Post on 23-Dec-2015
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.
0)(,0)(,)(
)()( sXsY
sX
sYsG
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.
ss YYYXXX ,
Example:
In the mixing tank system the following function was obtained. Evaluate the transfer function.
1
1
)(
)()(
)()1)((
)()()(
))()(()(
))()(()0()(
)(
0)(
ssC
sCsG
sCssC
sCsCssCq
V
sCsCqsVsC
sCsCqCssCV
CCqdt
dCV
i
i
i
i
i
tC
i
ot
q,Ci
q,C
Example:
Consider the blending system with two input units.
)()(
)()(
)(
)(
)(
)(
22
11
2211
221121
221121
2211
)(
2211
21
21
xxV
wxx
V
w
dt
dx
xxwxxwdt
dxV
wxxwxwxwxwxwdt
dxV
wxxwxwwwwxdt
dxV
wxxwxwdt
dVx
dt
dxV
dt
Vxd
wxxwxwdt
xVd
wwwdt
Vd
wwwx
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 ( )
3. Stream 2 is pure material A, x2=1
4. Process is initially at steady-state,
Since x1 and x are very small, required flow rate of pure component w2 will be also small.
w1=w=constant
11 ww
xwwxw )1(0 211
21 wwxwxdt
dxV
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;
021 xwwxw
Subtracting steady state equation from general equation gives;
xw
wx
dt
xd
w
V
wwwxxxxxx
xxwww
xxdt
xxd
w
V
xwxwwwxwxwdt
xxdV
xxwwwxxwdt
xxdV
21
222111
2211
2211
2211
1
,,
)()(1
)()(
)(
)()()()(
dividing both sides with givesw
defining the deviation variables;
: 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.
0)(,1
)()(
)(
0)(,1
1)(
)(
)(
)(1
1)(
1
1)(
)()()1)((
)()())0()((
122
211
21
21
21
21
sXs
KsG
sW
sX
sWs
sGsX
sX
sWKs
sXs
sX
sWKsXssX
sXWKsXXsXs
xwKxdt
xd
In transfer functions there can be a single output and a single output. However, in this equation there exists two inputs.
Properties of Transfer Functions
1. 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.
2. In any transfer function order of the denominator polynomial is the same as the order of the equivalent differential equation.
01
1
01
1
0
0
011
1
1
011
1
1
.........
.........
)(
)()(
.........
..........
asasa
bsbsb
sa
sb
sU
sYsG
ubdt
dub
dt
udb
dt
udb
yadt
dya
dt
da
dt
yda
nn
nn
mm
mm
n
i
ii
m
i
ii
m
m
mm
m
m
n
n
nn
n
n
st.st. gain is obtained by setting s to zero, therefore b0/a0
3. Transfer functions have additive property.
Y(s)
U1(s)
U2(s)
G1(s)
G2(s)
U3(s)
U4(s)
)()()()()(
)()()(
2211
43
sUsGsUsGsY
sUsUsY
X3 (s)
G1(s)
G2(s)
X1(s)
X2(s)
X0(s)
)()()(
)(
))()()(()(
)()()()()(
)()()(
210
3
2103
02013
213
sGsGsX
sX
sGsGsXsX
sXsGsXsGsX
sXsXsX
4. Transfer functions also have “multiplicative property”.
G1(s) G2(s)U(s)
Y1(s) Y2(s)
)()()()()()( 12122 sUsGsGsYsGsY
! Always from right to left
qi
q
R
A
h
R
hq
dt
dhA
R
hq
hAVdt
Vdqq
i
i
.,)(
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
q1
q2
R1
R2
A1
A2
h1
h2
1
11
,
,111
11
i1
1
,1,1
,11
si,
1
11
11
11
1
1q0
eqn.2 q
1 eqn. 1
R
hq
qqq
hhh
hR
qdt
hdA
R
hq
hR
R
h
hR
qdt
dhA
siii
s
ss
s
i
for tank 1
in order to convert variables into deviation variable form, steady state equations for eqn 1 and 2 should be written;
subtracting st.st. equations from general equation gives;
where
111
1
1
1
1
11
1
11
111
1111
1
1111
11
11
110
111
11
i1
11
)(
)(
11)(
)(
2,)(
)(
)()1)((
)(1()(
)()()(
)()(1
)(
)(1
)())0()((
1
1
KRsH
sQ
s
K
s
R
sQ
sH
eqnR
sHsQ
sQRssH
sQRsRAsH
sQR
sHssHRA
sQsHR
ssHA
sHR
sQHssHA
hR
qdt
hdA
i
i
i
i
i
i
taking Laplace transform of eqns 1 and 2 gives;
these two transfer functions give information about;
input:Qi, output;H1
and
input:H1, output:Q1
however, relationship between Q2 and Qi is required
)1)(1(
1
)(
)(
)(
)(.
)(
)(.
)(
)(.
)(
)(
)(
)(
11
)(
)(
11)(
)(
21
2
1
1
1
1
2
2
22
222
2
2
2
22
2
1
2
2
sssQ
sQ
sQ
sH
sH
sQ
sQ
sH
sH
sQ
sQ
sQ
KRsH
sQ
s
K
sRA
R
sQ
sH
i
ii
for tank 2
required ;
for interacting systems;
qi
q1q2
R1 R2
A1
A2
h1 h2
2
22
211
1
2221
111
)(1
R
hq
hhR
q
dt
dhAqq
dt
dhAqqi
s
s
s
s
siii
ss
sss
ss
ssi
hhh
hhh
qqq
qqq
qqq
R
hq
hhR
q
,222
,111
,222
,111
,
2
,2,2
,2,11
,1
,2,1
,1,
)(1
0
0
at st.st.
deviation variables
2
22
211
1
0
22221
0
1111
2
22
211
1
2221
111
)()(
))()((1
)(
))0()(()()(
))0()(()()(
)(1
R
sHsQ
sHsHR
sQ
HssHAsQsQ
HssHAsQsQ
R
hq
hhR
q
dt
dhAqq
dt
dhAqq
i
i
taking the Laplace transform of the equations;
)(
2
2221
2112111
)(
21111
1
21
2
1
)()()(
)()()()()(
))()(()()(
)(
)(.
)(
)(
sQ
i
sH
i
i
R
sHssHAsQ
ssHAsQsHsRAsQsQ
sHsQRsAsQsQ
sQ
sH
sQ
sQ
ssRAsssQ
sQ
ssRAss
R
sQ
sH
R
sRAsRAsRAsRAsRAsHsQ
ssHAR
sHssHAssHsARA
R
sHsRAsQ
ssHAR
sHssHAssHA
R
sHsRAsQ
i
i
i
i
i
2212
211
2
2212
211
22
2
22212211112
222
2212211
2
211
222
22122
2
211
1
1
)(
)(
1)(
)(
1)()(
)()(
)()()(
)(
)()(
)()()(
)(
2211