Transcript Document

A Study on Motor Speed Control
T  K t ia
Vb  K e
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System Equations
KVL and Newton’s 2’nd
law gives :
J  b  K t i a Tl
dia

K e  L
 Ria  va
dt
Let : Output : y = 
Input : u = va
Disturbance w = Tl
Laplace transforming, we obtain :


JL
JR  bL
2

s 
s  1Y ( s ) 
bR  K t K e
 bR  K t K e

Kt
1
U (s) 
W ( s)
bR  K t K e
bR  K t K e
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Transfer Functions
A
B
Y ( s) 
U ( s) 
W ( s)
( 1s  1)( 2 s  1)
( 1s  1)( 2 s  1)
Where:
Kt
A
bR  K t K e
RJ
1 
Kt Ke
1
B
bR  K t K e
: Mechanicaltimeconstant
L
2 
: Electricaltimeconstant
R
(Assumingb  0 and L is small)
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System Parameters
For this example, the following values for the physical parameters will
be assumed.
•moment of inertia of the rotor (J) = 0.0001 kg.m2/s2
* damping ratio of the mechanical system (b) = 0.0 Nms
* electromotive force constant (K=Ke=Kt) = 0.01 Nm/Amp
* electric resistance (R) = 0.1 ohm
* electric inductance (L) = 0.0005 H
* input (V): Source Voltage
* output ( ): angular velocity of shaft
* disturbance (Tl) : Load torque
* The rotor and shaft are assumed to be rigid
Reference : http://www.engin.umich.edu/group/ctm/index.html
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Numerical Values
J=0.0001;
b=0.0;
Ke=0.01;
Kt=0.01;
R=0.1;
L=0.0005;
A=Kt/(b*R+Kt*Ke); A=100
B=1/(b*R+Kt*Ke); B=10000
den=[J*L/(b*R+Kt*Ke) (J*R+b*L)/(b*R+Kt*Ke) 1];
r = roots(den);
tau1= -1/r(1); tau2=-1/r(2); tau1 = 0.0947 s.: mech. time constant
tau2 = 0.0053 s.: electrical time constant.
100
10000
Y ( s) 
U ( s) 
W ( s)
(0.095s  1)(0.005s  1)
(0.095s  1)(0.005s  1)
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Open Loop Step Response
step(A,den,0:.05:1.5)
title('Step Response for the Open Loop System')
Step Response for the Open Loop System
From: U(1)
100
90
80
60
To: Y(1)
Amplitude
70
50
40
30
20
10
0
0
0.5
1
Time (sec.)
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1.5
Steady State Model
yss = 100 u + 10000 w
where y is in rad/s,
u is in Volts
and w is in N/m.
For example, u = va = 1.5 Volts gives ss = 150 rad/s = 1432 rpm.
If w = Tl = -0.002 N m., then ss = 150 - 20 = 130 rad/s = 1241 rpm.
Result : Without control, the speed decreases (or increases)
proportional to w. Controller must increase (or decrease) u, to
compensate for the effect of disturbance.
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Feedback Control
100
4.2
Disturbance :
Load Torque Tl
Ref.
signal
(Volts)
Gain
1.5
PID
PID Controller
Steady state
Voltage input
x' = Ax+Bu
y = Cx+Du
Saturation
DC MOTOR
Motor
speed
Voltage input
0.028
TACHOMETER
Simulink block diagram
Input saturation : Input voltage va can be between 0 and 2.5 volts
Measurement : Tachometer gain : 0.028 Volts/rad/s
Reference signal : 150 rad/s * 0.028 Volts/rad/s = 4.2 Volts; is
compared with tachometer output.
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Proportional Control
yref is increased from 150 to 160 rad/s.
162
160
Kp = 5
158

(rad/s)
Kp = 0.5
156
154
152
150
0
0.1
0.2
0.3
0.4
0.5
time (s)
0.6
0.7
0.8
0.9
1
There is ss. error, ss. error decreases as KP is increased
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Disturbance Rejection
A step torque disturbance of Tl = -0.002 Nm. is applied
Disturbance rejection with P-control
150
Kp = 5
149
148

147
(rad/s.)
146
145
144
143
Kp = 0.5
142
141
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
time (s.)
Compared to no control (130 rad/s), there is improvement, still there
is ss. error.
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PID Controller
t

1
de(t ) 

u (t )  K  e(t )   e(t ) dt  Td

T
dt
i 0


1
K ( s )  K (1 
 Td s )
Ti s
Ki
or : K ( s )  K p 
 KDs
s
Control signal is a linear combination of error, integral of error
and time rate of change of error.
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PID Control of DC Motor
Kp  5
Ki  50
Kd  0.005
PID control servo and disturbance rejection
162
ref = 150 to 160 rad/s.
160
158

(rad/s)
156
154
152
Tl = 0 to -0.002 Nm.
150
148
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
time (s.)
No steady state error, good response.
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0.9
1