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Chapter 2 Chapter 2 Mathematical Mathematical BackgroundBackground
Professor Shi-Shang JangChemical Engineering DepartmentNational Tsing-Hua University TaiwanMarch, 2013
1
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2-1 The Essence of 2-1 The Essence of Process DynamicsProcess Dynamics
2
As the plant is not operated at its steady state as designed, the process is
in a dynamic state.
In many cases, small deviations of process steady state (noise) is
allowed, but keep monitoring on the system is very important.
In case of strong deviations (trend), process control becomes a must.
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Example: Thermal Example: Thermal ProcessProcess
Rate of Energy Input- Rate of Energy Output=Accumulation
i i i
vi p i i p
d V u tf h t f h t
dt
d V C T tf C T t f C T t
dt
Inputs: f(t), Ti(t),Ts(t)Output: T(t)
Ts
CV: T(t)MV: f(t)DV: Ti(t), Ts(t)
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Noise v.s. TrendNoise v.s. Trend
4
0 50 100 150 20069.9
70
70.1
70.2
70.3
70.4
70.5
70.6
70.7
Noise
Trend
T
time
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Conception of deviation Conception of deviation variablevariable
5
PlantManipulative
variablesControlled variables
Disturbances
H= hdeviation = h - hsteady-
state
tank
Valve
liquid level h
control stream
Fc
Wild stream
Fw
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The Essence of Process The Essence of Process Dynamics - Dynamics - ContinuedContinued
6
The feedback/feedforward process control needs to understand the
relationships between
◦CVs and MVs
◦DVs and CVs
the relationships are called process models.
For the ease of mathematical analyses, the process modeling implements
in Laplace transform instead of direct use of time domain process model.
Implementation of deviation variables is needed.
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Conception of deviation Conception of deviation variable ( variable ( =T-70=T-70))
7
0 50 100 150 200-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
Noise
Trend
time
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2-2 Linear Systems2-2 Linear Systems
8
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2-2 Linear Systems2-2 Linear Systems
Consider a process system with a CV, say y(t), and a MV,
say m(t). In process industries, it is traditional to assume
the system is lumped and linear. A lumped process
system is such that:
A lumped system is linear if
9
11 1 0 0'n n k
n ky a y a y a y b m b m c
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Linear Systems-Linear Systems-ContinuedContinued
A lumped process system is autonomous if
An autonomous lumped system is said to be stable at the origin if
10
0 0lim 0; for some (0) ; t
y t y y y S x x
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Linear Systems-Linear Systems-ContinuedContinued An autonomous lumped system is called asymptotic
stable if
Theory 2-1: A linear system is stable if the roots of
the characteristic equation all have negative real
part.
Corollary 2-1: A linear system is asymptotic stable if it
is stable.
11
0 0lim 0; for (0) ; t
y t y y y R
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Linear Systems-Linear Systems-ExamplesExamples
12
3
2
31 2 1 2
31 2 1 2
Example 1: " 4 ' 3 0; (0) 0, '(0) 2
: charac. eqn.: 4 3 0 3; 1
( ) 0
'( ) 3 3 2
Finally, we have ( ) t
t t
t t
t
y y y y y
Sol r r r r
y t c e c e c c
y t c e c e c c
y t e e
Note that no matter what the initial conditions is, the solution indicates :
lim 0t
y t
The system is hence asymptotic stable.
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Linear Systems-Linear Systems-ExamplesExamples
13
2
31 2
3
Example 2: " 2 ' 3 0; (0) 1; '(0) 4
2 3 0
From the inital conditions, we have
17 3
4
t t
t t
y y y y y
r r
y c e c e
y t e e
Note that the above system is unstable since: limt
y t
2
3 3
2 21 2
Example 3 : " 3 ' 8 0
3 8 0
3 23 / 2
1 1cos 23 sin 23
2 2
t t
y y y
r r
r i
y t c e t c e t
Note that the above system is also unstable since: limt
y t
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Linear Systems-Linear Systems-ContinuedContinued
Theory 2-2: A forced linear system is stable if the inputs of the system is bounded.
A process control system is basically assuming that the controlled variable (CV) is to be controlled to a certain set point (zero), thus it is assumed that the input (MV) is the forcing function to a linear system with a dependent variable of CV.
14
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Linear Systems-Linear Systems-ContinuedContinued A forced system is of the following form, for example:
y”+a1y’+a0y=bm(t)
In case of m(t) is in special functions such sint the
above equation can be solved by implementing
particular solutions i,e., y(t)=yh+yp
However, in case of feedback process control, m(t) is
most likely as a function of y(t), i.e.
15
1 0" ' ( )y a y a y bf y
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Linear Systems-Linear Systems-ContinuedContinued In most analysis of a feedback control systems, it is
our tradition to assume that the system performance
is in its steady state, i.e.
y”+a1y’+a0y=bm(t), y(0)=0; y’(0)=0
For the convenience of the analysis, a Laplace
transform expression of the above linear system
becomes necessary.
16
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2-3 Laplace Transform: 2-3 Laplace Transform: Definitions and PropertiesDefinitions and Properties
17
0
( ) ( ) ( )stf t f t e dt F s
L
0
1( ( )) 1 stf x e dt
s
L
Definition 2-1: Consider a function of time f(t), the Laplace transform of f(t)is denoted by
Property 2-1: If f(t)=1, t 0, f(t)=0, t<0, this is called a step function as shown below (to describe an abrupt event to cause the change of operation), then:
time0
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Scenario 1: Step function Scenario 1: Step function (shift)(shift)
18
-10 -8 -6 -4 -2 0 2 4 6 8 10-0.5
0
0.5
1
1.5
Time (seconds)
data
Time Series Plot:
The process endures an abruptlong term change, such a PM on a tool of manufacturingplant.
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Definitions and Properties- Definitions and Properties- ContinuedContinued
19
Corollary 2-1: Given the following step function:
0 0
0 )(
t
tCtu
Then: 0
st Cu t Ce dt
s
L
Property 2-2: Given a ramp function:
0 0
0 )(
t
tCttf
Then: 20
st Cf t Cte dt
s
L
time
time
0
0
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Scenario 2: Ramp function Scenario 2: Ramp function (drift)(drift)
20
-10 -8 -6 -4 -2 0 2 4 6 8 10
0
20
40
60
80
100
120
Time (seconds)
data
Time Series Plot:
The process endures slow change on the process such as tool aging of a manufacturing plant.
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Definitions and Properties Definitions and Properties - Continued- Continued
21
tc 0
0 1/c
0t 0
)( cxf
Property 2-3: delta function (t)
0 0
0 t)(
t
tCtf
Then: 0
stf t C t e dt C
L
Note that, a pulse function can also represent an abrupt event in plantoperation, but the event can be vanished very soon unlike step function.
Definition 2-2: A pulse function is denoted by:
Definition 2-3: A delta function (t)=limc0f(x)
time0 c
1/c
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Scenario 3: Pulse function Scenario 3: Pulse function (Excursion)(Excursion)
22
-10 -8 -6 -4 -2 0 2 4 6 8 10-0.5
0
0.5
1
1.5
Time (seconds)
data
Time Series Plot:
The process endures a sudden short change but returns to its originalstate.
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Definitions and Properties- Definitions and Properties- ContinuedContinued
23
0 0
0 )(
-at
t
tetf
0
1at stf t e e dts a
L
0 0
0 )(
-at
t
tCetf
Property 2-4: Exponential function
Corollary 2-2
0
at st Cf t C e e dt
s a
L
Then:
Then:
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Scenario 4: Exponential Scenario 4: Exponential function (Excursion)function (Excursion)
24
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (seconds)
data
Time Series Plot:
The process endures a sudden abrupt change, but returnto its original state gradually..
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Definitions and Properties- Definitions and Properties- ContinuedContinued
25
Sinusoildal Functions Property 2-5:
0 0
0 t Asin)(
t
ttf
2 20
sin st Af t A te dt
s
LThen:
Property 2-6:
0 0
0t Acos)(
t
ttf
Then: 2 2
0
cos st Asf t A te dt
s
L
Sinusoildal Functions
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Scenario 5: Sinusoidal Scenario 5: Sinusoidal functionfunction
26
-10 -8 -6 -4 -2 0 2 4 6 8 10-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Time (seconds)
data
Time Series Plot:
The process endures a sinusoidal wave input to force it response another periodical wave.
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TheoryTheory
27
Theory 2-1: LinearityConsider two functions f(t) and g(t) with their Laplace transform F(s) and G(s) exists, and let a and b be two constants, then
af t bg t aF s bG s L
Theory 2-2: DerivativeLet f(t) be differentiable, and its Laplace transform F(s) exist, then
( )(0 )
df tsF s f
dt
L
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Theory - Theory - ContinuedContinued
28
( ) ( )0 ( )d s
d
dy t dy t dysY s y sY s
dt dt
L L
Remark 2-1: Deviation variable (perturbation variable)Let y(t) has a steady state ys, then yd is called a deviation variable ifyd(t)=y(t)-ys
Remark 2-2: In the study of control theory, we all assume that the processis originally at its steady state, and then a change of the system starts. Therefore, y(0)= ys, or yd(0)=0.
Corollary 2-3: Derivative of a deviation variable
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Theory - Theory - ContinuedContinued
29
Theory 2-3: Final Value TheoremConsider a function f(t) with its Laplace Transform F(s), then:
ssFtfst 0limlim
Theory 2-4: Dead time (Time Delay, Translation in Time)Consider a function f(t) with its Laplace Transform F(s), then:
sf t e F s L
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General ProcedureGeneral Procedure
30
Time domain
Laplace domain
Step 1Take Laplace
Transform
ODE
Initial conditions
Step 2Solve for
sD
sNsY
Step 3Factor D(s)
perform partialfraction expansion
Step 4Take inverse
Laplace transform
Solution y(t)
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Solution of a Linear Solution of a Linear SystemSystem
Example: Solve the differentiation equation
1 0.8
5 4 2; (0) 1
5 4 2
25 4
25 ( ) 1 4 ( )
2 5 2( ) 5 4 5
5 2( )
5 4
5 2( ) 0.5 0.5
5 4t
dyy y
dt
dyL y L
dt
dyL L y
dt s
sY s Y sss
Y s ss s
sY s
s s
sy t L e
s s
Sol: Take the Laplace transform on both sides of the equation
or,
By Laplace Transform
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Solution of a Linear Solution of a Linear SystemSystem
t
s
s
etyssss
sY
s
ss
s
where
ssss
ssY
8.0
8.02
01
21
5.05.0)(8.0
5.05.0
45
5.25.0)(
5.28.0
2
8.0
28.0525
5.045
25
45)45(
25)(
By Laplace Transform – Partial Fraction Expansion
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2-4 Transfer Functions –2-4 Transfer Functions – Example: Thermal Example: Thermal ProcessProcess
Rate of Energy Input- Rate of Energy Output=Accumulation
i i i
vi p i i p
d V u tf h t f h t
dt
d V C T tf C T t f C T t
dt
Inputs: f(t), Ti(t),Ts(t)Output: T(t)
Ts
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2-4 Transfer Functions 2-4 Transfer Functions – – Cont.Cont.Let f be a constant V= constant, Cv=Cp
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Transfer Functions – Transfer Functions – Cont.Cont.
where, Gp(s) is call the transfer function of the process, in block diagram:
Gp(s)M(s) Y(s)
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Scenario Simulation (Step Scenario Simulation (Step Change)Change)
36
1
10s+1
Transfer Fcn
simout
To Workspace
Step Scope
0 5 10 15 20 25 30 35 40 45 500
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
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Homework and Homework and Reading AssignmentsReading Assignments
37
Homework – Due 3/19Text p572-1,2-2, 2-6 (a), (c)Reading Assignments:Laplace Transform(p11-26)
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2-52-5 Linearization of a Linearization of a FunctionFunction
X0X0 -△ X0+△
- 0 △ △
F(X)
X
aX+b
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Linearization of a Function Linearization of a Function (single variable)(single variable)
Example: Linearize the Arrhenius equation, at T=300C, k(300)=100 , E=22,000kcal/kmol.
T
1s
280 285 290 295 300 305 310 315 32040
60
80
100
120
140
160
180
200
1
1
2
2/
0
/0
7.13330031037.31003.139310
3.6630029037.310095.70290
..
30037.3100
37.3273300987.1
22000100
)()(
sk
sk
ge
TTk
TR
Eek
T
k
where
TTT
kTkekTk
TRETT
TTRTE
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Linearization of a Function Linearization of a Function (two variables)(two variables)
2, 1
, 2 2 2 1
if 2.2, 1.1 respectively
2.2 1.1 2.42 2 2.2 2 2 1.1 1 2.4
w l
a aa w t l t a w w l l w l
w lw l
a
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Linearization of Differential Linearization of Differential EquationsEquations
Definition 2-4: A linear function L(x) is such that for xRn, for all scalars a and b, L(ax1+bx2)= aL(x1)+bL(x2).
Definition 2-5: A Linear system is denoted as the following:
nnn
n
n
xxxLdt
dx
xxxLdt
dx
xxxLdt
dx
,,,
,,,
,,,
21
2122
2111
Provided that L1,…,Ln are linear functions
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Linearization - Linearization - ContinuedContinuedConsider a function f(x1,…,xn),
the linearization of such a function to a point
is defined by the first order Tylor expansion at that point, i.e.
nn
xx
xxxx
n
xx
xxxxn
nn
xxx
fxx
x
fxxxf
xxxLxxxf
nnnn
1
22
1
1
22
1 111
21
2121
,,,
,,,,,,
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Linearization of a single input Linearization of a single input single output systemsingle output system
0 0
0 0
0 0
0 0
,
( , )
0
Laplace Transform
( )
1
x x x xu u u u
f fx f x u x x u u
x u
f x u
X aX bU
sX s aX s bU s
or
X s b K
U s s a s
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Example – Level ProcessExample – Level Process
hV
f0
fCross-sectional=A
1/ 2f kh
0 0 0 ,dV dh
A f f f k h f f hdt dt
A=5m2
mh 9
f0=1m3/min
30
1/ 2
1 1 / min
39
f mk
mh
Abrupt change of f0
from 1 to 1.2m3/min
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Example – Level Example – Level Process - ContinuedProcess - Continued
0 0
1/ 3 15
182 9
dHF H F H
dt
sFsHs 0)()1815(
/90
/90
0.2 344 3.6( )
5 1/18 90 1
( ) 3.6 3.6
( ) 12.6 3.6
t
t
H ss s s s
H t e
h t e
0 0
0 2
d h hdh dH f f kA A A F H F H
dt dt dt F h h
Taking Laplace Transform on both sides:
Now, F0(s)=0.2/sTransfer Fcn
90 s+1
18
To Workspace
simout
Step Scope
Constant
9
Add
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[A B C D]=linmod('example_model');[b,a]=ss2tf(A,B,C,D);
example_model
Linearization
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47
Time(min.)
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Conclusive Remarks for Conclusive Remarks for Laplace TransformLaplace Transform
Laplace transform is convenient tool to represent the dynamics of a linear system.
Laplace transform is very easy to use to special inputs, especially in case of abrupt change of system inputs and time delays.
Linearization is a frequent approach for a nonlinear system, and it is a good approximation in small change of the inputs.
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Homework – Due 3/19Text p572-17,2-23Linearization of Function(p50-56)