Transcript 投影片 1 - National Tsing Hua University

Chapter 2
Mathematical Background
Professor Shi-Shang Jang
Chemical Engineering Department
National Tsing-Hua University Taiwan
March, 2013
1
2-1 The Essence of Process Dynamics
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.
2
Example: Thermal Process
Ts
Inputs: f(t), Ti(t),Ts(t)
Output: T(t)
Rate of Energy Input- Rate of Energy Output=Accumulation
f i hi  t   f i h  t  
d V  u  t  
dt
d V CvT  t  
f i C pTi  t   f iC pT  t   
dt
CV: T(t)
MV: f(t)
DV: Ti(t), Ts(t)
Noise v.s. Trend
70.7
70.6
70.5
70.4
T
70.3
Trend
70.2
70.1
70
69.9
0
Noise
50
100
150
200
time
4
Conception of deviation variable
Disturbances
Manipulative
variables
control stream
Plant
tank
Wild stream
Fw
H= hdeviation = h - hsteady-state
Valve
Fc
Controlled
variables
liquid level h
5
The Essence of Process Dynamics - Continued
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.
6
Conception of deviation variable ( =T70)
0.6
0.5
0.4

0.3
0.2
Trend
0.1
Noise
0
-0.1
0
50
100
150
200
time
7
2-2 Linear Systems
8
2-2 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
y   an1 y
n
n1

 a1 y ' a0 y  bk m  
k
 b0m  c
9
Linear Systems-Continued

A lumped process system is autonomous if

An autonomous lumped system is said to be stable at the origin if

lim y  t   0; for some y(0)  y0 ; y0  S  x x  
t 

10
Linear Systems-Continued

An autonomous lumped system is called asymptotic stable if
lim y  t   0; for  y(0)  y0 ; y0  R
t 

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
Linear Systems-Examples
Example 1: y " 4 y ' 3 y  0; y (0)  0, y '(0)  2
Sol : charac. eqn.: r 2  4r  3  0  r  3; r  1
y (t )  c1e 3t  c2e t  c1  c2  0
y '(t )  3c1e3t  c2e t  3c1  c2  2
Finally, we have y (t )  e 3t  e t
Note that no matter what the initial conditions is, the solution indicates :
lim y  t   0
t 
The system is hence asymptotic stable.
12
Linear Systems-Examples
Example 2:y " 2 y ' 3 y  0; y (0)  1; y '(0)  4
r 2  2r  3  0
y  c1et  c2 e 3t
From the inital conditions, we have
y t  
1
7et  3e 3t 

4
y t   
Note that the above system is unstable since: lim
t 
Example 3 : y " 3 y ' 8 y  0
r 2  3r  8  0


r  3  i 23 / 2
3
t
1

1

y  t   c1e cos 
23t   c2e 2 sin 
23t 
2

2

3
t
2
Note that the above system is also unstable since:
lim y  t   
t 
13
Linear Systems-Continued
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
Linear Systems-Continued

A forced system is of the following form, for example:
y”+a y’+a0y=bm(t)
1

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.
y " a1 y ' a0 y  bf ( y)
15
Linear Systems-Continued

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
2-3 Laplace Transform: Definitions and Properties
Definition 2-1:
Consider a function of time f(t), the Laplace transform of f(t)
is denoted by

L  f (t )    f (t )e st dt  F (s)
0
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:

L ( f ( x))  1e st dt 
0
1
s
0
time
17
Scenario 1: Step function (shift)
Time Series Plot:
1.5
data
1
0.5
0
-0.5
-10
-8
-6
-4
-2
0
Time (seconds)
2
4
6
8
10
The process endures an abrupt
long term change, such a PM
on a tool of manufacturing
plant.
18
Definitions and Properties- Continued
Corollary 2-1: Given the following step function:
C
u(t )  
0
t0
t0

Then:
L u  t    Ce st dt 
0
C
s
0
time
Property 2-2: Given a ramp function:
Ct
f (t )  
0

Then:
t0
t0
L  f  t    Cte st dt 
0
C
s2
0
time
19
Scenario 2: Ramp function (drift)
Time Series Plot:
120
100
data
80
60
40
20
0
-10
-8
-6
-4
-2
0
Time (seconds)
2
4
6
8
10
The process endures slow change on the process such as tool aging
of a manufacturing plant.
20
Definitions and Properties - Continued
Definition 2-2: A pulse function is denoted by:
0


f ( x)   1/c

0

t  0
1/c
0c
c  t
0
c
time
Definition 2-3: A delta function (t)=limc0f(x)
Property 2-3: delta function (t)
t 0
C t 
f (t )  
t0
 0

 st
Then: L  f  t    C  t  e dt  C
0
Note that, a pulse function can also represent an abrupt event in plant
operation, but the event can be vanished very soon unlike step function.
21
Scenario 3: Pulse function
(Excursion)
Time Series Plot:
1.5
data
1
0.5
0
-0.5
-10
-8
-6
-4
-2
0
Time (seconds)
2
4
6
8
10
The process endures a sudden short change but returns to its original
22
state.
Definitions and Properties- Continued
Property 2-4: Exponential function
e -at
f (t )  
0
Then:
t0
t0

L  f  t    e at e st dt 
0
1
sa
Corollary 2-2
Ce -at
f (t )  
 0
t0
t0
Then:

L  f  t   C  e at e st dt 
0
C
sa
23
Scenario 4: Exponential function
(Excursion)
Time Series Plot:
1
0.9
0.8
0.7
data
0.6
0.5
0.4
0.3
0.2
0.1
0
-10
-8
-6
-4
-2
0
Time (seconds)
2
4
6
8
10
The process endures a sudden abrupt change, but return
to its original state gradually..
24
Definitions and Properties- Continued
Property 2-5:
Sinusoildal Functions
Asin t
f (t )  
 0
Then:
t0
t0

L  f  t    A sin te st dt 
0
Property 2-6:
Sinusoildal Functions
Acost
f (t )  
0
Then:
A
s2   2

t0
t0
L  f  t    A cos te st dt 
0
As
s2   2
25
Scenario 5: Sinusoidal function
Time Series Plot:
1
0.8
0.6
0.4
data
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-10
-8
-6
-4
-2
0
Time (seconds)
2
4
6
8
10
The process endures a sinusoidal wave input to force it
response another periodical wave.
26
Theory
Theory 2-1: Linearity
Consider 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
L af t   bg t   aF  s   bG  s 
Theory 2-2: Derivative
Let f(t) be differentiable, and its Laplace transform F(s) exist, then
 df (t ) 

L

sF
s

f
(0
)



 dt 
27
Theory - Continued
Remark 2-1: Deviation variable (perturbation variable)
Let y(t) has a steady state ys, then yd is called a deviation variable if
yd(t)=y(t)-ys
Remark 2-2: In the study of control theory, we all assume that the process
is 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
 dyd (t ) 
 dy(t )  dys 


L


  sY  s   yd  0   sY (s)
dt
 dt 


L
28
Theory - Continued
Theory 2-3: Final Value Theorem
Consider a function f(t) with its Laplace Transform F(s), then:
lim f t   lim sF s 
t 
s 0
Theory 2-4: Dead time (Time Delay, Translation in Time)
Consider a function f(t) with its Laplace Transform F(s), then:
L  f t    e s F  s 
29
General Procedure
Time
domain
ODE
Initial
conditions
Laplace
domain
Step 1
Take Laplace
Transform
Step 2
Solve for
N s 
Y s  
Ds 
Solution y(t)
Step 4
Take inverse
Laplace transform
Step 3
Factor D(s)
perform partial
fraction expansion
30
Solution of a Linear System
By Laplace Transform
Example: Solve the differentiation equation
5
dy
 4 y  2;
dt
y (0)  1
Sol: Take the Laplace transform on both sides of the equation
 dy

L 5
 4 y   L 2
 dt

2
 dy 
5 L    4 L  y 
s
 dt 
2
s
2
5s  2
Y ( s ) 5s  4    5 
s
s
5  sY ( s )  1  4Y ( s ) 
or,
Y ( s) 
5s  2
s  5s  4 
 5s  2 


0.8 t
y (t )  L1 
  0.5  0.5e

 s  5s  4  

Solution of a Linear System
By Laplace Transform – Partial Fraction Expansion
Y (s) 
5s  2

2
 1
s (5s  4) s 5s  4
where
5s  2
1 
s  0  0.5
5s  4
5s  2
 5  0.8  2
2
2 


 2 .5
s  0.8
s
 0.8
0.8
0.5
2.5
0.5
0.5
Y (s) 



 y (t )  0.5  0.5e 0.8t
s 5s  4
s
s  0.8
2-4 Transfer Functions –
Example: Thermal Process
Ts
Inputs: f(t), Ti(t),Ts(t)
Output: T(t)
Rate of Energy Input- Rate of Energy Output=Accumulation
f i hi  t   f i h  t  
d V  u  t  
dt
d V CvT  t  
f i C pTi  t   f iC pT  t   
dt
2-4 Transfer Functions – Cont.
Let f be a constant V= constant, Cv=Cp
Transfer Functions – Cont.
where, Gp(s) is call the transfer function of the process, in block diagram:
M(s)
Gp(s)
Y(s)
Scenario Simulation (Step Change)
1
10s+1
Step
Scope
Transfer Fcn
simout
To Workspace
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
5
10
15
20
25
30
35
40
45
50
36
Homework and Reading Assignments
Homework – Due 3/19
 Text p57
 2-1,2-2, 2-6 (a), (c)
Reading Assignments：
Laplace Transform(p11-26)
37
2-5 Linearization of a Function
F(X)
aX+b
X0 -△ X0
-△
0
X0+△
△
X
Linearization of a Function (single variable)

Example: Linearize the Arrhenius equation, at T=300C,
k(300)=100 , E=22,000kcal/kmol.
k (T )  k0 e  E / RT  k (T ) 
k
T
T T
200
T  T 
where
k
E
 k0 e  E / RT
T T
T
RT 2
22000
 100
 3.37
2
1.987 300 273
 k T   100 3.37T  300
e.g .
180
s
1
k 290  70.95  100 3.37290 300  66.3s
160
140
120
100
1
k 310  139.3  100 3.37310 300  133.7 s 1
80
60
40
280
285
290
295
300
305
310
T
315
320
Linearization of a Function (two variables)
w  2, l  1
a
a
 w  w   l  l   2   w  2   2 l  1
w
l
if w  2.2, l  1.1 respectively
a  w  t  , l  t    a 
a  2.2 1.1  2.42  2   2.2  2   2 1.1  1  2.4
Linearization of Differential Equations


Definition 2-4: A linear function L(x) is such that for
xRn, for all scalars a and b, L(ax1+bx2)= aL(x1)+bL(x2).
Definition 2-5: A Linear system is denoted as the
following:
dx1
 L1  x1 , x 2 , , x n 
dt
dx2
 L2  x1 , x 2 ,  , x n 
dt

dxn
 Ln  x1 , x 2 ,  , x n 
dt
Provided that L1,…,Ln are linear functions
Linearization - Continued

Consider 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.
f x1 , x2 , , xn   Lx1 , x2 ,, xn 
f
 f x1 , x2 ,, xn  
x1
x1  x
x2  x2

xn  xn1
f
x1  x1    
xn
x1  x
x2  x2

xn  xn1
 xn  xn 
Linearization of a single input single output system
x  f  x, u  
f
x
x  x0  x  x0  
u u0
 f ( x0 , u0 )
X  aX  bU  0
Laplace Transform
sX  s   aX  s   bU  s 
or
X (s)
b
K


U  s s  a  s 1
f
u
x  x0
u  u0
 u  u0 
Example – Level Process
f0
A=5m2
V
f0=1m3/min
h
h  9m
f
Cross-sectional=A
dV
dh
A
 f0  f  f0  k h  f  f0 , h 
dt
dt
1 1  m3 / min 
k

 

1/ 2
9 3 m
h

f0
f  kh1/ 2
Abrupt change of f0
from 1 to 1.2m3/min
Example – Level Process - Continued
d h  h 
dh
dH
f
f
k
A
A
A

F0 
H  F0 
H
dt
dt
dt F0
h
2 h
5
dH
1/ 3
1
 F0 
H  F0  H
dt
18
2 9
Taking Laplace Transform on both sides:
9
(5s  1 18) H (s)  F0 s 
Constant
18
90 s+1
Step
Now, F0(s)=0.2/s
Transfer Fcn
Add
Scope
simout
0.2
344
3.6
H ( s) 


 5s  1/18 s 90s  1 s
H (t )  3.6  3.6e t /90
h(t )  12.6  3.6e t /90
To Workspace
example_model
Linearization
[A B C D]=linmod('example_model');
[b,a]=ss2tf(A,B,C,D);
Level
Time(min.)
47
Conclusive Remarks for Laplace 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.

Homework – Due 3/19
 Text p57
 2-17,2-23
Linearization of Function(p50-56)