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Overview and Control of DC and AC Motors
Brian Bouma

Abstract— an overview of the differences in form and function
between DC and AC motors. An in-depth analysis of the general
differential equations and transfer functions of DC and AC
motors. Strategies for controlling DC and AC motors using gain
and PID control are discussed and analyzed in depth. Methods
for determining the time constants of DC and AC motors are
discussed and analyzed.
Index Terms – AC motors, Control systems, DC motors,
Motor drives.
I. NOMENCLATURE
PID
Proportional Integral Derivative
II. INTRODUCTION
Motors are an integral part of engineering in today’s society.
They are used in a wide variety of applications, from running
fans to driving belts to turning wheels. Yet, despite their
prevalence in the designs of undergraduate engineering
students, most such students have very little idea of how
motors actually work, or of how to control them safely and
dependably. This paper describes both DC and AC motors,
analyzes them from a control standpoint, and determines
adequate strategies for controlling them in a manner which is
both safe and reliable. Stepper and servo motors will not be
discussed here, as their form, function, and application are
considerably different from that of DC and AC motors, and the
analysis of those four motor types would be too much
information to cover in this setting.
III. MOTOR OVERVIEW
To the uninformed observer, DC and AC motors appear to
be basically identical. Even though they seem to operate in
essentially the same way, their physical structures, and thus
their range of applications, vary significantly.
The brush DC motor is arguably the simplest variable-speed
DC motor design, in addition to being the most common. For
these reasons, the brush design is the one being described and
analyzed here. The brush DC motor (or all DC motors, for
that matter) is made up of a stator and a rotor (refer to Fig. 1
for all descriptions relating to the brush DC motor). As the
names suggest, the rotor (the circular portion of Fig. 1 made
This work was done for Engineering 315 at Calvin College in Grand
Rapids, Michigan, in the fall of 2004. All software used in this paper was
supplied by Calvin College. This project was supported financially by Calvin
College and Smiths Aerospace LLC.
Brian Bouma works for Smiths Aerospace LLC and attends Calvin
College in Grand Rapids, MI 49546 USA (e-mail: [email protected]).
up of eight “T”-shaped parts) is the part of the motor that
rotates during operation, while the stator (the dark blue block
and light blue “fingers” around the rotor in Fig. 1) remains
stationary during operation, relative to the motor’s casing and
mounting [3]. The stator is made up of either a winding or a
magnet, which creates magnetic flux in the magnetic field
formed between the stator and the rotor [3]. For the simple
analytical purposes herein, it makes no difference whether a
winding or a magnet is used in the stator, so the use of a
magnet will be assumed. The rotor has a winding on its
surface, termed the armature, in which electromotive forces
are induced by the magnetic field formed between the stator
and the rotor [3]. The armature winding is supplied current
through the collector (the yellow cylinder attached to the rotor
in Fig. 1), on which the brushes (the two brown tabs touching
the collector) apply pressure [3]. The collector is mounted on
the same shaft as the armature, and the fixed brushes are
connected to the armature terminals [3]. Thus, the power to
the motor runs through the brushes, into the collector, and
through the armature winding to produce the electromotive
forces between the stator and the rotor (the power cables are
the red lines attached to the brushes in Fig. 1). The brushcollector assembly provides current to the armature windings
in such a way that the current flows in one direction when the
windings are under a magnetic North pole (from the stator
magnet), and in the other direction when the windings are
under a magnetic South pole [3]. The rotor windings are made
up of coils, called sections, all sections being of an equal
number of turns (in Fig. 1, the eight colored sets of line
segments, two segments per set, on the “front” of the T-shaped
parts of the rotor are the ends of the coils; the coils run through
the length of the rotor and wrap around the opposite end of the
T-shaped parts of the rotor) [3]. Each section has two sides,
which are inserted into two slots spaced apart a distance equal
to the distance between the two field poles [3]. This way,
when the conductors of one side of a section are under the
North pole, the conductors of the other side of the same
section are under the South pole [3]. The sections of armature
winding are all connected together, in series, with the end of
the last section being connected to the beginning of the first
section, so that the winding as a whole is continuous, having
no particular start or finish [3]. For this to work, each slot
must contain two sides of sections (half each of two different
sections) [3]. As the rotor rotates, when a section changes
from being under the North pole to being under the South pole
(and hence the current in that section reverses direction), that
section commutates. Commutation of a section of the winding
is the changing of the section from being under one pole to
2
being under the other pole [3]. Two sections (two sections
that are opposite each other on the rotor) commutate at a time,
one switching from North to South pole, the other switching
from South to North pole. Because the two poles “swap”
sections simultaneously, half of the windings are under each
pole at all times. The North and South poles are an effect of
the flux in the armature created by the current flowing through
the two sets of windings (each pole containing one set). When
a section commutates, the brushes, applying pressure on the
collector, short-circuit the two ends of that section together, to
release the energy stored in the coils of the section before the
direction of current flow in the section is reversed [3]. Despite
this ingenious design, sparks are still produced between the
brushes and the collector [3].
Fig. 2. AC Motor Stator and Rotor
The inner face of the stator is made up of deep slots (or
grooves, depending on the exact design, but that distinction
makes no difference for this analysis), through which the
windings run (the red arrow in Fig. 3 indicates one such slot)
[3].
Fig. 1. Brush DC Motor
The primary advantage of a DC motor is that the magnitude
of the torque produced by the output shaft never changes [3].
This makes the DC motor perfect for applications that have
large startup loads, particularly automotive applications (like
the drive wheels in electric vehicles, for instance), where DC
power is readily available under most circumstances.
Today, the asynchronous (or induction) motor is the most
commonly used electric motor in industry and in household
devices [3]. For this reason, the asynchronous motor is the AC
motor design being described and analyzed here.
Electromagnetic induction is the creation of a current through
a conductor that that is within a magnetic field [3]. The
magnetic field is capable of generating a large current in the
conductor without requiring any physical contact whatsoever
[3]. It is this principle that allows the induction motor to
function without having any sliding electrical contacts (such as
the brushes in the brush DC motor) [3]. The stator (the hollow
cylinder in Fig. 2) is essentially a hollow cylinder with no
ends, and may be constructed out of either cast iron or
aluminum [3].
Fig. 3. Winding Slot in AC Motor Stator
The motor’s number of poles is determined by the layout of
the windings within the stator (the number of poles will always
be a multiple of two, but, again, that has no impact here) [3].
The rotor is also cylindrical, though this cylinder does have
a core. The rotor is made of steel disks (the long black
segments in Fig. 4) slotted around the exterior of the cylinder
[3]. A short-circuited winding is placed in the slots between
the steel disks, preventing the need for a supply to be
connected to the rotor [3]. The currents in the rotor are
induced by the interaction between the magnetic fields of the
stator and the rotor, leading to the name for this type of motor
[3]. A copper or aluminum bar (in the case of Fig. 4,
aluminum, judging by the color) is also placed in each slot
between the steel disks [3]. A circular conducting ring is
placed on the end of the cylinder (refer to Fig. 4), to connect
the ends of the bars together (because the conducting ring and
bars resemble a circular cage, this rotor design is called
“squirrel cage”) [3].
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viscous damping, and θ(t) is the motor’s angular position [2].
By assuming zero initial conditions and then Laplace
transforming each of these equations, s-domain equations are
reached. The developed torque is now
T( s ) K2 If( s )
,
(4)
the field voltage is now
Vf( s )
Lf s  Rf  If( s )
,
(5)
and the mechanical torque is now

Is2  Bs (s ),
T( s )
Fig. 4. Squirrel Cage and Steel Disks in AC Rotor
The interaction between the stator magnetic field and the rotor
magnetic field forces the rotor to spin, relative to the stator,
resulting in a functioning AC motor [3].
There are two varieties of asynchronous motors: three phase
and single phase. Three phase motors are used mainly in
industry, while single phase motors are more common
household appliances [3].
The primary advantage of an induction motor is that it
contains no sliding electrical contacts, resulting in a simple
robust design that is easy to manufacture and maintain [3]. A
secondary advantage is that the available range of induction
motors is from only a few watts to several megawatts, making
the use of induction motors in a wide variety of applications
physically feasible [3].
IV. TRANSFER FUNCTIONS
The differential equations and transfer functions of DC and
AC motors are crucial to the analysis of the control of these
machines. The transfer functions are derived from the
differential equations using Laplace transforms, a method all
too familiar to most engineers.
The differential equations and transfer functions for DC
motors are more complicated than those of AC motors, due to
the fact that DC motors have time lags because of both the
armature inductance and the winding, while AC motors have
only a single time constant. DC motors are described by three
differential equations: The developed torque (T(t)) is
described by
T( t) K2 if( t)
,
(1)
where if(t) is the current through the field and K2 is constant
[2]. The field voltage (vf(t)) is described by
vf( t)
d
Rf if( t)  Lf if( t)
dt
,
(2)
where Rf is the field resistance and Lf is the field inductance
[2]. Lastly, the mechanical torque (T(t)) is described by
I
d
2
d
 ( t)  B  ( t)
dt
dt
,
(3)
where I is the motor’s moment of inertia, B is the motor’s
T( t)
2

(6)
where all of the constants have the same meaning as in the
time-domain differential equations, If is used in place of if and
Θ is used in place of θ [2]. By substituting and solving, the
transfer function of the motor is found to be
Km
( s )
Vf( s )



s  m s  1  e s  1
,
(7)
where
m
I
B
is the mechanical time constant of the motor,
Lf
e
Rf
(8)
(9)
is the electrical time constant of the motor, and
K2
Km
B Rf
(10)
is another constant [2]. This is the transfer function that will
be used for the control analysis of the DC motor in the next
section.
The differential equations and transfer functions for AC
motors are considerably less complicated than those for DC
motors, owing to the fact that AC motors only have a single
time constant while DC motors have two. AC motors are
described by two differential equations: The torque (T(t)) is
described by
d
K v( t)  m  ( t)
dt
,
T( t)
(11)
where K is a constant, v(t) is the voltage provided to the
motor, θ(t) is the angular position of the motor, and m is
described by
"stall torque (at rated voltage)"
m
"no-load speed (at rated voltage)" ,
(12)
where stall torque (at rated voltage) and no-load speed (at
rated voltage) are characteristics of any specific AC motor [2].
The torque is also described by
I
d
2
d
 ( t)  B  ( t)
dt
dt
,
(13)
which is identical to the third differential equation that
describes DC motors, and has the same meaning [2]. By
equating the two AC motor equations, assuming zero initial
T( t)
2
4
conditions, and then taking the Laplace transform of the
resultant equation, the transfer function of an AC motor is
found to be
Km
( s )
V( s )
s    s  1 ,
(14)
where
Km
K
m B
PID controller only has two zeros, only two of the poles may
be eliminated. This raises the obvious question: Which poles
should be eliminated and which one should be left alone?
That depends on which pole, as the only pole in the system,
results in a system with the shortest rise time, the shortest
settling time, and the least overshoot. This system was
modeled and simulated in MATLAB/Simulink (see Fig. 5),
using a unit step input as a standard input.
(15)
is a constant, and

I
m B
(16)
is the time constant of the motor [2]. This is the transfer
function that will be used for the control analysis of the AC
motor in the next section.
V. CONTROL
Precise control of motors is vital to the use of motors in any
application. Without a system in place to prevent the motor
from operating unchecked, a step increase to the inputs of a
motor would result in the motor accelerating until it literally
broke apart, costing untold amounts of money to repair and
replace damaged equipment, and undoubtedly ruining
someone’s day.
The first step to safe control is the use of negative feedback,
a concept so common that it warrants no explanation here,
which has been employed in all of the simulations whose
results are shown here.
A control method which is popular because of its
robustness, its simplicity, and its reusability is PID control. A
PID controller contains a proportional gain, an integrator, and
a differentiator (hence its name), all of which are summed
together to produce the output of the controller. The transfer
function of a PID controller has the form
PID
KI
KP 
 KD s
s
2
KD s  KP s  KI
s
,(17)
where KP is the proportional gain coefficient, KI is the
integrator coefficient, and KD is the differentiator coefficient.
The proportional gain is used to amplify the input signal. The
integrator is used to improve the accuracy of the control
system, that is, to minimize the steady-state error (the
difference between the input value and the final output value)
as much as possible. The differentiator is used to increase the
damping in the system, which will decrease both the peak time
and the settling time of the system.
As can be recalled from above, the transfer function of the
DC motor is third order in the denominator, so it has three
poles (roots of the polynomial in the denominator). Likewise,
the transfer function of the PID controller is second order in
the numerator, so it has two zeros (roots of the polynomial in
the numerator). Thus, the PID values may be set so that the
zeros of the PID controller eliminate the poles of the DC
motor. However, since the DC motor has three poles and the
Fig. 5. Simulink Model of DC Motor Control System with
PID Control
The gain block in this system is used to dramatically speed up
the response of the system, although it does not effect the
location of any of the poles or zeros, so it does not impact the
“control” aspect of this analysis (even though it decreases the
peak and settlings times for the system). The electrical time
constant of the motor (τe) was assumed to be 1 ms, the
mechanical time constant of the motor (τm) was assumed to be
100 ms, and the motor constant (Km) was assumed to be 0.050
N*m/A, all of which are typical constants for a DC motor [1].
The first two poles to be eliminated were the poles that result
from the electrical and mechanical time constants, leaving the
third pole (at s = 0) alone. This was done by setting the
differentiator coefficient to the product of the electrical and
mechanical time constants, the proportional gain coefficient to
the sum of the electrical and mechanical time constants, and
the integrator coefficient to one. The result of this simulation
is so worthless that it will not be shown here. By eliminating
those two particular poles (hand calculations would confirm
this, but are not necessary here), the overshoot of the system
became 100% (the output was “2” while the input step was
only “1”), the peak time became 0.44 seconds (not terrible, but
not great either), and the settling time became infinite. The
system never settled. The output of the system was sinusoidal
and continued for all of eternity (until the simulation time of
ten seconds was reached). This is no good. In the next
simulation run, the two poles eliminated were the pole at s = 0
and the pole created by the electrical time constant. This was
done by setting the differentiator coefficient to the value of the
electrical time constant, the proportional gain coefficient to
one, and the integrator coefficient to zero. The output of this
simulation is shown in Fig. 6.
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Fig. 6. Time Response of System with Mechanical Time
Constant Pole
Fig. 6 shows the time response of the system when the
mechanical time constant and s = 0 poles were eliminated by
the settings of the PID controller. The horizontal axis shows
time, in seconds, and the vertical axis is the output of the
system, normalized to the magnitude of the input step. Closer
analysis of this simulation output shows a settling time (using
2% criterion) of approximately 0.76 seconds, a peak time of
approximately 0.14 seconds, and an overshoot of
approximately 49%. This simulation shows a better peak time
(0.14 seconds as opposed to 0.44 seconds), a better overshoot
(49% as opposed to 100%), and a much better settling time
(0.76 seconds as opposed to not settling at all). However, the
peak time could still improve (a little), the peak time could
improve quite a bit, and the overshoot has the most room for
improvement. The simulation was run a third time, this time
eliminating the pole at s = 0 and the pole created by the
mechanical time constant. This was done by setting the
differentiator coefficient to the value of the mechanical time
constant, the proportional gain coefficient to one, and the
integrator coefficient to zero. This simulation output is shown
in Fig. 7.
while the system corresponding to Fig. 6 was simulated for 1
second (with the step input occurring at time t = 0 seconds),
the system corresponding to Fig. 7 was simulated for only 0.25
seconds, because its time response was so fast that it could
scarcely be seen on a 1-second time plot.
From these findings, it appears as though the best way to
use a PID controller in conjunction with a DC motor is to set
the PID values so that the s = 0 pole and the pole created by
the larger of the two time constants (typically the mechanical
time constant) are eliminated.
The transfer function of an AC motor is simple enough that,
even though PID control may be implemented for an AC
motor (as it will be later), acceptable time response values may
be attained by simply adjusting the gain in the system (see Fig.
8). For all of the AC motor control system simulations, a time
constant (τ) of 0.1 seconds (100 ms) and a motor constant of
0.050 N*m/A have been assumed.
Fig. 8. Simulink Model of AC Motor Control System
By running this simulation many times, and adjusting the gain
value each time, it can be found that, between the gain values
of approximately 128.957 and 143.076, the overshoot is less
than (or equal to) 10% and the peak time is less than (or equal
to) 0.5 seconds. At a gain of 128.957, the overshoot is 8.21%
while the peak time is exactly 0.5 seconds. At a gain of
143.076, the peak time is 0.461 seconds while the overshoot is
exactly 10.0%. For gain values between the two just given,
the overshoot and peak time are between the two extremes
given above for overshoot and peak time. It just so happens
that the settling time, regardless of the gain value, is always
0.8 seconds (two opposing values, both functions of the gain,
cancel out in the calculation of the settling time). These values
are, of course, also dependent on the value of the motor’s time
constant, so the overshoot, peak time, and settling time will
change if the time constant is changed. For comparison
purposes, the model of Fig. 8 was simulated (see Fig. 9), using
a gain of 135, as that value is roughly midway between the two
gain values discussed earlier which denote the two endpoints
of the range in which optimum output values are produced.
Fig. 7. Time Response of System with Electrical Time
Constant Pole
The simulation output in Fig. 7 shows that the elimination of
the s = 0 and mechanical time constant poles produces the best
system performance yet. Since the output does not oscillate,
the peak time cannot be characterized as having a unique
value, so the peak time will be considered to be the same as
the settling time. The settling time is approximately 0.073
seconds (73 ms), and there is no overshoot. Note also that
Fig. 9. AC Motor System Time Response with Gain of 135
6
This simulation output shows a peak time of approximately
0.48 seconds, a settling time of approximately 0.73 seconds,
and an overshoot of approximately 9.0%.
Since the AC motor transfer function is only second order
(as opposed to the third order DC motor transfer function),
those values may be easily changed by adjusting the gain
value. However, better system response may be attained by
placing a PID controller in the control system with the motor
(see Fig. 10), and setting the PID values accordingly.
had to be used to attain a high enough resolution in the output
graph that the response could be seen without difficulty.
From these findings, it appears as though the best way to
control an AC motor is the use a PID controller and to set the
PID values so that the poles of the motor are eliminated.
When such equipment is not available, the gain can be
adjusted to optimize the response, although that approach
produces much less favorable results than those ensuing from
the use of a PID controller.
VI. TIME CONSTANTS
Fig. 10. Simulink Model of AC Motor Control System
with PID Control
Because the denominator of the AC motor transfer function is
second order, both poles may be eliminated by setting the PID
values strategically.
This was done by setting the
differentiator coefficient to the value of the time constant, the
proportional gain coefficient to one, and the integrator
coefficient to zero. The output of that simulation is shown in
Fig. 11.
Fig. 11. Time Response of System with Both Motor Poles
Eliminated
Closer inspection of Fig. 11 reveals a peak time of 0.14
seconds, a settling time of 0.065 seconds, and an overshoot of
0.4%. The settling time is less than the peak time because the
peak value is less than 2% greater than the final value. In
comparison to the simulation (of Figs. 8 and 9) which was
optimized by adjusting nothing other than the gain, this
simulation exhibited considerable improvement, decreasing
the peak time by approximately 0.34 seconds (from 0.48 to
0.14 seconds), decreasing the settling time by approximately
0.665 seconds (from 0.73 to 0.065 seconds), and decreasing
the overshoot by approximately 8.6% (from 9.0 to 0.4%). A
simpler way to compare the two responses by inspection is to
note the time scale in Figs. 9 and 11. The simulation
represented in Fig. 9 was run for 1.5 seconds, while the
response of the system whose simulation is represented in Fig.
11 was so much faster that a time scale of only 0.5 seconds
A method to control both DC and AC motors with PID
control has been determined. This method is heavily reliant on
the time constants of the motor, and assumes that anyone
trying to implement PID control for a motor knows the values
of the motor’s time constants. Since the nameplate on most
motors only provides the name of the manufacturer, the
motor’s serial number, the motor’s input voltage, the
maximum rotational speed of the motor’s output shaft (in
RPM), the power of the motor (in horsepower), and the weight
of the motor, another questions arises: How can the time
constants be determined for a motor that one already has in
hand? This is a very good question.
While the data sheets for some motors do contain the values
of the time constants, this is not true in every case, so those
values must be obtained experimentally.
In the case of a DC motor, this is easier said than done.
First, a system resembling that of Fig. 5 must be constructed,
where the output being measured is the rotational speed of the
output shaft of the motor. Keep in mind that, in this case, the
transfer function of the motor is really a black box; all that is
known about is for certain is that it has three poles, one of
them at s = 0, and the other two created by the two time
constants. Unfortunately, there is no cut and dried, step by
step process by which to determine the time constants of the
DC motor. As such, the system must be run many times,
adjusting the PID values each time. The integrator coefficient
should be set to zero, and the proportional gain coefficient
should be set to one, and both values should left there for all of
the system runs.
Only the value of the differentiator
coefficient should be changed between runs. The electrical
time constant will likely be around 1 ms, and the mechanical
time constant will be greater than the electrical. Recall from
earlier that the best system response will be attained when the
differentiator coefficient is equal to the mechanical time
constant. The mechanical time constant may range anywhere
from only a few milliseconds to hundreds of milliseconds, so
that full range should be tested. Thus, when a system response
closely resembling Fig. 7 (no oscillations, no overshoot, and a
very short settling time) is achieved, the value of the
differentiator coefficient for that run is the value of the
mechanical time constant.
To determine the time constants of an AC motor, a system
resembling that of Fig. 8 must be constructed, where the
output being measured is the rotational speed of the output
7
shaft of the motor. Choose an arbitrary gain value (it should
be somewhat high, to get a relatively fast time response, so a
run will take no more than a couple seconds, but not so high
that separate oscillations are indistinguishable), and then do
not change the gain value again. Keep in mind, again, that the
transfer function of the motor is really a black box, and the
only thing that is known about it for certain is that it has two
poles, one at s = 0 and the other created by the time constant.
The system must be run a single time (after running it several
times to determine a good gain value), and the peak time,
settling time, and percent overshoot all need to be observed.
Fig. 12 shows the output of the model of Fig. 8, using a gain of
1000 instead of the gain of 135 shown in Fig. 8.
Fig. 12. AC Motor System Response with Gain of 1000
Fig. 12 shows a peak time of approximately 0.14 seconds, a
settling time of approximately 0.76 seconds, and an overshoot
of approximately 49%. Next, several calculations must be
made, using the following equations:
   
PO 100 exp 
 1  2 

,
(18)
where PO is the percent overshoot of the response and ζ is the
damping ratio of the system,

Tp
2
n  1  
,
(19)
where Tp is the peak time of the response and ωn is the natural
frequency of system,
4
Ts
 n
,
(20)
where Ts is the settling time of the response, and
1
2  n
,
(21)
where τ is the time constant of the motor. Using the percent
overshoot (in this case, 49%) in conjunction with (18), solve
for the damping ratio, which in this case turns out to be
approximately 0.22. Next, use the value of the damping ratio
and the peak time (in this case, 0.14 seconds) in conjunction
with (19), and solve for the natural frequency, which in this
case turns out to be approximately 23.0 radians/second. If, for
some reason, the value of the peak time is not available or
cannot be gleaned from the time response graph, the damping
ratio and the settling time may be used in conjunction with
(20) to find the value of the natural frequency, but since the
peak time may generally be observed with greater accuracy
than the settling time, (20) should only be used as a last resort.
Similarly, if there is some problem with the value of the peak
time or percent overshoot, any two of (18), (19), and (20) may
be used to solve for the values of the damping ratio and natural
frequency, but since the percent overshoot and peak time are
the two that may be measured the most easily and with the
most accuracy, (18) and (19) should be used whenever
possible. Finally, using the values of the damping ratio and
natural frequency in conjunction with (21), the time constant
may be solved for, in this case turning out to be approximately
0.10 seconds. Considering that the value used for the time
constant in the simulation was 0.1 seconds, it appears as
though this method works with some degree of accuracy.
When attempting to determine the time constants of a DC
motor, the best method available is adjust the values of the
coefficients of a PID controller until the result of a run bears
resemblance to that of Fig. 7, and then to use those coefficient
settings from then on. When attempting to determine the time
constant of an AC motor, the system should be run, and the
peak time, settling time, and overshoot observed and used to
calculate the value of the time constant.
VII. ACKNOWLEDGMENTS
The author gratefully acknowledges the contribution of
Paulo Ribeiro, for his help in writing this paper. Without his
support, this process would have been much more difficult.
The author also acknowledges the contributions of John
Washburn, Frank Saggio, Paul Bakker, Matt Husson, and Nate
Studer for their help in the research process.
VIII. REFERENCES
[1]
[2]
[3]
Robert H. Bishop, Richard C. Dorf, Modern Control Systems, 9th ed.
Upper Saddle River, NJ: Prentice-Hall, 2001, p. 52-56, 227-231.
D.K. Anand, Introduction to Control Systems. New York: Pergamon
Press Inc., 1974, p. 34-37.
The ST Microcontroller Support Site Motor Control Tutorial,
http://mcu.st.com/contentid-7.html.
IX. BIOGRAPHY
Brian Bouma was born in Grand Rapids, Michigan
in the United States of America, on September 30,
1982. He graduated from Grand Rapids Christian
High School in 2001, and is currently pursuing his
undergraduate degree at Calvin College. He is
majoring in engineering, with a concentration in
electrical and computer engineering, and currently
holds a minor in mathematics.
His work experience includes Smiths Aerospace
LLC where he works in Digital Design.
He has been a student member of IEEE since 2001.