A A 1 c 1 2 1 If ; ; K ; K 1 2 1 2 c c - - PowerPoint PPT Presentation

Department of Chemical Engineering I.I.T. Bombay, India

) ( 1 ) ( ) ( ; 1 ; c A ; c A If

1 2 2 1 2 2 1 2 1 2 2 1 2 1 1 2 2 2 1 1 1

s u s K s K K s y c c K c K                

First Principles Model The roots of the denominator (poles of the transfer function) could be complex. They could give rise to oscillatory behaviour for different u.

Department of Chemical Engineering I.I.T. Bombay, India

Second Illustrative example: Control of a first order process

Ku y dt dy             



t d I d c

dt y y y y K t u ) ( 1 ) ( ) ( 

and

          



t d I d c

dt y y y y KK y dt dy ) ( 1 ) (  

Therefore,

d I c c

y y KK dt dy KK dt y d       ) 1 (

2 2 1

This is a second order system and will give rise to a second

• rder transfer function.

Plant controller yd +

• y

u

Department of Chemical Engineering I.I.T. Bombay, India

plane of initial rest when DP = 0 h h P1 P2 DP = P1 - P2

U-Tube Manometer

P g h dt dh R g L dt h d g L D       2 1 4 2

2 2 2

L = length of fluid in the manometer tube ,  = density and viscosity

• f manometer fluid

R = radius of manometer tube g = gravitation constant

Department of Chemical Engineering I.I.T. Bombay, India

General second order transfer function

 

1 2 ) ( ) ( ) (

2 2

    s s K s u s y s G   

The value of  (damping factor) determines the dynamic response of second order systems. It represents viscous or dissipative forces. If 0   < 1 (underdamped system) - oscillatory approach to steady state  1 (critically damped, overdamped systems) - non-oscillatory approach to steady state.  is also called the natural time constant of the system. For the manometer

3 2 g L   g L R 6

2

   

Department of Chemical Engineering I.I.T. Bombay, India

Step Response of Overdamped Second Order Systems

s A s u  ) (

         

  2 1 / 2 / 1

2 1

1 ) (    

  t t

e e KA t y

        

  2 1 / /

2 1

 

  t t

e e KA dt dy

= 0 at t= 0 Initial Slope = 0. This is in contrast to that of a first order system

Department of Chemical Engineering I.I.T. Bombay, India

Step Response of Critically Damped Second Order Systems

s A s u  ) (

= 0 at t= 0 Again, the initial Slope = 0

              

 



/

1 1 ) (

t

e t KA t y       

 2 /



 t

e t KA dt dy

Department of Chemical Engineering I.I.T. Bombay, India

Step Response of Underdamped Second Order Systems

s A s u  ) (

= 0 at t= 0 Again, the initial Slope = 0

                                           



t t e KA t y

t

     

  2 2 2 /

1 sin 1 1 cos 1 ) (

                              

 2 2 /

1 1 sin    

 

t e KA dt dy

t

Department of Chemical Engineering I.I.T. Bombay, India

tr

Rise time tr: Time at which the output first hits the steady state value

) cos ( 1

1 2

   



  

r

t

Step Response of Underdamped Second Order Systems

Department of Chemical Engineering I.I.T. Bombay, India

Time to first peak tp: Time at which the output hits the first maximum value

tp Peak time

Department of Chemical Engineering I.I.T. Bombay, India

a a = max(y) - D y b = D y b

) 1 exp(

2

      OS

Overshoot (OS) = a / b

Department of Chemical Engineering I.I.T. Bombay, India

ts b = D y

0.95 * b 1.05 * b Time taken to reach and remain within 5% of the total change in y (95% response time)

Settling time

Department of Chemical Engineering I.I.T. Bombay, India

a a = value of first peak - D y c = value of second peak - D y Decay ratio (DR) = c / a c

 

) 1 2 exp(

2 2

       OS DR

Department of Chemical Engineering I.I.T. Bombay, India

Period of Oscillation, P = time between successive peaks = time between successive valleys

P

                              

 2 2 /

1 1 sin    

 

t e KA dt dy

t

2

1 2      P

Department of Chemical Engineering I.I.T. Bombay, India

Frequency Response

 

1 2 ) ( ) ( ) (

2 2

    s s K s u s y s G   

with ) sin( ) ( wt A s u  It can be shown that the output y is also a sinusoid with the same frequency as the input and is given by, ) sin( ) 2 ( ) 1 ( ) (

2 2 2 2

       

  

wt w w AK t y

t

and

        



) 1 ( 2 tan

2 2 1

    w w