of the pendulumdifferential equation relating torque variation (t)and about operating point

- Equation (X) becomes

2

0

ML

T

L

goo

)cos()sin(

..

(XX)

)(t

(XX1)

(XX0)

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)()( sFtf

I.C Laplace Transform

0dttfetfF(s)

st )()]([L

L

Powerful method in analyzing/solving LTI differentialequations

I.C.1 Definition

F(s) complex variable function of complex variable s is such that

0000)( tfortf)(tf

Function of t

Function ofcomplex variable s

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Example:

s

s

e

s

e

dtetutfF

t

st

t

st

st

1

0

0

)()]([L(s)

Q: Laplace transform F(s) exists only for real s>0?

A: Analytic Extension theorem extends existence ofF(s) towhole s-plane except at some singular point(s), if any.

0atfunctionStep ttuf(t) )(

becomes)(sF

Complex planelocations where

u(t)

1

t

(This term is 0 with real s > 0 )

Same for all Laplace Tansforms:F(s) exists in part of s-plane existence in

whole s-plane except of a few singular points!

In this case, singular point at s=0

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I.C.2 Properties of Laplace Transform

(a) Superposition of functions )(and)( 21 tftf

)()()]()([ 2121 sFsFtftf L

(b) Time Delay (Function shift to right by duration )0

)()]([ sFetf s L

L )( sF)(tf

Lt

)( tft

)( sFe s

Given

Then

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(c) Time Scaling (Expansion/contraction of time axis)

01 aFatfas

a ,)()]([L

Given

Then

L )( sF)(tf

tot

L

t

)(atf

ato

ato

)(1as

aF

a >1Contraction

of time axis

a < 1Extension

of time axis

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(d) Modulation by Exponential factor

)( tfe at

L)( sF

)(tf

Lt

t

Given

Then

e at

)()]([ asFtfe at L

)( asF

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(e) Differentiation of a function

)()()]([ 0fssFtfdtdL

(f) Integration of a function

s

sFt

df)(

0])([ L

)()]([ sFstf ndt

dn

n

L

- For function and derivatives all starting at zero

)0()0()0()()]([)1(

21

n

ffsfssFstf nnn

ndt

ndL

- Generalization Value ofat )( tfdt

d

0t

)(11

tfnn

dt

d

Value ofat 0t

)( tfValue ofat 0t

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dtff

dftftftf

t

t

)()(

)()()()(

20

1

20

121

Laplace Transform of selected function in Table 2.3

Laplace Transform of complicated functions maybeobtained from Laplace Transforms of simpler functionsusing Transform properties

(g) Convolution Theorem

where convolution of functionf1(t) andf2(t) defined as:

)()()]()([ 2121 sFsFtftf L

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Then z1

, z2

, ., zm

are the m zeros of the F(s), and

p1, p2, ., pn are the npoles of the F(s)

))...()()(())...(()(

n

m

pspspspszszsCsF

3211

I.C.3 Laplace Transform Theorems

Poles and zeros of F(s): factorize F(s) to obtain

If some of the pi are the same Repeated pole case

Repeated zero case not of concern

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Final Value Theorem: If all poles of sF(s) inside lefthalf plane (LHP) of s-plane, then exists and

Example:

Initial Value Theorem: Given any F(s),

Example:

)(lim)(lim ssFftfst 0

)(lim)( ssFfs

0

)( f

imply

existenceof

)(f

(Later!)?)()(

)()(

f

ss

ssF

102

332

)( f

?)()()( ys

ssY 102

?)()(

)(

0102

ys

ssY

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I.C.4 Inverse Laplace Transform

Partial Fraction Expansion and term-by-term Inversion

)()( sFtf

)4)(2(

1)(

ss

G s

4

1

2

1

2

1

2

1

ss

ttt eeg

42)(

2

1

2

1

L-1 L-1

L-1

Partial FractionExpansion

Term-by-termLaplace Inversion

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(All poles different)

))...()((

))...(()(

21

1

n

m

pspsps

zszsksF

- Write

-

)(...

)()()(

2

2

1

1

n

n

ps

C

ps

C

ps

CsF

- Determine C1,,Cn as

ipsii sFpsC )()( i = 1,,n

Partial Fraction Expansion: Distinct Pole Case

nppp 21

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Partial Fraction Expansion: Repeated Pole Case

))...(()(

))...(()(

11

1

nr

r

m

pspsps

zszsksF

-

nr ppppp ...... 1111

- Write

)(...

)()(...

)()()(

n

n

r

r

r

r

ps

C

ps

C

ps

C

ps

C

ps

CsF

1

1

121

2

1

1

(first rpoles equal)

Cr+1,,Cn of non-repeatedpoles determined by formula

of distinct pole case

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1)]()[(

!

11 ps

r

i

i

ir sFpsds

d

iC

- Determine C1,,Cras

, i=0,,r-1

1

1

211

212

ps

ps

sFpsds

dC

sFpsC

)]()[(

)]()[(

* Special case: r =2

))...(()(

))...(()(

12

1

1

nr

m

pspsps

zszsksF

)(...

)()()()(

1

12

1

2

1

1

n

n

r

r

ps

C

ps

C

ps

C

ps

CsF

1)]()[(

!

1 212 psi

i

i sFpsds

d

iC

i=0,1

r-1

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* Special case:

1

1

1

)]()[(!2

1

)]()[(

)]()[(

312

2

1

312

313

ps

ps

ps

sFpsds

dC

sFpsds

dC

sFpsC

r =3

))...(()(

))...(()(

13

1

1

nr

m

pspsps

zszsksF

)(...

)()()()()(

1

13

1

32

1

2

1

1

n

n

r

r

ps

C

ps

C

ps

C

ps

C

ps

CsF

1)]()[(

!

1 313 psi

i

i sFpsds

d

iC

i=0,1,2r-1

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- Distinct Real Pole:

- Multiple Real Pole:

- Distinct Complex Pole:

ptCe

ps

C

1

L

ptr

r et

rC

ps

C 1

)!1(

1

)(

1

L

)cos(||21

*

*

CteCps

C

ps

C t

L

where jpeCC Cj ,||

Term-by-Term Laplace Transform Inversion

- Multiple Complex Pole: Complicated!

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Example 2.2 Application of Laplace Transform to solvingLTI differential equation

)()()()( trtytyty 34

)0()0( ,yy Initial conditions :

Apply Laplace transform :

))(())(()()(

)())(())()()(( )(

312

314

23400 02

sssssssY

ssYyssYysysYs

Excitation on RHS:r(t)= 2 (meaning 2u(t) )

Step function

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Taking Inverse Laplace Transform

Response due to InitialCondition

(zero if =0,=0)

)())(()( tetetetety 331

323

223

Response due toexcitation r(t)

(zero if r(t)=0)

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I.C.5 Pole Locations and Time Response

Partial Fraction Expansion yields

......)(*

*

1

1

ps

C

ps

C

ps

CsY

is real for real pole; and complex for complex poleiC

-- From pole locationpi of Y(s): more information on time behavior ofy(t)

Information abouty(t) from Y(s)?

-- From FVT and IVT:

information on final and initial values ofy(t)

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Inverse Laplace Transform Each pole contributing

a term toy(t) Contribution from a single real pole:

)(

)()(1

satpole

tyinesYin

s

t

Single real pole location and its contribution toy(t):

te

te

t

e

te te

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-- Graphic illustration: Location (of complex polepair) and time contribution

dj

dj

y(t)

t

y(t)

t

te

y(t)

te

Faster

decayenvelop

Fastergrowthenvelop

IncreasingFrequency

Growth Envelop

d

Increasing

Frequency

FrequencyDecay Envelop

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-Representation forp in LHP(b) ),( n

-- Conversion to/from CartesianRepresentation

-- Correspondingly,y(t)becomes sinusoidal signal with

tne

22

22

2

or1

d

dn

nd

n

,

,

cos

Frequency:

Decay Envelop:

21 n

21 nn jp

or

Distance between pole and origin

Cosine of pole angular positionrelative to ve real axis

:

:

n

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Fixed n , varying 10:

-- Pole location and time response in (n, ) representation

0

1

7070 .

090

0

Double real pole at

ns

nj

nj

y(t)

t

y(t)

t

Complex pole pair

Complex pole pair on j-axis

y(t)

t

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Example: Spring-Mass-Damper system

000 )(,)( yyy o

)()()()( tFtkytydt

dbty

dt

dM

2

2

k

M y

F

Wall friction, b

- With F(t)=0,

M

k

M

bnn 22 ,

oy

MksMbs

Mbs

sY

2)(

- Two independent design parameters b/Mand k/M toyieldand n

o

nn

n yss

ssY

22 2

2

)(

No externalexcitation!

kM

b

2

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where , and hence

*)(

*

)(*))((

)(

ps

C

ps

Cy

psps

ssY o

n

2

- Underdamped case: , i.e.,10

901212

122

2

CyCyj

jC oo ;

21 nn jp

cos

)sin()(

tey

ty nto n 2

21

1

kMb 20

representation!),( n

Y(s) has two complex poles:

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22

2

)()()()(

nno

n

n

s

B

s

Ay

s

ssY

- Critically damped case: or1 kMb 2

on

o

yB

yA

tno

netyty )()( 1

Response reaches its final value at the shortest timewithout oscillation

Y(s) has two overlapping real poles:

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)()())((

)(

2121

2

ps

B

ps

A

psps

yssY on

- Overdamped case: or1

oo yByA12

112

12

2

2

2

,

1221 nnpp ,

otptp yeety

21

121

121 2

2

2

2

)(

kMb 2

Y(s) has two distinct (non-overlapping) real poles:

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- Example of responses:

http://www.efunda.com/formulae/vibrations/sdof_free_damped.cfm

- For the case F(t)=0, , see000 vyyy o )(,)(

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Design Concept #1:

To properly place the poles of Y(s) so that the resultingy(t) yields satisfactory performance

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Example 2.15: Design of an accelerometer

- Idea: constant Force Fo on massMs

Wanty as measurement of acceleration:s

o MFay

x = case position

Fo

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ydt

dbkyxy

dt

dM )(

2

2

os Fxdt

dM

2

2

s

o

MFy

Mky

Mby

- Absolute movement of mass inside case:x+y

- Force acting on mass:

- Combining

- Apply Laplace Transform:

M

ks

M

bss

MF

M

ks

M

bs

yMbysy

sY so

22

000 )()()()(

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- Conduct PFE,

)()()()(

))(())((

)()()(

)(

2121

2121

000

ss

D

ss

C

s

E

ss

B

ss

A

sssss

MF

ssss

yMbysy

sY so

- Inverse Laplace transform,

Particularly,,

s

o

M

F

k

ME

tsts

s

otsts DeCeM

F

k

MBeAety 2121

)(

Due to initial value

M

ks

M

bs

221 and sswhere are roots of

Due to excitation

- ConstantsA,B, C,D andEcan be determined as in PFE

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- Acceleration measurement:

* As t large,

and

* Hence, plot verse tyield a as steady state value!

21 and ss

021 tsts ee ,

s

oss

MF

kMyty

)(

021 tsts ee ,

Fo/Ms=a is acceleration!

ssy acceleration!

)(tyM

k

* With b, k, M +ve, inside LHP21 and ss

Can also obtain from FVT

- The design issues

* How fasty(t) becomesyss depends on how fast

* Design b/Mand k/Msuch that are farther in theLHP so that and hence y(t)yss fast enough

* However, faster decay expensive components trade-off

021 tsts ee ,

- Specific numerical design in Example 2 15

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- Specific numerical design in Example 2.15

* What are ? How long to reach a reasonable acceleration measurement? Can we redesign to get as a faster measurement?

21 and ss

- Question: what if the applied force F, hence a, is time varying?

2.46Fig.3,2,3,2)0(,1)0( aM

F

M

k

M

b

yy s

o

)(tyM

k

2.46

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I.D. Transfer Function

I.D.1 Definition

Given

r(t) Plant y(t)

Transfer Function = L.T. of outputL.T. of input )(

)(sR

sY

(with zero initial conditions)

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u(t)

Laplace Transform converts differential equation description

to transfer function description (zero initial conditions)

Plant y(t)

cbsasd

2

)()()()( tudtcytybtya

U(s) Y(s)

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I.D.2 Block Diagram and Transformation Rules

Pictorical representation of system in termsof its components

- Tedious to obtain Transfer Function)()(0

sVsV

i

- Divide system into three components of BridgeCircuit

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)()()()()( sGsGsG

sVsV

i

o321

Complicated system?

- Then

Vi

Use Block Diagram transformation rules

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T-1

T-2

T-3

T-4

T-5

T-6

Same relationsbetweenX1,X2

andX3

Example 2.7

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p

T-4

T-1

T-6

T-6

T-6

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T 6

)(sGY(s)

U(s) )(sGY(s)

U(s)

Useful fact about pickoff point

Summarizing example of all things taught (so far)

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A water tank with input u and water levely

The differential equation:

(for simplicity, letA=1 m2

)

Laplace transform leads to transfer function representation

If u is a step function (constant water flowing in)

Summarizing example of all things taught (so far)

udt

dyu

dt

dyA

AreaA

y (m)

u (m3/sec)

s

1)(sU )(sY

ttys

sYs

sU )()()(2

11

t (sec)

y (m)

2

2

)0)0((assuming y

D i d f l l 2 ?

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Desired performance: water levely to stay aty=2 m?

Open loop control: knowing that water levely=2 at t=2,just stop water flow at t=2

Closed loop control: measurey, compare with the desiredperformance to determine the input u(t)=r(t)-y(t)

t (sec)

y (m)

2

2

t (sec)

u (m3/sec)

2

1

s

1)(sU)(sY)(sR

+_

s

sR2

)(Desired performance:water level =2 Feedback actualy to compare

with desired performance

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Closed loop transfer function using Transformation Rule #6:

With desired performance r=2,

Moreover, input u can be determine as

tetu

ssYsRsU

2)(

)1(

2)()()(

)1(2)(

)1(

2)(

2)(

tety

sssY

ssR

)()(

)()()( sRs

sYsRGH

GsY

1

1

1

t (sec)

u (m3/sec)

4

2

t (sec)

y (m)

4

2

Indeed,y(t)

stayed at y=2

I E Modeling of DC Motor

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I.E. Modeling of DC Motor

Mathematical model: Newtons Law; Kirchoffs Laws; and

laws describing the mechanical/electrical interactions:

- Magnetic field induced by current if:

if

Electromechanical System interacting mechanical andelectrical components

if

if

i i i i fi ld

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- Induced voltage on wire moving with velocity v in :

e vPolarity of e induced current to yield

force/torque against velocityv

Induced force F

ia

Wire movingwith velocity v

+

_

- Force on wire carrying currentia

in magnetic field :

F ia

Induced voltage e

E l 2 5 DC M t~

Vf

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Field circuit:- current ifto generate magnetic field

Armature circuit:- current ia flowing inside magnetic field

- TorqueTm on rotor:

- Induced back voltage e on armature circuit:

Example 2.5: DC Motor

fiKKve 32

ffiK

affam iiKKiKT 11

Rf

Lf

if

Magnetic field

Field

circuitArmature

circuit

+

-e

K1,Kf,K2,K3 someproportional

constants

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- Rotor equation (with friction term):

- With ia

constant: Torque

)()()( tTtbtJ L

- Possible existence of disturbance torque total torque on rotor:

dmL TTT

- Field circuit equation:

)()()( tVtiRtidt

dL fffff

afm iKKK 1

fmaffm

iKiiKKT 1

Case #1: Field current if

-controlled motor

- Armature current ia kept constant (can use currentgenerator instead of armature circuit)

Mechanical Component:i ( i h f i i )

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Rotor equation (with friction term)

Or, inserting the variable (t) asangular speed

Electrical component:

Field circuit equation

)()()( tTtbtJ L

Possible presence of

disturbance torquedmL TTT

)()()( tVtiRtidt

dL fffff

Torque applied on rotordue to current if :

Coupling electrical and

mechanical components

fmm iKT

)()( )()()( tt

tTtbtJL

- Block diagram transform

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- With

)()(

1)())((

)( sTbJss

sVbJsRsLs

Ks dfff

m

,,and,0 ff

fLd R

L

b

JT

)())((

)( sVsss

bRK

s fLf

f

m

11

fL - Typically,

)()(

)( sVss

bRKs f

L

f

m

1

Electric response

much faster thanmechanical response

3rdorder differential equationrelating input and outputsimplified to 2ndorder

- Block diagrams

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** After neglecting the electrical response by omitting f

)(sPosition

1

fR

Electrical component Mechanical component

Coupling Presence of disturbance torque

- Block diagrams

** Full dynamics

Figure 2.19

Electrical component Mechanical component

Coupling Presence of disturbance torque

)(s

Position

Case #2: Armature current (ia ) controlled motor

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- Rotor equation (with friction term):

- With if constant:

- Armature circuit equation:

Case #2: Armature current (ia ) controlled motor

)()()( tTtbtJ L

- Possible existence of disturbance torque total torque on rotor:

dmL TTT

)()()()( tKtVtiRti

dt

dL baaaaa

Back emf term on armaturecircuit introduces another

source of coupling between

electrical and mechanical

components

ffm iKKK 1

amaffm iKiiKKT 1

- Field current if kept constant constant magnetic field (or use permanent magnet)

fiKe 3

fb iKK 3

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- Block diagram: full dynamics

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- After transformation rules

)(

))((

)()(

))(()( sT

KKbJsRsLs

RsLsV

KKbJsRsLs

Ks

dbmaa

aa

abmaa

m

)(

))((

)( sV

bR

KKsss

bRKs a

a

bmLa

a

m

11

- Typically, ,,,a

aaLd R

Lb

JT withand0

Figure 2.20

)(sVa)(s

Position

L - Typically,Again, 3rdorder differentialequation simplified to 2ndorder

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s1)(s

Speed

)(s

Position

aL Typically,

)()(

)()(

)(

)( sVss

KKbRK

sV

bR

KKss

bRK

s abma

m

a

a

bmL

a

m

11 1

)( bma

a

KKbR

JR

1where

- Block diagram: neglecting electrical response (omitting )a

equation simplified to 2 order

Electrical

Coupling: current-

induced torque Mechanical

Coupling: angular speed-induced back emf

Presence of disturbance torque

Design Example: Disk Drive Read System y(t)= t)

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- Output = from DC Motor (Flexture flexibility ignored)

- Armature-current controlled motor with zero back emf:

- Controller = gain amplification transfer function=Ka- Perfect measurement sensor transfer functionH(s)=1

Kb=0

Figure 2.76 (a)

- Disturbance torque=0

Head

Flexture

Rotor

Magnets

Arm

2.75

r(t)

Armature-current controlledDC Motor/Read Arm:G(s)

Figure 2.76 (b)

with added details

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Va(s)

- Parameters:

)20)(1000(

5000

))(()(

sssbJsRsLs

KsG

aa

m

Kb=0

Ka

H(s)=1

0R(s)

- Armature-current controlled DC Motor Transfer Functionwith Kb=0:

a

Y(s)

= s)

-G(s) in terms of :L

aL and

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Figure 2.742.78

)105.0)(1001.0(

25.0

)1)(1(

)(

ssssss

bRK

sGLa

a

m

msbJmsRL

La

aa 50,1

aL - Reduced model with

)().(

.)(

20

5

1050

250

ssss

sG

- Closed loop transfer function usingreduced model:

a

a

Kss

K

sR

sY

520

52 )(

)(

40aK

- Closed loop with full model yield nearly the same response

(Full model)

% textbook design example (Reduced

model)

Ka=40;

- Matlab program to generate plot:

The case in Fig 2 78

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Ka=40;

num=5*Ka;den=[1 20 5*Ka];

t=[0:0.002:0.7];

y=step(num,den,t);

plot(t,y)

grid

xlabel('Time (s)')

ylabel('y(t) (rad)')

Title('Figure 2.74')

hold

% Do case for ka=20

Ka=20;

num=5*Ka;

den=[1 20 5*Ka];

t=[0:0.002:0.7];

y=step(num,den,t);

plot(t,y,'--')

% Do case for ka=60

Ka=60;

num=5*Ka;

den=[1 20 5*Ka];t=[0:0.002:0.7];

y=step(num,den,t);

plot(t,y,'-.')

% Label the cases

gtext('Ka=20')

gtext('Ka=40')

gtext('Ka=60')

Exercise: generating plot for full model?

The case in Fig. 2.78

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Third Reading Assignment:

PP. 256-276, PP. 295-301 of Textbook