Transcript sistem kendali - Gadjah Mada University

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 closedloop control system.
Self-regulating Processes
• The process outputs are not regulated, its value will easily change.
• following is an example of self regulating process
cold water in
heat exchanger
steam
in
hot water out
steam out
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
current to pneumatic
pressure converter
I/P
4-20 mA
set point
Controller
air supply
3-15 psi
4-20 mA
cold water in
steam in
temperature
transmitter
hot water out
steam
out
heat exchanger
Heat exchange process when under Automatic control.
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
Component
Diff. equ.
Laplace transform
i(t)
v(t )  Ri (t )
V (s)  RI (s)
v(t)
i(t)
v(t)
t
1
v(t )   i(t )dt  v(0)
C0
1
V (s)  I (s)
sc
di
v (t )  L
dt
V (s)  LsI (s)
i(t)
v(t)
Electrical circuit example modeling
i(t)
From the first Laplace transformed equation
R1
R2
v1(t)
v2(t)
C
•
•
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
1
R1i(t )  R2i(t )   i( )d  v1 (t )
C0
t
1
R2i(t )   i( )d  v2 (t )
C0
1
R1 I ( s )  R2 I ( s ) 
I ( s )  V1 ( s )
sC
1
R2 I ( s ) 
I ( s )  V2 ( s )
sC
R1 I ( s)  R2 I ( s) 
1
I ( s)  V1 ( s)
sC
we solve for I(s)
I ( s) 
V1 ( s)
R1  R2  (1 / sC )
Substituting I(s) to the second equation
R2 I ( s ) 
1
I ( s )  V2 ( s )
sC
we get
V2 ( s) 
R2  (1 / sC )
V1 ( s)
R1  R2  (1 / sC )
The transfer function of this system is
G( s) 
V2 ( s)
R2 sC  1

V1 ( s) ( R1  R2 ) sC  1)
Electrical circuit example modeling
• 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
Op-amp example modeling
Zf (s)
Zi (s)
Vi (s)
+
Vo (s) Z f (s)
G( s ) 

Vi (s) Zi (s)
Vo (s)
Block diagram and signal flow graph
E(s)
G(s)
C(s)
E(s)
G(s)
C(s)
C(s)= G(s)E(s)
R(s) +
E(s)
G(s)
C(s)
R(s) 1 E(s) G(s)

H(s)
 H(s)
G( s)
C ( s) 
R( s )
1  G( s) H ( s)
C(s)
Mechanical Translational System modeling
B
K
d 2x
dx
f (t )  M 2  B  Kx
dt
dt
C ( s) 
M
X ( s)
1

F ( s) Ms 2  Bs  K
F (s)  Ms2 X (s)  BsX (s)  KX (s)  (Ms2  Bs  K ) X (s)
Mechanical Rotational System modeling
K
 (s)
J
B
,
T ( s)

1
Js2  Bs  K
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
Let us start with an example
d 2 y (t )
dy(t )
M

f
(
t
)

B
 Ky (t )
dt 2
dt
G( s) 
Y ( s)
1

F ( s) Ms 2  Bs  K
For second order system we define
two state variables x1(t) and x2(t) as
x1(t) = y(t)
x2 (t ) 
dx1 (t )
 x1 (t )
dt
Then we may write
d 2 y (t ) dx2 (t )

 x 2 (t )
dt 2
dt
B
K
1

x2 (t ) 
x1 (t ) 
f (t )
M
M
M
Written in specific format, we have
x1  x2
K
B
1
x1 (t ) 
x2 (t ) 
f (t )
M
M
M
y (t )  x1 (t )
x2 (t )  
In matrix notation, we have
 x1   0
 x    K
 2   M
1  x   0 
B  1    1  f
   x2   
M
M 
 x1 
y (t )  1 0  
 x2 
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 = (nn) system matrix
–
B = (nr) input matrix
– u(t) = input vektor = (r1) vector composed of the system input function
– y(t) = output vektor = (p1) vector composed of the defined output
–
C = (pn) output matrix
–
D = (pr) 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
Example 3.1
Consider a D.E as follows:
y1  k1 y1  k2 y1  u1  k3u2
y 2  k4 y2  k5 y1  k0u1
(1)
where u1 and u2 are inputs and y1
and y2 are outputs. Let us define
the states:
x1  y1 x2  y1  x1
x3  y2
(2)
from (1) and (2) we write :
x1  x2
x2  k2 x1  k1 x2  u1  k3u2
x3  k5 x2  k4 x3  k0u1
with the outputs equation
y1  x1
y2  x3
this equation equations may be
written in matrix form
 0
x   k 2
 0
1 0
y
0 0
1
0  0
 k1
0  x   1
 k5  k 4  k6
0
x

1
0
k3 u
0 
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.
x(t)

y(t)
1
s
X(s)
Y(s)
s-1
X(s)
Y(s)
Symbols of integrator
B
y(t )  
M
We will construct simulation diagram of
f(t)
1
M
y(t )
+
-
-
B
M
K
M
1
s
y (t )

K

y
(
t
)




M
1
s

 1
y
(
t
)




M
y(t)

 f (t )

SIMULATION DIAGRAM
The transfer function of third order system is:
G(s) 
•There are two common form of simulation diagram
•The first one is the control canonical form
b2 s 2  b1s  b0
s 3  a2 s 2  a1s  a0
b2
b1
u(t)
1
s
+
-
x3
-
a2
a1
a0
1
s
x2
1
s
+ +
x1
b0
+
y(t)
SIMULATION DIAGRAM
The second one is the observer canonical form :
G(s)

b2 s 2  b1s b0
s 2  a2 s 2  a1s  a0
u(t)
b1
b0
+

a0
1
s
+
+
x3

a1
b2
1
s
+
+
x2

1
s
y(t)
x1
a2
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 integrator
2. 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)
b2
b1
1
s x3
u(t) +
-
-
1
s x2
1
s x1
+ +
b0
y(t)
+
 0
x   0
 a0
a2
a1
1
0
 a1
0  0 
1  x  0u
 a2  1
a0


y  b0 b1 b2  x


STATE EQUATION FROM SIMULATION DIAGRAM (observer canonical form)
u(t)
b1
b0
+

1
s
+
+
x3
1
s
a1
a0
  a2
x    a1
  a0

b2
1 0 b2 
0 1 x   b1 u
0 0 b0 
y  1 0 0x
+
+
x2

a2
1
s
y(t)
x1
Example
Consider the mechanical system with the following transfer function
1
Y ( s)
1
M
G( s) 


F ( s) Ms 2  Bs  K s 2  B s  K
M
M
The state model of control
canonical form is
 0
x (t )   K
 M
1 
0
B  x(t )    f (t )
 
1
M
1

y(t )  
0 x(t )
M 
The state model of observer
canonical form is
 K
 M
x (t )  
B

 M

1
0
 x(t )   1  f (t )
 M 
0

y(t )  1 0x(t )