and implementation of servo motor sliding mode control matlab … · 2008. 7. 5. · motivation •...
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Simulation and Implementation of Servo Motor Control
with Sliding Mode Control (SMC) using Matlab
and LabView
Bondhan
Novandyhttp://bono02.wordpress.com/
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Outline
• Background• Motivation• AC Servo Motor• Inverter & Controller• Hardware Block Diagram• Mathematical Modeling• Sliding Mode Controller Design• Simulation using Matlab• Simulation & Implementation using LabView• Several Ideas
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Motivation
• Sliding Mode Control is a robust control scheme based on
the concept of changing the structure of the controller in
response to the changing state of the system in order to
obtain a desired response • The biggest advantage of SMC is its insensitivity to variation
in system parameters, external disturbances and modeling
errors• This can be achieved by forcing the state trajectory of the
plant to the desired surface and maintain the plants state
trajectory on this surface for subsequent time• Because of these factors SMC is chosen as the controller
for our device
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AC Servo Motor
• The difference between AC Servo Motor and DC servo motor
is the design of the motor where in AC motor the permanent
magnet is on the rotor. The block diagram of an AC servo
motor is very similar to the block diagram of DC servo motor:
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AC Servo Motor
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Inverter and Controller
• Inverter is used to transform electricity from 1 single phase
into 3 phase
• It works by controlling the rotational speed of an AC motor by
controlling the frequency of the electrical power supplied to
the motor
• Our inverter the we use the linear V/F mode
Output max voltage is equalto max input voltage = 220 V RMS
The default frequency is 50 Hz
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Inverter and Controller Cont’d• Analog input voltage 0‐
10 V is used to change the volt/frequency of
the inverter output
• The scale between input : output = 10 : 220
depending on the configuration of V/F
• NI‐PXI 7358 is chosen as the controller and it has DAC 0‐10 output
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Hardware Block Diagram
PXI‐7358 InverterAC Servo
Motor
Encoder
DAC 0‐10 V 0‐220V/0‐50 Hz
Position, Speed, Acceleration
x22
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Mathematical Modeling
2
2
2
2
* * ...(1)
* ...(2)
1 ( )...(3)
Load Load
Load
dI dIa Ra La Kb Vdt dt
d dT T Kt Ia T Tj Tb J Bdt dt
d dIa J B TKt dt dt
θ
θ θ
θ θ
+ + =
− = − = + = +
= + +
Position control:
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Mathematical Modeling Cont’d
1 2
2 3
3 3 3 2 2* * 1 '
* * * *Load LoadKt B Ra Ra B Kb Kt La RaV T T
J La J La La J La J J Kt La J
θ θ
θ θ
θ θ θ θ θ
=
=
= − − − − − −
2 2
2 2
2 3 2
2 3 2
(3) (1)
( ) ( )
* * * * '
Load Load
Load Load
Ra d d La d d d dJ B T J B T Kb VKt dt dt Kt dt dt dt dtJ Ra d B Ra d Ra J La d B La d La dT T Kb V
Kt dt Kt dt Kt Kt dt Kt dt Kt dt
θ θ θ θ θ
θ θ θ θ θ
→
+ + + + + + =
+ + + + + + =
Finally we get (4):
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Sliding Surface
• We have 3rd
order system1( ) , choosing n=3 and is tracking errorn
ddsdt
λ θ θ θ θ−= + = −
2
22
2
2
2
( )
( 2 )
2
( ) 2 ( ) ( )d d d
ddtd ddt dt
λ θ
θ θλ λ θ
θ λθ λ θ
θ θ λ θ θ λ θ θ
= +
= + +
= + +
= − + − + −
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Equivalent Control
• We try to force the state trajectory to slide on our surface so that: 0
0ss==
2
3 3 2 2
23 2
3 3 2 2
( ) 2 ( ) ( )* * 10 '
* * *
2 ( ) ( )*
* * * *' *
d d d
Load
Load d d d
eq Load Load
sKt B Ra Ra B Kb Kt LaV T
J La J La La J La J J KtRa T
La JB La La Ra J Ra B Ra La JV T T Kb
Kt Kt Kt Kt Kt K
θ θ λ θ θ λ θ θ
θ θ θ θ
θ λ θ θ λ θ θ
θ θ θ θ
= − + − + −
= − − − − −
− − + − + −
= + + + + + +
23 2
* *2 ( ) ( )
d
d d
tLa J La J
Kt Kt
θ
λ θ θ λ θ θ− − − −
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SMC Controller
, * ( / )
and is the function: ( ) abs( ) / abs( )
eq switching switchingV V V where V K sat s
satsign s if ss if s
φ
φφ φ
= + = −
><
3 3 2 2
23 2
* * * *' *
* *2 ( ) ( ) * ( / )
eq Load Load d
d d
B La La Ra J Ra B Ra La JV T T KbKt Kt Kt Kt Kt Kt
La J La J K sat sKt Kt
θ θ θ θ θ
λ θ θ λ θ θ φ
= + + + + + +
− − − − −
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Lyapunov
Function
If there exist Lyapunov
function, so that
It is stable in the sense of Lyapunov
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Stability
We ensure the stability of our system choosing K to be large enough so that stable in the sense of Lyapunov.
Lyapunov
candidate:2
3 3 2 2
23 2
1 02
* * 1( '* * *
2 ( ) ( ) * ( / ))*( * ( / ))
*
Load
Load d d d
V s
Kt B Ra Ra B Kb Kt LaV ss s V TJ La J La La J La J J Kt
Ra T K sat sLa J
V s f K sat sf s K s
θ θ θ θ
θ λ θ θ λ θ θ φ
φ
=
= = − − − − −
− − + − + − −
= −
= − 0≤
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Simulation Using Matlab
Plant parameter:j=0.01;b=0.1;Kt=0.01;Kb=0.29098;Ra=1;La=0.5;
The unity feedback transfer function is:
2------------------s^2 + 12 s + 22.58
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1.4
-1.2
-1
-0.8
-0.6
-0.4
-0.2
0
x1
x2
phase portrait
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Transient Response Without Controller
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.02
0.04
0.06
0.08
0.1
0.12
0.14Impulse Response
Time (sec)
Ampl
itude
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09Step Response
Time (sec)
Ampl
itude
Stability Analysis (Linear System)Controllability:% Number of uncontrollable states>> unco = length(A)-rank(ctrb(A,B))unco =
0Observability:% Number of unobservable statesunob = length(A)-rank(Ob)unob =
0Thus the system is stable.
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Simulation Result With SMC Controller
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Step Response
K = 50; Lambda = 10; delta = 0.5;
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Step Response
K = 100; Lambda = 10; delta = 0.9;
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Trajectory Following
The trajectory is defined so that it will not produce shock while
moving because of discontinuity.
Quintic
polynomial:
2 3 4 50 0 1* 2* 3* 3* 5*:
0 start positionfinal position
0 start velocityfinal velociy
0 start accelerationfinal acceleration
q a a t a t a t a t a twhereqqfvfv
fαα
= + + + + +
======
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Quintic
polynomial trajectory for position, velocity and acceleration
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Trajectory Following Response K = 80; Lambda = 10; delta = 0.9;
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Error
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Simulation Using LabViewTrajectory Generator
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Simulation Using LabView
plantSMC controller
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Step Response
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Trajectory Response
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Implementation
• Implementation in NI Real‐Time can be done by replacing the runge‐kutta
ODE solver and
array composition with the real motor which is the SERVO Motor
• The feedback of the motor must be available (encoder)
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Problems and Discussion
• Real‐time clock generation ‐> Sampling time
• Motor parameters
• The feedback of the system are:– Position– Velocity– Acceleration *) In the simulation when acceleration feedback was defined
as 0, the error is still small
• Integral sliding mode control1
0
( ) , choosing n=4 and is tracking errort
nd
dsdt
λ θ θ θ θ−= + = −∫