Post on 04-Jan-2016
description
Teknik kendali
priyatmadi
FEEDBACK CONTROL SYSTEM
• Process control is methods to force process parameters to have specific values.
• Objective to maintain the value of some quantity at some desired level regardless of influences
Process Types
• Self-regulating Processes– These are uncontrolled processes. The process
variables are not regulated. Example!• Manual Controlled Processes
– These proceses are controlled by human Example!• Automatic Controlled Processes
– These process are controlled by automatic controller. There are 2 types:
• feed forward control system. Example!• feedback control system (closed-loop control system).
Example!– We concern with the analysis and design of closed-
loop control system.
Self-regulating Processes• The process outputs are not regulated, its value will easily change. • following is an example of self regulating process
steam out
cold water in
hot water outsteam in
heat exchanger
Manual controlled process
Heat exchange process when under Manual control. The dot line represent the closed loop of the controller and the process
Automatic controlled process
Heat exchange process when under Automatic control.
set pointController
3-15 psi
air supply
4-20 mA
4-20 mA
temperature transmitter
hot water out
steam out
cold water insteam in
heat exchanger
I/P
current to pneumatic pressure converter
MODELS OF PHYSICAL SYSTEM
• Mathematical model of a system is defined as a set of equation used to represent physical system.
• It should be understood that no mathematical model of physical system is exact, although we may increase the accuracy by increasing the complexity of the equations
• In this module we only concern with LTI system, whose equation can be solved using Laplace transform and can be represented by a transfer function.
models of electrical elements
)()( tRitv
dt
diLtv )(
t
vdttiC
tv0
)0()(1
)(
v(t)
i(t)
v(t)
i(t)
v(t)
i(t)
)()( sRIsV
)(1
)( sIsc
sV
)()( sLsIsV
Component Diff. equ. Laplace transform
Electrical circuit example modeling
• In this circuit we consider v1 to be the input and v2 to be the output.
• here we have to write a set equations whose solution will yield v2(t) as function of v1(t) or V2(s) as a function of V1(s)
t
tvdiC
tiRtiR0
121 )()(1
)()(
i(t)
v1(t)
R1
C
R2
v2(t)
)()(1
)(
)()(1
)()(
22
121
sVsIsC
sIR
sVsIsC
sIRsIR
t
tvdiC
tiR0
22 )()(1
)(
From the first Laplace transformed equation
)()(1
)()( 121 sVsIsC
sIRsIR
)()(1
)( 22 sVsIsC
sIR
we solve for I(s)
Substituting I(s) to the second equation
)/1(
)()(
21
1
sCRR
sVsI
we get
)()/1(
)/1()( 1
21
22 sV
sCRR
sCRsV
The transfer function of this system is
)1)(
1
)(
)()(
21
2
1
2
sCRR
sCR
sV
sVsG
• thus the circuit can be modeled by:– two differential equation– two equation in the LAPLACE transform
variable, or– a transfer function
• another model using state space will be discussed next time
Electrical circuit example modeling
Op-amp example modeling
-
+
Zi (s)
Zf (s)
Vi (s)Vo (s)
)(
)(
)(
)()(
sZ
sZ
sV
sVsG
i
f
i
o
Block diagram and signal flow graph
E(s) C(s)G(s)
E(s) C(s)G(s)
G(s)
H(s)
E(s) C(s)R(s) E(s) C(s)G(s)R(s) 1
H(s)
C(s)= G(s)E(s)
)()()(1
)()( sR
sHsG
sGsC
+
Mechanical Translational System modeling
KBsMssF
sXsC
2
1
)(
)()(
Kxdt
dxB
dt
xdMtf
2
2
)(
)()()()()()( 22 sXKBsMssKXsBsXsXMssF M
K B
J
K
B
,
KBsJssT
s
2
1
)(
)(
Mechanical Rotational System modeling
STATE VARIABLE MODELING
• Purpose: to develop presentation which preserves the input output relationship, but which is expressed in n first order equation
• Advantage: in addition to the input-output characteristic, the internal characteristic of the system is represented
• Computer aided analysis and design of state models are performed more easily
• We feedback more information (internal/state variable) about the plant
• Design procedure that result in the best control system are almost all based on state variable models.
STATE VARIABLE MODELING
)()(
)(1
)()()(
1
212
21
txty
tfM
txM
Btx
M
Ktx
xx
2
1
2
1
2
1
01)(
1010
x
xty
fMx
x
M
B
M
Kx
x
)(1
)()(
)()()(
12
22
2
2
tfM
txM
Ktx
M
B
txdt
tdx
dt
tyd
KBsMssF
sYsG
2
1
)(
)()(
)()(
)()(
2
2
tKydt
tdyBtf
dt
tydM
Let us start with an example
For second order system we define
two state variables x1(t) and x2(t) as
x1(t) = y(t)
)()(
)( 11
2 txdt
tdxtx
Then we may write
Written in specific format, we have
In matrix notation, we have
A second order D.E. has been modified into two first order D.E’s. we used two state variables x1 and x2. For one n order D.E there will be n first order D.E’s having n state variables.
STATE VARIABLE MODELING
Definition: The state of a system of any time t0 is the amount of information at t0 that together with all inputs for t t0, uniquely determines the behavior of the system for all t t0
• The standard form of the state equation is dx(t)/dt = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) where
– x(t) = state vector– A = (nn) system matrix– B = (nr) input matrix– u(t) = input vektor = (r1) vector composed of the system input function– y(t) = output vektor = (p1) vector composed of the defined output– C = (pn) output matrix– D = (pr) matrix to represent direct coupling between input and output
STATE VARIABLE MODELING
Recall that standard form of the state equation is dx(t)/dt = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) The first equation, called the state equation, is a 1st order D.E and x(t)
is the solution of the equation. The second one is the output equation. Given x(t) and u(t) y(t) can be
found. Usually matrix D is zero. Nonzero D indicates that there are some path
coupled input and output. On the first equ only the first derivatives of the state var may appear
on the left side of equation and no derivatives on the right side No derivatives may appear on the output equation. The standard format of the state equation valid for multiple input and
output system.
STATE VARIABLE MODELING
1015242
23112111
ukykyky
ukuykyky
1034253
23121122
21
ukxkxkx
ukuxkxkx
xx
32
11
xy
xy
Example 3.1 Consider a D.E as follows:
xy
uxx
100
001
0
1
00
0
0
010
6
3
45
12
k
k
kk
kk
where u1 and u2 are inputs and y1 and y2 are outputs. Let us define the states:
23
11211
yx
xyxyx
from (1) and (2) we write :
(1)
(2)
with the outputs equation
this equation equations may be written in matrix form
SIMULATION DIAGRAM We have presented the way of finding state model from differential Equation. We will present a method of finding state model form transfer function. The method is based on simulation diagram. It is a block diagram or flow graph consisted of gain, summing junction and integrator only.
Y(s)X(s) s-1
s
1Y(s)X(s)x(t) y(t)
Symbols of integrator
We will construct simulation diagram of )(1
)()()( tfM
tyM
Kty
M
Bty
s
1 y(t)
s
1 )(ty)(tyM
1f(t) +
M
K
-
M
B
-
SIMULATION DIAGRAM
s
1x2s
1u(t) +
1a
-2a
-
0a
s
1x3
-
b0
b1
b2
++
+
y(t)
x1
•There are two common form of simulation diagram
012
23
012
2)(
asasas
bsbsbsG
The transfer function of third order system is:
•The first one is the control canonical form
SIMULATION DIAGRAM
The second one is the observer canonical form :
012
22
012
2)(
asasas
bsbsbsG
u(t)
s
1x3
b1
a1
s
1x2
b2
a2
b0
a0
+
+
+
s
1+ + y(t)
x1
The diagrams can be easily expanded to higher order system
SIMULATION DIAGRAM
Once simulation diagram of transfer function is constructed, a state model of the system is easily obtained. The procedure has two step
1. Assign a state variable to the output of integrator2. Write an equation for the input of each integrator
and an equation for each system output . These equation are written as function of integrator outputs and the system inputs
This procedure yields the following state equation
STATE EQUATION FROM SIMULATION DIAGRAM (control canonical form)
s
1x2s
1u(t) +
1a
- 2a
-
0a
s
1x3
-
b0
b1
b2
++
+y(t)
x1
xy
210 bbb
uxx
1
0
0
100
010
210 aaa
u(t)
s
1x3
b1
a1
s
1x2
b2
a2
b0
a0
+
+
+
s
1+ + y(t)
x1
STATE EQUATION FROM SIMULATION DIAGRAM (observer canonical form)
xy 001
uxx
0
1
2
0
1
2
00
10
01
b
b
b
a
a
a
Example
)(01)( tty x
)(10
)(0
1)( tf
Mt
M
BM
K
t
xx
MK
sMB
s
MKBsMssF
sYsG
22
11
)(
)()(
)(1
0)(
10)( tft
M
B
M
Kt
xx
Consider the mechanical system with the following transfer function
)(01
)( tM
ty x
The state model of control canonical form is
The state model of observer canonical form is