View
237
Download
0
Category
Preview:
Citation preview
1
2
Outline1. Introduction2. Modeling3. Simulation4. Implementation5. Demo6. Conclusion
3
Outline1. Introduction2. Modeling3. Simulation4. Implementation5. Demo6. Conclusion
4
Introduction
• Rotating arm and inverted pendulum.• Rotating arm is actuated by a DC motor.• The angular disturbance will be sensed by the
potentiometer.
l1 length from the center of rotating arm to the pendulum.
l2 length of the inverted pendulum.
m1 mass of the rotating arm.
m2 mass of the inverted pendulum.
α The angular displacement of the rotating arm rotated.
θ The angular displacement of the inverted pendulum.
linear velocity of the mass center of the inverted pendulum.
Potentiometer
Motor
l 2
l1
m2
α
ө
X
Y
Z
m1
X
Y
r cv
5
Introduction
• The system is controlled by a PID control circuit.• Two equilibrium points existed.• Use a cut-off device to protect the system.
Sensor
Controller PlantDriver
Poweramplifier
DCmotor
PIDcontroller
Potentio-meter
Inverted pendulum
AngleVoltage signal
Disturbance
6
Outline1. Introduction2. Modeling
Find the transfer function of input voltage and the angle of inversed pendulum.
– Equation of motion.– Linearization– Laplace transform– Transfer function
3. Simulationment4. Implementation5. Demo6. Conclusion
7
Modeling -Equation of motion
2 2.2 2 22
2 2
1 1 1(cos ) (sin )
2 2 2
1cos 1
2
r cx y zT I m v I I I
V m gl
L T V
• Step 1 : Find the equation of motion by Lagrange equation
1 2 2
1 2 2 2
1 1( sin sin cos cos sin )
2 21 1 1
( cos sin sin cos cos ) ( sin )2 2 2
r cv l l l i
l l l j l k
8
Modeling -Equation of motion
2 2 22 2 1 2 2 2 2 2 2
2 2 2 2 22 1 2 2 1 2 2
1 2 2
0
1 1 1 1cos cos sin sin 0.......
4 2 4 2
1 1sin cos sin cos
4 2
1
2
x y z
y z
d L L
dt
I l m l l m m l I I m l g a
d L L
dt
I m l m l I I l l m
l l m
2 22 2
1sin 2 cos sin .......
4z yI m l I b
9
• Step 2 : Linearization – To do the linearization, we have to find the equilibrium points first.– Find the position where the extreme value of the potential energy exist.
Modeling -Linearization
2 2
2 2
1cos 1
21
sin 02
0 or 180
V m gl
dVm gl
d
Θ=180.0°
Θ=0.0°
10
• In this case, we set the equilibrium point at θ=0°• Expand the nonlinear terms in Taylor series.
•
Modeling -Linearization
2 4 3 51 1 1 1sin ... ; cos 1 1
2 24 6 120
2 22 2 1 2 2 2 2 2 2
22 2 1 2 2 2 2
2 22
2
2
22
21 2 2 1 2
1 1 1 10......(a)cos sin
1
cos c
4 2 4 2
1 1 1-
cos sin
sin sin
0......(a*)4 2
o1
4s
2
1
2
x y z
x
y z
I l m l l m m l I I m l g
I l m l l m m l g
I m l m l I I l l m
12
2 2
22 2
22 1 1 2 21 1
1
2
12 cos .......
4
1..
si
...
n
.. *2
sinz y
y
l l m
I m l I b
I m l I l l m b
11
System modeling -Linearization
• If the angle of disturbance is 5°, the max. error between linear and nonlinear model is 0.046°, less then 1%.
12
• Step 3 : Laplace transform of the motion equations
System modeling -Laplace transform
22 2 1 2 2 2 2
L 22 2 1 2 2 2 2
22 1 1 2 2
L 2
2 2
2 1 1 22
22
1 1 1- 0......(a*)
4 2 2
1 1 1- 0......(1)
4 2 2
1......(b
A
*)2
1......(2)
2A
x
x
y
y
I l m l l m m l g
I l m l l m m l g
I m l I l l m
I m l I
s s
s l sl m
13
System modeling -Transfer function
• Step 4 : Find the transfer function of a DC motor• According to Kirchhoff’s voltage law (KVL)
Where is the voltage of coil
is the induced voltage of the motor
is the torque generate by motorAe K
.....(3)
T A A A
A
L A
T
T
v
v e i R
s
K RK
RK
KV
Tv
AK i
Equivalent circuit of a DC motor
14
System modeling -Transfer function
22 2 1 2 2 2 2
22
2 2
12 2
1 2 2
1 1 1- 0......(1)
4 2 2
1......(2)
2
.....(3)
y
AT
xI l m l l m m l g
I m l I
s s
s s
V
l l m
RK
Ks
• Step 5 : Transfer function of the system
15
Symbol Value Unit
0.10 m
0.32 m
0.02841 Kg
0.046 Kg
9.81 m/s2
7.6707e-5 Kgm2
3.925e-4 Kgm2
3.925e-4 Kgm2
1
2
1
2
X
Y
l
l
m
m
g
I
I
I
Modeling -Transfer function
• Set the values we need
Symbol Value Unit
1 Ω
0.03 AR
K
• Assume the values we need but we don’t know
Ref. : Stephen J. Chapman “Electric Machinery Fundamentals” Chap. 9 McGraw. Hill
16
Modeling -Transfer function
2
4 3 2
2
4 3 2
4.71s -216.6
0.0918s 0.1413 6.7088 6.49
-2.208s
0.0918s 0.1413 6.7088 6.49
V s s s
V s s s
• Transfer function.
17
Modeling -Transfer function
• Unit step command test
Sensor
Controller Plantactuator
Poweramplifier
DCmotor
PIDcontroller
Potentio-meter
Inverted pendulum
AngleVoltage signal
Command
18
Modeling -Transfer function
• Command unit step and disturbance is zero to check transfer function.
19
Modeling –Routh-Hurwitz Stability
3 2
3
2
( ) 0.0718 (0.1413 20208 ) (2.208 6.7088) (2.208 6.49)
0.0918 2.208 6.7088
0.1413 20208 2.208 6.49
1 (0.1413 2.
0.0918
s s D s P s I
s P
s D I
s
0
3.038
2.93
0.06
0.3119 4.875 14.813 2.
208 ) (2.208 6.7088) 0.0918 (2.208 6.49)
026 0
2.208 6.
. 8
9
4 2
4
P
I
D
P PD D I
D P I
s I
• Using Routh-Hurwitz stability to find the stable range of the gain of PID or PD controller.
20
Modeling -Reference
• S. Awtar, N. king, T. Allen, I. Bang, M, Hagan, D.Skidmore, K. Craig, “Inverted pendulum systems: rotary and arm-driven- a mechatronic system design case study.” Mechatronic 12 (2002)
• Y. Yavin, “Control of a Rotary Inverted Pendulum.” Applied Mathematics Letters 12 (1999)
21
Outline1. Introduction2. Modeling3. Simulation
– Open loop– PD controller– PI controller– PID controller
4. Implementation5. Demo6. Conclusion
22
Simulation
• Use SimMechanics to build a nonlinear system model
23
Simulation
• Use Simulink to build a nonlinear system model
24
Simulation
• Use Simulink to build a linear system model
25
。 Simulation –open loop (angular V)
26
Simulation -PD controller
I controller ×0
P controller ×6
D controller ×22
10K
10KInverter ×20 Signal input
10K
6 20 120
0
22 20 440
P
I
D
27
Simulation -PD controller
28
Simulation -PD controller
• Response simulation.(PD controller)
• Absolute error between the simulation of SimMechanics and Simulink.
29
Simulation -PI controller
I controller ×2.5
P controller ×6
D controller ×0
10K
10KInverter ×20 Signal input
10K
6 20 120
2.5 20 50
0
P
I
D
30
Simulation -PI controller
31
Simulation -PI controller
• Response simulation.(PI controller)
• Absolute error between the simulation of SimMechanics and Simulink.
32
Simulation -PID controller
I controller ×1.5
P controller ×10
D controller ×11
10K
10KInverter ×15 Signal input
10K
10 15 150
1.5 15 22.5
11 15 165
P
I
D
33
Simulation -PID controller
34
Simulation -PID controller
• Response simulation.(PID controller)
• Absolute error between the simulation of SimMechanics and Simulink.
35
Outline• System introduction• System modeling• Simulation• Implementation
– Inversed pendulum– Control circuit
• Demo• Conclusion
36
Implementation
• System block diagram
Sensor
Controller Plantactuator
Poweramplifier
DCmotor
PIDcontroller
Potentio-meter
Inverted pendulum
AngleVoltage signal
Disturbance
37
• The length and mass of pendulum:32 cm and 28.41g
• The length and mass of rotating arm: 10 cm and 46 g
• Gear ratio: 5
Implementation -Inversed pendulum
38
Implementation -Control circuit
CommonCOM
PowerAmplifier
DCmotor
PIDController
Potentio-meter
Power supplyNO. 1
Power supplyNO. 2
AngleCut-offCircuit
• Circuit block diagram
39
Implementation -Control circuit
PID controller
Power amplifier
Cut-off circuit
Powersupply II
On/Off
Sensor
Signal light
Limit switch
Motor
• Circuit board
Powersupply I
40
Implementation -Potentiometer
• Use a variable resistor as a potentiometer.
-15V
+15V
Output voltage
Inverted pendulum
Potentiometer
41
Implementation - Potentiometer
-15V
+15V
Output voltage
-15V
+15V
Output voltage
0V
15k ohm
15k ohm
-15V
+15V
Output voltage
1V
14k ohm
16k ohm
• How does it work?
42
Implementation -PID controller
• Use 17741 operational amplifier
• Modes switch
• Elements shiftable
PID controller
43
Implementation -PID controller
P
Inverter
I
D
3.2kΩ 10ηF
500Ω 500Ω
200kΩ10ηF
500Ω
500Ω
500Ω500Ω
Input
Output
500Ω
500Ω
500Ω
44
Implementation -Cut-off circuit, signal light
NPN transistor
Relay5V 2 Form C Contact
500 ohm resistances
Resistance with signal
light
7404 NOT
7408 AND
74047408
45
Limit switch II
Limit switch I
Switch
Circuit for LED
500Ω
5V
500Ω
500Ω
500Ω
Output to relay
Implementation -Cut-off circuit, signal light
46
Implementation -Power amplifier
PNP TIP107
NPN TIP41
+15V
-15V
Input
B
C
E
B
C
EMotor
NPN TIP41
NPN TIP107
Diode
47
Implementation
• Why we use two power supply?
• The DC motor turns on, the voltage of power supply drops.
Input:
triangular
±200mV;2Hz
Output:
DC power supply
+15V port
The DC motor use the power from +15V port
normal
48
Outline1. Introduction2. Modeling3. Simulation4. Implementation5. Demo6. Conclusion
49
Demo -PD controller
• Steady state error exist
50
• Steady state error is zero
Demo -PID controller
51
Outline1. Introduction2. Modeling3. Simulation4. Accomplishment5. Demo6. Conclusion
52
Conclusion
• We use different ways to model the system by MATLAB.
• For a small disturbance, linearized model is reliable.
• The rotary inverted pendulum can be controlled by a PID controller.
• I controller can eliminate the steady state error.
Recommended