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matlab - servo motor transfer function matlab code |
Servo Motor transfer function Using in MATLAB software
Servo Motor
Servo Motor, we can say Control motors and used in feedback control systems (close loop) as output actuators. Its newer apply for continue energy transfer or conversion. The standard of the Servo motor is like that of the other electromagnetic motor, yet the development and the activity are extraordinary.
Their capacity rating differs from a small amount of a watt to a
couple hundred watts.
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The servo motor is most usually utilized for high innovation gadgets in the industrial applications like robotics technology.
It is an independent electrical gadget that turns portions of machine with high proficiency and incredible exactness.
Additionally
the yield shaft of this motor can be moved to a specific angle. Additionally
the yield shaft of this engine can be moved to a specific point.
The rotor inertia of the motors is low and have a high speed of reaction. The rotor of the Motor has the extensive length and lesser diameter. They work at low speed and now and then even at the zero speed.
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The servo motor is generally utilized in home
electronics gadgets, toys, cars, airplanes, radar and computers PCs, robot,
machine tool, tracking and guidance systems, processing controlling, and so on.
The Applications of the
Servomotor
- Radar system and process controller.
- Computers and robotics.
- Machine tools.
- Tracking and guidance systems.
- Robotics
- Conveyor belts
- Camera auto focus
- Solar tracking system
Type of Servo Motor
Ø DC Servo
Motor
Ø AC
servomotor
o Two Phase
AC Servo Motor
o Three
Phase AC Servo Motor
DC Servo Motor
Matlab Code-
clc
clear all
close all
J=0.01;
f=0.1;
K=0.01;
R=1;
L=0.5;
s=tf('s');
P_motor=K/(s*((J*s+f)*(L*s+R)+K^2))
bode(P_motor)
grid
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Result-
P_motor =
0.01
-------------------------------
0.005 s^3 + 0.06 s^2 + 0.1001 s
Continuous-time transfer function.
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AC Servo Motor
Matlab Code-
clc
clear all
close all
% Transfer function, Gm (s) = Km / (1+ s *Tm)
%% Find out Km
R = input ( 'radius of the rotor in m is R= ')
T = 9.81 * R
C=1
K = T / C
N = input ('rated speed in rpm is N = ')
FO = T / N
RC = 174 % resistance ohm
V = input ('supply voltage,V= ')
% IR is current through reference winding, Amp
IR = input ('current through reference winding IR= ')
% IC is current through control winding, Amp
IC = input ('current through control winding IC= ')
No_load_input = V *(IR+IC)
Copper_loss_in_watts = IC^2 * RC
Constant_loss_in_watts = No_load_input - Copper_loss_in_watts
% Frictional loss W is 30 % of constant loss in Watts
W = Constant_loss_in_watts *30/100
% Frictional_co-efficient
F = W / (2 *N / 60)^2
Km = K / FO + F
d =39.5e-3 % diameter of the rotor in m
LR =76e-3 % length of the rotor in m
rho=7.8*10^(-1) % density
J = pi*d*4* LR*rho / 32
tau_m=J / FO + F
% Gm (s) = Km / (1+ s m)
Gm = tf(Km, [tau_m 1])
bode(Gm)
grid
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Result-
radius of the rotor in m is R= 0.15
R =
0.1500
T =
1.4715
C =
1
K =
1.4715
rated speed in rpm is N = 3000
N =
3000
FO =
4.9050e-04
RC =
174
supply voltage,V= 220
V =
220
current through reference winding IR= 4
IR =
4
current through control winding IC= 3
IC =
3
No_load_input =
1540
Copper_loss_in_watts =
1566
Constant_loss_in_watts =
-26
W =
-7.8000
F =
-7.8000e-04
Km =
3.0000e+03
d =
0.0395
LR =
0.0760
rho =
0.7800
J =
9.1953e-04
tau_m =
1.8739
Gm =
3000
-----------
1.874 s + 1
Continuous-time transfer function.
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