05/05/2009 fall 2008 advanced topics (eeng 4010) control systems design (eeng 5310)

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Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

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What is a Control System? System- a combination of components that act together and perform a certain objective Control System- a system in which the objective is to control a process or a device or environment Process- a progressively continuing operations/development marked by a series of gradual changes that succeed one another in a relatively fixed way and lead towards a particular result or end.

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Page 1: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Fall 2008Advanced Topics (EENG 4010)Control Systems Design (EENG

5310)

Page 2: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

What is a Control System?

System- a combination of components that act together and perform a certain objective

Control System- a system in which the objective is to control a process or a device or environment

Process- a progressively continuing operations/development marked by a series of gradual changes that succeed one another in a relatively fixed way and lead towards

a particular result or end.

Page 3: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Control Theory

Branch of systems theory (study of interactions and behavior of a complex assemblage)

Control SystemManipulated Variable(s)

Control Variable(s)

Open Loop Control System

Control System

Manipulated Variable(s)

Control Variable(s)

Closed Loop Control System

Feedback function

Page 4: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Classification of Systems

Classes of Systems

Lumped ParameterDistributed Parameter (Partial Differential Equations, Transmission line example) Deterministic

Discrete TimeContinuous Time

NonlinearLinear

Time Varying

Stochastic

Constant Coefficient

Non-homogeneous Homogeneous (No External Input; system behavior depends on initial conditions)

Page 5: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Example Control Systems Mechanical and Electo-mechanical (e.g. Turntable) Control Systems Thermal (e.g. Temperature) Control System Pneumatic Control System Fluid (Hydraulic) Control Systems Complex Control Systems Industrial Controllers

– On-off Controllers– Proportional Controllers– Integral Controllers– Proportional-plus-Integral Controllers– Proportional-plus-Derivative Controllers– Proportional-plus-Integral-plus-Derivative Controllers

Page 6: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Mathematical Background

Why needed? (A system with differentials, integrals etc.)

Complex variables (Cauchy-Reimann Conditions, Euler Theorem)

Laplace Transformation– Definition– Standard Transforms – Inverse Laplace Transforms

Z-Transforms

Page 7: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Laplace Transform

Definition Condition for Existence Laplace Transforms of exponential, step, ramp, sinusoidal,

pulse, and impulse functions Translation of and multiplication by Effect of Change of time scale Real and complex differentiations, initial and final value

theorems, real integration, product theorem Inverse Laplace Transform

dtetfsFtf st 0

)()()]([L0|)(|0

tfe t

t

suchthat Limit

te)(tf

Page 8: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Inverse Laplace Transform

Definition Formula is seldom or never used; instead, Heaviside

partial fraction expansion is used. Illustration with a problem: Initial conditions: y(0) = 1, y’(0) = 0, and r(t) = 1, t >= 0. Find the steady state responseMultiple pole case with Use the ideas to find and

dsesFj

tfsFjc

ic

st

)(21)()]([1

L

)(2342

2

trydtdy

dtyd

22

1

)( as

L

22

1

)( asasL

3

2

)1(32)(

ssssF

Page 9: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Applications

Spring-mass-damper- Coulomb and viscous damper cases

RLC circuit, and concept of analogous variables Solution of spring-mass-damper (viscous case) DC motor- Field current and armature current

controlled cases Block diagrams of the above DC-motor problems Feedback System Transfer functions and Signal flow graphs

Page 10: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Block Diagram Reduction

Combining blocks in a cascadeMoving a summing point ahead of a blockMoving summing point behind a blockMoving splitting point ahead of a blockMoving splitting point behind a block Elimination of a feedback loop

G1 G2 G3 G4

H2

H1

H3

R(s)Y(s)

+ +

+ -

-+

Page 11: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Signal Flow Graphs

Mason’s Gain Formula

Solve these two equations and generalize toget Mason’s Gain Formula

r1

r2

x1

x2

a21 a12

a22

a11

11212111 xrxaxa 22222121 xrxaxa

k

ijkijk

ij

FG

G1 G2 G3 G4

G5G6 G7 G8

H2 H3

H7H8

R(s) Y(s)

Find Y(s)/R(s) using the formula

Page 12: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Control System Stability: Routh-Hurwitz Criterion

Why poles need to be in Right Hand PlaneNecessary condition involving Characteristic

Equation (Polynomial) Coefficients Proof that the above condition is not sufficient

Ex: s3+s2+2s+8.Routh-Hurwitz Criterion- Necessary & Sufficient

Page 13: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Routh-Hurwitz Criterion: Some Typical Problems

2nd and 3rd order systems q(s)=s5+2s4+2s3+4s2+11s+10 (first element of a

row 0; other elements are not) q(s)=s4+s3+s2+s+K (Similar to above case) q(s)=s3+2s2+4s+K (for k = 8, first element of a

row 0; so are other elements of the row) q(s)= s5+s4+2s3+2s2+s+1: Repeated roots on

imaginary axis; Marginally stable case

Page 14: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Root-Locus Method: What and Why?

Plotting the trajectories of the poles of a closed loop control system with free parameter variations

Useful in the design for stability with out sacrificing much on performance

Closed Loop Transfer Function

Let Open Loop Gain Roots of the closed loop characteristic equation depend on K.

G(s)

H(s)

Y(s)R(s)+ -

)()(1)(

)()(

sHsGsG

sRsY

n

jj

m

ii

n

m

ps

zsK

pspspszszszsKsHsG

1

1

21

21

)(

)(.

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

0)()(1 sHsG

Page 15: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Relationship between closed loop poles and open loop gain

When K=0, closed loop poles match open loop poles

When closed loop poles match open loop zeros.

Hence we can say, the closed loop poles start at open loop poles and approach closed loop zeros as K increases and thus form trajectories.

,K

Page 16: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Mathematical Preliminaries of Root Locus MethodComplex numbers can be expressed as (absolute value, angle) pairs.Now,

The loci of closed loop poles can be determined using the above constraints (particularly, the angle constraint) on G(s)H(s).

s=+j

|s|

s=|s|.ej

s

s=+j

-s1=-j

|s+s1|

s+s1=|s+s1|ej

1)()(0)()(1 sHsGsHsG

)12(180)()(&1|)()(| 0 ksHsGsHsG

Page 17: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Root Locus Method- Step1 thru 3 of a 7-Step Procedure

Step-1: Locate poles and zeros of G(s)H(s). Step 2: Determine Root Locus on the real-axis using angle constraint. Value of

K at any particular test point s can be calculated using the magnitude constraint.

Step 3: Find asymptotes by using angle constraint in . Find asymptote centroid

. This formula may be

obtained by setting

Illustrative Problem:

1)(;)2)(1(

)(

sHsss

KsG

)()(0

SHsGLimits

mn

zpn

j

m

i ijA

1 1

)()(

0-1-2

3j

3j

A

n

jj

m

ii

mnA ps

zsKsHsG

sK

1

1

)(

)(1)()(1

)(1

Page 18: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Root Locus Method- Step 4

Step 4: Determine breakaway points (points where two or more loci coincide giving multiple roots and then deviate). Now, from the characteristic equation

where , we get, at a multiple pole s1, , because at s1,

. Thus we get at s1,

Since at s = s1. Thus, we get break points by setting dK/ds=0. In the

example, we get s = -0.4226 or -1.5774 (invalid).

0)()(.1)()(1 sPsZKsHsG0)(.)()( sZKsPsf

0| 1ssdsdf

))...(()()( 21 nr sssssssf

0)()(')(')()(;

)(')('

sZsZsPsPsf

sZsPK 0)().(')(').( sZsPsZsP

0)(

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

2

sZ

sZsPsZsPdsdK

sZsPK

Page 19: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Root Locus Method- Step 5

Step 5: Determine the points (if any) where the root loci cross the imaginary axis using Roth-Hurwitz Stability Criterion.

Illustration with the Example Problem

Characteristic equation for the problem:s3+3s2+2s+KFrom the array, we know that the system is marginally stable at K=6. Now, we canget the value of (imaginary axis crossing) either by solving the second row3s2+6 =0 or the original equation with s=j.

S3 1 2S2 3 K

S1 (6-K)/3

S0 K

Page 20: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Root Locus Method- Step 6 and 7

Step 6: Determine angles of departure at complex poles and arrival at complex zeros using angle criterion.

Step 7: Choose a test point in the broad neighborhood of imaginary axis and origin and check whether sum of the angles is an odd multiple of +180 or -180. If it does not satisfy, select another one. Continue the process till sufficient number of test points satisfying angle condition are located. Draw the root loci using information from steps 1-5.

Page 21: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Root Locus approach to Control System Design

Effect of Addition of Poles to Open Loop Function: Pulls the root locus right; lowers system’s stability and slows down the settling of response.

Effect of Addition of Zeros to Open Loop Function: Pulls the Root Locus to Left; improves system stability and speeds up the settling of response

j

x

j

xx

j

x x x

j

xxxo

j

xxx o

j

xxx o

Page 22: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Performance Criteria Used In Design

We consider 2nd order systems here, because higher order systems with 2 dominant poles can be approximated to 2nd order systems e.g.

when

For 2nd order system

For unit step inputWhere . Two types of performance criteria (Transient and Steady State) Stability is a validity criterion (Non-negotiable).

))(2(1)( 22 asss

sTnn

||10|| na

)(2

)( 22

2

sRss

sYnn

n

)1sin(111)(;1)( 2

2

tety

ssR n

tn

1cos

Page 23: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Transient Performance Criteria

t

y(t)1.0

TR

ess

overshoot

TP

TS

)1sin(111)( 2

2

tety n

tn

•Rise Time TR= Time to reach Value 1.0

•Rise Time Tr1= Time from 0.1 to 0.9

Empirical Formula isfor• Settling time (Time to settle to within 98% of 1.0)=4/n

• Peak Time

Percentage Overshoot =

nrT

6.016.21

2.0

0.6

nTr1

8.03.0

21

n

PT

21/100 e

Page 24: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Series Compensators for Improved Design

RC OP-Amp Circuit for phase lead (or lag) compensator

Lead Compensator for Improved Transient Response; Example: Required to

reduce rise time to half keeping = 0.5. Lag Compensator for Improved steady-state

performance. Example:

)2(4)(

ss

sG

)2)(1(06.1)(

sss

sG

)3386.2)(5864.03307.0)(5864.03307.0(06.1

06.1)2)(1(06.1

)()(

sjsjsssssRsC

Page 25: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Frequency Response Analysis

Response to x(t) = X sin(t)G(s) = K/(Ts+1) and G(s)=(s+1/T1)/(s+1/T2) cases Frequency response graphs- Bode, and Nyquist

plots of Resonant frequency and peak valueNichols ChartNyquist Stability Criterion

1

])/()/(21[,)1(,)(, 211

nn jjTjjK

Page 26: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Control System Design Using Frequency Response Analysis

Lead Compensation Lag Compensation Lag-Lead Compensation

Page 27: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

State Space Analysis

State-Space Representation of a Generic Transfer Function in Canonical Forms:– Controllable Canonical Form– Observable Canonical Form– Diagonal Canonical Form– Jordan Canonical Form

Eigenvalue Analysis

Page 28: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Solution of State Equations

Solution of Homogeneous Equations Interpretation of Show that the state transition matrix is

given by Properties of Solution of Nonhomogeneous EquationsCayley-Hamilton Theorem

Axx

11 )()( AIA set t L

teA

)(t

Page 29: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Controllability and Observability

Definitions of Controllable and Observable Systems

Controllabililty and Obervability Conditions Principle of Duality

Page 30: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Control System Design in State Space

Necessary and Sufficient Condition for Arbitrary Pole Placement

Determination of Feedback Gain Matrix by Ackerman’s formula

Design of Servo Systems

Page 31: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Introduction to Sampled Data Control Systems

Z-transform and Inverse Z-transform Properties of Z-Transform and Comparison

with the Corresponding Laplace Transform Properties

Transfer Functions of Discrete Data Systems

Page 32: 05/05/2009 Fall 2008 Advanced Topics (EENG 4010) Control Systems Design (EENG 5310)

Analysis of Sampled Data Systems

Input and Output Response of Sampled Data Systems

Differences in the Transient Characteristics of Continuous Data Systems and Corresponding Discrete (Sampled) Data Systems

Root Locus Analysis of Sampled Data Systems