Chapter 4 Dynamic Systems: Higher Order Processes

Chapter 4 Dynamic Chapter 4 Dynamic Systems: Systems: Higher Order Higher Order ProcessesProcesses

Prof. Shi-Shang JangNational Tsing-Hua UniversityChemical Engineering Dept.

Hsin Chu, TaiwanMay, 2013

4-1 Non-interactive Systems – 4-1 Non-interactive Systems – Thermal TanksThermal Tanks

21 2 1

21 2 1

42 3 4 2

1

Let ;

2

A A

p v

A p A p v

A p B p A B p v

Tank

duf h t f h t V

dth C T u C T

dTf C T t f C T t V C

dtTank

dTf C T t f C T t f f C T t V C

dt

4-1 Non-interactive Systems –4-1 Non-interactive Systems –Thermal Tanks – Cont.Thermal Tanks – Cont.

1 22 1

1 2 1

2 44 2 3

2 4 1 2 2 3

1

1

2

1

v

A p

A p B pv

A B p A B p A B p

Tank

V C dt t

f C dt

s s s

Tank

f C f CV C dt t

f f C dt f f C f f C

s s K s K s

4-1 Non-interactive Systems4-1 Non-interactive Systems

1 1 11 1 1 1 1

1

2 2 22 2 1 2 2 1 2 1 2

2

;2

;2

i o i o

dh dH kA A f k h f F aH F a

dt dt h

dh dH kA A f k h F bH aH bH b

dt dt h

To Workspace

simout

Subtract 2Subtract 1

Subtract

Scope

MathFunction 1

sqrt

Math

Function

sqrt

Integrator 1

1s

Integrator

1s

Gain 3

1.5

Gain 2

0.1

Gain 1

0.1

Gain

1.5

Constant 1

1

Constant

4.5

4-1 Non-interactive Systems4-1 Non-interactive Systems

1 1 11 1 1 1 1 1 1

2 2 21 2 2 2 1 2 2 2 2 1

1

1

i oi o i o

F FA dH dHH K F F H s H s K F F

a dt a dtA dH dHa

H H K H H s H s K H sb dt b dt

Transfer Fcn 1

26 .667 s+1

1

Transfer Fcn

26 .667 s+1

2.667

SubtractScope

Constant 2

4

Constant 1

0

Constant

1

Add

NumericalNumerical ExampleExample

21 2 1 2

3 3

3 3

1 21/ 21/ 2

1

1

1

2

11 2

10 ; 4

4 / min; 1 / min

4 1 / min / min1.5

41.5

0.3752 22

12.667

1

1026.667 min

0.375

i o

A A m h h m

f m f m

m mk k

mmk

a bh

Kab

Ka

A

a

4-1 Non-interactive Systems4-1 Non-interactive Systems

1112 2

2

1

122

s

K

s

K

ss

K

sm

sy p

1 21 2

1 2

1 22 2

; ;2

;1 1

1222

ss

K

sm

sy p

An over-damped second order system has two negative real poles. Therefore, 2s2+2s+1=(1s+1)(2s+1); hence

such that

12

2

sK

= 11

1

sK

4-2 Interactive Systems – 4-2 Interactive Systems – Thermal Tanks with RecycleThermal Tanks with Recycle

4-2 Interactive Systems – Thermal 4-2 Interactive Systems – Thermal Tanks with Recycle- Cont.Tanks with Recycle- Cont.

21 4 2 1

1 22 1 4

1 2 1 1 2 4

42 3 4 2

2 44 2

Tank 1

1

Tank 2

A p R p A R p v

v A R

A R p A R A R

C p B p C B p v

C p Bv

C B p C B p

dTf C T t f C T t f f C T t V C

dtV C d f f

t t tf f C dt f f f f

s s K s K s

dTf C T t f C T t f f C T t V C

dtf C fV C d

tf f C dt f f C

3

2 4 3 2 4 31

p

C B p

Ct

f f C

s s K s K s

4-2 Interactive Systems – 4-2 Interactive Systems – Thermal Tanks with Recycle- Thermal Tanks with Recycle- Cont.Cont.

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

2-5 Step Response of Second-Order 2-5 Step Response of Second-Order Processes – Over-damped processProcesses – Over-damped process

1112 21

22

ss

K

ss

K

sm

sy pp

m(s)=A/s

12 /2

/1

21

ttpp ee

AKAKty

Inflection point

=3

=2

=11

3s+1

Transfer Fcn1

1

s+1

Transfer Fcn

simout

To Workspace

Scope

1

Constant

4-2 Interactive Systems4-2 Interactive Systems

1 1 11 1 0 1 1 2 0 1 2

1 2

2 22 2 1 1 2 2 2 1 2 2 1 2

2

2

;2

;

; ;2

i i

dh dH kA A f f k h h F F aH aH a

dt dt h h

dh dHA A k h h k h aH aH eH fH gH

dt dtk

e f a g a eh

f1

Cross-sectional=A2

h2V2

f2

h1V1

fi

Cross-sectional=A1

fo

4-2 Interactive Systems4-2 Interactive Systems

01 1 11 2 1 1 0 1 2

1 1 1 0 2

2 2 21 2 2 2 1 2 2 2 2 1

1

1

ii

i

F FA dH dHaH H K F F H H

a dt a a dts H s K F F H s

A dH dHfH H K H H s H s K H s

g dt g dt

H1(s)

H2(s)+

+

1

1

1 s

H2(s)

111

1/

2

212

2

21

221

sK

sK

KKK

F0(s)

Fi(s)+

-

F0(s)

Fi(s) +

-

11

1

sK

12

2

sK

Numerical ExampleNumerical Example

21 2 1 2

3 3

3 3

1 21/ 21/ 2

1

1 2

2

2

1

2

11

22

10 ; 8 ; 4

4 / min; 1 / min

4 1 / min / min1.5

8 41.5

0.3752 22

0.3752

0.75

12.667

0.5

1026.667 min

0.37510

13.333min0.75

i o

A A m h m h m

f m f m

m mk k

mmk

ah h

ke

h

g a e

Kaf

Kg

A

aA

g

180711s

667.2

15.01

333.13667.265.01

333.13667.265.01/5.0667.2

111

1/

)(

2

2

2

212

2

21

2212

s

sssK

sK

KKK

sF

sH

i

4-1 Second Order Systems4-1 Second Order Systems

2 1

pKY s

X s as bs

A second order system is of the following form:

2 2 2 1

pKY s

X s s s

Another form:

Kp is called process gain, is called time constant, is calleddamping factor. The roots of the denominator are the poles of the system.

4-1 Second-Order Processes - 4-1 Second-Order Processes - ContinuedContinued

Definition 4-1: A second order process is called over-damped, if >1; is called under-damped if <1; is called critical damped if =1.

Property 4-1: Consider the roots of denominator, in case of over-damped system, the poles of the system are all negative real numbers.

Property 4-2: The poles of a under-damped system are complex with negative real numbers.

Property 4-3: The pole of a critical damped system is a repeated negative real number.

State Space ApproachState Space Approach

112

1

21

22221

11211

2

1

21

22221

11211

2

1

KKNNNN

KNKNN

K

K

NNNNN

N

N

N

MBXA

M

M

M

bbb

bbb

bbb

X

X

X

aaa

aaa

aaa

X

X

X

dt

dX

Consider the following linear system with N differential equationsK inputs and P sensors

where X is termed the state vector and M is the input vector. The followingobservation equation is available:

112

1

21

22221

11211

2

1

21

22221

11211

2

1

KKPNNP

KPKPP

K

K

NPNPP

N

N

P

MDXC

M

M

M

ddd

ddd

ddd

X

X

X

ccc

ccc

ccc

Y

Y

Y

Y

State Space Approach _ State Space Approach _ Cont.Cont.

Assume that it is desirable to realize the input/output transfer functionsand neglecting the state variables.

sMsG

sMDBAsIC

sDMsBMAsICsY

sBMAsIsX

sBMsXAsI

sBMsAXssX

KP

1

1

1

State Space Approach _ State Space Approach _ ExampleExampleInteracting TanksInteracting Tanks

)(0

1.0

075.00375.0

0375.00375.0

)(

)(

)(0

1.0

)(

)(

075.00375.0

0375.00375.0

)(0

1.0

)(

)(

075.00375.0

0375.00375.0

)(

)(

10

01

)(0

1.0

)(

)(

075.00375.0

0375.00375.0

)(

)(

0

1.0

075.00375.0

0375.00375.0

75.0375.010

375.0375.010

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

212

211

sFs

s

sH

sH

sFsH

sH

s

s

sFsH

sH

sH

sHs

sFsH

sH

sH

sHs

FH

H

H

H

dt

d

HHdt

dH

FHHdt

dH

i

i

State Space Approach _ State Space Approach _ ExampleExampleInteracting Tanks – Cont.Interacting Tanks – Cont.

1s08711s

2.66667)(

0375.0075.00375.0

00375.0

)(0

0375.0075.00375.0

00375.00375.0075.00375.0

075.01.0

10

)(0)(

10

)()()()(

)(

375.075.00375.0

0375.00375.0075.00375.0

075.01.0

)(0

1.0

0375.0075.00375.0

0375.0

375.0075.00375.0

0375.00375.075.00375.0

0375.0

0375.075.00375.0

075.0

)(

)(

22

2

2

2

1

2

2

2

22

22

2

1

sFss

sF

ss

ss

s

sFsH

sH

sDMsCXsHsY

sF

ss

ss

s

sF

ss

s

ss

ssss

s

sH

sH

State Space to Transfer State Space to Transfer Function-MATLABFunction-MATLAB>> A=[-0.0375 0.0375;0.0375 -0.075]

A =

-0.0375 0.0375

0.0375 -0.0750

>> B=[1;0];C=[0 1];D=0;

>> [num,den]=ss2tf(A,B,C,D,1)

num =

0 -0.0000 0.0375

den =

1.0000 0.1125 0.0014

>> ss=den(3)

ss =

0.0014

>> num=num/ss

num =

0 -0.0000 26.6667

>> den=den/ss

den =

711.1111 80.0000 1.0000

>> tf(num,den)

Transfer function:

-9.869e-015 s + 26.67

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

711.1 s^2 + 80 s + 1

Rise time

A

B

C

time

Resp

ons

e

2-5 Step Response of Second-2-5 Step Response of Second-Order Processes – Under-damped Order Processes – Under-damped processprocess

1222

sss

AKsy p

/1sin1

2/

2te

AKAKty tp

p

21/ eA

B1. Overshoot=

21/2eB

C 2. Decay Ratio=

3. Rise time=tr=2

1

2

1tan

1

4. Period of oscillation=T=21

2

5. Frequency of oscillation (Natural Frequency)=1/T

Settling time

T

Example: Temperature Example: Temperature Regulated ReactorRegulated Reactor

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000100

100.5

101

101.5

102

102.5

103

time(seconds)

tem

pera

ture

(C)

t=3060st=1000sFeed flow rate0.4→0.5kg/sat t=0.

1. What is process gain?2. What is transfer func?3. What is rise time?

Textbook Reading Assignment Textbook Reading Assignment and Homeworkand Homework

Chapter 2, p41-49Homework p58, 2-9, 2-15, 2-16Due June 8th

Non-isothermal CSTRNon-isothermal CSTR

Non-isothermal CSTR- Non-isothermal CSTR- Cont.Cont.

20

M.B. for component A

where

E.B. for the tank

E.B. for the cooling jacket

AAi A A

ERT

A A

p i p A c v

cc c pc ci c c pc c c c c cv

dcf t c t f t c t Vr t V

dt

r t k e c

dTf t C T t f t C T t Vr t H UA T t T t V C

dt

dTf t C T t f t C T t UA T t T t V C

dt

Non-isothermal CSTR- Non-isothermal CSTR- Cont.Cont.

;56.1

);/(1;36);/(75

/000,12);/(88.0;/55

)/(987.1;/27820

min)/(1033.8;26.13

3

22

3

380

3

ftV

FlbBtuCpcftAFfthBtuU

lbmoleBtuHFlbBtuCpftlb

FlbmoleBtuRlbmoleBtuE

lbmoleftkftV

c

r

Process Information

Steady State Values

min/8771.0;540

9.678;/2068.0

min/3364.1;7.602

635;/5975.0

3

3

3

3

ftfRT

RTftlbmolec

ftfRT

RTftlbmolec

cci

A

c

iAi

Non-isothermal CSTR- Cont.Non-isothermal CSTR- Cont.Cooling flow rate Cooling flow rate 0.87710.87710.80.8

62.4*1

thou*cp2

62.4*1

thou*cp1

55*0.88

thou*cp

55*0.88

thou*cp

8.33e8

k0

0.8

fc(t)

1.3364

f(t)

12000

dH

0.5975

cAi(t)1s

cA(0.2068)

13.26

V1

13.26

V

75/60*36

UA

635

Ti(t)

1s

Temp (678.9)

540

Tci

1s

Tc(602.7)

Subtract

Scope2

Scope1

Scope

1.987

R

Product5

Product4

Product3

Product2

Product1

Product

eu

MathFunction

[jacket_energy_in]

Goto9

[heat_transfer_out]

Goto8

[reaction_rate]

Goto7

[Tc]

Goto6

[energy_out]

Goto5

[energy_in]

Goto4

[Temp]

Goto3

[mass_out]

Goto2

[jacket_energy_out]

Goto10

[cA]

Goto1

[mass_in]

Goto

[reaction_rate]

From9

[mass_out]

From8

[mass_in]

From7

[Tc]

From6

[Temp]

From5

[cA]

From4

[cA]

From3

[Temp]

From2

[heat_transfer_out]

From17

[jacket_energy_out]

From16

[jacket_energy_in]

From15

[Tc]

From14

[heat_transfer_out]

From13

[reaction_rate]

From12

[energy_out]

From11

[energy_in]

From10

[Temp]

From1

[cA]

From

-27820

EDivide1

Divide

Add2

Add1

Add

1/13.26

1/V

1/(1.56*55*1)

1/(Vc*thouc*Cv)1

1/(13.26*55*0.88)

1/(V*thou*Cv)

Energy Balance of the Reactor

Material Balance of the Reactor

Energy Balance of the Jacket

Inlet and Outlet Energy and Material fluxes of the Reactor

Inlet and Outlet Energy fluxes of the Jacket

Rate Constant Heat Exchange between Jacket and Reactor

Transfer Functions Derived Transfer Functions Derived by Linearizationby Linearization

DUCX

F

F

CC

y

BUAX

F

F

C

b

b

bbC

aa

aaa

aaC

dt

d

dt

dX

Fbaa

Ff

g

T

g

T

gfTTg

dt

d

dt

dT

bFbaaCa

T

gF

f

g

T

g

T

gC

c

gTfTTcg

dt

d

dt

dT

FbCbaCa

Cc

gF

f

g

T

gC

c

gcfTcg

dt

dC

dt

dc

c

i

Ai

c

A

c

i

Ai

c

A

c

A

cc

cc

cc

cccc

icA

ii

cc

AA

icA

AiA

AiAi

AA

AiAAA

0000010

0

0

000

b0

0

0

0

,,

,,,

,,

34

2322

1211

3332

232221

1211

343332

3333

2322232221

22222,2

12111211

1111,1

Linearization of the Reactor ExampleLinearization of the Reactor Example

48.026.13/2068.01033.826.1323364.1/2 9.678987.127820

80

1

1

111

eVceVkf

c

g

x

ga A

TRE

A

It can be shown as generated as above:

Transfer Functions Derived by Linearization – Cont.Transfer Functions Derived by Linearization – Cont.

[num, den]=ss2tf(A,B,C,D,4)num = 0 1.33226762955019e-015 -2.81748023 -1.356898478768den = 1 1.3804 0.3849816 0.038454805392

>> ss=den(4)>> den=den/ssden = 26.0045523519419 35.8966840666205 10.0112741717343 1>> num=num/ssnum = 0 3.46450233194353e-014 -73.2673121415962 -35.2855375273927

SIMULINK of Linear System - SIMULINK of Linear System - CSTRCSTR

Transfer Fcn

2.07s+1

26 .27s +36 .31s +10 .14s+13 2

To Workspace

simout

ScopeGain

-35 .77

Constant 1

678 .9

Constant

-0.0771

Add

Non-isothermal CSTR- Cont.Non-isothermal CSTR- Cont.Cooling flow rate Cooling flow rate 0.87710.87710.80.8

0 5 10 15 20 25 30 35 40 45 50678.5

679

679.5

680

680.5

681

681.5

682

Time (min)

Tank

temperature

Non-isothermal Non-isothermal CSTR- Cont.CSTR- Cont.Linearized Model Linearized Model (page 127)(page 127)

3 2

35.77 2.07 1

26.27 36.31 10.14 1c

s s

F s s s s

0 5 10 15 20 25 30 35 40 45 50678.5

679

679.5

680

680.5

681

681.5

682

Time(min)

Tan

kTem

pera

ture

The problem of The problem of nonlinearitynonlinearity

4-3 Step Response of the 4-3 Step Response of the High Order SystemHigh Order System

1 1 1

s s

a b

Ke Keor

s s s

systemsorder high theseeapproximat to

model timedead plusorder first

following theimplement wecases,many In

/1

1

1

( ) 1itnn

in

ii j

jj i

eY t K

1

( )

( ) 1n

ii

Y s K

X s s

X(s)=A/s

n=2

n=3

n=5

n=10

time

Response

s

4-3 Step Response of the High Order 4-3 Step Response of the High Order System- ContinuedSystem- Continued

11.41

1 3.2

5

s

e

s

s

Method of Reaction Curve:

time

Respons

e

inflection point

4-3 Step Response of the High 4-3 Step Response of the High Order System- ContinuedOrder System- Continued

time

Response

s

Real

Approximate

4-3 Response of Higher-4-3 Response of Higher-Order Systems – Cont.Order Systems – Cont.

1

2

1

1 2 1 3 1

0.5 1

1 2 1 3 1

Y s

X s s s s

Y s s

X s s s s

4-4 Other Types of Process 4-4 Other Types of Process ResponseResponse

( )op o p

i o

df tf t K m t

dtdh t

f t f t Adt

Integrating Processes: Level Process

4-4 Other Types of Process 4-4 Other Types of Process ResponseResponse

1

1

1

1

i o

po

p

pi

p

H s F s F sAsK

F s M ss

KH s F s M s

As As s

4-4 Other Types of Process 4-4 Other Types of Process ResponseResponse

4-4 Other Types of Process 4-4 Other Types of Process ResponseResponse The most general transfer function is as

the following:

p1, p2,…,pn are called the poles of the system, z1, z2,…,zm are the zeros of the system, Kp is the gain.

Note that nm is necessary, or the system is not physically realizable.

n

mp

nn

n

mm

mm

pspsps

zszszsK

asasas

bsbsbsb

sm

sy

21

21

011

1

011

1

4-4 Poles and Zeros - Example4-4 Poles and Zeros - Example

2

2

243

12

1

222

21

1,

/1

0

1

121

jss

s

s

ssss

KsG

Imaginary part

Real

part

2

21

2

21

1

1

2

1

0

0

Left Half PlaneLHP Right Half Plane

RHP

4-4 Poles and Zeros - 4-4 Poles and Zeros - ExampleExample

time

Respons

e

4-4 Location of the Poles and Stability in a Complex Plane

Re

Im

Purdy oscillatory

Purdy oscillatory

Fast Decay Slow Decay

Exponential Decay

Exponential Decay with oscillation

Slow growth

Fast Exponential growth

Exponential growthwith oscillation

Stable (LHP) Unstable (RHP)

4-4 The Stability of the 4-4 The Stability of the linear systemlinear system

)(lim tyt

Definition 4-2: A system is called stable for the initial point if given any initial point y0, such that ∣y0∣≦ε, there exists a upper bound , such that:

Definition 4-3: A system is called asymptotic stable if given any initial point y0, then

0)(lim

tyt

4-4 The Stability of the 4-4 The Stability of the linear systemlinear systemDefinition 4-4: A system is called input

output stable if the input is bound, then the output is bounded. (Bounded Input Bounded Output, BIBO)

Property 4-4: A linear system is asymptotic stable and BIBO if and only if all its poles have negative real parts.

4-4 Stability - Example4-4 Stability - Example

time

Resp

ons

e

G4

G2 G3G1

13011015

1)(

)(

)(1

sss

sGsm

sy

105.011.01

1

)(

)(2

sss

sGsm

sy

1

1)(

)(

)(233

sss

sGsm

sy

113

1

)(

)(24

ssssG

sm

sy

m(s)=1

Open Loop Unstable Open Loop Unstable Process- Process- Chemical Reactor (text page Chemical Reactor (text page 139)139)

0

M.B. for component A

where

E.B. for the tank

AAi A A

ERT

A A

p i p A c v

dcf t c t f t c t Vr t V

dt

r t k e c

dTf t C T t f t C T t Vr t H UA T t T t V C

dt

Open Loop Unstable Open Loop Unstable Process- Process- Chemical ReactorChemical Reactor

3

2 3

3

2 3

At 566

9.79 1;

9.75 1 6.6 1

9.75 1 6.6 1

At 620

7.47 1;

11.3 1 11.47 1

11.3 1 11.47 1

i

A

i

i

A

i

T R

s K s

s s s

C s K K

s s s

T R

s K s

s s s

C s K K

s s s

HomeworkHomeworkText p1484-4, 4-5, 4-7, 4-8, 4-10, 4-11,4-12Due April

Supplemental Material Supplemental Material

Development of Empirical Development of Empirical Models from Models from Process DataProcess Data

S-1 IntroductionS-1 IntroductionAn empirical model is a model that

is developed from experience and their parameters are found based on experimental tests.

The most frequent implemented empirical models are first order, second order and/or with time delays.

The input changes is basically a step or an impulse.

S-2 First Order without Time S-2 First Order without Time Delay Systems Using Step InputDelay Systems Using Step Input

/1)(

1)(

)(

1)(

)(

teKAty

ss

KAsy

s

Asm

s

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Consider a first order system with a output signal y(t) and input signal m(t), then:

First Order with Time Delay First Order with Time Delay Systems Using Step InputSystems Using Step Input

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Consider a first order system with a output signal y(t) and input signal m(t), then:

Example: A Typical Example: A Typical ExperimentExperiment

Time (second)

Y(temperature,oC,70-100oC)

Y(temperature,mA,4-20mA) Y (temperature, %) ln(1-Y)

0 70 4 0. 0

1 71.74 4.928 0.058 -0.0598

2 76.51 7.472 0.217 -0.2446

3 80.8 9.76 0.360 -0.4463

4 84.64 11.808 0.488 -0.6694

5 88 13.6 0.600 -0.9163

6 90.76 15.072 0.692 -1.1777

7 93.16 16.352 0.772 -1.4784

8 94.99 17.328 0.833 -1.7898

9 96.64 18.208 0.888 -2.1893

10 97.75 18.8 0.925 -2.5903

Graphical Fitting MethodsGraphical Fitting MethodsFit 1: Method of 63.2% ResponseFit 2: Method of initial slopeFit 3: Method of Log plot

Example: An Experiment Example: An Experiment PlotPlot

Fit 1Fit 2

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Consider a First-Order Plus Dead Time Model

Method of log plot - ContinuedMethod of log plot - Continued

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Fit 3

Method of log plot - ContinuedMethod of log plot - Continued

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Fit 3

real

Fit 2Fit 1

S-3 Over-damped Second Order S-3 Over-damped Second Order Systems Using Step InputSystems Using Step Input

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Smith’s Method for Second Smith’s Method for Second Order SystemsOrder SystemsStep 1: Get time delay by observing the response curve.Step 2: Find time t20 such thaty/y=0.2, find t60 such thaty/y=0.6Step 3: Get t20/t60, then and From the right figure.

ExampleExample

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t20=1.9

t60=5

t20/t60=0.38From the figuret60/=2.4 =5/2.4=2.1=1.22=4.32, 2 =5.04

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1