Prof.watanabe July28 APSS2010

8/20/2019 Prof.watanabe July28 APSS2010

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Introduction of Control Competition& Lecture on PID Control

Asia-Pacific Summer School on Smart Structure Technology

Toru WatanabeNihon University, JPN

Overview

Control object and tool for CompetitionIntroduction of DCMCT unit and QICii software

Introduction to classical control

Dynamical modelingTransfer FunctionBlock Diagram and FeedbackPoles and Zeros

Introduction to PID control

Control Object and Tools forCompetition

Control Object

DCMCT (DC Motor Control Trainer unit)Equipped with DC motor with flywheel, amplifier,microchip controller and digital/analog interface

QICii Interactive interface software Interactive modeling, feedback controller design andevaluation can be carried out

Able to upload controller to DCMCT / download timehistory

Outlook of DCMCT Unit

1. DC Motor 2. Flywheel3. Amp.4. Encoder 7. Analo I/O 9. Microchip10. USB port11. CTL Reset12. Interrupt SW15. Power port

Controller on 9 drives 1 through 3 based on signal from 4

Block diagram of the DCMCT

Schematic view of open loop

Controlcommand

DrivingCurrent

Error of angleor angular

l i Angle

FrictionTorque

Angle

Ampli-fier

Micro-chip

l iMo-tor Torque

Fly-wheel

En-coder

Controller is downloaded fromPC by using QICii


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Window of QICii

Universal Interface for the operation of DCMCT

Tasks in “Control Lab & Competition ”

Educational LabsIdentify parameters in the control objectStudy basic properties of P, I, and D controls

Supervisors add extra weights to flywheelDesign PID controller to above perturbed systemInvestigate robustness and evaluate performance

Why DCMCT & QICii?

Electromagnetic motor is commonly used instructural control to drive mechanisms such asmass dampers

,design procedure and characteristics of gains

are essentially identical over any dynamicalsystemsThey are so well-organized that fair competitioncan be done within such limited time

Introduction to ClassicalControl

What is “Classical Control ”?

In the field of “Control Engineering,” thereare two major stream of formulation

“State space equation” – essentially in time domain“Transfer function” – essentially in frequency domain

Both formulations are importantState space model is widely used in modern (or LQ)control, while transfer function is the basis ofclassical control.Post modern control (e.g. H-infinity) uses both.

State-Space vs. Transfer Function (1)

Example of State Space Formulationmx(t) cx(t) kx(t) u(t)

y(t) x(t)

Disp. x(t)

T

Springk

Dampingc

x v

0 1 0x(t) x(t)u(t)k c 1

v(t) v(t)m m m

x(t)y(t) 1 0

v(t)

u


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State-Space vs. Transfer Function (2)

Example of transfer function formulationmx(t) cx(t) kx(t) u(t)

Replacing “derivative” and “time”“ ”

Mass m

Springk

Disp. x(t)

Dampingc

2ms X(s) csX(s) kX(s) U(s)

2

X(s) 1G(s)

U(s) ms cs k

Take output/input ratio

u

Framework of “Classical Control ”

ModelingUtilizing transfer functions to describe thedynamics of control objects

Anal sisMainly in frequency domain – “s” domain

Controller DesignGenerally for single-input single-output systemManual calculation based design (practically bytry-and-error approach)

Advantage of “Classical Control ”

AdvantageSimple

Analysis and controller design can be done withoutcomputer, nor “MATLAB” – of course they are helpfulSimple controller such as “PID controller” is easy to tunemanually, and easy to understand intuitively

DisadvantageHard to deal with MIMO (multi-input multi-output) systemThe optimality of controller is not guaranteed

Introduction to Laplace Transform

Laplace Transform

Identification -st0

F s = f t e dt = L f t

Laplace transform of f t ( ) (0)d

L sF s f er ve unc on t

Laplace transform ofintegrated function 0 1f ( ) L d F ss

Neglecting initial value … f t ( ) L sF s

Laplace Transform and DifferentialEquation

Differential equation in time domain1

0 1 11

( ) ( ) ( )( )

n n

n nn n

d x t d x t dx t u t

dt dt dt

1

1 11

( ) ( )( )

n

n nn

d x t dx t y t

dt dt

Transformed differential equation

10 1 1

11 1

X U( )

Y( ) X

n nn n

nn n

s s s s s

s s s s

Transformed equation is no longer differential, but algebraic


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Laplace Transform to Solve DifferentialEquation – “Operator Method ”

DifferentialEquation

t-domainsolution

(t-function)

Tough to solve

s -domain AlgebraicEquation

Algebraicsolution

(s-function)

Laplace transform Inverse transform

Easy to solve

Laplace Transform and Fourier Transform

Identification of Fourier Transform

-= ( ) j t F j f t e dt Infinite function (e.g.unit step function)cannot be transformed

Introduce “convergence factor”

- -= ( ) j t t F j f t e e dt Replacing “ +j ” with “s,” Laplace transform

- -= ( ) = ( ) j t st F j f t e dt f t e dt F s

Infinite function (e.g.unit step function) canbe transformed!

Representing Operator “s”

In Fourier transform, operator “j ” can besimply understood as “frequency ”

Therefore, the transformed function must becomplex to denote damping as Real/Imaginal ratio

In Laplace transform, operator “s” can be

understood as “frequency and damping ”Therefore, the transformed function simplybecame rational function because the existence ofdamping is already taken into consideration as “s”

Modeling of DynamicalSystems as Transfer Function

Modeling Dynamical System (1)

First, identify all elements concerning dynamicsof control object

Electric Circuit includin DC motor

Flywheel including rotor

Moment ofElectromotivecoefficient Kml i i i i l i

VoltageVm(t)

Resistance Rm

Current Im(t)

Inductance Lm Armature (rotor)rotation speed

m(t)

Motortorque

Tm(t)

Inertia J eq(t)

Friction Torque Td(t)


Describe all elements as differential equationsEach element is supposed to possess single input andsingle output

Electric Circuit includin DC motor

m m m m m m mV (t) R I (t) L I (t) K (t)t

l i i i i l i

Flywheeleq m m m d J (t) K I (t) T (t)t

m m m be m mT (t) K I (t), V (t) K (t) Electromotive / Backelectromotive Force


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Transform all equations into “s” domain

m m m m m m mV (s) R I (s) L sI (s) K (s) Electric Circuit including DC motor

Flywheel

eq m m m d J s (s) K I (s) T (s)


Erase non-essential variables

m m m m m mR L s I (s) V (s) K (s) Electric Circuit including DC motor

Flywheel

eq m m m d J s (s) K I (s) T (s)

m

m m mm m m m

1 K I (s) V (s) (s)

R L s R L s

Substituted in …


Transform the equation into transfer function

2m m

eq m m m d K K

J s (s) V (s) (s) T (s)

Flywheel & Electric Circuit with motor

m m m m

Here, Td(s)is neglected

2m m

eq m m d m m m m

K K J s (s) V (s) T (s)R L s R L s

m m2

m mm m eq

m m

(s) K V (s) K

R L s J sR L s

TransferFunction

Transfer Function & State Equation (1)

You may be able to obtain transfer functionfrom state equation

m m m m m m mV (t) R I (t) L I (t) K (t)t

eq m m m d J (t) K I (t) T (t)t

m m

m mm m

m

eq

R K 1

L L I(t) I(t)I(t)L V (t) , y(t)= 0 1

K (t) (t)(t) 0 0J

Here, Td(s)is neglected

Transfer Function & State Equation (2)

Matrix equation can be dealt with as … ,t t t t t X AX Bu y = CX

Transform all equations into “s” domain

,s s s s s s X AX Bu y = CX

1 1( ), G s( )

ss s s s

s

yy = C I - A Bu C I - A Bu

Transpose terms,

m m

2m eq m m m

(s) K V (s) J s R L s K

TransferFunction

NOTE: Denominator is reduced and compound fraction is turned to simple

Transfer Function andFrequency Response Function


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Analogy: Frequency Response Function

Transfer Function is quite similar to“Frequency Response Function”

mx(t) cx(t) kx(t) u(t) Externalforce u

Suppose

2

X( ) 1G( )

U( ) m cj k

and take output/input ratio

t

j t

u(t ) U( )e

x(t ) X( )e

Mass m

Springk

sp. x

Dampingc

Difference between TF and FRF

As shown, operator “s” is a complex variables j

,transform (basis of FRF) is quite similar toLaplace transform (basis of TF).FRF can be obtained by simply replacing “s”of TF to “j ”

Laplace Transform and FrequencyResponse Function

Systemt-domain Frequency ResponseFunction ( -function)

Tough to derive

s -domain System

(s -function)

Transfer Function

(s-function)

Laplace transform Replace “s” with “j ”

Easy to derive

Block Diagram andConnection

Block Diagram – Graphical Description

Each block denotes transfer function

Transfer functionInput Output

Addition

Subtraction

Input A

Input BOutputC=A-B

Distribution

Input A Output B

Output C A=B=C

,

Parallel Connection = Addition orSubtraction

In parallel connection, total transfer function isequal to the sum or difference of two functions

G(s)=G1(s) - G2(s)

InputInput

2(s)

OutputOut-put


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Cascade Connection = Multiplication

In cascade connection, total transfer function isequal to the product of two functions

Input Output

G(s)=G1(s)G2(s)Input Output

Feedback Connection = FractionalFunction

Feedback connection makes fractional function

G1(s)Input u Output ysignal e

2(s)

G(s)=G1(s)/{1+G1(s)G2(s)}Input Output

signal c

Why Feedback Connection makesFractional Function?

G1(s)Input u

Output ysignal e

signal c

Proof y = G1(s) e = G1(s) { u – c } = G1(s) { u – G2(s) y }{ 1+G1(s) G2(s) } y = G1(s) uTherefore, y / u = G1(s) / {1+G1(s) G2(s) }

s

DCMCT ’s Block Diagram

Element-by-element block diagram

1 Vm Km

Im

1 1

Tdm

m

m m mm m m m

1 K I (s) V (s) (s)

R L s R L s

Tm

Km

eq m m m d J s (s) K I (s) T (s) =Tm

DCMCT ’s Transfer Function accordingto Block Diagram

Ignore TdLet G1(s)= Km / { S Jeq (S Lm +Rm)}

= According to “Feedback Connection,” G(s) =G1(s) / { 1+G1(s)G2(s) }, Then

m m

2m eq m m m


TransferFunction

Feedback Compensation andCascade Compensation


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Where to Put Controller?

Feedback compensation (common in modern control)

Control ObjectInput u

Output y

Controller

u

yError eControl ObjectController

Cascade compensation (common in classical control)

Difference in Transfer Function

Suppose control object as G1(s), whilecontroller as G2(s)Feedback Compensation

=

Cascade CompensationG(s) = G1(s) G2(s) / { 1+G1(s)G2(s) }

Looks not a big difference – just in numerator

No, it is a big difference – “zeros” are different!

Poles and Zeros

Definition – Poles and Zeros

Suppose G(s) = N(s)/D(s)N(s): Numerator polynomialD(s): Denominator polynomial

= “ ” = On poles, G(s) possess “infinite transmission”

Zeros (Zi) = Operator “s” that gives N(s)=0On zeros, G(s) possess “zero transmission”

Poles and Eigenvalues

Poles in classical control is identical to eigenvaluesin modern control

According to transfer function, poles of DCMCT aregiven by solving the following equation

2

According to state equation, eigenvalues of DCMCTare given by solving the following equation

eq m m ms s

2

det det 0

m m

mm m m

m m eq m

eq

R K s

L L R K s s s

K L J Ls J

I A

Poles and Behavior of System

Poles denotes the behavior of the systemIf any Re(Pi) > 0, the system possesses instableMagnitude of Pi (=Re (Pi)* Re (Pi) + Im(Pi)*Im(Pi))is identical to “natural frequency”

Angle of Pi ( = arctan(Re (Pi)/Magnitude of Pi) ) isidentical to “damping ratio”

See Prof. R. Christenson’s slide No. 30-34.


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Zeros and Root Loci

We know that the position of poles of the system ischanged according to controller gain.The map that shows the trace of poles according tothe increase of gain is called “Root Loci”

to zeros or toward infinite as the gain is increasedto infinity.Therefore, position or location of zeros matters tothe characteristics of control systems

Introduction to PID Control

Concept of PID Controller

Prefix the structure of controller as combinationof Proportional-Integral-Derivative gains

ControlObject

(target)

K p

K I

1s

errorou pu

Why PID? How PID used?

Advantage of PID controller Simple – only three gains are required to tuneEasy to understand – roles of each gains can berepresented intuitivelyPractically useful –PID controller generally

satisfies requirements.

Application Any simple control problem (e.g., AMD for SDOF)Local feedback for actuator (e.g., servomotor)

The Role of P, I and D gains (1)

Proportional gain – equivalent to “spring”

Output = displacement

If the feedback gain isproportional to displacement

to "Spring"ControlForce

F kx


Derivative gain – equivalent to “damper”

E uivalent

Output = displacementDerivative = velocity

If the feedback gain isproportional to velocity

to "Damper"ControlForce F cx


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Integral gain – “Stationary disturbance canceller”

Output = displacementIntegral = sum of drift

If the feedback gain isproportional to integral

Suppose stationaryforce is subjected

ControlForce

D gain does not work.P gain reduces drift, butnot perfectly cancel it.

Control force isgradually increased

until drift is converged

F I xdt

Aim of PID Controller

To let the output (rotational angle or speed)follow the target

Output Y(s)

Command

DCMCT

R(s)

K p

K I

1s

rrorEm(s)

m m (s))

m s

Transfer Function of DCMCT and PID

DCMCT’s transfer function

m m

2m eq m m m


Y(s) = m(s)

Output : rotational speed

2m P I D

P I Dm

V (s) 1 sK K s K K K sK

E (s) s s

PID controller’s transfer function

Y(s) = m (s)

m m2

m eq m m m

(s) K V (s) s J s R L s K

Output : rotational angle

Transfer Function of DCMCT with PID

Cascade CompensationG(s) = G1(s) G2(s) / { 1+G1(s)G2(s) }

Y(s) = (s)

m P I D

2 2

eq m m m m P I D

K sK K s K G(s)

s J s R L s K K sK K s K

Y(s) = (s)

2m P I D

22 2eq m m m m P I D

K sK K s K G(s)

s J s R L s K K sK K s K

Effect of PID on DCMCT (1)

Y(s) = (s) case, No controlLm is neglected for brevity

2m mK K G s s

eq meq m m J R J R s K

Effect of PID on DCMCT (2)

Y(s) = (s) case – Lm is neglected for brevity

2m P I D

2 2eq m m m P I D

K sK K s K G(s)

s J sR K K sK K s K

m P I D

22D m eq m m m P m I

22 2m m P m m P D m eq m m I

D m eq m

K sK K s K

K K J R s K K K s K K

K K K K K K 4 K K J R K K s

2 K K J R

Poles become complex by introducing PID feedback


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How to Design PID gains?

Each gains possess their own “trade offs” Any excessive gain makes system unstableP gain as increased

as er u v ra ng response

D gain as increasedmore damping but too sensitive to noise and error

I gain as increasedresistant to drift, but slower response

Design “ Theories ” for PID Controller

Several design procedures for PID controllerare already presented

Ziegler-NicholsCritical sensitivitStep response

All these methods are “procedures,” not“theories” because they do not guarantee theoptimality

Documents

Prof.watanabe July28 APSS2010