Process Control Systems (2) C. Siettos N.T.U.A.

Process Control Systems

(2)

C. Siettos

N.T.U.A.

Transfer Function

differential equation

r(t) y(t)

transfer function

r(s) y(s)

A Transfer Function is a relation which

connects the output with the input of a linear systems in the s-domain (SISO)

H(s) = y(s) / r(s)

Transfer Function

H(j w )

( ) ( ) ( ) Y s  H s X s

est

s X )(

   

0 0

i i

n m

i i i i

i i

d y t d u t

a b

dt dt

 





Let

   

 

 

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

i i i

i i i

b s u a s y

b s

Y s U s

a s a s a s

 

   

   



  





 

 

  

Then the LaplaceTransformation is

Initial Conditions ⁱ

 

 

d y y i n

dt   

From the Transfer Function it can be derived the differential equation by substituting

s d

 dt

Transfer Function

H(j w )

( ) ( ) ( ) Y s  H s X s

est

s X )(

The Transfer Function is the Laplace (Fourier) Transformation of the Response in dirac function : H(s) (H(jw))

The poles and roots are given by H(s)Pi(s-zi)/Pj(s-pi).

It allows the computation and (from differential equations) and the analysis of the response

f(t)

t

 (t

) F(s) = e

 (t

) dt = e

0



MIMO Transfer Functions

   

 

     

     

     

   

 

1 11 12 1 1

2 21 22 2 2

1 2

...

...

...

...

m m

n n n nm m

Y s G s G s G s U s

Y s G s G s G s U s

Y s G s G s G s U s

     

     

     

     

     

     

     

    

Set of differential equations U1

Um

Y1

Yn

1 ^st Order System



c(t) 1 e



C(s)  R(s) G(s)  a

s(s  a)

1 1

( ) ( )

( )

C s a

s a s s a

 

  

Response of 1 ^st order system

Transient Response:

1. Time-constant, 1/a 2. Rise time, T_r

3. Settling time, T_s

Experimental Construction of Transfer Function of 1

order



G(s)  K (s  a) C(s)  K

s(s  a)  K a

s  K a (s  a)

The time that the system nees to reach 63% of its final value:

63 x 0.72 = 0.45  0.13 sec , a = 1/0.13 = 7.7

Steady State value K/α =0.72  K= 0.72 x 7.7= 5.54

G(s) = 5.54/(s+7.7) .

( ) / ( / )

c t  K a  K a e

2 ^nd order System

R Y

2

2

2

( )

( )

n n

Y s R s

s s

w

w w

  

2

2 2 2

( ) ⁿ

n n

G s s s

w

w w

  

Response inunit step

) 2

) (

( ₂

2

2

n n

n

s s

s s

Y w w

w



 

) 1 sin(

1 )

( w  



w





 e

t

t

y

_

_

2 ^nd order system

) 1 sin(

1 )

( w  



w





 e

t

t

y

^

^

frequency w

– the frequency of oscillations for two complex poles if the dissipation was zero

Dissipation ratio  - a measure of the dissipation for the second order characteristic equation



s

 2 w

s  w

 0 s

  w

 w



1 s

  w

 w



1

2

2

2

( )

( )

n n

Y s R s

s s

w

w w

  

2 ^nd order system

) 1 sin(

1 )

( w  



w





 e

t

t

y

_

_

Transfer Function of a motor of continuous current

e : electromotive force

Analogous to the angular velocity of the shaft

The torgue T is analogous to the intensity of the current i: T=Kⁱi

i

Inertia J is the inertia of the motro plus the inertia of the load bdθ/dt: the total friction

     

 

   

i 2

di t d t

Ri L V t K

dt dt

d t d t

K i t J b

dt dt



 

  

 

Τhe model:

Transfer Function of a motor of continuous current

i

2

2

, _i

di d d d

Ri L V K K i J b

dt dt dt dt

  

    

The model:

Laplace transformation:

       

     

i

R sL I s V s K s K I s Js b s

   

  

       

i i

i i

s K K JL

V s Js b R sL KK b R bR KK

s s

J L JL JL

  

         

Transfer Function

0 0.5 1 1.5 2 2.5 3 0

0.05 0.1 0.15 0.2 0.25

time

angular velocity

Use of Matlab for computing the Dynamic Response of a motor of continuous current

 

  

 ^{ }

i i

i i

s K K JL

V s Js bs R sL KK s s s b R s bR KK

J L JL JL

  

 

         

Ki=1;

J=0.01;

L=0.5;

b=1;

R=1;

K=0.01;

num=Ki/J*L;

den=[1 (b/J+R/L) b*R/(J*L)+Ki*K/(J*L)];

sys=tf(num,den);

t=[0:0.1:3];

[y,t]=step(sys,t);

plot(t,y),grid xlabel('time')

ylabel('angular velocity')

Transfer Function of Car Suspension

Mass of Car body

Suspension mass

Road

1 1 1

x x x





2 2 2

x x x





k1

k1

b1

b2

xr

The model

   

       

1 1 1 2 1 1 2 1

2 2 2 r 2 2 r 2 1 2 1 1 2 1

m x k x x b x x

m x k x x b x x k x x b x x

   

       

  

    

Input

Transfer Function of Car Suspension

Mass of Car body

Suspension mass

Road

1 1 1

x x x





2 2 2

x x x





k1

k1

b1

b2

xr

The model

   

       

1 1 1 2 1 1 2 1

2 2 2 r 2 2 r 2 1 2 1 1 2 1

m x k x x b x x

m x k x x b x x k x x b x x

   

       

  

    

Input

2

1 1 1 1 1 1 1 2 1 21

2

2 2 2 2 1 2 2 2 2 r 2 r 1 1 1 1

m s X b sX k X b sX k X

m s X b sX b X k X k X b sX k X b sX

   

      

Laplace transform

  ^{ }

  ^{   }

2

1 1 1 1 1 1 2

2

2 2 1 2 2 2 2 r 1 1 1

m s b s k X b s k X

m s b s b k X b s k X b s k X

   

      

Transfer Function of Car Suspension

Mass of Car body

Suspension mass

Road

1 1 1

x x x





2 2 2

x x x





k1

k1

b1

b2

xr

The model

  ^{ }

  ^{   }

2

1 1 1 1 1 1 2

2

2 2 1 2 2 2 2 r 1 1 1

m s b s k X b s k X

m s b s b k X b s k X b s k X

   

      

 

    ^{ }

2

1 1 1 1 1 1

2 2 2 2

1 1 2 2 1 2

0

r

m s b s k b s k X

b s k X b s k m s b s b k X

        

           

     

 

 

    ^{ }

2 1

1 1 1 1 1

1

2 2 2

2 1 1 2 2 1 2

0

r

m s b s k b s k

X X

b s k

X b s k m s b s b k

       

 

 

  

         

     

Use of Matlab for computing the response of the suspension system

Mass of Car body

Suspension mass

Road

1 1 1

x x x





2 2 2

x x x





k1

k1

b1

b2

xr

To μοντέλο

 

    ^{ }

2 1

1 1 1 1 1

1

2 2 2

2 1 1 2 2 1 2

0

r

m s b s k b s k

X X

b s k

X b s k m s b s b k

       

 

 

  

         

     

syms m1 b1 k1 m2 b2 k2 s m1=2500;

m2=320;

k1=80000;

k2=500000;

b1 = 350;

b2 = 15020;

A=[m1*s^2+b1*s+k1 -(b1*s+k1);b1*s+k1 - (m2*s^2+b2*s+b1+k2)];

B=inv(A);

C=[0;b2*s+k2];

num1=B(1,1)*C(1)+B(1,2)*C(2);

num2=B(2,1)*C(1)+B(2,2)*C(2);

G=(num1-num2);

Matlab code

Output: X1-X2

Use of Matlab for computing the response of the suspension system

Mass of Car body

Suspension mass

Road

1 1 1

x x x





2 2 2

x x x





k1

k1

b1

b2

xr

The model

syms m1 b1 k1 m2 b2 k2 s m1=2500;

m2=320;

k1=80000;

k2=500000;

b1 = 350;

b2 = 15020;

A=[m1*s^2+b1*s+k1 -(b1*s+k1);b1*s+k1 - (m2*s^2+b2*s+b1+k2)];

B=inv(A);

C=[0;b2*s+k2];

num1=B(1,1)*C(1)+B(1,2)*C(2);

num2=B(2,1)*C(1)+B(2,2)*C(2);

G=(num1-num2);

[num, den]=numden(G) num1=sym2poly(num);

den1=sym2poly(den);

step(num1,den1) Matlab code

Use of Matlab for computing the response of the suspension system

step(num1,den1) : response in unit step change of the input

Αποκρίσεις

0 5 10 15 20 25 30 35 40 45

-1 -0.5 0 0.5 1

1.5 Step Response

Time (sec)

Amplitude

step(0.1*num1,den1) : response in 0.1 step change of the input

0 5 10 15 20 25 30 35 40 45

-0.1 -0.05 0 0.05 0.1

0.15 Step Response

Time (sec)

Amplitude