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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion1
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-1
054461 Process Control Laboratory
LECTURE 1:
COURSE INTRODUCTION
Daniel R. Lewin
Department of Chemical EngineeringTechnion, Haifa, Israel
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-2
Lecture Objectives
Be familiar with the course structure andobjectives.
Have recalled all of the material you learned inthe introductory control course, and inparticular, be able to:
a. Formulate a linear process model
b. Sketch the response of a linear system
c. Design a simple (PID) feedback controllerusing the Root Locus method
On completing this section, you should:
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion2
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-3
Course Objectives
Derive an empirical linear model for a realprocess by generating experimental data and itsanalysis.
Design, tune and implement SISO controllers fora real process. The types of controllers that youshould be able to implement are: simple feedback(e.g. PID), cascade controllers, IMC and FF.
Design and implement control system for MIMOprocesses.
On completing this course, you should:
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-4
Linear Systems – Review + More
Modeling process transient behavior
Linearization
Laplace transforms
Linear system response Root locus design of FB controllers
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion3
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-5
Modeling - Dead Sea Pond (1)
Prepare a model describing an evaporating pond in the DeadSea Works.
Feed brine [T/h]
Brine conc. [kg salt/kg]
Evap. Rate [T/h]
Solution.
rate of accumu- rate of input rate of out- rate of lossby evaporationlation w ith feed put w ith effluent
f0 q q E
= − − = − −
Overall mass balance:
fThus, q q E= −
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-6
Modeling - Dead Sea Pond (2)
( )
rate of accumu- rate of salt rate of saltout-
lation of salt input with feed put with effluent
f f
dVc q c qc
dt
= −
ρ = −
The balance for salt gives:
fUsing q q E, and noting that and V are constant:= − ρ
( )1 , 0ff
dc E 1 c c c 0 cqdt
+
= − =τ
( )ff f c q Eq cdcV Vdt
−−
=ρ ρ
fDefining V q :τ = ρ What is the sign of
(E/qf – 1) ?
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion4
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-7
Linearization (1) Consider the general SISO nonlinear process:
( )dx
f x,udt
=
Linearization around the stationary point (x0, u0) givesthe linear equation:
( ) ( ) ( )0 0 0 0
0 0 0 0x ,u x ,u
dx f ff x ,u x x u u
dt x u
∂ ∂+ − + −
∂ ∂
But: ( )0 0 0dx
f x , udt
=
Hence: ( ) ( ) ( )
0 0 0 0
0 0 0 0
00 0
x ,u x ,u
x ,u x ,u
d x x f fx x u udt x u
dx f fx u
dt x u
− ∂ ∂− + −∂ ∂
∂ ∂+
∂ ∂
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-8
Linearization (2) Extension to MIMO nonlinear process:
( )
( )
dxf x,u
dt y g x,u
=
=
Linearization around the stationary point (x*, u*) gives
the linear system:
dxAx Bu
dt y Cx Du
= +
= +
ii,j
fA a
x
∂≡ =
∂
The matrices are Jacobian matrices, e.g.A,B,C and D
*
*
x x x
u u u
= −
= −
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion5
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-9
Linearization – Example (1)
( ) ( )1 , 0f f ff
dc Ef c;q ,c ,E 1 c c c 0 cqdt
+
= = − =τ
Continuing the Dead Sea Ponds example, we have:
Data: qf = 10 T/h, E = 5 T/h, cf = 0.1 kg salt/kg brine, ρV = 100 T
At steady state:
( )= ⇒ = − =f fdc
0 c c 1 E q 0.2 kg salt/kg brinedt
Stationary point: c* = 0.2; qf* = 10; cf* = 0.1; E* = 5; τ* = 10 h.
Linearization:
ffss ssf fss ss
dc f f f fc q c E
dt c q c E
∂ ∂ ∂ ∂= + + +
∂ ∂ ∂ ∂
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-10
Linearization – Example (2)
-1*
*ss f*
Ef 11 0.05 h
c q
∂= − = −
∂ τ -1*f*
f ss
c cf0.001 T
q V
−∂= = −
∂ ρ
-1
*f ss
f 10.1 h
c
∂= =
∂ τ-1*
ss
cf0.002 T
E V
∂= =
∂ ρ
( )ffdc
0.05c 0.001q 0.1c 0.002E, c 0 0dt
= − − + + =
ff
ss ssf fss ss
dc f f f fc q c E
dt c q c E
∂ ∂ ∂ ∂= + + +
∂ ∂ ∂ ∂
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion6
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-11
Laplace Transforms General MIMO linear process is:
dxAx Bu
dt y Cx Du
= +
= +
*
*
x x x
u u u
= −
= −
Example: [ ]
( )f
f
qdc
0.05 c 0.001 0.1 0.002 c ,c 0 0dt
E
= − + − =
Taking Laplace Transforms (around steady state):
( ) ( ) ( )
( ) ( ) ( )
sX s AX s BU s
Y s CX s DU s
= +
= +
Hence: ( ) ( ) ( )1
Y s C sI A B D U s− = − +
P(s) –transferfunction matrixrelating all inputsto outputs
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-12
Laplace Transforms
A strictly proper system has n > m (physical system).A proper system has n = m (e.g. PI controller) Roots of numerator are zeros and roots of denominatorare poles:
Analysis of transfer functions: ( ) ( )1
P s C sI A B D−
= − +
The transfer function matrix is composed of elements:
( )m m 1
m m 1 1 0i,j n n 1
n n 1 1 0
b s b s b s bp s
a s a s a s a
−−
−−
+ + + +=
+ + + +…
…
zi are zeros
( ) ( ) ( ) ( )
( )( ) ( )1 2 m
i,j1 2 n
s z s z s zp s
s p s p s p
− − −=
− − −
pi are poles
For stability ALL poles must have a negative real part. Complex poles give oscillatory response. Zeros shape the response (see later)
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion7
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-13
Response to step
in OP: --OP —PVClassification
Transfer
Function
Stable self-
regulatings
p
p e1s
K θ−+τ
Non self-regulating
sp es
K θ−
Unstable sp
p e1s
K θ−+τ−
Linear System Response
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-14
Identification by Step Test First order response of process, y, to a step change in input, u:
Dynamics are approximated by the FOPTD model:p s
Kp(s) e
s 1−θ=
τ +
pwith K y u, and θ and estimated from the trajectory= ∆ ∆ τ
u(t)
y(t)
∆u
0.623∆ y ∆ y
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion8
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-15
Higher Order Responses (1)
p
2 2
Kp(s)
s 2 s 1=
τ + τξ +
Response of 2nd order transfer function to a unit step:
( ) ( )( )= =
− −τ + τξ +p p 1 2
2 21 2
K K p p y(s)
s s p s ps s 2 s 1ξ −ξ
= − ±τ τ
2
1,2
1p
Need to differentiate between three cases:
static gain
damping coefficient
ξ >ξ =ξ <
A) 1B) 1C) 1
natural period
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-16
Higher Order Responses (2)ξ > 1 2A) For 1, p and p are real negative roots:
( )− −= − −1 2p t p tp 1 2 y(t) K 1 A e A e
ξ = = = − τ1 2B) For 1, p p 1/ :
( )( )− τ
= − + τt
p y(t) K 1 1 t e
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion9
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-17
Higher Order Responses (3)ξ < ξ −2 1 2C) For 1, 1 is imaginary, and p and p
are complex roots:
( )tp 21
y(t) K 1 e sin t1
−ξ τ
= − ω − φ − ξ
= − ξ τ ± − ξ τ21,2p i 1
− ξω =
τ
21
( )−
φ = − ξ ξ
1 2
tan 1
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-18
Higher Order Responses (4) Example 1: A processing system with its controller
( ) ( )( ) ( )
p
1 2
KCp s s
F s 1 s 1= =
τ + τ +
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( )s
C s P s B s e s
P s B s C s C s
=
= −
( ) ( ) ( )
( ) ( )
c p
c p
1 2 1 2
c p c p
s 2
K K
K K 1
K K 1 K K 1
P s B sCs
C P s B s 1s s 1
+
τ τ τ + τ
+ +
= =+
+ +
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion11
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-21
Root Locus for Control Design
1. The open-loop transfer function is expressed in the form: The Rules:
2. The root loci start (at K = 0) at the poles of pc(s) and end(K = ∞) at the zeros of pc(s) or at ± ∞
3. On the real axis of the complex plane, the RL isconstructed from right to left, according to the total polesand zeros (Σpz) met along the way:
If K > 0, draw the RL if Σpz is odd
If K < 0, draw the RL if Σpz is even
4. The number of loci = order of the system = number ofpoles of pc(s). The complex loci always appear as complexconjugates (mirror image either side of the real axis).
( ) ( ) ( ) ( )
( ) ( ) ( )1 2 m
1 2 n
s z s z s zpc s K
s p s p s p
− − −=
− − −
…
…
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-22
Root Locus for Control Design
5. The angle of the asymptotes of the loci (as K →∞) is:
The Rules (Cont’d):
( )
( )
180 2k 1 k 0,1, ,n m 1 (K 0)
n m180 2k
k 0,1, ,n m 1 (K 0)n m
−= − − >
−
= − − <−
…
…
where n and m are the number of poles and zeros of pc(s).
6. The asymptotes are centered atwhere pi and zi are the locations of poles and zeros of
pc(s), respectively.
( ) ( )i ip z n mσ = − −∑ ∑
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion12
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-23
Root Locus for Control Design( )
( ) ( ) ( ) c
1 p s ,c s K
s 1 s 3= =
+ +Example 3
1 2-1-2-3-4
-1
-2
1
2
Re(s)
Im(s)3
-3
-5 ××××××××
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-24
Root Locus for Control Design
( )( )( )
( ) ( )c i21
p s ,c s K 1 1 Tss 1 s 3
= = ++ +
Example 4
1 2-1-2-3-4
-1
-2
1
2
Re(s)
Im(s)3
-3
-5 ××××××××
2 ××××
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054461 PROCESS CONTROL LABLECTURE ONE
Daniel R. Lewin, Technion13
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-25
Root Locus for Control Design( )
( ) ( ) ( )s c i
1 p s e ,c s K 1 1 Ts
s 1−= = +
+
1 2-1-2-3-4
-1
-2
1
2
Re(s)
Im(s)3
-3
-5
Example 5
×××××××× ××××
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-26
Design Specifications for RL The desired closed loop response is:
2 2s
y 1(s)
y s 2 s 1=
τ + τξ +
( )t
2
1
y(t) 1 e sin t1
−ξ τ
= − ω − φ− ξ
The response to a unit step in ys is:
The analytical response is used to estimate desired overshootand settling time.
Overshoot: 2OS exp 1 = −πξ − ξ
Thus: ( )
( )
2
e2 2
e
log OS
log OSξ ≥
+ π
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054461 PROCESS CONTROL LABLECTURE ONE
Course IntroductionPROCESS CONTROL LAB - (c) Daniel R. Lewin1-29
Summary
Linearization
Laplace transforms
Linear system response
Root locus design of FB controllers Use of approximate specifications in design
( ) ( )= =x f x,u , y g x,u
Modeling process transient behavior Generating models in standard form: