C5-Control_Srategy_process_control_etc_etc

Process

1. Process Control
Chapter 5: Control Strategies

2. Content
3.1 Objective
3.2 Feed- forward Control
3.3 Feedback Control
3.4 Cascade Control
3.5 Ratio Control
3.6 Selective Control
3.7 Split-range control
3.8 MIMO system control structure
3.9 MIMO system control structure design
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3. 3.1 Objective
Process Control Problem: maintain y ≈r while
–Set point r can be changed
–Exist disturbance d
–Exist measurement noise n
–Process Model is not accurate
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Process
Disturbances d
Controlled
Variables y
Manipulated
Variables u
Note:
u, r, y, d arevariables
of G(s)

4. Specific Control Objectives
Stabilizing system
High response speed and good response quality
–Set point changing response
–Process disturbance response
–Low sensitivity with measurement noise
The values of manipulated variables are slow/insignificantly
change.
Persistent:
–Persistent stability
–Persistent quality
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5. Complex problems
Processes are complex, hard to control (multi-dimension
interactive, non-minimum phase system, constraints on
manipulated variables’ values and rate of change, constant on
allowed changing range of manipulated variables...)
Process Model is hard to build accurately
Disturbances are hard to measure/predict
There is limits in implement & installing control law
The operators’ knowledge about control theory are limited
Etc.,
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6. Control Strategies
Control Strategy/Structure:principles about structure in using the
information of process variables to give the control actions
Control Strategy
–Single or multiple variables
–Coordinate using what input variables and how to
control output variables?
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Process
Disturbances d
Measurable
variable y
m
Manipulated
variables u
Set point r

7. Basic control strategies
SISO system
–Feed-forward Control
–Feedback control
–Cascade control
–Ratio control
–Selective control
–Split-range control
MIMO system
–Centralized control
Decoupling control
Multivariable control
–Decentralized control
–Hierarchical control
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8. 3.2 Feed-forward Control
Example: The heat exchanger
–Control the steam inlet flow F
sto maintain the output temperature T
2at
desired set point
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Steam
Input Temp
Flow
Output Temp
Flow
Oil

9. Basic structure
Principle:
–Assumption: Model is accurate, disturbances are measurable
–Measure the disturbance d , then calculate u in order to y ≈r:
–Do not measure y
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1
()( ())
() ()
r d
r
u K s r G sd
K s Gs
−
= −
≈
(3.1)
Controller

10. control deviant
Quality analysis
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1
1
1
()()
()
r
rd d
d dd
K Gd
u K r Gd G r Gd
y Gu Gd GG r Gd Gd r
−
−
−
=
=−= −
⇒=+= − +=
b) Processmodel deviant (assumed= 0):
(3.2)
(3.4)
a) Ideal control
1
()
realG GG
G
y G GG r r r
G
−
= +∆
∆
⇒ = +∆ = +
(3.3)
c) Disturbance model deviant
real
()
d dd
d dd d
G GG
y r Gd G G d r Gd
= +∆
⇒ = − + +∆ = +∆

11. d) Unmeasurable disturbance existence
→System is unstable when G
d2is unstable
22 22dd dyGuGdGd rGd=++ =+ (3.5)
control deviant
e) The model have zero point on the right haft of complex plane
→The ideal controller is unstable, needed to approximate!
Example: System model:
The ideal controller is unstable → The system is internal unstable!
The approximate controller (for steady state):
1
()
1
s
Gs
s
−
=
+
1
11
() ()
1
r
s
K s Gs
s
−+
= =
−
1
2
(0) 1
r
KG
−
= =
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u (K1 –unstable) y (K1 –unstable)
y (K1 –unstable)u (K2 –approximate)
y (K2 –approximate)
Time Time

13. Quality analysis (continuous)
f) Model has delayed time or the degree of denominator is higher than the degree of
numerator
→ The ideal controller is non-causal
Example: System model:
The ideal controller is non-causal
G) Unstable process: The ideal controller eliminates the unstable pole → The system is
not internal stable, if there is a little input noise, the system may become unstable.
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2
1
()
1
ss
Gs e
ss
−+
=
−+
2
1
()
1
s
r
ss
Ks e
s
−+
=
+
()
ud u
y G u d G d r Gd= ++ =+ (3.6)

14. Example: Level control
Control principle: Input flow must equal to output flow
Problem: If there is a small error in measurement flowrate value or
in control valve model, the tank can be overflow or empty
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15. Summary of Feed-forward control
Advantages:
–Simple
–Fast response (Compensate the disturbance before it affects the output)
Disadvantages:
–Must use disturbance measurement device
–Can not eliminate the effect of unmeasurable disturbance
–Sensitive with model deviant (process model and disturbance model)
–The ideal controller may be unstable or unusable → Approximate
–Unable to stabilize an unstable process.
Main applications
–Simple problems, non optimum phase, low quality requirement
–Implemented in conjunction with feedback control to improve the response
speed of close- loop system
–Compensate measurable disturbance (mainly static compensation), pre-filter
(pretreatment) the dominant signal
–Ratio control
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16. Static compensator design steps
1.Determine controlled variables, choose the manipulated variables and
measurable disturbance.
2.Construct the process model, write down equilibrium equations and
constitutive equations
3.Substitute controlled variables by the set point, solve the equilibrium
equations of manipulated variables according to set point and
disturbances.
4.Analyze and assess the impact of model deviant to control quality
5.Eliminate the disturbances that have less impact in order to reduce the
sensor installation cost.
6.Correct the parameters of feed-forward controller for the working
point to compensate the model deviant and eliminated disturbances
7.Add feedback controllers to eliminated the steady- state error, reduce
the effect of model deviant and unmeasurable disturbances
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17. Example: Heat exchanger control
Skip the heat loss, we have energy
conservation equation:
(3.7)
where:
C
p–oil heat capacity
λ–thermal parameter in condensation
process
Substitute T
2with Set point :
(3.8)
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21
()
op s
FC T T Fλ−=
1
()
p
so
C
F F SP T
λ
= −
Steam
Input Temp
Flow
Output Temp
Flow
Oil

18. Example: Stirred-tank System
Assumption:
-c
1andc
2are constants
-Output flowwis random (free
flowing)
Equilibrium equation:
Substitute c with Set point:
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1122 12 ()wc wc w w c+=+
11
2
2
()w SP c
w
c SP
−
=
−
(3.9)
(3.10)

19. 3.3 Feedback Control
Example: Example: Heat exchanger control
Control principle: Control the steam flow (manipulated variable) based on the
deviant between the output oil temperature (controlled variable) and the set
point
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Steam
Feed Oil

20. Acting direction of Feedback Control
Direct acting (DA):The controller outputs increase when the
controlled variables increase and vice versa
Reverse acting (RA) The controller outputs decrease when the
controlled variables increase and vice versa
The selection of acting direction is depend on:
–Process characteristic: The relation between controlled variables and
manipulated variables
–The action type of controlled valve (Notice the arrow direction on the
controlled valve symbol)
Fail close, air-to-open: direct acting
Fail open, air-to-close: reverse acting
In example: Reverse acting
–Process: More steam → more temperature
–Controlled valve: fail close
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21. Feedback Control Configuration
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The response with set point and
the response with disturbance are
bonding →impossible to design a
complete independent controller.
Have the ability to design the
controller K
r(s) to improve the
response with set point
a) First degree of freedom configuration
CONTROLLER
b) Second degree of freedom configuration
CONTROLLER

22. Why using Feedback Control?
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()
dy GK r y n G d= −− +
(1 )
dGK y GKr G d GKn+ =+−
111
d
dLG L
y r d n Tr G Sd Tn
LLL
= + − =+−
+++
1
111
d
d
G GK
e r y r d n Sr G Sd Tn
GK GK GK
=−= − + = − +
+++
Consider First degree of freedom configuration:
(3.11)
(3.12)
Close-loop response:
(3.13)
Control deviant
(3.14)

23. The role of feedback control
1.An unstable process can be stabilized by using a feedback
controller to move the poles to the left half of the complex
plane (see the denominator polynomial
1 +GKin 3.13 and
3.14)
2.When the disturbance is unmeasurable or the model response is
uncertain, the disturbance effect can only be eliminated by using
feedback method:
3.The process model is inaccurate, so the steady-state error
elimination can only be happened when observing the output
state:
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10
1
d
G
GK
GK
⇒≈
+

0, 1 0n GK e≈ ⇒≈

24. The problems of feedback control
A close- loop contains a stable subject can become unstable
Feedback control needs addition sensors
Measurement noise can affect the control quality (notice the last
term in 3.13 and 3.14) → the need to have a good noise filter
method and a good measurement data processing
It's hard to have a good feedback controller if we don't have a
good model.
In some processes that have reverse acting or delayed (non
optimum phase), a feedback controller that is careless designed
may even worsen the response characteristic
The feedback controller have slow response with load noise and
set point change
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25. Combined with Feed-forward control
Example: Levelcontrol
–Feed-forwardNoisecompensator
–FeedbackStabilizingsystemandeliminatingsteady-stateerror
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26. Cascade control
Question: The effect of disturbances on processes that have slow
dynamic (temperature, level and concentration) or high delayed →
single-loop controller can not bring the desired fast rate of response and
small overshoot
Typical example: with the same opening of the valve, the pressure
change of steam/oil has high impact to flowrate
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Steam
Feed Oil

27. Example: Heat exchanger control
Solution:Eliminatingearlytheeffectofdisturbancesusingan
insidecontrolloop,useanothermeasurementvariable.
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Steam
Feed Oil

28. Tow basic structures
Classical structure (serial structure): Add an measurement variable
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Parallel structure: Add a manipulated variable
Subject
Subject

29. Quality analysis (classical structure)
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212
2 11 2 1 1 1 2 2 2
2 12 2 12
2 1 1212
1 12
2 12 2 12 2 12
2 1 12 21
12
21 21 21
() , ,
11
(1 )
11 1
(1 1/ ) /
1/ 1 1/ 1 1/ 1
d
d
dd
dd
GKL
y r G d d L KG L KG
L LL L LL
L G GGLL
y rd d
L LL L LL L LL
L G GG LL
rdd
LL LL LL
= −+ = =
++ ++
+
=++
++ ++ ++
+
=++
++ ++ ++
With in the consider frequency range (of L
1), we have:
2
1L
1 121
1 12
1 1 21
1 1 (1 )
dd
G GGL
y r d ds
L L LL
≈+ +
++ +
The effect of d2 has
been reduced
(3.15)
(3.16)
Close-loop response:

30. When using cascade control?
Single-loop controller doesn’t meet the quality requirements.
It’s easy to measure and control another process variable
(that have relation with the first variable)
The second manipulated variable shows the clear effect of the
noise that is hard to measure.
There is a casual relationship between controlled variables
and the second manipulated variable (maybe the same
variable)
The dynamic characteristic of the second manipulated
variable is faster than of the first variable.
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31. 3.5 Ratio control
Ratio control is maintaining a same ratio between two variables to
control indirectly a third variable → in fact it is a type of Feed-forward
control.
Example: Heat exchanger control
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Steam
Oil
Steam
Oil
a) Detailed Implementation Diagram b) Combined Diagram

32. Tow basic structures
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Uncontrolled Flow
Set point
Real Ratio
Controlled Flow
a) Feedback Ratio
Calculation
Uncontrolled Flow
Controlled Flow
Set point
b) Set point Flow
Calculation

33. Example: Stirred-tank System Ratio Control
(combined with Feedback control)
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34. 3.6 Selective Control
Use a selector to select signal: A manipulated variable (an
actuator)
Select a measurement signal: Limit control
–A controlled variable
–Measurement variables (measured at different location)
–A control loop
Select a control signal: Override control
–Two (or more) controlled variables, tow (or more) measurement
variables
–Two (or more) control loop
→ Ensure safety
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35. Limit control
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Control configurator
Example: Reactor temperature control
FC
100
TT
117
TT
116
TT
115
UC
101
Cold water
T
z
t
max
Process
Selector
(max, min,...)
-
Controller

36. Override Control
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Control configurator
Process
Controller 2
Controller 1
-
-
Example: Boiler control
Steam
LC
101
PC
100
FY
102

37. Override Control Application
Avoid overflow in a distillation column by limiting the steam
flow or fuel flow.
Prevent the state that the level is too high or too low by win
the right to intervene in control valves (feed valves and
exhaust valves)
Prevent overpressure or over temperature in a reactor by
reducing feed heat
Reduce feed fuel in a combustor to reduce the low oxygen
content in exhaust gas.
Prevent the overpressure in a pipe (steam or water) by
opening the by-pass valve
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38. 3.7 Split-range control
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Split-Range-
Controller
Process
+
-
•A controlled variable
•Lots of manipulated variables or
actuators
Example: Reactor temperature control
TT
100
TC
100
Cold water
Steam
Valve opening
100%
Control signal

39. 3.8 MIMO system control structure
Centralized control, Multivariable control:
–A multiple-input, multiple-output
–Design using channel separating or multi-variable method
Decentralized control, multi-loop control:
–Divide the system into smaller problems that are easy to be solved
(single-variable or multi-variable)
–Implemented by many independent controller
Hierarchical control
–Divide the system into partial control loops and master control
loops
–Utilizing both decentralized and centralized controller
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40. Centralized controller
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Process
Controller
SP
Example: Stirred-tank control
–Controlled variable level
and output concentration
(c)
–Manipulated variable: Input
flow w1 and w2

41. Centralized controller
Example: Two product
distillation tower
–Control-needed
variables: Distillate &
bottom product
composition XD and
XB,
–Controlled variables:
Distillate and bottoms
product temperature
–Manipulated variable:
Reflux and feed flow
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TT
100
TT
101
X
D
X
B
Reflux
Heating medium
Feed
UYC
102

42. Decentralized control
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K3
SP1
K2
K1
Process
SP2
SP3
Definition
A control system that consist of many
independent feedback controller, each
connects to a set(not sharing) of output
variables (measurable) and set points
with a set of manipulated variables

43. Decentralized control
Example: Two product
distillation tower
–Control-needed
variables: Distillate &
bottom product
composition X
Dand
X
B,
–Controlled variables:
Distillate and bottoms
product temperature
–Manipulated variable:
Reflux and feed flow
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TC
100
TC
101
X
D
X
B
Reflux
Heating medium
Feed fuel

44. Comparisons
Centralized control
+High quality (if the model is accurate)
+Lots of modern design methods and tools
–Complex model construction
–Complex digital controller implementation (lack of available block library,
various sampling cycle...)
–Difficult for user to follow → Difficult to accept
–Low reliability
Decentralized control
+Traditionally industrially used approach, therefore high acceptance
+Presumably high transparency
+Easy re-adjustment of the control parameters (also auto-tuning)
+High reliability
–Structure design is complex (one has to select the right coupling of control and
controlled variables)
–(Usually) low performance, not always available
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45. Decentralized control design
Design steps:
–Select controlled variables, manipulated variables, and measurable variables
–Coupling controlled variables and manipulated variables
–Apply control strategy for each SISO system
The arisen problems:
–When to use decentralized control?
Assess the level of interaction between the control channels
Assess the control quality that can be archived (the difficult of problem)
–Coupling input/output variables
Each manipulated variable affects different controlled variables → choose the
input/output couple that have the best relation
Simple situation → can draw conclusions by analyzing physical processes
High input/output quantity → great coupling abilities, need a systematic method
–Assessing system quality and stability
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46. 3.9.1 Relative Gain Array (RGA)
RGA concept
–Purposed by Bristol in 1966 (AC-11) → indicator assessing
interaction between the input/output channels in a MIMO system
–Help selecting the pairs of controlled and manipulated variables in
designing the decentralized control structure
–Have many great characteristics in analysis the control stability and
quality of decentralized controller
RGA is a m x m nondegenerate complex square matrix:
With x is Hadamard product
Example:
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1
RGA() () ( )
T
G GGG
−
≡Λ × (3.17)
1
1 2 0.4 0.2 0.4 0.6
, , ()
3 4 0.3 0.1 0.6 0.4
GG G
−
−    
= = Λ=    
−
    

47. Explain the properties of RGA by the 2x2 example.
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Consider a 2x2 system at steady-state
11 12
21 22
11 12 11 11
11
21 22 11 11 12 21 11 22
(0)
1 1
() ,
1 1/
gg
G
gg
G
gg gg
λλ λ λ
λ
λλ λ λ

=


−  
Λ= = =
  
− −
  
(3.19)
2
1 11 1
0u
y gu
=
∆=∆
When u
2= 0:
When u
2≠0 to remainy
2=0:
2
12 21
1 11 1 11 11 1 11 1
0
22
ˆ (/) ( )
y
gg
y gug ug u
g
λ
=
∆ ≈ ∆= ∆= − ∆
(3.20)
(3.21)
Indirect action
If
→There is no two-way interaction, easy to decentralized control.
12
11 12 21
21
0
10
0
g
gg
g
λ
=
=⇒=⇒ 
=


48. Generalization for m x n system
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11 12 1
21 22
1() () ()
() ()()
()
()
() ()
m
m mmgs gs g s
gs gsys
Gs
us
gs g s



= =





 

Consider the sensitivity between the input variables u
jand the output variables y
iin
two cases:
-Without others input variables, means:
-Have others input variables, as long as other output variables are constant, means:
The ration factor between two values shows the interacting level between u
jand y
i:
0,
[]
k
i
ij ij
j
u kj
y
gG
u
= ≠
∂
=
∂

0,
k
u kj= ∀≠
11
ˆ/ () ( )[ ][ ]
T
ij ij ij ij ji
g g GG GGGλ
−−
= ⇒Λ = ×
(3.18)
0,
k
y ki= ∀≠
1
0,
ˆ1/[ ]
k
i
ij ji
j
y kiy
gG
u
−
= ≠∂
=
∂


49. Properties of RGA matrix
Sum of the elements in each row or column is equal to one
The RGA matrix is not affected by choice of units or scaling of variables
Exchanging two rows (columns) of G leads to exchanging two rows
(columns) of Λ(G)
Λ(G)is an identity matrix if G is a Upper or Lower triangular matrix
(one-way interactive)
If G(s) is a transfer function matrix, then Λ (G(jω))is calculated
corresponding with each frequency ωin considered frequency range.
The number is an index that show the interactive
level of the process (the most important ones are around the tần số cắt
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1.0
ij ij
ij
λλ= =∑∑
12 1 1 2 2( ) ( ), diag( ), diag( )
iiG DGD D d D dΛ =Λ ∀= =
sum
()
ij
ij
RGA G I g
≠
Λ− = ∑

50. The pairing method of input/output variables
based on RGA
Law 1: The pair input/output (j, i) which corresponds with
the element
λ
ijhas the value near 1 around the system desired
tần số cắt, ưu tiên the number bigger than 1
–The more the range in which λ
ij≈1 is bigger, the better
–In simple cases, we can choose the transfer function at the steady-
state (s=0)
Law 2: Avoid choose λ
ij<< 1 or λ
ij< 0 for the system at
steady-state.
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12 34
1
2
3
4
0.931 0.150 0.080 0.164
0.011 0.429 0.286 1.154
0.135 3.314 0.270 1.910
0.215 2.030 0.900 1.919
uu uu
y
y
y
y
−

−−
Λ=
− −−

−

Example 1:
1
2
3
1 23
1.98 1.04 2.02
0.36 1.10 0.26
0.62 1.14 2.76
u uu
y
y
y
−

Λ= −

−−

Example 2:

51. The stability of decentralized control system
With the process G(s) stabilizes
1.IfeachsingleloopisstablewhenothersisopenandthematrixΛ(G)=I
∀ω,thenthewholesystemisstable→SelectthepairsothatΛ(G)≈I
aroundtầnsốcắt
2.Ifcontrollerusetheintegrateelementandthepairthatiscorresponding
withΛ(G(0))hasthenegativevalue,then:
Thewholesystemisunstable,or
Thecorrespondingsingleloopisunstable,or
Thewholesystemisunstablewhenthecorrespondingsingleloopisopen.
3.Ifcontrollerusetheintegrateelementandisstablewhenallothersloop
areopen,andtheNiederlinskiindex:
Thentheicontrolloopisunstable.Withn=2,theaboveconditionis
necessaryandsufficient
10/18/2017 Chapter 5: Control Strategies HMS 2006-2017
51
1
det (0)
0
(0)
n
ii
i
G
NI
g
=
= <
∏

52. Consider the stability: The distillation column
example
Steady-state model:
10/18/2017 Chapter 5: Control Strategies HMS 2006-2017
52
12,8 18,9
6,6 19,4
2,01 1,01
1,01 2,01
D
BxL
xV
−    
=
    
−
    
−
Λ≈

−

2
1
det (0)
NI
(0)
12,8 19,4 18,9 6,6
0,498
12,8 19,4
ii
i
G
g
=
=
−⋅+⋅
= =
−⋅
∏
Process
K
1
K
2
r
D
x
D
x
B
r
B
L
V
K
1
K
2
r
B
x
D
x
B
r
D
L
V
Process
2
1
det (0)
NI
(0)
18,9 6,6 12,8 19,4
0,991
18,9 6,6
ii
i
G
g
=
=
− ⋅+ ⋅
= =−
−⋅
∏
18,9 12,8
19,4 6,6
1,01 2,01
2,01 1,01
D
BxV
xL
−    
=
    
−
    
−
Λ≈

−


53. 3.9.2 Singular Value Decomposition (SVD)
Singular value and the singular value decomposition have
many uses in analyzing system quality
In process control, beside the RGA analysis, the SVD is a
useful tool:
–Select controlled variables, manipulated variables, and control
variables
–Assess the persistent of a control strategy/structure.
–Determine the best decentralized control structure
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53

54. Singular variables
Each singular variable σof a complex matrix A (m x n) is defined
as eigenvalues of A
H
A → the distance ruler of the decomposition
of A
Consider the second norm of A:
Explanation:
–With the input vector x, the matrix A maps to y = Ax with the maximum
gain factor is and the minimum gain factor is
–The gain factor depends on the direction of vector x
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54
2
2
2
2
2
01
2
2
2
01
2
max ( ) max ( ) ( )
( ) max max
( ) min min
H
ii
ii
xx
xx
A AA A A
Ax
A Ax
x
Ax
A Ax
x λ σσ
σ
σ
≠=
≠=
= =
= =
= =

(3.24)
(3.25)
(3.26)
σ σ
/σσ

55. The SVD analysis and the direction dependent
SVD analysis
From the system theory point of view, if consider G(jω) is matrix
A and x is input signal vector, we can conclude some similar and
deeper explanations:
–The input vectors x have direction same with the first column of V will be
gained the most → the result is that vector y has the same direction with the
first column of U
–The input vectors x have direction same with the last column of V will be
gained the least → the result is that vector y has the same direction with the
last column of U
10/18/2017 Chapter 5: Control Strategies HMS 2006-2017
55
(3.27)
1
12
11
,,
... , min( , )
,,
T TT T
k
k
kk
A UV U V UU IVV I
k mn
AV U Av u Av u
σ
σ
σσσ σ σ
σσ


=Σ= = =



= ≥ ≥≥ = =
⇒=Σ = =

(3.28)

56. Selecting the controlled variables
Example:Selectwhichtrayinthedistillationcolumnthattemperature
iscontrolled(ThemanipulatedvariablesistherefluxflowLand
heatingmediumpowerQ)
Fromchapter2:Selecttheoutputvariablethatisheavyeffectedbythe
controlledvariables→correspondingwiththeelementthathave
highestvalueineachcolumnofU
9 0.00773271 0.0134723
8 0.2399404 0.2378752
7 2.5041590 2.4223120
6 5.9972530 5.7837800
5 1.6773120 1.6581630
4 0.0217166 0.0259478
3 0.1976678 0.1586702
2 0.1289912 0.1068900
1 0.0646059 0.0538632
−
−
−
−
−
−
−
−
0.0015968 0.0828981
0.0361514 0.0835548
0.3728142 0.0391486
0.8915611 0.1473784
0.2523673 0.6482796
0.0002581 0.6482796
0.0270092 0.4463671
0.0178741 0.2450451
0.0089766 0.1182182
U
−−

−−

−−

−

−−=

−−

−

−

−







10/18/2017 Chapter 5: Control Strategies HMS 2006-2017 56
Tray DT/DL DT/DQ
0.7191691 0.6948426
0.6948426 0.7191691
T
V
−
=

−−

9.3452 0
0 0.052061

Σ=


6
nd
tray

57. Condition number
Condition number
In linear algebra, cond(A) represents the sensitive of system with the
error in A or in y, or the ability to find exactly the solution of Ax = b,
the bigger cond(A), the lower ability
Example:
If A
12changes from 0 to 0.1, then A becomes degenerate
In system theory, cond(G(jω)) is related to the ability to control, limits
the control quality.
–The bigger the condition number, the more sensitive the system is with the
model error.
–The condition number is related to the reachable quality criteria (frequency
domain)
–The condition number depends on the scale selection/normalization.
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57
(3.29)cond() () /AA γ σσ= =
10
10 1
A

=


10.1 0
( ) = , cond( ) = 101
0 0.1
AA

Σ



58. Note: Frequency dependent
10/18/2017 Chapter 5: Control Strategies HMS 2006-2017 58
Example:Two-product distillation tower

59. Reduce the input/output variables number
Based on Seborg et. al., 2000
–After model normalization, the SVD analysis and arranging the singular
variables in descending order, it’s possible to eliminate some input/output if:
Example:
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59
1
/10
ii
σσ
+
<
0.48 0.90 0.006
(0) 0.52 0.95 0.008
0.90 0.95 0.020
G
 −

=

 −

0.5714 0.3766 0.7292
0.6035 0.4093 0.6843
0.5561 0.8311 0.0066
U


= −

−

1.618 0 0
0 1.143 0
0 0 0.0097


∑=



0.0541 0.9984 0.0151
0.9985 0.0540 0.0068
0.0060 0.0154 0.9999
V


= −−

−−

2.4376 3.0241 0.4135
1.2211 0.7617 0.5407
2.2165 1.2623 0.0458
− 

Λ= −

 −

→May consider to remove a input/output pair (u
3andy
1ory
2)