Automatic Control System

Modelling

Modelling dynamical systemsEngineers use models which are based upon mathematical relationships between two variables. We can define the mathematical equations:• Measuring the responses of the built process (black model)Not interested in the number and the real value of the time constants of the process, only it is enough that the response is similar sufficient accuracy• Using the basic physical principles (grey model). In order to simplification of mathematical model the small effects are neglected and idealised relationships are assumed.

Developing a new technology or a new construction nowadays it’s very helpful applying computer aided simulation technique.This technique is very cost effective, because one can create a model from the physical principles without building of process.

Grey box model

Modelling mechanical systems

;2

2

dt

sdmF ;

dt

dsCvCFD ksFs

Newton’s law one-dimensional translation and rotational systems.

;2

2

dt

dIM

adt

sd

2

2

2

2

dt

d

velocity

acceleration

F: force [N]; FD: absorber’s force; Fs: spring’s force M: moment about centre of mass of body [Nm]I: the body’s moment of inertia [kgm2]s, : displacement [m; rad]v, ω: velocity [m/sec; rad/sec]a, : acceleration [m/sec2; rad/sec2]C: friction constant [Nsec/m]k: spring constant [N/m]

•Assign variables and sufficient to describe an arbitrary position of the object.•Draw a free-body diagram of each components and indicate all forces acting.•Apply Newton’s law in translation and rotational form.•Combine the equations to eliminate internal forces.

adt

ds

dt

d

Modelling mechanical systems

m2

m1

k2

k1

C

t

y

t

x

t

r

The car’s wheel vertical motion is assumed one-dimensional and the mass hasn’t got extension. The sock absorber is represented by a dashpot symbol with friction constant C.The force from the spring acts on both masses in proportional their relative displacement with spring constant k.The equilibrium positions of the two mass are offset from the spring’s unstretched positions because of the force of gravity.Nowadays the mathematical programs allow one to create a better model.The system can be approximated by the simplified system shown in left.

Road surface

Modeling mechanical systems

m2

m1

k2

k1

C

t

y

t

x

t

r

2

2

dt

sdmF

dt

dsCFD ksFs

The positive displacements or velocity of mass, and so the positive force are signed by up arrows.

2

2

112 dt

xdm)rx(k)xy(k

dt

)xy(dC

2

2

22 )()(

dt

ydmxyk

dt

xydC

Simulating the system by MATLAB

dt

tdy )(

dtdt

dt dt2

1

m

1

1

m1k

)(tr

1k

)(tx

C 2k

)(ty

2k

C

2

2 )(

dt

txd

2

2 )(

dt

tyd

dt

tdx )(

Difference equations

0 T0 2T0 3T0 4T0 5T0 6T0 7T0

y7

y6

y5

y4

y3

y2

y1

y0

0

)1()()(

T

iyiy

dt

tdy

20

2

2 )2()1(2)()(

T

iyiyiy

dt

tyd

30

3

3 )3()2(3)1(3)()(

T

iyiyiyiy

dt

tyd

)()( 0iTyiy A block input is energised by x(t), and the responseis y(t). Because of the response needs any time or insimulation the calculation of the response also needs any time, and so y(iT0) can be calculated from previous value than x(iT0). The sampled block has a delay time (T0). Example: )()(

)(tbxty

dt

tdyT

)1()()1()(

0

ibxiyT

iyiyT

)1()1()(00

0

ibxiyT

Tiy

T

TT

iiT

iyTdtty0

0

0

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

0

0

0

ibxTT

Tiy

TT

Tiy

Using difference equationrkyk

dt

dyCxkxk

dt

dxC

dt

xdm 12122

2

1

)()}1()({

)1()}()1({)}1()(2)1({

2

222

ixkixixT

C

iykiyiyT

Ciyiyiy

T

m

0)1(,)1(,)1(,0)0()0(,)0( 110 yxxrrandyxrr

xkdt

dxCyk

dt

dyC

dt

ydm 222

2

2

)1()()}1()({

)()()}1()({)}2()1(2)({

12

2121

irkiykiyiyT

C

ixkkixixT

Cixixix

T

m

Using Laplace transform to model mechanical systems

m2

m1

k2

k1

C

2

2

112 )()()(

dt

xdmrxkxyk

dt

xydC

2

2

22 )()(

dt

ydmxyk

dt

xydC

)()()()()()()( 211122 sxsmsrksxksxksyksCsxsCsy

)()()()()( 2222 sysmsxksyksCsxsCsy

)()()()( 2111

22 sxsmsrksxksysm

Block modeling mechanical system

sm2

1

sm1

1

s

1

1k

2k

C

)(ssy

)(22 sysm

)(sy)(21 sxsm

)(sr

)()()()()( 2222 sysmsxksyksCsxsCsy

)()()()( 2111

22 sxsmsrksxksysm

2ks

1)(ssx

C

)(sx

Block modeling mechanical systemBlock reduction

sm2

1

sm1

1

s

1

21

1

sm

k

2k

C

)(sy

)(sr2k

s

1

Csm1

1s

22sm

Modeling electromechanical systems

;dt

dKU eE

;

dt

dILU a

L

DC motor’s voltages.

;dt

dCT f

The electromotive force based on:

DC motor’s tongues.

B: magnet field [T: Tesla]; Ta armature torqueIa: armature current Tf friction torqueUE: electromotive force [V] TL load torqueUL: voltage on inductance [V] C: friction constant [Nsec/m]UR: voltage on resistance [V]

aaa IKT

lBUE

;aR RIU

Modeling DC motor

M

aT LTfTaLaR

eUaI

dcU

aJ armature inertia

angular displacementTa armature torqueTf friction torqueTL load torqueIa armature currentRa armature resistanceLa armature inductance

Ka motor torque constantC rotational friction constant

dt

dILIR

dt

dKUUUUUU a

aaaedcLRedc 0

Generated electromotive force (emf) against the applied armature voltage

LaaLfaa Tdt

dCIKTTT

dt

dJT

2

2

Ke electromotive force constant

Simulating the system by MATLAB

dtdtaJ

1)(tTL

C

)(teK

dcU

aR

dtaL

1 )(tIa

aK

dt

dIa

2

2

dt

d dt

d

Example: Block modeling DC motor

M

aLaR

eUaI

dcU

aT LTfT

aJ

dtLa

1dt

aR

aK

eK

C

dt

d2

2

dt

dJa

dt

dIL aa

LT

fT

dcU aI aT

Laaa Tdt

dCIK

dt

dJ

2

2

dt

dILIR

dt

dKU a

aaaedc

dtJa

1

In time domain using difference equations

LaaaLaaa TIKdt

dC

dt

dJT

dt

dCIK

dt

dJ

2

2

2

2

)1()}1()({)()}1()({ iUiIiIT

LiIRii

T

Kdcaa

aaa

e

)}1()({)1()1()(}{ iiT

KiI

T

LiUiI

T

LR e

aa

dcaa

a

dca

aaae Udt

dILIR

dt

dK

)1()1()}1()({)}2()1(2)({2

iTiIKiiT

Ciii

T

JLaa

a

)}2()1({)2()2()1(

iiTRL

KiI

TRL

LiU

TRL

TiI

aa

ea

aa

adc

aaa

In operator frequency domain

M

aT LTfTaLaR

eUaI

dcU

aJ

Assuming than Udc is constant andthe system is steady-state when onechange the value of Udc

Examination of dynamic behaviourcan be used the Laplace transform.

Laaa Tdt

dCIK

dt

dJ

2

2

dt

dILIR

dt

dKU a

aaaedc

)()()()( ssILsIRssKsU aaaaedc

)()()()(2 sTsCssIKssJ Laaa

G(s))(sx )(sy )()()()(

)(

)(sxsGsysG

sx

sy

Block modeling DC motor

M

aLaR

eUaI

dcU

aT LTfT

aJ

)()()()( ssILsIRssKsU aaaaedc

)()()()(2 sTsCssIKssJ Laaa

sJa

1

sLa

1

s

1

aR

aK

eK

C

s2sJa aasIL

LT

fT

dcUaI aT

a

aa

K

sLR

Simpler block model of DC motor

s

1

sLR

K

aa

a

sJC a1

dcU

LT

eK

sJCsLR

K

aaa

a

1

dcU

eK

LT

s

1

sKsJCsLR

K

eaaa

a 1

))((

a

aa

K

sLR LT

dcU

Models of electronic circuitKirchhoff’s current law: The algebraic sum of current leaving a junction or node equals the algebraic sum of the current entering that node.Kirchhoff’s current law: The algebraic sum of all voltages taken around a closed path in a circuit is zero.

Resistor

)()( tRitu

InductorCapacitor

)()( sRIsU )()( sLsIsU

dt

tdiLtu

)()(

dt

tduCti

)()(

)()( sCsUsI

Models of electronic circuit

U1 U2

I1

I2

I3

A B

All resistance equal R and all capacitanceequal C. Point “A” is nearly ground and such as a summing junction for the currents.“B” is a take off point for U2.The OpAmp amplitude gain is Au(s)

)(1 sG

)(2 sG

)(3 sG

)(sAu

U2U1 I1

I2

I3

U1 I1

U2I2

U2I3

21

1

2

1

)(

)(

1

1

RsCRsU

sI

sCR

sC

sU

sI

1)(

)(

2

2

21

1

2

1

)(

)(

2

3

RsCRsU

sI

Modeling heat flow

);(1

21 TTR

q

vcmC

Heat energy flow.

Specific heat:

Thermal conductivity:

;1q

Cdt

dT

q: heat energy flow J/sec R: thermal resistance °C/J T: temperature °CC: thermal capacity J/°C

l

kA

R

1

A: cross-sectional area l: length of the heat-flow pathk: thermal conductivity constant

Temperature as a function of heat-energy flow:

Heat flow models

q1

q2

room T1

T0

R1

R0

The heat energy flows through substances (across the room’s wall):

))(11

()(1

1011

2110 TTRR

qqTTR

q

The net heat-energy flow into a substance:

qCdt

dT 1

vmcC m: the mass of the substancecv: specific heat constant

qdt

dTC

The heat can also flow when a warmer mass flow into a cooler mass or vice versa: )( 01 TTc

dt

dmq v

Heat flow models

))(()()( 101

11

2

222121

11 TT

l

Ak

l

AkTTc

dt

dmqqq

dt

dTC vm

)}()()()()()()( 20111 sTcssmsTsconstTsTcssmssT vv

The total heat-energy flow: )(1

)( 1021 TTR

TTcdt

dm

dt

dTC v

qdt

dTC )(

110 TT

Rq

It’s non-linear, except T1=T2.

Modelling a heat exchanger

)(1

)( wsssivssss

s TTR

TTcKdt

dA

dt

dTC

water steam

siT dt

dAK

dt

dm ss

s www AK

dt

dm

sT

wiT

wT realT

)(1

)( wswwivwwww

w TTR

TTcKAdt

dTC

The time delay between the measurementand the exit flow of the water:

)( dwreal tTT

As: area of the steam inlet valve, Aw: area of the water inletKs: flow coefficient of the inlet valve, Kw: flow coefficient of the water inletcvs: specific heat of steam, cvw: specific heat of waterTsi: temperature of inflow steam, Twi: temperature of inflow waterTs: temperature of outflow steam, Ts: temperature of outflow waterCs=mscvs thermal capacity of the steam, Cw=mwcvw thermal capacity of the waterR: thermal resistance (average over the entire exchanger)

Simulating heat exchanger

)(1

)( wsssivssss

s TTR

TTcKdt

dA

dt

dTC )(

1)( wswwivwww

ww TT

RTTcKA

dt

dTC

dt)(tTs

vsscK

sC

1

dt

dTC s

s

wC

1dt

dTC w

w

dt)(tTw

R

1

)(tTsi

)(tTwivwww cKA

)(tAs

dt

d

Laplace form heat exchanger

)(1

)( wsssivssss

s TTR

TTcKdt

dA

dt

dTC )(

1)( wswwivwww

ww TT

RTTcKA

dt

dTC

dswreal esTsT )()(

The equation is nonlinear because the state variable Ts is multiplied by the the controlinput As. The equation can be linearized at the working point Ts0, and so Tsi-Ts0=Ts

nearly constant. To measure all temperature from Twi, it’s eliminated Twi=0.

))()((1

)()( sTsTR

TcKsAssTC wssvsssss

))()((1

)()( sTsTR

sTcKAssTC wswvwwwww

)(1

1)(

1)( sT

sRCsA

sRC

TcRKsT w

ss

s

svsss

)(

1

1)( sT

sRCcKRAsT s

wvwwww

)(1

1

1

1)(

11

1)( sT

sRCsRCcKRAsA

sRC

TcRK

sRCcKRAsT w

swvwwws

s

svss

wvwwww

)(})(){( 2 sATcKCRCsCcKsRACCscKAsT ssvssswsvwwwswvwwww

Block model of heat exchanger

vwww cKA1

sRCs

1)(ssTC ss)(sTw

R

1

)(sRTssvss TcK

)(sAs

))()((1

)()( sTsTR

TcKsAssTC wssvsssss

))()((1

)()( sTsTR

sTcKAssTC wswvwwwww

dswreal esTsT )()(

sCw

1)(ssTC wwse

Black box model

Modeling by reaction curve

Process field

plant

controller

Feedback control

A/M

GC(s) GA(s)

GT(s)

GW(s)

GP1(s) GP2(s)

When auto / manual switch is manualposition (open), then GC(s)=1

A/M

GC(s) GA(s)

GT(s)

GW(s)

GP(s)R0+r U0+u

YM0+yM

W

Modelled the process from reaction curveby dead-time proportional first order transfer function HPT1

t

,My u

uMy

uT gT

1( ) ;

1

usT

p Pg

G s K esT u

yK M

p

The error of the model

0

2mod })()({

ielmeasured ixix

The principle of the less squares:

The better model the less sum of value of the squares.

The error of the model:

%100|)(|

|)()(|

0

0mod

N

imeasuredM

N

ielMmeasuredM

iy

iyiy

A better model of process from reaction curveby dead-time second order transfer function HPT2

t

,my u

umy

1 2

1 1( )

1 1

tsT

p PG s K esT sT

It needs computer! The beginning parameters: tf T

TTT ,..

221

Modelled the process from reaction curveby “n” order transfer function PTn

t

,my u

umy

30t

npp sTKsG

)1(

1)(

70t

%30

%70

%10

10t

Modelled the process from reaction curveby dead-time integral first order transfer function HIT1

t

u

gT

1 1( )

1pi g

G sT sT

iT

,my u