EEE2012 Control Notes

EEE2012: Control.

1st and 2nd order systems page 1

Chapter 1. Dynamic Systems and Control

Dynamic systems and their control can be found in any branch of engineering or science; electrical, mechanical, chemical, biological, ecological and financial systems to name a few. In all cases one aims to simplify the system to allow its’ representation in a mathematical form. A set of common terms can be found, regardless of the branch of science; Controlled Variable: is the output of the system that is measured and controlled. Control Signal: is the signal or quantity that is varied by the system controller to affect the value of the output. Plant: The plant is the piece of equipment which performs a particular operation. Any physical object to be controlled is called a plant.

System: is a set of objects/elements that are connected or related to each other in such a way that they create and hence define a unity that performs a certain objective.

Disturbance: A disturbance is a signal that adversely affects the value of the output of a system. This may be internal or external.

Feedback control: is an operation that reduces the difference between the output of a system and the reference input.

Dynamics: Properties describing the behaviour of the system in response to an input.

It is our aim to study, analyse and ultimately to control a system to produce a “satisfactory” performance. We would like the system to be controlled automatically, that is without the involvement or intervention of a human beyond giving an input demand.

EEE2012: Control.


a) System model

The model is derived from the set of Ordinary Differential Equations (ODE) describing the system. The differential equations are obtained by analyzing the physical processes of the system. This applies equally well to electrical, mechanical, thermodynamic, fluid dynamic……. systems. In this course we will concentrate on electrical and mechanical system modeling and analysis. A couple of examples are given.

Electrical system

For the electrical system below:

Applying kirchoff’s law gives the equation;

∫

And as ∫

So the electrical system is defined by a 2nd order differential equation.

V

L R

C

EEE2012: Control.


Mechanical system

For a mass being pushed along a surface;

f(t)

x(t)

friction

The sum of forces = mass x acceleration

∑

Applied force – Force due to friction = Mass x Acceleration

and

Therefore the mechanical system can be expressed by the 2nd order differential

equation;

EEE2012: Control.


Tutorial 1 An electrical machine is modeled by the system diagram show below;

V=voltage, i=current, R=resistance, L=inductance, E=motor EMF, T=torque, ω=angular velocity,

J=inertia, B=friction coefficient, Kt=motor torque constant, Ke=motor EMF constant.

1. Express the electrical half of the system mathematically using Kirchoff’s law.

2. Express the mechanical half of the system by summing the moments of force

(consider torque T, Friction B, inertia J)

3. If the motor back EMF E=keω and the torque produced T=Kti combine both

systems into one single differential equation

1.

2.

3. Rearrange 2:

Sub into 1:

V

i

ω T

R L

B

J E

Mechanical system Electrical system

EEE2012: Control.


b) First order Ordinary Differential Equations

First derivative of x is a function of x and time

),( txfdt

dx

General form of an ODE:

A(t)x’+b(t)x=c(t) Non autonomous (time dependent)

Ax’+bx=c Autonomous (independent of time)

Both forms can be solved using the same methods.

Analytic methods to study first order ODEs:

There is an explicit formula for x(t) (a solution which is obtained by the separation of variables or by the use of an integrating factor) which satisfies

),( txfdt

dx

Separation of variables;

Cattxadtdxadt

dx => INFINITE curves (for all Initial

Conditions (ICs)).

x(t) is called solution of the system which is described by adt

dx .

EEE2012: Control.


Integrating factor:

ktconstk

kdtee

uexeuekxxe ktktktkt ''

udtedtxe ktkt '

ceudteexcudtexe ktktktktkt

or:

tktktkt udteexex

0

110

Assuming that k>0 the first part is called transient and the second is called

steady state solution.

EEE2012: Control.


Tutorial 2a.

Solve

using the separation of variables

SOLUTION:

In standard form:

Let ⁄

⁄

Then we have:

Separating the variables: ∫

∫ and solving gives;

( ) (equ 1)

Substituting in the initial conditions, :

( )

Therefore

( ) (equ 2)

Substituting equation 2 into equation 1 and rearranging gives;

( )

( ) or (

)

Raising to the power e;

(

) so

Making i the subject of the equation;

(

) and simplifying

( )

Substitute back in ⁄

⁄ :

⁄

⁄(

⁄ )

⁄ so;

(

⁄ )

⁄

EEE2012: Control.


Tutorial 2b.

Solve

using an integrating factor

SOLUTION;

In standard form:

Let ⁄

⁄

Using the integrating factor;

∫ where ⁄ , substitute this into the 1st ODE

⁄ (

)

⁄ (

)

as

(

⁄ )

⁄

then

(

⁄ )

⁄ (

)

and integrating both sides;

(

⁄ ) (

) ∫

⁄ then;

(

⁄ )

⁄ +C multiplied by ⁄ on both sides allows simplification to;

⁄ (equ 1)

then substitute in the initial conditions :

⁄

substituting back into equation 1;

(

)

⁄

So

(

⁄ )

⁄

EEE2012: Control.


c) Numerical solution of a first order ODE

Numerical solution of using Simulink;

Rearrange making the highest differential the subject, i.e.

Define x’ and integrate to give x.

Connect blocks with summation and correct gains.

Tutorial 3 Using Simulink to solve each of the following;

1.

2.

3.

4.

For each sketch the output over time and describe the system behaviour (stability

and steady state value).

SOLUTIONS: (1) Stable and steady state value = 1/6 (2) Unstable. (3) Stable and

steady state value = -1/6 (4) Zero input therefore zero output

Scope

1

s

Integrator

k/A

Gain

u/A

Constant

x' x

EEE2012: Control.


d) Response to a sinusoidal input

When a first (or 2nd) order system is exposed to a periodic input, the output will clearly not be constant. It will follow the same periodic nature. If a first order system is exposed to a periodic input;

( )

The solution to the 1st order differential equation takes the form;

ktj

t

k

e

k

yk

1

2

2

tan

2

2tancos

1

1

1

1Re~Re

1

Where the periodic input is still evident

But altered by a magnitude change and a phase shift

EEE2012: Control.


Tutorial 4

Using Simulink show the effect of a periodic input on a first order system;

( )

Compare the numerical solution of the above to a Simulink model of the equation

kt

k

1

2

2tancos

1

1

Let k=1 and y(0)=0.1

The model below will help you

Sine Wave

Scope3

Scope2

Scope1

Scope

Product1Product

eu

Math

Function1

eu

Math

Function

1

s

Integrator1

1

s

Integrator

0.1

Gain3

k

Gain2

-k

Gain1

k

Gain

Clock

total

total

steady state

Transient

0 2 4 6 8 10-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

TransientSteadystate Overall

EEE2012: Control.


0 1 2 3 40

0.02

0.04

0.06

0.08

0.1

e) First order ODE summary

First order systems have an underlying exponential growth or decay

The system is considered stable if the output reaches a steady state.

Dynamics described by time to reach a percentage of steady state and stability or lack of.

Output follows input after transient with change in magnitude (and phase)

Steady state value

↑

50% rise time

EEE2012: Control.


f) Second order Ordinary Differential Equations

Second derivative of x is a function of the first derivative of x, x and time

),,'(2

2

txxfdt

xd

Second order linear ODEs with constant coefficients: uBxAxx '''

u=0 => Homogeneous ODE; I need two “representative solutions”

0''' BxAxx , assume rtex => rtrex ' & rterx 2'' =>

00''' 2 rtrtrt BeAreerBxAxx

02 BArr ; Characteristic equation => Check its roots.

2

42 BAAr

EEE2012: Control.


Roots are real and unequal roots r1 and r2 (A2>4B Overdamped system)

trex 1

1 and trex 2

2 are solutions of the ODE =>

trtreCeCxCxCx 21

212211 . If r1 and r2<0 then 0x .

Example 01303403'4'' 2 rrrrxxx

tt eCeCx 23

1 . Assume that 10 x and 00' x :

10 21 CCx and tt eCeCx 23

13' 030' 21 CCx => C1=-0.5,

C2=3/2 => tt eex 2

35.0 3 :

Scope3

Scope2

Scope1

Scope

eu

Math

Function1

eu

Math

Function

1

s

Integrator1

1

s

Integrator

3/2

Gain5

-1

Gain4

-0.5

Gain3

-3

Gain2

-4

Gain1

-3

Gain

0

Constant

Clock1

Clock

xDxD2x

x1

x2

EEE2012: Control.


Roots are real and equal: r1=r2 (A2=4B Critically damped system)rtex 1

and rttex 2 => rtrt teCeCxCxCx 212211

Example:

A=2, B=1, x(0)=1, x’(0)=0 => c1=c2=1

Scope3

Scope2

Scope1

Scope

Product

eu

Math

Function

1

s

Integrator1

1

s

Integrator

-1

Gain2

2

Gain1

Clock

t

t exp(-t)texp(t-)

0 1 2 3 4 5 6 -0.5

0

0.5

1

1.5 Overall solution

x 2

x 1

EEE2012: Control.


Roots are complex: r=a+bj (A2<4B Under damped system A0)

So )sin()cos( btjbteeeeeex atjbtatbjtattbjart =Re+jIm.

Theorem: If x is a complex solution to a real ODE then Re(x) and Im(x) are the real

solutions of the ODE:

)sin(),cos( 21 btexbtex atat =>

btGebtcbtce

btecbtecxcxcx

atat

atat

cos)sin()cos(

)sin()cos(

21

212211

where

1

21

1

21

1 tan&,

tancosc

c

cc

cG

A=1, B=1, x(0)=1, x’(0)=0 => c1=1, c2=1/sqrt(3)

0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

e-t

te-t

overall

EEE2012: Control.


cos

Trigonometric

Function1

sin

Trigonometric

Function

Scope5

Scope4

Scope3

Scope2

Scope1

Scope

Product

eu

Math

Function

1

s

Integrator1

1

s

Integrator

1/sqrt(3)

Gain5

sqrt(3)/2

Gain4

1

Gain3

-0.5

Gain2

1

Gain1

Clockt

bt sin(bt)

cos(bt)

exp(-at)

0 2 4 6 8 10-1.5

-1

-0.5

0

0.5

1

1.5

C1cos(bt)

C2sin(bt)

C1cos(bt)+C

2sin(bt)

0 2 4 6 8 10-1.5

-1

-0.5

0

0.5

1

1.5

C1cos(bt)+C

2sin(bt)

eat

Overall

EEE2012: Control.


Undamped system (A=0)

BrBeerBxx rtrt 22 0000'' => Imaginary roots (If B<0 then I

would have two equal real roots).

So jbr => btGbtcbtcx cos)sin()cos( 21

A=0, B=1, x(0)=1, x’(0)=0 =>c1=1, c2=0:

cos

Trigonometric

Function1

sin

Trigonometric

Function

Scope5

Scope4

Scope3

Scope2

Scope1

Scope

Product

eu

Math

Function

1

s

Integrator1

1

s

Integrator

0

Gain5

1

Gain4

1

Gain3

0

Gain2

0

Gain1

Clockt

bt sin(bt)

cos(bt)

exp(-at)

0 2 4 6 8 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Overall

EEE2012: Control.


In all previous cases if the real part is positive then the solution will diverge to

infinity and the ODE (and hence the system) is called unstable.

Space of roots

jb

a

Critical or

overdamped

underdamped

jb

a

Stable

Unstable

EEE2012: Control.


g) Natural frequency, damping frequency, damping factor

2nd order systems very important with rich dynamic behaviour

So 2,2 nn BA => 0'2'' 2 xxx nn

is the damping factor and n is the natural frequency of the system.

2

422

2

4222nnnBAA

r

222nnnr

1. Real and unequal 110

2222

nn => Overdamped system

implies that 1 ; 2222,1 nnnr => replace at

trtreCeCx 21

21 .

2. Real and equal 112222 nn => Critically damped system

implies that 1 ; nr => tt nn teCeCx

21 .

3. Complex 110

2222

nn => Underdamped systems implies

1 ; 22222,1 1 nnnnn jjr

dn jnd

21

=>

tGex d

tn cos ; d is called damped

frequency or pseudo-frequency.

4. Imaginary roots 0 and therefore the solution is njr =>

tGx ncos ; so when there is no damping the frequency of the

oscillations = natural frequency. tGx ncos

5. In all the previous cases if 0 then the transient part tends to zero. If

0 then the system will diverge to infinity with or without oscillations.

EEE2012: Control.


Oscillations? Name Components of solution

1 No Overdamped Two exponentials:

tktkee 21 , , 0, 21 kk

1 No Critically

damped

Two exponentials:

ktkt tee , , 0k

1 Yes Underdamped One exponential and one

cosine tekt cos, , 0k

0 Yes Undamped one cosine tcos

If 0 then cases 1-3 are the same but with 0k

dj

n

Critical or

overdamped

underdamped

Stable

Unstable

dj

n

EEE2012: Control.


h) NonHomogeneous (NH) differential equations

uBxAxx '''

u=0 => Homogeneous => x1 & x2.

Assume a particular solution of the nonhomogeneous ODE: xp

o If u(t)=R=cosnt => B

RxP

Then all the solutions of the NHODE are 2211 xcxcxx P

So we have all the previous cases for under/over/un/critically damped

systems plus a constant R/B.

If complementary solution is stable then the particular solution is called

steady state.

Example:

22''' Pxxxx

)sin()cos(22 212211 btcbtcexcxcx at

x(0)=1, x’(0)=0 => c1=-1, c2=-1/sqrt(3)

EEE2012: Control.


cos

Trigonometric

Function1

sin

Trigonometric

Function

Scope5

Scope4

Scope3

Scope2

Scope1

Scope

Product

eu

Math

Function

1

s

Integrator1

1

s

Integrator

-1/sqrt(3)

Gain5

sqrt(3)/2

Gain4

-1

Gain3

-0.5

Gain2

1

Gain12

Constant

Clockt

bt sin(bt)

cos(bt)

exp(-at)

0 2 4 6 8 10-1

-0.5

0

0.5

1

1.5

2

2.5

Overall

Homogeneous solution

Particular solution

EEE2012: Control.


Tutorial 5

State whether each of the following systems are stable/unstable, oscillatory/non-oscillatory.

1.

+3

+2y=1

2.

+2

+2y=1

3.

-5

+6y=1

4.

-2

+104y=1

5.

y=1

Using Simulink numerically solve each and sketch the output. Solutions

1. Stable, non-oscillatory

2. Stable, oscillatory

3. Unstable

4. Unstable, oscillatory

5. Marginally stable

0 1 2 3 4 5 6 70

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Step Response

Time (sec)

Am

plit

ude

0 1 2 3 4 5 6 70

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Step Response

Time (sec)

Am

plit

ude

0 2 4 6 8 10 12 14 16 18 200

1

2

3

x 1025 Step Response

Time (sec)

Am

plit

ude

0 10 20 30 40 50 60-10

-8

-6

-4

-2

0

2

4

6

8x 10

23 Step Response

Time (sec)

Am

plit

ude

10 20 30 40 50 60 70-0.5

0

0.5

1

1.5

2

Step Response

Time (sec)

Am

plit

ude

EEE2012: Control

Transfer functions and the s-plane

Chapter 2. Transfer functions

A Transfer function is the ratio of the Laplace transform of the output over the Laplace transform of the input.

The Laplace Transform

Used only on Linear Time Invariant systems Convert the Differential expression into a Polynomial expression;

{ ( )

( ) } ( ) ( )

The LT is transforming a DE from the time domain (domain is a set of values that describe a function, in that case the variable is the time) to another complex domain (i.e. the variable has a real and imaginary part).

dttfetfLsF st )()()( , js

Can perform this from first principles, but the formula tables are easier.

Differentiation

{ ( )

} ( ) ( )

{ ( )

} ( ) ( )

( )

Where f(0) and df(0)/dt are the initial conditions.

EEE2012: Control


Example 1

A mechanical system with force as the input and position as the output:

xout

(t)xin(t)

Define the system equations;

2

2

dt

xdmfmaf out

Spring

2

2

dt

xdmxxK out

outin

Perform the Laplace Transform;

KmssXsKX

sXmssKXsKX

dt

xdmxxK

outin

outoutin

LT

IC

out

outin

2

2

02

2

And express the transfer function;

Kms

K

sX

sX

in

out

2

This result is compared to the Characteristic Equation (CE) of differential equation;

outout

inout

outin Kxdt

xdmKx

dt

xdmKxKx

2

2

2

2

The homogeneous system is: outout Kx

dt

xdm

2

2

0 and therefore the CE is

02 Kmr i.e. exactly the same as the denominator of TF.

EEE2012: Control


Example 2 An electromechanical system with a voltage input, an intermediate force and position as the output:

Define the system equations. The force of attraction of the mass to the coil is non-linear, therefore cannot be transformed to the Laplace domain, therefore linearise this over a small region of operation;

2

2

x

ikf A => ikf A

Define the remaining linear system equations – both electrical and mechanical and perform the Laplace transform on each set;

)()()()(

)()(1

)(1

2 sXmssBsXskXsIk

RsIsVL

ssI

xmxBkxik

iRvLdt

di

AA

Separate both sets with Output=input – (Current and voltage, forces and acceleration);

)()()()(

)()()(

2 sXmssBsXskXsIk

sVRsIssLI

A

As current is in both the electrical and mechanical equations isolate and substitute to simplify;

)()()()(

)()(

2 sXmssBsXskXRsL

sVk

RsL

sVsI

A

v(t)

R

L

f

Km

x

B

EEE2012: Control


Then simplify the combined equation;

)()( 2 sXkBsmsRsL

sVkA

And finally express as a transfer function;

kBsmsRsL

k

sV

sX A

2)(

)(

Example 3 A motor connected via an ideal gear box to an inertial load with friction.

Define the physical equations;

0

1

2

0

1

20

1

20

000

n

nK

dt

diLRivK

dt

diLRiv

BiKn

nJ

iKT

Tn

nT

BTJ

Ta

aaaamTa

aaaa

aT

aTm

m

Combine and simplify;

0

1

2

0

2

10

n

nK

dt

diLRiv

BiKn

nJ

T

a

aaaa

aT

v(t)

R L

EMF

Tm,θ

m

r1

r2

N1

N2

θo

J2

B

Tm,θ

m

To,θ

o

EEE2012: Control


Now perfume the Laplace transform;

sKssILRsIsV

sBsIKsJs

aaaaa

aLT

n

nKK

n

nKK TT 01

020

2

,2

12

1

21

Isolate the component in both equations;

sKsIsLRsV

sK

BJssI

aaaa

a

01

0

2

2

And simplify;

1

2

2

0

010

2

2

1

KK

BJssLR

sV

s

sKsK

BJssLRsV

aa

a

aaa

Now express the transfer function in its most simple form;

21

2

20

KKBJssLR

K

sV

s

aaa

EEE2012: Control


a) Characteristic Equation and Order

The denominator of Transfer Function is the characteristic Equation of ODE, this is the characteristic Equation (CE) of the system. The order of the ODE is the order of the system. 1. Perform the Laplace transform on the following system equation and rearrange to give the transfer function. What is the system order?

( )

( )

( )

2. What is the order of this system?

( )

( )

( )(

)

4th order

EEE2012: Control


b) Pole location in the s-plane

sD

sNsG )( : the roots of the numerator are called zeros whilst the roots of

the denominator are called poles.

Points on the S plane where G(s) is analytic are called ordinary points.

Points where G(s) is not analytic are called singular points.

Singular points approaching infinity are called poles.

Singular points approaching zero are called…………………..zeros!

Example 1 ( )

( )

plot the real and imaginary part of the poles and zeros on the

s-plane; ( )

( )

( )

( )( )

( )( ) {

0

j

-1-2 3

EEE2012: Control


Example 2

1

1)(

2

sssG

Poles at -0.5+/- 0.866j ;

To find the roots of a polynomial use the command roots( ):

>> num=[1 4];

>> den=[1 3 -6];

>> roots(num)

ans =

-4

>> roots(den)

ans =

-4.3723

1.3723

0

j

-0.5

+0.866j

-0.866j

EEE2012: Control


To find and plot directly the poles/zeros use the command pzmap( ):

>> num=[1 4];

>> den=[1 3 -6];

>> g=tf(num,den);

>> pzmap(g)

>> [p,z]=pzmap(g)

p =

-4.3723

1.3723

z =

-4

Useful commands: zgrid, sgrid.

-5 -4 -3 -2 -1 0 1 2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1 Pole-Zero Map

Real Axis

Imagin

ary

Axis

EEE2012: Control


c) Final value theorem

The final value theorem is used to calculate the steady state value – If

one exists.

( ) where fss is the value of f(t) after infinite time.

Performing the Laplace transform gives;

( ) ( ) ( ) ( )

As a transfer function this is;

( )

( )

( )

Rearranging to express the output, X(s) as a function of the input;

( ) ( )

( )

And applying the final value theorem ( ( )

( )) with

step input of magnitude r; ( )

gives

(

( ))

ONLY VALID WHEN A STEADY STATE EXISTS.

EEE2012: Control


d) Transfer functions in Matlab

Define the numerator

Define the denominator

Use the command tf( numerator, denominator )

Example 1

2

5

ssG

>> num=5;

>> den=[1 2];

>> g=tf(num,den)

Transfer function:

5

-----

s + 2

Example 2

63

42

ss

ssG

>> num=[1 2];

>> num=[1 4];

>> den=[1 3 -6];

>> g=tf(num,den)

Transfer function:

s + 4

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

s^2 + 3 s - 6

EEE2012: Control


e) Block diagrams

System

X(t) Y(t)

Input OutputIn the time domain

G(s)

X(s) Y(s)

And in the s domain

Block Diagram Algebra

1. To sum (or subtract) two signals, we use a summing point:

2. To “distribute” a signal, we use a branch point:

X1(s)

X2(s)

X1(s)+X

2(s) X

1(s)

X2(s)

X1(s)-X

2(s)

X1(s) X

1(s)

X1(s)

EEE2012: Control


3. Series connection:

4. Parallel connection

G1(s)

Y(s)

X(s)

G2(s)

Z(s)

X(s)

X(s)

C(s)

G1(s)+G

2(s)

X(s) C(s)

G1(s)

Y(s)X(s)G

2(s)

Z(s)

G1(s) G

2(s)

Z(s)X(s)

EEE2012: Control


Block diagrams in Matlab:

>> num1=1;

>> den1=[1 1];

>> num2=[1 0];

>> den2=[1 5 6];

>> g1=tf(num1,den1)

Transfer function:

1

-----

s + 1

>> g2=tf(num2,den2)

Transfer function:

s

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

s^2 + 5 s + 6

>> g=series(g1,g2)

Transfer function:

s

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

s^3 + 6 s^2 + 11 s + 6

>> g=parallel(g1,g2)

Transfer function:

2 s^2 + 6 s + 6

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

s^3 + 6 s^2 + 11 s + 6

EEE2012: Control


Tutorial

1. Find the order and plot the poles and zeros of;

( )

, ( )

( ) ( )

( )( ) ( )

2. If Y(s)=R(s)G(s), for the stable systems in Q2 find the final value Y

ss in

response to a unit step input ( )

3. Using Matlab find the transfer function of the following;

a) G1

and G2

are in parallel connection

b) G1

and G2

are in series connection

c) G1

and G2

are in series and this is in parallel with G2

connection

d) G1

and G2

are in parallel and this is in series with G1

connection

Where ( )

( )

4. Simplify the system below:

G2(s)X(s)

G1(s) G3(s)

G4(s)

G5(s)

Y(s)

EEE2012: Control


Q1. G1, 1st order, No zeros, Pole@s=-4 G2, 2nd order, No zeros, Pole@4±j8 G3, 3rd order, zero@s=-4, pole@s= 2±j6 G4, 2nd order, no zeros, pole @ s=±j6 Q2.

Q3.

>> ga=parallel(g1,g2) Transfer function: 2 s^2 + 9 s + 11 --------------------- s^3 + 4 s^2 + 9 s + 6

>> gb=series(g1,g2) Transfer function: s + 5 --------------------- s^3 + 4 s^2 + 9 s + 6

>> gc=parallel(gb,g2) Transfer function: s^4 + 10 s^3 + 37 s^2 + 72 s + 60 ----------------------------------------- s^5 + 7 s^4 + 27 s^3 + 57 s^2 + 72 s + 36

>> gd=series(ga,g1) Transfer function: 2 s^2 + 9 s + 11 ------------------------------- s^4 + 5 s^3 + 13 s^2 + 15 s + 6

Q4. Done in class

( ( )

) ( )

(

)

EEE 2002: Control

Chapter 3 – Time domain characteristics 41/132

Chapter 3. Time domain characteristics

You have derived the mathematical model of the system. Now apply various methods to analyse the performance of the system. All systems have a forcing signal which we want to control, four main types of input are given below, all are simple functions of time (and s)

;

The Dirac function The step or the pulse function

The Ramp function The parabolic function

“Once a control system is designed on the basis of test signals, the performance of the system in response to actual inputs is generally satisfactory” [Ogata p.220] Gradually changing function of time → Ramp Sudden disturbance → Step Shock → Impulse Acceleration → Parabolic Allows for comparison of all systems on the same basis

0

r(t)

t

ttr

1sR

0

r(t)

t

Atr

s

AsR

0

r(t)

t

Attr

2s

AsR

0

r(t)

t

2Attr

3s

AsR

0

r(t)

t

ttr

1sR

0

r(t)

t

Atr

s

AsR

0

r(t)

t

Attr

2s

AsR

0

r(t)

t

2Attr

3s

AsR

EEE 2002: Control


a) First order systems

Remember: baxx ' ;

Homogeneous equation: 0' axx

Characteristic equation; 0 am A solution of the HE: ateCx 1 hence this is a stable solution if a<0 (or the

pole is at the left hand side)

The same in the s-plane: LsRsV

sI

1

s

K

sV

sI

1, K=1/R and τ=L/R. =>

1

s

KsVsI

Step response: 1

1

ss

VKsI

=>

t

eR

Vti 1

Steady state value;

R

V

R

Ve

R

Vi

t

tF

01lim

And the same in the s domain;

R

VVK

s

K

s

VsI

sss

1lim

0

V(s) I(s)

LsR

10

j

LR

EEE 2002: Control


To find the step response directly in Matlab (apart from using Simulink), use

the command:

step( , ):

>> num=1;

>> den=[1 1];

>> step(num,den)

Step Response

Time (sec)

Am

plit

ude

0 T 2T 3T 4T 5T 6T0

0.1(V/R)

0.2(V/R)

0.3(V/R)

0.4(V/R)

0.5(V/R)

0.6(V/R)

0.7(V/R)

0.8(V/R)

0.9(V/R)

1 (V/R)

Step Response

Time (sec)

Am

plit

ude

0 T 2T 3T 4T 5T 6T0

0.1(V/R)

0.2(V/R)

0.3(V/R)

0.4(V/R)

0.5(V/R)

0.6(V/R)

0.7(V/R)

0.8(V/R)

0.9(V/R)

1 (V/R)

0.632 (V/R)

EEE 2002: Control


Or to extend the plot to 15 seconds; >> step(num,den,15)

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1Step Response

Time (sec)

Am

plitu

de

0 5 10 150

0.2

0.4

0.6

0.8

1Step Response

Time (sec)

Am

plitu

de

EEE 2002: Control


Ramp response;

1

12

ss

VKsC

=>

t

etR

Vtc

The final value of the error is;

R

Ve

R

Vive

t

xF

1lim

Step Response

Time (sec)

Am

plit

ude

0 T 2T 3T 4T 5T 6T 0 T 2T 3T0

0.1(V/R)

0.2(V/R)

0.3(V/R)

0.4(V/R)

0.5(V/R)

0.6(V/R)

0.7(V/R)

0.8(V/R)

0.9(V/R)

1 (V/R) VT----- R

EEE 2002: Control


Tutorial

A first order system is given by ( )

( )

1. Write the transfer function as ( )

( )

and hence find the value

of the DC gain ( ) and of the system’s time constant ( ). 2. For k=1 and a=5 repeat the previous step. 3. Plot a pole zero map for this transfer function, and based on this

predict the dynamic behaviour of the system in response to a (unit) step input.

4. Using the final value theorem calculate the steady state value Yss in response to a (unit) step input.

5. Confirm your findings using Matlab or Simulink 6. Repeat the previous steps and comment on your results for:

1. k=0.1 and k=10 (keep a fixed to 5). 2. a=0.5 and a=50 (keep k fixed to 1).

7. Using the appropriate Matlab commands find and plot the error for the system in question 2. What is the steady state error?

EEE 2002: Control


b) Second order systems

Physical equations;

xout(t)xin(t)

Friction

LT

IC

outoutoutin

dt

xdm

dt

dxBxxK

02

2

In the s domain; sXmssBsXsKXsKX outoutoutin

2

sXmsBsKsKX outin

2

And simplified to give a transfer function;

KBsms

K

sX

sX

in

out

2

mKs

mBs

mK

sX

sX

in

out

2

Expressed in terms of damping factor and natural frequency n; 22

2

2 nn

n

sssR

sC

,

mK

mB

n 2,2

The roots of the characteristic equation are given by;

12 nns

EEE 2002: Control


Case 1 >1:

Then the system has two negative real roots and is called overdamped:

212

21

121

s

e

s

etc

tsts

n

1 => tn

etc

12

1

An overdamped system response is similar in form to a slow first order system (long decaying exponential).

0

j

s1

s2

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Step Response

Time (sec)

Am

plit

ude

Overdamped system

EEE 2002: Control


Case 2 =1:

The system has two equal real roots at ns :

0

j

n

tetc n

tn 11 :

Step Response

Time (sec)

Am

plit

ude

0 2 4 6 8 10 120

0.5

1

1.5

Overdamped system

Critically damped system

nnn ss 12

EEE 2002: Control


Case 3 0<<1:

21 nn js or dn js

The line between the origin and the pole is:

222

ndd nnn 2222 1

And the angle is;

1coscos

n

n

In the time domain the response expressed in terms of damping factor and

natural frequency;

21

2

1tansin

11 t

etc d

tn

0

j

n

dj

dj

EEE 2002: Control


The envelope that will be created from the exponential terms is:

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

1

Sinusoidal term

Exponential term

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

1

EEE 2002: Control


And their product:

And by adding the constant factor:

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

1

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

-0.5

0

0.5

1

1.5

2

EEE 2002: Control


The three stable second order system responses are compared below;

And with reducing further toward zero the damping reduces and the system oscillates for longer;

Step Response

Time (sec)

Am

plit

ude

0 2 4 6 8 10 120

0.5

1

1.5

Overdamped system

Critically damped systemUnderdamped system

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8Step Response

Time (sec)

Am

plit

ude

EEE 2002: Control


Case 4 =0:

Note: The system is called marginally stable because the solutions do not diverge to infinity. Hence if the previous four cases are combined to one graph:

0

j

nj

nj

0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plit

ude

Marginally stable

Underdamped

Critically damped

Overdamped

EEE 2002: Control


Case 5 <0 (negative damping i.e. positive feedback):

21

2

1tansin

11 t

etc d

tn

Since 0 then angle defined in the s-plane

(

1coscos

n

n ) has to be greater than 90o:

By combining the previous s-planes we have:

0 1 2 3 4 5-10

-5

0

5

10

0

j

n

dj

dj

EEE 2002: Control


0

j

n

dj

dj

nj

nj

1,0cos

1

0

1

0,1cos

1

1

EEE 2002: Control


0

j

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

0 2 4 6 8 100

0.5

1

1.5

2

0 0.25 0.5 0.75 1 -4

-3

-2

-1

0

1

2x 10

8

0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

0 0.2 0.4 0.6 0.8 10

20

40

60

80

100

120

140

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

EEE 2002: Control


A general response in the time domain is:

This response can be described by a number of parameters;

Time that the system needs to reach half of its final value:

Rise time (10%-90% or 5%-95% or 0%-100%) d

rt

Peak time: d

pt

Maximum overshoot:

"1

eMp

Settling time: n

st

3%5 and

n

st

4%2

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plit

ud

e

EEE 2002: Control


0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plit

ude

td

tr

tp

Mp

1.05

0.95

ts

EEE 2002: Control


c) Time domain characteristics in Matlab

>> num=1;

>> den=[1 4 13];

>> g=tf(num,den);

>> step(g)

Right click on figure and choose the characteristics that you would like to see.

0 0.5 1 1.5 2 2.5 30

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09Step Response

Time (sec)

Am

plitu

de

EEE 2002: Control


EEE 2002: Control


d) Extra poles and zeros

Systems with an order higher than 2; General form of a Transfer function;

nn

nn

mm

m

asasasa

bsbsb

sR

sC

1

1

10

10

...

...

)(

)(

For a step input:

r

k kkk

kkkkkkq

j j

j

ss

csb

ps

a

ssC

122

1 2

11

nrq 2

i.e. combination of first and second order systems.

Example – 3rd order system

Express the TF as a product of 1st and 2nd order systems;

edssfscbsass

223

11=>

Equate the like terms;

edssfscbsass 223

fesfdesfdscbsass 2323

Form a set of simultaneous equations and solve;

fec

fdeb

fda

11

The response of a higher order system is the sum of exponential and damped sinusoidal curves.

Assuming that all poles are at the left hand side then the final value of the output is “1” since all exponential terms will converge to 0.

Let’s assume that some poles have real parts that are far away from the imaginary axis=>

EEE 2002: Control


21

2

1tansin

11 t

etc d

tn

=> 0 tne

Overall performance is characterised by the isolated (far away from zeros) poles that are close to the imaginary axis.

If we have only one pole (or a pair for complex roots) that is closed to the real axis then we say that this pole (or pair of poles) is (are) the DOMINANT pole(s) for the system.

A simple rule is that the dominant poles must be at least five to ten times closer to the imaginary axis than the other ones.

r

k kkk

kkkkkkq

j j

j

ss

csb

ps

a

ssC

122

1 2

11

r

k

kk

t

k

r

k

kk

t

k

q

j

tp

j tectebeatc kkkkj

1

2

1

2

1

1sin1cos1

The values of b (numerator coefficients) determine the amplitude of the oscillations of the system.

EEE 2002: Control


Tutorial

1. A second order system is given by cbsas

ksG

2; write the

transfer function as: 22 2

'

nnss

ksG

.

2. For k’=1 and sradn /5,5.0 define the transfer function.

3. Enter this into Matlab and use the command “damp” to confirm the damping factor and natural frequency.

4. Based on the previous answer predict the behavior of the system for a (unit) step input.

5. Find the step response of that system using Matlab AND Simulink and hence crosscheck your previous answer. Also plot the system’s error.

6. Use the formulae to calculate overshoot, MP, time to peak, TP, settling time to within 5%, ts5%, and confirm these findings using your Matlab plot.

7. Using the final value theorem, calculate the steady state value in response to a unit step input. Confirm this finding with your Matlab plot.

8. The mechanical system in Figure 1 is described by a differential equation expressing the sum of forces;

a) Where F is the applied force, k is the spring constant, is the

coefficient of friction and m is the mass. are position, velocity and acceleration respectively.

b) Assuming zero initial conditions calculate the Laplace transform of this

system, expressing your result as a transfer function relating output X and input F.

m

←

k F→

FIGURE 1.

EEE 2002: Control


c) Express the damping factor ζ and the natural frequency ωn as a function of m, and k.

d) Assuming the mass m=0.1kg and the friction coefficient =0.1 what spring constant k will result in an under-damped system?

e) For a spring constant k ten times greater than your calculated value in part iii, calculate the position of the system poles and the damping factor. Can you confirm that you have invoked a stable, under-damped system?

9. A plant has the transfer function ( )

( )( )( ) and forms

part of an open loop system where the output Y(s)=X(s)G(s). X(s) is the system input. a) Calculate the position of any poles and zeros for this system. Plot

these on an s-plane diagram. b) Considering the location of the poles and zeros, what do you think will

be the system response to a unit step input? c) Calculate the final value Yss(s) in response to a step input. d) Use matlab to plot the system step response and confirm your answer

to 4. e) Approximate this system by considering the position of any dominant

poles. f) Take account of the final value of your approximated system and

modifying the system gain (numerator value). g) Hold your previous step response and superimpose your simplified

system step response. Has your simplification worked?

EEE 2002: Control

Chapter 4 –Closed loop systems 66/132

Chapter 4. Closed loop systems

Assume the following electrical system:

R L

U

Change or control V so the output current i is equal to the demanded current i*. Assume R=2 Ohms, L=1 H and i*=2 A:

Ri* U

RLs

1 i

Expected response:

i

t

2A

Assume that you want a faster system (same steady state):

V

V

EEE 2002: Control


t

2A

L & R = constant hence we can ONLY change U but then iss i* Ideally we would like:

t

2A

U

=i* x R

I

So V=f(i*,i)

As ii* Vi*R

And as i-i*0 Vi*R The difference i-i* is called ERROR (e)

V

EEE 2002: Control


So we would like:

Ri* U i

e

+

+RLs

1

Where e0 ii*

Ri* U i

e

+

+

+ -

RLs

1

Crosscheck:

RLsG

1

iGGRii

RiiiRieV

VGi1*

***

11

1

1

1

*1*1

sR

LG

RG

i

iGRiGi

Which implies that 1R

Lnew

As new> the system is now faster. At the steady state:

2

11

12lim

11

1*

0

s

RLs

sis

RL

iis

ss .

Hence the input to the system sees the output. This is called Feedback and is the cornerstone of control systems. A system with feedback is called a CLOSED LOOP system.

V

V

EEE 2002: Control


a) General closed loop systems

A more general form for a closed loop system:

HG

G

R

C

'1

'

where GGcG ' Gc= Transfer function of the controller G = TF of plant (or the open loop transfer function)

UC

GGc+

-

H

Feedback

Feedback

TF

Controller System

RE

Your task is to design the controller Gc and feedback H (if applicable). Assume H=1 and Gc=K (a constant value gain).

KR

K

KRLS

K

ssC

KRLS

K

KG

KG

R

C

sss

2lim

1 0

If K>>R then Css=2, also:

1

SKR

L

KRK

KRLS

K

R

C

So KRL

new

As new then we have a faster system.

A closed loop can also achieve a predefined steady state:

OLFT: 2

2

sG

Open loop response (desired current=1A): >> num=2;

>> den=[1 2];

>> g=tf(num,den);

EEE 2002: Control


>> step(g)

CLTF: K

KCss

Ks

K

R

CGCL

22

Closed loop response (k=10): >> numc=10;

>> denc=1;

>> gc=tf(numc,denc);

>> gol=series(g,gc);

>> h=tf(1,1);

>> gcl=feedback(gol,h);

>> hold on

>> step(gcl)

0 1 2 30

0.2

0.4

0.6

0.8

1Step Response

Time (sec)

Am

plitu

de

EEE 2002: Control


And for K=100:

Step Response

Time (sec)

Am

plitu

de

0 1 2 30

0.2

0.4

0.6

0.8

1

Step Response

Time (sec)

Am

plitu

de

0 1 2 30

0.2

0.4

0.6

0.8

1

EEE 2002: Control


Ri* U i

e

+

+

+ -

RLs

1

Tutorial

1. The RL circuit described during the lectures is to be controlled by

Simulate the system for a unit step and ramp response and crosscheck the

results using the final value theorem. Monitor the value of the error and

the control signal U.

2. Change the control strategy to:

with ksGsHRLs

sG c

,1,1

For k=1, 10 and 100 simulate the system for a unit step and ramp response

and crosscheck the results using the final value theorem. Monitor the value

of the error and the control signal U.

UC

GGc+

-

H

Feedback

Feedback

TF

Controller System

RE

EEE 2002: Control


b) Higher order systems

Until now we have seen only 1st order systems with feedback. In these systems, the feedback and the controller influence the steady state and the time constant. The decrease of the time constant means that we moved the pole further into “minus infinity” area. Hence we changed the s-plane of the system. What is it going to happen if we use a 2nd order system in a feedback system?

21

1

ssG

>> num=1;

>> den=conv([1 1],[1 2]);

>> g=tf(num,den);

>> step(g)

>> hold

Current plot held

>> numc=10;

>> denc=1;

>> gc=tf(numc,denc);

>> gol=series(g,gc);

>> h=tf(1,1);

>> gcl=feedback(gol,h);

A 2nd order system with feedback becomes faster, but note that oscillations are introduced.

kss

k

kss

kGCL

2321 2

CE: 0232 kss but the general from: 02 22 nnss

Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

EEE 2002: Control


so 32 n and kn 22

for k=10 12122 nn and hence 433.03122 :

>> [wn,z]=damp(gcl)

wn =

3.46410161513775

3.46410161513775

z =

0.43301270189222

0.43301270189222

So the feedback and the controller can completely change the location of the poles in the s-plane.

Example: 321

sss

KG

>> num=10;

>> den=conv(conv([1 1],[1 2]),[1 3]);

>> g=tf(num,den);

>> step(g)

>> hold

Current plot held

>> num=50;

>> g=tf(num,den);

>> step(g)

The open loop response for various gains is:

EEE 2002: Control


The open loop system will be stable for all values of K since they do not influence the poles of the system. The response of closed loop system for K=1, 10, 100 is:

Hence the feedback may introduce instability.

0 1 2 3 4 5 60

1

2

3

4

5

6

7

8

9Step Response

Time (sec)

Am

plitu

de

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-4

-3

-2

-1

0

1

2

3

4

5

Step Response

Time (sec)

Am

plit

ude

EEE 2002: Control


To understand why we have these changes solve the characteristic equation. We will extensively study this at the “Root Locus” chapter. Properties of feedback systems:

Minimise steady state error.

Faster system.

Less sensitive to system uncertainties.

Introduce instability (even for negative feedback).

Expensive (we need to feedback the signal, i.e. use a sensor).

EEE 2002: Control


c) Systems classification

Types We already have seen some of these categories:

Electrical

Mechanical

Hydraulic

1st order

2nd order

Higher order

Overdamped

Underdamped

Critically damped

Stable

Unstable

Marginally stable and others Another way to classify control systems is to use the number of the poles at the origin of the OPEN LOOP system:

m

N

nOL

cscscss

bsbsbsKsG

21

21

This implies a pole multiplicity N at the origin. This system is called type N.

43

10

ss

ssGOL => Type 0

43

51

sss

sssGOL => Type 1

43

12

sss

sGOL => Type 2

1005

122

sss

sGOL => Type 2

1

1105

ss

sGOL => Type 5

Note: THE TYPE OF THE SYSTEM IS DIFFERENT FROM THE ORDER OF THE SYSTEM!!!

EEE 2002: Control


Error constants

Consider a unity feedback system with an open loop transfer function G(s), hence:

sG

sG

sR

sCsGCL

1

sCsRsE , this implies that the transfer function between the error

and the input is:

sGsG

sG

sR

sC

sR

sE

1

1

111

And therefore sG

sRsE

1

1

From the final value theorem:

sG

ssRE

sss

1lim

0

For a unit step input:

01

1

1

1

lim0 GsG

ss

Es

ss

We define as static position error constant as Kp: 0lim0

GsGKs

p

Hence p

ssK

E

1

1

For a unit ramp:

ssGssGssG

ss

Esss

ss

1lim

1lim

1

1

lim00

2

0

We define as static velocity error constant as Ku: ssGKs

u0

lim

And therefore u

ssK

E1

For a unit parabolic input:

sGssGsssG

ss

Esss

ss 20220

3

0

1lim

1lim

1

1

lim

We define as static acceleration error constant as Ka: s GsKs

a

2

0lim

EEE 2002: Control


And therefore a

ssK

E1

Types and error constants

Another way to classify control systems is by their static error with a combination of their type. Error and unit step 1tr :

Type 0: n

mOL

cscscs

bsbsbsKsG

21

21 so

'limlim21

21

00K

cscscs

bsbsbsKsGK

n

m

ssp

So p

ssK

E

1

1

Type N>0: m

N

nOL

cscscss

bsbsbsKsG

21

21

m

N

n

ssp

cscscss

bsbsbsKsGK

21

21

00limlim

So 0ssE

Error and unit ramp ttr :

Type 0: m

nOL

cscscss

bsbsbsKsG

21

21 so

0limlim21

21

00

m

n

ssu

cscscs

bsbsbsKsssGK

So ssE

Type 1: m

nOL

cscscss

bsbsbsKsG

21

21

'limlim21

21

00K

cscscss

bsbsbsKsssGK

m

n

ssu

So v

ssK

E1

Type N>1: m

N

nOL

cscscss

bsbsbsKsG

21

21

m

N

n

ssa

cscscss

bsbsbsKsssGK

21

21

00limlim

So 0ssE

EEE 2002: Control


Error and unit parabolic input 2

2

1ttr :

Similarly we find that: Type 0: ssE

Type 1: ssE

Type 2: a

ssK

E1

Type N>2: 0ssE

From the above the following table of steady state error can be derived:

Input STEP: r(t)=1

RAMP: r(t)=t

Parabola: r(t)= ½

t2

Type 0 pK1

1

Type 1 0 vK

1

Type 2 0 0 aK

1

Hence if we want to decrease the error we have to increase the type of the system.

EEE 2002: Control


Tutorial 1. Find the Closed Loop (CL) (with unity feedback, and GC(s)=1) Transfer

Function (TF) when the Open Loop (OL) TF is:

a. 1

1

sGA

b. 1

12

ss

GB

c. 11

12

sss

GC

2. In each case find the CL pole and zero location.

3. Using the final value theorem find the steady state value of the CL

response for a unit step and ramp input.

4. Find the error TF in each case.

5. Using the final value theorem find the steady state value of the error for

a unit step and ramp input.

6. Using Simulink crosscheck your previous answer.

7. State the order and type of each system.

8. Find the error constants Kp and Kv.

9. For each closed loop system; A,B,C double the gain GC(s) and assess the

response to a unit step input for each system.

10. Using Matlab plot the pole-zero map of the CLTF and crosscheck this

with your findings in 9.

11. Repeat 9 and 10 for another doubling in gain, which systems can be

forced into instability?

12. Find the gain for which the 3rd order system is marginally stable.

EEE 2002: Control

Chapter 5 –PID Control 82/132

Chapter 5. PID control

Proportional Integral Differential. A PID Controller modifies the error signal fed to the plant. From previous table, if the type is 0 then we have a steady state error for a step input. Also by increasing K, we increase the oscillations and we may cause instability to the system. Hence we need another solution.

Consider the system: 321

sss

KsG for K=10:

To increase the type of the OLTF (which is GCxG) we add an integrator:

sKKsG ipc

1)( =>

321

1)(

ssss

KsKsG

ip

OL

This is the so-called PI controller (proportional + integrating).

0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

error

output

EEE 2002: Control


Further increase of Kp, Ki:

1

s +6s +11s+63 2

Transfer Fcn

StepScope

10

Kp

1

Ki

1

s

Integrator

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Ki=0

Ki=2

Ki=6

Proportional gain

Integrator gain

EEE 2002: Control


0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Kp=10, K

i=6

Kp=20, K

i=10

EEE 2002: Control


Check the derivative of error, i.e. the rate of change of e:

Maximum value of De just before e=0. So De can control the oscillations:

sKs

KKsG dipc 1

)( => 321

1)(

2

ssss

KsKKssG

ipd

OL

0 2 4 6 8 10-1.5

-1

-0.5

0

0.5

1

De

e

10

ki

1

kd

1

s +6s +11s+63 2

Transfer FcnStepScope

20

Kp

1/s

Integrator

du/dt

Derivative

EEE 2002: Control


Since I have no oscillations I can increase Kp a little bit more to make the system faster:

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Kd=0

Kd=5

Kd=10

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Kp=20

Kp=50

EEE 2002: Control


0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

0 2 4 6 8 10-20

-10

0

10

20

30

40

50OutputP output

0 2 4 6 8 100

0.5

1

1.5

0 2 4 6 8 10-20

-10

0

10OutputD output

EEE 2002: Control


Another way to write the PID controller:

s

K

K

sK

KKsK

s

KKsG

p

d

p

ipd

ipCL

11)(

sT

sTKsG d

i

pCL

11)(

0 2 4 6 8 100

0.5

1

1.5

0 2 4 6 8 100

2

4

6

OutputI Output

EEE 2002: Control


a) Tuning of PID controllers

Trial and error.

Ziegler Nichols I

Ziegler Nichols II

Root locus

Frequency response

Other advanced control methods

Trial and error:

P: Faster system, in some cases reduces the error (can cause instability). I: Reduces the steady state error, increases the number of oscillations. D: Reduces the oscillations.

Ziegler Nichols I

Assume a system with no delays (we do not study these systems) and with no-complex conjugate poles.

u(t) c(t)Plant

Its open loop step response may look like (obtained experimentally of from simulations):

EEE 2002: Control


This can be modelled as:

1)(

)(

Ts

Ke

sU

sC Ls

Based on that we have the following table:

Type of controller

Kp Ti Td

P LT 0

PI LT9.0

3.0L 0

PID LT2.1 L2 L5.0

Ziegler Nichols II

1. Initially assume Ki=Kd=0. 2. Increase Kp until the system is marginally stable.

3. Record the Kp=Kcr and the frequency of oscillations: 4. Set the controller gains as follows based on the critical gain and

frequency of oscillation;

Type of controller

Kp Ti Td

P crK5.0 0

0

t,s

c(t

)

Pcr

EEE 2002: Control


PI crK45.0 crP

2.1

1 0

PID crK6.0 crP5.0

crP125.0

These methods aim at achieving an overshoot of 25%.

EEE 2002: Control


Zeigler Nichols 2 Class example

A PID controller ( )

is connected in series with a plant

whose transfer function is

( )

( )( ) and a unity feedback loop is imposed.

Using the Zeigler Nichols 2 method determine the values of Kp, Ti and Td

ZN2 STEPS

1. Set ( ) ( )

2. Find closed loop transfer function;

( )

( )

( )( )

( )

( )( )

3. Find the critical gain at which the system is marginally stable;

Marginally stable when poles lie on the imaginary axis so

substitute into C.E;

4. Separate variables in

√

EEE 2002: Control


=

As √

Critical gain for marginal stability Kcr=120

5. Substitute KCR into ZN2 table for PID controller

6. Proportional gain:

KP=0.6 x KCR = 0.6 x 120 = 72

7. Period of natural oscillation

So

√

8. Integral and differential gain

Ti=0.5 x PCR = 0.8112 seconds

Td=0.125 x PCR = 0.2028 seconds

EEE 2002: Control


Zeigler Nichols 2 Tutorial

A plant is described by the transfer function;

( )

( )( )

This system is to be controlled by a PID controller and a unity feedback loop implemented. Tune the PID controller using the Zeigler Nichols II method and state your calculated values of Kp, Ti and Td.

EEE 2002: Control


b) Root locus method

With the root locus method we specifically target a pole location at the s-plane, i.e. we target a desired damping factor, natural and damped frequency.

Root locus class example

The OLTF is 3411

1)(

2

sssG use a PI controller.

34113411

1)(

22

sss

KsK

sss

KKsG

ipipOL

ip

ip

CLKsKsss

KsKsG

3411)(

2

This is a 3rd order system = 2nd order x 1st order: CE: 222 23411 nnip ssasKsKsss

222323 223411 nnnnip asasasKKsss

2

2234

211

ni

nnp

n

aK

aK

a

Assume that the design specs are:

5.0

6

n

So:

180

100

5

36

36634

611

i

p

i

p

K

K

a

aK

aK

a

Matlab: CL response, no controller: >> num=1;

>> den=[1 11 -34];

>> g=tf(num,den);

>> gcl=feedback(g,1);

>> step(gcl)

EEE 2002: Control


CL, response, PI controller: >> kp=100;

>> ki=180;

>> gc=tf([kp ki],[1 0]);

>> gol=series(gc,g);

>> gcl=feedback(gol,1);

>> step(gcl)

Homework:

Find the PID gains : the CLTF of 166

1)(

2

sssG has

5.0

6

n and a real pole

at a=-5. Solution: Kp=50, Ki=180, Kd=5.

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35Step Response

Time (sec)

Am

plitu

de

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.5

1

1.5

2Step Response

Time (sec)

Am

plitu

de

EEE 2002: Control


Root Locus Tutorial

A plant is controlled by a proportional controller with gain K and a unity

gain feedback loop. The plant transfer function is ( )

( )( )

If a ramp input where r(t)=5t is used, what is the steady state error with a proportional gain K=1.5? What value of gain K will ensure an error ESS<0.1?

15.01

sss

KsGOL , Unity feedback and input: r(t)=5t

a) If K=1.5, find the steady state error b) The system must have steady state error, Ess<0.1 find the value of K

C(s)

C(s)

15.01 sss

K

2

5

s sR

Ksss

K

sR

sC

15.01

Ksss

K

ssC

15.01

52

sCsRsE

Ksss

K

ssE

15.011

52

Ksss

ss

ssE

15.01

15.015

Ksss

ss

ssE

sss

15.01

15.015lim

0

KEss

5

a) ...33.35.1

5ssE

b) 1.0ssE 1.05

K50K

EEE 2012: Control

Chapter 6 – The Root Locus Method of 132

Chapter 6. The Root Locus Method

“The Root Locus is the path (locus) of the roots of the characteristic equation traced out in the s-plane as a system parameter is changed” The dynamic response depends on system’s and controller’s parameters , therefore the closed loop pole location depends on system’s and controller’s parameters. The system designer will find the controller gain(s) such as we have a satisfactory performance, this equates to finding the controller gain(s) such as we have a satisfactory pole location. To do that we need a good understating of how poles move in the s-plane as the controller parameters are changed.

EEE 2012: Control


a) Root locus on a 1st order system

First order system with a proportional controller with gain k and a unity gain feedback loop;

( )

Gc(S)=K H(s)=1

Forward path: ( )

Closed loop: ( )

The pole location s + 1 + k = 0

then s = -1 -k As k increases the pole moves further in the negative x direction

Properties:

Root locus starts from the open loop pole location.

Ends at .

This implies that as K is increased the system becomes faster (we

already know this from before)

-10

-8

-6

-4

-2

0

2

4

6

8

10

-10 -8 -6 -4 -2 0 2 4 6 8 10

jω

σ

X

EEE 2012: Control


Example 2 Second order system with a proportional controller with gain k and a unity gain feedback loop;

( )

Gc(S)=K H(s)=1

Forward path: ( )

Closed loop: ( )

√ ( )

Two real poles until ( ( )) i.e. k > 0.25 After this we have

√ hence a complex pole

Properties:

Starts from the OL pole location.

Collides at a point.

Moves on a straight line perpendicular to the real axis.

Symmetrical with respect to the real axis

Ends at -1.5 ±

-6

-4

-2

0

2

4

6

-2.5 -2 -1.5 -1 -0.5 0 0.5

jω

σ

X X

X X

EEE 2012: Control


Root locus tutorial

For each the following open loop systems;

( )

( )

( ) ( )

a) Input the open loop transfer function into Matlab

hint: >>g1=tf(1,[1 2 2])

b) Use the Matlab command >>rlocus(g1) to plot the open loop poles,

zeros and root locus for each.

c) Sketch each root locus and discuss the general patterns found, i.e

direction of travel of roots, asymptotes, effect of zeros, collision

points….

EEE 2012: Control


G1(s)

Second order system, unity feedback loop, proportional controller gain K;

kss

kG

ss

kG CLOL

2222 22 so pole location: 0222 kss :

Properties:


Starts already with complex poles so no collision.

Moves on a straight line perpendicular to the real axis.


Ends at 1 .

-1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0-5

0

5Root Locus

Real Axis

Imagin

ary

Axis

EEE 2012: Control


G2(s)

Third order system with a unity gain feedback loop;

ksss

kG

sss

kG CLOL

61166116 2323 so pole location:

06116 23 ksss :

Properties:


Two poles collide and then they follow asymptotically a line which is

not perpendicular to the real axis.


The other pole goes to .

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

-4

-3

-2

-1

0

1

2

3

4

5Root Locus

Real Axis

Imag

inar

y A

xis

EEE 2012: Control


G3(s)

Third order system with a unity gain feedback loop;

226116

22

6116

22223

2

23

2

ssksss

sskG

sss

sskG CLOL so pole location:

02621160226116 23223 ksksksssksss

Properties:


Two poles collide and then they follow asymptotically a line which is

not perpendicular to the real axis.

They do not diverge to infinity but they converge to the two

complex zeros.


The other pole goes to .

-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0

-1

-0.5

0

0.5

1

Root Locus

Real Axis

Imagin

ary

Axis

EEE 2012: Control


b) Summary

Hence it can be seen that the root location depends on the order and the number of zeros of the system. The procedure of finding the closed loop pole location for various values of the proportional gain k is known as root locus method. We can find that location numerically or graphically (W. R. Evans). Matlab provides the command rlocus(num_OL,den_OL) but it is better to have a deeper understanding of the graphical method. This method can also be used for other parameters (apart from the proportional controller) but is rather tricky. Steps;

1. To include the effect of the feedback TF we name as OLTF the :)(sHsG

G(s)

H(s)

We always start from the location of the OL poles (i.e. poles of G(s)H(s)):

HGHG

HG

H

HGG

G

H

H

G

G

G

G

CLNkNDD

DkN

D

NkND

kN

D

N

D

Nk

D

Nk

GH

GG

11

0: HGHG NkNDDCE . For k=0 (first value): 0: HGDDCE , so poles of

G(s)H(s). We use the symbol “x” to denote OL poles. We use the symbol “o” to denote OL zeros. We use the symbol “□” to denote CL poles. No need for CL zeros as they do not move.

EEE 2012: Control


c) Angle and magnitude conditions

1)()(0)()(1:)()(1

)(

)(

)(

sHsGsHsGCE

sHsG

sG

sR

sC

This implies that 1)()( sHsG and ,12180)()(arg nsHsG ...2,1,0n

For example:

321

1

pspsps

zsksHsG

1

1p

3

2

1

3p

2p

1z

s

1B

1A

2A

3A

EEE 2012: Control


321

1

1321)arg(

AAA

kBsHsG

s

Every point that belongs onto the Root Locus (RL) must satisfy the magnitude and angle criteria.

For the system that we studied before 232

ss

kGOL

kss

kGCL

232

We will analyse the root locus for two test points; js 5.18.1 and js 5.15.1 .

The OL case is 232

ss

kGOL so I have two poles at -1 and -2 and no zeros:

1

1p

2

jxs 5.1

2A

1A

2p

1

jxs 5.1

1A

xs 1s

1x

5.1

EEE 2012: Control


1

5.1atan180

1

5.1180tan 11

xx

22

1 15.1 xA

2

xjs 2

2A

2s

xs

5.1

x2

xx 2

5.1atan

2

5.1tan 22

22

2 25.1 xA

For the point js 5.18.1 :

o

1 07.11818.1

5.1atan180

o

2 40.828.12

5.1atan

So 18047.20021 so that point does not belong to the root locus.

For the point js 5.15.1 :

o

1 43.10818.1

5.1atan180

o

2 56.718.12

5.1atan

So 18021 so that point does belong to the root locus.

By using the magnitude condition we can find the gain:

EEE 2012: Control


58.15.125.125.1

58.115.15.115.1

2222

2

2222

1

xA

xA

5.21 21

21

AAkAA

k

Analytically: For the test point s=-1.8±1.5j

023

112

ss

kGH 09.04120232 kj- .-kss

This is not valid as -2.41 + 0.9j 0 For the test point s=-1.5±1.5j

023

112

ss

kGH 05.20232 k-kss

This is a valid point, therefore s=-1.5±1.5j belongs on the root locus.

EEE 2012: Control


d) Simple root locus

For the open loop transfer function;

21

sss

ksG .

1. The number of branches equals the number of poles. The root locus is

always symmetrical with respect to the real axis and the branches start from poles and end up to zeros or infinity (infinite zero). In this specific case I have three branches, which will diverge to imaginary zeros at infinity.

2. There are three real OL poles with angles: 2arg,1arg,arg sss .

Applying the angle criteria; If s>0 then ,0arg s ,01arg s 02arg s and therefore that area

does not belong to the root locus as 02arg1argarg sss .

If s<0 and s>-1 then ,180arg s ,01arg s 02arg s and therefore

that area belongs to the root locus as 180001802arg1argarg sss .

If s<-1 and s>-2 then ,180arg s ,1801arg s 02arg s and

therefore that area does not belong to the root locus as 36001801802arg1argarg sss .

If s<-3 then ,180arg s ,1801arg s 1802arg s and therefore that

area belongs to the root locus as 1121801801801802arg1argarg sss with 1n .

So the root locus on the real axis is:

EEE 2012: Control


0s1s2s

3. We have seen that when we numerically calculate the root locus at

some point the poles from -1 and -2 will collide. To find this break-out point we rearrange the CE as:

sA

sBkskAsBsfCE 00)(: and then we find the 1st

derivative with respect to s and we equate the result with zero:

0ds

dk. Hence in this case: 21021 ssskksss and

hence

42.0

5.1026321 2 ssssss

ds

dk. Since the 2nd

point only belongs to the root locus the other one (-1.5) is ignored:

0s1s2s

Break away point

The same procedure can be used to find break-in points.

EEE 2012: Control


4. To find the asymptote that the locus of these two poles will follow:

The angle of the asymptote is:

zp nn

n

12180, where pn is the number of

finite poles and zn is the number of finite zeros. Hence in this case

0,3 zp nn and hence their angle is 180/3=60. The point of

intersection with the real axis is:

zp

n

i

i

n

i

i

nn

zp

s

zp

11 => 1

3

210

s :

0s1s2s

Asymptote

5.1

5. The point of intersection of the root locus with the imaginary axis can

be calculated by using the CE at js . So in this case

023023021 2323

kjjksssksssjs

So

4.1

0023 and 60603 2 kkk .

6. Now use this information to sketch the root locus;

EEE 2012: Control


0s1s2s

5.1

EEE 2012: Control


In Matlab: >> num=1; den=[1 3 2 0]; rlocus(num,den)

-6 -4 -2 0 2-4

-3

-2

-1

0

1

2

3

4Root Locus

Real Axis

Ima

gin

ary

Axis

EEE 2012: Control


Example: Find the root locus of: 32

)2(2

ss

sksG , H(s)=1:

1. I have two complex poles and 1 real zero.

2. Hence I have two branches that are symmetrical with respect to the

real axis.

3. Place the OL poles and zeros:

1

j4.1

j4.1

2

4. Find the loci on the real axis:

a. For s>-2 I have no poles or zeros so even number and hence

that area does not belong to the locus.

b. From to -2 I have 1 zeros, i.e. odd number and hence this

belongs to the locus:

EEE 2012: Control


1

j4.1

j4.1

2

5. Determine break in and out points:

2

320)2(32:

32

)2( 22

2 s

sskskssCE

ss

sksG

268.0

73.30142 sss

ds

dk. The acceptable value is 73.3s

EEE 2012: Control


6. Determine behaviour of complex poles:

These poles will move towards the real axis by increasing k. After the collision at -3.73 one will move towards to the zero at -2and the

other will diverge to -. The angle of departure of the complex poles is calculated by; 180 – (angle with respect to other poles) + (angle with respect to other zeros);

180 :

1

j4.1

j4.1

2

1

1

902

So 1455590180180 121 . The angle of departure for

-1-1.4j is -145 since the locus is symmetric.

EEE 2012: Control


1

j4.1

j4.1

2

Now we simply have to draw the locus from the complex poles to the real axis:

1

j4.1

j4.1

2

With Matlab: >> num=[1 2]; den=[1 2 3]; rlocus(num,den)

145°

-145°

EEE 2012: Control


-4 -3 -2 -1 0-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Root Locus

Real Axis

Ima

gin

ary

Axis

EEE 2012: Control


Design procedure summary

1. The branches are always symmetric with respect to the real axis.

2. The number of the branches equals the number of the OL poles.

3. Every branch starts from an OL pole (k=0) and ends at an OL zero

(k=). If we have more poles than zeros then we assume that there

are imaginary zeros at .

4. There are zp nn asymptotes where they intersect on the real axis at:

∑ ∑

5. The angle between them is ( )

6. According the angle criterion, a root locus exists on the real axis if the number of poles and zeros to the right is odd. Complex poles do not contribute.

7. The break in and break out points occur when

8. The intersection with the imaginary axis occurs at . Substitute this into the CE.

9. Departure angle from a complex pole is found by using the angle

criterion: ∑ ∑ In matlab the command rlocus(g) is very useful!

EEE 2012: Control

Chapter 7 – Design Based Root Locus of 132

Chapter 7. Design Based Root Locus

Lead Control (equivalent to PD control)

Used when the steady state properties of the system are ok but there is poor performance, i.e. we want to move the poles of the system without greatly altering the gain.

General form: abbs

asKsG cc

,

Other forms of Lead controller: 1,1

1

1

1

Ts

Ts

KTsa

TsKsG ccc , so:

bTTa /1,/1

Design strategy: 1. Express the design requirements as a desired pole location. 2. Plot the uncompensated root locus and check if there is a gain that

brings the system to that location, i.e. if a P controller is enough. 3. If not, place the zero at the real part of the desired pole location, i.e.

na .

4. Find the value of the pole by using the angle criterion:, i.e. zerosrealpolesreal argarg

5. Find the value of Kc by using the magnitude criterion, i.e. 1GH .

6. Check to see if the new compensated system satisfied the design specifications.

7. If not repeat by placing the zero at a different location.

EEE 2012: Control


Example – Lead Control

Assume the system:

2

4

sssG

Create a closed loop (unity feedback strategy) such as the closed loop system will have sradn /4 and 5.0 ;

EGc(s)

RG(s)

U C

If a simple proportional controller is used then the CLTF is:

'2

'

42

42 Kss

K

Kss

KsGCL

0'2

21lim

'2

2

'2

22

2

02

2

2

2

Kss

ss

ssE

Kss

sssRsE

Kss

ss

sR

sE

sss

And the root locus is: >>num=[1]

>>den=[1 2 0]

>>rlocus(num,den)

EEE 2012: Control


Since the required damping factor is 0.5 this implies an angle of 60o which

will give 73.15.0122 2 dn and this is achieved for K’=2:

-2.5 -2 -1.5 -1 -0.5 0-1.5

-1

-0.5

0

0.5

1

1.5Root Locus

Real Axis

Ima

gin

ary

Axis

Root Locus

Real Axis

Ima

gin

ary

Axis

-2.5 -2 -1.5 -1 -0.5 0-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

System: sysGain: 4

Pole: -1 + 1.73iDamping: 0.5

Overshoot (%): 16.3Frequency (rad/sec): 2

EEE 2012: Control


But for this gain and damping factor the natural frequency is not the required, i.e. the system is slower. So we replace the proportional controller with a more general:

The new open loop TF is: 2

4

ssbs

asKG

The location of the open loop poles and zeros and the desired closed loop

poles 322 j is:

0-2

3.464

-3.464

One strategy is to place the open loop zero such as it will be equal to the real part of the desired closed loop, i.e. a=-2 (in this example it just happens to coincide with the open loop pole). This will leave us with the location of the pole. Let’s assume that the controller pole is placed at x=-b and create an angle with the real axis:

EEE 2012: Control


0-2

3.464

-3.464

-b

o90

In order for the closed loop pole to be on the desired point the new characteristic equation must satisfy the angle condition:

301801201809090 and hence b=4.

So the OLTF is:

4

4

2

4

4

2

ssK

sss

sKG

Now we can use the magnitude criterion to find the gain:

1

4

4

322

jsss

K (as 01 GH )

EEE 2012: Control


4116

4

1644

4

1322322

4

KK

K

jj

K

Root Locus

Real Axis

Ima

gin

ary

Axis

-4 -3 -2 -1 0-5

0

5

System: gol

Gain: 4

Pole: -2 + 3.46i

Damping: 0.5

Overshoot (%): 16.3

Frequency (rad/sec): 4

EEE 2012: Control


a) Lag Control

Used when the steady state properties of the system are not satisfactory but the pole location is ok.

General form: babs

asKsG cc

,

Other forms of Lead controller: 1,1

1

1

1

Ts

Ts

KTs

TsKsG ccc , so:

bTTa /1,/1

Design strategy: 1. Express the design requirements as gain (usually as Kv), pole location

(damping factor and natural frequency). 2. Check if it is possible to achieve these with a simple proportional

controller. 3. If the location is possible but at that point Kv is too small (i.e. big

error for a ramp input) then add a lag controller such as the controller pole will be close to the origin (for example at -0.001) and the zero to the left of the pole as many times as it is required to achieve the desired Kv.

4. Use the magnitude criterion to find the necessary controller gain. 5. Check to see if you have achieved the design specifications.

EEE 2012: Control


Example – Lag Control

An plant transfer function is 21

06.1

ssssG . Create a control structure

that will achieve the following: 673.0,491.0 n

5vK

EGc(s)

RG(s)

U C

If a simple proportional controller is used then the CLTF is:

'21

'

06.121

06.1

Ksss

K

Ksss

KsGCL

Hence the CE is:

0'23

0'21

23

Ksss

Ksss

As this is a 3rd order system the generic CE ca be written as:

022

02

2223

22

nnnn

nn

asasas

ssas

By equating the corresponding coefficients:

34.2,06.1'

'

22

32673.0

491.02

2

aK

Ka

a

an

n

nn

n

Hence remarkably the desired pole location is achieved even without any extra proportional controller. But:

53.0

21

06.1lim

0

ssssK

sv

Hence we need to increase Kv 10 times (more or less). We use a Lag controller: We place the pole at 005.0b and hence the zero at 05.0a (as we need to increase the gain 10 times).:

EEE 2012: Control


0181.1

121

06.1

005.0

05.0

21

06.1

005.0

05.0

58.033.0

c

js

c

cOL

K

ssss

sK

ssss

sKsG

The new Kv is:

396.5

21

0792.1

005.0

05.0lim

0

ssss

ssK

sv

And the new pole location is: >>num=1.0792*[1 0.05]

>>den=conv(conv(conv([1 0.005],[1 0]),[1 1]),[1 2])

>>gcl=feedback(g,1); damp(gcl)

Eigenvalue Damping Freq.

(rad/s)

-5.46e-002 1.00e+000 5.46e-

002

-3.06e-001 + 5.73e-001i 4.71e-001 6.50e-

001

-3.06e-001 - 5.73e-001i 4.71e-001 6.50e-

001

-2.34e+000 1.00e+000

2.34e+000

EEE 2012: Control


b) Lead - Lag Control

This is a combination of a Lead and a lag controller (similar to PID). General form:

2

2

1

1

/1

/1

/

/1

cTs

Ts

Tcs

TsKsG cc

1

1

/

/1

Tcs

Ts

is the lead controller and

2

2

/1

/1

cTs

Ts

is the lag controller.

Example – Lead - Lag Control

An plant transfer function is 5.0

4

sssG . Create a control structure

that will achieve the following: 5,5.0 n

80vK

EGc(s)

RG(s)

U C

0-0.5

Desired pole

location

4.33

-2.5

EEE 2012: Control


By using the lead controller design steps:

0-0.5

Desired pole

location

4.33

-2.5

o120

o90

2.5

4.33

o652

33.4tan 1

o351809012065180

35

4.33

X-2.5

68.85.2

33.435tan

x

x

Hence 47.368.8

1 c

Tc

By choosing now T2=10 so that the second pole will be close to the origin:

10

805.0

4

805.0

4

/1

/1

/

/1lim

2

2

1

1

0

c

c

cs

v

K

K

sscTs

Ts

Tcs

TssKK

And hence:

EEE 2012: Control


5.0

1

0288.0

1.0

68.8

5.240,

0288.0

1.0

68.8

5.210

sss

s

s

ssG

s

s

s

ssG OLc

>>num=40*conv([1 2.5],[1 0.1]);

den=conv(conv(conv([1 8.68], [1 .0288]),[1 0]),[1

0.5]);

>> g=tf(num,den)

Transfer function:

40 s^2 + 104 s + 10

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

s^4 + 9.209 s^3 + 4.604 s^2 + 0.125 s

>> gcl=feedback(g,1)

Transfer function:

40 s^2 + 104 s + 10

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

s^4 + 9.209 s^3 + 44.6 s^2 + 104.1 s + 10

>> damp(gcl)

Eigenvalue Damping Freq. (rad/s)

-1.00e-001 1.00e+000 1.00e-001

-4.34e+000 1.00e+000 4.34e+000

-2.39e+000+4.16e+000i 4.97e-001 4.80e+000

-2.39e+000-4.16e+000i 4.97e-001 4.80e+000

80

5.0

1

0288.0

1.0

68.8

5.240

5.0

1

288.0

1.0

68.8

5.240lim

0

sss

s

s

ssKv

s

EEE2012 Control Notes

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