control theory eee322 - national university theory.pdf · – determine the stability of a...

Control Theory

EEE322

Dr. Mudathir Fagiri

Introduction to

Control Theory

Prerequisite by topics

• Knowledge and proficiency in MATLAB

• Concept and solution of linear Ordinary

Differential Equations (ODE)

• Laplace Transform and its applications

• Vectors and Matrices

• Complex Numbers

OBJECTIVES On completion of this subject, the student will be able to do the following

either by hand or with the help of computation tools such as MATLAB:

– Define the basic terminologies used in controls systems.

– Explain advantages and drawbacks of open-loop and closed loop control systems.

– Obtain models of simple dynamic systems in ordinary differential equation, transfer

function, state space, or block diagram form.

– Obtain overall transfer function of a system using either block diagram algebra, or signal

flow graphs, or Matlab tools.

– Compute and present in graphical form the output response of control systems to typical

test input signals.

– Explain the relationship between system output response and transfer function

characteristics or pole/zero locations.

– Determine the stability of a closed-loop control systems using the Routh-Hurwitz criteria

– Analyze the closed loop stability and performance of control systems based on open-

loop transfer functions using the frequency response techniques.

Topics Covered • Introduction to open loop and closed loop control systems.

• Review of signal systems concepts and techniques as applied to control

system.

• Block diagrams and signal flow graphs.

• Modeling of control systems using ODE, block diagrams, and transfer

functions.

• Modeling and analysis of control systems using state space methods.

• Analysis of dynamic response of control systems, including transient

response, steady state response, and tracking performance.

• Closed-loop stability analysis using the Routh-Hurwitz criteria.

• Stability and performance analysis using the Root Locus techniques.

• Control system design using the Root Locus techniques.

• Stability and performance analysis using the frequency response

techniques.

Textbook

• Automatic Control Systems, Golnaraghi

and Kuo, ninth edition, Wiley, 2009

http://www.amazon.com/gp/reader/0470048964/ref=sib_dp_pt

Other References • Feedback Control of Dynamic Systems (6th Edition) by

Gene Franklin, J.D. Powell, and Abbas Emami-Naeini

(Hardcover - Oct 11, 2009)

• Modern Control Engineering (5th Edition) by Katsuhiko

Ogata (Paperback - Aug 30, 2009)

http://www.amazon.com/Modern-Control-Engineering-Katsuhiko-Ogata/dp/0136156738/ref=sr_1_5?ie=UTF8&s=books&qid=1251076006&sr=1-5

Other References • Modern Control Systems (11th Edition) (Pie) by Richard

C. Dorf and Robert H. Bishop (Hardcover - Aug 10,

2007)

• Control Systems Engineering, Just Ask! Package by

Norman S. Nise (Hardcover - Jun 21, 2004)

http://www.amazon.com/Control-Systems-Engineering-Just-Package/dp/047167589X/ref=sr_1_8?ie=UTF8&s=books&qid=1251075560&sr=1-8

http://www.amazon.com/Modern-Control-Systems-11th-Pie/dp/0132270285/ref=sr_1_1?ie=UTF8&s=books&qid=1251075560&sr=1-1

Other References • Modern Control Theory (3rd Edition) by William L.

Brogan (Paperback - Oct 11, 1990)

• Feedback Control Systems (4th Edition) by Charles L.

Phillips and Royce D. Harbor (Hardcover - Aug 19, 1999)

http://www.amazon.com/Feedback-Control-Systems-Charles-Phillips/dp/0139490906/ref=sr_1_21?ie=UTF8&s=books&qid=1251075811&sr=1-21

http://www.amazon.com/Modern-Control-Theory-William-Brogan/dp/0135897637/ref=sr_1_6?ie=UTF8&s=books&qid=1251075560&sr=1-6

History of Control Engineering

18th Century James Watt’s centrifugal governor for the speed control of a steam engine.

1920s Minorsky worked on automatic controllers for steering ships.

1930s Nyquist developed a method for analyzing the stability of controlled systems

1940s Frequency response methods made it possible to design linear closed-loop control systems

1950s Root-locus method due to Evans was fully developed

1960s State space methods, optimal control, adaptive control and

1980s Learning controls are begun to investigated and developed.

……………………….

……………………….

……………………….

Earlier Control Systems?

Water-level float regulator (before BC)

Human System

i. Pancreas Regulates blood glucose level

ii. Adrenaline

Automatically generated to increase the heart rate and oxygen in times of flight

iii. Eye

Follow moving object

iv. Hand

Pick up an object and place it at a predetermined location

v. Temperature

Regulated temperature of 36°C to 37°C

Earlier Control Systems?

A manual level control system

A modern high voltage tranformator

A wind farm

Transportation

Car and Driver

• Objective: To control direction and speed of car

• Outputs: Actual direction and speed of car

• Control inputs: Road markings and speed signs

• Disturbances: Road surface and grade, wind, obstacles

• Possible subsystems: The car alone, power steering system,

breaking system

Control System Terminology

• Input - Excitation applied to a control system from an external source.

• Output - The response obtained from a system

• Feedback - The output of a system that is returned to modify the input.

• Error - The difference between the reference input and the output.

Negative Feedback Control System

CONTROLLER CONTROLLED

DEVICE

FEEDBACK

ELEMENT

+ + +

-

Input

Output

Feedback

Error

• Control – is the process of causing a system variable to conform to some desired value.

• System – An interconnection of elements and devices for a desired purpose.

• Control System – An interconnection of components forming a system configuration that will provide a desired response.

• Process – The device, plant, or system under control. The input and output relationship represents the cause-and-effect relationship of the process.

• The interaction is defined in terms of

variables.

i. System input

ii. System output

iii. Environmental disturbances

Types of Control Systems

Open-Loop – Simple control system which performs its

function with-out concerns for initial conditions or external inputs.

– Must be closely monitored.

Closed-Loop (feedback) – Uses the output of the process to modify

the process to produce the desired result.

– Continually adjusts the process.

Advantages of a Closed-Loop

Feedback System Increased Accuracy

– Increased ability to reproduce output with varied input.

Reduced Sensitivity to Disturbance

– By self correcting it minimizes effects of system changes.

Smoothing and Filtering

– System induced noise and distortion are reduced.

Increased Bandwidth

– Produces sat. response to increased range of input changes.

Major Types of Feedback Used Position Feedback

– Used when the output is a linear distance or angular measurement.

Rate & Acceleration Feedback

– Feeds back rate of motion or rate of change of motion (acceleration)

– Motion smoothing

– Uses a electrical/mechanical device call an accelerometer

Present

Position

Future

Position

Ship’s

Heading Range Change Bearing

Change

Fire Control Problem

Fire Control Problem

• Input

– Target data

– Own ship data

• Computations

– Relative motion procedure

– Exterior ballistics procedure

Fire Control Problem • Solutions

– Weapons time of flight

– Bearing rate

– Line of Sight(LOS): The line between the target

and the firing platform

– Speed across LOS

– Future target position

– Launch angles

• Launch azimuth

• Launch elevation

– Weapon positioning orders

• The above determines weapon trajectory: The line

the weapon must travel on to intercept the target.

The Iterative Process to the

Fire Control Solution

Step 1

Step 2

Step 3 Last Step

A 3-Dimensional Problem

Horizontal Reference Plane

Line of Sight

Target

Elevation

Gun

Elevation

Solving the Fire Control Problem

Continuously Measure

Present Target Position Stabilize Measured

Quantities

Compute Relative

Target Velocity

Ballistic

Calculations

Relative

Motion

Calculations

Time of

Flight

Future

Target

Position

Prediction Procedure

Unstabilized

Launch

Angles

Environmental Inputs

Launch Angles

(Stabilized)

Weapons Positioning orders

Idle-speed control system.

Figure 1-5 (p. 5) Solar collector field.

Conceptual method of efficient water

extraction using solar power.

Important components of the sun-

tracking control system.

a. system concept;

b. detailed layout;

c. schematic;

d. functional block diagram

Antenna azimuth position control system:

(a)

(b)

(c)

a. Video laser disc player;

b. objective lens reading

pits on a laser disc;

c. optical path for playback

showing tracking mirror

rotated by a control system

to keep the laser beam

positioned on the pits.

Computer hard

disk drive,

showing disks

and read/write

head

Courtesy of Quantum Corp.

Response of a

position control

system showing

effect of high

and low

controller gain

on the output

response

High gain; fast but oscillating

Control goal; fast reaction, lower overshoot, less settling time

The control system design process

Aircraft attitude defined

Winder

© J. Ayers, 1988.

Control of a nuclear reactor

Grinder system

© 1997, ASME.

High-speed proportional

solenoid valve

© 1996, ASME.

High-speed rail system showing

pantograph and catenary

© 1997, ASME.

Control Theory

EEE322

Dr. Mudathir A. Fagiri

Mathematical Modeling

and Representation of

Physical Systems

Types of Systems

Static System: If a system does not change

with time, it is called a static system.

Dynamic System: If a system changes with

time, it is called a dynamic system.

Dynamic Systems

• A system is said to be dynamic if its current output may depend on the past history as well as the present values of the input variables.

• Mathematically,

Time Input, ::

]),([)(

tu

tuty 0

Example: A moving mass

Model: Force=Mass x Acceleration

uyM

M

y u

Ways to Study a System

Model

• A model is a simplified representation or

abstraction of reality.

• Reality is generally too complex to model

exactly.

What is Mathematical Model?

A set of mathematical equations (e.g., differential eqs.) that

describes the input-output behavior of a system.

What is a model used for?

• Simulation

• Prediction/Forecasting

• Prognostics/Diagnostics

• Design/Performance Evaluation

• Control System Design

Basic Elements of Electrical Systems

• The time domain expression relating voltage and current for the resistor is given by Ohm’s law

Rtitv RR )()(

• The Laplace transform of the above equation is

RsIsV RR )()(


• The time domain expression relating voltage and current for the Capacitor is given as:

dttiC

tv cc )()(1

• The Laplace transform of the above equation (assuming there is no charge stored in the capacitor) is

)()( sICs

sV cc

1


• The time domain expression relating voltage and current for the inductor is given as:

dt

tdiLtv L

L

)()(

• The Laplace transform of the above equation (assuming there is no energy stored in inductor) is

)()( sLsIsV LL

V-I and I-V Relations

Component Symbol V-I Relation I-V Relation

Resistor

Capacitor

Inductor

dt

tdiLtv L

L

)()(

dttiC

tv cc )()(1

Rtitv RR )()( R

tvti R

R

)()(

dt

tdvCti c

c

)()(

dttvL

ti LL )()(1

Kirchhoff’s voltage law:

The algebraic sum of voltages around any closed loop in an

electrical circuit is zero.

Kirchhoff’s current law:

The algebraic sum of currents into any junction in an

electrical circuit is zero.

Laplace Transform

Name Time function f(t) Laplace Transform

Unit Impulse (t) 1

Unit Step u(t) 1/s

Unit ramp t 1/s2

nth-Order ramp t n n!/sn+1

Exponential e-at 1/(s+a)

nth-Order exponential t n e-at n!/(s+a)n+1

Sine sin(bt) b/(s2+b2)

Cosine cos(bt) s/(s2+b2)

Damped sine e-at sin(bt) b/((s+a)2+b2)

Damped cosine e-at cos(bt) (s+a)/((s+a)2+b2)

Diverging sine t sin(bt) 2bs/(s2+b2)2

Diverging cosine t cos(bt) (s2-b2) /(s2+b2)2

Find the inverse Laplace transform of

F(s)=5/(s2+3s+2).

Solution:

Find inverse Laplace Transform of

Find the inverse Laplace transform of

F(s)=(2s+3)/(s3+2s2+s).

Solution:

Laplace Transform Theorems

Transfer Function

Transfer Function

After Laplace transform we have

X(s)=G(s)F(s)

We call G(s) the transfer function.

System interconnections

Series interconnection

Y(s)=H(s)U(s) where H(s)=H1(s)H2(s).

Parallel interconnection

Y(s)=H(s)U(s) where H(s)=H1(s)+H2(s).

Feedback interconnection

Example 1

The two-port network shown in the following figure has vi(t) as

the input voltage and vo(t) as the output voltage. Find the transfer

function Vo(s)/Vi(s) of the network.

C i(t) vo(t)

dttiC

Rtitvi )()()(1

dttiC

tvo )()(1

vi( t)

Example 1

Taking Laplace transform of both equations, considering initial

conditions to zero.

Re-arrange both equations as:

dttiC

Rtitvi )()()(1

dttiC

tvo )()(1

)()()( sICs

RsIsVi

1 )()( sI

CssVo

1

)()( sIsCsVo ))(()(Cs

RsIsVi

1

Example 1

Substitute I(s) in equation on left

)()( sIsCsVo ))(()(Cs

RsIsVi

1

))(()(Cs

RsCsVsV oi

1

)()(

)(

CsRCs

sV

sV

i

o

1

1

RCssV

sV

i

o

1

1

)(

)(

Models of Electrical Systems

R-L-C series circuit, impulse voltage source:

Model of an RLC parallel circuit:

Models of Mechanical Systems

Mechanical translational systems. Newton’s second law:

Device with friction (shock absorber):

B is damping coefficient.

Translational system to be defined is a spring (Hooke’s law):

K is spring coefficient

Model of a mass-spring-damper system:

Note that linear physical systems are modeled by linear

differential equations for which linear components can be

added together. See example of a mass-spring-damper

system.

Differential equations as mathematical models of physical

systems: similarity between mathematical models of

electrical circuits and models of simple mechanical

systems (see model of an RCL circuit and model of the

mass-spring-damper system).

CONTROL SYSTEMS

Dr. Mudathir A. O. Fagiri

lecture 3

Dr. Ali Karimpour Feb 2013

2

Lecture 3

Different representations of

control systems Topics to be covered include:

High Order Differential Equation. (HODE model)

State Space model. (SS model)

Transfer Function. (TF model)

State Diagram. (SD model)

lecture 3


3

High Order Differential Equation (HODE).

inputtheisu

outputtheisy

bdt

dub

dt

udb

dt

uda

dt

dya

dt

yda

dt

ydm

m

mm

m

n

n

nn

n

011

1

1011

1

1 ...........

HODE model

The study of differential equations of the type described

above is a rich and interesting subject. Of all the methods

available for studying linear differential equations, one

particularly useful tool is provided by Laplace Transforms.

lecture 3


4

Example 1: A high order differential equation (HODE).

HODE model

lecture 3


5

State Space Models (SS)

For continuous time systems

For linear time invariant continuous time systems

SS model

SS model

lecture 3


6

General form of LTI systems in state space form

pnpnn

p

p

nnnnn

n

n

n u

u

u

bbb

bbb

bbb

x

x

x

aaa

aaa

aaa

x

x

x

.

.

..

.....

.....

..

..

.

.

..

.....

.....

..

..

.

.

2

1

21

22221

11211

2

1

21

22221

11211

2

1

SS model

State Space Models (SS)

lecture 3


7

Example 2: A linear time invariant continuous time systems

SS model

output

lecture 3


8

Example 2: Continue

The equations can be rearranged as follows:

We have a linear state space model with

)()(

)(1

)(11

)(1)(

)(1)(

121

tvtc

tvCR

tvCRCR

tiCdt

tdv

tvLdt

tdi

f

lecture 3


9

A demonstration robot containing several servo motors

lecture 3


10

Different Representations

SD model

HODE model

TF model

SS model

lecture 3


11

Transfer Function Model (TF)

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

...........

01

1

101

1

1

011

1

1011

1

1

subssubsusbsussyassyasysasys

conditioninitialzerotransformlaplaceTaking

ubdt

dub

dt

udb

dt

udya

dt

dya

dt

yda

dt

yd

m

m

mn

n

n

m

m

mm

m

n

n

nn

n

)(

)(

)(

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

sA

sB

su

sysGsusBsysA TF model

HODE model TF model

Or

Input-output model

lecture 3


12

Different representations

SD model

HODE model

TF model

SS model

lecture 3


13


SD model

HODE model

TF model

SS model

lecture 3


14

TF Model Properties

1- It is available just for linear systems.

2- It is derived by zero initial condition.

3- It just shows the relation between input

and output so it may lose some information.

4- It can be used to show delay systems but

SS can not.

lecture 3


15

State Diagram Model (SD)

)0()()( 2212

1 xssxsxdt

dxx

)0()()( 21

1

2 xsxssx

s -1

s -1

x2(0)

x1(s) x2(s)

SD model

lecture 3


16

State Diagram to State Space

1

212

21

54

xc

rxxx

xx

SD model SS model

lecture 3


17


SD model

HODE model

TF model

SS model

lecture 3


18


1

212

21

54

xc

rxxx

xx

SS

SD

TF

45

1)(

2

sssG

HODE

rccc 45

Discussed in this lecture

Will be discussed in next lecture

Mason’s rule

CONTROL SYSTEMS


Lecture 4

Reduction of Multiple Systems

Block Diagram

Figure 4.1

Components of a block diagram for a linear, time-invariant system

Figure 4.2

a. Cascaded subsystems;

b. equivalent transfer function

Figure 4.3

a. Parallel subsystems;

b. equivalent transfer function

Figure 4.4

a. Feedback control system;

b. simplified model;

c. equivalent transfer function

Figure 4.5: Block diagram algebra for summing junctions

equivalent forms for moving a block

a. to the left past a summing junction;

b. to the right past a summing junction

Figure 4.6: Block diagram algebra for pickoff points

equivalent forms for moving a block

a. to the left past a pickoff point;

b. to the right past a pickoff point

Block diagram reduction via familiar forms for Example 4.1

Problem: Reduce the block diagram shown in figure to a single transfer

function

Steps in solving Example 4.1:

a. collapse summing junctions;

b. form equivalent cascaded system

in the forward path

c. form equivalent parallel system in

the feedback path;

d. form equivalent feedback system

and multiply by cascadedG1(s)

Block diagram reduction via familiar forms for Example 4.1 Cont.

Problem: Reduce the block diagram shown in figure to a single transfer

function

Block diagram reduction by moving blocks Example 4.2

Steps in the block diagram reduction for Example 4.2

a) Move G2(s) to the left past of

pickoff point to create parallel

subsystems, and reduce the feedback

system of G3(s) and H3(s)

b) Reduce parallel pair of 1/G2(s)

and unity, and push G1(s) to the right

past summing junction

c) Collapse the summing junctions,

add the 2 feedback elements, and

combine the last 2 cascade blocks

d) Reduce the feedback system to

the left

e) finally, Multiple the 2 cascade

blocks and obtain final result.

CONTROL SYSTEMS


Lecture 5

Signal-flow graph

and Mason’s Rule

Signal-flow graph components:

a. system;

b. signal;

c. interconnection of systems and signals

a. cascaded system nodes

b. cascaded system signal-flow

graph;

c. parallel system nodes

d. parallel system signal-flow

graph;

e. feedback system nodes

f. feedback system signal-flow

graph

Building signal-flow graphs

Converting a block diagram to a signal-flow graph

Problem 5.1: Convert the block diagram to a signal-flow graph.


Signal-flow graph development:

a. signal nodes;

b. signal-flow graph;

c. simplified signal-flow graph


Problem 5.2: Convert the block diagram to a signal-flow graph.


Mason’s Rule - Definitions

Mason’s gain rule is as follows: the transfer function of

a system with signal-input, signal-output flow graphs is

332211)(ppp

sT

Δ=1-(sum of all loop gains)+(sum of products of gains of all

combinations if 2 nontouching loops)- (sum of products of gains of

all combinations if 3 nontouching loops)+…

A path is any succession of branches, from input to output, in the

direction of the arrows, that does not pass any node more than once.

A loop is any closed succession of branches in the direction of the

arrows that does not pass any node more than once.

Mason’s Rule - Definitions

Loop gain: The product of branch gains found by traversing a path that starts at a node and ends at the

same node, following the direction of the signal flow, without passing through any other node more than

once. G2(s)H2(s), G4(s)H2(s), G4(s)G5(s)H3(s), G4(s)G6(s)H3(s)

Forward-path gain: The product of gains found by traversing a path from input node to output node

in the direction of signal flow. G1(s)G2(s)G3(s)G4(s)G5(s)G7(s), G1(s)G2(s)G3(s)G4(s)G5(s)G7(s)

Nontouching loops: loops that do not have any nodes in common. G2(s)H1(s) does not touch

G4(s)H2(s), G4(s)G5(s)H3(s), and G4(s)G6(s)H3(s)

Nontouching-loop gain: The product of loop gains from nontouching loops taken 2, 3,4, or more at a

time.

[G2(s)H1(s)][G4(s)H2(s)], [G2(s)H1(s)][G4(s)G5(s)H3(s)], [G2(s)H1(s)][G4(s)G6(s)H3(s)]

Mason’s Rule

The Transfer function. C(s)/ R(s), of a system represented by a signal-flow graph is

Where

K = number of forward paths

Tk = the kth forward-path gain

= 1 - loop gains + nontouching-loop gains taken 2 at a time - nontouching-loop

gains taken 3 at a time + nontouching-loop gains taken 4 at a time - …….

= - loop gain terms in that touch the kth forward path. In other words, is formed by eliminating from those loop gains that touch the kth forward path.

( )( )

( )

k k

k

TC s

G sR s

k k

Transfer function via Mason’s rule Problem: Find the transfer function for the signal flow graph

Solution:

forward path G1(s)G2(s)G3(s)G4(s)G5(s)

Loop gains G2(s)H1(s), G4(s)H2(s), G7(s)H4(s),

G2(s)G3(s)G4(s)G5(s)G6(s)G7(s)G8(s)

Nontouching loops 2 at a time

G2(s)H1(s)G4(s)H2(s)



3 at a time G2(s)H1(s)G4(s)H2(s)G7(s)H4(s)

Now

= 1-[G2(s)H1(s)+G4(s)H2(s)+G7(s)H4(s)+ G2(s)G3(s)G4(s)G5(s)G6(s)G7(s)G8(s)] +

[G2(s)H1(s)G4(s)H2(s) + G2(s)H1(s)G7(s)H4(s) + G4(s)H2(s)G7(s)H4(s)] – [G2(s)H1(s)G4(s)H2(s)G7(s)H4(s)]

= 1 - G7(s)H4(s)

[G1(s)G2(s)G3(s)G4(s)G5(s)][1-G7(s)H4(s)]

1

1 1( ) T

G s

CONTROL SYSTEMS


Lecture 9


2

Lecture 6

Stability of Linear

Control Systems

Lecture 9


3

The response of linear systems can always be decomposed as the

zero-state response and zero-input response. We study

1. Input output stability of LTI system is called BIBO (bounded-input bounded-output) stability ( the zero-state response )

2. Internal stability of LTI system is called Asymptotic stability ( the zero-input response )

Stability analysis

Lecture 9


4

Input output stability of LTI system

tt

dtugdutgty00

)()()()()(

Consider a SISO linear time-invariant system, then the

output can be described by

where g(t) is the impulse response of the system and

system is relaxed at t=0.

)()()()()()(00

Idtugdutgtytt

Lecture 9


5


Definition: A system is said to be BIBO stable (bounded-input

bounded-output) if every bounded input excited a bounded

output. This stability is defined for zero-state response and is

applicable only if the system is initially relaxed.

Lecture 9


6


Theorem: A SISO system described by (I) is BIBO stable if

and only if g(t) is absolutely integrable in [0,∞), or

For some constant M.

0)( Mdttg

Lecture 9


7


then)(so boundedbe)(Let mututu

Proof: g(t) is absolutely integrable system is BIBO

0

)()()( dtugty

So the output is bounded.

Mum dtug )()(0

dgum

0

)(

Now: System is BIBO stable g(t) is absolutely integrable

If g(t) is not absolutely integrable, then there exists t1 such that:

Let us choose

dgt1

0)(

1

011 )()()(

t

dtugty 1

0)(

t

dg

0)( if1

0)( if1)( 1

g

gtu

So it is not BIBO

Lecture 9


8

Theorem: A SISO system with proper rational transfer

function g(s) is BIBO stable if and only if every pole of g(s)

has negative real part.


Lecture 9


9

Example 1: Discuss the stability of the system .

9

1

1

s

)(sC

2

1

s

s)(sR

2

1

)(

)()(

ssR

sCsG

There is no RHP root , so system is

BIBO stable.

BIBO stability:

Lecture 9


10

Different regions in S plane

RHP plane LHP plane Unstable Stable

Lecture 9


11

Example 5: Check the BIBO and internal stability of the

following system.

s

s 1+

- 1

1

s

1x2x cr

BIBO stability

1

1

11

1

)(

)(

s

s

s

sr

sc 1p We have BIBO stability

CONTROL SYSTEMS


Lecture 9


2

Lecture 7

Stability of linear control systems

through Routh Hurwitz criterion

Lecture 9


3

Routh Hurwitz Algorithm

The Routh Hurwitz algorithm is based on the following

numerical table.

Routh’s table

ns

1ns

1

1na2na ..............

3na4na

5na ..............

2ns

3ns...

0s

1,2

1,3 2,3

2,2 3,2

3,3

..............

..............

1,n

Lecture 9


4

Routh’s table

ns

1ns

1

1na2na ..............

3na4na

5na ..............

2ns

3ns...

0s

1,2

1,3 2,3

2,2 3,2

3,3

..............

..............

1,n


1

3121,2

1

n

nnn

a

aaa

1

5142,2

1

n

nnn

a

aaa

1

7163,2

1

n

nnn

a

aaa

Lecture 9


5

Result

Consider a polynomial p(s) and its associated table.

Then the number of roots in RHP is equal to the number

of sign changes in the first column of the table.

Lecture 9


6

Routh’s table

ns

1ns

1

1na2na ..............

3na4na

5na ..............

2ns

3ns...

0s

1,2

1,3 2,3

2,2 3,2

3,3

..............

..............

1,n


Number of sign changes=number of roots in RHP

Lecture 9


7

Example 1: Check the number of zeros in the RHP

010532)( 234 sssssp

051

1032

3

4

s

s

Two roots in RHP

01072 s

007

451s

00100s

Lecture 9


8

Routh Hurwitz special cases

Routh Hurwitz special cases

1- The first element of a row is zero. (see example 2)

2- All elements of a row are zero. (see example 3)

Lecture 9


9


0322)( 234 sssssp

021

321

3

4

s

s

0302s

00321

s

0030s

3

Two roots in RHP

for any

Lecture 9


10


047884)( 2345 ssssssp

484

781

4

5

s

s

0663s

0442s

0001s

044)( 2 ssq

0081s

0040s

No RHP roots + two roots on imaginary axis

08)( ssq

Auxiliary Polynomial

Lecture 9


11

Remarks:

If all elements of the row s2n-1 are zero, there are 2n roots with

same magnitude that they are symmetrical to the center of the s

plane. It means they can be real roots with different signs or

complex conjugate roots.

If there are no sign changes in first column, all roots of the

auxiliary polynomial lie on the jω axis.

However, one important point to notice is that if there are repeated

roots on the jω axis, the system is

actually unstable.

022)( 45 ssssp

022)( 45 ssssp

02422)( 2345 ssssssp

084632)( 2345 ssssssp

Lecture 9


12

Example 4: Check the stability of following system

for different values of k

42

)13(3

ss

ssk

+

-

42

)13(1

42

)13(

)(

3

3

ss

ssk

ss

ssk

sM)13(42

)13(3

sksss

sks

04)2(3 23 skkss

To check the stability we must check the RHP roots of

43

21

2

3

ks

ks

03

4)2(31

k

kks

040s

03

463

03

2

k

kk

kWe need k>0.528

for stability

Lecture 9


13

Example 6: The block Diagram of a control system is depicted in the

following figure. Find the region in K-α plane concluding the system stable.

s

s

1

)2(2

s

sK+

-

R(s) Y(s)

Lecture 9


14

Example 6: The block Diagram of a control system is depicted in the

following figure. Find the region in K-α plane concluding the system stable.

s

s

1

)2(2

s

sK+

-

R(s) C(s)

Lecture 9


15

Exercises

1- Are the real parts of all roots of following system less than -1.

42654)( 2345 ssssssp

Lecture 9


16

Exercises (Cont.)

2- The open-loop transfer function of a control system

with negative unit feedback is:

)21)(1(

)2()()(

sTss

sKsHsG

Find the region in K-T plane concluding the system stable.

Answer:

0.33

0.66

Stable

Unstable

Unstable

Un

stab

le

T

K

Lecture 9


17

Exercises (Cont.)

3- Find the number of roots in the region [-2,2].

015115 23 sss

4- The closed loop transfer function of a system is :

ksksss

sksG

2)25(157

)2()(

234

For the stability of the system which one is true? ( k>0 )

1) 0 ≤ k ≤ 28.12 2) 0 < k < 28.12

3) 0 ≤ k < 28.12 4) 0 < k ≤ 28.12

Answer: (4)

CONTROL SYSTEMS


Lecture 10


2

Lecture 8

Time domain analysis of

control systems

Lecture 10


3

Some test signals

R

t

R

t

1

R

R

t

Step input

Velocity input

Acceleration input

)()( tRutr s

RsR )(

)()( tRtutr 2

)(s

RsR

)(2

)(2

tuRt

tr 3)(

s

RsR

Lecture 10


4

Error in control systems

G K

+

-

c e r

Error signal input output e(t)=r(t)-c(t)

E(s)=R(s)-C(s)

Loop

transfer function G(s)K(s)

sT

dndd

j

m dessss

sssk

)1).......(1)(1(

)1).......(1)(1(

21

21

j=2 type two system

j=0 type zero system

j=1 type one system

Lecture 10


5

1


G K

+

-

c e r

)()()()( sRsTsRsE

)()()(1

1)( sR

sKsGsE

)()(1

)()()(

sKsG

sKsGsT

)(lim)(lim0

ssEteest

ssIf the system is

stable:

(very important)

)()()(lim0

sRsTsRss

)(lim)(lim0

ssEteest

ss

2

)()()(1

lim0

sRsKsG

s

s

Lecture 10


6

Error in control systems for step input

)(lim0

ssEes

ss

)(lim0

ssEes

ss

s

RsR )(

)(lim1)(lim)(lim000

sTRs

RsT

s

RsssEe

sssss

ps

ssss

K

R

sKsG

R

s

R

sKsG

sssEe

1)()(lim1)()(1lim)(lim

0

00

)()(lim0

sKsGKs

p

Position constant

1

)()()(lim0

sRsTsRss

2

)()()(1

lim0

sRsKsG

s

s

Lecture 10


7

Error in control systems for velocity input

2)(

s

RsR

vs

ssss

K

R

sKssG

R

s

R

sKsG

sssEe

)()(lim0)()(1lim)(lim

0

200

)()(lim0

sKssGKs

v

Velocity constant

)(lim0

ssEes

ss

)(lim0

ssEes

ss

1

)()()(lim0

sRsTsRss

2

)()()(1

lim0

sRsKsG

s

s

s

sTR

s

RsT

s

RsssEe

sssss

)(1lim)(lim)(lim

02200

Lecture 10


8

Error in control systems for parabolic input

3)(

s

RsR

as

ssss

K

R

sKsGs

R

s

R

sKsG

sssEe

)()(lim0)()(1lim)(lim

2

0

300

)()(lim 2

0sKsGsK

sa

Acceleration constant

)(lim0

ssEes

ss

)(lim0

ssEes

ss

1

)()()(lim0

sRsTsRss

2

)()()(1

lim0

sRsKsG

s

s

203300

)(1lim)(lim)(lim

s

sTR

s

RsT

s

RsssEe

sssss

Lecture 10


9


sT

dndd

j

m dessss

sssksKsG

)1).......(1)(1(

)1).......(1)(1()()(

21

21

position velocity acceleration

Type Kp Kv Ka ess ess ess

0

1

2

3

k k

R

10 0

k

k

0 k

R0

0 0 k

R

0 0 0

Lecture 10


10

Example 1: Find the different errors in following system.

+

-

c e r

1

9

s1

9)()(

ssKsG

10

9)(

ssT

9)()(lim0

sKsGKs

p 1.091

1

sse

1.09.01)(lim11or 0

sTes

ss

0 av KK sse

Error for unit step input

Errors for unit velocity and unit parabolic input

System is stable so we continue

s

sTe

sss

)(1lim1or

0

Lecture 10


11

Example 1: Step response

1

9

s

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

Step Response

Time (sec)

Am

plit

ude

+

-

c e r

10

9)(

ssT

step(9,[1 10])

hold on;step(1,1)

Lecture 10


12

Example 1: Velocity response

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

3velocity response

Time (sec)

Am

plit

ude

1

9

s

+

-

c e r

10

9)(

ssT

step(9,[1 10 0])

hold on;step(1,[1 0])

Lecture 10


13

Example 2: Find the different errors in following system

+

-

c e r )10(

100

ss

)10(

100)()(

sssKsG

10010

100)(

2

sssT

)()(lim0

sKsGKs

p0

1

1

sse 011)(lim11or

0

sTe

sss

0aK sse

Error for step input

Error for parabolic input

10)()(lim0

sKssGKs

v1.0

10

1sse

Error for velocity input


1.0)(1

lim1or 0

s

sTe

sss

20

)(1lim1or

s

sTe

sss

Lecture 10


14

Example 2: Step response

0 0.2 0.4 0.6 0.8 1 1.20

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plit

ude

+

-

c e r )10(

100

ss

10010

100)(

2

sssT

step(100,[1 10 100])

hold on;step(1,1)

Lecture 10


15

Example 2: Velocity response

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

3velocity response

Time (sec)

Am

plit

ude

step(100,[1 10 100 0])

hold on;step(1,[1 0])

+

-

c e r )10(

100

ss

10010

100)(

2

sssT

Lecture 10


16

Example 3: The closed loop transfer function of a system

is given. Determine a such that the system error to step

input is zero.

23612)(

23

sss

asT

First of all check the stability:

23

012

49

2312

61

0

1

2

3

s

s

s

s

Clearly it is

stable

0)(lim10

sTRes

ss

1)(lim0

sTs

23123

aa

Lecture 10


17


is given. Determine a and b such that the system error to

velocity input is zero.

23612)(

23

sss

bassT


s

sTRe

sss

)(1lim

0

s

sss

bas

Rs

236121

lim23

0

0

23612

)23()6(12lim

23

23

0

ssss

bsassRe

sss

It is stable by example 3

23

6

b

a

Lecture 10


18


is given. Find the system error to unit step.

23612)(

23

sss

bassT


It is unstable so the error is infinity

Lecture 10


19

Example 6: Find the different errors in following system

+

-

c e r

1s

k

1)()(

s

ksKsG

ks

ksT

1)(

ksKsGKs

p

)()(lim0 k

ess

1

1

0 av KK sse

Error for step input

Errors for velocity and parabolic input


Note that the above method doesn’t say anything about how

the errors go to infinity

0k

Error series can explain the matter.

Lecture 10


20

Error series

G K

+

-

c e r

dtrwtesRsWsEt

ee )()()()()()(0

: taround)( Expand tr

....!3

)(

!2

)()()()( 32

trtrtrtrtr

....)(!2

)()()()()()(0

2

00

dwtrdwtrdwtrte

t

e

t

e

t

e

Lecture 10


21

Error series

....)(!2

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

0

2

00

dw

trdwtrdwtrte

t

e

t

e

t

e

Now consider steady value for r

....)(!3

)(!2

)()()( 3210 tr

Ctr

CtrCtrCte sssss

....)(!2

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

0

2

00

dwtr

dwtrdwtrte es

esess

0C2C1C

Lecture 10


22

Error series coefficients

....)(!3

)(!2

)()()( 3210 tr

Ctr

CtrCtrCte sssss

dwC e

0

0 )( dwC e

0

1 )( dwC e

0

2

2 )( .........

Calculation of coefficients

dewsW s

ee

0)()(

)(lim0

0 sWC es

ds

sdWC e

s

)(lim

01

n

e

n

sn

ds

sWdC

)(lim

0........ ........

Lecture 10


23

Example 7: Determine the error series coefficients for system

in example 1

ksWC e

s

1

1)(lim

00 20

1)1(

)(lim

k

k

ds

sdWC e

s

........

+

-

c e r

1s

k

)(1

1)( sR

ks

ssE

)(sWe

32

2

02

)1(

2)(lim

k

k

ds

sWdC e

s

Lecture 10


24

Example 8: Determine the error series for step and velocity

inputs in example 1.

,.......)1(

2,

)1(,

1

122210

k

kC

k

kC

kC

+

-

c e r

1s

k

Step:

ktr

Ctr

CtrCtrCte sssss

1

1....)(

!3)(

!2)()()( 32

10

0...)()(,1)( trtrtr sss

Ramp:

2

3210

)1(1....)(

!3)(

!2)()()(

k

k

k

ttr

Ctr

CtrCtrCte sssss

0...)()(,1)(,)( trtrtrttr ssss

Lecture 10


25

Example 9: Compare the result of example 1 and 7

+

-

c e r

1s

k

ktes

1

1)(

2)1(1)(

k

k

k

ttes

kess

1

1

sse

Step:

Ramp:

Example 1 Example 7

They have similar result but error series show how

the error go to infinity

Lecture 10


26

Example 10: Determine the error for following input

,.......)1(

2,

)1(,

1

1 have we8 exampleBy

32210k

kC

k

kC

kC

+

-

c e r

1s

k ttr 0sin)(

This problem can just be solve with error series!

.......cos)(,sin)(,cos)(,sin)( 0

3

00

2

0000 ttrttrttrttr ssss

tC

CtC

Ctes 0

3

03

010

2

02

0 cos...)!3

(sin...)!2

()(

k=100

20

Lecture 10


27

Example 10: Determine the error for following input

(continue)

+

-

c e r

1s

k

tC

CtC

Ctes 0

3

03

010

2

02

0 cos...)3.2.1

(sin...)2.1

()(

6

4332210 10766.5)101(

600,000194.0

)101(

200,0098.0

)101(

100,0099.0

101

1

CCCC

tttes 2cos)83.2.1

10766.520098.0(2sin)4

2.1

000194.00099.0()(

6

)3.622sin(02213.02cos019592.02sin010288.0)( ttttes

1002sin)( 00 kttr

Lecture 10


28

Example 11: Determine the exact value of error in example 8.

+

-

c e r

1s

k 1002sin)( 00 kttr

)(1

1)( sR

ks

ssE

4

2

101

12

ss

s

jsjsssE

22101)(

4101

2002

)4)(1012(

24

jj

j

)4)(1012(

24

jj

j

jtjt

s ejejte 22 )0047.00098.0()0047.00098.0()(

)4.642sin(0217.02cos0196.02sin0094.0)( ttttes

jtjtt ejejete 22101 )0047.00098.0()0047.00098.0(0196.0)(

Lecture 10


29

Example 12: Determine the response example 5.

+

-

c e r

1s

k1002sin)( 00 kttr

101

100

)(

)(

ssr

sc

[u,t]=gensig('sin',2);

T1=tf(100,[1 101])

lsim(T1,u,t);

hold on;

T2=tf(1,1)

lsim(T2,u,t);

0 1 2 3 4 5 6 7 8 9 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Linear Simulation Results

Time (sec)

Am

plit

ude

Lecture 10


30

Exercises

2005015

200)()

23

ssssMb

6005015

500)()

23

ssssMc

1- Find the error of the following systems to step input.

20005015

2000)()

23

ssssMa

2- Find the error of the following systems to velocity input.

2005015

20050)()

23

sss

ssMb

20005015

200050)()

23

sss

ssMa

CONTROL SYSTEMS


Lecture 11


2

Lecture 9

Transient response of a

prototype second order system

Lecture 11


3

Introducing a prototype second order system.

+

-

c e r

)2(

2

n

n

ss

c r 22

2

2 nn

n

ss

)(2

)(22

2

sRss

sCnn

n

)2()(

22

2

nn

n

ssssC

Step

response

10 if1- :are Poles 2 nn j

2222

22 1)(1)(

1)(

nn

n

nn

n

ss

s

ssC

Lecture 11


4


)1sin(

1)1cos(1)()( 2

2

2 ttetutc nn

tn

)1sin(

1

11)()( 2

2

tetutc n

tn

)1sin()1cos(11

1)()( 222

2tt

etutc nn

tn

)1sin(cos)1cos(sin1

1)()( 22

2tt

etutc nn

tn

1cos

2222

22 1)(1)(

1)(

nn

n

nn

n

ss

s

ssC

Lecture 11


5

Step response

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

13 n8.0,13 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

6.0,8.0,13 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

4.0,6.0,8.0,13 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

2.0,4.0,6.0,8.0,13 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

0,2.0,4.0,6.0,8.0,13 n

Lecture 11


6

Step response

13.0 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

2,13.0 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

3,2,13.0 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

4,3,2,13.0 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

28.6,4,3,2,13.0 n

Lecture 11


7

Specifications of a prototype second order system.

)1sin(

1

11)()( 2

2

tetutc n

tn 1cos

Lecture 11


8

Rise time

:rt The time elapsed up to the instant at which the step

response reaches, for the first time, the value kry. The

constant kr varies from author to author, being usually either

0.9 or 1.

Lecture 11


9

Settling time

:st The time elapsed until the step response enters (without

leaving it afterwards) a specified deviation band, ±, around

the final value. This deviation , is usually defined as a

percentage of y, say 2% to 5%.

Lecture 11


10

Overshoot

:pM The maximum instantaneous amount by which the step

response exceeds its final value.

Lecture 11


11

Peak time

:pt The time at which corresponding to maximum

instantaneous amount by which the step response exceeds its

final value.

Lecture 11


12

If the closed loop system includes an RHP zero

Lecture 11


13

Undershoot

:uM The (absolute value of the) maximum instantaneous

amount by which the step response falls below zero.

Lecture 11


14


+

-

c e r

)2(

2

n

n

ss

c r 22

2

2 nn

n

ss

10 if-1- :are Poles 2 dnnn jj

n-

21 nj

d

n

j

j

21

n

frequency Naturaln

ratio Damping

frequency damped Natural1 2 nd

factor Dampingn

Lecture 11


15

Percent Overshoot

..OP 100%y

M p

How can we find

P.O. ?

Lecture 11


16

Calculation of Percent Overshoot and Peak Time

)1sin(

1

11)()( 2

2

tetutc n

tn 1cos

0)1cos()1sin(1

)( 22

2

tete

t

tcn

t

nn

tn nn

tan

1)1tan(

2

2

tn ntn 21

Lecture 11


17

Peak time

ntn 21

21

n

pt

21

n

nt

Let n=1

n=1

n=3

n=4

n=5

n=2

Lecture 11


18

Calculation of Percent Overshoot

)1sin(

1

11)()( 2

2

tetutc n

tn 1cos

ntn 2121

n

pt

)( pp tcMy

211

e 100%..

y

MOP

p21

100%..

eOP

Lecture 11


19

Percent Overshoot

21100%..

eOP ..OP

0 100%

0.100 73%

0.200 53%

0.300 37%

0.400 25%

0.500 16%

0.707 4.3%

1 0% 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Perc

ent

Overs

hoot

Lecture 11


20

Rise time

rt

How can we find

rise time ?

yktc r)(

ytc 9.0)(Let

n

rt

5.28.0 10

917.24167.01 2

n

rt

Lecture 11


21

Settling time

st

How can we find

settling time ?

ytcy )(

%5for 2.3

n

st %2for 4

n

st 7.00

Lecture 11


22

Exercises

1 – Consider following system.

+

-

c e r

)2(

2

n

n

ss

a) Find the step response of the system for

b) Find the rise time, settling time, overshoot, and percent overshoot.

3.0,56.12 n

+

-

c e r

)2(

2

n

n

ss

2 – Consider following system.

a) Find the step response of the system for

b) Find the rise time, settling time, overshoot, and percent overshoot.

9.0,56.12 n

Lecture 11


23

Exercises (Cont.)

+

-

c e r

)36( ss

k

3- In the following system set k such that the percent overshoot of system be 4.3%

+

-

c e r

))((

100

bsas

4- In the following system set a and b such that the percent overshoot of system be 4.3%

And the steady state error to step input be 0.

5- In the system of problem 1 set k such that

a) The error to ramp input be 0.01

b) The percent overshoot of system be 4.3%

c) The error to ramp input be 0.01 and the percent overshoot of system be 4.3%

Lecture 11


24

Exercises (Cont.)

+

-

c e r

)25( ss

k

6- In the following system

a) For k=200 derive settling time, rise time and percent overshoot.

Confirm your result with step response.

b) For k=1000 derive settling time and percent overshoot.

Confirm your result with step response.

+

-

c e r

)10)(5( sss

k

7- In the following system set the k such that the imaginary poles

have 0.707 damping ratio.

CONTROL SYSTEMS


Lecture 14

Dr. Ali Karimpour May 2013

2

Lecture 10

Root Locus Technique

Lecture 14


3

0)(1 skf

Root locus

Root loci (RL)

Complement root loci (CRL)

Complete root loci

Rk

Rk

Rk

Root locus, shows the position of roots of the following equation

for different values of k

Lecture 14


4

Root locus

Root locus, shows the position of roots of above equation for different

values of k

+

-

c1 e r )(skG

Suppose:

Closed loop system is:

)(1

)()(

skG

skGsM

0)(1 skG

Characteristic equation is:

Lecture 14


5

The Root Locus procedure

0)(1 skf

)(

1

sfk

0)( kRsf

Condition of magnitude

Condition of angle

0)( kRsf

Which points lie on the root loci?

Rsf )(

Lecture 14


6


0)(1 skf

Rule 1: Specify the equation exactly in the following form.

063 23 kskss 06

131

3

2

ss

sk

How many branches in root loci?

It is : ),max( branches of No. nm

Lecture 14


7


0)(1 skf

Rule 2: Specify the poles and zeros of f(s). The root loci lie on

the poles of f(s) for k=0 and lies on the zeros of f(s) for k=±∞

-20 0

0k0k kk

100k

k

k

0k0k

0)20(

11

ssk

Lecture 14


8

The Root Locus procedure 0)(1 skf

Rule 3: Define the real axis section for positive and negative value of k.

-20 0

0k0k kk

100k

k

k

0k0k

0)20(

11

ssk

Lecture 14


9


Rule 4: Find the asymptotes and centered of asymptotes for positive and

negative values of k.

,...2,1,02

0k

,...2,1,0)12(

0k

mnn

m

mnn

m

zp

zp

zp

n

i

n

i

ii

nn

zpp z

1 1

center Asymptotes

-20 0

0k0k kk

100k

k

k

0k0k

0)20(

11

ssk

zp nn asymptotes ofnumber

Lecture 14


10


Rule 5: Find the break point.

pointbreak 0)(

1

s

s

k

sfk

-20 0

0k0k kk

100k

k

k

0k0k

0)20(

11

ssk

ssk 202 0202

s

s

k10s

Lecture 14


11


Rule 6: Find the cross of root locus with imaginary axis by Routh

Hurwitz criteria.

-20 0

0k0k kk

100k

k

k

0k0k

0)20(

11

ssk

0202 kss

0)(s axisimaginary and locusroot ofpoint cross theis 0 k

Lecture 14


12

0)10)(5(

11

sssk

Example 2: Draw the complete root loci of the following

system.

Rule 1: Specify the equation exactly in the standard form.

Clearly it is ok

Lecture 14


13

0)10)(5(

11

sssk


the poles of f(s) for k=0 and lie on the zeros of f(s) for k=±∞

0k

10 5 0

0k 0k

Lecture 14


14

0k

10 5 0

0k 0k

0)10)(5(

11

sssk


0k

0k0k

0kk

k

Lecture 14


15

0)10)(5(

11

sssk

Rule 4: Find the asymptotes and centered of asymptotes .

3

4,

3

2,0

2 0k

3

5,

3

3,

3

)12( 0k

zp

zp

nn

m

nn

m

53

510

center Asymptotes

1 1

zp

n

i

n

i

ii

nn

zpp z

0k

10 5 0

0k 0k

0k0k0k0k kk

Lecture 14


16


0)(1 skf


pointbreak 0)(

1

s

s

k

sfk

Lecture 14


17

0k

10 5 0

0k 0k

0k0k0k0k kk

0)10)(5(

11

sssk


sss

ssssf

k

5015

)10)(5()(

1

23

50303 2

ss

s

k11.2,89.7 s

?

?

Lecture 14


18


0)(1 skf

Rule 6: Find the cross of root locus with imaginary axis by Routh

Hurwitz criteria.

Lecture 14


19

0)10)(5(

11

sssk

Rule 6: Find the cross of root locus with imaginary axis

0)10)(5( ksss

ks

ks

ks

s

0

2

3

015

750

15

501

075015 2 s 07.7js

0k

10 5 0

0k 0k

0k0k0k0k kk

j07.7

j07.7

0750 k 750k

Lecture 14


20

0k

10 5 0

0k 0k

0k0k0k0k kk

07.7

07.7

0)10)(5(

11

sssk

Lecture 14


21

-30 -20 -10 0 10-20

-10

0

10

20Root Locus

Real Axis

Imagin

ary

Axis

-40 -20 0 20-40

-20

0

20

40Root Locus

Real Axis

Imagin

ary

Axis

rlocus(1,[1 15 50 0]); hold on; rlocus(-1,[1 15 50 0])

-30 -20 -10 0 10-20

-10

0

10

20Root Locus

Real Axis

Imagin

ary

Axis

-40 -20 0 20-40

-20

0

20

40Root Locus

Real Axis

Imagin

ary

Axis

0)10)(5(

11

sssk

CONTROL SYSTEMS


Lecture 14


2

Lecture 11

Property and Construction

of Complete Root loci

Lecture 14


3

Example 1: Draw the complete root loci of following system.

Rule 1: Specify the equation exactly in the standard form.

0)1(

)3)((101

2

ss

sks

0)3)((10)1( 2 sksss 0)2910(

)3(101

2

sss

sk

Lecture 14


4


the poles of f(s) for k=0 and lie on the zeros of f(s) for k=±∞

0)2910(

)3(101

2

sss

sk

0k

5

0k

0k

2

3

Example 2:

Lecture 14


5


0)2910(

)3(101

2

sss

sk

0k

5

0k

0k

2

3

0k0k0k

kkk

Example 2:

Lecture 14


6

Rule 4: Find the asymptotes and centered of asymptotes .

2

2,0

2 0k

2

3,

2

)12( 0k

zp

zp

nn

m

nn

m

5.32

)3(10

center Asymptotes

1 1

zp

n

i

n

i

ii

nn

zpp z

0)2910(

)3(101

2

sss

sk

0k

5

0k

0k

2

3

0k0k0k kk k

Example 2:

Lecture 14


7


)3(

)2910(

10

1

)(

1 23

s

sss

sfk

0)3(

)2910()3)(29203(

10

12

232

s

ssssss

s

k

0)2910(

)3(101

2

sss

sk

0k

5

0k

0k

2

3

0k0k0k kk k

47.5s

Example 2:

Lecture 14


8

Rule 6: Find the cross of root locus with imaginary axis

0)2910(

)3(101

2

sss

sk

0k

5

0k

0k

2

3

0k0k0k kk k

030)1029(10 23 kskss

ks

ks

ks

ks

30

0729

3010

10291

0

1

2

3

0 k

?

We need another rule.

Lecture 14


9

Exercises

1- The transfer function of a single-loop control system are given as:

sTsHsss

sG d

1)()3)(1(

10)(

2

.

Construct the root loci of the Zeros of 1+G(s)H(s)=0 for -∞<Td<∞

2- The open loop transfer function of a unity-feedback (negative sign)

system is:

)2)(1()(

1.0

sss

eKsG

s

Constract the complete root loci of the characteristic equation.

Lecture 14


10

Exercises

Construct the complete root loci of the characteristic equation for

Let n=1, n=2 , n=3 and n=4


system is:

nps

KsG

)5()(


system is:

a) Construct the root loci for -∞<K<∞ , with α=5 .

b) Construct the root loci for -∞< α <∞ , with K =5 .

)1(

)3)(()(

2

ss

ssKsG

Lecture 14


11

Exercises


system is:

Construct the root loci for 0<p<∞

))(10(

500)(

psss

psG

6- Consider following system

Construct the root loci for 0<k1<∞

01

131

s

sk

Lecture 14


12

Exercises

7- Consider following system

For k1=0, k1=1 1nd k1=10 construct the root loci for 0<k2<∞ .

0111

3

2

2

ksks

sk

8- Construct the root loci of the closed loop poles of the following

system for 0<a<∞(Midterm spring 2008).

s

1

2

1

s

a

-

+ + R(s) C(s)

-

Lecture 14


13

Exercises

9- Find the root-locus graph for the following system.

The Answer :

Lecture 14


14

Exercises


system is:

Construct the root loci for 0 < p < ∞

)(

10)(

psssG

control theory eee322 - national university theory.pdf · – determine the stability of a...

Documents