THE APPLICATION OF STATE FEED BACK CONTROL SYSTEM FOR STABILIZE INVERTER PENDULUM.
No. 27 Vol.3 Thn. XIV April 2007
ISSN: 854-8471
THE APPLICATION OF
STATE FEED BACK CONTROL SYSTEM
FOR STABILIZE INVERTER PENDULUM
Syarkawi Syamsuddin,Darmawan
Andalas University, Padang
ABSTRACT
Actually, inverter pendulum position is unstable. It is possible to find equilibrium condition but there
is unstable. Under position control system it is feasible to find a stable condition. So we can
overcome the unstable problem. Even though the system was getting any disturbances but under
stability control the system look for its equilibrium. The control position also consists of DC motor
and other devices. Analyzing the system we applied state feed back for stabilize inverter pendulum.
The concept of this system can be developing for unstable object like a rocket position. Many
unstable The Control System of this “Inverter Pendulum” is realized under digital signal
processing controlling DC motor.
INVERTER PENDULUM SYSTEM
Inverter pendulum absolutely unstable. It
derived by movable cart at the rail. By using a
belt the pendulum have fix connection to DC
motor and encoder-1 as in figure 1 below.
As a instrumentation measurement, it read both
cart horizontal position and angle pendulum position.
Calculating a value of voltage parameter input. More
friendly as a signal processing, we used personal
computer.
Physical and Mathematical Model
Let we consider a physical and mathematical
model of the pendulum control system as a figure 2
below.
Fig.1. Pendulum Model
There are 2 encoder, it called encoder-l and
encoder-2. Encoder-l is used for measure
horizontal position of cart, and encoder-2 is
used for measure angle position. In this control
system the problem was :
How to control inverter pendulum position
in stability?
In the method of this research it need a
signal processing. The function signal
processing are:
• instrumentation measurement,
• signal processing, and
r
• control voltage input parameter of DC
motor.
TeknikA
Fig.2. Physical Model of Pendulum
From Newton’s Law, we derived the dynamic
equations :
M&r& = ku − cc r& − H
M
u(t)
(1)
= mass of cart [kg]
= horizontal displacement of cart [m]
=control voltage of motor [volt]
57
No. 27 Vol.3 Thn. XIV April 2007
k
cc
H
ISSN: 854-8471
= torque gain [N/volt]
= bisques friction coefficient [kg/sec]
= force at pendulum base in horizontal
direction [N]
= moment inertia of pendulum [kgm2]
= angle of pendulum [rad]
= force of to pendulum in vertical
direction [N]
= force of to pendulum in horizontal
direction [N]
= distance between joint and C.G [m]
= bisques frictional coefficient of joint
[kg2/sec]
= mass of pendulum [kg]
= gravitation acceleration [m/sec2]
J
θ
V
H
L
cp
m
g
(8)
( ml cos θ )&r& + ( J + ml 2 )θ&& + c pθ& − mgl sin θ = 0
Equation (7) and (8) can be changed to matrix mode
like:
2
⎡ M + m ml cos θ ⎤ ⎡ &r&⎤ ⎡− c c r& + ml sin θθ& + ku ⎤ (9)
⎢ml cos θ J + ml 2 ⎥ ⎢θ&&⎥ = ⎢ − c θ& + mgl sin θ ⎥
⎦ ⎣ ⎦ ⎣⎢
⎣
p
⎦⎥
⎡ J + ml 2
⎡ &r&⎤
1
⎢
⎢θ&&⎥ =
2
2 2
2
⎣ ⎦ ( M + m)( J + ml ) − m l cos θ ⎢⎣− ml cos θ
Because the equilibrium state is at r = 0 and θ = 0. Also
θ quite small and so assume that sin θ = θ, cos θ = θ
2
and θ& = 0 . So equation (9) can be linierized as:
From Newton’s Law, we derived the dynamic
equations :
M&r& = ku − cc r& − H
u(t)
k
cc
H
J
θ
V
H
L
cp
m
g
Jθ&& = Vl sin θ − Hl cosθ − C pθ&
Moment equation of pendulum at joint to CG:
(2)
Horizontal CG pendulum acceleration equation:
H =m
d
(t + l sin θ )
dt 2
2
(3)
d
(l cos θ ) = V − mg
dt 2
2
mr + ml cos θ − ml sin θθ 2 = H
− ml cos θθ 2 − ml sin θθ = V − mg
⎤
− ml ⎤ ⎡ − c c r& + ku
⎥⎢
⎥
&
M + m ⎥⎦ ⎣− c pθ + mgl sin θ ⎦
(10)
=
2 2
1 ⎡ −m l g
⎢
α 0 ⎣⎢( M + m) mgl
− c c ( J + ml 2 )
mlc c
( J + ml 2 ) k ⎤ ⎢⎢ r& ⎥⎥
⎥
− c p ( M + m)
− mlk ⎦⎥ ⎢θ& ⎥
⎢ ⎥
mlc p
(11)
where
α 0 = ( M + m) J − Mml 2
State Vector and State Equation
There are four states variable in the pendulum control
system, it consist of:
r as displacement of cart
•
• θ as angle of pendulum
•
r& as velocity of cart of pendulum
• θ& as angular velocity of pendulum
Let the four states:
⎡ x1 ⎤ ⎡ r ⎤
⎢ x ⎥ ⎢θ ⎥
x = ⎢ 2⎥ = ⎢ ⎥
⎢ x3 ⎥ ⎢ r& ⎥
⎢ ⎥ ⎢ &⎥
⎣ x 4 ⎦ ⎣θ ⎦
(12)
(4)
from equation 3 and 4:
(5)
(6)
from equation 1, 2, 5 and 6:
( M + m)&r& + ml cos θθ&& + c c r − ml sin θθ& 2 = ku
(7)
TeknikA
⎡ J + ml 2
⎢
⎢⎣ − ml
we call this x vector state as a state vector. Accordance
equation (11) it have can be find vector equation like:
Vertical CG pendulum acceleration equation:
m
⎡ &r&⎤
1
⎢θ&&⎥ =
2
⎣ ⎦ ( M + m) J − Mml
(1)
= mass of cart [kg]
= horizontal displacement of
cart
[m]
=control voltage of motor [volt]
= torque gain [N/volt]
= bisques friction coefficient [kg/sec]
= force at pendulum base in horizontal
direction [N]
= moment inertia of pendulum [kgm2]
= angle of pendulum [rad]
= force of to pendulum in vertical
direction [N]
= force of to pendulum in horizontal
direction [N]
= distance between joint and C.G [m]
= bisques frictional coefficient of joint
[kg2/sec]
= mass of pendulum [kg]
= gravitation acceleration [m/sec2]
M
r
2
− ml cos θ ⎤ ⎡− cc r& + ml sin θθ& + ku ⎤
⎥
⎥⎢
&
c
mgl
θ
sin
θ ⎦⎥
−
+
M + m ⎦⎥ ⎣⎢
p
⎡0 0
⎢0 0
x& = ⎢
⎢0 a32
⎢
⎣0 a 42
1
0
a33
a 43
0 ⎤
⎡0⎤
⎥
⎢0⎥
1 ⎥
x + ⎢ ⎥u (13)
⎢b3 ⎥
a34 ⎥
⎥
⎢ ⎥
a 44 ⎦
⎣b4 ⎦
58
No. 27 Vol.3 Thn. XIV April 2007
and in matrix :
x& = Ax + Bu
ISSN: 854-8471
(14)
We have already know state equation:
⎡0 0
A = ⎢0 0
⎢
⎢0 a32
⎢
⎣0 a 42
where :
⎡0 0
⎢0 0
A= ⎢
⎢0 a32
⎢
⎣0 a 42
1
0
a33
a 43
0 ⎤
1 ⎥⎥
and B =
a34 ⎥
⎥
a 44 ⎦
⎡0⎤
⎢0⎥
⎢ ⎥
⎢b3 ⎥
⎢ ⎥
⎣b4 ⎦
The elements of matrix A and matrix B will be
founded from equation (11) as like :
⎡ a32
⎢a
⎣ 42
a33
a 43
a 34
a 44
2 2
b3 ⎤
1 ⎡ −m l g
=
⎢
b4 ⎥⎦ α 0 ⎢⎣( M + m)mgl
− c c ( J + ml 2 )
mlc c
mlc p
− c p ( M + m)
The output vector is :
⎡ y ⎤ ⎡ r ⎤ ⎡1 0 0 0⎤
y = ⎢ 1⎥ = ⎢ ⎥ = ⎢
⎥ x (15)
⎣ y 2 ⎦ ⎣θ ⎦ ⎣0 1 0 0⎦
and in matrix
(16)
where :
C =
y=Cx
1 0 0 0
0 1 0 0
Open Loop Pendulum
And by using
And then :
⎡ a 32
⎢a
⎣ 42
a 33
a 43
a 34
a 44
⎡0⎤
⎢0⎥
and B = ⎢ ⎥
⎢b3 ⎥
⎢ ⎥
⎣b4 ⎦
0 ⎤
1 ⎥⎥
a34 ⎥
⎥
a 44 ⎦
1
0
a33
a 43
α 0 = ( M + m) J − Mml 2
b3 ⎤ 1
=
b4 ⎥⎦ α 0
⎡ − m 2l 2 g
⎢
⎣⎢( M + m) mgl
− c c ( J + ml 2 )
mlc c
mlc p
− c p ( M + m)
( J + ml 2 ) k ⎤
⎥
− mlk ⎦⎥
( J + ml 2 )k ⎤
⎥
− mlk ⎥⎦
And by using all pendulum parameter
constants, we can find that :
⎡ a 32
⎢a
⎣ 42
a 33
a 43
a 34
a 44
b3 ⎤ ⎡− 0.3027 − 11.71 0.00378 0.884 ⎤
=
b4 ⎥⎦ ⎢⎣ 53.14
61.71 − 0.814 4.714⎥⎦
0
1
0 ⎤
⎡0
⎥
A = ⎢0
0
0
1
⎥
⎢
⎢0 − 0.3027 − 11.71 0.00378⎥
⎥
⎢
53.14
61.71 − 0.814 ⎦
⎣0
⎡ 0 ⎤
⎢
⎥
and B = ⎢ 0 ⎥
⎢ 0.884 ⎥
⎥
⎢
⎣ − 4.17 ⎦
BLOK DIAGRAM WITH STATE FEED
BACK CONTROL
By using a state feed back control, the block diagram
system as below:
Accordance with equation (14) and
equation (16), we make a open loop block of
pendulum as:
Fig.4. Block Diagram with State Feed Back Control
Fig.3 Block Diagram Pendulum
PARAMETERS AND STATE VALUES
As a result of measurement (except
gravitation acceleration), it find that the values
of parameter needs as below:
F = ( f1
F is a state feed back, and
f2
f3
f4 )
In order to find a best response, we have to design
feed back parameter. One of concepts is adjust the
system's poles location. It should be done too many
times until we find a suitable poles location which
produce a best response. It can be done by using
simulation program.
A suitable poles are:
p1 = −400 + j 20
p 2 = −400 − j 20
p3 = −50
p4 = −2.9
Base
on
poles
above
( p1 = −400 + j 20 ;
p2 = −400 − j 20 ; p3 = −50 ; p 4 = −2.9 ), we can
TeknikA
59
No. 27 Vol.3 Thn. XIV April 2007
find a “feed back state”. And it's feed back state
value is:
F = ( f1 f 2 f 3 f 4 ) = (− 141 − 189 − 80.9 − 29.4)
Where
0
1
0 ⎤
⎡0
⎥
0
0
0
1
⎥
⎢
⎢0 − 0.3027 − 11.71 0.00378⎥
⎥
⎢
53.14
61.71 − 0.814 ⎦
⎣0
A= ⎢
TeknikA
⎡ 0 ⎤
and B = ⎢ 0 ⎥
⎢
⎥
⎢ 0.884 ⎥
⎢
⎥
⎣− 4.17⎦
ISSN: 854-8471
SUMMARY
1.
2.
3.
4.
The system will stable when we use a suitable
value of a state feed back.
The output response system still have a small over
shot.
Because of measuring parameter case so it's
slightly difference between simulation and
experiment.
By adding one more direction (x-y direction), the
concept of this control system can be developed
for control vertically position of rocket.
60
ISSN: 854-8471
THE APPLICATION OF
STATE FEED BACK CONTROL SYSTEM
FOR STABILIZE INVERTER PENDULUM
Syarkawi Syamsuddin,Darmawan
Andalas University, Padang
ABSTRACT
Actually, inverter pendulum position is unstable. It is possible to find equilibrium condition but there
is unstable. Under position control system it is feasible to find a stable condition. So we can
overcome the unstable problem. Even though the system was getting any disturbances but under
stability control the system look for its equilibrium. The control position also consists of DC motor
and other devices. Analyzing the system we applied state feed back for stabilize inverter pendulum.
The concept of this system can be developing for unstable object like a rocket position. Many
unstable The Control System of this “Inverter Pendulum” is realized under digital signal
processing controlling DC motor.
INVERTER PENDULUM SYSTEM
Inverter pendulum absolutely unstable. It
derived by movable cart at the rail. By using a
belt the pendulum have fix connection to DC
motor and encoder-1 as in figure 1 below.
As a instrumentation measurement, it read both
cart horizontal position and angle pendulum position.
Calculating a value of voltage parameter input. More
friendly as a signal processing, we used personal
computer.
Physical and Mathematical Model
Let we consider a physical and mathematical
model of the pendulum control system as a figure 2
below.
Fig.1. Pendulum Model
There are 2 encoder, it called encoder-l and
encoder-2. Encoder-l is used for measure
horizontal position of cart, and encoder-2 is
used for measure angle position. In this control
system the problem was :
How to control inverter pendulum position
in stability?
In the method of this research it need a
signal processing. The function signal
processing are:
• instrumentation measurement,
• signal processing, and
r
• control voltage input parameter of DC
motor.
TeknikA
Fig.2. Physical Model of Pendulum
From Newton’s Law, we derived the dynamic
equations :
M&r& = ku − cc r& − H
M
u(t)
(1)
= mass of cart [kg]
= horizontal displacement of cart [m]
=control voltage of motor [volt]
57
No. 27 Vol.3 Thn. XIV April 2007
k
cc
H
ISSN: 854-8471
= torque gain [N/volt]
= bisques friction coefficient [kg/sec]
= force at pendulum base in horizontal
direction [N]
= moment inertia of pendulum [kgm2]
= angle of pendulum [rad]
= force of to pendulum in vertical
direction [N]
= force of to pendulum in horizontal
direction [N]
= distance between joint and C.G [m]
= bisques frictional coefficient of joint
[kg2/sec]
= mass of pendulum [kg]
= gravitation acceleration [m/sec2]
J
θ
V
H
L
cp
m
g
(8)
( ml cos θ )&r& + ( J + ml 2 )θ&& + c pθ& − mgl sin θ = 0
Equation (7) and (8) can be changed to matrix mode
like:
2
⎡ M + m ml cos θ ⎤ ⎡ &r&⎤ ⎡− c c r& + ml sin θθ& + ku ⎤ (9)
⎢ml cos θ J + ml 2 ⎥ ⎢θ&&⎥ = ⎢ − c θ& + mgl sin θ ⎥
⎦ ⎣ ⎦ ⎣⎢
⎣
p
⎦⎥
⎡ J + ml 2
⎡ &r&⎤
1
⎢
⎢θ&&⎥ =
2
2 2
2
⎣ ⎦ ( M + m)( J + ml ) − m l cos θ ⎢⎣− ml cos θ
Because the equilibrium state is at r = 0 and θ = 0. Also
θ quite small and so assume that sin θ = θ, cos θ = θ
2
and θ& = 0 . So equation (9) can be linierized as:
From Newton’s Law, we derived the dynamic
equations :
M&r& = ku − cc r& − H
u(t)
k
cc
H
J
θ
V
H
L
cp
m
g
Jθ&& = Vl sin θ − Hl cosθ − C pθ&
Moment equation of pendulum at joint to CG:
(2)
Horizontal CG pendulum acceleration equation:
H =m
d
(t + l sin θ )
dt 2
2
(3)
d
(l cos θ ) = V − mg
dt 2
2
mr + ml cos θ − ml sin θθ 2 = H
− ml cos θθ 2 − ml sin θθ = V − mg
⎤
− ml ⎤ ⎡ − c c r& + ku
⎥⎢
⎥
&
M + m ⎥⎦ ⎣− c pθ + mgl sin θ ⎦
(10)
=
2 2
1 ⎡ −m l g
⎢
α 0 ⎣⎢( M + m) mgl
− c c ( J + ml 2 )
mlc c
( J + ml 2 ) k ⎤ ⎢⎢ r& ⎥⎥
⎥
− c p ( M + m)
− mlk ⎦⎥ ⎢θ& ⎥
⎢ ⎥
mlc p
(11)
where
α 0 = ( M + m) J − Mml 2
State Vector and State Equation
There are four states variable in the pendulum control
system, it consist of:
r as displacement of cart
•
• θ as angle of pendulum
•
r& as velocity of cart of pendulum
• θ& as angular velocity of pendulum
Let the four states:
⎡ x1 ⎤ ⎡ r ⎤
⎢ x ⎥ ⎢θ ⎥
x = ⎢ 2⎥ = ⎢ ⎥
⎢ x3 ⎥ ⎢ r& ⎥
⎢ ⎥ ⎢ &⎥
⎣ x 4 ⎦ ⎣θ ⎦
(12)
(4)
from equation 3 and 4:
(5)
(6)
from equation 1, 2, 5 and 6:
( M + m)&r& + ml cos θθ&& + c c r − ml sin θθ& 2 = ku
(7)
TeknikA
⎡ J + ml 2
⎢
⎢⎣ − ml
we call this x vector state as a state vector. Accordance
equation (11) it have can be find vector equation like:
Vertical CG pendulum acceleration equation:
m
⎡ &r&⎤
1
⎢θ&&⎥ =
2
⎣ ⎦ ( M + m) J − Mml
(1)
= mass of cart [kg]
= horizontal displacement of
cart
[m]
=control voltage of motor [volt]
= torque gain [N/volt]
= bisques friction coefficient [kg/sec]
= force at pendulum base in horizontal
direction [N]
= moment inertia of pendulum [kgm2]
= angle of pendulum [rad]
= force of to pendulum in vertical
direction [N]
= force of to pendulum in horizontal
direction [N]
= distance between joint and C.G [m]
= bisques frictional coefficient of joint
[kg2/sec]
= mass of pendulum [kg]
= gravitation acceleration [m/sec2]
M
r
2
− ml cos θ ⎤ ⎡− cc r& + ml sin θθ& + ku ⎤
⎥
⎥⎢
&
c
mgl
θ
sin
θ ⎦⎥
−
+
M + m ⎦⎥ ⎣⎢
p
⎡0 0
⎢0 0
x& = ⎢
⎢0 a32
⎢
⎣0 a 42
1
0
a33
a 43
0 ⎤
⎡0⎤
⎥
⎢0⎥
1 ⎥
x + ⎢ ⎥u (13)
⎢b3 ⎥
a34 ⎥
⎥
⎢ ⎥
a 44 ⎦
⎣b4 ⎦
58
No. 27 Vol.3 Thn. XIV April 2007
and in matrix :
x& = Ax + Bu
ISSN: 854-8471
(14)
We have already know state equation:
⎡0 0
A = ⎢0 0
⎢
⎢0 a32
⎢
⎣0 a 42
where :
⎡0 0
⎢0 0
A= ⎢
⎢0 a32
⎢
⎣0 a 42
1
0
a33
a 43
0 ⎤
1 ⎥⎥
and B =
a34 ⎥
⎥
a 44 ⎦
⎡0⎤
⎢0⎥
⎢ ⎥
⎢b3 ⎥
⎢ ⎥
⎣b4 ⎦
The elements of matrix A and matrix B will be
founded from equation (11) as like :
⎡ a32
⎢a
⎣ 42
a33
a 43
a 34
a 44
2 2
b3 ⎤
1 ⎡ −m l g
=
⎢
b4 ⎥⎦ α 0 ⎢⎣( M + m)mgl
− c c ( J + ml 2 )
mlc c
mlc p
− c p ( M + m)
The output vector is :
⎡ y ⎤ ⎡ r ⎤ ⎡1 0 0 0⎤
y = ⎢ 1⎥ = ⎢ ⎥ = ⎢
⎥ x (15)
⎣ y 2 ⎦ ⎣θ ⎦ ⎣0 1 0 0⎦
and in matrix
(16)
where :
C =
y=Cx
1 0 0 0
0 1 0 0
Open Loop Pendulum
And by using
And then :
⎡ a 32
⎢a
⎣ 42
a 33
a 43
a 34
a 44
⎡0⎤
⎢0⎥
and B = ⎢ ⎥
⎢b3 ⎥
⎢ ⎥
⎣b4 ⎦
0 ⎤
1 ⎥⎥
a34 ⎥
⎥
a 44 ⎦
1
0
a33
a 43
α 0 = ( M + m) J − Mml 2
b3 ⎤ 1
=
b4 ⎥⎦ α 0
⎡ − m 2l 2 g
⎢
⎣⎢( M + m) mgl
− c c ( J + ml 2 )
mlc c
mlc p
− c p ( M + m)
( J + ml 2 ) k ⎤
⎥
− mlk ⎦⎥
( J + ml 2 )k ⎤
⎥
− mlk ⎥⎦
And by using all pendulum parameter
constants, we can find that :
⎡ a 32
⎢a
⎣ 42
a 33
a 43
a 34
a 44
b3 ⎤ ⎡− 0.3027 − 11.71 0.00378 0.884 ⎤
=
b4 ⎥⎦ ⎢⎣ 53.14
61.71 − 0.814 4.714⎥⎦
0
1
0 ⎤
⎡0
⎥
A = ⎢0
0
0
1
⎥
⎢
⎢0 − 0.3027 − 11.71 0.00378⎥
⎥
⎢
53.14
61.71 − 0.814 ⎦
⎣0
⎡ 0 ⎤
⎢
⎥
and B = ⎢ 0 ⎥
⎢ 0.884 ⎥
⎥
⎢
⎣ − 4.17 ⎦
BLOK DIAGRAM WITH STATE FEED
BACK CONTROL
By using a state feed back control, the block diagram
system as below:
Accordance with equation (14) and
equation (16), we make a open loop block of
pendulum as:
Fig.4. Block Diagram with State Feed Back Control
Fig.3 Block Diagram Pendulum
PARAMETERS AND STATE VALUES
As a result of measurement (except
gravitation acceleration), it find that the values
of parameter needs as below:
F = ( f1
F is a state feed back, and
f2
f3
f4 )
In order to find a best response, we have to design
feed back parameter. One of concepts is adjust the
system's poles location. It should be done too many
times until we find a suitable poles location which
produce a best response. It can be done by using
simulation program.
A suitable poles are:
p1 = −400 + j 20
p 2 = −400 − j 20
p3 = −50
p4 = −2.9
Base
on
poles
above
( p1 = −400 + j 20 ;
p2 = −400 − j 20 ; p3 = −50 ; p 4 = −2.9 ), we can
TeknikA
59
No. 27 Vol.3 Thn. XIV April 2007
find a “feed back state”. And it's feed back state
value is:
F = ( f1 f 2 f 3 f 4 ) = (− 141 − 189 − 80.9 − 29.4)
Where
0
1
0 ⎤
⎡0
⎥
0
0
0
1
⎥
⎢
⎢0 − 0.3027 − 11.71 0.00378⎥
⎥
⎢
53.14
61.71 − 0.814 ⎦
⎣0
A= ⎢
TeknikA
⎡ 0 ⎤
and B = ⎢ 0 ⎥
⎢
⎥
⎢ 0.884 ⎥
⎢
⎥
⎣− 4.17⎦
ISSN: 854-8471
SUMMARY
1.
2.
3.
4.
The system will stable when we use a suitable
value of a state feed back.
The output response system still have a small over
shot.
Because of measuring parameter case so it's
slightly difference between simulation and
experiment.
By adding one more direction (x-y direction), the
concept of this control system can be developed
for control vertically position of rocket.
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