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Experiment # 7
Servo Motor Position Control
1. Objective:
ο‚·
To become familiar with the fundamentals of servo technology.
ο‚·
To become familiar with position control.
ο‚·
Investigate the effect ofPID-Controller on the performance of Servo Motor system.
2. Theory and Background:
Servo system is a system to control mechanical instruments in compliance with variation of
position or speed target value. The word “servo” derives from Greek “servant”. The system is
called “servo system” as it responds faithfully to command. The major difference of servo motor
compared with general use motors is that it has a detector to detect rotation speed and position.
DC servo motors have beenwidely used as an actuator for motion control anddirect-drive
applications. Examples are as roboticand actuator for automation process, mechanicalmotion and
others. The DC servo motors have beenextensively applying in many servome chanisms.
Therefore, it is very important to study about theposition control of the DC servo
motors.Consider the servo system shown in Figure 1. The objectiveof this system is to control
the position of the mechanical load in accordance withthe reference position.
Figure 1: Schematic diagram of servo system.
The operation of this system is as follows: A pair of potentiometersacts as an errormeasuring device. They convert the input and output positions into proportional electric signals.
The command input signal determines the angular positionr of the wiper arm of the input
potentiometer.The angular position r is the reference inputto the system, and the electric
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potential of the arm is proportional to the angularposition of the arm. The output shaft position
determines the angular position c of thewiper arm of the output potentiometer. The difference
between the input angular positionr and the output angular position c is the error signal e, or
𝑒 =π‘Ÿ−𝑐
The potential difference π‘’π‘Ÿ − 𝑒𝑐 = 𝑒𝑣 is the error voltage, where e, is proportional to rand 𝑒𝑐 is
proportional to c; that is, π‘’π‘Ÿ = 𝐾0 π‘Ÿ and 𝑒𝑐 = 𝐾0 𝑐, where 𝐾0 is a proportionalityconstant. The
error voltage that appears at the potentiometer terminals is amplified bythe amplifier whose gain
constant is 𝐾1 .The output voltage of this amplifier is appliedto the armature circuit of the dc
motor.A fixed voltage is applied to thefield winding. If an error exists, the motor develops a
torque to rotate the output loadin such a way as to reduce the error to zero. For constant field
current, the torque developedby the motor is
𝑇 = 𝐾2 π‘–π‘Ž
where𝐾2 is the motor torque constant and π‘–π‘Ž is the armature current.When the armature is rotating,
a voltage proportional to the product of the flux andangular velocity is induced in the armature.
For a constant flux, the induced voltage 𝑒𝑏 is directly proportional to the angular velocity π‘‘πœƒ/𝑑𝑑,
or
𝑒𝑏 = 𝐾3
π‘‘πœƒ
𝑑𝑑
where𝑒𝑏 is the back emf, 𝐾3 is the back emf constant of the motor, and πœƒ is the
angulardisplacement of the motor shaft.
The transfer function between themotor shaft displacement and the error voltage can be
simplified and get the following:
𝐺(𝑠) =
𝐾
𝐽𝑆 2 + 𝐡𝑠
Where 𝐽is the moment of inertia referred to the output shaft.𝐡 = [𝑏0 + (𝐾2 𝐾3 /π‘…π‘Ž )]/𝑛2is the
viscous-friction coefficient referred to the output shaft.
𝐾 = 𝐾0 𝐾1 𝐾2 /π‘›π‘…π‘Ž .
Figure 2: block diagram for the servo system; (c) simplified block diagram.
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3. Components Required:
1 Power Supply +/- 15 V
72686
1 PID Controller
734064
1 Power Amplifier
73413
1 Servo Set point potentiometer 73410
1 DC-Servo
73414
1 CASSY– interface with Computer
4. Procedures:
a) Characterization of the Set point potentiometer

Connect the Set point potentiometer to the DC power supply 72686.

Adjust the set point integrator to ∞.

Adjust the setpoint value to 0° (360° ), and the output voltage to 0.0 v using the
“zero” adjustor.

Adjust the setpoint value to 90° (360° ), and the output voltage to 5.0 v using the
“scale” adjustor.

Start changing the set value clock-wise and counter clock-wise and record the output
voltages as in table 1.
Table 1: Angular displacement πœƒ ° of the set point potentiometer vs. the output voltage 𝑉1.
°
𝜽
π•πŸ

Counter Clock-wise
Clock-wise
0 10 20 30 60 90 120 150 0 10 20 30 60 90 120 150
0
-5
0
5
Draw the input output characteristic curve. Is it linear? Comment on your results.
b) Step response of servo system:

Set the experiment according to the connection diagram shown in Figure 3.

With the Switch S1 kept off, repeat the 2nd, 3rd and 4th steps in part a.

Adjust 𝐾𝑃 = 1 and switch off Ki and Kd.

Set the angular displacement of the set point potentiometer πœƒ ° = 90° .

Set Channel input B in CASSY to the output of system (i.e. o/p of DC servo block).

At the same time, pushthe switch S1 (ON) and start the CASSY, F9 key.
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c) P- Control:

[switch off Ki and Kd]
For 𝐾𝑃 = 1, 𝐾𝑃 = 5, 𝐾𝑃 = 10, change the set value of the potentiometer (i.e. system
input) as indicated in table 2 and record the output angle of the DC servo motor.
Table 2: Input and output values of servo system in different cases for P-controller.
𝑲𝑷 = 𝟏
°
πœƒπ‘–π‘›
°
πœƒπ‘œπ‘’π‘‘
Counter Clock-wise
0
50
90
0
°
πœƒπ‘–π‘›
°
πœƒπ‘œπ‘’π‘‘
Counter Clock-wise
0
50
90
0
150
Clock-wise
0
50
0
90
150
Clock-wise
0
50
0
90
150
Clock-wise
0
50
0
90
150
𝑲𝑷 = πŸ“
150
𝑲𝑷 = 𝟏𝟎
°
πœƒπ‘–π‘›
°
πœƒπ‘œπ‘’π‘‘

Counter Clock-wise
0
50
90
0
150
Plot the step response, fix the input angle to πœƒ ° = 90°, and change 𝐾𝑃 = 1, 𝐾𝑃 =
5, 𝐾𝑃 = 10.

Comment, analyze and discuss the effects of each response and choose the best one.

Evaluate the performance in terms of speed, steady state error, overshoot and
oscillation.
d) PI- Control:[Fix 𝑲𝑷 = 𝟏𝟎 and switch off Kd]

Repeat the first step in part (b) when π‘²π’Š = 𝟐 𝒂𝒏𝒅 π‘²π’Š = πŸ“ and complete table
𝟏
3.Note:π‘²π’Š = 𝑻
π’Š

Plot the above two responses and put them on the same graph.
Table 3: Input and output values of Servo system in different cases for PI-controller.
°
πœƒπ‘–π‘›
°
πœƒπ‘œπ‘’π‘‘
0
0
𝑲𝑷 = 𝟏𝟎 and π‘²π’Š = 𝟐
Counter Clock-wise
50
90
150
0
0
4
Clock-wise
50
90
150
°
πœƒπ‘–π‘›
°
πœƒπ‘œπ‘’π‘‘
0
0
𝑲𝑷 = 𝟏𝟎 and π‘²π’Š = πŸ“
Counter Clock-wise
50
90
150
0
0
Clock-wise
50
90
150
e) PID Control design:

Repeat the first step in part (b) when 𝑲𝑷 = 𝟏𝟎, π‘²π’Š = 𝟐 𝒂𝒏𝒅 𝑲𝑫 = 𝟎. πŸ“ and complete
table 4.
Table 4: Input and output values of Servo system for PID-controller cases.
°
πœƒπ‘–π‘›
°
πœƒπ‘œπ‘’π‘‘

0
0
𝑲𝑷 = 𝟏𝟎, π‘²π’Š = 𝟐 𝒂𝒏𝒅 𝑲𝑫 = 𝟎. πŸ“
Counter Clock-wise
Clock-wise
50
90
150
0
50
90
0
Plot the above response and put them on the same graph.
Switch
S1
Figure 3: Experiment connection diagram.
5
150
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