Control servo motors

1. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 1 | P a g e
[control Laboratory II]
University of Baghdad
Name: - Saif Al-din Ali -c-

2. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 2 | P a g e
TABLE OF CONTENTS
ABSTRACT......................................................I
INTRODUCTION............................................II
THEORY.......................................................III
APPARATUS..................................................V
Calculations and results…………..................VI
DISCUSSION...............................................VII

3. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 3 | P a g e
Experiment Name
Quick Return Mechanism
1. Abstract
Servo motors (or servos) are self-contained electric devices that rotate
or push parts of a machine with great precision. Servos are found in
many places: from toys to home electronics to cars and airplanes. If you
have a radio-controlled model car, airplane, or helicopter, you are using
at least a few servos. In a model car or aircraft, servos move levers back
and forth to control steering or adjust wing surfaces. By rotating a shaft
connected to the engine throttle, a servo regulates the speed of a fuel-
powered car or aircraft. Servos also appear behind the scenes in devices
we use every day. Electronic devices such as DVD and Blu-ray Disc
players use servos to extend or retract the disc trays. In 21st-century
automobiles, servos manage the car's speed: The gas pedal, similar to
the volume control on a radio, sends an electrical signal that tells the
car's computer how far down it is pressed. The car's computer calculates
that information and other data from other sensors and sends a signal to
the servo attached to the throttle to adjust the engine speed.
Commercial aircraft use servos and a related hydraulic technology to
push and pull just about everything in the plane.
2. Introduction
Principle of working :
Servo motor works on the PWM ( Pulse Width Modulation ) principle, which means its angle
of rotation is controlled by the duration of pulse applied to its control PIN. Basically servo
motor is made up of DC motor which is controlled by a variable resistor (potentiometer) and
some gears
Mechanism of servomotor :
Basically a servo motor is a closed-loop servomechanism that uses
position feedback to control its motion and final position. Moreover the
input to its control is a signal ( either analogue or digital ) representing
the position commanded for the output shaft .

4. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 4 | P a g e
The motor is incorporates some type of encoder to provide position and
speed feedback. In the simplest case, we measure only the position.
Then the measured position of the output is compared with the
command position, the external input to controller. Now If the output
position differs from that of the expected output, an error signal
generates. Which then causes the motor to rotate in either direction, as
per need to bring the output shaft to the appropriate position. As the
position approaches, the error signal reduces to zero. Finally the motor
stops.
The very simple servomotors can position only sensing via a
potentiometer and bang-bang control of their motor. Further the motor
always rotates at full speed. Though this type of servomotor doesn’t
have many uses in industrial motion control, however it forms the basis
of simple and cheap servo used for radio control models.
Servomotors also find uses in optical rotary encoders to measure the speed of output shaft
and a variable-speed drive to control the motor speed. Now this, when combined with a PID
control algorithm further allows the servomotor to be in its command position more quickly
and more precisely with less overshooting.
Working of servomotors :
Servo motors control position and speed very precisely. Now a
potentiometer can sense the mechanical position of the shaft. Hence it
couples with the motor shaft through gears. The current position of the
shaft is converted into electrical signal by potentiometer, and is
compared with the command input signal. In modern servo motors,
electronic encoders or sensors sense the position of the shaft .
We give command input according to the position of shaft . If the
feedback signal differs from the given input, an error signal alerts the
user. We amplify this error signal and apply as the input to the motor,
hence the motor rotates. And when the shaft reaches to the require
position , error signal become zero , and hence the motor stays standstill
holding the position.
The command input is in form of electrical pulses . As the actual input to
the motor is the difference between feedback signal ( current position )
and required signal, hence speed of the motor is proportional to the

5. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 5 | P a g e
difference between the current position and required position . The
amount of power require by the motor is proportional to the distance it
needs to travel .
Controlling of servomotors :
Usually a servomotor turns 90 degree in either direction hence
maximum movement can be 180 degree . However a normal servo
motor cannot rotate any further to a build in mechanical stop.
We take three wires are out of a servo : positive , ground and control
wire. A servo motor is control by sending a pulse width
modulated(PWM) signal through the control wire . A pulse is sent every
20 milliseconds. Width of the pulses determine the position of the shaft .
for example ,
A pulse of 1ms will move the shaft anticlockwise at -90 degree , a pulse
of 1.5ms will move the shaft at the neutral position that is 0 degree and
a pulse of 2ms will move shaft clockwise at +90 degree.
variable pulse width control
servo motor
When we command a servo motor to move by applying pulse of
appropriate width, the shaft moves to and holds the require position of
the shaft. However the motor resists to change . Pulses need repetition
for the motor to hold the position .

6. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 6 | P a g e
3. Theory

7. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 7 | P a g e
Theoretical Analysis: -
𝜔 𝛼 𝑉
ω = Kg Vg ; ϴ =Kg Vg
T(t) = K Ф Ia
T(t) = K Ia
Lb = Kb ω
L
𝑑𝐼𝑎
𝑑𝑡
+ 𝐼𝑎 𝑅 + 𝐿𝑏 = 𝑉𝑖 ---- (*)
𝐽
𝑑𝜔
𝑑𝑡
+ 𝐵 𝜔 = T(t) → 𝐽
𝑑𝜔
𝑑𝑡
+ 𝐵 𝜔 =K Ia → Ia=
1
𝐾
[ 𝐽
𝑑𝜔
𝑑𝑡
+ 𝐵 𝜔 ]
𝑑𝐼𝑎
𝑑𝑡
=
1
𝐾
[ 𝐽
𝑑2𝜔
𝑑𝑡2
+ 𝐵
𝑑𝜔
𝑑𝑡
]
𝐿
1
𝐾1
[ 𝐽
𝑑2𝜔
𝑑𝑡2
+ 𝐵
𝑑𝜔
𝑑𝑡
] +
1
𝐾
[ 𝐽
𝑑𝜔
𝑑𝑡
+ 𝐵 𝜔] R + eb = Vi
For small L ;-
𝑑𝜔
𝑑𝑡
(
𝑅𝐽
𝐾1
) + 𝜔 (
𝐵𝑅
𝐾1
+ 𝐾𝑏 ) = 𝑉𝑖 * K1/R and *
1
𝐵+
𝐾𝑏 𝐾1
𝑅
(
𝐽
𝐵+
𝐾𝑏𝐾1
𝑅
)
𝑑𝜔
𝑑𝑡
+ ω = (
𝐾1
𝐵𝑅+𝐾𝑏𝐾1
) 𝑉𝑖
Let τ = (
𝐽
𝐵+
𝐾𝑏𝐾1
𝑅
) ; Ks =(
𝐾1
𝐵𝑅+𝐾𝑏𝐾1
)
Thus ; τ
𝑑𝜔
𝑑𝑡
+ ω = 𝐾𝑠 𝑉𝑖
ω( τ𝐷 + 1 )= 𝐾𝑠 𝑉𝑖
𝝎
𝑽𝒊
=
𝑲𝒔
𝝉𝑫+𝟏
where τ : time constant ; Ks: gain
For steady state:
𝑑𝜔
𝑑𝑡
= 0 → ω = 𝑲𝒔 𝑽𝒊

8. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 8 | P a g e
4. APPARATUS

9. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 9 | P a g e
5. Calculations and results
NO Vout Vin time (sec)
1 0 0 0
2 4.4 3.5 14.75
3 5 4 6.45
4 5.9 4.5 4.88
5 6.6 5 3.56
6 7.2 5.5 2.12
ω =
𝟓
𝒕𝒊𝒎𝒆
× 𝟑𝟎
• ω =
5
𝟏𝟒.𝟕𝟓
× 30 → ω =10.17rad/s
• ω =
5
𝟔.𝟒𝟓
× 30 → ω = 23.26rad/s
• ω =
5
𝟒.𝟖𝟖
× 30 → ω = 30.74rad/s
• ω =
5
𝟑.𝟓𝟔
× 30 → ω = 42.13rad/s
• ω =
5
𝟐.𝟏𝟐
× 30 → ω = 70.75rad/s
NO Vout Vin time (sec) w rad/s
1 0 0 0 0
2 4.4 3.5 14.75 10.17
3 5 4 6.45 23.26
4 5.9 4.5 4.88 30.74
5 6.6 5 3.56 42.13
6 7.2 5.5 2.12 70.75

10. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 10 | P a g e
K =
𝜔
𝑉𝑖
= slope =8.1115
K =
𝜔
𝑉𝑖
= slope =10.467

11. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 11 | P a g e
Discussion
Q1 / Find ks, an estimable of the DC motor steady state gain as follow: -
-Take the mean of the values of Ks -Find the standard deviation of the estimate, AKs
• ω =
5
𝟏𝟒.𝟕𝟓
× 30 → ω =10.17rad/s
KS motor =w/vi = 10.1695/3.5 =2.90557
KS generator = w/Vo = 10.1695/4.4 = 2.3112
KS motor = 2.9055+5.8139+6.8306+8.4269+12.34 / 5 =7.36356 rpm/v
KS generator= 2.31125+4.65116+5.20977+6.38406+9.42684 / 5 5.596616 rpm/v
Q2/ determine KS by taking the mean of value of KS
KS motor = 2.9055+5.8139+6.8306+8.4269+12.34 / 5 =7.36356 rpm/v
KS generator= 2.31125+4.65116+5.20977+6.38406+9.42684 / 5 5.596616 rpm/v
Q3/ discuss the relationship obtained from the plot
Note the proportionality directly between v and w
As the voltages increase, the number of turns increases
Q4/ develop a block diagram for dc motor modeling
Q5/ comment on whether or not the system behaver linearly as the model
predicts
Note from the drawing a non-linear relationship subject to the influence of the load and
operating conditions