Automatic controlAutomatic control

Introduction to automatic control

Dr Ing Mohamed Saber SokarDr.-Ing. Mohamed Saber Sokar

▐ Objectives Review on elementary function

▐ Objectivesy

Complex numbers

Laplace transform Examples propertiesproperties Inverse transform

Partial fraction e pansion Partial fraction expansion Partial fraction Common test input signals

U d t d th C t f T f F ti Understand the Concept of Transfer Functions.٢ Dr.-Ing. Mohamed Saber Sokar

▐ Elementary functionsExponential Function xe

▐ Elementary functionsExponential Function

, /x y x y x y x y

ee e e e e e

0( ) , 1L i h i F i l ( )

x y xye e e Logrithmic Function ln( ) ln ln ln

x (xy) x y

ln ln ln

( y) yx x y

ln ln , ln(1) 0a

y (x ) a x

1lnln ln

, ( )

1x x x

( )

e x, e e ,x

٣ Dr.-Ing. Mohamed Saber Sokar

▐ Elementary functionsEven Function: f( x) f(x) x

▐ Elementary functionsEven Function: ,Odd Function: ,

f( x) f(x) xf( x) f(x) x

2 2

sin sin cos cos( x) x, ( x) x 2 2sin cos 1

sin sin cos cos sinx x

(x y) x y x y sin sin cos cos sin

sin sin cos cos sin (x y) x y x y (x y) x y x y

sin 2 2sin cos x x xcos c(x y) os cos sin sincos cos cos sin sin

x y x y(x y) x y x y

cos cos cos sin sin(x y) x y x y

٤ Dr.-Ing. Mohamed Saber Sokar

▐ Elementary functions▐ Elementary functionsxxxxx 1cos2sin21sincos2cos 2222

x)(x

xxxxx

2cos121cos

1cos2sin21sincos2cos

2

x)(x 2cos121sin

22

-x)π ()π( x-x2

cos2

cossin

2

-x)π ()π( xx2

sin2

sincos

22

xx)(π x x) (πcoscos

sinsin22

x, x, xx, xx

x x)(π

i1csc1sec

icoscotsintan

coscos

xxxx sincossincos٥ Dr.-Ing. Mohamed Saber Sokar

▐ Elementary functions▐ Elementary functions

gSimplifyin B A sincosD

gSimplifyin

AB, BAC tanthen 22

)(BAxBxAA

cossincos 22 )(

phasemagnitudeC :: phase, magnitudeC ::

٦ Dr.-Ing. Mohamed Saber Sokar

▐ Complex Numbers▐ Complex NumbersyixzandyixzIf

)()()()( 2121221121 yyixxiyxiyxzzz adding

222111 .,. yixzandyixzIf

)()( iyxiyxzzz Multiplying

)()()()( 2121221121 yyixxiyxiyxzzz adding

)()(.)(.

2112212121

221121

yxyxiyyxxzzziyxiyxzzz

Multiplying

))(( iyxiyxiyxz Multiplying by conjugate of the denominator

2222

2211

22

11

2

1

))(())((

iyxiyxiyxiyx

iyxiyx

zzz

222112

222121

yxyxyxi

yxyyxxz

2222 yxyx

٧ Dr.-Ing. Mohamed Saber Sokar

▐ Polar form of Complex Numbers▐ Polar form of Complex Numbers

z = x + i . y , if x=r . cos, y= r . sin

Th ( +i i Then z = r(cos +i . sin = r

Absolute value (modulus) r2=x2+y2 Absolute value (modulus) r =x +y

Magnitude 22 yxr g

Argument = tan-1(y/x)

Example z=2+i.420

4tan

20

1

z

2tan

٨ Dr.-Ing. Mohamed Saber Sokar

▐ Laplace Transform▐ Laplace TransformD fi iti d l Definition and examples

)()()( dttfefLsF st

)()(LapalceInverse

)()()(

10

FLtf

dttfefLsF

ExampleUnit Step Function u(t)

)()(LapalceInverse FLtf

tutf

111

1)()(

Unit Step Function u(t)

es

es

es

dteLsF sstst 1111)1()( 0

00

ssF 1)(

s٩ Dr.-Ing. Mohamed Saber Sokar

▐ Laplace Transform (Example)▐ Laplace Transform (Example)ttf 3)(Fi d L l t f f th f ll i tetf 3)( Find Laplace transform for the following

function

dtedteeeLsF tststt

)()( )3(33

ts

11 )3(

00

se

sts

33

0

)3(

ssF

31)(

aseL at

1)(as

١٠ Dr.-Ing. Mohamed Saber Sokar

▐ Laplace Transform▐ Laplace Transformsin)sin(

tdteeteL atstat

)()(

0

cos)1(sin1

)(

tdteas

teas

tastas

0

)(

00

cos

tdteas

asas

tas

0

2)(

0

)(2

0

sin)1(cos)(

tdte

asaste

astastas

2

0

)(2

2

2 sin)()(

tdte

asastas

20

2

2

)(sin)

)(1(

astdte

asst

220

cos

)(sin

astdte

astdte

st

st

220 )(

cos

astdte

١١ Dr.-Ing. Mohamed Saber Sokar

▐ Laplace transform table▐ Laplace transform table

١٢ Dr.-Ing. Mohamed Saber Sokar

▐ Laplace transform theorem-1▐ Laplace transform theorem-1

١٣ Dr.-Ing. Mohamed Saber Sokar

▐ Laplace transform theorem-2▐ Laplace transform theorem-2

١٤ Dr.-Ing. Mohamed Saber Sokar

▐ Laplace Transform▐ Laplace Transform Find Laplace transform Find Laplace transform

y”+9y=0, y(0)=0, y’(0)=2y y y( ) y ( )1- L(y”)=s2Y(s)-sy(0)-y’(0)= s2Y(s)-22- L(y)=Y(s)

Sol tionSolution(s2+9)Y(s)-2=0(s +9)Y(s) 2 0Y(s)=2/ (s2+9) = (2/3)[3/(s2+9)]y(t)=(2/3) sin 3t

١٥ Dr.-Ing. Mohamed Saber Sokar

▐ Laplace Transform▐ Laplace Transform

Solve: y”+2y’+5y=0 y(0)=2 y’(0)= 4Solve: y”+2y’+5y=0, y(0)=2, y’(0)=-4 L(y”)=s2Y(s)-sy(0)-y’(0)= s2Y(s)-2s+4(y ) ( ) y( ) y ( ) ( ) L(y’)=sY(s)-y(0)=sY(s)-2 L(y)=Y(s) (s2+2s+5)Y(s)=2s (s2+2s+5)Y(s)=2s Y(s)=2s/

(s2+2s+5)=2(s+1)/[(s+1)2+22]-2/[(s+1)2+22]2/[(s+1) +2 ]

y(t)= e-t(2cos 2t –sin 2t)

١٦ Dr.-Ing. Mohamed Saber Sokar

▐ Inverse Laplace Transform▐ Inverse Laplace Transform3i1)31()1()1( 111 tLLL

123

3sin3

)33

()3

()9

( 221

221

21

ss

ts

Ls

Ls

L

)3)2(

13)2(

2()3)2(

3( 22221

221

sssL

ssL

)3sin313(cos3sin

313cos 222 ttetete ttt

sin2cos3)1

21

3()123( 2222

122

1

tt

sssL

ssL

23

111 Find

sss

?)52

23( 21

sssL

١٧ Dr.-Ing. Mohamed Saber Sokar

▐ Partial Fraction▐ Partial Fraction1s

11

y(t)?is What ,6

1)( 23

CBAsss

ssY

32)3)(2(1

61)( 23

sC

sB

sA

ssss

sssssY

0sput and s,by multiply ))((

61

32)3)(2(1

00

AsCs

sBsA

sss

2sput and 2,-sby multiply 632)3)(2( 000

ssss sss

103

3)2()2(

)3(1

B

ssC

ssAB

sss

103)3( 222 ssss sss

١٨ Dr.-Ing. Mohamed Saber Sokar

▐ Partial Fraction▐ Partial Fraction

AB )3()3(1

3sput and 3,sby multiply

sss ssA

ssBC

sss

333

)3()2()3(

)2(1

ss

C

33

152

ssY 1213111)(

15

tt

sssssssY

32

23

231)(

31521066)(

tt eety 32

15106)(

١٩ Dr.-Ing. Mohamed Saber Sokar

▐ Partial fraction; repeated factor▐ Partial fraction; repeated factor-1(0)y'1,y(0),42'3" 3 tetyyy

142)1(31

( )y,y( ),

22 YsYsYs

yyy

)(124137)(

3)(

234

2

sGssssss

)()(

)23)(3(124137)( 22

sHsG

sssssssssY

123)( 1

22

sD

sC

sB

sA

sAty

?Abut before, as obtained becan D C, B, ,A123

12

sssss

2)23)(3(124137)(A 2

234

02

s ssssssssQ

)23)(3( 0ssss

٢٠ Dr.-Ing. Mohamed Saber Sokar

▐ Partial fraction; repeated factor▐ Partial fraction; repeated factor124137 234 ssss)23)(3(124137)( 22

ssss

sssssY

1231

22

sD

sC

sB

sA

sA

obtainedbecan A 0,sput and ,sby Multiply 123

22

sssss

124137

,p,yp y222234

2

DsCsBssAAssss

0sputandswithateDifferenti123)23)(3( 122

sss

sAAsss

3)('A0sput ands, with ateDifferenti

1

sQ 3)(A01 s

sQ٢١ Dr.-Ing. Mohamed Saber Sokar

▐ Test signals used in control systems▐ Test signals used in control systems

٢٢ Dr.-Ing. Mohamed Saber Sokar

▐ Test signals used in control systems▐ Test signals used in control systems

٢٣ Dr.-Ing. Mohamed Saber Sokar

▐ Free & Forced Responses▐ Free & Forced Responses Forced Response [zero ICs & nonzero u(t)]Forced Response [zero ICs & nonzero u(t)]

The response of a system to nonzero input and zero initial conditions. Can be obtained byCan be obtained by

Assume zero ICs and use LT to solve for the forced response {replace differentiation with (s) in theresponse {replace differentiation with (s) in the Input/Output (I/O) ODE model}.

Free Response [u(t) = 0 & nonzero ICs] The response of a system to zero input and nonzero initial conditions.The response of a system to zero input and nonzero initial conditions. Can be obtained by

Let u(t) = 0 and use Laplace transform (LT) consideringLet u(t) = 0 and use Laplace transform (LT) considering Initial condition (IC) values to solve for the free responseresponse.

٢٤ Dr.-Ing. Mohamed Saber Sokar

▐ Transient and Steady State Response▐ Transient and Steady State ResponseEx: Find the total response of a stable first order system:

5u t t

5 10y y u

to a ramp input:

p y

with I.C.: 0 2y

- Total response

20

10 5 1 25 5 y

Y ss s s

Transfer FunctionFree Response

Forced Response

U s

- Partial fraction eq.2 31 2

2

25 5

aa aY ss s s s

2 1

21 50 50( ) 2d da s Y s

22 forced ( ) 10a s Y s

0

1 forced 22 1 00

( ) 22 1 ! 5 5

s

ss

a s Y sds ds s s

3 forced5 ( ) 2a s Y s 2 forced 0( )

s 3 forced 5( )

s

5 5

3T i td

2 10 2t ty t t a e e

Transient responseSteady state response Transient response free responsefrom Forced response from Forced response

٢٥ Dr.-Ing. Mohamed Saber Sokar

▐ Transient and Steady State Response▐ Transient and Steady State ResponseEx: Find the response of a given second order system:

5sin 3u t t4 3 6y y y u

to a sinusoidal input:

p g y

with I.C.: 0 0, 0 2y y

- Total response 2 2 2 2

6 5 3 2 4 24 3 3 4 3

sY ss s s s s

Forced Response Free Response

- Partial fraction equation

31 2 4 1 2aa a a b bY s

3 1 3 3 3 1s

s s s j s j s s

292

a 152

a 3112

a j 4112

a j 2 3b 1 1b 2 22 3 2

3 33 1 1 2 22 Re jt t ty t a e a b e a b e

4 2 1

Steady state response Transient response

1 37 155 sin 3 tan 22 2

t tt e e

2 2

٢٦ Dr.-Ing. Mohamed Saber Sokar

▐ Transfer functionThe transfer function H() of a system is the frequency

▐ Transfer functionThe transfer function H() of a system is the frequencydependent ratio of the system output Y(s) to a systeminput U(s)input U(s).

H(s) = )()(

sUsYU(s) Y(s)

)(V

)(sU

)(V)(V gain Voltage )(H

i

o

)(I)(I gain Current )(H o

)(Ii

٢٧ Dr.-Ing. Mohamed Saber Sokar

٢٨ Dr.-Ing. Mohamed Saber Sokar