Transcript Process Control

Process Control
CHAPTER IV
INPUT-OUTPUT MODELS AND
TRANSFER FUNCTIONS
Transfer Functions
Control systems are based on a single output
and a few input variables. For this reason
solution of model equation for all input
variables is usually not required.
“We need a method for “compressing” the
model”
For linear dynamic models used in process
control, it’s possible to eliminate intermediate
variables analytically to yield an input-output
model.


The Transfer Function, is an algebraic
expression for the dynamic relation
between a selected input and output
of the process model. A transfer
function can only be derived only for a
linear differential equation model
because Laplace transforms can be
applied only to linear equations.

The transfer function is a model, based on,
Laplace transform of output variable y(t),
divided by the Laplace transform of the
input variable with all initial conditions
being equal to zero.
Y ( s)
G( s) 
, Y ( s)  0, X ( s)  0
X (s)
The assumptions of y(0)=0 and x(0)=0 are
easy to be achieved by expressing the
variables in the transfer function as deviations
from the initial conditions.
 Thus all transfer functions involve variables
that are expressed as deviations from an initial
steady state.
Deviation variables; difference between variables
and their steady state values.

X   X  X s , Y   Y  Ys
Example:
In the mixing tank system the following function
was obtained. Evaluate the transfer function.
dC
V
 q (Ci  C )
dt


V  sC ( s )  C (0)   q (Ci ( s )  C ( s ))




C ( t ) t o  0 

VsC ( s )  q (Ci ( s )  C ( s ))
V
sC ( s )  C ( s )  Ci ( s )
q


C ( s )(s  1)  Ci ( s )
G (s) 
C (s)
1

Ci ( s ) s  1
q,
Ci
q,C
Example:
Consider the blending system with two input units.
d (V )
 w1  w2  w
dt
d (Vx)
 w1 x1  w2 x2  wx
dt
d (Vx )
dx
dV

 V
 x
 w1 x1  w2 x2  wx
dt
dt
dt

x ( w1  w2  w )
dx
V
 x( w1  w2  w)  w1 x1  w2 x2  wx
dt
dx
V
 xw1  xw2  xw  w1 x1  w2 x2  wx
dt
dx
V
 w1 ( x1  x)  w2 ( x2  x)
dt
dx w1
w

( x1  x)  2 ( x2  x)
dt V
V
Output:x
Inputs:x1,x2,w1,w2
One input-one output ?
Assumptions:
1.
Both feed and output compositions are dilute
(x1<x < <1)
2.
Feed flow rate w1 is constant ( w1  w1 )
3.
Stream 2 is pure material A, x2=1
4.
Process is initially at steady-state, 0  w1 x1  w2 (1)  w x
Since x1 and x are very small, required flow
rate of pure component w2 will be also small.
w1=w=constant
dx
V
 wx1  wx  w2
dt

In the definition of transfer function it was
indicated that input and output variables
should be zero at the initial conditions. In this
example, the variables have initial steady
state values different than zero. In order to
define deviation variables we should subtract
steady state equation from the general
equation.
at steady state;
w x1  w2  w x  0
Subtracting steady state equation from general equation gives;
d (x  x)
V
 w ( x1  x1 )  ( w2  w2 )  w ( x  x )
dt
d (x  x)
V
 w x1  w x1  w2  w2  w x  w x
dt
dividing both sides with w gives
V d ( x  x )
1
 ( x1  x1 )  ( w2  w2 )  ( x  x )
w
dt
w
defining the deviation variables;
x  x  x , x1  x1  x1 , w2  w2  w2
V dx
1
 x1  w2  x
w dt
w



: Time Constant
It is an indication of the speed of response of the
process. Large values of τ mean a slow process
response, whereas small values of τ mean a fast
response.
K : Steady-state gain
The transfer function which relates change in input to
change of output at steady state conditions.
The steady state gain can be evaluated by setting s
to zero in the transfer function.
dx 

 x1  Kw2  x
dt
 ( sX ( s )  X (0))  X 1( s )  KW2  X ( s )
X ( s )(s  1)  X 1( s )  KW2( s )
1 
1 


X ( s )  
 X 1( s )  
 KW2( s )
 s  1 
 s  1 
In transfer functions there can be a single output and a single output.
However, in this equation there exists two inputs.
X ( s )
1
 G1 ( s ) 
, W2( s )  0
X 1( s )
s  1
X ( s )
K
 G2 ( s ) 
, X 1( s )  0
W2( s )
s  1
Properties of Transfer Functions
1.
2.
By using transfer functions the steady state output
change for a change in input can be calculated directly.
(i.e., simply setting s→0 in transfer function gives the
steady state gain.
In any transfer function order of the denominator
polynomial is the same as the order of the equivalent
differential equation.
dny
d n 1
dy
an

a

..........

a
 a0 y 
n 1
1
n
n 1
dt
dt
dt
d mu
d m 1u
du
bm

b

.........

b
 b0u
m 1
1
m
m 1
dt
dt
dt
m
Y (s)
G (s) 

U ( s)
i
b
s
 i
i 0
n
a s
i 0
i
bm s m  bm 1s m 1  ......... b0

an s n  an 1s n 1  ......... a0
i
st.st. gain is obtained by setting s to zero, therefore b0/a0
3.
Transfer functions have additive property.
U3(s)
U1(s)
G1(s)
Y(s)
U2(s)
G2(s)
U4(s)
Y ( s)  U 3 ( s)  U 4 ( s)
Y (s)  G1 (s)U1 (s)  G2 (s)U 2 (s)
X1(s)
G1(s)
X3 (s)
X0(s)
G2(s)
X2(s)
X 3 (s)  X 1 (s)  X 2 (s)
X 3 ( s )  G1 ( s ) X 0 ( s )  G2 ( s ) X 0 ( s )
X 3 ( s )  X 0 ( s )(G1 ( s )  G2 ( s ))
X 3 (s)
 G1 ( s )  G2 ( s )
X 0 ( s)
Transfer functions also have
“multiplicative property”.
4.
Y2(s)
Y1(s)
U(s)
G1(s)
G2(s)
Y2 (s)  G2 (s)Y1 (s)  G2 (s)G1 (s)U (s)
! Always from right to left
qi
h
R
q
A
d ( V )
qi  q 
,V  A.h
dt
h
q
R
dh
h
A
 qi 
dt
R
Example:

Consider two liquid surge tanks that are placed in
series so that the output from the first tank is an
input to the second tank. If the outlet flow rate
from each tank is linearly related to the height of
the liquid (head) in that tank, find the transfer
function relating changes in flow rate from the
second tank to changes in flow rate into the first
tank.
qi
h1
R1
q1
A1
h2
A2
R2
q2
for tank 1
A1
d h1
1
 qi 
h1
dt
R1
q1 
h1
R1
eqn .1
eqn .2
in order to convert variables into deviation variable form, steady state equations for
eqn 1 and 2 should be written;
0  q i,s 
q1, s 
1
h1, s
R1
h1, s
R1
subtracting st.st. equations from general equation gives;
A1
dh
1


 qi 
h1
dt
R1
where

h1  h1  h1, s
qi


q1
 qi  qi , s

h1

R1
taking Laplace transform of eqns 1 and 2 gives;
A1
dh
 1 
 qi 
h1
dt
R1
1




A1 ( sH 1 ( s )  H 1 (0))  Qi ( s ) 
H1 ( s)

R1
0
1



A1sH 1 ( s ) 
H 1 ( s )  Qi ( s )
R1


A1 R1sH 1 ( s )  H 1 ( s )

 Qi ( s )
R1



H 1 ( s ) ( A1 R1 s  1  R1Qi ( s )





 1



H 1 ( s )( 1s  1)  R1Qi ( s )

H1 ( s)

Q1 ( s ) 
, eqn2
R1

H1 ( s )
R1
K1


Qi ( s )
 1s  1  1s  1

Q1 ( s )
1
1



K1
H 1 ( s ) R1

these two transfer functions give information
about;
input:Qi, output;H1
and
input:H1, output:Q1
however, relationship between Q2 and Qi is
required
for tank 2

H 2 (s)
R2
K2



A2 R2 s  1  2 s  1
Q1 ( s) 
2

Q2 ( s) 1
1



H 2 ( s ) R2 K 2
required ;





Q2 ( s) Q2 ( s ) H 2 ( s ) Q1 ( s ) H1 ( s)

.
.
.





Qi ( s ) H 2 ( s ) Q1 ( s ) H1 ( s ) Qi ( s )

Q2 ( s)
1


Qi ( s ) ( 1s  1)( 2 s  1)
for interacting systems;
qi
h1
R1
q1
A1
d h1
qi  q1  A1
dt
d h2
q1  q 2  A2
dt
1
q1 
( h1  h2 )
R1
h2
q2 
R2
h2
A2
R2
q2
at st.st.
qi , s  q1, s  0
q1, s  q 2 , s  0
q1, s 
1
( h1, s  h2 , s )
R1
q2 , s 
h2 , s
R2
deviation variables
qi

 q i  qi , s

q1  q1  q1, s
q2

 q2  q2 , s

h1  h1  h1, s
h2

 h2  h2 , s


qi  q1

d h1
 A1
dt

d h2
q1  q2  A2
dt
1



q1 
( h1  h2 )
R1

h2

q2 
R2
taking the Laplace transform of the equations;




Qi ( s )  Q1 ( s )  A1 ( sH 1 ( s )  H 1 (0))



0




Q1 ( s )  Q2 ( s )  A2 ( sH 2 ( s )  H 2 (0))

0
1



Q1 ( s ) 
( H 1 ( s )  H 2 ( s ))
R1

H 2 (s)

Q2 ( s ) 
R2


Q1 ( s ) H 2 ( s )
.


Qi ( s ) Q1 ( s )




Qi ( s )  Q1 ( s )  A1s ( R1Q1 ( s )  H 2 ( s ))

H1 ( s )





Qi ( s )  Q1 ( s )  A1sR1 H 2 ( s )Q1 ( s )  A1sH 2 ( s )

H 2 ( s)


Q1 ( s )  A2 sH 2 ( s ) 
R2

Q2 ( s )






H
(
s
)
H




2
2 ( s)
Qi ( s )  A1 R1s 
 A2 sH 2 ( s)  A1sH 2 ( s )  
 A2 sH 2 ( s)
 R2

 R2



H
(
s
)
H




2
2 ( s)
Qi ( s )  A1 R1s
 A1 R1sA2 sH 2 ( s )  A1sH 2 ( s ) 
 A2 sH 2 ( s)
R2
R2
1
1
2
2






  A1 R1 s  A1 R1 s A2 R2 s  A1 R2 s  1  A2 R2 s 
Qi ( s )  H 2 ( s ) 

R
2





H 2 (s)
R2

2

1


s



s
 A1 R2 s   2 s
1
1 2
Qi ( s )

Q2 ( s )
1

2

1


s



s
 A1 R2 s   2 s
1
1 2
Qi ( s )