1. REPORT
SELF BALANCING ROBOT
May 3, 2016
Teru Komal Kumar, 12EE35018
Guide : Prof. Siddhartha Sen
Indian Institute of Technology, Kharagpur
Department of Electrical Engineering
1
3. Electrical Engineering Indian Institute of Technology, Kharagpur
1 INTRODUCTION
In this report, we discuss the process of building the self balancing robot from the scratch.
This robot uses the model of inverted pendulum as its reference model. The cart in this case
is made up of two wheeled differential drive. The robot that is built is inherently unstable
and without the action of an external controller, it would fall down. To stabilise it, the motors
need to driven in the appropriate direction.
The report is majorly divided into three sections. In the first section, we look at the system
overview i.e., different modules involved and their functioning. In the second section, we
worked on finding the parameters related to motor dynamics and the robot dynamics that
would help us in getting the state space equations. In the next section, we have discussed
about two control methods that would help us in stabilising the system namely the PID
control and the state space control.
Figure 1: Model of the robot developed for this project
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2 NOMENCLATURE OF SYMBOLS USED
τm Motor Torque
τa Applied Torque
Va Applied Voltage
Ve Back emf
R Armature resistance
IR Rotor Inertia
Km Torque Constant (N.m/A)
Ke Back emf Constant (V.s/rad)
i Armature current
Mw Mass of the wheel
Mp Mass of the chasis
Ip Robot Chasis Inertia
CL,CR Torque exerted by motors on the wheels
Hf Frictional force between the wheel and the ground
HL,HR,PL,PR Reaction forces between wheels and chasis
Kp Proportional Gain
Kd Differential Gain
Ki Integral Gain
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3 SYSTEM OVERVIEW
3.0.0.1 Mechanical system The basic structure of the robot is built from wood. The
wooden structure has two plates where the components like the motors, sensors and com-
munication module can be mounted.These plates can be moved up and down which would
help us in manipulating the centre of gravity of the bot. Two wheels of radii of about 5 cm are
used for the ’cart’ part of the system.
3.0.0.2 Electronic System The system contains 4 different modules which include Proces-
sor module, Sensor module, Actuator module and Communication module.
Figure 2: System Overview
3.1 Processor Module
Owing to our experience in working with Atmega processor as part of our coursework, we
have chosen Arduino Mega 2560 microcontroller as the processor for the robot. Some of its
desirable qualities include
1. It has 4 serial ports and hence can be used to communicate with multiple devices
simultaneously.
2. It has a onboard TTL-to-USB serial chip that allows for ease in communication with
the PC.
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Figure 3: Arduino MEGA 2560
3. It has 15 PWM pins which are helpful in controlling the actuators.
4. It also has a onboard FTDI chip that smoothens the process of burning the code on to
the processor.
Microarchitecture AVR
Clock frequency 16 MHz
Flash Memory 256 KB (8 KB bootloader)
SRAM 8 KB
Microprocessor ATmega2560
Word Size 8 bit
Operating Voltage 5V
EEPROM 4 KB
3.2 Sensor Module
In the system, we need two sensing elements. One is for determining the orientation of the
robot and another for sensing the angular position of the motor which effectively gives us the
linear displacement.
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3.2.1 Inertial Measurement Unit
For the orientation feedback of the robot, a 9 DOF (Degrees of Freedom) IMU was used. This
has 9 DOF through the combination of a Gyroscope, an Accelerometer and a Magnetometer.
The outputs of all sensors are processed by an on-board ATmega328 and transmitted over a
serial interface. It operates in different modes configured by the firmware uploaded in the
on-board Atmega 128. When placed over the bot, the IMU gives us the yaw, pitch and roll
(called as YPR from here on) of the system. The sensor encodes the values of YPR as double
precision floating point format and sends it through the serial output. The Rx and Tx pins are
connected to one of the serial ports of Arduino Mega and thus the data is transmitted.
3.2.1.1 Testing The IMU was tested and calibrated. For testing purpose, the IMU was
connected to the PC through a FTDI USB2Serial converter chip. A graphical interface was de-
veloped using software called Processing to visualize the results. Shown below is a snapshot
of the said visualization.
From the above tests, the geometric conventions of the IMU were found to be as below:
• X axis pointing forward (towards the short edge with the connector holes)
• Y axis pointing to the right
• Z axis pointing down.
• Positive yaw : clockwise
• Positive roll : right wing down
• Positive pitch : nose up
3.2.1.2 Calibration Process The precision and responsiveness of IMU can be improved a
lot by calibrating the sensors. If not calibrated there might be effects like
• Drifts in yaw when you apply roll to the board.
• Pointing up does not really result in an up altitude.
All three component sensors namely, gyroscope, accelerometer and magnetometer were
calibrated individually. The accelerometer was calibrated by moving the sensor slowly about
all the three axes and noting the extreme values of the sensor readings along them (this
accounts for the acceleration due to gravity). Gyroscope was calibrated by holding the
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Figure 4: IMU Testing in Processing software
sensor still for about 10 seconds. Magnetometer was calibrated by noting the maximum and
minimum values of the earth s magnetic field along each axis. The values obtained above,
referred as offset values, were incorporated in the firmware of the on-board processor of the
IMU.
3.2.2 Incremental Quadrature encoder
A 900 ppr incremental optical encoder was used for DC motor position feedback. It gives
two outputs (Channel A and Channel B) with phase difference depending on the direction of
the rotation. Direction of rotation can be identified by checking if a channel’s waveform is
leading or lagging the other. The Figure 5 demonstrates the waveforms.
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Figure 5: Quadrature signals of motor encoders
3.2.2.1 Testing The motor position was decoded from the quadrature signals using inter-
rupts triggered at rising edge of the two signals. The interrupt service routine was written
in assembly language for optimized execution. After successful position feedback from the
motors, P controller was implemented for position control of the motor with voltage as the
control input. The voltage was varied using pulse width modulation (PWM) at a frequency of
490 Hz. The result was imported in MATLAB and plotted as shown below.
Figure 6: Response of P controller on the motor position
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3.3 Actuator Module
The actuator module consists of two brushed DC motors and a motor driver IC.
3.3.0.2 DC Motor Two 12V, 170 RPM gear (1:30) motors with position encoders (with
specification as mentioned in the previous section) were used to drive the system. Each
motor has a stall torque of 15 kg/cm @ 12 V and 7.5 kg/cm @ 6 V. The stall current required is
1.8A @ 12V and 0.8 A @ 6 V. The motors are specified to run smoothly from 4V to 12 V.
Note: The above specifications are according to the datasheet provided by the vendor.
3.3.0.3 Motor Driver Motor driver board based on L298 was used. It can provide current
up to 2A. It can operate with PWM frequency as high as 20 kHz.
3.4 Communication Module
All the on-system communications are done through various Rx and Tx pins connected
across different modules. Wireless communication is used to control the motion of the robot
remotely and to track the motion of the robot in real time on our Personal Computer. For
this purpose, we have chosen to use Zigbee wireless protocol. It is mainly useful for short-
range low-rate wireless transfer. Its low power consumption limits transmission distances
to 10 to 100 meters. ZigBee has a defined rate of 250 kbit/s, best suited for intermittent data
transmissions from a sensor or input device.
Using this, the real time sensor data and the robot position could be sent to our Personal
computer and this data could be processed by some application like Labview or MATLAB
and plot the robot s movement. For this, we need a pair of Zigbee modules. One is connected
on board to the Arduino and the other is connected to our PC and they both are coupled.
Through this module, we could also control the movement our robot by giving the appropriate
commands through the keyboard. Thus the wireless module serves a dual purpose.
3.4.0.4 Testing Each module was connected to two different PCs and they were configured
with appropriate source and destination addresses using a software called XCTU. One of the
modules was connected to the Arduino communicating through the USART protocol. This
module then communicates with the other Zigbee connected to the PC.
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4 POWER CIRCUIT DESIGN
The current and power ratings of the various components of the system have been tabulated
below.
Components Voltage Ratings (volts) current rating (mA)
Arduino Mega 5 200
IMU Razor 3.3 10
Encoder 5 20
DC Motor 12 2*1800
Motor Driver 5 110
Zigbee 5 50
Total Current Requirement 3990
Total Power Requirement 45.133 W
As can be seen from the above table, the current requirements and the power requirements
are around 4 Amps and 45 Watts. A Li Po battery rated at 18.5 V, 2000mAH, 20C is used as
the power source for the entire system. Drawing the required current of 4 A will drain the
supply in about half an hour which is acceptable for our application. We use 7812 IC (voltage
regulators) to step down the 18.5 voltage to the required 12V. This reduces the efficiency of the
system but increases the stability in the supply voltage. Appropriate power distribution circuit
was developed with LEDs, diodes, decoupling capacitors and ON/OFF switches. Shown below
is the schematic of the same.
Figure 7: Power circuit schematic in Proteus
All the modules connect to the central processor, in this case, the Arduino MEGA. A shield
specific for this application has been developed for an easy-to-plug system. Shown below is
the labelled picture of the completed shield.
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5 DYNAMIC MODELLING
To develop a robust control system, we need to know the dynamics of the system. So, now we
derive the linear model of the DC motor and the equations of motion for the robot which
would help us in writing the state space equations of the system.
5.1 DC Motor Model
We are using two DC motors to drive the robot. The input to these motors is the voltage
applied which we are controlling in our system by varying the PWM output from the Arduino.
From this model, we get the relation between voltage applied and the controlling torque
required to balance the system.
In the DC motor, the armature current is developed as we connect a voltage source across
it. The torque produced τm is directly proportional to the this current i.
τm = kmi (1)
Also, as the motor rotates, a back emf Vem f is generated. This can be approximated to be
directly proportional to shaft velocity ω
Ve = keω (2)
We can ignore motor inductance for the purpose of modelling. Now, applying Kirchoff’s law
to the motor circuit, we have
Va −Ri −Ve = 0 (3)
The sum of all torques produced on the shaft is linearly related to the moment of inertia of
the rotor Ir i.e.,
τm −τa = IR
dω
dt
(4)
Substituting the value of Ve from Eq 2 in Eq 3 and rearranging the terms, we get
i =
−Keω
R
+
Va
R
(5)
Substituting the value of τm from Eq 1 in Eq 4 and rearranging the terms, we get
dω
dt
=
Kmi
IR
+
−τa
IR
(6)
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Substituting the value of i from Eq 5 in Eq 6 and rearranging the terms, we get
dω
dt
=
−KmKeω
IRR
+
KmVa
IRR
+
−τa
IR
(7)
The motor dynamics can be represented with a state space model, it is a system of first
order differential equations with parameters position θ, and velocity ω, that represents its
operation. The inputs to the motor is the applied voltage and applied torque.
˙θ
˙ω
=
0 1
0 −KmKe
IR R
θ
ω
+
0 0
Km
IR R
−1
IR
Va
τa
(8)
y = 1 0
θ
ω
+ 0 0
Va
τa
(9)
5.2 Model of the robot
We are equating our robot to a inverted pendulum. The two wheels act as the cart and the
robot chassis acts as the inverted pendulum. We first analyse the wheel dynamics and then
analyse the body dynamics which will eventually get us the state equations that govern the
motion of the system.
5.2.1 Equations for the wheels
We are writing the equations of motion with respect to two wheels. The different forces
acting on one of the wheels could be shown in the below figure. Since, both the wheels are
analogous, we could write for one of them and get the other easily.
From the Figure 8, the forces acting on the right wheel could be seen. Equating the forces
in the horizontal x direction, we get
Mw ¨x = Hf − HR (10)
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Figure 8: Free body diagram of the right wheel
Sum of moments around the centre of wheels is given by
Iw ¨ωw = CR − Hf r (11)
We know CR is nothing but the τm which we derived in the previous section Eq 4.By substitut-
ing Eq 7 in Eq 4 and rearranging the terms, we have
CR =
−KmKeω
R
+
KmVa
R
(12)
Substituting the value of CR in Eq 11, we get
Iw ¨ωw =
−KmKeω
R
+
KmVa
R
− Hf (13)
Rearranging the above equation to get Hf as the subject
Hf =
−KmKeω
Rr
+
KmVa
Rr
−
−Iw ¨ωw
r
(14)
Substituting this value of Hf in Eq 10, we get
Mw ¨x =
−KmKeω
Rr
+
KmVa
Rr
−
−Iw ¨ωw
r
− HR (15)
We can similarly find this equation for the left wheel which is
Mw ¨x =
−KmKeω
Rr
+
KmVa
Rr
−
−Iw ¨ωw
r
− HL (16)
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Since the centre of the wheel is moving in a straight line, we can approximate the angular
rotation into linear motion by simple transformations,
r ¨θ = ¨x (17)
r ˙θ = ˙x (18)
Using this transformations and rewriting the equations 15 and 16, we get the following
equations for the left and the right wheel respectively
Mw ¨x =
−KmKe x
Rr2
+
KmVa
Rr
−
−Iw ¨x
r2
− HL (19)
Mw ¨x =
−KmKe x
Rr2
+
KmVa
Rr
−
−Iw ¨x
r2
− HR (20)
Adding the above two equations and rearranging the terms, we get
2 Mw +
Iw
r2
¨x =
−2KmKe x
Rr2
+
2KmVa
Rr
−(HL + HR) (21)
5.2.2 Equations for the chassis
Figure 9: Free body diagram of the chasis
We are approximating the chassis to be the inverted pendulum in the above figure which
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shows us all the forces acting on it
By summing all the forces in the horizontal direction, we get
(HL + HR) = Mp ¨x + Mpl ¨θcos(θ)− Mpl ˙θ2 sin(θ) (22)
Now, summing the forces in the direction perpendicular to the pendulum, we have
(HL + HR)cos(θ)+(PL +PR)sin(θ)− Mp g sin(θ)− Mpl ¨θ = Mp ¨x cos(θ) (23)
Now, summing the moments around the centre of mass of pendulum, we get
−(HL + HR)I cos(θ)−(PL +PR)l sin(θ)−(CL +CR) = I ¨θ (24)
From Eq 12, we have
(CL +CR) = 2
−KmKeω
R
+2
KmVa
R
(25)
Substituting the value of (CL +CR) in Eq 24, we get
−(HL + HR)l cos(θ)−(PL +PR)l sin(θ) = I ¨θ +2
−KmKeω
R
+2
KmVa
R
(26)
Multiplying the Eq 23 by -l and equating its RHS with the RHS of Eq 26, we get
−Mp gl sin(θ)− Mpl2 ¨θ − Mpl ¨x cos(θ) = I ¨θ +2
−KmKeω
R
+2
KmVa
R
(27)
Substituting the value of (HL + HR) from Eq 22 in Eq 21 gives
2 Mw +
Iw
r2
¨x =
−2KmKe x
Rr2
+
2KmVa
Rr
− Mp ¨x − Mpl ¨θcos(θ)+ Mpl ˙θ2 sin(θ) (28)
The above 2 equations can be linearised by assuming θ = π + φ where φ is the small angular
deviation from the vertical axis. With this approximation, we have cos(θ) = -1, sin(θ) = −φ
and ¨θ = 0
Using the above approximations in the equations 27 and 28 and rearranging the terms,
we have
¨φ =
Mpl ¨x
Ip + Mpl2
+
−2KmKe
˙(x)
Rr Ip + Mpl2
+
−2KmVa
Rr Ip + Mpl2
+
Mp glφ
Ip + Mpl2
(29)
¨x =
2KmVa
Rr 2Mw + 2Iw
r2 + Mp
−
2KmKe ˙x
Rr2 2Mw + 2Iw
r2 + Mp
+
Mpl ¨φ
2Mw + 2Iw
r2 + Mp
(30)
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After substituting the above two equations in each other and some manipulation of terms,
we get the state space equations given by
˙x
¨x
˙φ
¨φ
=
0 1 0 0
0
2KmKe (Mp lr−Ip −Mp l2
)
Rr2α
M2
p gl2
α 0
0 0 0 1
0
2KmKe (rβ−Mp l)
Rr2α
M2
p glβ
α
0
x
˙x
φ
˙φ
+
0
2Km(Ip +Mp l2
−Mp lr)
Rrα
0
2Km(Mp l−rβ)
Rrα
Va
(31)
where α and β are given by
α = Ipβ+2Mpl2
Mw +
Iw
r2
(32)
β = 2Mw +
2Iw
r2
+ Mp (33)
x
φ
=
1 0 0 0
0 0 1 0
x
˙x
φ
˙φ
(34)
Thus we found the matrices A, B, C for the state space model of our system based on the
system parameters. We can find the transfer function for the input and output using the
equation below and the resulting transfer function is shown in the next section.
Y (s) = C (sI − A)−1
BU (s) (35)
5.3 Approximate values of system parameters
Here, we are approximately finding the values of system parameters to get a rough idea by
simulating the resulting transfer functions.
1. The mass of the wheel has been given in the buyer’s manual and the weight of the robot
is approximated to 4 kg.
2. Radius of the wheel and the length of the bot have been measured with the help of a
ruler
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3. The wheel is assumed to be a solid cylinder and its moment o inertia is thus found from
its mass and radius.
4. The moment of inertia of the robot is measured by approximating it to a rod with the
axis of rotation being perpendicular to it through one of the ends which gives us Ml2
3
5. The armature resistance is assumed to be 1 ohm.
6. The Km and Ke values are approximated to be 0.01 N.m/Amp and 0.01 V/rad/sec
respectively
Mp 4 kg
Ip 0.12 kg.m2
Mw 0.123 kg
Iw 0.00173 kg.m2
r 0.05 m
l 0.3 m
R 1 ohm
Km 0.01 N.m/Amp
Ke 0.01 V/rad/sec
6 CONTROL SYSTEM DESIGN
In this section we describe the detailed analysis of the dynamic model of the system formu-
lated in the previous section. An appropriate control system for the system is suggested by
various techniques. Hardware implementation of the same has also been commented on by
the end of the section.
The transfer function of the pendulum’s angular position, φ, to the input voltage, Va, can
be obtained from Eq 35. Solving it we get the transfer function as follows.
T (s) =
φ(S)
V (s)
=
αzS +γz
s3 −αp s2 −βp s +γp
(36)
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where,
αz =
2Km(Mpl)−2r Mw −2Iw
r − Mpr
Rr(Ipβ+2Mpl2(Mw + Iw
r2 ))
γz =
−4K 2
mKe(Ip + Mpl2
− Mplr)2
+4K 2
mKe(Mplr − Ip − Mpl2
)(Mpl −rβ)
R2r3(β)
αp =
2KmKe(Mplr − Ip − Mpl2
)
Rr2(Ipβ+2Mpl2(Mw + Iw
r2 )
βp =
Mp gl(Mp −β)
β
γp =
2KmKe Mp gl(Mplrβ− Ipβ− Mpl2
β+ Mpl2
−rlβ)
Rr2(β)2
β = 2Mw +
2Iw
r2
+ Mp
The open loop impulse response and step response of the system are shown in Figure 10
and Figure 11 respectively.
Figure 10: Open loop impulse response
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Figure 11: Open loop step response
6.1 PID Control
The control system problem is to find an appropriate compensator which places the closed-
loop poles at desired locations to achieve required transient response. Figure 12 shows the
structure of the compensated system with compensator transfer function given by C(s).
Figure 12: System Structure
The generic model of the PID compensator is given by Eq 37.
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C(s) =
Kd s2
+Kp s +Ki
s
(37)
Tuning the three gains, Kd, Kp and Ki, we can place the closed-loop poles at the desired
locations. Further analysis in terms of root-locus of the system can yield a detailed insight
into the system behaviour.
The root-locus of the plant without any compensator, plotted in MATLAB (with sample
system parameters for the purpose of qualitative analysis), is shown in Figure 13. The poles
of T (s) are located at 5.5651, -5.6041 and -0.1428. One zero is present at origin. The same can
be seen in the root-locus plot.
Figure 13: Root locus of the plant with varying gain
As can be seen, one branch (coloured blue) is always on the right side of the imaginary
axis. This establishes that for any value of the gain K chosen, the system can not be stabilized
without a compensator. To solve this, we add a pole (integrator) at origin to cancel the zero at
that location. This generates two poles branching out to the right-side of the s-plane as shown
in Figure 14. From the properties of root-locus we know that we can pull the poles to the
left-half plane by addition of zeroes. We observe that addition of one zero does not pull the
poles enough to the left-side to meet acceptable transient behaviour. Thus we add two zeroes
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at −3 and −4. The resulting root-locus is shown in Figure 15. Choosing the gain at appropriate
closed loop pole-zero location on the root-locus will result in a fully compensated system
meeting the transient response requirements. The system impulse response of pendulum
angular position with gain at 20 is shown in Figure 16. The poles for the final compensated
closed loop transfer function are as follows.
p1 = −84.0439+0.0000i
p2 = 0.0000+0.0000i
p3 = −3.5235+0.7157i
p4 = −3.5235−0.7157i
All poles are located on the left side of the imaginary axis indicating the stability of the
system. The response of the cart’s position is shown in Figure 17.
Figure 14: Root locus of the plant with integrator
This method allows for control of the pendulum position but provides no means for the
control of the cart position, as evidenced by Figure 17. The state-space method described in
the next section caters to this need.
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Figure 15: Root locus of the plant with PID compensator
Note that in the current hardware implementation, these gains were tuned using the
Ziegler-Nichols PID tuning algorithm. This is because of the unavailability of the value of
various system parameters.
6.2 State Space Control
The state-space control technique can be applied to Multi-Input, Multi-Output (MIMO) sys-
tems. Our system, when considering cart position control, is a SIMO (Single-Input, Multiple-
Output) system.
The system dynamics can be represented by
˙X = AX +BU
Y = C X
where matrices A, B and C are given Eq 31 and Eq 34. X is the state vector with four states
as cart position: x, cart velocity: ˙x, pendulum’s angular position from the vertical: φ and
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Figure 16: Root locus of the plant with integrator
pendulum’s angular velocity: ˙φ. The outputs of the system are cart position: x and pendulum
angular position from the vertical: φ
The state-space control uses full state feedback. The schematic is shown in Figure 18.
The state-space equations of the above system configuration is given as follows.
˙X = AX +BU = AX +B(−K X +r) = (A −BK )X +Br
˙X = (A −BK )X +Br
Y = C X (38)
where K = [k1k2 ···k3]
The feedback gain vector K can be easily found by representing the system in Phase-
Variable form. The characteristic equation of the system is given by
sn
+ an−1sn−1
+···+ a1s + a0 = 0
The state-space representation of the open-loop system in this form is given by Eq 39.
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Figure 17: Root locus of the plant with integrator
Figure 18: State feedback control loop structure
A =
0 1 0 ... 0
0 0 0 ... 0
. . . . .
. . . . .
. . . . .
−a0 −a1 −a2 ... −an−1
B =
0
0
.
.
.
1
C = c1 c2 c3 ... cn (39)
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The closed loop form is given by Eq 40.
A =
0 1 0 ... 0
0 0 0 ... 0
. . . . .
. . . . .
. . . . .
−(a0 +k1) −(a1 +k2) −(a2 +k3) ... −(an−1 +kn)
(40)
The characteristic equation of the closed loop system is given by Eq 41.
sn
+(an−1 +kn)sn−1
+(an−2 +kn−1)sn−2
+···+(a1 +k2)s +(a0 +k1) = 0 (41)
Let the desired characteristic equation be as follows,
sn
+dn−1sn−1
+dn−2sn−2
+···+d1s +d0 = 0
In the above two equations, by equating the coefficients to the desired coefficients we get
the values of Ki .
Ki+1 = di − ai
After placing the poles at the following locations, the system step response is shown in
Figure 19. The input is the step input for target cart position.
p1 = −8.4910+7.9283i
p2 = −8.4910−7.9283i
p3 = −4.7592+0.8309i
p4 = −4.7592−0.8309i
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Figure 19: Response of Cart’s position and Pendulum’s angular position under state space control
7 MATLAB CODE
The MATLAB code used for obtaining the above simulation results is shown from here on for
the reference of the reader.
clear all;
%%%%%%% System Parameter Values %%%%%%%%
Mp = 4;
Ip = 0.12;
Mw = 0.123;
Iw = 0.00173;
r = 0.05;
l = 0.3;
R = 1;
Km = 0.01;
Ke = 0.01;
g = 9.8;
b = 2*Mw + 2*Iw/(r^2) + Mp;
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29. Electrical Engineering Indian Institute of Technology, Kharagpur
% legend('x', 'phi')
%
% step_info = lsiminfo(y, t);
% cart_info = step_info(1)
% pend_info = step_info(2)
%%%%%%%%% PID Control %%%%%%%%
% Kp = 100;
% Ki = 1;
% Kd = 0;
%
% C = pid(Kp, Ki, Kd);
% T = feedback(P_pend, C);
%
% T2 = feedback(1,P_pend*C)*P_cart;
%
% t=0:0.01:10;
% impulse(T,t)
% title('Response of Pendulum Position to an Impulse Disturbance under PID Control: Kp =
% t = 0:0.01:5;
% impulse(T2, t);
% title('Response of Cart Position to an Impulse Disturbance under PID Control: Kp = 100
%%%%%%% Root locus analysis %%%%%%%
% rlocus(P_pend)
% title('Root locus of the plant with vaying gain')
% C_int = 1/s;
% rlocus(C_int * P_pend)
% title('Root locus of the plant with integrator')
%
% % zeros = zero(C_int * P_pend)
% % poles = pole(C_int * P_pend)
%
% z = [−3 −4];
% p = 0;
% k = 1;
% C_pid = zpk(z, p, k);
% rlocus(C_pid*P_pend)
% title('Root locus of the plant with PID compensator')
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30. Electrical Engineering Indian Institute of Technology, Kharagpur
% % %
% [k, target_pole] = rlocfind(C_pid*P_pend)
%
% k = 35;
% T = feedback(P_pend, k*C_pid);
% figure(1)
% impulse(T)
% title('Impulse response of the plant with PID compensator')
%
% T3 = feedback(1,P_pend*C_pid)*P_cart;
% figure(2)
% impulse(T3)
% title('Impulse response of the cart position with PID compensator')
% axis([0 10 −200 200])
%%%%%%% State−space control %%%%%%%
% poles = eig(A)
P = [−8.4910 + 7.9283i −8.4910 − 7.9283i −4.7592 + 0.8309i −4.7592 − 0.8309i];
K = place(A, B, P)
Ac = [(A − B*K)];
Bc = [B];
Cc = [C];
Dc = [D];
p = eig(Ac)
states = {'x' 'x_dot' 'phi' 'phi_dot'};
inputs = {'r'};
outputs = {'x' 'phi'};
sys_cl = ss(Ac, Bc, Cc, Dc, 'statename', states, 'inputName', inputs, 'outputName', outp
t = 0:0.01:5;
r = 0.2*ones(size(t));
[y, t, x] = lsim(sys_cl, r, t);
[AX, H1, H2] = plotyy(t, y(:,1), t, y(:, 2), 'plot');
set(get(AX(1), 'Ylabel'), 'String', 'Cart Position (m)');
set(get(AX(2), 'Ylabel'), 'String', 'Pendulum angle (radians)');
title('Step Response with State Space Control')
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31. Electrical Engineering Indian Institute of Technology, Kharagpur
REFERENCES
1. The Control of an Inverted Pendulum,[manual] Available at:<https://engineering.
purdue.edu/AAE/Academics/Courses/aae364L/2011/Fall_2011/lab3/manual>
2. Self-balancing two-wheeled robot,[report] Available at:<http://sebastiannilsson.
com/wp-content/uploads/2013/05/Self-balancing-two-wheeled-robot-report.
pdf>
3. Control Systems Engineering - Norman S. Nise
4. Exploring XBees and XCTU,[manual] Available at:<https://learn.sparkfun.com/
tutorials/exploring-xbees-and-xctu>
5. 9 DOF razor IMU calibration manual,[manual] Available at:<https://github.com/
ptrbrtz/razor-9dof-ahrs/wiki/Tutorial>
6. Arduino tuitorials on www.arduino.cc
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