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1 Introduction

what is a control system?

what are the basic components of a control system?

some examples of control system applications

open-loop and closed-loop systems

effects of feedback

1 Introduction 2

Control systems and models

Acontrol systemis a [combinationorsetorgroup] of [componentsorelementsordevices] [connectedoracting] together under a set of[regulationsorrules] to perform a [certain objectiveoruseful task].

Fig. 1-1: Basic components of acontrol system.

System model

inputs

outputs

boundarysystem

{law rule}

environment = components

inputs outputssystem

c2005 Dept. of Electrical Engineering, Chulalongkorn Univ., based on B. C. Kuo & F. Golnaraghi,Automatic Control Systems, 7th ed.

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1 Introduction 3

A Control System

contains a component called controller whose role is to control thesystem output so that an objective is achieved.

Example: A simple control system

r controllingcomponent

controller

A control system

componentcontrolled

plant

u y

This type of control systems is called open-loop.


1 Introduction 4

An Example: Open-Loop Temperature Control

Consider as an example

r

component

air-conditioner

A room with air-conditioning system

componentcontrolled

a.c. room

u ycontrolling

where r fan speed control (LO, MED, HI)u amount of cool airy room temperature

Open-loop controller neglects information about room-size, numberof persons in room, inside- and outside-temp., time of day, etc.


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1 Introduction 5

Closed-Loop Temperature Control

Consider theclosed-loopconfiguration

room

u

a.c.thermostat

componentcomparing e

ry

d

wheredrepresents the information being neglected. The controller(block a.c.) called closed-loop controller can bedesignedto makeyless insensitive tod.

Benefits

more comfortable

save energy, etc.


1 Introduction 6

Feedback Control

Also known as Automatic ControlCan reduce the effect of disturbancedonybecause

d affectsyin the feedforward manner,

andyisfed backto the feedforward path for making decision.

In this way, the closed-loop control makes the cause-and-effectrelationship a complete loop.


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1 Introduction 7

Control system example: Sun-tracking solar collectors

Fig. 1-5: Solar collector field.

the collector dish must trackthe sun accurately

a predetermined desired rateis modified or trim by actualposition errors determined bythe sun sensor

the controller constantlycalculates the suns rate forazimuth and elevation

Fig. 1-7: Important components of

the sun-tracking control systems.


1 Introduction 8

Open-loop and closed-loop control systems

Open-loop or nonfeedback control

good for plant without disturbances

good for plant with accurate model

require less equipment (e.g. sensor, comparing device)

no stability problem

Closed-loop or feedback control

reduce the effect of unknown disturbances

tolerate model inaccuracy

require more equipment

need good design to avoidinstability


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1 Introduction 9

Effect of feedback

complete the cause-and-effect relationships

increase the gain of a system in one frequency range but decrease itin another

can improve stability or be harmful to stability

can increase or decrease the sensitivity of a system

can reduce the effect of noise

can affect bandwidth, impedance, transient response, steady-stateresponse, and frequency response


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2 Mathematical Foundation

Laplace transform

inverse Laplace transform by partial-fraction expansion

application of Laplace transform to the solution of linear ODE

impulse response and transfer functions of linear systems

MATL AB tools and case studies

2 Mathematical Foundation 2

Laplace transform

Definition Given a real functionf(t)satisfying

0 |f(t)et| dt

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Impulse function

An impulse (or Dirac delta) function is defined as the limit of arectangular pulse function:

(t) =

1, 0 t

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Important theorems of Laplace transform

Linearity

L[f1(t) +f2(t)] =F1(s) +F2(s)

DifferentiationL[f(t)] =sF(s) f(0)

L[f(t)] = sL[f

(t)] f(0)

= s2F(s) sf(0) f(0)Integration

L

t0

f() d

=

1

sL[f(t)] =

F(s)

s

L t

0

1

0

f(2) d2 d1=F(s)

s

2

Shift in time

L[f(t T)us(t T)] =eT sF(s)

Initial value theoremlimt0

f(t) = lims

sF(s)

if the limit exists.

Final value theoremlim

tf(t) = lim

s0sF(s)

ifsF(s)is analytic inRe s 0.

Complex shifting

L[etf(t)] =F(s )

Real convolution

F1(s)F2(s) =L[f1(t) f2(t)]=L

t0

f1()f2(t) d



0+ and 0

If we define

F+(s) =L+[f(t)] =

0+

f(t)est dt and F(s) =L[f(t)] =

0

f(t)est dt

we must replace0in the theorems by0+or0 accordingly, e.g.,

L+[f(t)] =sF+(s) f(0+), L+

t0+

f() d

=

F+(s)

s , f(0+) = lim

ssF+(s)

Then 0+ or 0? Both work fine. For example,consider du(t)

dt =(t)

and the differentiation formula,

1

0 1 2

u(t)

u(0) = 0u(0+) = 1

L+[u(t)] = sL+[u(t)] u(0+)= s

1

s 1 =L+[(t)] = 0

L[u(t)] = sL[u(t)] u(0)= s

1

s 0 =L[(t)] = 1


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Inverse Laplace transform

The inverse Laplace transform ofF(s)can be found from

f(t) =L1[F(s)] = 1

2j c+j

cjF(s)est ds

wherecis a real constant chosen so thatF(s)is analytic in regionRe s c.

The method of partial-fraction expansion can also be used to findL1[]ofrational functions.

Rational function Ifan= 0we call the function

P(s) =ansn +an1sn1 + +a1s+a0

apolynomialofdegreenwitha1,a2, . . . ,anas itscoefficients.

Rational functionsare fractions of polynomials (c.f. rational numbersare fractions of integers).



Partial-fraction expansion: Real roots

Ifis a real root ofP(s)withmultiplicitym, i.e.,P(s) = (s )mR(s),then we write

Q(s)

P(s)=

A1s

+ A2

(s )2+ +

Am(s )m

+ {terms ofR(s)}

Multiplying both sides by(s )mQ(s)

R(s)=Am+ +A2(s )

m2 +A1(s )m1 + {terms ofR(s)}(s )m

Letting(s) = (s )mQ(s)

P(s)=

Q(s)

R(s), we can findAkfrom

Ak = 1

(m k)!(mk)(), k= 1, 2, . . . , m

and its correspondingL1[]term is

L1

Ak(s )k

= et

tk1

(k 1)!


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Partial-fraction expansion: Complex conjugate roots

If+jis a complex root (multiplicitym) ofP(s)with , i.e.P(s) = (s j)mR(s), then we write

Q(s)P(s)

= A1+jB1s j

+ + Am+jBm(s j)m

+ A1 jB1s +j

+ + Am jBm(s +j)m

+ {terms ofR(s)}

Letting(s) = (s j)mQ(s)

P(s), we have

Ak+jBk = 1

(m k)!(mk)(+j), k= 1, 2, . . . , m

and its L

1

[](with its conjugate pair) isL1

Ak+jBk

(s j)k+

Ak jBk(s +j)k

= et

tk1

(k 1)!

(Ak+jBk)e

jt + (Ak jBk)ejt

= et tk1

(k 1)!(2Akcos t 2Bksin t)



Partial-fraction expansion: Examples

Example 1 5s+ 3

(s+ 1)(s+ 2)(s+ 3)= 1

s+ 1+

7

s+ 2+ 6

s+ 3

Example 2 2

s(s+ 1)3(s+ 2)=

1

s+

1

s+ 2

2

s+ 1

2

(s+ 1)3

Example 3 2(s + 1)

s2(s2 + 2s+ 2)=

1

s2+

j/2

s+ 1 +j+

j/2

s+ 1 j

Example 4 G(s) = 2n

s2 + 2ns+2n=

n1 2

j/2

s++j

j/2

s+ j

where = nand =n1 2 or

g(t) =L1[G(s)] = n1 2

ent sin nt

1 2


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Application of Laplace transform to solution of linear ODE

Example Solvey(t) + 3y(t) + 2y(t) = 5us(t)wherey(0) =1,y(0) = 2.

Take the Laplace transform of both sides of the ODE to get

s2Y(s) sy(0) y(0) + 3sY(s) 3y(0) + 2Y(s) =5

s

Substituting initial conditions and solving forY(s), we get

Y(s) = s2 s+ 5

s(s+ 1)(s+ 2)

which can be expanded by partial-fraction expansions to give

Y(s) =

5/2

s

5

s+ 1+

3/2

s+ 2

Taking the inverse Laplace transform, we get the complete solution

y(t) =5

2 5et +

3

2e2t, t 0



Impulse response

Consider alinear time-invariantwith inputu(t)and outputy(t). Thesystem can be characterized by itsimpulse responseg(t)which is definedas the output when the input is a unit-impulse function(t).

1

0

(t)

(t) =

1, 0 t

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Transfer function

Thetransfer functionof a linear time-invariant system isdefinedas theLaplace transform of the impulse response, withall the initial conditions

set to zero. G(s) =L[g(t)]

Letnandmbe the order of the denominator and the numeratorpolynomials ofG(s), respectively. The transfer functionG(s)is said to be[strictly]properwhen [n > m]n m. Ifn < m, we call itimproper.

We can show that the transfer function is also theratiobetween theLaplace transform of the output and the Laplace transform of the input

G(s) =Y(s)

U(s)

by the use of convolution, linearity, and time-invariance properties.



Approximation of input signal

We can depict the approximation of input signal by

aweighted sumof sequence of pulse functions(t n)

as follows:

u(t) = lim0

n=0

u(n)(tn)

1

2

2 3 410 n

u(t)

u(n)(tn)

. . .. . .

u(n)


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Interpretation of convolution

Defineg(t)to be the systems response to(t).

input output(t) g(t) by definition

(tn) g(tn) time invarianceu(n)(tn) u(n)g(tn) scaling

n=0

u(n)(tn)

n=0

u(n)g(tn) superposition 0

u()(t ) d

0

u()g(t ) d

dn ,

By sifting property of the impulse function

LHS=

0

u()(t ) d=u(t)

andg(t)on the RHS is the response to(t)or impulse response.



MATLABtools and case studies

>> [r,p]=residue([5 3],conv([1 1],conv([1 2],[1 3])))

r = -6.0000 7.0000 -1.0000

p = -3.0000 -2.0000 -1.0000

>> [r,p]=residue([2],conv([1 2],[1 3 3 1 0]))

r = 1.0000 -2.0000 -0.0000 -2.0000 1.0000

p = -2.0000 -1.0000 -1.0000 -1.0000 0

>> syms s>> G=2/s/(s+1)^3/(s+2)

G =2/(s*(s + 1)^3*(s + 2))

>> ilaplace(G)

ans =1/exp(2*t) - 2/exp(t) - t^2/exp(t) + 1


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3 Block Diagrams and Signal-Flow Graphs

block diagramsblock diagrams of control systems

block diagrams of multivariable systems

signal-flow graphs (SFGs)basic properties of SFG

SFG algebra

gain formula for SFG


3 Block Diagrams and Signal-Flow Graphs 2

Block diagrams

is a simple pictorial representation of a system basic language of control engineers describescompositionandinterconnectionof a system

describes the cause-and-effect relationships throughout the system

Fig. 3-1: (a) Block

diagram of a dc-motor

control system.

(b) Block diagram with

transfer functions and

amplifier characteristics.


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Block diagrams of control systems

Basic block-diagram elements are

blocksrepresent system components, and lines with arrowrepresent signals and showhow signals flow from one block to another.

Y(s) =G(s)U(s)

G(s) Y(s)U(s)

Some special blocks are usedto represent sensing devices

that perform simplemathematical operations suchasadditionandsubtraction.



Drawing a block diagram

Consider a voltage divider as shown in the circuit below.

+ +

R1

Vi Vo

R2I

Vo= R2I R2VoI

I=Vi Vo

R1

+ 1R1

Vi I

Vo

Combining both equations, we have

+Vi I

R2Vo1

R1 or Vo

Vi=

R2R1+R2


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Basic block diagram of a feedback control system

r(t), R(s) =reference input (command)

y(t), Y(s) =output (controlled variable)

b(t), B(s) =feedback signalu(t), U(s) =actuating signal

=error signale(t), E(s), ifH(s) = 1

H(s) =feedback transfer function

From the figure above, we can write

Y(s) =G(s)U(s), B(s) =H(s)Y(s), and U(s) =R(s) B(s)Solving these forY(s)in terms ofR(s), we have

Y(s)

R(s)=

G(s)

1 +G(s)H(s)



Block diagrams of multivariable systems

Y(s) = M(s)R(s)

M(s) = [I + G(s)H(s)]1G(s)= G(s)[I + H(s)G(s)]1


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Block diagram algebra: Combining two blocks

original diagram equivalent diagram

uG1

yG2

u yG2G1

+u y

G2

G1u y

G1 G2

+u y

G1

G2

u yG1

1 G1G2



Block diagram algebra: Moving one block

+u1

Gy

u2

+

u1 y

G

u2G

+u1 y

G

u2

+u1 y

G

u21

G

u yG

y

u yG

Gy

u yG

u

u yG

1

G

u


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Block diagram reduction

Example Consider a multi-loop feedback control system as shownbelow

+

+ + +

G3 G4G2G1

H2

H1

H3

YR

Answer:Y

R=

G4G3G2G11 G4G3H1+G3G2H2+G4G3G2G1H3



Reduction difficulty?

Wrong way: move the summing point behind the branching point

+

+ +

+Y

G2

G4

G1

G3

R

Right way:

+

+ +

+Y

G2

G4

G3

R

G1

G1

Answer: Y

R=G2G1+G4G1+G2G3


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Signal-flow graphs (SFGs)

may be regarded as a simplified version of a block diagram

developed by S. J. Mason in 1950s

can be used to determine relationship between two signals incomplex systems

a small circle callednodedepicts a signal (line in block diagram) and a line calledbranchwhose arrow and number show direction

and gain (multiplier) for signal traveling from one node to another

G(s)

Y(s)U(s)

G(s)

Y(s)U(s)



Signal-flow graph: Examples

Example 1 The voltage divider can be viewed as aunity negativefeedback whose SFG graph is shown below.

VoI

R21

R1

11

Vi

Example 2

R 1 G1 G2 G3 G4

H2

H3

H1

Y


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Definition of SFG terms

Aninput node (source)is a node that has only outgoing branches. Anoutput node (sink)is a node that has only incoming branches. Apathis any collection of a continuous succession of branches

traversed in the same direction.

Aforward pathis a path that starts at an input node and ends at anoutput node, and along which no node is traversed more than once.

Aloopis a path that originates and terminates on the same node andalong which no node is traversed more than once.

Apath gainor aloop gainis the product of the branch gainsencountered in traversing that path or loop.

Two parts of an SFG arenontouchingif they do not share a commonnode.



Gain formula for SFG

The gain between the input nodeyinand output nodeyoutis

M=yout

yin=

1

Nk=1

Mkk

where = 1 (all individualloops)

+

(gain products of all combinations oftwonontouching loops)

(gain products of all combinations ofthreenontouching loops)+

(. . . fournontouching loops) (. . . fivenontouching loops) + N = total number of forward paths betweenyinandyout

Mk = the gain products of thekth forward path

k = thefor that part of the SFG that is nontouching with thekth forward path

Theis also known as thedeterminantof SFG.


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Gain formula: An example

r 1 a b c d

e

gh

f

y

k

l

has 7 loops and 4 paths:

Loop: L1= l, L2= bh, L3= dg, L4= abck, L5= eck, L6= abfgk, L7= efgkPath: P1=abcd, P2= abf, P3= ecd, P4=efNontouching loops: L1L3, L2L3, L1L5, L1L7

= 1 l bh dg abck eck abfgk efgk+ldg +bhdg+leck+lefgk

y

r =

1

(abcd+abf+ecd(1 l) +ef(1 l))



Gain formula: Another example

Y(s)

R(s)=

G1G2G3+G1G41 +G1G2H1+G2G3H2+G1G2G3+G4H2+G1G4

, Y(s)

E(s)=?


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MATL ABtools and case studies (1)

Transfer functions can be easily handled using Control System Toolbox.

>> s=tf(s);

>> G1=1/(s+1);>> G2=tf(1,[1 2])

Transfer function:1

-----s + 2

>> G1*G2

Transfer function:1

-------------

s ^ 2 + 3 s + 2

>> G3=feedback(G1,G2)

Transfer function:s + 2

-------------s ^ 2 + 3 s + 3

>> G4=G1/(1+G2*G1)

Transfer function:s ^ 2 + 3 s + 2

---------------------s ^ 3 + 4 s ^ 2 + 6 s + 3

The commandtfis used to specify the model in transfer function form.Then normally arithmetics +, -, *, and / can be used with transferfunctions.




Commandsseries,parallel,feedbackcan be used to make morecomplicated models.

>> zpk(G4)

Zero/pole/gain:(s+2) (s+1)

---------------------(s+1) (s^2 + 3s + 3)

>> minreal(G4)


-------------s ^ 2 + 3 s + 3

The commandzpkis used to specify the model in zero-pole-gain formwhile the commandminrealcan be used to simplify model by pole-zerocancellation.


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More complicated connection can be done using commandsappendandconnect.

>> G5=append(G1,G2);>> connect(G5,[1 -2;2 1],1,1)


-------------s ^ 2 + 3 s + 3

Theappend(G1,G2)command creates the structure y(1) = G1 u(1)andy(2) = G2 u(2)while theconnectcommand has the syntax

sys = connect(blksys,Q,input,output)

Theindex-based interconnection matrixQ=[1 -2;2 1]feeds-y(2)intou(1)andy(1)intou(2)andinput=1andoutput=2indicate thatrdrivesu(1)andyisy(1).




Multivariable systems can also be handled.

>> G=[1/(s+1),-1/s;2,1/(s+2)]; H=[1 0; 0 1];>> feedback(G,H)

Transfer function from input 1 to output...

3 s^2 + 9 s + 4#1: ----------------------s ^ 3 + 7 s ^ 2 + 1 2 s + 4

2 s ^ 3 + 6 s ^ 2 + 4 s#2: ----------------------

s ^ 3 + 7 s ^ 2 + 1 2 s + 4

Transfer function from input 2 to output...- s ^ 2 - 3 s - 2

#1: ----------------------s ^ 3 + 7 s ^ 2 + 1 2 s + 4

3 s^2 + 8 s + 4#2: ----------------------

s ^ 3 + 7 s ^ 2 + 1 2 s + 4


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4 Modeling of Physical Systems

modeling of electrical networks modeling of mechanical systems elements

translational motion & rotational motion

friction, backlash and dead zone (nonlinear characteristics)

equations of mechanical systems sensors and encoders in control systems DC motors in control systems linearization of nonlinear systems systems with transportation lags (time delays) MATL AB tools and case studies

4 Modeling of Physical Systems 2

Introduction

two most common models: transfer function and state-variable transfer functions are valid only for linear time-invariant systems

state equations can be applied to linear as well as nonlinear systems analysis and design for nonlinear systems are usually quite complex linear models are only approximations must determine how to accurately describe a system mathematically

and how to make proper assumptions so that the system may berealistically characterized by a linear mathematical model

the objective of this chapter is to demonstrate mathematicalmodeling of control systems and components

the coverage here is by no means exhaustive


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Ideal v.s. real

Example 1 A resistor V

I

+

ideal: Ohms law V =I R real: over voltage and current, parasitic capacitance and inductance

Example 2 A mechanical spring

ideal: F =kxmasslessfrictionless

real:

F

F

x

slope= k

free length

solidheight

What is real is that everything is subject to certainconditions.To describe it, we need someassumptions.



Modeling of electrical systems elements

component voltage-current current-voltageC

capacitore(t) =

1

C

t0

i() d i(t) =Cd

dte(t)

R

resistore(t) =Ri(t) i(t) =

1

Re(t)

L

inductore(t) =L

d

dti(t) i(t) =

1

L

t0

e() d

Capacitor current-charge: i(t) = d

dtq(t)

Inductor voltage-flux: e(t) = d

dt(t)


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Modeling of electrical networks

Use Kirchhoffs voltage and current laws: loop and node methods.

Usingi1(t),i2(t), andec(t)as states, we can write

L1di1dt

=R1i1 ec+e

L2di2dt

=R2i2+ec

Cdecdt

=i1 i2

=L1L2Cs3 + (R1L2+R2L1)Cs

2 + (L1+L2+R1R2C)s+R1R2

I1(s)

E(s)=

L2Cs2 +R2Cs+ 1

,

I2(s)

E(s)=

1

,

Ec(s)

E(s) =

L2s+R2



Operational amplifiers

Operational amplifiers or OpAmps areactivecomponents which can beused inrealizationof wider class of transfer functions.

+

+

C R1

R2

R3

R

+

EO(s)

+

EI(s)

EO(s)

EI(s) =

R3R1R2R

(RCs+ 1)

Note that this transfer function isimproper.


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Modeling of mechanical systems elements: Translational

component force-displacement force-velocityk

springf(t) =kx(t) f(t) =k

t0

v() d

b

damperf(t) =b

d

dtx(t) f(t) =bv(t)

m

massf(t) =m

d2

dt2x(t) f(t) =m

d

dtv(t)



Modeling of mechanical systems elements: Rotational

component torque-angulardisplacementtorque-

angular velocityK

springT(t) =K (t) T(t) =K

t0

() d

B

damper

T(t) =Bd

dt(t) T(t) =B(t)

J

inertia

T(t) =Jd2

dt2(t) T(t) =J

d

dt(t)


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Modeling of friction

Viscous frictionis a linear relationship between force and velocity.

Static frictiontends to prevent motion from the beginning. Once the

motion begins, static friction vanishes and other frictions take over.Coulomb frictionis a retarding force having a constant amplitude with

respect to the direction of velocity.

f(t) =Bdx(t)

dt f(t) =(Fs)|x=0 f(t) =Fc

dx(t)

dt

dx(t)dt



Conversion between translational and rotational motions

The mechanisms ofbelt-and-pulleyand rack-and-pinioncan be used tochange between translational and rotational motions. LetW =M g.

Belt and pulley Ifris the

radius of the pulley. Then theequivalent inertiathat the motorsees is

Jeq= M r2

Rack and pinion IfLis thescrew lead, i.e., the lineardistance thatMtravels perrevolution of the screw. Then

Jeq= M L

2

2


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Gear trains

IfTi,i, andNi(i= 1,2) are torque applied,angular displacement, and teeth number of

the geari, then we haver1r2

=N1N2

, 1r1= 2r2, T11= T22

by considering the teeth ratio, distance alongsurface, and work done, respectively.

In practice, considering inertia loads andfrictions at both sides, the equivalenttorque seeing from gear 1 is given as

T =J1ed21dt2

+B1ed1dt

+Fc11|1| +

N1N2

Fc12|2|

where J1e= J1+

N1N2

2J2, B1e= B1+

N1N2

2B2.



Backlash, dead zone, and saturation


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Equations of mechanical systems: Translational

x(t)

forcef(t)

bfriction

k

mmass x(t)

f(t)

bx kx

m

Applying Newtons law of motionto the free-body diagram, we have

f(t) =md2

dt2x(t) +b

d

dtx(t) +kx(t)

Assuming zero initial conditions

(ms2 +bs+k)X(s) =F(s) = X(s)F(s)

= 1

ms2 +bs+k



What about the gravity?

b

k

x(t)

f(t)

y(t)

x0

m

Whenf(t) = 0the spring stretches by theweight of the mass (mg) alone to the length

x0=mg

k

Let y(t)be the displacement measured fromthe free length of the spring, and

x(t)be the displacement measured fromx0= mg/k

Now we havey(t) =x0+x(t). With reference from the free length

f(t) +mg =md2

dt2y(t) +b

d

dty(t) +ky(t)

However, fromy(t) =x0+x(t)

y(t) = x(t) and y(t) = x(t) = f(t) =md2

dt2x(t) +b

d

dtx(t) +kx(t)


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Equations of mechanical systems: Rotational

K BJ

(t)T(t)

From Newtons law of motion,

T(t) =Jd2

dt2(t) +B

d

dt(t) +K(t)

Assuming zero initial conditions,

T(s) = (Js2 +Bs+K)(s)(s)

T(s) =

1

Js2 +Bs+K



Sensors and encoders in control systems (1)

Potentiometer mechanical displacement electrical voltage

e(t) =Ksc(t) or e(t) =Ks[1(t) 2(t)]c2005 Dept. of Electrical Engineering, Chulalongkorn Univ., based on B. C. Kuo & F. Golnaraghi,Automatic Control Systems, 7th ed.

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Sensors and encoders in control systems (2)

Tachometer mechanical velocity electrical voltage

et(t) =Ktd(t)

dt =Kt(t) = Et(s)

(s) =Kts

Incremental encoder mechanical movement electrical pulse



DC motors in control systems

The dc motor is a torquetransducerconverting electrical energy tomechanical energy.

The relationship among the developedtorqueTm, the flux, and the armaturecurrentiais

Tm= KmiawhereKmis a proportional constant.

When motor rotates, the voltage acrossits terminals calledback emf ebisdeveloped

eb=Kmm


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Mathematical modeling of PM DC motors (1)

Consider a dc motor system with the following variables andparameters:

= magnetic flux in the air gapJm = rotor inertia

Bm = viscous-friction coefficientm(t) = rotor displacementm(t) = rotor angular velocity

Ra = armature resistanceLa = armature inductanceKi = torque constantKb = back-emf constant

ia(t) = armature currenteb(t) = back emf

TL(t) = load torqueTm(t) = motor torqueea(t) = applied voltage



Mathematical modeling of PM DC motors (2)

ea(t) eb(t) =Ladia(t)dt

+Raia(t)

Tm(t) =Kiia(t) [ :=Kmia(t)], eb(t) =Kbdm(t)

dt =Kbm(t)

Tm(t) TL(t) =Jmd2m(t)dt2

+Bmdm(t)dt

m(s)

Ea(s)

TL=0

= Ki

LaJms3 + (RaJm+BmLa)s2 + (KbKi+RaBm)s


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Linearization of nonlinear systems

Consider a nonlinear system

x(t) = f(x(t), r(t)) where x(t) =

x1(t)

x2(t)...xn(t)

, r(t) = r1(t)

r2(t)...rp(t)

, f(, ) = f1(, )f2(, )...fn(, )

The linearization of the system about an operating condition(x0, r0),e.g., f(x0, r0) = 0, is given by

x(t) = Ax(t) + Br(t) where x(t) = x(t) x0, r(t) = r(t) r0and

A= [aij] where aij =

fi

xjx0,r0 and

B= [bij] where bij =

fi

rjx0,r0



Magnetic-ball-suspension system: Nonlinear model

Consider a magnetic-ball-suspension system

e(t) =input voltagei(t) =winding currenty(t) =ball position

L=winding inductanceR=winding resistance

M=mass of ball

Md2y(t)

dt2 =M g i2(t)

y(t) , e(t) =Ri(t) +Ldi(t)

dt


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Magnetic-ball-suspension system: Linearization

Define the state variables as

x1(t) =y(t)

y0, x2(t) = y(t), and x3(t) =i(t)

i0

around the nominal positiony0= constant, we havei0=

M gx01andthe linearized model is

d

dt

x1(t)x2(t)

i(t)

=

0 1 0g

y002

g

M y0

0 0 RL

x1(t)x2(t)

i(t)

+

001

L

e(t)



Systems with transportation lags (time delays)

There are many ways to approximate time delays using Maclaurin series

eTds = 1 Tds+12

T2d s2

or

eTds = 11 +Tds+T2d s2/2

A better approximation is to use the Pade approximation

eTds =1 Tds/21 +Tds/2


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5 Poles and Zeros of Transfer Functions

poles and zeros residue and system response first- and second-order systems pole-zero cancellation


5 Poles and Zeros of Transfer Functions 2

Linear time-invariant systems

Consider a linear constant coefficient ODE

y(n)(t) +1y(n1)(t) +2y(n2)(t) + +n1y(t) +ny(t) =f(t)

whose solution can be found substitutingy(t) =emt.

If the coefficientsiare not constant, but depend on time asi(t), wehave alinear time-varyingsystem.

If the coefficientsidepend onyasi(y, y,...,y(n1)), we have anonlinear time-invariantsystem.


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Time-domain response

The time-domain response is given by

y(t) = k=1

ck

emkt +Yp

(t) =c1em1t +c

2em2t +

+c

nemnt +Y

p(t)

wherem1,m2, . . . ,mnare (real or complex) roots of the characteristics(orauxiliary) equation

mn +1mn1 +2mn2 + +n1m+n= 0

andYp(t)is the particular solution depending onf(t).

If the initial conditionsy(0),y(0), . . . ,y(n1)(0)are given, we candeterminec1,c2, . . . ,cn.



Poles and zeros

In more general cases, the forcing functionf(t)is related to the inputu(t)by

f(t) =1u(n1)(t) +2u(n2)(t) + +n1u(t) +nu(t)

or

y(n) +1y(n1) + +n1y+ny=1u(n1) + +n1u+nu

and the transfer function is given by

G(s) = 1s

n1 +2sn2 + +n1s+nsn +1sn1 +2sn2 + +n1s+n =:

n(s)

d(s)

We call the roots ofn(s) = 0zeros andd(s) = 0poles ofG(s).

Example G(s) = 26.25(s+ 4)

s(s+ 3.5)(s + 5)(s+ 6)

has a zero at 4and four poles at0, 3.5, 5, and 6.c2005 Dept. of Electrical Engineering, Chulalongkorn Univ., based on B. C. Kuo & F. Golnaraghi,Automatic Control Systems, 7th ed.

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Residues and system response

The system response can be found fromG(s)by taking the inverseLaplace transform. To do that, we need to find the partial fraction

G(s) = Ai

s+pi+ Bk

s+k+jk+

Bks+k jk

In complex analysis,AiandBkare known as theresidueofG(s)ats= piand ats= k jk, respectively.In this case, the system response is given by

g(t) =

Aiepit +

ekt [2Re(Bk)cos kt+ 2Im(Bk)sin kt]



First- and second-order systems

Hence, many system responses can be considered as summation of thefirst-order system

G1(s) = A

s+p

and the second-order system

G2(s) = K2

s2 + 2s+2

We will consider the behavior of their system responses in more detail.


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Pole-zero cancellation

Consider, for example, two transfer functions with their partial fractions

C1(s) =

26.25(s+ 4)

s(s+ 3.5)(s + 5)(s+ 6) =

1

s 3.5

s+ 5+

3.5

s+ 6 1

s+ 3.5

and

C2(s) = 26.25(s+ 4)

s(s+ 4.01)(s+ 5)(s+ 6)=

0.87

s 5.3

s+ 5+

4.4

s+ 6+

0.033

s+ 4.01

We can see that the residue of the pole at 4.01, which is closer to thezero at 4, is smaller. Hence we can approximateC2(s)by neglecting theresponse corresponding to the pole at 4.01with the second-order

C2(s) 0.87

s 5.3

s+ 5+ 4.4

s+ 6

orc2(t) 0.87 5.3e5t + 4.4e6t.However, the unstable polecannotbe cancelled by the unstable zero.




Use commandsresidue,roots,pole, andzero.

>> [r,p,k]=residue(26.25*[1 4],poly([0,-4.01,-5,-6]))

r = p = k =

4.3970 -6.0000 []-5.3030 -5.0000

0.0332 -4.01000.8728 0

>> p=[1 7 9 23 10]>> roots(p)>> s=tf(s); G=(s^2-2*s+2)/(s^4+7*s^3+19*s^2+23*s+10)>> pole(G)>> zero(G)


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6 Stability of Linear Control Systems

bounded-input bounded-output (BIBO) stability

relationship between charactersitic equation roots and stability

zero-input and asymptotic stability

Routh-Hurwitz criterion

Rouths tabulation

special cases when Rouths tabulation terminates prematurely

MATL ABtools and case studies

6 Stability of Linear Control Systems 2

Introduction

absolute stability v.s. relative stability

Absolute stability refers to the condition whether the system isstable or unstable; it is ayesornoanswer.

Once the system is found to be stable, to determine how stable it

is, we need a measure of relative stability.

zero-state v.s. zero-input responses

total response= zero-state response+zero-input response

c2005 Dept. of Electrical Engineering, Chulalongkorn Univ., based on B. C. Kuo & F. Golnaraghi, Automatic Control Systems, 7th ed.

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Bounded-input bounded-output stability

y(t)u(t)

G(s)

g(t)

U(s) Y(s)

Definition With zero initial conditions, thesystem is said to be(BIBO) stableif its output

y(t)is bounded to any bounded inputu(t).

From the convolution integral relatingu(t),y(t), andg(t)

y(t) =

0

u(t )g() d

we have

|y(t)| =

0

u(t )g() d

0

|u(t )| |g()| d

Suppose the input is bounded, i.e., |u(t)| Mfor some positiveM, thenthe output is bounded if

0

|g(t)| dt Q <



Relationship between charactersitic eqn. roots & stability

The transfer functionG(s), by the Laplace transform definition, is

G(s) = L[g(t)] =

0

g(t)est dt

For anys= s0= 0+j0, we have

|G(s0)| =

0

g(t)es0t dt

0

|g(t)|es0t dt=

0

|g(t)|e0t dt

since |es0t| = |e0t|. If there is a poles0=0+j0ofG(s)with0 0,using the fact that |e0t| 1for allt 0,

= |G(s0)|

0

|g(t)|e0t dt

0

|g(t)| dt

which violates the BIBO stability requirement. The system isunstable.

Thusfor a stable systems, the roots of the characteristic equation, orthe poles ofG(s)cannot be located in the closed right half s-plane.


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Zero-input and asymptotic stability

Let the input of an nth-order system be zero, the output due to the initialcondition can be expressed as

y(t) =n1k=0

y(k)(t0)gk(t) =y(t0)g0(t) +y(t0)g1(t) +. . .+y

(n1)(t0)gn1(t)

If the zero-input responsey(t), subject to the finite initial conditionsy(k)(t0)reaches zero astapproaches infinity, the system is said to bezero-input stableorasymptotic stable.

Let thencharacteristic equation roots be expressed assi=i+ji,i= 1, 2, . . . , n.

y(t) =m

i=1

Kiesit +nm1

i=0

Litiesit

Fory(t)to approach0ast , the real parts ofsimust be negative.



Stability for linear time-invariant systems

Hence, for linear time-invariantsystems, BIBO stability, zero-input

stability, and asymptotic stability allhave the same requirementthat theroots of the characteristic equationmust all be located in the left-halfs-plane.

If the characteristic equation hassimple roots on thej-axis and nonein the right-half plane, we say that thesystem ismarginally stable.


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Routh-Hurwitz criterion

The Routh-Hurwitz criterion is a method of determining the location ofzeros of a polynomial whether they are in the left- or right-half planes.

Consider that the characteristic equation of a linear time-invariant SISOsystem is of the form

F(s) =ansn +an1s

n1 + +a1s +a0= 0

In order thatF(s)does not have roots with non-negative real parts, it isnecessary (but not sufficient)that

1. All the coefficientsa0,a1, . . . ,an1,anhave the same sign.

2. None of the coefficients vanishes.

an(s z1)(s z2) (s zn) =ansn an(

zi) s

n1 + +an(1)n zi



Rouths tabulation or Rouths array (1)

sn p11= an p12= an2 p13= an4 p14= an6 sn1 p21= an1 p22= an3 p23= an5 p24= an7 sn2 p31 p32 p33 p34 sn3 p41 p42 p43 p44 ...

...

...s0 pn+1,1

where

p31= 1

p21

p11 p12p21 p22 , p32= 1p21

p11 p13p21 p23 , p33= 1p21

p11 p14p21 p24 ,

p41= 1

p31

p21 p22p31 p32 , p42= 1p31

p21 p23p31 p33 , p43= 1p31

p21 p24p31 p34 ,

...

The roots are all in the LHP ifp11,p21,p31, . . . ,pn+1,1have the same sign.


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Rouths tabulation or Rouths array (2)

The number of changes of signs in the elements of the first columnsequals the number of roots with positive real parts or in the RHP.

Example 1 P(s) =s4 + 2s3 + 3s2 + 4s + 5 = 0

s4 1 3 5s3 2 4 0s2 23142 = 1

25102 = 5

s1 14251 = 6 0s0 5

There are two sign changes: 1 6 and 6 5. HenceP(s)has two

RHP roots.

Example 2 2s4 +s3 + 3s2 + 5s + 10 = 0



Proof of Rouths criterion (1)

Supposenis even. The polynomialP(s) =ans

n +an1sn1 + +a1s +a0can be divided into real and

imaginary parts asP(s) =P1(s) +P2(s)where

Even part: P1(s) = ansn +an2s

n2 + +a2s2 +a0

Odd part: P2(s) = an1sn1 +an3sn3 + +a3s3 +a1s

Definition If two polynomials have the same numbers of roots in LHP,RHP, and onj-axis, we say that they areequivalent.

Fact 1 Ifj0is a root ofP(s), then it is also a root ofP1(s)andP2(s).

Fact 2 For real constantnear zero,P(s)is equivalent to (as long as ithasndegree)Q(s, ) :=P(s) sP2(s) = (an an1)s

n +an1sn1 + .

Proof: Note thatQ(s, 0) =P(s). Now from Fact 1,P(s)and

Q(s, ) =P1(s) +P2(s) sP2(s) = [P1(s) sP2(s)] +P2(s)

have the same number ofj-axis roots for small.


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Proof of Rouths criterion (2)

Fact 3 If anan1

,Q(s, )drops degree as one of its roots approaches

an1an an1

=

an

an1

1

Now we can prove the Rouths criterion by noting that

sum of the first and second rows of Rouths array isP(s),

sum of the second and third rows isQ(s, an/an1).

Whenan1= 0, the construction of Rouths array fails. To work aroundthis problem, we use the following result.

Fact 4 If(s)is a polynomial such that(j)> 0for all, anddeg P2< deg P1, thenP1(s) +P2(s)is equivalent toP1(s) +(s)P2(s).Proof: LetQ(s, ) =P1+ [(1 ) +]P2where0 1. We can seethatQ(s, 0) =P1+P2,Q(s, 1) =P1+P2, and(1 ) + >0for all.



Rouths array: Special case 1

Case 1 Somep,1= 0but there is a nonzero memberp,kin the row.

Let(s) = 1 + (s2)k1 and use Fact 4.

Example 1 p(s) =s5 + 2s4 + 3s3 + 6s2 + 5s + 3

s5

1 3 5s4 2 6 3s3 0 7

P2(s) = 7s P2(s) =s

Let(s) = 1 s2.

(s)P2(s) = s3 +ss

3 1 1s2 8 3s1 118s0 3

There are two sign changes inthe first column sop(s)hastwo RHP roots.

Example 2 s4 +s3 + 2s2 + 2s + 3 = 0


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Rouths array: Special case 2 (1)

Case 2 The entire row is zero.

ThenP(s)is even or odd and is also a factor of the original polynomial.

Fact 5 Even or odd polynomials have as many roots in LHP as roots inRHP. Proof: Ifsis a root, then sis also a root.

To remedy the situation, considerP(s) =P(s+)for some small >0.

P(s) = (P+2

2!P + )

even

+ (P +2

3!P + )

odd

is equivalent to

(P+22!

P + ) + (P +23!

P + ) P(s) +P(s) as 0

In this case, substituteP(s)withP(s) +P(s)and count only RHP roots.



Rouths array: Special case 2 (2)

Example P(s) =s5 +s4 + 6s3 + 6s2 + 8s + 8

s5 1 6 8s4 1 6 8s3 0 0

P(s) =P1(s) =s4 + 6s2 + 8

P(s) = 4s3 + 12s

Use 14(4s3 + 12s) =s3 + 3s.s3

1 3s2 3 8s1 13s1 1s0 8

Hence,P(s)has one (1)LHP root and four (4)

j-axis roots.

Remark For design purposes, we can use the all-zero-row condition tosolve for the parameter that causes the systems marginal stability.


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A design example and relative stability

Consider a negative feedback system shown in the block diagram below.

Given

G(s) = s2 2s + 2

s4 + 7s3 + 19s2 + 23s + 10

Find ranges ofKsuch that

1. the closed-loop system is stable, and

+G(s)

K

yu

2. all the closed-loop system poles have real part less than 1.

Solution The characteristic equation is1 +KG(s) = 0or

s4 + 7s3 + 19s2 + 23s + 10 +K(s2 2s + 2) = 0

s

4

+ 7s

3

+ (19 +K)s

2

+ (23 2K)s+ 10 + 2K = 0The regionRe s < 1is equivalent toRe (s + 1)< 0. So lets= s + 1ors= s 1and findKsuch that1 +KG(s) = 1 +KG(s 1) = 0is stable inthe sense thatRe s

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7 Time-Domain Analysis of Control Systems

time response and test signals for continuous-data control systems the unit-step response and time-domain specification steady-state error: unity feedback v.s. nonunity feedback time response of a first-order system transient response of a prototype second-order system

damping ratio and damping factor, natural undamped frequency

maximum overshoot, delay time, rise time, and settling time

time-domain analysis of a position-control system dominant poles of transfer functions MATL ABtools and case studies

7 Time-Domain Analysis of Control Systems 2

Time response and typical test signals

Time response of a control system can be divided into two parts:

transient response: yt(t) goes to zero ast , i.e., limt

yt(t) = 0

steady-state response: yss(t) is what remains afteryt(t)has died out

Typical test signals

step-function input:r(t) =Rus(t)

ramp-function input:r(t) =Rtus(t)

parabolic-function input:r(t) = 1

2

Rt2us(t)


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Unit-step response and time-domain specification



Time-domain specification: Definition

1.Maximum overshoot=ymax yssis used to measure relative stability.

percent maximum overshoot=maximum overshoot

yss 100%

2.Delay time,td, is defined as the time required for the step response toreach50%of its final value.

3.Rise time,tr, is defined as the time required for the step response torise from10%to90%of its final value.

4.Settling time,ts, is defined as the time required for the step responseto decrease andstay withina specified percentage of its final value.

5.Steady-state error,ess, is defined as the difference between theoutput and the reference input when the steady state (t ) isreached.

Here we assume that the system has unityd.c. gain, i.e.,Gc(0) = 1.


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Steady state error: Unity feedback

Consider a feedback system withclosed-loop transfer function

M(s) =Y(s)

R(s)=

G(s)

1 +G(s)H(s)

+ G(s)

H(s)

R(s) U(s) Y(s)

IfH(s) = 1, we have aunity feedbacksystem and define the error of thesystem ase(t) =r(t) y(t). Hence the steady-state error is given by

ess= limt

e(t) = lims0

sE(s) = lims0

sR(s)

1 +G(s)



System types and error constants

An open-loop transfer function

G(s) =K(Tas+ 1)(Tbs+ 1) sn(T1s+ 1)(T2s+ 1)

is called oftypenwhich is the order of the pole ofG(s)ats= 0.

The steady-state error due to various kinds of inputs is shown below:

Step (position) error constant:Kp= lim

s0G(s)

Ramp (velocity) error constant:Kv = lim

s0sG(s)

Parabolic (acceleration) error constant:

Ka= lims0 s2

G(s)

type step ramp parabolic

0 R

1 +Kp

1 0 R

Kv

2 0 0

R

Ka


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Steady state error: Unity feedback examples

GivenH(s) = 1, determine system type ofG(s)and findessfor three basictypes of inputs.

G(s) = K(s+ 3.15)s(s+ 1.5)(s+ 0.5)

G(s) = Ks2(s+ 12)

G(s) = 5(s+ 1)s2(s+ 12)(s+ 5)



Steady state error: Nonunity feedback

IfH(s)is nonunity, the outputY(s)and the reference signalR(s)havedifferent units. So we cannot define the error byr(t) y(t). Differentdefinitions oferrorare needed.

Case H(0) = 0 Thed.c. gainof sensorH(s)is given byKH=H(0).

E(s) = 1KH

R(s) Y(s)

Case H(0) = 0 orH(s)has zero(s) at the origin, i.e.,H(s) =sNH1(s).

E(s) = 1

KHsNR(s) Y(s) where KH= lim

s0H(s)

sN =H1(0)

Examples G(s) = 1

s2(s+ 12)and

H(s) =5(s+ 1)s+ 5

H(s) = 10ss+ 5


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Time response of a first-order system

Consider a first-order system (T >0)

G(s) =

Y(s)

R(s)=

K

T s+ 1

Impulse response r(t) =(t) R= 1

y(t) = L1[G(s)] = KT

et/T

whent 0 yss(t) = 00.0

0.2

0.4

0.6

0.8

1.0

0 1 2 3 4 5

63.2%

t/T

y(t)/K

Step response r(t) =u(t) R(s) = 1/s

y(t) = L1 G(s)s

= L1 K

s(T s+ 1)

= L1 K

s KT

T s+ 1

= K(1 et/T)

whent 0 yss(t) =K

Notice that step response=

impulse response



A prototype second-order system

Consider the closed-loop system shown inthe block diagram. We have

Y(s)

R(s)=

K

s2 +ps+K

which is asecond-ordertransfer function.

+ K

s(s+p)

yr e

We write this as acanonicalorprototypesecond-order transfer function

n

n

n

G(s) = 2n

s2 + 2ns+2n

where >0andn>0whose poles are

s2 + 2ns+2n = s

2 + 2ns+22n+ (1 2)2n

= (s+n)2 +22

n

where2 = 1 2.


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Transient response of a second-order system (1)

Whenr(t) = 1 R(s) = 1/s, we have

y(t) = L1 2n

s(s2 + 2ns+2n)

= L1 1

s

+ 2 Re L1 B

s+ (+ j)n

where

B = 2n

s(s+ (j)n)

s=(+j)n=

j2(+j)

Hence

y(t) = 1 1

ent sin(nt+)

where= (+j) = cos1 andt 0. damping ratio natural (undamped) frequency n damping factor := n conditional (damped) frequency :=n



Transient response of a second-order system (2)


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Pole location and response



Maximum overshoot

peak time tmaxortptmax=

n1 2

=

maximum overshoot Mp:=y(tmax) y()

Mp= e /

12

percent (maximum) overshoot %Mpor P.O.

P.O.=y(tmax) y()

y() 100%

P.O.= 100e /

12


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Delay time and rise time

Delay time 0<

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Effects of adding a pole to transfer function (1)

Add a pole ats= 1/Tpto aforward-path transfer function of aunity-feedback system as

G(s) = 2n

s(s+ 2n)(1 +Tps)

The closed-loop transfer functionM=G/(1 +G)becomes

2n

Tps3 + (1 + 2nTp)s2 + 2ns+2n

= 1, n= 1 = 0.25, n= 1



Effects of adding a pole to transfer function (2)

Add a pole ats= 1/Tpto the closed-loop transfer function as

M(s) =Y(s)

R(s)=

G(s)

1 +G(s)=

2n(s2 + 2ns+2n)(1 +Tps)


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Effects of adding a zero to transfer function (1)

Add a zero ats= 1/Tpto the closed-loop transfer function as

M(s) = G(s)

1 +G(s)= 2n(1 +Tps)

(s2 + 2ns+2n) or y(t) =y1(t) +Tz

dy1(t)

dt



Effects of adding a zero to transfer function (2)

Consider, for example, adding a zero ats= 1/Tzto a forward-path of athird-order system as

G(s) = 6(1 +Tzs)

s(s+ 1)(s+ 2) so that

Y(s)

R(s)=

6(1 +Tzs)

s3 + 3s2 + (2 + 6Tz)s+ 6


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Non-mininum phase (NMP) zeros and undershoot

IfG(s)has unity d.c. gain and a real NMP zero ats= z0, then we have

e(t) = 1

y(t)

E(s) = (1

G(s))1

s E(z0) =

1

z0=

0

e(t)ez0t dt

Assuming its step responsey(t)has a-settling timetsand splittingintegral into two parts[0, ts] (ts, ), we have ts

0

e(t)ez0t dt+

ts

|e(t)|ez0t dt 1z0

IfMuis the minimum undershoot, we havemax

t0e(t) =Emax= 1 +Mu> 0

and from the definition of-settling time

|e(t)

| ,

t

ts, we have

Emax1 ez0ts

z0+

ez0ts

z0 1

z0or Mu 1

ez0ts 1If 1andz0ts 1, this meansMu> 1z0ts .

fast settling time large undershoot



Dominant poles

For analysis and design, it isimportant to sort out polesthat have a dominant transient

effect, calleddominant poles.

Dominant poles controldynamic performance,

whereasinsignificant polesensure realizable controller.

It is widely accepted that if the magnitude of the real part of a pole isat least510times that of a dominant pole or a pair of complex

dominant poles, then the pole may be regarded as insignificant.


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Relative damping ratio

When the dynamic of a system of higher-order can be accuratelyrepresented by a pair of complex-conjugate dominant poles, then we

can still useandnto indicate the transient dynamics. The damping ratio in this case is referred to as the relative damping

ratio.

Example

M(s) = 20

(s+ 10)(s2 + 2s+ 2) 2

s2 + 2s+ 2




Use commandsstep,ltiviewandsisotool.SIMULINKis also useful.


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8 Basic Control Actions

introductiondesign specifications

controller configurations

PID controller

proportional action

integral action

derivative action

tuning of PID controllerclosed-loop method

open-loop method

8 Basic Control Actions 2

Introduction

Design specifications

relative stability, steady-state accuracy (error), transient response:maximum overshoot, rise time, settling time

frequency-response characteristics: gain margin, phase margin,Mr

Controller configurations

series (cascade), feedback, and feedforward compensations

one degree-of-freedom, two degrees-of-freedom


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PID control

PID controllers a.k.a. three-term or three-mode control

one of the oldest controller types and the most widely used in

industries

pulp & paper 86%

steel 93%

oil refineries 93%

only textbook version is introduced

differs a lot from the industrial version



Text-book PID controllers

PID controller:

Integral

Proportional

Derivative

Plant


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PID controllers

PID controller

u(t) = KP

e(t)

P

+ KI

t

0

e() d

I

+ KD

d

dte(t)

D

U(s) =

KPE(s) +

KI

s E(s) +

KDsE(s)

or

Gc(s) = Kp+KI

s +KDs

= Kc 1 + 1TIs+ TDs



P action

LetKI= 0andKD= 0, i.e.

Gc(s) =KP U(s) =KPE(s)

u=

umax e > e0KPe+u0 e0 < e < e0umin e < e0


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P control

0 5 10 15 20

0

0.5

1.0

1.5

set point, measured variable

0 5 10 15 20

2

0

2

4

6control variable

KP = 5

KP = 2

KP = 1

KP = 5

KP = 2

KP = 1

stationary error (offset) largeKP fast response,small offset,

worse stability



I action

+

e

t

u = KPe(t) +u0

= KP

e(t) +

1

TI

t0

e() d

PI control

offset exists e dincreases u0increases yincreases reduce offset


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PI control

0 5 10 15 20

0

0.5

1.0

1.5


0 5 10 15 20

0

1

2

control variable

TI = 1TI = 2

TI = 5

TI =

TI = 1

TI = 2

TI = 5

TI =

removes stationary error

largeKI fast offset removal,worse stability



D action

A PI-controller contains no prediction

The same control signal is obtained for both these cases:

time

II

P

time

P

e

t

e

t


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D actionPrediction

error

time

e(t)e(t+TD)

e(t) +TDde(t)

dt

P control:u(t) =KPe(t)

PD control:u(t) = KP

e(t) +TD

de(t)

dt

KPe(t+TD)

TD= prediction horizon (time)



PD control

0 5 10 15 20

0

0.5

1.0


0 5 10 15 20

2

0

2

4

6 control variable

TD = 0.1

TD = 0.5

TD = 2

TD = 0.1

TD = 0.5

TD = 2

TDis used to reduce oscillation

TDtoo small, no influence

TDtoo large, decrease performance

In industrial practice, especially in noisy environment, D-control isoften turned off.


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PID control

In brief, we use

P control to adjust overall dynamic (increase P to get faster response)

I control to adjust steady-state part (remove offset)

D control to adjust transient part (reduce oscillation)

PID tuning How to tune? Ziegler & Nichols (1942) suggested twomethods:

open-loop method (or step response method)

closed-loop method (or ultimate cycle method)



PID tuning: Open-loop method

or step response method or reaction curve method: Draw a straight linetangent to the step response at thepoint of inflection. Suppose that linecross 0%-output and 100%-output at A and B respectively. LetLbe thetime measured from step input to the point A and Tbe the time

measured from A to B.

P: KP=T

L

PI: KP= 0.9T

L, TI=

L

0.3

PID: KP= 1.2T

L, TI= 2L, TD= 0.5L


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PID tuning: Closed-loop method

or ultimate cycle method: Apply P-control and gradually increase gainKuntil the system oscillate at marginal stability. Call thatKthe ultimate

gainKuand the period of oscillationthe ultimate periodPu.

P: KP= 0.5Ku

PI: KP= 0.45Ku, TI= Pu

1.2

PID: KP= 0.6Ku, TI=Pu

2 , TD=

Pu

8


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9 Root-Locus Technique

basic properties of the root loci

properties of the root lociK= 0andK= points, number of branches, symmetryroot loci on the real axis

angles and intersect of asymptotes: behavior at |s| angles of departure and angles of arrival

intersection with imaginary axis

breakaway points (saddle points)

root contours: multiple-parameter variation design gain from root locus shaping the root locus lead and lag compensators

9 Root-Locus Technique 2

Introduction

location of closed-loop poles (roots of characteristic equation) relatesclosely tostabilityandtime-domaincharacteristics

root locus: a systemetic contruction of the trajectories of the rootsof the characteristic equation

proposed by Walter R. Evans (1920-99) in 1948 use poles and zeros of open-loop (or equivalents) to determine thetrajectories of closed-loop poles whenoneparameter is changing

when more than one parameter varies, we called themroot contours give rough and quick graphical feeling of how to compensate for

simple SISO system

for root contours and more accurate root loci, computer tool(MATL AB) can be used


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Basic properties of the root loci: Changing parameter

Suppose the closed-loop transfer function of a control system is

Y(s)

R(s)=

G(s)

1 +G(s)H(s)

Then the roots of characteristic equation must satisfy1 +G(s)H(s) = 0.Suppose thatG(s)H(s)contains a real variable parameter in the form

G(s)H(s) =KB(s)

A(s) 1 + K B(s)

A(s) =

A(s) +KB(s)

A(s) = 0

Example Consider the characteristic equation

s(s+ 1)(s + 2) + s2 + (3 + 2K)s+ 5 = 0

which can be written as

1 +K 2s

s3 + 4s2 + 5s + 5 or

B(s)

A(s)=

2s

s3 + 4s2 + 5s + 5



Basic properties of RL: Magnitude & angle conditions

For simplicity, let us express the characteristic equation as

1 +KG(s) = 0 G(s) = 1K

where

G(s) =B(s)

A(s)=

mi=1(s zi)n

j=1(s pj)=

sm +bm1sm1 + +b1s +b0sn +an1sn1 + +a1s +a0

To satisfy1 +KG(s) = 0, the following conditions must be satisfiedsimultaneously:

Magnitude condition |G(s)| = 1|K|, < K <

Angle condition G(s) =

(2i + 1) =odd multiples of, K 0

2i =even multiples of, K

0


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Graphical interpretation of magnitude & angle conditions

Example

G(s) = s z1

(s p1)(s p2)The figure shows thesituation whens0is on theroot loci.

Re-axis0

j-axis

2 13

s0p1

s0p2

s0

s0z1

Onlythe angle condition is used to construct the root loci.

Only magnitude condition hasKinvolved. So, to find Kfor a specificlocation on a root locus we must use magnitude condition.



Properties of the root loci

the number of root loci equals the order of characteristic equation root loci begin (K= 0) from open-loop polespjand end (K= ) at

open-loop zeros (zi)

the poles and zeros include those at infinity, if any

they are symmetrical with respect toRe-axis and with respect to theaxes of symmetry of the pole-zero configuration.

the entire real axis is occupied by the root loci for all values ofK forK 0, the root loci are found on a given section onRe-axis, if the

total number of (real) poles and zeros to the right of the section is odd

Example G(s) = s + 1

s2

+s 2=

s + 1

(s 1)(s + 2)


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Multiple roots

ForK 0, root loci for

G(s) = 1s2

can be found from

s2 +K= 0

Re-axis0

j-axis

and for

G(s) = 1

(s )3from

(s )3 +K= 0j-axis

Re-axis0



Asymptotic properties

When the magnitude ofsis large (|s| ),1 +KG(s) = 1 +KB(s)A(s)

can

be approximated using long division

A(s)

B(s)

=snm + (an1

bm1)snm1 +

as1 +KG(s) = 1 +K

B(s)

A(s) 1 + K

(s )nmwherean1 bm1= (n m)or

= 1

nm

n

j=1

pj m

i=1

zi

where bm1=

mi=1

zi, an1= n

j=1

pj

In summary, the large-scale behavior of root locus is asymptotic to the

lines (akaasymptotes) with center (centroid) atand anglesigiven by

i=(2i+ 1) 180

n m for K 0


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Examples

Example 1

G(s) =

s + 1

s(s+ 2)(s + 3)

Example 2

G(s) = 1

s(s+ 3)2

Example 3

G(s) = 1

s(s+ 1)(s + 3)(s + 4)



Angles of departure and angles of arrival

Angle of the tangent of root locus at a points0can be found using theangle conditionas

m

i=1

(s0 zi) n

j=1

(s0 pj) = 180

with the desired angle replaced byDfor the angle of departure and byAfor the angle of arrival.

Example 4 G(s) = s + 2

s2 + 2s + 2

In case of multiple roots, use2,3, ... for sum of angles of multipleroots.

Example 5 G(s) = s + 2

(s2 + 2s + 2)2


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Breakaway points (saddle points)

At the point where more than one roots meet, a break-away occurs. Wecan find that point from thenecessarycondition for multiple roots that

dds

(1 +KG(s)) = 0

ord

dsG(s) = 0 or

d

ds

1

G(s)= 0

or

A(s)d

dsB(s) =B(s)

d

dsA(s)

orn

j=1

1

s pj =m

i=1

1

s zi



Intersection with imaginary axis

of root loci should be found using Routh-Hurwitz test because of thestability of the closed-loop system.

The values of

Kthat causes thej-axis crossing and the crossing locationshould also be determined.

Example 2 [continued] The root loci crossj-axis atK= 54and theauxiliary equation iss2 + 9 = 0or at = 3rad/sec.

Example 3 [continued] The root loci crossj-axis atK= 26.25and

the auxiliary equation iss2

+ 1.5 = 0or at = 1.225rad/sec.


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More examples

Example 6

G(s) =(s+ 1)(s + 2)

s(s2 + 2s + 2)

Example 7

G(s) = 1

s(s2 + 8s + 32)

Example 8

G(s) = 1

s(s+ 2)(s2

+ 2s + 2)

Example 9

G(s) = s2 + 4

s(s2 + 9)



. . . and their root loci

3

2

4

1

1

1

j

2

24

4

26 4

j

1

1

2

j

2 134

2

1

12

3

3j


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Multiple roots: An example

G(s) = s + 1

s2(s+ 9)

j

8 6 4 2

2

4

2

4



Root locus construction: Summary

1. mark poles () & zeros () ons-plane2.nbranches, symmetrical aboutRe-axis, start (K= 0) from and end

(K= ) at 3. draw loci forK

0onRe-axis where the total count of real

&

to

the right is odd; the remaining parts ofRe-axis are the loci forK 04. draw asymptotes (angles and centroid)

5. find breakaway points and multiple roots (and gain)

6. findj-axis crossing points (and gain) by Routh-Hurwitz criterion

7. compute angles of departure & arrival

8. complete the root locus


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How to solve f(s) = 0

bisection search: Find range[a, b]such thatf(a)f(b)< 0, i.e.,f(a)andf(b)has different sign. Withc= a+b2

testf(c)

iff(a)f(c)< 0, thenb cor use[a, c]iff(b)f(c)< 0 , thena cor use[c, b]

as next range. Repeat until |a b| < . Newton-Raphson method: x x+h

h= f(x)f(x)

or 1

h= f

(x)f(x)

+ f(x)2f(x)

iteration: Substitute a good guesssinn

j=1

1s pj =

mi=1

1s zi

leaving only two unknowns. Solve and repeat. Good for findingbreakaway points.



Design aspect of the root loci

+ yr e

K G(s)u

where

G(s) = 1

s(s+ 2)

Find the range ofKso that

the closed-loop is stable, has overshoot less than10%, has settling time less than

1.5s,

etc.

Effects of adding poles and zeros to G(s)

adding a pole toG(s)has the effect of pushing the root loci towardthe right-halfs-plane

adding left-half plane zeros to

mws slide

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