[A305] Otomatik Kontrol Ders Notu (Slayt)

8/12/2019 [A305] Otomatik Kontrol Ders Notu (Slayt)

1/27

LAPLACE TRANSFORMS

Definition of the Laplace transform:

0

[ ( )] ( ) ( )stf t f t e dt F s

0

0 0

a tU t

t

0

[ ( )] ( ) ( )stu t u t e dt U s

00

( )st

st ae aU s ae dt

s s


2/27

( )( )

0 0 0

1[ ]

s a tat at st s a e

L e e e dt e dts s

Aadaki rampa (ramp) fonksiyonu analitik yntemle znz

0

0 0

bt tf t t

0 0

( ) ( ) st st F s f t e dt bte dt

0

0 0

1(1)

stst steb te dt bt b e dt

s s

200

0( ) ( )

ststb b e b be dt

s s s s s s


3/27

Properties of Laplace transforms:

1) Linearity : a sabit bir say veya s ve t den bamsz iseL[af(t)]=aL[f(t)]=aF(s)

2) Sperpozisyon : her iki fonksiyonunda laplace dnmalnabiliyorsa

1 2 1 2 1 2[ ( ) ( )] [ ( )] [ ( )] ( ) ( )f t f t L f t L f t F s F s


4/27

3)Translation in time:

[ ( )] ( )asf t a e F s

4)Complex Differention:

[ ( )] ( )d

tf t F sds

5)Translation in the s domain:

[ ( ) ( )ate f t F s a


5/27

6)Real differantiation:

2 2

[ ( )] ( ) (0 )

[ ( )] ( ) (0) (0)

L Df t sF s f

D f t s F s sf Df

7)Final value Theorem:

0

( ) ( )lim lims s

sF s f t


6/27

Example:

3( )

( 2)Y s

s s

Solution:

0 0 0

3 3 3( ) ( ) ( )

( 2) 2 2lim lim lim lims s s sy t sY s s

s s s

8)Initial value Theorem:

0

( ) ( )lim lims s

F s f t


7/27

Laplace Transforms of Most Common Functions of TimeContinuous Function Laplace Transform

Impulse 1

Steps

1

t2

1

s

2t 32

s

ate as1

atte 2)(1

as

Sin(wt))( 22 ws

w

Cos(wt) )( 22 wss


8/27

rnek:2

3( )

( 2 5)f s

s s s

1 2 32 2

3

( 2 5) 2 5

K K s K

s s s s s

1 2 32 2

3 ( )

( 2 5) 2 5

K K s Ks s s

s s s s s s

1

3

5K

22 3

3 63 ( ) ( ) 3

5 5K s K s


9/27

1 2 32 2

3

( 2 5) 2 5

K K s K

s s s s s s

2 21 1 1 2 33 2 5K s K s K K s K s

1

3

5K idi.

22 3

3 33 ( ) 3 (2 )

5 5K s x K s

22 3

3 63 ( ) 3 ( )

5 5K s K s


10/27


11/27

2 2

( )[ cos sin ]

( )at at A s a Bw

L Ae wt Be wts a w

2 2

23 ( 1)35 2( )

5 ( 1) 2

sF s

s s

3 3 1( ) (cos2 sin 2 )

5 5 2t

t e t t

rnek:2

2( )

( 1)( 2)f s

s s

1 2 32 2

2( )

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

K K Kf s

s s s s s


12/27

2 12 3

2( 2) ( 2)

1 1

Ks K s K

s s

1 21

2 22

( 1)( 2) ( 1 2)sK

s s

1 2K

2s 2 2K

1 32 2

2 ( 2)

( 1) ( 1)

s sK K

s s


13/27

2 2 2 21 2 32 2

2( 2) ( 2) ( 2) ( 2)

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

K K Ks x s s s

s s s s s

12 32 ( 2) ( 2)

( 1) 1Ks K s K

s s

leminin trevi alndnda

s = -2ye yaklar.

3 2K


14/27

1 2 32 2

2( 2) ( 2) ( 2) ( 2)

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

K K Ks s s s

s s s s s

2s iin ; 30 0 K

2 12 3

2( 2) ( 2)

1 ( 1)s K s K

s s

3 12 2

(0)( 1) (1)(2) [0( 1) 1]0 (2 4)

( 1) ( 1)

s sK K s

s s

2 2

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

s s s s

s s


15/27

=2 2

2

2( 2 2) ( 2)

( 1)

s s s s

s

=

2 2

2

2 2 2 4 4

( 1)

s s s s s

s

2( 2)

( 1)

s

s

=

2

2

2

( 1)

s s

s


16/27

Bir Fonksiyonun Tekil Noktalar ve Kutuplar

S dzleminde tekil noktalar, fonksiyonun yada trevinin bulunmadnoktalardr.Kutup, tekil noktadr.

G(s) s civarnda analitik ve tek deerlidir.

[( ) ( )]limi

r

is s

s s G s

2

10( 2)( )

( 1)( 2)

sG s

s s s

fonksiyonunun sfrlar s=-2 de bir sonlu ve

sonsuzda 3 sfr vardr. s=-3 de katl, s=0 da ve s=-1 de katsz kutbu

vardr.G(s) fonksiyonu bu noktalar dnda analitiktir denir.


17/27

3

10( ) 0lim lim

s s

G ss

Adi Dorusal Diferansiyel Denklemler:

Seri RLC devresini ele alalm;

( ) 1( ) ( ) ( )

di ti t L id t e t

dt C .()

kinci mertebeden bir diferansiyel denklem:

11

11

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

n n

n nn n

d y t a d y t dy t a a y t f t

dt dt dt

( )


18/27

Katsaylar y(t)nin bir fonksiyonu olmad srece dorusal adidiferansiyel denklemdir.

()da 1( ) ( )x t i t dt

ve 12( )

( ) ( )dx t

x t i tdt

21 2

( ) 1 1( ) ( ) ( )

dx t Rx t x t e t

dt LC L L


19/27

1. mertebeden durum deikenleri;

1

2

( ) ( )

( )( )

x t y t

dy tx t y

dt

( ) ..

.1

1

1

( )( )

nn

n n

d y tt y

dt


20/27

1 2

2 3

x

x

.

.

.

1n nx

1 1....n n na x a x u

Dinamik Sistemlerin Matematiksel Modeli

Lineer Sistemler: Bir sisteme sperpozisyon teoremi uygulanyorsasistem lineerdir.


21/27

1 1( ) ( )t y t se 1 2 1 2( ) ( ) ( ) ( )x t x t y t y t

2 2( ) ( )t y t

Lineer zamanla deimeyen ve lineer zamanla deien sistemler:

Bir diferansiyel denklemin katsaylar sabit ise veya fonksiyonlarbamsz deikenlerden oluuyorsa lineerdir.( Zamanla deiensistemlere rnek:Uzay arac kontrol sistemidir.Yakt tketimindendolay uzay aracnn ktlesi deiir.)

Dorusal olmayan sistemler:Bir sisteme sperpozisyon teoremi

uygulanamyorsasistem nonlineerdir.


22/27

22

2sin

d x dxx A wt

dt dt

22

2 ( 1) 0

d x dx

x xdt dt

23

20

d x dxx x

dt dt


23/27

Dinamik Sistemlerin Durum Uzay Gsterimi

1( )t ve 2 ( )x t durum deikenleri olsun;

u(t); Giri, 11 12 21 22 11 21, , , , ,a a a a b b ise sabit katsaylar:

111 1 12 2 11

( )( ) ( ) ( )

dx ta x t a x t b u t

dt

221 1 22 2 21

( )( ) ( ) ( )

dx ta x t a x t b u t

dt

1

2

( )( )

( )

x tx t

x t


24/27

Durum denklemleri;

( )( ) ( ) ( )

dx tx t Ax t Bu t

dt

ile ifade edilir.

1

2

n

x

x

x

,


25/27

A =

1 2

0 1 0 0

0 0 1 0

0 0 0 1

n n n n xa a a a


26/27

B =

0

0

0

1

k ( y= Cx) Y =

1

2

1 0 0

n

x

x

x

x


27/27

Filename: kon_sis_tem_2.docDirectory: C:\Documents and

Settings\Administrator\Desktop\FUNDAMENTALS OF CONTROLSYSTEMS\kontrol_temelleri

Template: C:\Documents and Settings\Administrator\Application

Data\Microsoft\Templates\Normal.dotmTitle: LAPLACE TRANSFORMSSubject:Author: hpKeywords:Comments:Creation Date: 09.10.2009 11:01:00Change Number: 39Last Saved On: 08.07.2010 15:34:00Last Saved By: PERFECTTotal Editing Time: 541 Minutes

Last Printed On: 08.07.2010 15:40:00As of Last Complete Printing

Number of Pages: 26Number of Words: 752 (approx.)Number of Characters: 4.293 (approx.)