An introduction to process control systems Constantinos Siettos

(1)

An introduction to process control systems

Constantinos Siettos

School of Applied Mathematics and Physical Sciences

National Technical University of Athens

(2)

Control Systems

Linear, Non Linear and Adaptive

Control: Sensing + Computation + Actuation

(3)

Open and Closed Loop Systems

(4)

First Applications

Watts Engine (1788)

(5)

Εφαρμογές • Navigation systems

• Robotics

• Chemical Reaction Control

• Communication

• Engines

• Biologocial

• Food Engineering

• Economic

(6)

Applications : Flying of insects

(7)

Outline

• Laplace transformation

• Solution of linear differential equations with Laplace

• Transfer Functions

• Dynamic responses of first and second order

• Intro to Feedback control systems

• Stability Ruth-Hurwitz, Root-Locus

• Feedback controllers

• Digital controllers

(8)

Laplace transformation

Laplace

transform Solution in s domain

inverse Laplace transform

Solution in time Signal in

time

• Other transformations

• Fourier

• z-transform

• wavelets

 ^ ^



 ( ) 0 ( )

)]

(

[ f t F s f t e ^st dt L

s =σ+jw

(9)

Laplace Transfromations of known functions

Function f(t) F(s)

Impulse Step

Ramp Exponential

Sine

1

s 1

2

1 s

a s 

1

2 s2



 

1 ) ( t  f

t t

f ( )  e

at

t f ( ) 

) sin(

)

( t t

f  

 





 

0 0

0 ) 1

( t

t t f

Cosine ^{f t} ^{( ) cos(} ^ ^ ^t ⁾

2 2

s

 s

(10)

Laplace Transformation-Properties

0 t 0

( ) ( ) ( ),

L d f t sF s f dt

     



 

 

 

0

)

t

F s

L f( dτ

  s



Linearity

L a f t [

_{1 1}

( )  a f t

_{2 2}

( )]  a F s

_{1 1}

( )  a F s

_{2 2}

( )

Derivative

Integral

1 2 1 2

0

) ( ) ( )

t

L f (tτ)f (τ dτ F s F s   

Time delay

^{Lf t}  ^ ^  ^ ^{e F s}

^^s^

 

Transformation of

^{e f t}

^^^t

  ^{L e f t} ^ _

^^^t

  ^{ } _ ^{F s a}  ^ 

Convolution

(11)

Laplace Transformation-Properties

Final Value Theorem

^f   ^{ } ^lim

_t_

^{f t}   ^ ^lim

_s_₀

^{sF s}  

       

0 0 0

0

0 0 0

st st st

st

df t df t

L e dt e f t se f t dt

dt dt

f s f t e dt sF s f

 

   





 

           

 

      

 



     

0 0

0

0 lim

^st

lim

s s

df t e dt sF s f

dt

 

 

 

 

   

 

  

Taking the limit

 

₀

   

0

0 lim

s

df t dt sF s f

dt





 

   



or

     

0 0

0

0 lim

^st

lim

s s

df t e dt sF s f

dt

 

 

 

 

   

 

  

PROOF

or ή

^{f t}  ^{  }  ^f   ⁰ ^ ^lim

_s_₀

^ _ ^{sF s}   ^ _ ^ ^f   ⁰

(12)

Laplace Transformation-Properties

Find the value of the function

^{f t}  

^for

^t ^{ }

     

5 1

2 3 4

s s

F s s s s s

 

   

  ^lim

_t

  ^lim

_s ₀

 

f f t sF s

 

  

       

     

0 0

5 1 5

2 3 4 24

lim lim lim

t s s

s s

f t sF s s

s s s s

  

 

  

  

when

(13)

Laplace Transformation-Properties

       

0 0 0

0

0 0 0

st st st

st

df t df t

L e dt e f t se f t dt

dt dt

f s f t e dt sF s f

 

   





 

           

 

      

 



     

0

0 lim

^st

lim

s s

df t e dt sF s f

dt

 

 

 

 

   

 

  

Taking the limit

   

0 lim 0

s

sF s f



 

   

or

     

0

0 lim

^st

lim

s s

df t e dt sF s f

dt

 

 

 

 

   

 

  

PROOF

or

  ⁰ ^lim

_t ₀

^{ } ^lim

_s

^{ } ^, ⁰

f

^

f t sF s t

 

  

(14)

Laplace Transformation-Properties

Initial Value Theorem

Find the initial value of a function

^f   ⁰

     

5 1

2 3 4

s s

F s s s s s

 

   

       

   

0

2 2

5 1

0 2 3

4 5 5 6 1

lim lim lim

lim

t s s

s

s s

f sF s s

s s s

s s

  



 

  

 

 

 

 

when

  ⁰ ^lim

_t ₀

^{ } ^lim

_s

^{ } ^, ⁰

f

^

f t sF s t

 

  

(15)

Laplace Transformations

 

0

)

t

F s

L f( dτ

  s



PROOF

0 0 0

) )

t t

L f( dτ  _{ }

^

^ _ f( dτ e dt  ^ _ _

^st

 

  

We set

u e 

^^st

 

0 t

v   f t dt

και

du  se dt

^st

^{dv f t dt} ^  

 

   

     

0

0 0 0 0

0 0 0

1 1

1 1 1 1

0 0 )

t

st

t

st st

f( dτ e dt vdu vu udv

s s

f t dt e e f t dt s

F s F s

s s s





 

 

 

 

         

   

   

   

   

 

      

 

 

 

   

   

 ^ 

(16)

Laplace Transformations

   

L e  

^at

f t    F s a 

PROOF

       

0 0

( )

at at st a s t

L e f t e f t e dt e f t dt F a s

 

    

     

   

at

1 L e s a





 

  

PROOF

 

0 0 0

( )

1

s a t

at at st a s t

e

L e e e dt e dt

s a s a

  

 

    

 

       

         

(17)

Laplace Transformations

       

2

2 2

0 0

d f t df

L s F s sf

dt dt

 

  

 

 

PROOF

   

         

2 2

0 0 0 0

2 0

0 0

at st st st st

st

d f t d f t d df df df

L e e dt e dt e e dt

dt dt dt dt

dt dt

df df t df

s e dt s F s sf

dt dt dt

   

    





          

         

 

 

 

     

 

 

  



         

   

1

1 2

2 1

1 1 0 1

0 0

...

n n

n n n

n

n n i

n i

i n i

d f t df d f

L s F s s f s

dt

dt dt

d f

s F s s

dt



 



  

  

 

     

 

 

 

IN GENERAL

(18)

Laplace Transformations with Matlab

Use of Symbolic toolbox of Matlab Commands:

L = laplace(F)

is the Laplace transform of the scalar symbol F with default independent variable t. The default return is a function of s.

L=laplace(F,t)

makes L a function of t instead of the default s.

L = laplace(F,w,z

) makes L a function of z and F a function of w instead of the default variables s and t, respectively

>> syms f t

>> f=t^4

>> laplace(f) ans =

24/s^5

syms w t

>> laplace(sin(w*t)) ans =

w/(s^2+w^2)

>> laplace(exp(-w*t)) ans =

1/(s+w)

(19)

Solution of D.Ε.with Laplace

Two kinds of linear differential equations

   

0 n i

i i

i

d y t

a u t



dt

  ^{ } ^{ }

0 0

i i

n m

i i i i

i i

d y t d u t

a b

dt dt

 

  

 

u t

Input signal

^{y t}  

Output signal

Initial Conditions ⁱ

 

i

⁰

i

  ^{0 ,} ⁰ ^,...., ¹

d y y i n

dt   

The Laplace transformation of (1) is given by

 

¹ ¹

   

0 0

n i

0

i i j

i j

a s Y s

^

s

^{ }

y U s

 

 

 

 

 

 

We solve for

^{Y s}      

¹ ¹

 

0 0

0

n i

i j

i j

n n

i i

a s y Y s U s

a s a s

  

 

   

 

(1) (2)

1

1 1 0

0

....

n i n n

i n n

i

a s a s a s

_ ^

a s a



    

 Characteristic Equation

n

1 a 

(20)

Solution of D.E. with Laplace

   

¹ ¹

 

0 0

n i

0

i j

i j

n n

i i

a s y Y s U s

a s a s

  

 

   

 

   

¹ ¹

 

0 0

1 1

0 0

0

n i

i j

i j

n n

i i

a s y

y t L U s L

a s a s

  

 

 

 

 

 

 

 

   

 

For (1)

(21)

Solution of D.Ε. with Laplace

   

¹ ¹

 

¹ ¹

 

0 0 0 0

0

0 0 0

0 0

m i n i

m i i j i j

i j i j

i i j i j

i

n n n

i i i

b s u a s y

b s

Y s U s

a s a s a s

 

   

   



  

       

  

For (2)

   

0 0

i i

n m

i i i i

i i

d y t d u t

a b

dt dt

 

  

(22)

Inverse Laplace Transformation with expansion in partial fractions

   

 

⁰

0

m j

j j

n i

i i

Q s b s F s P s

a s



  

 ^a

ⁿ

^ ¹

Let

Let the characteristic equation has n roots(poles) so that the pole s₁has multiplicity r₁, pole s₂ multiplicity r₂…… pole s_kmultiplicity r_k:

1 k

i i

r n



 

^Then

^ ^

0 1

i

k

n i r

i i

a s s s

 

 

 

Hence

 

0

1

i

m j

j j

k r

i i

b s F s

s s











The expansion in partial fractions is given by

 

¹ ¹

  ^, ^{0 iff}

ri

k ij

n j n

i j i

F s b c b m n

s s

 

   

 

 ¹     

!

i

i i

i

r j r

ij r j i

s s

i

c d s s F s

r j ds



 

 

    

(23)

Example 1

 

¹ ¹

  ^, ^{0 iff}

ri

k ij

n j n

i j i

F s b c b m n

s s

 

   

  _ ¹ _ _!

ⁱi

^ ^{  }

ⁱ

i

r j r

ij r j i

s s

i

c d s s F s

r j ds



 

 

    

      ¹ ⁵ ^s ² ³ ³

F s s s s

 

  

 

¹¹ ²¹ ³¹

1 2 3

c

c c

F s  s  s  s

  

     

   

     

   

     

   

11 1

21 2

31 3

5 1 3

1 1

2 1 3 1

5 2 3

2 7

1 2 3 1 5 3 3

3 6

1 3 2 3

s

c s F s



   

        

   

       

   

        

(24)

Example 2

 

¹ ¹

  ^, ^{0 iff}

ri

k ij

n j n

i j i

F s b c b m n

s s

 

   

  _ ¹ _ _!

ⁱi

^ ^{  }

ⁱ

i

r j r

ij r j i

s s

i

c d s s F s

r j ds



 

 

    

    

³



1 1 2

F s  s s s

   

¹¹



¹²

¹  

¹³

1  

² ¹⁴

1  

³ ²¹

² 

c c c c c

F s  s  s  s  s  s

   

 

   

     

11 0

21 2

3

14 1

3

13 1

1

2 2

3

12 2 2

1 1

1 2 2 1

2 1 1

1 1 0

2 1 1 1

1 1

2 ! 2 2

s

s s

c sF s

c s F s

d d

c s F s

ds ds s s

d d

c s F s

s s

ds ds





 

 

   

 

    

 

     

 

            

 

             

(25)

Use of Matlab για for expansion in partial function of the inverse Laplace

Residue: Convert between partial fraction expansion and polynomial coefficients Syntax

[r,p,k] = residue(b,a) [b,a] = residue(r,p,k)

      ¹ ⁵ ^s ² ³ ³

F s s s s

 

  

>> num=[5 3];

>> den=poly([-1 -2 -3]);

>> [r,p,k] = residue(num,den)

r =

-5.99999999999999 7

-1.00000000000001

p =

-3 -2 -1 R: Column vector of residues P: Column vector of poles K: Row vector of direct terms

(26)

Use of Matlab για for expansion in partial function of the inverse Laplace

Residue: Convert between partial fraction expansion and polynomial coefficients Syntax

[r,p,k] = residue(b,a) [b,a] = residue(r,p,k)

 

₂

¹

2 F s s

s s

 



>> num=[1 1];

>> den=[1 2 0];

>> [r,p,k] = residue(num,den)

R: Column vector of residues P: Column vector of poles K: Row vector of direct terms

r =

0.5 0.5 p =

-2 0

k = []

(27)

Inverse Laplace Transformation with the use of partial functions

       

1 1 1

1 1 1 1

1 !

i i

i

r r

k k

ij ij j s t

n j n

i j i i j

c c

L F s L b b t t e

s s  j

^

  

   

 

 

             

Example

^{F s}      ^ ^s ^s ^

²

^ ² ² ^s ^s ^ ^ ¹ ³

     

1 1

1 1 1 3 2

4 1

1 3 2

4 1

1 4

3 2

t t

L F s L

s s

L L L t e e

s s 

 

    

  

     

 

 

 

   

         

 

   

   

(28)

Use of Matlab for the inverse Laplace Transformation

Use of Symbolic toolbox of Matlab Commands:

L = ilaplace(F)

is the Laplace transform of the scalar symbol F with default independent variable t. The default return is a function of s.

L=ilaplace(F,t)

makes L a function of t instead of the default s.

L = ilaplace(F,w,z

) makes L a function of z and F a function of w instead of the default variables s and t, respectively

>> syms F,s;

>> F=(s^2+2*s+1)/((s+2)*(s+3));

>> ilaplace(F) ans =

dirac(t)-4*exp(-3*t)+exp(-2*t)

(29)

Example of the solution of D.Ε. With Laplace

0 )

0 ( ' )

0 ( 2

8

2

6

2

  y  y  y 

dt dy dt

y

d • ODE / initial conditions

• Laplace Transformation

• Solution for Y(s)

• partial fraction expansion

• Inverse Laplace Tranformation

s s

Y s

Y s s

Y

s

²

( )  6 ( )  8 ( )  2 /

) 4 (

) 2 (

) 2

(   

s s

s s Y

) 4 (

4 1 )

2 (

2 1 4

) 1

(  



 

 s s s

s Y

4 2

4 ) 1

(

4 2t

e

t

t e y









(30)

Example of Solution of D.Ε. with Laplace- illustration with Matlab

0 )

0 ( ' )

0 ( 2

8

2

6

2

  y  y  y 

dt dy dt

y

d ODE / αρχικές συνθήκες

4 2

4 ) 1

(

4 2t

e

t

t e y









>> t=0:0.01:10;

>> y=(1/4)-(1/2)*exp(-2*t)+(1/4)*exp(-4*t);

>> plot(t,y)

>> xlabel('Time')

>> ylabel('System Response')

0 1 2 3 4 5 6 7 8 9 10

0 0.05 0.1 0.15 0.2 0.25

Time

System Response

An introduction to process control systems Constantinos Siettos