Wireless Communication Electronics by Example

advertisement
Robert Sobot
Wireless
Communication
Electronics
by Example
Wireless Communication
Electronics by Example
Robert Sobot
Wireless Communication
Electronics by Example
123
Robert Sobot
Electrical and Computer Engineering
Western University
London, ON
Canada
ISBN 978-3-319-02870-5
DOI 10.1007/978-3-319-02871-2
ISBN 978-3-319-02871-2
(eBook)
Springer Cham Heidelberg New York Dordrecht London
Library of Congress Control Number: 2013951765
Springer International Publishing Switzerland 2014
This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of
the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations,
recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or
information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar
methodology now known or hereafter developed. Exempted from this legal reservation are brief
excerpts in connection with reviews or scholarly analysis or material supplied specifically for the
purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the
work. Duplication of this publication or parts thereof is permitted only under the provisions of
the Copyright Law of the Publisher’s location, in its current version, and permission for use must
always be obtained from Springer. Permissions for use may be obtained through RightsLink at the
Copyright Clearance Center. Violations are liable to prosecution under the respective Copyright Law.
The use of general descriptive names, registered names, trademarks, service marks, etc. in this
publication does not imply, even in the absence of a specific statement, that such names are exempt
from the relevant protective laws and regulations and therefore free for general use.
While the advice and information in this book are believed to be true and accurate at the date of
publication, neither the authors nor the editors nor the publisher can accept any legal responsibility for
any errors or omissions that may be made. The publisher makes no warranty, express or implied, with
respect to the material contained herein.
Printed on acid-free paper
Springer is part of Springer Science+Business Media (www.springer.com)
To Allen
Preface
This tutorial book comes as a supplement to my ‘‘Wireless Communication
Electronics, Introduction to RF Circuits and Design Techniques’’ textbook, which
resulted from my lecture notes in ‘‘Communication Electronics I’’ undergraduate
course that I was offering over the last 7 years to students at Western University in
London, Ontario, Canada.
My main inspiration to write this tutorial book again came from my students
who would always ask ‘‘How to I practice for this course?’’ while being frustrated
for having to browse through large number of books in order to find a few
applicable examples for practice. In addition, most of the modern undergraduate
textbooks follow the same approach of giving a number of solved examples,
accompanied by a number of problems with no solutions. Feedback that I received
from my students is that without being able to verify both the final results and the
methodology, it is very discouraging to practice the unsolved problems from most
textbooks. Consequently, many students become discouraged and never take that
first most critical step.
In this tutorial book, I choose to give not only the complete solutions to the
given problems, but also to give detailed background information about the relevance of the problem and the underlying principles. By doing so, my hope is that
my students will be able to make their first steps with my help, and then to develop
their own confidence and understanding of modern electronics. In order to solve
engineering problems, one must first understand the underlying principles, and one
must know the basic set of methods of how to solve the problems.
The intended audience of this book are primarily senior engineering undergraduate students who are just entering the field of wireless electronics after only
first courses in electronics. At the same time, my hope is that graduate engineering
students will find this book useful reference for some of the details that have been
either only touched upon in the previous stages of their education, or explained
from a different point of view. Finally, the practicing junior RF engineers may find
this book handy source for quick answers that are routinely omitted in most
textbooks.
Paris, France, Winter, Spring and Summer, 2013
Robert Sobot
vii
Acknowledgments
Writing a textbook is an undertaking that, by default, relies on ‘‘standing on the
shoulders of giants.’’ I would like to acknowledge all those wonderful books that I
used as the references and the source of my knowledge, and to say thank you to
their Authors for providing me with the insights that otherwise I would not have
been able to acquire. Under their influence, I was able to synthesize my own
picture of reality, which is what acquiring of the knowledge is all about. My hope
is that their guidance and shaping of my own understanding of the topics in this
book are clearly visible, hence I do want to acknowledge their contributions, which
are now being passed to the readers.
In the professional life, one learns both from the written sources and from the
‘‘experience.’’ The experience comes from the interaction with people that we
meet and projects that we work on. I am grateful to my former colleagues who I
was fortunate to have as my technical mentors on really inspirational projects, first
at the Institute of Microelectronic Technologies and Single Crystals, University of
Belgrade, former Yugoslavia, then at PMC–Sierra Burnaby BC, Canada, where I
gained most of my experiences about ‘‘the real world.’’
I would like to acknowledge contributions of Prof. John MacDougall who
initialized and restructured the course into the form of ‘‘design and build’’ concept,
and of Profs. Alan Webster, Zine Eddine Abid, and Serguei Primak who taught the
course at various times.
I would like to thank all of my former and current students who relentlessly
keep asking ‘‘Why?’’ and ‘‘How did you get this?’’ with hope that material
compiled in this book contains answers to at least some of those questions, and that
it will encourage them to keep asking questions with unconstrained curiosity about
all phenomena that surround us.
Most of all, I want to thank my wife for her unconditional and continuous
support, and our son for always being around my writing desk and for making me
laugh.
ix
Contents
Part I
Problems
1
Introduction: Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
2
Basic Terminology: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
3
Electrical Noise: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
4
Electronic Devices: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
5
Electrical Resonance: Problems . . . . . . . . . . . . . . . . . . . . . . . . .
25
6
Matching Networks: Problems . . . . . . . . . . . . . . . . . . . . . . . . . .
29
7
RF and IF Amplifiers: Problems . . . . . . . . . . . . . . . . . . . . . . . . .
33
8
Sinusoidal Oscillators: Problems . . . . . . . . . . . . . . . . . . . . . . . . .
37
9
Frequency Shifting: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
10 Modulation: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
11 Signal Demodulation: Problems . . . . . . . . . . . . . . . . . . . . . . . . .
49
12 RF Receivers: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
Part II
Solutions
13 Introduction: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
14 Basic Terminology: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
15 Electrical Noise: Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
107
xi
xii
Contents
16 Electronic Devices: Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . .
143
17 Electrical Resonance: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . .
161
18 Matching Networks: Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . .
183
19 RF and IF Amplifiers: Solutions . . . . . . . . . . . . . . . . . . . . . . . . .
201
20 Sinusoidal Oscillators: Solutions . . . . . . . . . . . . . . . . . . . . . . . . .
215
21 Frequency Shifting: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .
225
22 Modulation: Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
235
23 Signal Demodulation: Solutions. . . . . . . . . . . . . . . . . . . . . . . . . .
243
24 RF Receivers: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
251
Appendix A: Physical Constants and Engineering Prefixes . . . . . . . . .
267
Appendix B: Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . .
269
Appendix C: Second Order Differential Equation. . . . . . . . . . . . . . . .
271
Appendix D: Complex Numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
273
Appendix E: Basic Trigonometric Identities . . . . . . . . . . . . . . . . . . . .
275
Appendix F: Useful Algebra Equations . . . . . . . . . . . . . . . . . . . . . . .
277
Appendix G: Bessel Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . .
279
Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
281
Bibliographies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
291
Acronyms
A/D
AC
ADC
AF
AFC
AGC
AM
BiCMOS
BJT
BW, B
CMOS
CRTC
CW
D/A
DAC
dB
dBm
DC
ELF
EM
eV
FCC
FET
FFT
FM
GaAs
GHz
HBT
HF
Hz
IC
IF
InGaAs
InP
Analog to digital
Alternate current
Analog to digital converter
Audio frequency
Automatic frequency control
Automatic gain control
Amplitude modulation
Bipolar–CMOS
Bipolar junction transistor
Bandwidth
Complementary metal–oxide semiconductor
Canadian Radio–Television and Telecommunication Commission
Continuous wave
Digital to analog
Digital to analog converter
Decibel
Decibel with respect to 1 mW
Direct current
Extremely low frequency
Electromagnetic
Electron volts
Federal Communication Commission
Field effect transistor
Fast Fourier transform
Frequency modulation
Gallium arsenide
Gigahertz
Heterojunction bipolar transistor
High frequency
Hertz
Integrated circuit
Intermediate frequency
Indium gallium arsenide
Indium phosphide
xiii
xiv
I/O
IR
JFET
KCL
KVL
LC
LF
LNA
LO
MMIC
MOS
MOSFET
NF
PCB
PLL
PM
pp
ppm
Q
RADAR
RF
RMS
SAW
SHF
SINAD
S/N
SNR
SPICE
TC
THD
UHF
UV
VCO
V/F
VHF
V/I
VLF
VSWR
Acronyms
Input output
Infrared
Junction field–effect transistor
Kirchhoff’s current law
Kirchhoff’s voltage law
Inductive–capacitive
Low frequency
Low noise amplifier
Local oscillator
Monolithic microwave integrated circuit
Metal oxide semiconductor
Metal oxide semiconductor field effect transistor
Noise figure
Printed circuit board
Phase locked loop
Phase modulation
Peak to peak
Parts per million
Quality factor
Radio detecting and ranging
Radio frequency
Root mean square
Surface acoustic wave
Super high frequency
Signal to noise plus distortion
Signal to noise
Signal to noise ratio
Simulation Program with Integrated Circuit Emphasis
Temperature coefficient
Total harmonic distortion
Ultra high frequency
Ultraviolet
Voltage controlled oscillator
Voltage to frequency
Very high frequency
Voltage current
Very low frequency
Voltage standing wave ratio
Part I
Problems
Chapter 1
Introduction: Problems
Electromagnetic (EM) waves and their transmission through space are possible due to
several relatively simple but fundamental physical phenomena. In modern theoretical
physics, EM wave is an abstract concept that epitomizes transfer of energy through
the space, which is equivalent to say that information is transferred through the space.
At the same time, the information itself is embedded into the wave in the form of the
energy variation in time. In this chapter we review some of the fundamental concepts
from physics related to energy, matter, EM waves, EM fields, propagation of energy
through matter and space, and basic interaction between two waveforms.
Problems:
1.1. An average sized snowflake consists of approximately n = 6.68559 × 1019
molecules. Assuming the complete matter of the snowflake is converted into energy,
estimate for how long a laptop computer whose average power consumption is
P = 25 W could be powered?
1.2. Our Sun produces energy in form of light by continuous nuclear fusion reaction
in its core where hydrogen (H) atoms are fused together to form atoms of helium
(He). Activity of the Sun is monitored by satellites positioned around Earth that
measure the solar constant k S , i.e. average radiated power per square metre at the
distance of one astronomical unit ( AU ) from the Sun. Estimated average value of
the solar constant is k S = 1366 W/m2 .
Knowing the average radius of the Earth r , the average distance between the Earth
and the Sun l, and estimated mass of the Sun m S , estimate:
1. what is the total mass (M) of matter at rest that the Sun converts into energy per
second?
2. what percentage of the total converted mass M at rest is used to supply energy to
the Earth?
3. what is the total mass (M H ) at rest of hydrogen (H) used in the fusion process
per second?
4. ignoring all other causes, and assuming that the Sun will collapse after converting
10 % of its current mass at rest, estimate how long the Sun will last?
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2_1, © Springer International Publishing Switzerland 2014
3
4
1 Introduction: Problems
5. what is the total power radiated by the Sun and how much is received at the Earth?
1.3. Sketch an EM sinusoidal wave that propagates in x–direction with clearly marked
all three axes, as well as the other relevant wave vectors.
1.4. Sketch a rough drawing of EM wave as being generated by a dipole antenna.
1.5. Show that expression E = E m sin(kz − ωt) describes a wave moving in the
positive z direction. In addition, express k and ω parameters in terms of the corresponding wavelength λ, the frequency f , then find relationship to the wave velocity
v = f (ω, k), i.e. as function of angular frequency ω and wave number k.
1.6. Starting with classical approximation of the wave equation of a sinusoidal EM
wave E = E m sin(k z − ω t) derive expression and then calculate the speed of light
in vacuum with no electric charges.
1.7. Find the ratio of electric and magnetic field amplitudes of a sinusoidal EM wave
whose components are E = E m sin(kz − ωt) and B = Bm sin(kz − ωt).
1.8. We note that vector product of electric and magnetic fields E × B points in the
same direction as the wave velocity vector. Find expression for magnitude of a vector
S that is defined as S = 1/μ0 (E × B) and determine its measurement unit. Comment
on the nature of these results.
1.9. Estimate peak values of electric E m and magnetic Bm components of EM wave
generated by the Sun, as measured close to the Earth.
1.10. After learning that Poynting vector represents flow of energy per unit area,
and recognizing that energy is related to work, that work is related to force, and the
force is related to pressure, we realize that EM wave hitting a perfectly reflective
surface must exert “radiation pressure” on the surface. Derive expression for average
Poynting vector √S→ as function of the radiation pressure Pr (pressure is measured in
N/m2 ).
1.11. A solar sail that is made of ideally reflective light material called Kapton is
used to propel a probe away from the Sun by relying only on the pressure generated
by the Sun light. Ignore influence of all planets in the solar system, and use results
from the previous problems as needed. Assuming that the total mass of the sail plus
payload is m = 2 kg, end the specific mass of the used Kapton is m k0 = 768.5 µg/m 2 ,
estimate mass of the payload.
1.12. The electrical and magnetic components of a composite electromagnetic wave
are described as follows,
E = E m cos(kz − ωt) y − E m cos(kz + ωt) y
B = Bm cos(kz − ωt) z + Bm cos(kz + ωt) z
(1.1)
(1.2)
Find expression for the average value of its Poynting vector S and comment on the
result.
1 Introduction: Problems
5
Fig. 1.1 Problem 1.16: a long
conductor (relative to the λ)
is divided into infinitesimally
short sections z ∞ λ,
where each of the sections
is modelled with distributed
circuit elements R, L, C, and G
1.13. Calculate the intrinsic wave impedance, phase velocity, and wavelengths of an
electromagnetic wave in free space oscillating at the following frequencies: f 1 =
10 MHz, f 2 = 100 MHz, f 1 = 10 GHz.
1.14. Plot the graph of the radial magnetic field H (r ) inside and outside of infinitely
long wire in air of radius a = 5 mm aligned along the z–axis and carrying a DC
current of I = 5 A.
1.15. Find the induced voltage of a thin wire loop of radius a = 5 mm in air subjected to a time–varying magnetic field H = H0 cos ω t, where H0 = 5 A/m and the
operating frequency is f = 100 MHz.
1.16. Starting from an electrical line section model in Fig. 1.1 derive, first, expression
for general characteristic line impedance, then the expression for lossless characteristic line impedance Z 0 .
Calculate characteristic impedance of a two–wire transmission line that is characterized as L = 378 nH/m and C = 150 pF/m.
1.17. Instantaneous voltage of EM wave is described as v(t) = Vm cos (2π f t + φ0 )
where, ω = (2π 100) rad/s and φ0 = π/4. Calculate: (a) EM wave’s phase and the
front–end’s distance from the starting point at t = 15 ms; (b) wavelength λ of this
wave; and (c) amplitudes V of the instantaneous voltages at t = 0 s and t = 15 ms.
1.18. Instantaneous voltage of a waveform is described as
v(t) = sin (2π t) +
1
1
1
sin (6π t) + sin (10π t) + sin (14π t) + · · ·
3
5
7
Using plotting software of your choice, create the following plots (show the time
axis over at least two periods of the slowest tone):
1.
2.
3.
4.
on the same graph, plot v(t) and the first three terms terms;
plot v(t) using the first ten terms;
plot again the case 2. but this time remove the first two terms of the polynomial;
try removing some other combinations of terms from the case 2. polynomial.
6
1 Introduction: Problems
Fig. 1.2 Problems 1.21 and 1.22: two single tone signals and passive network (left), and time
domain plot of an AM modulated waveform (right)
Observe time–domain differences among waveforms in terms of their amplitude,
rising and falling edges, period, especially relative to the ideal waveform that v(t) is
expected to emulate.
1.19. Using the same time–domain square waveform function as in Problem 1.18 analyze the frequency content of waveforms in cases 2 and 3 by creating their respective
frequency–domain plots, i.e. signal power versus frequency.
1.20. Given three waveforms, v1 = 1 + 0.5 sin(ω t), v2 = 1 + 0.5 sin(ωt − π ), and
v3 = 0.5 + 0.5 sin(ω t − π ), without using any plotting software sketch by hands in
correct scale the following plots:
1.
2.
3.
4.
v1 ,
v1 ,
v1 ,
v1 ,
v2 , and v1 (t) − v2 (t);
v2 , and v1 (t) + v2 (t);
v3 , and v1 (t) − v3 (t);
v3 , and v1 (t) + v3 (t);
Comment on the created waveforms.
1.21. Two single tone signals, v1 = f (ω 1 , t) and v2 = f (ω 2 , t) are applied at the
input nodes of a passive network bellow, Fig. 1.2 (right). Assuming that the two
signals are drawn in scale, and v1 (t) has period of T1 = 1 µs, find the following:
1. expression for the output signal, vo (t) = f (v1 (t), v2 (t)), assume R1 = R2 ;
2. sketch to scale the signal frequency spectrum at the output node Vo when R1 = R2 ,
and clearly label frequencies on the horizontal axis;
3. expression for the output signal, vo (t) = f (v1 (t), v2 (t)), assume R2 = 2 R1 ;
4. sketch to scale the signal frequency spectrum at the output node Vo when R2 =
2 R1 , and clearly label frequencies on the horizontal axis.
1.22. For the given amplitude modulated (AM) signal vin , Fig. 1.2 (left), sketch shape
of the envelope waveform.
Chapter 2
Basic Terminology: Problems
As electrical engineers we often take for granted scientific concepts related to
potential and flow of current in conductive and semi–conductive materials.
Nevertheless, our understanding of phenomena related to the flow of electricity inside
electrical circuit is based on very few basic concepts studied in the previously taken
courses.
In this chapter we review some of the relevant elements from physics, terms and
definitions, and techniques commonly used by electrical engineers while either analyzing or synthesizing electrical circuits. Intuitive understanding of classic concepts
related to structure of an atom, internal forces, movement of charges, electrical field,
and electric power is important and the following problems serve the purpose of
refreshing the prior knowledge.
Problems:
2.1. A hydrogen atom consists of one proton (m p , q p ) and one electron (m e , qe )
separated by distance r , where (m, q) pair corresponds to their respective mass and
charge. Ignoring all other particles and forces and assuming the hydrogen atom to
be at rest, find the ratio R of electrostatic FC and gravitational FG forces between
these two particles.
2.2. A system of three charges, initially at rest, is located at three vertices of a square,
Fig. 2.1. Sketch direction in which the charge q3 will begin to move assuming that
charge q1 is a proton fixed in space, charge q2 is an electron fixed in space, and: (a)
the charge q3 is a proton free to move in space; or, (b) the charge q3 is an electron
free to move in space.
2.3. Derive expression for electric field around a single charged particle q.
2.4. In an experimental setup, two oil drops whose respective radiuses are measured as
r1 = 1.670 µm and r2 = 1.775 µm while the oil’s mass density is ωoil = 837 kg/m3 .
The two oil drops are first electrically charged and then suspended in the air by the
means of electric field of E = 2.0 × 105 N/C that is acting in direction opposite to
the gravity force.
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2_2, © Springer International Publishing Switzerland 2014
7
8
2 Basic Terminology: Problems
Fig. 2.1 Problem 2.2: a
system of three charges
located in space
For the given data, calculate the difference in electric charges stored on these two oil
drops.
2.5. Derive expressions for radial Er and the perpendicular Ep electric fields of an
electric dipole at point r √ a, where distance between the two charges making
the dipole is 2a. What happens if the derived results are applied for the case when
r → 0?
2.6. Derive expressions for radial component of electric field Er at distance r from
the centre of a very long metallic wire whose length is l and it is charged with a
uniform linear charge density ql [C/m/m].
2.7. Using a graphing tool of your choice, plot on the same graph various pairs of
the following single tone signals at f = 10 MHz along with their respective sums:
S1 = 2.0 sin (λt), S2 = 2.0 sin (λt + π/3), S3 = 2.0 sin (λt + π/2)
S4 = 1.0 sin (λt + 3π/4), S5 = 2.0 sin (λt + 2π ), S6 = 3.0 sin (λt + 4π/3)
For example, plot together S1 , S5 , and (S1 + S5 ) or S2 , S3 and (S2 + S3 ), and so
on. In particular, observe their relationships in respect to their phase differences, and
at the same time observe how the two amplitudes are related to each other at any
given point in time, i.e. at some of the typical instantaneous phase points. For a given
frequency, practice mental calculations of conversing various phase differences into
the units of time.
2.8. Using a plotting software, create plot that includes the following five single tone
signals (assume f = 10 MHz):
S1 (t) = 2 sin (λ t), S2 (t) = − sin (2λ t), S3 (t) =
1
2
S4 (t) = − sin (4λ t), S5 (t) = sin (5λ t), · · ·
2
5
and then plot their sum
2
sin (3λ t)
3
2 Basic Terminology: Problems
9
S(t) =
5
Sk
k =1
Note that aside from the fundamental tone S1 at λ this particular waveform also
contains both even and odd harmonics, i.e. both 2λ , 4λ , 6λ ,... and 3λ , 5λ , 7λ ,...
terms.
How the function Sk looks like? If number of the single tone signals Sk increases
indefinitely, i.e. if k → ∞, what waveform shape S(t) is synthesized?
2.9. For a highly simplified case where the noise waveform is described as follows
Sn (t) =
1
2
3
4
1
sin (λ t) − sin (2λ t) + sin (3λ t) + sin (5λ t) + sin (9λ t)
5
2
3
4
3
1. Find noise cancellation function Sm that should be used to remove this noise
completely;
2. Using plotting software, plot together Sn , Sm and (Sn + Sm ).
3. Assuming realistic signal processing system, how much signal processing delay
could you allow in the system, so that the cancellation is still satisfactory? (For
the sake of argument, let us assume that “satisfactory” noise reduction is when
the maximum noise amplitude is reduced about ten times.)
Fig. 2.2 Problem 2.12: schematic diagram of circuit network
Fig. 2.3 Problem 2.13: time domain plot of current flow
10
2 Basic Terminology: Problems
2.10. Calculate average power of a square pulse whose amplitude is v = 2V and the
pulse width is t = 1 ms and its energy is dissipated in a resistor R = 100 π. What
happens if the pulse duty cycle is changed to 25 or 80% ?
2.11. A current flowing in positive direction through an electronic component is
defined as:
−2t, if (t ≤ 0);
i(t) =
(2.1)
+3t, if (t ≥ 0);
1. Calculate values of the current at the following two time instances i(−2.2 s) and
i(+2.2 s);
2. Calculate the total charge g that has flown through the component within the time
interval of (−2s ≤ t ≤ 3s);
3. and, find the average value of i(t) within the same time interval.
2.12. With reference to Fig. 2.2 calculate power absorbed/generated by each element
in this circuit if: (vs , i s ) = (8 V, 7 A), v1 = 12 V, i2 = 2 A, i3 = 8 A. In this
example, a resistor symbol is used to show an element capable of both absorbing and
generating power.
2.13. Calculate the average power Pavg delivered/absorbed by a load R = 5 π and
i rms for the current whose time domain waveform is shown in Fig. 2.3.
Chapter 3
Electrical Noise: Problems
Most humans do not even realize the existence of a phenomena known as a “cocktail
party effect” where our brain is capable of filtering out sounds in a room full of people
and tuning in to a single voice. From the perspective of wireless communication
technology the pre twentieth century Earth was similar to an empty room, none to talk
and none to listen. Except for the always existing cosmic radiation, the humans did not
have a single device that is capable of using EM radiation for wireless communication.
In the modern world, the air is always filled in with EM waves generated by
countless wireless devices that we have. Now, the situation is similar to a room fool
of people, all talking loud all the time. That constant din sets the minimum energy
level of “noise”, thus any signal that is to be received must have energy level higher
than the noise. In this section we review basic definitions, the theoretical background,
and some of the basic techniques that help us design electronic equipment capable
of extracting only the wanted signal from the background din.
Problems:
3.1. Using plotting software of your choice, sketch two time–domain graphs:
(a) plot an example of a voltage noise signal n(t, ω 1 , ω 2 , ω 3 , ...) next to an ideal
single–tone voltage signal a(t, ω a ), so that the two voltage amplitudes are related
as SNR = 20dB; and
(b) plot a noisy single–tone signal A(t, ω) = a(t, ω a ) + n(t, ω), where the noise
component is described as n(t, ω 1 , ω 2 , ω 3 , ...).
For the purpose of this problem, experiment with the number of noise harmonics
used to create the n(t, ω) and their frequencies relative to the single tone frequency
so that the graphs are visually illustrative. Also, zoom–in the two graphs to one to
two periods of the single–tone signal.
Repeat this example for other SNR values (both positive and negative).
3.2. Using plotting software of your choice that is capable of performing FFT, sketch
frequency domain plot of the noisy voltage signal from Problem 3.1. What is your
observation?
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2_3, © Springer International Publishing Switzerland 2014
11
12
3 Electrical Noise: Problems
3.3. Using plotting software of your choice, sketch a quantitative time–domain graph:
(a) of two single–ended signals
a(t) = VCM + A0 sin(ω t) + n(t)
b(t) = VCM + A0 sin(ω t + λ ) + n(t)
i.e. after each of the two single tone signals was affected by the same noise n(t);
(b) of d(t) that is calculated as the difference of the two single–ended signals, i.e.
d(t) = a(t) − b(t)
Without affecting generality of the example, and for the simplicity of the plot, assume
that the noise signal is of the form n(t) = ( A0/10) [sin(10ω t) + cos(20ω t)]. Choose
values of the maximum amplitude A0 , the common mode DC voltage VCM , and
frequency ω so that the graph is zoomed-in to one or two periods of the signal.
Comment on the properties of signal d(t) relative to the signals a(t) and b(t).
3.4. For the given RLC network, Fig. 3.1, derive expression for its transfer function
H ( jω) = vout /vin .
Then, using plotting software of your choice, plot amplitude of the network transfer function |H ( jω)|, and find its bandwidth BW by using graphical method. Next,
suggest the equivalent ideal (i.e. “brick–wall”) approximation of the transfer function
and overlay the two transfer functions. For this exercise, use the following component
values: R = 0.1 , 1 , 10 , L = 1 H, and C = 1F. What difference the value of
the resistor makes? Explain.
It is useful to explore influence of all the components on the centre frequency f 0
and bandwidth BW of the transfer function. For instance, how would you scale the
component values so that the centre frequency of the BPF is set to f 0 = 10 MHz?
To f 0 = 2.4 GHz? Explore how the behaviour of this RLC network changes if
the components change their places inside the network by trying all other possible
topologies.
3.5. Sketch a qualitative frequency–domain graph that shows: (a) a white noise power
spectral density spectrum; (b) an ideal “brick–wall” bandpass (BP) filter transfer
function; and (c) the white noise spectrum after being filtered by the BP filter.
3.6. Using the graph from Problem 3.5, show how the filtered white noise power
is estimated graphically. Assume that the white noise PSD = 1 µW/ Hz, while
the system is intended to amplify Ps = 100 mW electrical signal whose frequency
Fig. 3.1 Problem 3.4:
schematic diagram of RLC
network
3 Electrical Noise: Problems
13
content is within the standard range of the human hearing. Calculate SNR for both
cases, i.e. without and with BP filter applied.
3.7. Two random signal voltage generators are connected in series. The two voltage
generators have maximal rms amplitudes e1 = 1 V and e2 = 10 V respectively.
Derive the expression for the rms voltage amplitude at the terminals of this
two source network. Comment on the result if the specific numerical values in this
example are used.
3.8. Resistors R1 = 20 k and R2 = 50 k are at room temperature T = 290 K.
For a given bandwidth of BW = 20 kHz find the thermal noise voltage for the three
cases shown in Fig. 3.2, i.e.: (a) each resistor separately; (b) their serial combination;
(c) their parallel combination; (d) the total available noise power Pn in the cases (a)
to (c).
3.9.Two resistors R1 = 20 k at temperature T1 = 285.5K and R2 = 50 k at
temperature T2 = 190K are connected in series. For a given bandwidth of BW =
20 kHz first find relevant expressions for the total output thermal noise voltage en ,
and the total available noise power at the output Pn of these two resistors, Fig. 3.2
(middle), and then, for the given data, calculate their respective numerical values.
The term “available power” refers to the power delivered to the load if the load
impedance were conjugately matched to the source impedance for the maximum
power transfer. However, if the source and load impedances are not matched, then
delivered power is obviously less than the available power, which still does not change
the amount of the available power (i.e. in this case it would be only partially used).
3.10. Assuming that a realistic source resistor R1 at T1 drives a realistic load resistor
R2 at T2 , find expression for noise figure NF at the output of the resistive networks,
Fig. 3.2, in the following two cases:
(a) the source resistance R1 and the load resistance R2 are connected in series,
Fig. 3.2 (middle), and then comment on the conditions required to achieve minimum noise figure;
(b) the source resistance R1 and the load resistance R2 are connected in parallel,
Fig. 3.2 (right), and then comment on the conditions required to achieve minimum noise figure.
3.11. In the radio astronomy, the local calibration reference resistor is cooled to,
for example, TR = 2.5K and its thermal noise power PR serves as the absolute
Fig. 3.2 Problems 3.8, 3.9, and 3.10: schematic diagrams of a single realistic resistor R (left),
series combination of resistors R1 , R2 (middle), and parallel combination of resistors R1 , R2 (right)
14
3 Electrical Noise: Problems
Fig. 3.3 Problems 3.12, 3.13
and 3.14: Schematic diagram
of an inverting operational
amplifier
reference level for the background radiation noise. Assuming an ideal amplifier
and the overall measurement setup to be ideal, calculate the equivalent antenna
temperatures Ta , if signal power levels measured at the antenna’s output terminals
are PS = 3.01dB, 6.02dB, 9.03dB, ... above the background radiation power level?
3.12. Simplified schematic diagram of an inverting amplifier is shown in Fig. 3.3.
For this problem assume that:
(a)
(b)
(c)
(d)
voltage source v S is ideal;
the operational amplifier has very high open–loop gain;
no current flows into the two input nodes of operational amplifier and
thermal noise is the only type of noise present in the system.
Derive expression for rms value of the output noise voltage eout if the only noise
source in the system is due to the total thermal noise voltage en at the input terminals
of the operational amplifier (en is usually specified in the accompanying datasheets).
The two resistors R1 , R2 are assumed ideal, i.e. noiseless.
Then, calculate rms value of the amplifier’s noise voltage√eout , datasheets for the
operational amplifier specify the input noise as en = 5 nV/ Hz, and the amplifier
parameters are R1 = 1 k, and R2 = 10 k.
3.13. Use the same circuit and the same set of assumptions from Problem 3.12.
However, this time the noisy operational amplifier is specified in terms of: (a) its
total input noise voltage en ; and (b) its total input noise currents i n+ and i n− , where
the two respective noise currents are associated with the positive and negative input
nodes of the operational amplifier, (i n+ and i n− noise currents are also usually equal
and specified in the accompanying datasheets). In addition, within the total frequency
bandwidth of the system the two noise currents are assumed to be non–correlated.
Again, the two resistors R1 , R2 are assumed ideal, i.e. noiseless.
First, derive expression for rms value of the total output noise voltage eout , then
calculate its numerical value. √
Datasheets for the operational amplifier specify
the
√
input noise as en = 5.082 nV/ Hz, the input noise current is i n = 5 pA/ Hz, and
the amplifier parameters are R1 = 1 k, and R2 = 10 k.
3.14. Use the same circuit and the same set of assumptions from Problem 3.13.
However, this time assume that the two resistors R1 , R2 are not ideal, i.e. they also
contribute to the total noise power in the system.
First, derive expression for rms value of the total output noise voltage eout . Second,
calculate its numerical value. Datasheets for the operational amplifier specify the
3 Electrical Noise: Problems
15
Fig. 3.4 Problem 3.15:
schematic diagram of an
operational amplifier
√
√
input noise as en = 5 nV/ Hz, the input noise current is i n = 5 pA/ Hz, and
the amplifier parameters are R1 = 1 k, R2 = 10 k, and T = 3.75 ◦ C. Third,
calculate the total noise voltage within frequency bandwidth BW = 20 Hz to 20 kHz.
3.15. An amplifier circuit, Fig. 3.4, is connected to the input signal source v S through
node 1iso that the source resistance R S is matched to the amplifier’s input resistance
R3 . The operational amplifier is assumed to have very high gain and infinite input
impedance.
Derive expression for the noise figure NF of this amplifier, then, calculate numerical value of the NF. Datasheets
for the operational amplifier specify
√
√ the total
input noise as en = 5 nV/ Hz, the input noise current is i n = 5 pA/ Hz, and the
amplifier parameters are R1 = 1 k, R2 = 10 k, R3 = 1 k, and T = 95 ◦ C.
3.16. An amplifier circuit, Fig. 3.5, is connected to the input signal source v S through
node 1iso that the source resistance R S is matched to the amplifier’s input resistance. The operational amplifier is assumed to have very high gain and infinite input
impedance.
Derive expression for the noise figure NF of this amplifier, then, calculate numerical value of the NF. √
Datasheets for the operational amplifier specify
√ the total
input noise as en = 5 nV/ Hz, the input noise current is i n = 5 pA// Hz, and the
amplifier parameters are R S = 800 , R1 = 1 k, R2 = 10 k, R3 = 1 k, R4
is adjusted so that the amplifier’s input resistance is matched with the source resistance, and temperature inside the amplifier box is measured as T = 80.8 ◦ C.
Fig. 3.5 Problem 3.16:
schematic diagram of
an inverting operational
amplifier
16
3 Electrical Noise: Problems
3.17. A realistic voltage source with the internal resistance is used to drive an ideal
amplifier whose noise figure is NF and the power gain G.
Derive expression for noise power No in dBm/Hz at the output terminal of the
amplifier, and then calculate it by using the following numerical data: NF =
3dB, G = 12dB, and the environment temperature is T = 15 ◦ C.
3.18. Power spectrum graph of a signal and the noise floor power levels at the input
side of an amplifier is shown in Fig. 3.6. What is the input side SNR I ? Assuming this
signal is amplified by an amplifier whose gain is G = 10dB and NF = 3dB, redraw
the power spectrum graph and comment on the result. What is the output side SNRo ?
3.19. Power levels of the space radiation routinely detected by Earth based radio
telescopes are comparable, for instance, with power level of a signal generated by a
cell phone placed on the Moon. In order to illustrate how weak this signal really is,
relative to the surrounding EM radiation that is perceived as the noise, let us consider
the following numerical example.
Assume that an astronaut on the Moon is trying to contact the Earth by using his
cellphone. At the same time, a bird that flies above the receiving radio antenna dish
generates thermal noise that interferes with the phone signal, Fig. 3.7.
A typical cell phone radiates signal at Pcell = 1W level over bandwidth of
B = 5 MHz,the bird’s body temperature is T = 37 ◦ C, and average distance to the
Moon is D = 384.4 × 106 m. Aside from the cell phone signal and the thermal noise
generated by the bird, ignore all the other signal and noise sources, and assume that
the signals are received uniformly over the whole surface of the antenna.
Fig. 3.6 Problems 3.18:
frequency domain of a signal
spectrum relative to the noise
floor
Fig. 3.7 Problem 3.19:
illustration for cellphone
on the Moon example
3 Electrical Noise: Problems
17
Calculate:
(a) Hight H above the antenna’s dish surface where the bird is located at the moment
when SNR = 0dB, i.e. the cell phone and the thermal noise powers are equal?
(b) SNR when the bird lands at centre of the dish cover, i.e. when H = h = 10m?
3.20. A signal whose amplitude is vin = 1 µV, bandwidth is B, and the source
resistance Rin = 50 , passes through an ideal noiseless “brick-wall” type BP filter
whose input impedance and bandwidth are always matched to the source. Derive
expression and then calculate
(a) SNR in dB at the BP filter output node, if bandwidth of the filter is B1 = 20 kHz
and T1 = 88 ◦ C;
(b) the environment temperature T2 in K if the same SNR as found in (a) needs to
be maintained while using the input signal source whose bandwidth is B2 =
82.080 kHz.
3.21. Starting with the known expressions for thermal noise voltage generated by a
resistor, for dynamic resistance R D of an R LC resonator at resonance ω 0 , and for
effective bandwidth Beff of an R LC resonator, derive the well known expression
for noise voltage squared en2 = kT/C . As a numerical example calculate the noise
voltage en if C = 10 pF and T = 17 ◦ C.
3.22. An oscilloscope probe’s RC equivalent network model is specified as R =
1 M and C = 1 pF. Determine:
(a) useful frequency bandwidth B of this probe;
(b) the lower limit of voltage that can be measured by this probe at T = 33 ◦ C if all
other effects and the probe load are ignored;
3.23. A television set consists of the following chain of five consecutive sub–blocks:
two RF amplifiers with 20dB gain and 3dB noise figure each, a mixer with −6dB
gain and 8dB noise figure, and two additional amplifiers with 20dB gain and noise
figure of 10dB each. Calculate:
(a) the system noise figure;
(b) the system noise temperature Tn at T = 26 ◦ C.
3.24. An amplifier with the input signal power of 5 × 10−6 W, noise input power
1 × 10−6 W, has output signal power of 50 × 10−3 W and the output noise power
40 × 10−3 W. Calculate the noise factor F and the nose figure NF of this amplifier.
3.25. Calculate noise current and equivalent noise voltage for a diode biased with
IDC = 1 mA at the room temperature 300K and within the bandwidth of 1 MHz.
3.26. The equivalent input resistance of an amplifier is Rin = 1 k, while its equivalent shot noise biasing current is I D = 1 µA at T = 70 ◦ C. The signal generator has
internal resistance R S = 1 k and it provides a signal of at least v S = 10 µVrms .
Calculate maximal bandwidth B of this amplifier.
18
3 Electrical Noise: Problems
Fig. 3.8 Problem 3.28: noise
frequency spectrum
3.27. A front end RF amplifier whose gain is A = 50dB and noise temperature
TnA = 90 K provides signal to a receiver that has a noise figure of NF = 12dB.
First, calculate the noise temperature TRx of the receiver itself, then calculate the
overall noise temperature Tsys of the amplifier plus the receiver system at the room
temperature T = 300 K.
3.28. A graph of a typical BJT input noise current power spectral density is shown in
Fig. 3.8. Using approximative methods estimate the total input noise current of this
BJT within the frequency bandwidth or 1 kHz to 100 MHz.
Chapter 4
Electronic Devices: Problems
Time domain electrical signals are processed using both linear and non–linear
devices. Rapid changes of the voltage/current signal levels, however, cause wide
range of different responses while being processed by electronic components and
devices. In this chapter we review important practical aspects of setting up right
conditions that enable active and passive devices to produce the desired response.
Variations of variables external to the devices themselves, such as the power supply level and the environmental temperature, have great impact on the overall circuit
behaviour. Thus, we practice to predict the overall circuit behaviour while accounting
for the environment imperfections.
Problems:
4.1. By definition, time domain voltage function v(t) across an inductor is related to
the time varying current i(t) flow as
v(t) ≡ L
di(t)
dt
(4.1)
where, L is the proportionality constant defined as inductance. If the current plot in
time domain is shown in Fig. 4.1, sketch the voltage graph across the inductor.
4.2. A square pulse voltage source v(t) , capacitor C, and a resistor R are connected
in a series circuit. At time t0 = 0 s the capacitor was completely discharged. Plot
a graph of the voltage vC across the capacitor over the next 5 ms. Numerical data:
pulse waveform frequency f = 1 kHz, its maximum amplitude is v(max) = 10 V,
capacitor value C = 1 µF, and the resistor value is R = 1 kω.
4.3. Assuming VD D = 3.0 V power supply, design a Vref = 1V voltage reference
using a resistive divider. Assume that there is no any other branching current and that
the equivalent Thévenin resistance of this voltage reference is Rth ≤ 100 ω. How
much the reference voltage changes if the power supply voltage variation is ±10 %?
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2_4, © Springer International Publishing Switzerland 2014
19
20
4 Electronic Devices: Problems
Fig. 4.1 Problem 4.1: time
domain diagram of current
waveform
4.4. A diode model is given as
VD
q VD
VD
− 1 = I S exp
(4.2)
− 1 ≈ I S exp
I D = I S exp
n VT
nkT
n VT
where,
I D – is current flowing through the diode
I S – is the diode leakage current
VD – is voltage across the diode, i.e. biasing voltage
VT – is the thermal voltage (VT = kT /q)
k – is Boltzmann constant
T – is temperature in degrees Kelvin [K]
q – is the elementary charge
n – is the emission coefficient, usually between 1 and 2
Typical 1N4004/1A diode has the following parameters: I S = 76.9 nA and the
emission coefficient n = 1.45. At the junction temperature of T = 28 ◦ C calculate
the diode current I D both using the exact and approximated model, then find the
calculation error in percents if:
1. the forward biasing voltage is VD = 600 mV ; and
2. the forward biasing voltage is VD = 50 mV.
4.5. The same 1N4004/1A diode from problem 4.4 is connected in series with an
ideal current source I = 1 A. At junction temperature T = 28 ◦ C calculate voltage
VD across the diode terminals as well as the voltage range if the current source varies
by ±10 %.
4.6. For a BJT transistor whose I S = 5 × 10−15 A, and at room temperature
VT = 25 mV, the biasing current is IC = 1 mA, and assume the emission
coefficient n = 1. Calculate the base emitter voltage VB E . Now, for VB E =
0.50, 0.55, 0.60, 0.65, 0.70, 0.75, and 0.80 V calculate the collector current IC . Also,
calculate gm of this BJT.
4.7. For the four resistive networks Fig. 4.2 calculate the equivalent impedances R AB
at the following frequencies: DC, 1 Hz, 1, 10 kHz, 1, 100 MHz, and at ∞ frequency.
When appropriate, instead of doing the exact calculations, estimate the results by
using reasonable engineering approximation. Assume R = 1 kω, and C = 1 nF.
4 Electronic Devices: Problems
21
Fig. 4.2 Problem 4.7:
schematic diagram of four
resistive networks
(a)
Fig. 4.3 Problem 4.8:
resistive networks
(a)
(b)
(c)
(d)
(b)
Fig. 4.4 Problems 4.9 (left)
and 4.10 (right): schematic
diagrams of networks
4.8. For the two networks given in Fig. 4.3, estimate maximum and minimum voltage
gains V A,B /Vout,B . First assume impedance ratio of Z 1 /Z 2 = R1 /R2 = 10 : 1, and
then assume Z 1 /Z 2 = R1 /R2 = 1 : 10.
4.9. For a network shown by schematic diagram in Fig. 4.4 (left),
1. assuming the base–emitter diode threshold voltage is Vth (B E) = 0 V, i.e. ideal
BE diode, find value(s) of R2 so that the transistor Q 1 is turned on. What potential VC is required at the collector node C to maintain the saturation mode of
operation?
2. assuming the base–emitter diode threshold voltage is Vth (B E) = 1 V, i.e. realistic
BE diode, find value(s) of R2 so that the transistor Q 1 is turned on. What potential
is required at the collector node VC to maintain the saturation mode of operation?
4.10. What is the required resistor ratio R1 /R2 for the network in Fig. 4.4 (right), so
that the transistor Q 1 operates in saturation, if VCC = 10 V and R E = 1 kω:
22
4 Electronic Devices: Problems
(a)
(b)
(c)
Fig. 4.5 Problem 4.11: schematic of resitive network
1. assuming the base–emitter diode threshold voltage is Vth (B E) = 0 V, i.e. ideal
BE diode, find value(s) of R2 so that the transistor Q 1 is turned on. What potential
is required at the collector node VC to maintain the saturation mode of operation?
2. assuming the base–emitter diode threshold voltage is Vth (B E) = 1 V, i.e. realistic
BE diode, find value(s) of R2 so that the transistor Q 1 is turned on. What potential
is required at the collector node VC to maintain the saturation mode of operation?
4.11. Estimate impedances looking into networks (a), (b), and (c), Fig. 4.5. For the
sake of argument, assume that the small signal internal base resistance rπ is very
large, and the internal emitter resistance is re is a small non zero value.
4.12. Calculate thermal voltage VT under the following conditions: (a) T = −55 ◦ C,
(b) T = 25 ◦ C, and (c) T = 125 ◦ C. As a side note, these three temperatures are
commonly used to characterize military grade electronic equipment.
4.13. A simple voltage reference is built using a resistor and 1N4004 diode as in
schematic Fig. 4.6 (left). Calculate voltage across the diode at room temperature
25 ◦ C, under the following conditions: VCC = 9 V, R = 1 kω, IS = 18.8 nA.
Express the result using engineering number notification with three decimal places.
4.14. Find biasing voltage VB E at T = −55 ◦ C, T = 25 ◦ C, and T = 125 ◦ C, if
BJT collector current is set to IC = 1 mA and I S = 100 fA, Fig. 4.6 (right). Repeat
the calculations for I S = 200 fA.
4.15. For BJT transistor in Fig. 4.6 (right), assuming the base current to be negligible,
estimate biasing voltage VB required at the base node so that the collector biasing
current is set to IC = 1 mA ≈ IE . Data: I S = 100 fA, RE = 100 ω, T = 25 ◦ C.
Fig. 4.6 Voltage reference
network for problems
4.13 (left) and 4.14 (right)
4 Electronic Devices: Problems
23
4.16. Estimate the input impedance Z in looking into the base node, Fig. 4.6 (right), if
emitter resistor is R E = 100 ω and the forward gain β F is assumed as: (a) β F = 99,
and (b) β F → ∞.
4.17. Design preliminary resistive voltage divider to set the base biasing voltage for
the circuit in Fig. 4.6 (right). Use your own engineering judgement and constrains
for the design. Assume power supply voltage VCC = 3.3 V, emitter resistor R E =
100 ω, and β F = 99.
4.18. For a BJT transistor with I S = 100 fA, VBE = 591.6 mV at temperature
T = 25 ◦ C calculate: (a) the collector current IC ; (b) the transistor’s transconductance
gm ; (c) the intrinsic emitter resistance r E ; and (d) r E if IC = 2, 3 mA, ....
What happens if the temperature changes to, for example, T = 30 ◦ C? Briefly
state your observations.
Chapter 5
Electrical Resonance: Problems
Resonance in general is one of the fundamental phenomena in the universe. Electrical
resonance in particular is so important for our telecommunication technology, that
if by some weird coincidence all LC circuits were dissolved and if we lost the
knowledge of its workings, our marvellous electronic technology would have become
totally unusable. It is not difficult to imagine consequence of such unfortunate event
for our technological civilization.
And yet, the seemingly trivial parallel connection of two passive components, an
inductor and capacitor, is all that is needed to create the electronic resonance. In this
chapter we study various important aspects of this tiny circuit, both in frequency and
time domain.
Problems:
5.1. Derive expressions for impedance Z ab (ω) of a series RLC network and expression for admittance Yab (ω) of a parallel RLC network, Fig. 5.1.
5.2. Knowing that in case of an ideal RLC network its impedance Z ab (ω0 ) at resonance must be real, derive expressions for resonant frequencies f 0 of the series and
parallel networks, Fig. 5.1.
5.3. Find Z ab (ω0 ) for ideal LC series and parallel networks, Fig. 5.2.
5.4. For a more realistic parallel LC network, Fig. 5.3, find expression for the resonant
frequency ω p0 . The resistance R accounts for all thermal losses, i.e. it represents
combined resistance of the inductor, the wires, and effective series resistance (ESR)
of the capacitor. Then, using the result find expression for the resonant frequency
ω p0 under condition that the “realistic” parallel LC networks degenerates into “ideal”
parallel LC network, i.e. R → 0. What is the conclusion?
5.5. At the resonance ω p0 : (a) find expression for dynamic resistance R D = Rab (ω p0 )
of the LC network in Fig. 5.3.
5.6. One possible definitions for quality factor Q of a series resonant LC circuit is
formulated as the ratio of the total energy stored the inductor (or the capacitor) versus
the energy dissipated in the resistive part of the network, Figs. 5.1 and 5.3.
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2_5, © Springer International Publishing Switzerland 2014
25
26
5 Electrical Resonance: Problems
Fig. 5.1 Problems 5.1, 5.2:
schematic diagrams of ideal
series and parallel RLC networks
Fig. 5.2 Problem 5.2:
schematic diagram of ideal
series and parallel LC network
Fig. 5.3 Problems 5.4, 5.5,
5.6: schematic diagram of
realistic parallel LC network
Derive the three versions of expressions for Q s factor of a series RLC network,
i.e.: (a) Q s (L , R), (b) Q s (R, C); and (c) Q s (R, L , C).
5.7. Derive relationships between series impedance Z s = Rs + j X s and its equivalent
parallel admittance Y p = 1/Z s = 1/R p + 1/ j X p . What is the conclusion?
5.8. For a series RLC network, Fig. 5.1 (left), derive expression for bandwidth BW
at the resonant frequency ω0 as a function of the Q. What is the conclusion?
5.9. Truly realistic model of an LC resonator must include losses of both the inductor
and capacitor, modelled by effective series resistance ESR and resistor r1 , Fig. 5.4.
Derive expressions for the resonant frequency ω0 and dynamic resistance R D of
this circuit as the function of Q 1 and Q 2 factors of the inductor and capacitor. Finally,
transform resonator in Fig. 5.4 into its equivalent parallel RLC network, assuming
that the capacitor is lossless, i.e. ESR = 0.
5.10. For a given coil, L = 2 µH, Q = 200, f 0 = 10 MHz, calculate:
1. its equivalent series resistance,
2. its equivalent parallel resistance,
3. the value of the resonating capacitor C,
5 Electrical Resonance: Problems
27
Fig. 5.4 Problem 5.9: schematic diagram of realistic parallel LC network
4. resistance R which, when added in parallel expands the bandwidth to B =
500 kHz.
5.11. In RF circuits it is common practice to use RLC circuit to design a bandpass
(BP) filter that is needed to isolate a single frequency carrier signal from the rest of
EM radiation that is constantly present in the air. For the sake of argument, compare
a series RLC resonating circuit with the combination of a lowpass (LP) and (HP) RC
filters used one after another, which effectively also creates BP filter.
Plot normalized transfer characteristics of RLC bandpass, LP, HP, and LPHP
bandpass filters. In the plot, normalize resonant frequencies of both RLC and LPHP
filters, as well as pole locations of LP and HP filters to 1Hz. Also, normalize all four
output amplitudes to 1 V. In addition, try to estimate Q factor of RLC resonator, so
that its bandwidth is equal to bandwidth of RC based LPHP bandpass filter.
5.12. For the given inductor L = 2.533 nH and a trimming capacitor whose range is
C = 1.3 nF to 857.345 pF calculate the tuning range of this LC resonator.
5.13. Design an LC resonator whose resonant frequency is f 0 = 10 MHz if only
L = 2.533 nH inductor and the following components are available:
Fig. 5.5 Problem 5.17: Illustration of a BP transfer function’s frequency bandwidth
28
5 Electrical Resonance: Problems
(a) C1 = 10 nF, C2 = 40 nF, and C3 = 50 nF.
(b) C1 = 200 nF, C2 = 400 nF, C3 = 0.5 µF, and C4 = 2 µF.
(c) C1 = 70 nF, C2 = 60 nF, C3 = 60 nF, and C4 = 100 pF.
5.14. Calculate the Q–factor of a serial RLC network if inductor L = 2.533 nH
and the lumped wire resistance r = (λ )m, at: (a) f 1 = 10 MHz; and (b) f 2 =
100 MHz.
5.15. A 1mH inductive coil has wire resistance of R = 5 and self–capacitance
of 5 pF. The inductor is used to create LC resonator at f 0 = 25 MHz. Calculate the
effective inductance and effective Q factor.
5.16. Calculate resonant frequency of a serial RLC network with R = 30 , L =
3 mH, and C = 100 nF. Calculate its impedance at f = 10 kHz and f = 5 kHz.
5.17. A frequency response curve of an LC resonator looks as in Fig. 5.5. Assuming,
f 1 = 450 kHz, f 2 = 460 kHz, and the resonance frequency f 0 = 455 kHz. Determine the resonator bandwidth, Q factor, inductance L if capacitance is C = 100 nF,
the total internal circuit resistance R.
5.18. A series RC branch consists of R S = 10 and C S = 7.95 pF. Convert it into its
equivalent parallel RC network form assuming operating frequency of f = 1GHz.
Chapter 6
Matching Networks: Problems
Inside electronic circuits, signals are processed either in the form of voltage or in
the form of current. In the section related to voltage and current dividers we learned
that the ratio between the signal source impedance and load impedances is extremely
important because it directly affects how much the signal amplitude is reduced at the
interface node between the source and load. In the ideal case, voltage amplitude stays
unchanged if the source impedance is zero, and/or the load impedance is infinite (i.e.
the load current is zero). Similarly, current amplitude stays unchanged if the load
impedance is zero (i.e. the load voltage is zero) and/or source impedance is infinite.
As a consequence, depending whether the goal is to pass voltage or current signal
from one stage of the system to another, there are two opposing requirements. Most
important conclusion, however, is that even if it were possible to achieve the ideal
either voltage or current signal transmission from source to load, in both of these
two ideal cases the transmitted signal power is zero because electric signal power is
the product of voltage and current.
In this section we practice Q–matching technique to design passive LC networks
that serve the purpose of maximizing power transfer of voltage signals delivered by
a source to the load, which implies that both voltage and current amplitudes of the
signal must take non–zero values. We design the matching network circuits assuming
complex–conjugate matching.
Problems:
6.1. Design a single stage LC matching network in between source with R S = 5 ω
and load R L = 50 ω termination at f = 10 MHz, Fig. 6.1 (left). Additional design
constrain is that the matching network must maintain DC connection between the
source and load.
6.2. Design a single stage LC matching network in between source with R S = 50 ω
and load R L = 5 ω termination at f = 10 MHz, , Fig. 6.1 (right). Additional design
constrain is that the matching network must maintain AC connection between the
source and load.
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2_6, © Springer International Publishing Switzerland 2014
29
30
6 Matching Networks: Problems
Fig. 6.1 Problems 6.1 and 6.2: typical cases of non matched sources and load resistances, R S < R L
(left), and R S > R L (right)
Fig. 6.2 Problems 6.7 and 6.8: Inductive source driving resistive load (left), and resistive load
driving capacitive RC load (right)
6.3. Using Q–matching technique, calculate the equivalent parallel network to series
connection of R S = 5 ω and L S = 238.732 nH at f = 10 MHz.
6.4. Using results from problem 6.1, find reflection coefficient λ and mismatch loss
ML at the interface between the serial and parallel sections of the matching network.
6.5. Using results from problem 6.1, estimate the 3 dB bandwidth, assuming that the
overall matching network’s frequency characteristics is symmetrical.
6.6. Following up on results from problem 6.4, if the input signal occupies frequency
bandwidth B from f = 8 MHz to f = 12 MHz then recalculate λ and ML. Briefly
comment on the results.
6.7. Design single–stage LC matching network at 10 MHz when the source impedance
is inductive, Z s = Rs + j L s , while the load is purely resistive, Fig. 6.2 (left). It is
required that the matching network maintains DC connection between its input and
output terminals. Data: R S = 5 ω, L S = 138.732 nH, R L = 50 ω.
6.8. Design single–stage LC matching network when parasitic capacitance C L exists
in parallel with the load resistance R L , where, R S = 5 ω, R L = 50 ω, C L = 1 nF,
f = 0.01 GHz. Assume DC connection inside the matching network.
6.9. Match 5 ω source resistance to 50 ω load resistance at 10 MHz by designing
a general two stage matching network. In this example, the goal is to decrease the
bandwidth relative to the single LC matching network solution. Make your own
choice of the ghost resistance R I N T and briefly explain your reasoning.
6.10. Match 5 ω source resistance to 50 ω load resistance at 10 MHz. Design two
stage LC matching network, where the first stage behaves as a LP filter, while the
6 Matching Networks: Problems
31
Fig. 6.3 Problem 6.12: Schematic diagram of RF amplifier connected to antenna through matching
network
second stage behaves as a HP filter. One of the goals of this matching network design
is to increase the bandwidth relative to the equivalent single LC matching network
solution.
6.11. An antenna impedance is assumed to be resistive R A = 50 ω. Tuneable RF
amplifier is set to f 0 = 665 kHz and its input impedance is Rin = 2 kω. Design
two possible matching networks using Q–matching technique and comment on differences between the two solutions.
6.12. For the given circuit, Fig. 6.3, RF antenna has internal resistance R A = 50 ω,
frequency of the RF signal is f R F = 10 MHz, C2 = 11.213 pF, C0 → ∞, R1 =
5200 ω, and DC voltage at node 1iis set to V (1) = 1/4VD D . Design matching
network that provides DC coupling to the RF antenna. In your solution use minimal
number of components to design the matching network.
6.13. Using the parasitic absorption method at 10 MHz match source impedance of
Z S = (5 + j10) ω to a load impedance of Z L = (50 − j30) ω (the capacitor is in
parallel to the load resistance R L ).
Chapter 7
RF and IF Amplifiers: Problems
The transition from the classic linear low–frequency amplifiers that are found in
literally in all undergraduate textbooks to RF amplifiers suitable for radio equipment
can be made gradual if, before even moving into the higher frequency regions, first
we review again important aspects of the circuit that become more protruding as the
frequency increases.
In the following examples we practice to recognize and mentally evaluate
parameters such as input/output impedance, biasing points of BJT devices, Miller
capacitance, and amplifier bandwidth. Instead of focusing at all details at once, and
risking not to see the forest for the trees, we study each of the amplifier aspects in
isolation while accepting numerical inaccuracies for the sake of developing intuition
and sense for the circuit behaviour.
Problems
7.1. Estimate impedances as seen by looking into the respective network nodes, as
indicated in Figs 7.1b, c, and d.
7.2. Resistance seen by looking into a BJT emitter is Rout = 100 ω. Resistance
looking into the base is Rin = 100 kω. For λ = 100, find reflected resistance
at the base node R B and R E ? (Ignore base internal resistance and small emitter
resistance re .)
7.3. For network shown in schematic diagram, Fig. 7.1 a,
1. assuming the base-emitter diode threshold voltage is Vth (B E) = 0 V , i.e. ideal
BE diode, find value(s) of R2 so that BJT transistor is turned on. What potential VC
is required at the collector node C to maintain the saturation mode of operation?
2. assuming the base-emitter diode threshold voltage is Vth (B E) = 1 V , i.e. a
realistic BE diode, find value(s) of R2 so that BJT transistor is turned on. What
potential is required at the collector node VC to maintain the saturation mode of
operation?
7.4. What is the required resistor ratio R1 /R2 for network in Fig. 7.1e, so that BJT
transistor operates in saturation, if VCC = 10 V and R E = 1 kω:
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2_7, © Springer International Publishing Switzerland 2014
33
34
7 RF and IF Amplifiers: Problems
(a)
(b)
(c)
(d)
(e)
Fig. 7.1 Problems 7.1, 7.3, and 7.4: network examples
(a)
(b)
(c)
Fig. 7.2 Problems 7.5, 7.6, 7.10, and 7.13: network schematics
1. assuming the base–emitter diode threshold voltage is Vth (B E) = 0 V, i.e. ideal
BE diode, find value(s) of R2 so that the transistor is turned on. What potential is
required at the collector node VC to maintain the saturation mode of operation?
2. assuming the base–emitter diode threshold voltage is Vth (B E) = 1 V, i.e. realistic
BE diode, find value(s) of R2 so that the transistor is turned on. What potential is
required at the collector node VC to maintain the saturation mode of operation?
7.5. If the only given data are RC = 10 kω and R E = 100 ω resistor values, estimate
the upper bound of voltage gain Av for circuit in Fig. 7.2c and express the estimated
gain in [dB].
7.6. Given data for circuit in Fig. 7.2c is: RC = 10 kω, R E = 100 ω, I S = 100 fA,
and VB E = 768.78 mV at temperature T = 25 √ C, and the input signal frequency
is f = 10 MHz. In addition, there is also capacitor C connected in parallel with the
emitter resistor R E that is not shown in the schematic.
Estimate the circuit voltage gain Av if: (a) C = 1 µF, and (b) C → ∞. How
large is the estimated gain difference between these two cases? How large is the gain
difference in comparison with the gain calculated in Problem 7.5? Briefly comment
on the results.
7.7. For grounded emitter amplifier powered from VCC = 10 V, assume VT =
25 mV, then estimate voltage gain A V in the following cases: (a) Vout = 7.5 V; (b)
Vout = 5 V; and (c) Vout = 0.2 V.
7 RF and IF Amplifiers: Problems
(a)
35
(b)
(c)
Fig. 7.3 Problems 7.9, 7.11, 7.12, and 7.14: network schematics
7.8. Assuming VT = 25 mV and that VB E ≤ VT , if VB E voltage of a BJT transistor
changes by 18 mV how much is the correspond change of IC as expressed in dB?
Repeat the calculations for VB E = 60 mV?
7.9. For a CE amplifying circuit in Fig. 7.3 estimate the Miller capacitance C M if
RC = 9.9 kω, R E = 100 ω, CC B = 1 pF.
7.10. Estimate the inherent (i.e. due to the amplifier topology itself) input side
bandwidth of CE amplifier in Fig. 7.2b if RC = 9.9 kω, R E = 100 ω, CC B =
(1/π )pF (not shown in the schematic), λ = 190, R1 = 2 kω, R2 = 2 kω.
7.11. Estimate the range of frequencies where CE amplifier in Fig. 7.3b should be
used, if the base side inductor L = 2.533 µH, RC = 9.9 kω, R E = 100 ω, CC B =
1 pF.
7.12. For amplifier in Fig. 7.3b , the small signal voltage gain is A V = −100. Estimate
value of the inductor L so that the input stage resonates at f 0 = 15.915 MHz. Data:
C = 1 pF. Assume the base current to be zero.
7.13. A signal generator with low internal resistance is coupled with the CE amplifier
in Fig. 7.2c through a series capacitor C = 1 µF. Estimate the range of frequencies
where the CE amplifier should be used, if RC = 9.9 kω, R E = 100 ω, CC B =
1 pF, R1 = 2 kω, R2 = 2 kω.
7.14. For circuit shown in Fig. 7.3c first suggest its application, and then estimate:
(a) DC voltage at BJT collector node; (b) transconductance gain of BJT gm (Q 1 ); and
(c) suggest the circuit’s input vi and output terminals vo , then estimate small signal
AC voltage gain, if the gain is defined a A V = vo/vi .
Data: assume λ = ∞, RC = 7.5 kω, I = 0.5 mA, C = ∞, VCC = 5 V, and
VT = 25 mV.
Chapter 8
Sinusoidal Oscillators: Problems
Transmission of wireless signals is based on the idea of existence of a sinusoidal
waveform, which in return is modulated in some way that encodes the intended
message. In mathematical terms, a sinusoidal function is required for several key
mathematical operations that we meet in the following chapters. Physical embodiment of a sinusoidal waveform is known as an oscillator, whose the only role is
to deliver clean predefined either voltage or current waveform that is accurately
described by a sinusoidal function. What is more, instead of designing a separate
oscillator for each desired frequency, in practical realizations we prefer to design
tuneable oscillators so that a single circuit delivers a sinusoidal waveform that can
take a range of frequencies.
In this chapter we review basic properties of closed loop systems, which are
fundamental to oscillator operation. Instead of taking classical approach of studying
specific oscillator topology, in this book we study oscillators from the perspective
of closed loop systems, thus simplifying the analysis by looking at the forward path
(i.e. amplifier) separately from the feedback path (i.e. passive feedback LC network).
By doing so we can combine three separate concepts that we already know about,
i.e. concept of a signal amplifier, concept of resonant LC circuit, and concept of
feedback systems into the new concept of oscillators.
Problems:
8.1. Derive expression for the general case of loop gain if the feedback loop consists
of forward path amplifier with gain A and the feedback circuit path with gain β.
8.2. One of several versions of a phase oscillator, Fig. 8.1 (left), is based on a CE
amplifier and three RC stages in the feedback loop. Derive expression for the minimal
transistor gain factor βmin (not to be confused with the feedback loop parameter),
and the resonant frequency ω0 , under the following assumptions: (a) the transistor’s
output resistance ro is infinite; (b) all capacitors have the same value; (c) all resistors
have the same values, while the transistor’s base resistance rb is absorbed in the left
most resistor R; (d) for simplicity, details of biasing network are not shown; and
(e) all elements are ideal, ignore the small emitter resistance re and base collector
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2_8, © Springer International Publishing Switzerland 2014
37
38
8 Sinusoidal Oscillators: Problems
Fig. 8.1 Problems 8.2, 8.3, and 8.6: simplified schematics of oscillator networks
capacitance C BC . Then, calculate values for resistors R and capacitors C, if RC =
10 kω.
8.3. Estimate the resonant frequency ω0 of an oscillator whose feedback network is
shown in Fig. 8.1(centre), if L 1 = 0.5 μH, L 2 = 1.5 μH and C = 126.65 pF.
8.4. Using the same example circuit and data as in Problem 8.3 estimate the feedback
network’s gain factor β.
8.5. Assuming the same data as in Problems 8.3 and 8.4, estimate the effective
resistance Reff that this feedback network presents to the output of the oscillator’s
amplifier whose input impedance is Rin = 10 kω, if the effective inductor’s Q factor
is Q(L eff ) = 50.
8.6. For the circuit shown Fig. 8.1 (right) derive: (a) expression for the resonant
frequency ω0 ; and (b) expression for gm of BJT transistor. Then, use the following
numerical data to calculate the resonant frequency ω0 and gm : RC = 10 kω, BJT
output resistance rc = 10 kω, L = 2 μH, C1 = C2 = 253.30 pF, and Q L → ∞.
Details of biasing network are omitted for simplicity.
Repeat the same problem assuming finite Q L = 50, derive and recalculate the
new equation(s) at the resonant frequency ω0 and gm .
Fig. 8.2 Problem 8.7: simplified schematic of Clapp oscillator with varicap diode
8 Sinusoidal Oscillators: Problems
39
8.7. For the Clapp oscillator shown Fig. 8.2 calculate the oscillating frequency at:
(a) zero bias VD = 0 V of the varicap diode, and (b) at VD = −7 V. Data: L =
100 μH, C1 = C2 = 300 pF, and C0 = 20 pF.
Chapter 9
Frequency Shifting: Problems
There are at least two fundamental reasons for the need of a practical technique to
move (i.e. shift) frequency of a signal in the frequency domain while still preserving
the embedded message.
First, there are many radio transmitters in existence. Television, satellite, cell
phone and other similar communication systems should be looked at as being nothing
more than a sophisticated radio. Thus, from the perspective of wireless carrier signals,
the space around us is equivalent to a room full of people, all talking loud and at the
same time. As we easily imagine, it is difficult to hear message of a person standing
on the other side of the room. In principle, there are two basic methods to enable a
message to be transmitted between transmitter and the receiver. More obvious method
is to apply the time sharing principle, i.e. to apply the rule that only one transmitter
is active at any given moment. This is equivalent to a conference setup where one
speaker after another gives a lecture, while all attendees quietly listen. Less obvious
method is to apply the frequency sharing principle. Now, all transmitters are active
all the time, however, each transmits its signal at different frequency bandwidth that
equals to the resonant frequency of the receiver. This is equivalent to a room full of
opera singers, all with perfect pitch, and all singing in non–overlapping frequency
regions, e.g. bass voice is very easily distinguished from soprano voice. Visually,
each frequency bandwidth serves role of a wire directly connecting transmitter and
the receiver. As long as the receiver is “deaf” for all other frequencies except the one
being used by the transmitter, the system works perfectly. In practice, the use of a
narrowband LC resonator in the receiver’s front end sections makes the receiver able
to “hear” only one transmitter, while all other active transmitters are effectively not
heard.
Second, conversion of an electrical signal into EM radiation happens in the antenna. In order to effectively produce EM radiation the antenna length should be at
least as long as a few of the waveform’s wavelengths. As soon as we realize that,
for example 3 kHz signal (which is in the range of a typical human voice) has wavelength of approximately 100 km it becomes obvious why direct wireless transmission
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2_9, © Springer International Publishing Switzerland 2014
41
42
9 Frequency Shifting: Problems
of voice frequencies is not practical. Actually, we already use 100 km “antennas”,
except that they are laid in trenches and called communication cables, phone lines.
In this chapter we learn how to actually implement frequency sharing mechanism
that has enabled our modern wireless communication systems.
Problems:
9.1. There are four single–tone signals:
S1 (t) = 3V sin(2π × 1 MHz × t)
S2 (t) = 4V sin(2π × 20 MHz × t)
S3 (t) = 12V cos(2π × 19 MHz × t)
S4 (t) = 12V cos(2π × 21 MHz × t)
For the given waveforms: (a) find expression for S(t) = S1 (t) × S2 (t). Then, using
a graphing software plot waveforms of [S(t)], [S1 (T )], and [−S1 (t)] in the same
window. Observe time domain relationships among these signals; and (b) then plot
So (t) = 1/2 (S3 (t) − S4 (t)). Briefly comment on the results.
9.2. Starting from f L O = sin(2π × 10 MHz × t), find two other single–tones that
could be used to generate a single tone at f L F = 1 kHz. Comment on the process
and the result.
9.3. There is a large number of radio stations transmitting their respective programs at
various carrier frequencies, all the same time. For example, a radio receiver is tuned
to receive AM modulated wave transmitted at carrier frequency of f R F = 980 kHz.
Local oscillator inside the receiver is set at f L O = 1435 kHz.
For the given data:
1.
2.
3.
4.
calculate frequencies at the output of the receiver’s mixer,
which frequency should be chosen as the intermediate frequency (IF),
at what frequency the ghost radio station would operate, and then
plot frequency domain graph of the frequencies involved in this problem.
9.4. Tuned RF amplifier with LC tank whose Q = 20 is tuned at RF frequency f 0 .
Estimate attenuation of the image signal, if the image frequency is 10 % higher then
RF signal.
9.5. Two voltage signals
v1 = V1 cos(ω1 t)
v2 = V2 cos(ω2 t)
are first added and then passed through an ideal diode, Fig. 9.1, whose voltage/current
function is given as
vD
−1
(9.1)
i D = I S exp
Vt
9 Frequency Shifting: Problems
43
Fig. 9.1 Problem 9.5:
Simplified schematic of a
diode mixer
First, derive expression for the current signal at the output of the diode as a
function of the two input voltages i D = f (v1 , v2 ), and comment on its frequency
spectrum; second, assume that V1 = V2 = 1 V, f 1 = 10.000 MHz, f 2 =
10.001 MHz, Vt = 25 mV, and I S = 1 µA, then write expression for the low–
frequency current tone coming out of this network.
9.6. Two voltage signals
v1 = V1 cos(ω1 t)
v2 = V2 cos(ω2 t)
are either first added and then applied to the gate of an ideal BJT transistor Q 1 ,
Fig. 9.2 (left), or v1 is applied to the gate of Q 2 and v2 is applied to the emitter node
by means of a 1:1 ratio transformer, Fig. 9.2 (right).
Find expressions for frequency spectrum of the collector current in both cases.
Assume ideal transistors with the current gain of β.
9.7. Two voltage signals
v1 = V1 cos(ω1 t)
v2 = V2 cos(ω2 t)
are applied to JFET transistors J1 and J2 that are used instead of Q 1 and Q 2 respectively, in the same topology as in Fig. 9.2.
Find expression for the square term in the output frequency spectrum.
Fig. 9.2 Problem 9.6 and 9.7: Simplified schematic of BJT mixer
44
9 Frequency Shifting: Problems
Fig. 9.3 Problem 9.8: Simplified schematic of a dual–gate FET mixer and its equivalent circuit
diagram, where M1 and M2 are assumed identical
9.8. Two voltage signals
v1 = V1 sin(ω1 t)
v2 = VDC2 + V2 sin(ω2 t)
are applied to dual–gate FET transistor used as a mixer, Fig. 9.3 (left). Note that a
dual gate FET transistor M1,2 is equivalent to cascode configuration of two stacked
transistors M1 and M2 , Fig. 9.3 (right). Assume an ideal FET transistor.
Find expression for the square term in the output frequency spectrum.
Chapter 10
Modulation: Problems
For the purpose of creating wireless communication system, a simple high frequency
sinusoidal waveform is actually not much useful. By itself, the only information it
carries is that there is an active source in existence. In order to carry any more
complicated message, a music or speech for example, the single tone waveform must
have some time variation that can be controlled and used to embed the message.
By inspection of a simple sinusoidal function
S(t) = A sin(ω t + λ)
we conclude that there are three possible parameters available for manipulation in
time, amplitude A, frequency ω , and phase λ.
Naturally, amplitude A is the first obvious choice, simply because we can choose
to use a switch and turn on and off the sinusoidal source. That switching action
translates into a binary type of modulation (i.e. time change) of the amplitude. If
periods of the silence and the signal activity are timed to “sound” as series of long
and short beeps, then transmission technique known as Morse code is created. Or, if
the instantaneous amplitude of the sinusoidal waveform is somehow controlled by
the instantaneous amplitude of a speech, then we say that the sinusoidal waveform
is amplitude modulated (AM).
Second choice would be to somehow control the instantaneous frequency ω of
the high frequency sinusoidal waveform by the instantaneous amplitude of a speech,
then we say that the sinusoidal waveform is frequency modulated (FM). In this
scenario, deviation of the instantaneous frequency relative to a no–signal frequency
f 0 is proportional to the speech amplitude.
Third choice would be to control the instantaneous phase λ and make it proportional to the speech amplitude, in which case we say that phase modulation (PM)
has been implemented.
Each of those there modulation techniques has its advantages and disadvantages
depending on the intended application. In this chapter we practice some of the basic
principles related to signal modulation.
R. Sobot, Wireless Communication Electronics by Example,
45
DOI: 10.1007/978-3-319-02871-2_10, © Springer International Publishing Switzerland 2014
46
10 Modulation: Problems
Problems:
10.1. For given audio signal and its carrier waveform
Aa (t) = 3 + 1.5 sin (2π × 1500 × t)
Ac (t) = 6.0 sin (2π × 50, 000 × t)
do the following:
1.
2.
3.
4.
5.
Plot both these two waveforms in a single plot;
Construct AM modulated wave;
Determine the modulation factor and percent modulation;
What are the frequencies of the audio signal and the carrier?
What tones are visible in frequency spectrum of the modulated wave?
10.2. How many AM broadcasting stations can be accommodated in a 100 kHz
bandwidth if the highest frequency used to modulate these carriers is 5 kHz?
10.3. Determine the power content of each of the sidebands and of the carrier of an
AM signal whose total power is 1,200 W while the modulation index is set at 85 %.
10.4. An AM signal whose modulation index is 70 % contains 1500 W at the carrier
frequency. Determine the power content of the upper and lower sidebands for this
percent modulation. Then, calculate the power at the carrier and the power of each
of the sidebands when the percent modulation drops to 50 %.
10.5. For FM waveform whose frequency modulation index of m f = 1.5 and the
modulation signal’s frequency is f b = 10 kHz:
1. estimate the required bandwidth BFM (using Carson’s rule);
2. calculate ratio of the total power PT relative to the power in the FM unmodulated
waveform;
3. find which harmonics has the highest amplitude.
10.6. Recommend a design parameters of an AM standard broadcast receiver that is
to operate with an intermediate frequency (IF) of 455 kHz, thus
1. Calculate the required frequency for the local oscillator f LO when the receiver is
tuned to f c = 540 kHz, assuming that frequency of the local oscillator is above
the frequency of the received signal.
2. Calculate the required frequency for the local oscillator f LO when the receiver is
tuned to f c = 540 kHz, assuming that frequency of the local oscillator is below
the frequency of the received signal.
10.7. A carrier waveform whose frequency is f c = 107.6 MHz is frequency modulated by a baseband sine wave whose frequency is f m = 7 kHz. The output FM
signal has frequency deviation of πf s = 50 kHz. For the given data,
10 Modulation: Problems
47
Fig. 10.1 Problem 10.1:
schematic of reactance
modulator
Fig. 10.2 Problem 10.2:
illustration of a phase modulation circuit with varicap diode
1. Calculate frequency swing of the carrier in this FM signal;
2. Determine the highest and the lowest frequencies attained by the modulated signal; and
3. Calculate the modulation index of the output FM wave?
10.8. The total signal power of FM transmitter is PT = 100 W while modulation
index is m f = 2.0.
1. Calculate power levels contained in all frequency components; and
2. if frequency of the used modulation signal is f m = 1.0 kHz estimate the FM
signal bandwidth requirement.
10.9. For the given data in circuit Fig. 10.1 find value of capacitor C so that frequency
of the output waveform is f out = 3.5 MHz. Data: C T = 83.4 nF, L T = 20 nH, R =
100 φ, gm (M1 ) = 10 mS.
10.10. For phase modulation circuit, Fig. 10.2, that consists of an inductor in parallel
with series combination of capacitor C and varicap diode Cd0 , find the phase deviation
constant K . Data: Q = 70, C = 10 Cd0 , V0 = 15 V.
Chapter 11
Signal Demodulation: Problems
Modulated waveform must be received and the message recovered. The message
recovery in AM communication systems requires a circuit that is capable to extract
the carrier envelope. A diode–capacitor based circuit known as envelope detector (or,
peak detector) is essential part of both AM and FM receivers, as well as many other
analog signal processing systems. In this chapter we familiarize with the envelope
detector operational principle and operation.
Problems:
11.1. With reference to the simplified schematic Fig. 11.1 and if R = 2 kω, estimate:
1.
2.
3.
4.
5.
Detector input impedance;
Total signal power delivered to the detector;
Voltages v0 (max), v0 (min), and V0 (DC);
Average output current I0 (DC); and
Appropriate capacitor value C to prevent diagonal clipping distortion for maximal
modulation frequency f m (max) = 5 kHz and maximal modulation index m a =
0.9.
11.2. Assume that the input signal Vin of an AM diode detector, Fig. 11.2, is a 665 kHz
IF carrier modulated with a 5 kHz tone. Component values are: C1 = 220 pF, C2 =
22 pF, R1 = 470 ω, R2 = 4.7 kω, R L = 50 kω. Diode I D versus VD characteristics
is shown in the graph.
1. Sketch qualitatively the detector output tones along an λ axis showing relative
amplitudes of the tones;
2. Sketch AM waveform shape at nodes 1–5;
3. Sketch equivalent circuit at 5 kHz. Calculate amplitude ratio of the input signal
and signal at node 3;
4. Sketch equivalent circuit at 665 kHz. Calculate amplitude ratio of the input signal
and signal at node 3;
5. Briefly comment on the above results.
R. Sobot, Wireless Communication Electronics by Example,
49
DOI: 10.1007/978-3-319-02871-2_11, © Springer International Publishing Switzerland 2014
50
11 Signal Demodulation: Problems
Fig. 11.1 Problem 11.1: simplified schematic of peak detector
Fig. 11.2 Problem 11.2: simplified schematic of peak detector, and voltage current characteristics
the diode
Fig. 11.3 Block diagram for Problem 11.3
11.3. Voltage signal received by a 50 ω antenna has amplitude of 10 µV. Gain
contributions are noted next to each block of the system (see Fig. 11.3). Estimate:
1. input signal power in W and dBm units,
2. power delivered to the speaker in dBm and W.
11.4. A signal waveform used to modulate RF carrier has symmetrical triangular
shape with zero DC = 1 V component and amplitude of Vb = 2 Vpp while the
carrier wave has amplitude of VC = 2 Vp . Sketch these two waveforms before and
after the mixer circuit. Estimate modulation index m by inspection of the plots.
11.5. For unmodulated signal, AM current in the antenna is I0 = 1 A, while sinusoidal modulation wave causes the antenna current to be Im = 1.1 A. Calculate the
modulation index m if the antenna impedance is R.
11 Signal Demodulation: Problems
11.6. An FM signal S FM (t) is described as
S FM (t) = 2,000 sin 2π × 108 + 2 × π × 104 cos (π × 104 t) t
as it arrives at a 50 ω antenna.
Determine:
1.
2.
3.
4.
5.
6.
the carrier frequency,
the transmitted power,
modulation index m f ,
the message signal frequency f b ,
required signal bandwidth B by using Bessel functions (not Carson’s rule),
power in the largest and smallest sidebands as predicted by Bessel functions.
51
Chapter 12
RF Receivers: Problems
System level analysis of an RF receiver includes characterization of its components,
the internal blocks, tuneability range, image frequencies, signal to noise ratio, and
dynamic range. In this chapter we practice some of the key system level concepts.
By all means, the reader is advised to take other textbooks and, with the acquired
introductory knowledge from this book, to keep developing knowledge of RF communication circuits and high frequency concepts and design techniques.
Problems:
12.1. Sketch a block diagram of a heterodyne AM receiver, and illustrate waveform
shapes at output of each stage. The transmitted message waveform is shown in
Fig. 12.1.
12.2. An AM receiver is designed to receive RF signals in the in the 500 kHz to
1600 kHz frequency range with the required bandwidth of B = 10 kHz at the resonant
frequency f 0 = 1050 kHz. The RF amplifier uses inductor L = 1 µH.
1. calculate bandwidth at f max = 1600 kHz and capacitance C;
2. calculate bandwidth at f min = 500 kHz and capacitance C; and
3. comment on the results.
12.3. An AM receiver is designed to receive RF signals in the f min = 500 kHz to
f max = 1600 kHz frequency range. All incoming RF signals are shifted to intermediate frequency f I F = 465 kHz. AM receiver tuning is commonly done by a knob
that simultaneously tunes resonating capacitors in the RF amplifier and LO oscillator
sections. For the receiver architecture in Fig. 12.2 (matching network not shown),
1. calculate tuning ratio C R F (max)/C R F (min) of the resonator capacitor in the RF
amplifier,
2. calculate tuning ratio C L O (max)/C L O (min) of the resonator capacitor in the
local oscillator LO,
3. recommend the resonating frequency for the local oscillator.
R. Sobot, Wireless Communication Electronics by Example,
53
DOI: 10.1007/978-3-319-02871-2_12, © Springer International Publishing Switzerland 2014
54
12 RF Receivers: Problems
Fig. 12.1 Problem 12.1: time
domain plot of baseband
modulating signal
Fig. 12.2 Problem 12.3: AM
receiver block diagram
Fig. 12.3 Problem 12.5:
input–output characteristics
of an RF amplifier
12.4. The LO oscillator frequency is 11 MHz, and RF signal frequency is 10 MHz.
What is the image frequency?
12.5. Input–output power characteristics of an amplifier is given in Fig. 12.3. Estimate
the gain, 1dB compression point, and the third order intercept point IIP3.
12.6. A receiver whose IF frequency is f I F = 455 kHz is tuned to an RF signal
with f R F = 950 kHz. No other transmission frequency is allowed within the RF
band f R F = 950 kHz ± 10 kHz. However, for the sake of argument, let us imagine
existence of a nearby non–linear transmitter whose emitting frequency spectrum
consists of its both first and the second harmonics.
Aside from the obvious 950 kHz ±10 kHz frequency range, what other frequency
range(s) should also be prohibited for the external transmitter?
12 RF Receivers: Problems
55
Fig. 12.4 Problem 12.9: block diagram of an FM transmitter
12.7. Medium wave AM transmitters operate in the 540–1610 kHz range with 10 kHz
spacing while using 455 kHz IF frequency. Estimate the range of the local oscillator
frequencies and suggest bandpass filter(s) suitable for use with AM medium wave
receiver?
12.8. A double conversion receiver architecture is based on two IF frequencies,
I F1 = 10.7 MHz and I F2 = 455 kHz. If the receiver is tuned to a f R F = 20 MHz
signal, find frequencies of the local oscillators and the internal frequencies.
12.9. An FM transmitter, Fig. 12.4, used the local VCO controlled by audio signal to
generate waveform whose frequency is centred at f0 = 3.5 MHz. The oscillator shifts
its frequency by ωf 0 = ±1.6 kHz when the audio signal amplitude is Vin = 3.6 V pp .
Each FM station is assigned a frequency channel that is B = 150 kHz wide.
At the point of antenna find: (a) the carrier rest frequency f c ; (b) the carrier
frequency deviation ωf c ; (c) the FM modulation percentage; and (d) peak–to–peak
voltage of the message signal Vin needed to cause 100 % of modulation.
12.10. An FM signal whose centre frequency is f 0 = 200 kHz and its frequency
deviation is ωf 0 = ±200Hz is passed through two frequency multipliers, Fig. 12.5.
The two newly created waveforms are then added together by a summing block.
Signal coming out of the summing block is passed through LC resonator whose
centre frequency and bandwidth are aligned with the lower–side–band (LSB) part of
the output spectrum. At the output of the LC resonator, power of the LSB waveform
is measured as Pout = 1 mW at the edges of its bandwidth.
Assuming an ideal system:
1. calculate f 1 ± ωf 1 ;
2. calculate f 2 ± ωf 2 ;
Fig. 12.5 Problem 12.10:
block diagram of an RF
section
56
12 RF Receivers: Problems
Fig. 12.6 Problem 12.11: simplified schematic diagram of a three stage RF amplifier
3. calculate value of Q factor in the LC resonator;
4. sketch detailed plot of the output power spectrum. For the vertical axis use [dBm]
units, and for the horizontal axis use [Hz] units. Clearly show power levels of all
relevant tones.
12.11. Conceptual and simplified schematic diagram of a three stage RF amplifier
is in Fig. 12.6. Power level Pin of the RF source signal that corresponds to the 1dB
compression point, signal–to–noise ratio SNR measured at the internal node 1i,
signal power level P2 measured at the internal node 2i, and the current i out of the
signal delivered to the load R L are shown explicitly. In addition, bandwidth of the
LC resonator is B is set by the design of RF amplifier’s LC resonator.
1. Data: R L = 100 λ, temperature is T = 26.85 √ C, power gain of the first stage is
A1 = 10 dB, noise figure of the second stage is N F2 = 6.02 dB, and noise figure
of the third stage is N F3 = 9.0309 dB,
2. Assumptions: the total input side noise is generated only by the equivalent input
resistance Rin while each gain stage generates its own noise, the load resistor
is ideal, all base currents are ignored. Using the given data and assumptions,
estimate:
a.
b.
c.
d.
power gain of the second stage only A2 in [dB];
noise voltage vn at the output;
noise figure NF of this amplifier;
dynamic range DR of this amplifier.
12.12. A cell phone receiver operates at room temperature, T = 16.85 √ C, and it
has the following specifications: noise figure NF = 20 dB, bandwidth B = 1 MHz,
signal to noise ratio SNR = 0 dB. Also, non–linearity of the receiver circuit can be
described by the following output signal y(x) function:
y(x) = 2x − 0.267x 3
12 RF Receivers: Problems
57
which applies to the desired incoming signal x1 = 0.1 V sin ω1 t. At the same time
when the x1 signal is being received, there is another cell phone nearby transmitting
a signal whose function is x2 = A2 sin ω2 t, i.e. (A2 → 0.1 V). For the described
situation calculate:
1.
2.
3.
4.
the 1dB compression point in [dB];
the input intercept point IIP3 in [dB];
receiver sensitivity S in [dBm];
estimate maximum dynamic range DR in [dB], and effective dynamic range DR
in [dB] (assuming 2/3 of the maximum DR);
5. amplitude of the second signal A2 in [V] that would cause the “signal blocking”
effect. In that case it is assumed that output of the receiver, when both x1 and x2
signals are present, can be described as
3
2
y(x1 , x2 ) ∞ a1 + a3 A2 A1 cos ω1 t.
2
Part II
Solutions
Chapter 13
Introduction: Solutions
Examples of physical phenomena given in Chap. 1 are related to fundamental
concepts in physics related to energy, matter, EM waves, EM fields, propagation
of energy through matter and space, and basic interaction between two waveforms.
In this chapter we find detailed, tutorial like numerical solutions to the given problems as well as comments related to historical developments of the theory and some
of the relevant practical applications.
Solutions:
1.1. Einstein discovered the mass–energy equivalence concept that is famously
formulated as
E = (mc2 )2 + ( pc)2
(13.1)
where, m is the mass of object whose equivalent energy is calculated, c is the velocity
of light in vacuum, and p is momentum of the mass m in motion. However, if the
mass is not moving then p = 0 and (13.1) reduces into the well known form
E = mc2
(13.2)
This equitation is one of the fundamental natural principles, and it is used to describe,
for example, the hydrogen fusion process in the starts as well as the energy conversion
process in nuclear reactors. In this example we quantify amounts of energy potentially
available to us through exploits of this principle.
A snowflake consists of water molecules (H2 O), thus, first we find its total
mass (m S ) by adding masses of all n individual molecules that constitute the
snowflake. We already know that each water molecule consists of two hydrogen
atoms (atomic weight H = 1.00794 g/mol) and one atom of oxygen (atomic weight
O = 15.9994 g/mol). Thus, a single water molecule has atomic weight of:
R. Sobot, Wireless Communication Electronics by Example,
61
DOI: 10.1007/978-3-319-02871-2_13, © Springer International Publishing Switzerland 2014
62
13 Introduction: Solutions
H2 O = 2 × H + O = 18.01528 g/mol
(13.3)
or, if the mass is expressed in g then
H2 O =
H2 O
18.01528 g/mol
=
= 2.99151 × 10−23 g
NA
6.0221415 × 1023 g/mol
(13.4)
then the complete snowflake has mass m S of
m S = n × H2 O = 6.68559 × 1019 × 2.99151 × 10−23 g
= 2 mg = 2 × 10−6 kg
(13.5)
Velocity of the snowflake is negligible relative to the speed of light, which leads into
the equivalent energy of:
E = m c2 = 2 × 10−6 kg × 2.99792458 × 108 m/s = 1.79751036 × 1011 J
(13.6)
Power is defined as the energy transfer rate, therefore in order to provide the average
power of P = 25W the total energy E must be distributed over the following time
t=
1.79751036 × 1011 J
E
=
= 7.190 × 109 s √ 228 years
P
25 W
(13.7)
We conclude that if we were able to completely convert, for instance, only 2 mg of any
matter (including a snowflake) into energy we would be able to provide power to our
hand-held electronic equipment for many years. Following this line of thought, we
already developed nuclear power plants that are based on this fundamental principle
and used to provide power to whole cities. However, we still need to learn how (and
if) this principle could be scaled down and used to provide energy to our smaller
everyday’s gadgets, e.g. electronic equipment, houses, and cars.
1.2. For all practical purposes the Sun is a nuclear reactor that, from our perspective,
represents an infinite source of energy. The Sun energy is radiated continuously into
the space, and even though it enabled the life on Earth, huge percentage of this freely
available energy is still not harnessed by the humans. In this example we find how
much of this energy is really available and for how long it is expected to last.
We keep in mind that the solar constant k S is measured at the Earth distance from
the Sun, and that by definition it represents the radiated energy flux per time, i.e.
flow of energy per second as measured over a square metre surface. For purposes of
this problem, it us handy to first express the solar constant in terms of the converted
mass, as
k S = 1366
J/s
kg m/s2
kg
W
Nm
= 1366 3
= 1366 2 = 1366 2 = 1366
2
m
m
m s
ms
s
(13.8)
13 Introduction: Solutions
63
and to use it to find the equivalent amount of “mass flow” m → per square metre per
second, which is calculated from E → = m → c2 as
m→ =
E→
1366 kg/s3
−14 kg
=
2 = 1.51988 × 10
2
8
c
m2 s
m
2.997924580 × 10 /s
(13.9)
where m → and E → are respective mass and energy flow per metre square per second.
The calculated mass m → is equivalent to the radiated energy by the Sun per square
metre per second in all directions, as measured at 1AU distance (i.e. at the Earth,
because that is how k S is measured).
1. To find the total mass per second (M) that is being converted into energy, we
integrate m → over the spherical surface A whose radius is l = 1 AU , thus
M = A × m → = 4ω l 2 × m →
2
kg
= 4ω × 1.495978707 × 1011 m × 1.51988 × 10−14 2
m s
kg
(13.10)
= 4.27435 × 109
s
2. This total mass M is converted into energy that is being dissipated from the Sun
in all directions over every second. However, by looking from the Sun the Earth
is visible only as a circular “shield” whose diameter is r and whose area S is
2
S = ω × r 2 = ω × 6.3568 × 106 m = 1.26948 × 1014 m2
(13.11)
therefore, the energy reaching the Earth is equivalent to mass flow m E ,
m E = m → × S = 1.51988 × 10−14
kg
kg
× 1.26948 × 1014 m2 = 1.92946
2
m s
s
∴
mE √ 2
kg
s
(13.12)
which, in comparison with the total converted mass M = 4.27435 × 109 kg/s,
represents only a minor amount of about 4.5 × 10−8 % , i.e. for all practical
purposes it is negligible relative to the total converted mass being dissipated in
all directions of the space.
3. The hydrogen fusion process occurring in the stars takes several steps, however
in principle it involves conversion of four hydrogen atoms (or, equivalently, two
deuterium atoms) into one helium-4 atom plus some amount of energy. This extra
energy is direct consequence of the mass difference between the starting hydrogen
mass and the resulting helium mass. By experiments we found that atomic mass
of the four hydrogen atoms and one helium atom are respectively
64
13 Introduction: Solutions
H = 4 × 1.00794 g/mol ∞ 6.694896 × 10−27 kg
He = 4.002602 g/mol ∞ 6.646476 × 10−27 kg
therefore, the mass λm that is actually converted into energy is difference between
the ending and the starting mass, i.e.
λm = H − H e = 4.842 × 10−29 kg
(13.13)
which is relative to the mass of four H atoms as
R=
6.694896 × 10−27 kg
√ 138
4.842 × 10−29 kg
(13.14)
where, R is the ratio of the mass of 4H atoms initially available before the fusion
starts and the equivalent mass converted during the fusion process. In other words,
only 1/138 of the initial H mass is actually converted into the radiated energy
while the rest of the initial mass is still preserved in the form of helium, which
means that the Sun must start the fusion process with the total of
M H = R × M = 138 × 4.27435 × 109 kg/s √ 5.9 × 1011 kg/s
(13.15)
of hydrogen in order to sustain the solar constant.
4. The Sun’s current mass is estimated as m S = 1.9891 × 1030 kg, thus to convert
10% of m S at the estimated mass conversion rate M H it will take time t that is
found as
t=
0.1 × 1.9891 × 1030 kg
0.1 × m S
=
= 3.371 × 1017 s
MH
5.9 × 1011 kg/s
∴
t √ 10.7 × 109 years
(13.16)
5. The total power PS generated by the Sun is calculated by using the complete
surface A of the sphere surrounding the Sun with diameter 1 UI and the solar
constant, i.e.
PS = k S × 4ω × UI 2
2
= 1366W/m2 × 4ω × 1.495978707 × 1011 m = 3.842 × 1026 W
It is interesting to see how power generated by the Sun compares to the World energy
consumption. According to publicly available statistical data, in the year 2012 the
World energy consumption reached E = 550 × 1018 J, which is approximately
equivalent to
13 Introduction: Solutions
P=
65
550 × 1018 J
E
=
√ 17.3 × 1012 W
t
365 × 24 × 3600 s
(13.17)
In comparison, power from the Sun that is reaching the Earth (that its, only the Sun
facing side), after using (13.11), is found as
P = k S × S = 1366
W
× 1.26948 × 1014 m2 √ 173 × 1015 W
m2
(13.18)
which is to say that, if we could capture all power delivered by the Sun we would
have had approximately 10,000 times more power than what we currently need.
1.3. Maxwell formalized symmetrical relationship between spatially varying electric
and magnetic fields. That is, a changing electric filed is always accompanied with
the changing magnetic field. Through his work Maxwell also calculated the speed of
light and, therefore, concluded that light must be a form of electromagnetic radiation
that propagates through space with finite velocity. Important note about these two
fields is that they propagate together through space, always being in phase, and most
importantly once the energy started moving through the space it will keep moving
even if the energy source is removed. For instance, that is why we receive light from
the distant stars even though they may not be there anymore.
A common way to illustrate an EM wave is to show two sinusoidal waveforms
as being perpendicular and propagating in the same direction, one for its electrical
and one for its magnetic component, Fig. 13.1. If the waveform propagates in the
x-direction, then the electric vector E and magnetic vector B must follow rules for
the vector product E × B in the right-handed coordinate system, which then dictates
that the electric vector must be in y-direction, while the magnetic vector B must
point in the z-direction. It is important to notice that in this interpretation, Fig. 13.1,
even though only a single sinusoidal functions are shown in the x and y-directions,
the respective electric and magnetic wave components extend infinitely in these two
directions.
1.4. A dipole antenna is simplest geometrical form of antenna that is derived from a
capacitive structure. Therefore, it consists of two symmetrical conductive elements,
for example two simple metallic rods. Once the antenna is stimulated with an AC
Fig. 13.1 Problem 1.3:
illustration of an EM sinusoidal wave propagating in the
time-direction
66
13 Introduction: Solutions
Fig. 13.2 Problem 1.4: illustration of a three step capacitor morphing into dipole antenna
Fig. 13.3 Problem 1.4:
illustration of an EM wave
generated by a dipole antenna
signal generator source, it creates EM filed that keeps propagating through the space
at the speed of light even when the source is turned off.
In order to demonstrate how a dipole antenna structure is derived, let us take a
look at Fig. 13.2. We already know that a simple capacitor stimulated by AC signal
generator creates changing electric field in between the two capacitive electrodes,
Fig. 13.2 (left). However, even if the two electrodes are physically separated at the
end that is further away from the source, the electric field does not break, Fig. 13.2
(centre). Instead, once the two electrodes are vertically aligned with each other, the
field morphs into radial shape, Fig. 13.2 (right). In other words, we can imagine that
a dipole antenna is only a fully “open” capacitor.
As the AM current is continuously changing its direction along the antenna
electrodes (as shown by green arrow), the electric and magnetic fields also keep
changing their respective directions, Fig. 13.3. In this graphical quasi 3D representation of EM field moving radially away from the dipole antenna, we should actually
imagine a torus shaped electrical field lines orthogonally wrapped around circular
magnetic field lines. In reality, the whole space is filled with the alternating fields,
not only the indicated vector lines. In order to better visualize this dynamic process,
watch some of very good 3D animations available on the web that show how this
radiation pattern is created and how it propagates through the space.
1.5. In principle, there are at least two ways to estimate number of participants in
a parade marching along the street. We either take a fixed position at the street side
and watch the parade passing in front of our eyes while counting the participants in
13 Introduction: Solutions
67
Fig. 13.4 Problem 1.5: illustration of an EM sinusoidal wave and its space–time parameters
time, or we make aerial photo of the whole street and count the participants in space.
In the former case we keep fixed position of observation, while in the latter case we
keep fixed time of observation. We take similar approach if we want to observe a
wave propagation.
From a strictly mathematical point of view, equation E = E m sin(k z − ω t) is
function of both time and space coordinates, i.e. E = f (z, t). That is to say, by
setting one of the two coordinates constant we are able to observe how E changes
relative to the other coordinate. In the given equation, both the wave number k and
angular frequency ω can be also seen as “scaling” or “gain” parameters for their
respective coordinates.
Let us first consider spatial wave movement by “freezing” the time, i.e. by setting
t = const, and by observing only E = f (z). If that is the case, and if without loosing
generality we set t = 0, then
E = E m sin(k z)
(13.19)
which is a simple sinusoidal function in z direction, Fig. 13.4 (top). As for any other
sinusoidal function, it is easy to identify its period T . We only must keep in mind the
physical meaning of the coordinate z and visualize this sinusoidal function in space.
If that is the case then the spatial period T must be measured in units of meters, thus
to emphasize this detail instead of using the general letter T we use letter π when
referring to the spatial period. This graph is equivalent to a photo of the sinusoidal
function, i.e. it shows amplitude at various spatial points at the same moment in time.
It is useful to take a closer look at the spatial “gain” factor k. At the point in space
where one full spatial period π is measured, i.e when z = π (keep in mind that time
is constant and set to t = 0, and that a sinusoidal function has period of 2ω ), we
write
68
13 Introduction: Solutions
kz − ωt = 2ω
∴
kπ − 0 = 2ω
∴
k=
2ω
π
(13.20)
where we used the fact that a full period of a sinusoidal function is equivalent to 2ω
angular measure. In literature k is often referred to as the “wave number.”
Similarly, we can set the space coordinate to a constant, e.g. z = 0, and observe
time variant function
E = E m sin(−ω t)
(13.21)
Again, (13.21) represents a simple sinusoidal function whose initial phase is negative,
Fig. 13.4 (bottom). Because the point of observation is fixed, this graph shows how
amplitude of the sinusoidal function changes in time, i.e. it shows the amplitude at
various time points as seen from the same point in space. As we are familiar with
using time variable when describing sinusoidal functions, we keep using letter T do
denote period of this function and we measure it in seconds.
Again, it is useful to take a closer look at the time “gain” factor ω. At the moment
in time when one full time period T is measured, i.e when t = T (keep in mind that
space is constant and set to z = 0), we write
kz − ωt = 2ω
∴
0 − ωT = 2ω
∴
|ω| =
2ω
= 2ω f
T
(13.22)
where the “–” sign indicates the initial phase.
In both cases shown in Fig. 13.4 the wave propagates in positive directions of its
respective coordinates. Now, it is very useful to determine velocity of the wave’s
front-end propagation through space by using the well known classical Newtonian
equation for the relationship among distance z, time t, and velocity v, i.e.
z = vt
(13.23)
Because it takes time t = T to travel distance in space of one wavelength π, by
definition and after substituting (13.20) and (13.22) we write
v∞
π
2ω ω
ω
λz
=
= πf =
=
λt
T
k 2ω
k
(13.24)
With (13.24) we have a tool to calculate time–space coordinates of a propagating
wave whose frequency is f and the wave number k.
1.6. Theoretical and independent derivation of the speed of light from Maxwell’s
equations was one of the most convincing confirmations that light is indeed a form
of EM radiation. By following same basic idea used in Example 1.5 that gave us
connection between time and space propagation of waves, we can derive and exactly
calculate the speed of light starting from Maxwell’s classical equations assuming
vacuum and no free electric charges (φ = 0 and J = 0).
13 Introduction: Solutions
69
Starting with three-dimensional Maxwell’s equations in form of
≤ ×E=−
∂B
∂t
≤ × B = μ0 ε0
(13.25)
∂E
∂t
(13.26)
First, we find second derivatives of E and B by finding space derivative ∂/∂z of
(13.25) and then after substituting the space derivative into (13.26) we write
≤(≤ × E) = −
∂(≤ × E)
∂ 2E
= −μ0 ε0 2
∂t
∂t
(13.27)
At this point we need a theorem from calculus that states that
≤(≤ × E) = −≤ 2 E + ≤ · (≤ · E)
(13.28)
where, if φ = 0 then ≤ · E = 0, which further leads into one dimensional form of
the wave equation as
≤ 2 E = μ0 ε0
∂2 E
∂t 2
∴
∂2 E
∂2 E
= μ0 ε0 2
2
∂z
∂t
(13.29)
Now we can find one dimensional second derivatives in space and time as
∂E
= k E m cos(kz − ωt)
∂z
∂2 E
= −k 2 E m sin(kz − ωt) = −k 2 E
∂z 2
(13.30)
∴
∂E
= −ω E m cos(kz − ωt)
∂t
∴
∂2 E
= −ω2 E m sin(kz − ωt) = −ω2 E
∂t 2
(13.31)
therefore, the wave equation (13.29) becomes
−k 2 E = −μ0 ε0 ω2 E
∴
v∞
ω
1
=≥
k
μ0 ε0
=
(13.32)
1
4ω
× 10−7 N/A2
× 8.854, 187, 817 × 10−12 F/m
∴
v = c = 299, 792, 458
m
s
(13.33)
70
13 Introduction: Solutions
where, value of vacuum permittivity ε0 constant was independently established
experimentally by measuring electrostatic force between capacitor plates. However,
value of magnetic constant (vacuum permeability) in SI system was postulated as
μ0 = 4ω × 10−7 N/m to help define the unit for electric current . Therefore, in SI
units, the speed of light is c = 299, 792, 458 m/s.
1.7. As a side note, starting with one dimensional form of (13.25), which is simply
∂B
∂E
=−
∂z
∂t
(13.34)
after substituting the two derivatives into (13.34), we write
k E m cos(kz − ωt) = ω Bm cos(kz − ωt)
∴
Em
ω
= =c
Bm
k
(13.35)
which gives us another interesting relationship that involves EM field components
and the speed of light. Thus, knowing amplitude of one EM wave component we
easily calculate amplitude of the second.
1.8. We continue discussion started in Problem 1.7, by saying that magnitude of the
S = 1/μ0 (E × B) vector product is found as
|S| =
1
1
1
1 E m2
|E × B| =
=
E m Bm =
c Bm2
μ0
μ0
μ0 c
μ0
∴
S=
1
N
A2
V 2
W2
V2 A 2 s
m
=
=
m
N m3
m
s
(13.36)
that is, this vector product has the unit of power per square metre, thus it represents
density of the energy being transferred. In addition, by applying the right hand rule
for the vector product, we find that the resulting vector S points in the same direction
as the wave velocity. In other words, it points in direction of the energy flow. We
already encountered the special case of this vector in Problem 1.2 when we called
it the solar constant. The energy flow vector is very important parameter because
it not only quantifies the amount of energy but also contains information about the
flow direction. In the literature, vector S is referred to as Poynting vector and it is
explicitly used for the purpose of estimating an EM wave energy flow.
1.9. Now that we are familiar with the nature of Poynting vector and its relationship
with the solar constant, we can perform various calculations, including calculations
of EM wave component amplitudes as the wave arrives to the Earth.
Electrical and magnetic field components of EM radiation generated by the Sun
are given as E = E m sin(ky −ωt) y, and B = Bm sin(kz −ωt) z. Therefore, average
13 Introduction: Solutions
71
value of Poynting vector ∗S∓ is calculated as
∗S∓ =
1
c Bm2
1 c Bm2
∗E × B∓ =
∗sin2 (kx − ωt)∓ x =
x
μ0
μ0
2 μ0
∴
c Bm2
⇒
2 μ0
2 μ0 k S
2 × 4 ω × 10−7 N/A2 NA2 × 1366 W/m2
=
Bm =
= 3.384 µT
c
2.99792458 × 108 m/s
kS =
∴
E m = c Bm = 2.99792458 × 108 m/s × 3.384 × 10−6 T = 1.014 × 103
V
m
1.10. Afer realizing that energy density can be also expressed as a function of
pressure, the whole discussion about EM waves becomes really interesting. Electromagnetic waves are assumed to be a non-material form of existence, which is defined
through very abstract concept of energy. At the same time, one of the implications of
Poynting vector is that interaction of EM wave with material object results in physical pressure. This statement by itself sounds almost as science fiction. However,
as we will see shortly, this concept of non-material with material interaction opens
completely new ways of thinking about, for instance, propulsion systems in space.
To see how the pressure comes into the equation, let us take another look at
Problems 1.8 and 1.8, where we already established the nature of Poynting vector
∗S∓, and rearrange the measuring units as
∗S∓ =
c Bm2
2 μ0
(13.37)
∴
∗S∓ ∼
∴
P∼
W
J s−1
N m s−1
N m
=
=
= 2
2
2
m
m
m2
m s
N
W m −1
=
m2
m2 s
(13.38)
By simple rearrangement of (13.37) to match the units in (13.38) we write expression
for EM wave pressure as
P=
B2
∗S∓
= m
c
2 μ0
72
13 Introduction: Solutions
Fig. 13.5 Solar sail concept
which can be then interpreted as the pressure on both sides of a surface immersed in
the EM field. However, if the EM wave is perfectly reflected, then the wave pressure
is distributed over only one side of the surface, that is, the pressure on that single
side increases by the factor of two (because area of two sides of a surface is twice
the area of one side). Thus, we conclude that radiation pressure Pr on one side of a
perfectly reflective surface must be
Pr = 2 P =
B2
2 ∗S∓
= m
c
μ0
N/m2
which is to say that by knowing either of the two components of EM wave we are
able to calculate mechanical pressure caused by the light. Practical consequences of
this conclusion we illustrate in the following examples.
1.11. First, Maxwell’s theory predicted the light pressure, then in 1899 Pyotr Lebedev
experimentally demonstrated the concept. Thereafter, a number of science fiction
writers used the concept, most notably Arthur Clarke in his 1964 novel “Sunjammer”,
which inspired naming of the next solar sail experimental space flight demonstration
planned for 2014. In this example we try to numerically illustrate feasibility of the
concept (Fig. 13.5).
In order to neutralize gravitational pull from the Sun, force produced by solar
pressure must be at least as equal. From the basic physics we know that pressure P
is defined as force F over the surface A, thus radiational force Fr is expressed as
Fr = Pr A. In addition, gravitational force between two masses is defined by the
Newton’s gravitational law. In addition, assuming a space probe of mass m, the Sun
mass m S , area of the solar sail A, and the distance of the probe from the Sun a, we
write expression for the balance between radiational and gravitational forces as
G
2S
mS m
A
≤ Fr = Pr A =
2
a
c
We already know that Poynting vector ∗S∓ represents power per surface, therefore
the total power of the Sun PS is distributed over spherical surface 4 ω a 2 . Thus,
13 Introduction: Solutions
73
G
mS m
2 PS
≤
A
2
a
c 4 ω a2
which, for the given numerical data, leads into
c 4 ω G mS m
2 PS
3 × 108 m s−1 × 4 ω × 6.674 × 10−11 N m2 kg−2 × 1.989 × 1030 kg × 2 kg
=
2 × 3.842 × 1026 W
3 2
A ≥ 1.302 × 10 m
A≥
This solar sail can be made of a square shaped area whose side is 36.079 m. Sail of
this size must weight
m k = m k0 A = 768.5 µg/m2 × 1.302 × 103 m2 = 1 kg
(13.39)
therefore, mass of the useful payload is 2 kg − 1 kg = 1 kg.
1.12. Waves are described by ordinary trigonometry equations, therefore in our
analysis of multi-tone waves we can assume that they have been created by superposition (i.e. interference) of two or more single-tone waves. Consequently, the overall
analysis is simplified because more complicated equations are derived by following
the elementary trigonometry algebra rules. While keeping in mind the two space–
time variables, there is a number of interesting situations that arise that depend on
the phase relationship between the two super-positioning waves. In this example
introduce the concept of “standing waves”, which is mathematically described as the
superposition of two waves traveling in opposite directions.
In this case, the electric filed is described as if it were composed of the sum of two
waves, one traveling in positive z-direction, and one traveling in negative z-direction.
Therefore, we write
E = E m (cos(kz − ωt) − cos(kz + ωt)) y = 2E m sin(kz) sin(ωt) y
B = Bm (cos(kz − ωt) + cos(kz + ωt)) z = 2Bm cos(kz) cos(ωt) z
(13.40)
(13.41)
Then, by definition,
1
E×B
μ0
2E m
1
cos(kz) cos(ωt) z
=
(2E m sin(kz) sin(ωt) y) ×
μ0
c
S=
=
E m2
sin(2kz) sin(2ωt) x
c μ0
and now we can calculate average value ∗S∓ by definition,
(13.42)
74
13 Introduction: Solutions
T
1
∗S∓ =
S dt
T 0
T
E m2
1
=
sin(2kz)
sin(2ωt) dt x
c μ0
T 0
E m2
T =2ω/ω
sin(2kz) cos(2ωt)|0
=−
x
c μ0 2ω T
=0
(13.43)
Let us take a closer look at the above results. Equation (13.40) shows that the
electric field becomes zero when sinkz = 0, in other works when
kz = nω
∴
z=
nω
π
nω
=
=n
2ω
k
2
π
(13.44)
where, n = 0, 1, 2, . . .. As the time advances, these points in space are not moving.
Similarly, the points of maximum amplitude E max = 2E m are also not moving in
space. We conclude that at distances equal to multiples of the half–wavelength the
total amplitude of the electrical component is zero. At the same time, from (13.41)
we see that Bm = 0 if coskz = 0, i.e.
1
kz = n +
ω
2
∴
z=
n
1
+
2 4
π
(13.45)
where, n = 0, 1, 2, . . .. Again, as the time advances, at these points in space the
corresponding total amplitude of the magnetic field is always zero, which coincides
with the maximum amplitudes of the total electric field.
Result found in (13.43) is also very interesting. We already concluded that
amplitude of Poynting vector represents rate of energy flow in the vector’s direction, thus saying the ∗S∓ = 0 is to say that the energy of the wave used in this
example does not propagate in space. This case is known as a “standing wave”,
it is very important and it can be created, for example, by using two ideal mirrors
positioned exactly at the distance equal to an integer number of wavelengths that
effectively create a resonant cavity, Fig. 13.6. The wave energy is oscillating, however it is confined within the cavity. For example, this phenomena forms the basis
for design of lasers. It is important to us to recognize that because standing waves
do not propagate in space, they can not be used transmission of radio signals. Therefore, we must design transmission circuits so that reflection, and therefore creation
of standing EM waves, is avoided.
1.13. We have already learned that SI unit for electric field E is V/m, while SI unit
for magnetic filed H is A/m. It is then easy to realize that the ratio between the two
fundamental fields is
13 Introduction: Solutions
75
Fig. 13.6 Problem 1.12: example of a standing wave formed between two perfectly reflective
surfaces at distance of d = 3/2π
E
∼
H
V
m
A
m
=
V
=α
A
(13.46)
which is, naturally, known as the free space impedance Z 0 . We keep in mind that in
free space B = μ0 H (i.e. for all practical purposes they are interchangeable), thus
after substituting (13.32) and (13.35), we derive several alternative expressions of
the free space impedance. Relative permeability μr and permittivity εr of free space
are defined as unity, while free space permeability μ0 and permittivity ε0 are defined
relative to the free space, resulting in the intrinsic impedance as
E
=c
B
∴
∴
Z 0 = μ0 c =
E
=c
μ0 H
1
ε0 c
=
∴
E
= μ0 c
H
μ0
= 4ω c × 10−7 m/s √ 120 ω √ 377 α
ε0
(13.47)
By definition, phase velocity was derived from (13.34) in (13.35), Therefore, the EM
wave phase velocity in free space is, obviously, the velocity of light
νp = c =
π
1
ω
=
=π f = ≥
k
T
μ0 ε0
∴
ν p = 2.99792458 × 108 m/s √ 300, 000 km/s
The wavelength then is calculated as
(13.48)
76
13 Introduction: Solutions
⎧
⎪
⎨30 m for f 1 = 10 MHz
νp
2ω ν p
π=
=
= 3m
for f 2 = 100 MHz
⎪
f
ω
⎩
3 mm for f 3 = 10 GHz
(13.49)
As this example shows, wavelength of EM waves oscillating at some of the commonly
used frequencies are comparable with physical size of our equipment. That fact is
important for design engineers when considering, for instance, the antenna size (e.g.
the dipole length), or the length of conductive wires on printed circuit boards (PCB).
As the EM wave frequency increases, its wavelength becomes comparable first with
the size of PCB, then with the sizes of discrete components themselves. We note that
at typical RF signal frequencies used in this book (i.e. f = 10 MHz), size of PCB
(which is in order of several centimetres per side) is much less then the estimated
30 m calculated wavelength of a 10 MHz signal. Therefore, the use of approximated
Maxwell’s equations is justified.
On the other hand, if the signal frequency is in order of multiples of GHz, even
integrated circuit (IC) designs (which typically occupy only a few millimetres per
side) must account for the phase differences along the signal path and use the transmission line theory in the design circuit calculations.
It is now good time to comment on the terms of “high frequency” (HF) and “low
frequency” (LF). As we see, the high frequency term is relative to the length of
the conductive medium (e.g. wire). If along the conductive path we can measure
significant variation of potential, or even full positive–zero–negative swing, then
we say that the signal frequency is high. However, if along the conductor length we
measure only negligible variations in the potential, we say that the signal frequency is
low and that the conductor represents good approximation of an equipotential surface.
It should be obvious that even 60 Hz signal (π =) is considered “high frequency” if
transmitted over distances of several thousands kilometres. Naturally, there is wide
“grey area” between HF and LF regions, thus it is up the designer to decide what
approximation is acceptable for the particular design.
1.14. In this example we revisit applications of Ampère’s law by looking at magnetic
field distribution within and outside of a cylindrical conductive wire. Straightforward
implementation of Ampère’s law,
d
H · d l = Ifree,enc +
dt
L
∂D
∇ × H = Jfree +
∂t
D · ds integral form
(13.50)
differential form
(13.51)
S
for case of constant current I (t) = const and all the charges contributing to the
current flow, the current density J is uniform through any cross-section area of radius
r inside the conductor, up to the radius a at the conductor’s surface. Hence, portion
of the total current Ir flowing thorough any area with radius r (0 ≤ r ≤ a) inside
the conductor is determined by the ratio between the full cross-section area ωa 2 and
the inside area ωr 2 , i.e. Ampère’s law can be written as
13 Introduction: Solutions
77
180
160
H, A/m
140
120
100
80
60
40
20
0
0
5
10
15
20
25
30
r, mm
Fig. 13.7 Problem 1.14: Magnetic field distribution inside and outside of an infinitely long wire or
radius a = 5 mm carrying a current of 5 A
H · 2ωr =
I
ωr 2
ωa 2
∴
H=
Ir
2ωa 2
(13.52)
where 0 ≤ r ≤ a. Outside of the conductor the current density is equal to zero, which
simplifies (13.50) (or (13.51)) so that the magnetic filed H outside of the conductor
is calculated as
H · 2ωr = I
∴
H=
I
2ωr
(13.53)
where r ≥ a. The total magnetic field outside and inside of the infinitely long
conductive wire is
⎧
Ir
⎪
⎪
= 31.810 × 103 r A/m
r ≤ 5 mm
⎪
⎨ 2ωa 2
H (r ) =
⎪
⎪
⎪
⎩ I = 0.798 A/m
r ≥ 5 mm
2ωr
r
It is important to notice that the magnetic field is affected by the surrounding material,
it increases linearly inside the conductor because more current contributes to the
magnetic field, while outside of the wire the magnetic field strength is inversely
proportional to the distance because the whole current has been accounted for and
there is no external contributors to the field, Fig. 13.7.
1.15. Free charge inside a time varying magnetic field B start to move in space,
which is by definition an electric current. If the charge is contained inside a metallic
conductor, then the moving charges create induced electrical field, which by definition creates voltage potential along a path. The newly induced potential is equal
78
13 Introduction: Solutions
Fig. 13.8 Problem 1.15: the
time rate of change of the
magnetic flux density induces
a voltage
to the loop line integral of the electrical filed E around the conductor. It the same
time, this potential is also connected with the B and its flux calculated across the
associated surface. Straightforward implementation of Faraday’s law
d
E · dl = −
B · ds integral form
dt S
L
∂B
∇×E=−
differential form
∂t
(13.54)
(13.55)
results in expression for potential V as
d
d
V =−
E · dl =
B · ds =
μ0 H · ds
dt S
dt S
L
d
d
=
μ0 H0 cos(ω t)n · ds = μ0 H0 cos(ω t) ωa 2 n
dt S
dt
= −ωa 2 ωμ0 H0 sin(ω t)
= −0.031 sin(2ω × 108 t) V
where, B = μ0 H is the magnetic flux density, n is the unity vector in the same
direction as magnetic field H vector. Maximum amplitude of the induced potential
is approximately 31 mV when measured at the ring terminals.
This problem example is a typical demonstration of Maxwell’s equation in a form
of Faraday’s law, also known as the transformer law, where the time-varying magnetic
field induces a voltage response in a conductive loop located nearby, Fig. 13.8. The
voltage induction principle is one of the most important discoveries that enabled
radio communication, electric motors, and generation of electric energy.
1.16. Low frequency approximation is a very handy tool in cases when wavelength
of of a signal is much longer than length of the conductive media (e.g. a wire) that is
used to transmit the said signal. In the low frequency approximation the assumption
is that the conductive surface is equipotential, i.e. at any given moment in time the
potential is equal at all points in space. Or, in other words the conductor is ideal, that
is without any parasitic RLC components. However, when the signal wavelength is
comparable with the conductor’s length non equipotential surfaces become reality.
In Problem 1.13 we could see a range of wavelengths used in telecommunications,
13 Introduction: Solutions
79
Fig. 13.9 Problem 1.16:
section of a long conductor
(relative to the π) located
between z and (z + λz)
while in this example we take a first look at what is known as transmission line
model. A long connector is cut into infinitesimally short pieces, and each section
is modelled as RLC network, Fig. 13.9. We keep in mind that in this case units are
normalized per length.
With the help of Kirchhoff’s voltage law around circuit loop that includes V (z)
and V (z + λz), Fig. 13.9,
V (z + λz) − V (z) = −(R + jωL) I (z) λz
(13.56)
at the same time, by definition, potential difference across the conductor section λz
is written as
V (z + λz) − V (z)
d V (z)
= lim −
(13.57)
−
λz∼0
dz
λz
which, after substitution of (13.56) and after applying the limit λz ∼ 0 leads into
−
d V (z)
= (R + jωL) I (z)
dz
(13.58)
a we derive expression
Similarly, applying Kirchhoff’s current law to the node for change of the current as function of the wire length as
I (z + λz) = I (z) − V (z + λz)(G + jωC) λz
(13.59)
and, by definition, we write
I (z + λz) − I (z)
d I (z)
= lim
λz∼0
dz
λz
∴
d I (z)
= −(G + jωC) V (z)
dz
(13.60)
Equations (13.58) and (13.60) are coupled first-order differential equations. Solution
to this system of equation is found by decoupling the two equations, which is
accomplished by differentiating both sides, first (13.58) and then (13.60). By doing
80
13 Introduction: Solutions
so, explicit solutions for V (z) and I (z) are found. Therefore, starting with spatial
differentiation of (13.58) and substituting (13.60),
−
d 2 V (z)
d I (z)
= (R + jωL)
dz
dz
∴
d 2 V (z)
= (R + jωL)(G + jωC) V (z)
dz
d 2 V (z)
− (R + jωL)(G + jωC) V (z) = 0
dz
d 2 V (z)
− k 2 V (z) = 0
dz
(13.61)
where, a complex propagation constant k is defined as
k = (k) + j(k) =
(R + jωL)(G + jωC)
(13.62)
and it is function of the transmission line geometry (keep in mind that the R, L, C,
and G are all distributed parameters, i.e. functions of length, calculated separately
for specific shape of the conductor).
Repeating the same procedure, while starting with spatial differential of (13.60)
instead, similar solution to (13.61) is found for the current spatial dependance, i.e.
d 2 I (z)
− k 2 I (z) = 0
dz
(13.63)
Equations (13.61) and (13.63) show explicitly, for a given moment in time, the
voltage/current spatial dependance along the transmission line. Solutions to these two
decoupled equations are already well known to be in the following form (assuming,
of course, that the transmission line is aligned with the z-axis):
V (z) = V + e−kz + V − e+kz
+ −kz
I (z) = I e
− +kz
+I e
(13.64)
(13.65)
As a convention, each of these two equations is interpreted as combination of two
waveforms, one propagating in the positive z-direction, while the second one propagates in the negative z-direction.
The two equations (13.64) and (13.65) are correlated because they describe the
same waveform, which means that they are connected through the transmission line
impedance. For example, substituting (13.64) back into (13.58) results in the explicit
relationship between the voltage V (z) and current I (z) as
13 Introduction: Solutions
−
81
d V (z)
= (R + jωL) I (z)
dz
∴
1
d V (z)
I (z) = −
(R + jωL) dz
d(V + e−kz + V − e+kz )
1
=−
(R + jωL)
dz
k
+ −kz
− V − e+kz )
=
(V e
(R + jωL)
k
V (z)
=
(R + jωL)
(13.66)
which is to say that the expression connecting current I (z) and voltage V (z) must be
impedance, and because it is important parameter of a transmission line it is named
characteristic line impedance Z 0 , i.e. after substituting (13.62)
(R + jωL)
V (z)
=
=
Z0 =
I (z)
k
(R + jωL)
(G + jωC)
(13.67)
which, in ideal lossless case, i.e. no thermal dissipation R = G = 0, simplifies as
Z0 =
L
C
(13.68)
Characteristic impedance of a lossless transmission line (13.68) is not function of
frequency. That fact should be contrasted to the general definition (13.67) that is a
complex quantity and does take into account always present thermal losses (but not
inter-component interference). However, the characteristic impedance is very strong
function of the line geometry (through the distributed values L and C), hence must be
calculated for each type of transmission line, for example for two-wire line, coaxial
line, parallel-plate line, etc.
Transmission line model emphasizes the fact that waves propagate through the
media with finite velocity, which causes different potentials between points where
the wave has “arrived” and the points that are still “waiting” for the waves to arrive.
Thus, the importance of a transmission line characteristic impedance is in the fact
that, from strictly electrical perspective, each section of the line represents resistance.
Therefore, any subsequent section or component that is connected to a particular
section should have the same impedance for maximum power transfer, otherwise,
at the interfacing point some of the transmitted energy will be “reflected back” and
perceived as the signal loss. Calculation of the reflection coefficient at the interface
of two impedances gives us very useful design parameter.
For the given numerical example, we calculate
82
13 Introduction: Solutions
Z0 =
L
=
C
378 nH/m
= 50.2 α
150 pF/m
(13.69)
which implies that this two-wire connection it is a good choice as the interface to a
50 α antenna, or a 50 α input of an oscilloscope.
1.17. Instantaneous values of a sinusoidal wave are very often expressed in their
respective interchangeable units. For example, inquiring about a wave’s instantaneous phase in radians is equivalent to asking about location of the wave’s front end
in time. A fine point to note, though, is that it is customary that phase of a periodic
waveform is folded back into the first period, while the wave’s front end position is
expressed in absolute time. Similar comment applies when comparing the waves’s
wavelength and the wave’s front end position in space. In this example we illustrate
these relations by using example of a waveform whose frequency is f = 100 Hz.
By observation of the given waveform expression
v(t) = Vm cos (ω t + φ0 )
(13.70)
where, radial frequency was given as ω = (2ω 100) rad/s and the initial phase φ0 =
then by definition it follows that
ω/4,
ω = (2ω 100) rad/s
⇒
f = 100 Hz
∴
1
T =
= 10 ms
f
∴
π = c T = 2.99792458 × 108 m/s × 10 ms √ 3, 000 km
where correlation between frequency and the wavelength is achieved through velocity
of light c and the wave period T . Similarly, correlation between phase and the absolute
time is established by using relationship for radial period
T = 2ω
∴
2ω ∞ 10 ms
∴
15 ms ∞ 3ω =
12ω
4
which is phase of a waveform whose initial phase is φ0 = 0, for example light grey
sinusoidal waveform in Fig. 13.10. However, in this example the zero moment in
time is associated with the initial phase of φ0 = ω/4. Thus, at the moment t = 15 ms
the phase is shifted to
φ=
ω
13ω
12ω
+ =
4
4
4
13 Introduction: Solutions
83
Fig. 13.10 Problem 1.17: unit conversion
As already mentioned, it is customary to express all phase results relative to the
period, i.e. after removing integer multiples n × 2ω , (n = 0, 1, 2 . . .), which implies
that phase takes only values within 0 ≤ φ ≤ 2ω range, Fig. 13.10, which in this
example means that φ = 5ω/4.
Instantaneous voltage amplitude relative to the maximum amplitude Vm at t = 0 s
is calculated directly from (13.70) as
Vm
V = Vm sin φ0 = ≥
2
while at t = 15 ms we find, for example, by inspection of Fig. 13.10 that
V = Vm sin
Vm
13ω
= −≥
4
2
Familiarity with various views and units related to sinusoidal waveforms are very
important for practicing engineers. As a side note, we observe huge value of wavelength for a 100 Hz waveform. Indeed, waveforms of LF signal are comparable with
the Earth’ dimensions. Knowing that practical length d of an antenna intended for
reception/transmission of a signal should be comparable with the signal’s wavelength
(for example, d = π/4), it should be obvious that wireless transmission of LF signals
is simply not practical. That is, we already use long-distance cables as telephone
lines laid in trenches, instead of using them as the antennas.
1.18. Fourier gave us a tool that is equivalent to X-ray generator used by medical
professionals. The X-ray machine enables us to see inside of a living being, without
the need to physically open the body. Similarly, Fourier transform and polynomials
enable engineers to “see” inside a complicated waveform and to find what is was
made from. Picture that shows content of a waveform is called “frequency spectrum”
13 Introduction: Solutions
1.0
1.0
vΣ10(t), vHP(t)
vΣ3(t), v1(t), v3(t), v5(t)
84
0.5
0
-0.5
vΣ3
v1
v3
v5
-1.0
0
0.5
0
-0.5
vHP
vΣ10
-1.0
0.5
1.0
time
1.5
2.0
0
0.5
1.0
1.5
2.0
time
Fig. 13.11 Problem 1.18: a complicated “squarish” looking waveform v3 (t) on the left,
synthesized by a simple addition of three single-tone functions v1 (t), v3 (t), v5 (t). Better approximation of a square waveform v10 (t) (right) is achieved by adding first ten terms of (13.71). Waveform
v H P (t) is created by removing the first two terms from v10 (t), which is equivalent to applying a
HP filter. Observe that sharp edges are preserved due to HF content (horizontal axis is in units of
period T )
and it serves the same purpose as X-ray pictures to doctors. In this example we learn
how to synthesize a square pulse waveform by using addition of several sinusoidal
functions. In that respect, Fourier polynomials are our recipes for how to make
complicated periodic waveforms.
Function v(t) in (13.71) represents Fourier polynomial of a square wave function
(actually it should have been 4/ω v(t), to be mathematically correct).
v(t) = sin (2ω t) +
1
1
1
sin (6ω t) + sin (10ω t) + sin (14ω t) + · · · (13.71)
3
5
7
By increasing the number of terms the square wave approximations becomes
better, with the ideal square function requiring infinite number of sinusoidal terms.
However, if some of the terms are missing, the shape of square waveform becomes
distorted when we observe it in time domain, Fig. 13.11 (right). In electrical engineering terms, the process of removing for instance low-frequency terms (e.g the
terms associated with ω and 2 ω, i.e. low-frequency terms) is equivalent of applying
high-pass (HP) filter. Similarly, when only high-frequency terms are removed, in
electrical terms it is equivalent to application of a low-pass (LP) filter. If both lowfrequency and high-frequency terms are removed and only the “middle” terms are
left in the equation, it is equivalent of applying a band-pass (BP) filter.
It is useful exercise to remove one or more neighbouring terms from the middle
of the polynomial and observe the time domain distortion. In addition, find in literature similar Fourier polynomials that describe triangular, sawtooth, or any other
“standard” waveform and repeat the exercise.
1.19. While time domain plot in Fig. 13.11 is very useful to compare signal
nonlinearities, i.e. distortion, it still does not give us clear picture of what v10 (t) is
made from. For that purpose we need to use Fourier transform and convert v10 (t)
13 Introduction: Solutions
85
Signal Power, dB
0
HPF
-20
-40
-60
-80
-100
-120
0
5
10
15
20
25
frequency, kHz
Fig. 13.12 Problem 1.19: frequency spectrum of square waveform v10 , Fig. 13.11 (right) (the
vertical axis is scaled relative to the first harmonic’s power)
into its equivalent frequency domain plot P( f ) to show power content of v10 (t) at
various frequencies.
In the frequency spectrum plot, Fig. 13.12, the ten single-tone signal terms that are
used in the synthesis of v10 (t) with their respective frequencies and relative powers
are clearly visible. Red dash-dot line (HPF) shows profile of a HP filter that would be
used if the first two terms are to be removed, v H P (t), for example if only crossover
points of the waveform need to be detected. Of course, in this example we started
from a signal whose frequency content was already known, however, this example
illustrates usefulness of Fourier transform to examine content of an unknown signal,
and it is indispensable to circuit designers.
1.20. It is very often needed to remove unwanted components in the frequency
spectrum of a signal. For example, if a noise becomes embedded into the information carrying tone becomes, then some form of filtering must be applied in order
to recover the information. Application of LP, HP, or BP filters is obvious and most
often used solution to the problem. With this approach all tones outside of the filter’s pass-band will be indiscriminately suppressed. However, it is not difficult to
imagine a case where a single component in the signal frequency spectrum needs
to be removed while leaving the rest of the spectrum intact. Fortunately, with a bit
of understanding how sinusoidal waveforms interact with each other, we can design
circuits to perform mathematical computations and to achieve the goal. In this example we learn how two or more sinusoidal signals can be added together, so that the
resulting waveform is derived either by “constructive” or “destructive” action. In
other words, instead of trying to suppress a certain tone by the filtering action, it
is always possible to add another tone that is “negative” and therefore cancels the
targeted term.
For example, with reference to (13.71), if the goal is to produce a waveform y(t)
that is identical to v(t), except for only the third term v3 (t) = 1/5 sin(10ω t), then we
86
13 Introduction: Solutions
can write the following:
y(t) = v(t) + (−v3 (t))
(13.72)
1
1
1
1
= sin (2ω t) + sin (6ω t) + sin (10ω t) + sin (14ω t) · · · − sin (10ω t)
3
5
7
5
1
1
1
= sin (2ω t) + sin (6ω t) + sin (14ω t) + sin (18ω t) + · · ·
3
7
9
As trivial as it may seem, (13.72) gives us direction of how to design a system that
can perform cancelation of v3 (t) tone. Aside from physical connection to v(t), we
need to design signal generator whose output equals −v3 (t), which is a sinusoidal
waveform. In the following chapters we will learn how to design a circuit that generates sinusoidal output with specific amplitude, phase and frequency. For the time
being we assume that we have such circuit. Lastly, we need an adding circuit (keep
in mind that the operation of subtraction may be looked at as the addition of negative
number). Depending whether the information signal is in form of current or voltage,
then the addition is performed either by connecting two wires together (and therefore
summing the two currents), or by using an operational amplifier whose two inputs
perform subtraction of the two voltage signals.
In this example we learn a special very important case of “constructive” addition
of two waveforms, that is, we find out how to create “differential” signal out of two
single-ended signals.
1. Before we proceed to performing signal operations, let us first take a closer look
at the given signals
v1 (t) = 1 + 0.5 sin(ω t)
v2 (t) = 1 + 0.5 sin(ω t − ω )
Both of these two sinusoidal signals contain a constant term equal to one. This
is important observation, because in electrical engineering, a constant current or
voltage is referred to as DC. That means, value of every point of the sinusoidal
function is increased by, in this case, one. Therefore, this DC value takes the role
of average value for the sinusoidal signal (because without it, average value of
zero is assumed in all sinusoidal waveforms). To emphasize this role, we refer
to the average value of sinusoidal signal as the common mode (CM). We quickly
realize that in this case CM = 1 is greater than maximum amplitude Vm = 0.5,
therefore v(t) is always positive, which is important information for the circuit
designer.
Next observation we make is that maximum amplitudes of v1 (t) and v2 (t) are
equal. Following that, the same comment applies to the two frequencies ω 1 and
ω 2 , their values are also equal. Finally, because the two frequencies are equal, we
can look at phase difference. In this case, phase difference λφ = ω .
Let us now see significance of these observations. Subtraction of v1 (t) and v2 (t)
leads into
13 Introduction: Solutions
87
1.5
v1(t), v2(t), v(t)
1.0
0.5
0
-0.5
v
v1
v2
-1.0
-1.5
0
π
2π
3π
4π
time
Fig. 13.13 Problem 1.20: a differential waveform v(t) (black line), created by subtracting of two
waveforms v1 (t) and v2 (t) with the same common mode voltage VCM = 1V , same frequency, and
the opposite phase
1
1
v(t) = v1 (t) − v2 (t) = 1 + sin(ω t) − 1 + sin(ω t − ω )
2
2
1
= [sin(ω t) − sin(ω t − ω )]
2
1
= [sin(ω t) + sin(ω t)]
2
= sin (ω t)
(13.73)
The resulting sinusoidal waveform v(t) has CM = CM1 = CM2 = 1 − 1 = 0,
which is logical because the two initial CM values were equal. Maximum amplitude of the resulting sinusoid is the sum Vm = Vm1 −(−Vm2 ) = 0.5−(−0.5) = 1.
It is important to understand that the sign of a sinusoidal amplitude is assigned
relative to the CM (just remember a simple sinusoid where CM = 0). In other
words, the two sinusoids v1 (t) and v2 (t) always have the opposite signs, which
is caused by the phase difference of exactly λφ = ω .
To summarize, when two sinusoids have same CM, same frequency, same
amplitude, and phase difference of ω (i.e. they are inverted relatively to each
other), their difference is a sinusoid with the same frequency whose amplitude is
doubled and CM is zero, Fig. 13.13. Important to notice, by doing the mathematical operation of subtraction we achieved factor of two gain. Differential signal
processing is most important in modern electronics and we will see it again in the
following chapters.
2. As counterintuitive it is, the subtraction of two inverted sinusoids results in
amplified sinusoid, let us now see what is the result if the same two sinusoids are
added, i.e.
88
13 Introduction: Solutions
1
1
v(t) = v1 (t) + v2 (t) = 1 + sin(ω t) + 1 + sin(ω t − ω )
2
2
1
= 2 + [sin(ω t) + sin(ω t − ω )]
2
1
= 2 + [sin(ω t) − sin(ω t)]
2
=2
(13.74)
This time, two time varying components (AC) canceled each other because they
are inverted (we can imagine that each waveform was pulling in the opposite
direction with the same force, thus going nowhere), and we are left only with a
DC signal the is result of the sum of two CM levels.
3. When two waveforms are not perfectly symmetrical (which is more realistic
case), then resulting signal differs from the previous two cases, for example, in
this exaggerated case of CM misalignment
1 1
1
+ sin(ω t − ω )
v(t) = v1 (t) − v2 (t) = 1 + sin(ω t) −
2
2 2
1 1
= + [sin(ω t) − sin(ω t − ω )]
2 2
1 1
= + [sin(ω t) + sin(ω t)]
2 2
1
= + sin (ω t)
(13.75)
2
the two CM levels did not cancel, therefore, the resulting differential signal does
not have zero CM and the sinusoid takes asymetrically positive and negative
values, Fig. 13.14, which may or may not be what we wanted to achieve.
4. Addition of two inverted sinusoids with misaligned CM results in
1 1
1
+ sin(ω t − ω )
v(t) = v1 (t) + v2 (t) = 1 + sin(ω t) +
2
2 2
3 1
= + [sin(ω t) + sin(ω t − ω )]
2 2
3 1
= + [sin(ω t) − sin(ω t)]
2 2
3
=
(13.76)
2
which means that two inverted sinusoids cancel even if their CM are different,
while the outcome is DC level that is the sum of two initial CM levels.
As already mentioned, differential signal processing is extremely important in
modern signal processing, and it is, therefore, very important to quantify all possible
13 Introduction: Solutions
89
1.5
v1(t), v2(t), v(t)
1.0
0.5
0
-0.5
v
v1
v3
-1.0
-1.5
0
π
2π
3π
4π
time
Fig. 13.14 Problem 1.20: a differential waveform v(t) (black line), created by subtracting of two
waveforms v1 (t) and v3 (t) with non-equal common mode voltage, same frequency, and the opposite
phase
imperfections due to the signal misalignments, or departures of any kind from the
ideal symmetrical case.
This example also illustrates how important it is for engineers to develop “mental
math” skills that enable us to imagine that a sinusoid is created by adding two
components, namely the varying AC and the static DC, and to separately manipulate
the two of them.
1.21. Analog signal processing is nothing more than direct implementation of mathematical equations by means of transfer functions that are used to describe electronic
components, i.e. R, L, and C (and from now on M, a memristor). That is, each component by itself embodies one or more specific mathematical operations, for example
operation of addition is implemented by using a resistive divider for two voltage
signals, while operation of the first derivative is implemented by knowing that first
derivative of a current signal is proportional to the voltage across inductor whose
inductance value serves as the proportionality constant. Therefore, it is possible to
piece together any equations by means of combining various analog components.
The equation synthesis is achieved either as direct one-to-one matching, for example exponential function is embedded in a BJT transistor’s operation, or by close
approximation, for example nth root function can be implemented by using Taylor
expansion and exponential function, and so on. In this example we learn how simple
operation of linear addition is implemented by using a voltage divider.
1. Because the two resistors are equal R = R1 = R2 = 1 kα the two input voltages
are added and divided equally. Thus, after passing through the voltage divider
network, the two signals are attenuated in the same proportion, which follows
from Kirchhoff’s current law (regardless whether the input voltage signal is AC
or DC) as
90
13 Introduction: Solutions
Fig. 13.15 Problem 1.21:
block diagram of an analog
voltage adder circuit
v2 − vo
v1 − vo
+
=0
R
R
⇒
vo =
v1 + v2 = 2vo
1
(v1 + v2 )
2
∴
vsum = v1 + v2 = 2 vo
(13.77)
which further illustrates operation of addition. Side effect of this approach is that
the addition operation is accompanied with the 1/2 scaling factor, which implies
that correct implementation of voltage signal addition as in (13.77) requires two
equal resistors and an amplifier whose gain is G = 2, Fig. 13.15.
2. From given data T1 = 1 µs it follows that f 1 = 1/T1 1 = 1MHz. Further, by
inspection of Fig. 1.2 we write T2 = 1/2 T1 = 0.5 µs, therefore f 2 = 1/T2 =
2MHz. Additional two pieces of information we can extract from Fig. 1.2 is that
the two maximum amplitudes are related as V2m = 2 V1m and that the phase
difference is zero (even though one frequency is double of the other, the two are
synchronized at all times and have the same zero-amplitude crossing points in
time). Therefore,
v1 (t) = 1 sin[(2ω 1MHz) t]
v2 (t) = 2 sin[(2ω 2MHz) t]
∴
vsum = v1 (t) + v2 (t) = sin[(2ω 1MHz) t] + 2 sin[(2ω 2MHz) t]
(13.78)
From the graph given in Fig. 1.2 and (13.78) we conclude that the linear addition
of two single tone signals does not produce any new frequency tones in the output
spectrum (we still have the 1 and 2 MHz single tones at the output), and that the
output amplitude of v2 is still twice the amplitude of v1 .
Knowing the two voltages (v1m = 1), (v2m = 2) and resistance R, we can correlate
the two corresponding powers (P1 , P2 ) as
13 Introduction: Solutions
91
Fig. 13.16 Problem 1.21: relative power spectrum of two waveforms v1 (t) and v2 (t)
P1avg =
2
v2
v2
v1rms
v2
1
4
= 1m =
and P2avg = 2rms = 2m =
R
2R
2R
R
2R
2R
∴
P1avg
1
=
P2avg
4
(13.79)
which is to say that 1 : 4 is relative power ratio between the two signals, both at
the input and the output nodes. Compact way of presenting normalized power vs.
frequency relationship is to use a frequency spectrum graph, Fig. 13.16 (left).
3. When the two resistors are not equal, i.e. R2 = 2 R1 = 2R, the two input voltages
are added and divided in the same ratio as the two resistors, thus we write
v1 − vo v2 − vo
1
+
= 0 ⇒ 2v1 − 2vo + v2 − vo = 0 ∴ vo = (2 v1 + v2 )
R
2R
3
Therefore, in this case the output voltage vo is not just a simple sum of the two
input signals, instead voltage v1 is multiplied by factor two before it is added to
v2 , which is then alltogether multiplied by 1/3 factor, thus we write
vo =
1
{2 sin[(2ω 1 MHz) t] + 2 sin[(2ω 2 MHz) t]}
3
(13.80)
We observe that the output frequency spectrum still did not change, while the
relative amplitudes of the two output tones are now equal.
4. In this case, after passing through the voltage divider network, the two signals
are not attenuated in the same proportion. Signal v2 passes through twice the
resistance relative to the resistance of the path that v1 takes. Consequently, at
the output side of the voltage divider network the proportion between v1 and v2
changed by factor two relative to the input side, and therefore became equal.
Now, the two signal powers contribute equally to the power of the output signal
vo , Fig. 13.16 (right).
This example illustrates linear operation of addition, which may change relative
powers of the internal tones in the output signal, however it does not change frequency
content. This is very important observation, because in the following chapters we
92
13 Introduction: Solutions
Fig. 13.17 Problem 1.22:
signal envelope sketch
will realize that design of wireless circuits is based on using non-linear operations,
such as multiplication, to produce output tones at different frequency relative to
the input tones. This “frequency shifting” operation is fundamental for wireless RF
communication systems and must be recognized as different than linear addition.
1.22. A continuous single tone waveform by itself does not convey any useful information, aside from revealing its own existence. Nevertheless, even that single bit of
information, i.e. existence of a waveform, is used to design first forms of communication systems for transmitting information over distances greater than the distance
achievable by a human voice. If time multiplexed, combination of “short” and “long”
beeps enables creation of Morse code where each letter of alphabet is encoded as a
unique combination of beeps. For instance, “RF” would be encoded as “· − · · · − ·”
and transmitted wirelessly by well timed turning ON and OFF power button of a
transmitter that emits a single tone.
In this example we learn about a bit more sophisticated encoding principle, where
we exploit fact that a single tone sinusoidal waveform can change either its amplitude, frequency, or phase. If amplitude of the waveform is changed in accordance
with the information signal, we refer to this technique as amplitude modulation.
The time varying maximum amplitude points generate a curve known as the “envelope waveform” and its shape is exact replica of the information being transmitted,
Fig. 13.17.
Because a sinusoid is symmetrical waveform, its amplitude change is also
symmetrical. Consequently, we note existence of two identical half-envelopes venv+
= |venv− | whose absolute values are equal (which is always the case). In this particular case only, the information is a low-frequency sinusoid, thus each half-envelope
has a shape of a sinusoidal signal. Also, the two half-envelopes have common modes
at CM+ for the positive, and CM− for the negative half-envelope. By using appropriate signal processing electronics, either of the two half-envelopes can be extracted,
which is to say that the information can be extracted from any of the half-envelopes.
Chapter 14
Basic Terminology: Solutions
Concept of an electric field and forces among charged particles at the atomic level
help us model macro-behaviour of materials and devices. The macro-models based
on these concepts enable us to predict outcome of an experiment, or to design a
device to perform a certain (mathematical) function. Duality of matter and waves
turns out to be fundamental for macro operation of these devices, thus in this section
we review some of the mathematical techniques commonly used in the wireless
electronic circuit design.
Solutions:
2.1. In classical physics model we use concept of the four fundamental forces in the
universe that are responsible for all non-contact interactions. Namely the four forces
are gravity, electromagnetism, strong nuclear force, and weak nuclear force. The two
nuclear forces are assumed to be “short distance” forces responsible for holding the
matter together. At the same time, gravity and electromagnetism are assumed to be
“long distance” forces that greatly influence bio-chemical, electrical and electronic
processes. In this example we review basic definitions and compare electrostatic and
gravitational forces.
Direct application of Coulomb’s and gravitational law gives
q q
ke e 2 p
k e qe q p
FC
R=
= m erm p =
FG
Gm
em p
G 2
r
2
(1/4ω λ0 ) Nm2/C2 1.602176565 × 10−19 C
= 6.67384 × 10−11 N m2/kg2 9.10938291 × 10−31 kg 1.672621777 × 10−27 kg
= 2.26882 × 1039
That is, for all practical purposes force between an electron and proton is determined
only by the electrostatic attraction, while the gravitation force is safely neglected
because it is approximately 1039 times smaller. Therefore, although we think about
R. Sobot, Wireless Communication Electronics by Example,
93
DOI: 10.1007/978-3-319-02871-2_14, © Springer International Publishing Switzerland 2014
94
14 Basic Terminology: Solutions
the charges as material particles, in electrical engineering we neglect the effects of
gravity.
2.2. As we already know, scalar variables are determined only by a number that quantifies their magnitude (e.g. temperature), while vector variables are the ones that need
both magnitude and direction for the complete description (e.g. force). Electrostatic
force is responsible for the movement of free electrons, which by definition is electrical current. In this example, we review vector arithmetics of forces among electrical
charges.
r23 = a, while distance from q1 to q3 equals to
Distance from q2 to q3 equals
√
the square diagonal r13 = a 2. Due to the fact that only charge q3 is mobile, there
is no need to calculate force between the two fixed charges (q1 , q2 ). Indeed, charge
q3 is pulled by (q1 , q2 ), each pooling in its respective direction, with Couloumb’s
law stating that electrostatic pulling/pushing force is inversely proportional to the
distance. Therefore, magnitude of force F12 relative to F13 stands as the ratio
ke
F23
=
F13
ke
q2 q3
2
r23
q1 q3
2
r13
=
r13
r23
2
∴
F23
√ 2
a 2
=
F13 = 2 F13
a
(14.1)
which helps us to sketch correctly scaled force vector diagram. Direction of the
vectors is determined by polarity of the there charges.
1. when q3 is a proton, i.e. positive charge, it is pushed away from q1 and pulled by
q2 . Thus, vector plot is constructed as in Fig. 14.1 (left). From 14.1 and the simple
geometric rules of similar and right-angled triangles, Pythagoras’ theorem, and
rules for squares we conclude (after normalizing F13 = 1) that
√ 2 √ 2 ⎧
√
2
2
F3 =
+ 2−
= 5−2 2
2
2
Fig. 14.1 Solution 2.2: Electrostatic forces among three electric charges in space
(14.2)
14 Basic Terminology: Solutions
95
while, horizontal and vertical projections of F3 are
√
√
2
2
F3x = 2 −
and F3y =
2 √
2
F3y
2
tan φ =
=
→ φ = 28.675∞
√
F3x
4− 2
(14.3)
which, as referenced in Fig. 14.1 (left), fully defines vector F3 .
2. when q3 is an electron, i.e. negative charge, it is pulled by q1 and pushed away by
q2 . Thus, vector plot is constructed as in Fig. 14.1 (right). Following exactly the
same reasoning as in the previous case we find magnitude and direction of F13 as
⎧
F3 =
√
5 − 2 2 and π = 28.675∞
(14.4)
This exercise should serve a purpose of refreshing vector algebra, basic geometry,
and basic physics.
2.3. Using a “test charge” q0 at distance r from charge q we measure Coulomb’s
force Fc acting at the test charge, and we use it to define electric field around the
charge q as
F
q
1 q
E = lim
= ke 2 r =
r
q0 ≤0 q0
r
4ω λ0 r 2
We observe that strength of electric field around a point charge is not function of the
used test charge, and the field decreases with square of the distance.
2.4. Developing ability to connect seemingly unrelated events and concepts in a
logical order is essential skill. In this example, we practice how to put together a few
very basic concepts and how to derive very important scientific conclusion.
The elementary electric charge constant was determined by famous Millikan’s
oil-drop experiment. In the experiment, Fig. 14.2, a large number of very fine drops
of oil whose mass density is known with high precision were electrically charged
and suspended in the air by the means of electric field acting in direction opposite
to the gravity force. By repeating the measurement many times for various sizes of
the oil-drops and magnitude of the electric field, Millikan noticed that all calculated
charges stored at the oil drops were multiples of a certain “unity charge”. He therefore
concluded that the total charge comes in discrete quanta, thus, the minimal possible
charge must be the one of a single electron.
Knowing mass density and radius of the two given oil drops and by assuming
that the oil drops take ideal spherical shape (which is reasonable assumption if the
drops are in the suspended, i.e.weightless, state) it is straightforward to calculate
their respective masses m 1 and m 2 as
96
14 Basic Terminology: Solutions
Fig. 14.2 Solution 2.4:
Simplified diagram of
Millikan’s experimental setup
m 1 = φoil V = φoil
4ωr 3
kg 4ω(1.670µm)3
= 837 3
= 1.6329146 × 10−14 kg
3
m
3
and
m 2 = 1.9606909 × 10−14 kg
These two masses are pulled down by gravitational force and at the same time
pulled up by the electrical force. When the oil drops are suspended, the two forces
are in balance, i.e.
mg
mg + q E = 0 ∴ q = −
E
which leads into
m
1.6329146 × 10−14 kg × 9.806650 2
m1g
s
q1 = −
= −8.0067111 × 10−19 C
=−
E
2.0 × 105 CN
and
q2 = −9.6139048 × 10−19 C
Therefore, the difference in electrical charge is
q = q2 − q1 = −1.6071937 × 10−19 C
(14.5)
which is about 0.3 % higher then the accepted value of of electron charge qe =
1.602176565 × 10−19 C. As we already know, the electric charge is one of the
fundamental constants in physics and it is important that it is determined by high
accuracy. As a side note, in his real experiment, Millikan apparently measured the
elementary charge with less than 2 % measuring error.
2.5. In reality, all natural processes are fundamentally non-linear. That is, strictly
speaking, a linear function is just a mathematical idealization that does not exist
in nature. However, within a relatively small enough window of observation any
nonlinear function can be linearized to a various degree. Therefore, by approximating
a nonlinear function around the point of interest with its small linear section, we
consciously trade accuracy for simplicity. Typical example of this approach is that
14 Basic Terminology: Solutions
97
Fig. 14.3 Solution 2.5:
Geometrical view of an electic
dipole
while we walk we do not think about the Earth as a sphere, instead within our close
proximity we perceive it as a plain.
Over long period of time, mathematicians worked out linearization methods for
many non-linear functions even before they became commonly used to model various
physical phenomena. For example some of commonly used mathematical approximations used by engineers are based on Taylor expansion series. In this example, we
use Taylor expansion of inverse square root function, which replaces the nonlinear
inverse square root with a “nice” polynomial function that is much easier to deal
with. Similarly to the Fourier polynomials, the first two polynomial terms are the
linear part (i.e. the constant and the first order variable (a + bx)), while the higher
order terms add more and more non-linearity while at the same time reducing the
approximation error.
By definition (14.6), the electric dipole moment is a vector quantity whose
magnitude equals to the product of positive and negative electrical charges, while
the vector direction is from the negative to the positive charge. In simpler words, it
is a measure of the separation between positive and negative electrical charges or its
polarity.
We already know that a potential V at any point in space is calculated using the
superposition principle, thus we write
V =
k=n
⎪
k=1
1
Vk =
4ω λ0
q
q
−
r1 r2
Using Pythagoras’ theorem, Fig. 14.3, we write
r12 = (r sin ∂ )2 + (r cos ∂ − a)2
∴
⎨
⎨
r1 = r 2 + a 2 − 2ra cos ∂ and r2 = r 2 + a 2 + 2ra cos ∂
98
14 Basic Terminology: Solutions
Assuming r ≥ a, we use only the linear terms of the Taylor series1 to simplify
expressions for r1 and r2 as
1
1
= ⎩
r1
r
1+
a 2
r
1
−2
a r
a cos ∂
1 a 2 a 1
∗ r 1 − 2 r + r cos ∂
1 a 2
1
1
cos ∂
1−
∗
−
r2
r
2 r
r
which leads into
1
q
1
V =
−
4ω λ0 r1 r2
1 a 2 a 1 a 2 a q
cos ∂ − 1 +
cos ∂
1−
∗
+
+
4ω λ0 r
2 r
r
2 r
r
p cos ∂
p·r
2aq cos ∂
=
=
(14.6)
V =
4ω λ0 r 2
4ω λ0 r 2
4ω λ0 r 2
where, p = 2 pq is known as the electric dipole moment. Electric dipole moment
concept enables us to treat the dipole as a single charge. Using E = −∓V in spherical
coordinates we write
p cos ∂
εV
=
Er = −
εr
2ω λ0 r 3
p sin ∂
1 εV
=
E∂ = −
r ε∂
4ω λ0 r 3
For small distances r the electric field becomes infinite, which is mathematical
consequence of the initial assumption that r ≥ a. However, from large enough
distance, dipole appears as a simple charge, which simplifies further analysis. In
engineering jargon, in most cases results within 10 % of the correct result are considered “good enough”. In the case of mental calculations sometimes even being within
the order of magnitude of the correct result is close enough because the estimate is
achieved fast and without the use of calculating equipment. We keep in mind that
mental calculations skills are extremely useful to practicing engineers because, even
with all involved inaccuracies, the engineer develops intuitive feeling for the problem
at hand.
2.6. In order to practice new principles, it is natural to start with simplest possible
cases and progressively work out more complicated ones. For instance, in this example instead of considering only point charges, we introduce multiple point charges
distributed over a cylindrical object. Geometry of the charged objects determines
shape of the surrounding electric field, thus in our work we progress from a point, to
1
√
Taylor’s series around x = 0 is: 1/
1+x
∗ 1 − (1/2) x + (3/8) x 2 − (5/16) x 3 + · · ·
14 Basic Terminology: Solutions
99
a line, a cylinder, a sphere, and then to non-regular shapes. It should be obvious by
now that non-regular objects may be “dissected” and considered simply as a sum of
basic regular objects, which in itself can be seen as being equivalent to the non-liner
to linear function approximation.
The total charge uniformly distributed on the given wire is found simply as q = ql l,
thus direct implementation of Gauss’s Law in cylindrical coordinates leads into
D · ds = qfree,enc
→
S
λ0 E · ds = qfree,enc
S
∴
Er · 2ωrl =
ql l
λ0
→
Er =
ql
r
2ωr λ0
(14.7)
where, r is the unity vector perpendicular to the wire surface.
2.7. Duality between matter and energy is still one of the greatest mysteries that we
are facing in modern science. Conclusion that particles and waves are interchangeable came from the famous double–slit experiment. Basic property of waves, i.e.
interference, is perfectly well modelled by interaction of sinusoidal functions. Thus,
mathematics of classic waveforms is based on trigonometry algebra. In this example we review some of basic definitions related to sinusoidal functions and their
relationships, namely addition of two sinusoidal function and their phase difference.
It is a simple matter to reconstruct a sinusoidal function if its zero–crossing points
(i.e. its period) and the amplitude are known in advance. For example, phase difference between S1 and S2 waveforms, Fig. 14.4, is ω/3. In the time–domain, for a
10 MHz waveform, the ω/3 phase difference is expressed in units of time as follows.
First, the waveform period is calculated as T = 1/ f = 100 ns ⇒ 2ω , then from a
Fig. 14.4 Solutoin 2.7: Illustration of a relationship between sinusoidal waveforms
100
14 Basic Terminology: Solutions
simple proportions between ω/3, 2ω , and 100 ns it follows that ω/3 is equivalent to
16.667 ns.
Similarly, every time when S1 and S2 cross their respective amplitudes are equal,
thus amplitude of S1 + S2 at the same instance must be twice the amplitude of either
waveform. Because their phase difference is ω/3, the two forms cross for the first time
16.667 ns. At that moment S1 = 2 sin (ω/3) = 1.732, thus the instantaneous sum of
the two sinusoids is 3.464. Similarly, when S2 = 0 then S1 = 1.732 the sum must
be 1.732 as well. We observe that addition of these two sinusoids produced another
sinusoid whose amplitude is greater than any of the two individual amplitude. In this
case we say that this was “constructive interference” because the resulting sinusoid
is “stronger” then each of the two sinusoids.
By careful engineering, that is, by understanding how the phase difference and
amplitude of two or more sinusoids add together, we are able to manipulate waveforms in many ways. Some of the practical uses of wave interference are shown in
the following examples.
2.8. Very important application of constructive addition principle is to use a carefully
chosen set of sinusoids to synthesize other periodic functions. In this example we
use a set of five sinusoidal functions to synthesize a sawtooth waveform.
If the given set of sinusoids is written as
2
S1 (t) = (−1)
1
2
S2 (t) = (−1)2+1
2
3+1 2
S3 (t) = (−1)
3
2
S4 (t) = (−1)4+1
4
2
S5 (t) = (−1)5+1
5
1+1
sin (1 α t)
sin (2 α t)
sin (3 α t)
sin (4 α t)
sin (5 α t)
then it is easy to see that all functions have the same general form, thus their sum is
S(t) =
5
⎪
k=1
Sk =
5
⎪
k=1
(−1)k+1
2
sin (k α t)
k
where, term (−1)k+1 alternates between −1 and +1 and therefore controls the initial
phase, term 2/k controls amplitude of its corresponding term, and term (k α t) sets
the harmonic frequency. When only the first five terms are used, then the sawtooth
function is still visibly distorted, Fig. 14.5 (left) while, for instance, 25 terms produce
much more realistic sawtooth function, Fig. 14.5 (right). Needles to say, an infinite
number of terms is required to synthesize mathematically perfect sawtooth function.
101
4
3
2
1
0
-1
-2
-3
-4
S1
S2
S3
S4
S5
S
0
0.5
1.0
time
1.5
amplitude
amplitude
14 Basic Terminology: Solutions
2.0
4
3
2
1
0
-1
-2
-3
-4
S(t)
0
0.5
1.0
1.5
2.0
time
Fig. 14.5 Soluton 2.8: Illustration of sawtooth waveform synthesis using five sinusoids (left), and
twenty five (right)
2.9. Noise cancellation concept is based on the idea that a multitone noise waveform
Sn is added to its “mirror image” waveform Sm so that Sn + Sm = 0, and is therefore
cancelled. The cancelation action is much easier executed in frequency domain, by
simply adding negative version of each component in the noise frequency spectrum.
We duly note that even though we used term “noise” in this example, any other signal
tone can be cancelled using the same principle.
In an ideal case, the cancelation function’s spectrum must contain inverted, i.e.
“mirrored” images of all single tone components found in the noise function. Thus,
each single tone noise component would be added with its mirrored tone, therefore
each of the sums would be zero. In other words, the mirrored tone is shifted by T /2
(i.e. by ω ) relative to the original tone. Therefore,
1. Noise Sn (t) and the corresponding cancelation function Sm (t) must be related as
follows,
1
2
3
4
1
sin (α t) − sin (2α t) + sin (3α t) + sin (5α t) + sin (9α t)
5
2
3
4
3
1
2
3
4
1
Sm (t) = − sin (α t) + sin (2α t) − sin (3α t) − sin (5α t) − sin (9α t)
5
2
3
4
3
Sn (t) =
or, equivalently
Sm (t) =
1
2
1
sin (α t + ω ) − sin (2α t + ω ) + sin (3α t + ω )+
5
2
3
3
4
sin (5α t + ω ) + sin (9α t + ω )
4
3
2. In this idealized case, the cancelation is perfect, Fig. 14.6 (left) because each tone
in the spectrum is perfectly cancelled in each instance of real time.
3. However, in a realistic case there is always a finite delay time, i.e. “processing
time”, between the moment when the noise value presents itself and the cancelation value is created. That is, the sequence of events in time is as follows.
14 Basic Terminology: Solutions
4
3
2
1
0
-1
-2
-3
-4
0
Sn(t)
Sm(t)
Sn(t)+Sm(t)
amplitude
amplitude
102
0.5
1.0
time
1.5
4
3
2
1
0
-1
-2
-3
-4
Sn(t)
T/10
T/20
T/100
0
0.5
1.0
1.5
time
Fig. 14.6 Solution 2.9: One of the possible illustrations of destructive interference
First, amplitude of the incoming noise signal is measure at a certain instance in
time. Then, signal whose amplitude is same but with the opposite sign is created.
Finally, the two signals are added so that their sum is a zero value signal. Naturally, execution of these actions takes finite amount of time, which means that by
the moment the correct cancelation value is created, the original instantaneous
value of the noise signal has already changed and the sum is not zero (except by
a chance, of course).
In practice, the delay is measured relative to period T of the first harmonic (whose
period is the longest, of course) because the absolute value of the period is not
relevant at all. In Fig. 14.6 (right) there are three cases shown, i.e. when the
cancelation signal was delayed relative to the noise signal by: (a) T/10 or 10 %
of the period; (b) T/20 or 5 % of the period; and (c) T/100 or 1 % of the period. A
rough comparison of the noise and the produced “cancelled” signal’s amplitudes,
Fig. 14.6 (right), shows that in this particular case we must finish the complete
signal processing in less then 1% relative to the period T , if we are to suppress
the noise signal by factor of ten.
By choosing higher operating frequencies we effectively reduce the allowed
processing time, i.e. the intended electronics must operate under much harder
time constrains. That is why, for example, it is not difficult to design noise cancellation headphones by using the state of the art technology that enables the
design of electronics operating in multi gigahertz range, while the audio signal
is in kilohertz range. For example, a 1 kHz signal’s period is T = 1 ms, thus one
percent of T is 10 µs. By the standards of modern technology, within a 10 µs
window it is possible to take millions of samples of the signal and perform large
number of numerical operation while the signal itself barely changed its amplitude. However, the situation is is completely different if we try to cancel a 10 GHz
tone. One percent of its period is only 1 ps, therefore even the fastest processors
of the day can not to much within this time window.
2.10. By definition, electrical power P is product of the two fundamental electrical
variables, voltage and the current, measured at the same time instance. Due to Ohm’s
law, there are three equivalent forms that describe instantaneous electric power as
14 Basic Terminology: Solutions
103
p = vi =
v2
= i2 R
R
(14.8)
where, for simplicity the resistance R is assumed constant in time. A square pulse
waveform is characterized by only two amplitude levels; first half of its period the
amplitude takes value of zero, while over the second half of its periods the amplitude
constantly holds its maximum value. In accordance to (14.8), if at any given moment
either voltage or current takes zero value, the power must take zero value too.
A periodic function is often described by duration of its duty cycle. By definition,
a square wave is said to work on 50 % duty cycle, which means that half of its period
the square wave holds its high value, and half of the period it holds zero. Because
the pulse amplitude is constant over its corresponding half–period, in this example
given as 1ms (i.e. the full period is T = 2ms), we simply write
 2
(2 V)2
v (t)


=
= 40 mW for 0 ∼ t ∼ 1 ms

 R
100 ν
p(t) =

2
2


 v (t) = (0 V) = 0 mW for 1 ms ∼ t ∼ 2 ms
R
100 ν
hence, by definition, average power Pavg over the full period T = 2 ms is
Pavg =
1
T
0
T
p(t) dt =
1
2 ms
1 ms
40 mW dt +
0
1
× 40 mW × 1 ms = 20 mW
=
2 ms
2 ms
0 mW dt
1 ms
(14.9)
which is the constant power level that needs to be continuously held over the full
time period so that it dissipates as same amount of energy as power “burst” of the
square wave that lasts only half–period but has double amplitude. In other words,
the total energy dissipated within the pulse period is due to the energy only within
the high voltage half–period, i.e.
W = Pt = 40 mW × 1 ms = 40 mJ
which is equivalent to the total energy dissipated by constant power of 20 mW
continuously delivered over the time period of 2 ms.
In order to be able to mentally evaluate or estimate integrals, it is very useful to
keep in mind that a definite integral represents area of the region bounded by the
function (where area below the horizontal axis is assumed negative and taken away
from the sum).
2.11. Whether an element serves the role of a power source or a power load depends
on the flow of current in and out of that element. Positive power implies that the
energy flows into the element where it is dissipated, while the negative power value
104
14 Basic Terminology: Solutions
Fig. 14.7 Solution 2.11:
Graph of a piecewise
linear function in current-time
coordinating system
that is easily integrated
geometrically
implies that the energy is leaving the element while being delivered to its load. In
this example we review simple definitions related to the current flow and charges.
1. After straightforward implementation of (2.1) and with reference to Fig. 14.7, we
write i(−2s) = −2A/s (−2s) = 4A; and i(3s) = 3A/s (3s) = 9A;
2. By direct implementation of definition for current, and in the case of linear current
change in time, definite integral is trivial to solve from the graph in Fig. 14.7 by
adding areas of right angled triangles bounded by the i(t) functions, thus,
i(t) ⇒
dq
dt
∴
q=
3s
−2 s
i(t)dt =
0
−2 s
3s
(−2t)dt +
(3 t) dt = 17.5 C
0
In this trivial case of linear i(t) function the same result is derived by inspection
of Fig. 14.7 and calculation of surface under the function, i.e. by adding the two
triangular areas, as (4 × 2)/2 + (9 × 3)/2 = 17.5 C, where the two catheti of each
right angled triangle are in the units of amperes and time respectively, therefore
the product (i.e. area) is in coulombs.
3. Knowing that the total charge moved in within the time interval of (−2 s, 3 s)
equals 17.5 C, the average current is easily found by calculating the hight of a
rectangle whose area is 17.5 C and horizontal side is 5 s long in current-time
coordinating system, therefore the constant current level that would move the
same amount of charge is found as i avg = 17.5 C/5 s = 3.5 A, marked as red
dash–dot line.
2.12. Understanding that definition of power source is based on the flow of current
relative to potential at the device’s terminal helps us to track the total energy in the
system. As we remember, the law of conservation of energy must apply to closed
systems.
By inspection of Fig. 2.2 and from the given data, it is straightforward to write:
(vs , i s ) = (8 V, 7 A) with the current leaving positing terminal of the element, therefore Ps = −56 W. Voltage v2 = vs = 8 V, the current entering R2 , therefore,
P2 = 16 W.
In accordance to Kirchhoff’s current law (KCL), i s = i 1 + i 2 , therefore, i 1 =
7 A − 2 A = 5 A leaving R1 , which leads into P1 = −60 W. This result is followed
14 Basic Terminology: Solutions
105
by conclusion that v3 = vs + v1 = 20 V, which means that P3 = 160 W. Voltage
controlled current source gm forces current of i gm = 0.25A/V × 12 V = 3 A leaving
the device, while v(gm ) = v3 = 20 V. Thus, we reach conclusion that P(gm ) =
−60 W.
It is simple to verify the conservation of energy law by adding all powers in the
system as
Psystem = Ps + P1 + P2 + P3 + P(gm )
= −56 W − 60 W + 16 W + 160 W − 60 W = 0
which illustrates behaviour of ideal voltage/current sources in a multi-device closed
system. As ideal elements, they are capable of absorbing/generating infinite amount
of power in order to hold the required voltage/current at its terminals.
2.13. By inspection of Fig. 2.3, and from definition of electric power P = i 2 R, we
conclude that during the first time period power absorbed by the load is P1 = 500 W,
during the second time period dissipated power is P2 = −125 W, during the third
time period net power is P3 = 0 W, and during fourth time period absorbed power
is P4 = 500 W. Therefore, the average power is found as
Pavg =
P1 + P2 + P3 + P4
= 218.75 W
4
(14.10)
In this case of constant current sections it is easy to geometrically solve the definition
integral, thus RMS current is
⎩
i rms =
(10 A)2 × 1 ms + (−5 A)2 × 1 ms + (0 A)2 × 1 ms + (10 A)2 × 1 ms
4 ms
= 7.5 A
Chapter 15
Electrical Noise: Solutions
Signal to noise ratio (SNR) is one of the most basic specifications used to characterize
quality of the received signal, as well as (implicitly) the complexity of the required
electronics. In the following examples, we develop a feeling for relationships between
the noise and signal, we learn about differential signals, thermal noise, and we practice
to use frequency spectrum plots to help us in the design process.
Solutions:
3.1. Saying that two voltage amplitudes are related as SNR = 20 dB is just another
way of saying that amplitude of the one assumed to be “signal” is ten times greater
than amplitude of the one assumed to be “noise”. For instance, for two given voltage
amplitudes A1 = A0 (i.e. “signal”) and A2 = A0/10 (i.e. “noise”), we write
SNR := 10 log
= 10 log
= 20 log
∴
= 20 log
P1
P2
(15.1)
A21/R
A2/R
A1
A2
A0
= 10 log
A1
A2
2
(15.2)
A0
10
= 20 log 10 = 20 dB
(15.3)
We keep in mind that SNR is defined as ratio of two powers (which is unitless quantity,
e.g. P1/P2 ). Often, it is more practical to use units of [dB] by using log function as
in (15.1), while SNR of the two respective voltage (or current) amplitudes is derived
and expressed in units of [dB] as in (15.2). Therefore, if ratio of two powers is, for
example, P1/P2 = 10 we can express it as power SNR P = 10 dB by using (15.1), or
equivalently we can use the respective two voltage amplitudes and say that voltage
R. Sobot, Wireless Communication Electronics by Example,
107
DOI: 10.1007/978-3-319-02871-2_15, © Springer International Publishing Switzerland 2014
108
15 Electrical Noise: Solutions
15
15
a(t)
n(t)
a(t)+n(t)
10
amplitude, V
amplitude, V
10
5
0
-5
-10
-15
5
0
-5
-10
0
50
100
150
200
-15
time, µs
0
50
100
150
200
time, µs
Fig. 15.1 Solution 3.1: time domain waveforms
SNR V = 20 dB as in (15.3). All three expressions are equivalent. Main reason for
using dB units is due to properties of the log function,1 which enable us to use the
simple operations of addition and subtraction when comparing two signals that are
already expressed in decibels. At the same time, by using the log function we can
plot both weak and strong signals in the same plot, even if their powers are orders of
magnitudes apart (which is, again, the case for transmitted and received EM signals
in radio communications). In addition, it turns out that B is too large unit for modern
wireless communication applications, thus we use dB.
For the sake of creating meaningful visualization of an ideal single tone signal
a(t, ω a ) and a noise signal n(t, ω 1 , ω 2 , ω 3 ) we arbitrary choose three random tones
and add them to the main tone as
a(t) = A0 sin ω a t
n(t) = n 1 sin(ω 1 t + λ1 ) + n 2 sin(ω 2 t + λ2 ) + n 3 sin(ω 3 t + λ3 )
where A0 = 10(a1 +a2 +a3 ) so that we set the required SNR = 20 dB (Fig. 15.1). For
purposes of this illustrative plot, the quasi–random “noise” function n(t) is created
simply by adding three random sine/cosine terms with various frequencies. Also,
frequencies of the noise harmonics are set a bit higher than the signal frequency
(which is not to say that it can not be the other way around). It is clearly visible from
the graph that the two signals, a(t) and n(t), have the maximum amplitudes ratio of
approximately ten. However, when noise affects the wanted signal, mathematically
speaking, that means the two functions are simply added together in time–domain
as a(t) + n(t) (Fig. 15.1, right). This graph helps us visualize how an example of
a single–tone voltage signal that contains high frequency noise with SNR = 20 dB
may look like.
As the additional exercise, plot more examples with various SNR to develop your
own sense of noisy signals and to practice estimating the associated SNR from the
resulting time–domain plots (which is what we do when we use an oscilloscope). For
1
log(x y) = log x + log y and log(x/y) = log x − log y
15 Electrical Noise: Solutions
109
example, find what is SNR in [dB] if amplitude of the noise function is |n(t)| = A0
while amplitude of the signal function is |a(t)| = A0/10 ?
3.2. In these two examples the “noisy” signal s(t) is synthesized as sum of a 10 kHz
single tone and three randomly chosen tones as
s(t) = 10V sin(2π × 10 kHz × t)
+ 0.3V sin(2π × 601 kHz × t + 13)
+ 0.5V sin(2π × 713 kHz × t − 42)
+ 0.2V sin(2π × 907 kHz × t + 87)
(15.4)
Note that the maximum amplitude of the three noise components added together
is one, which is ten times smaller relative to the main 10 kHz tone amplitude. We
observe that time domain plot (i.e. by using an oscilloscope), Fig. 15.1 (right), shows
10 kHz noisy signal with approximately SNR = 20 dB, however, we have no way
of determining the frequency content of the noise itself, and subsequently how to
design a low-pass (LP) filter to remove the noise. Here is where the frequency plot
(i.e. the use of spectrum analyzer) becomes indispensable. In the frequency plot,
Fig. 15.2, it becomes obvious that only three single tones are present in the signal.
On the horizontal axis we clearly read frequency values of the frequencies, while
the vertical axis shows powers relative to the 10 kHz tone that was normalized to
valued of 0 dB. By zooming on this plot we can read relative signal powers in units
dB, which we already know from (15.4). We keep in mind that in this simulated
plot the noise floor (measured as below −100 dB) is due to numerical noise of the
simulator tool. It goes without saying that it is extremely important for an engineer
to become proficient in using both time domain and frequency domain graphs to
evaluate signals. For example, from Fig. 15.2 we easily conclude that if we design a
LP filter whose frequency slope starts at just above 10 kHz and extends all the way
0
Signal Power, dB
-20
-40
LP filter
-60
-80
-100
-120
-140
0
0.2
0.4
0.6
frequency, MHz
Fig. 15.2 Solution 3.2: frequency spectrum plot
0.8
1.0
110
15 Electrical Noise: Solutions
Fig. 15.3 Solution 3.3: time domain differential signal
to 600 kHz (dash–dot red line) it will be sufficient to suppress all three noise tones
down into the noise floor. Naturally, this is much easier design requirement relative
to the case where the first noise tone is somewhere much closer to the 10 kHz signal
(because steeper filter slope requires higher order, i.e. more complicated, filter).
3.3. When two single–ended signals a(t) and b(t) that have their amplitudes equal,
their common–modes equal, their frequencies equal, and their phases inverted, are
subtracted from each other we refer to the resulting signal d(t) = a(t) − b(t) as differential signal. It is important to realize that the two waveforms are two separate real
physical signals, while the d(t) is only measured result of the applied mathematical
operation.
One way of looking at single–ended versus differential signals is that amplitude
of a single–ended signal is measured relative to a stable reference level (to which we
arbitrary assign value of “zero” and for the convenience we name it GND), while
“differential” measurement implies that the desired signal and its reference level
are phase inverted versions of each other, for instance a(t) and b(t) in Fig. 15.3
(left). Mathematically speaking, both measurements are relative, thus we can always
declare instantaneous value of, for instance, b(t) = 0 and imagine that at every
moment a(t) is measured relative to the “instantaneous zero level”. Also, one can
always imagine to be “walking along” the a(t) path while measuring the “distance”
to b(t). If one needs to “look down” (Fig. 15.3, green arrow (1)) then the distance
is positive, and “looking up” (Fig. 15.3, green arrow (2)) means the negative result
(Fig. 15.3, right).
Let us see again how this case looks from strictly mathematical point of view,
where the two complementary general signals a(t) and b(t) have common mode
levels CMa and CMb , have same frequency ω a = ω b = ω , and are subjected to the
interfering noise signals n a (t) and n b (t), i.e.
15 Electrical Noise: Solutions
111
a(t) = CMa + Aa sin ω t + n a (t)
b(t) = CMb + Ab sin(ω t + π ) + n b (t)
∴
a(t) = CMa + Aa sin ω t + n a (t)
b(t) = CMb − Ab sin ω t + n b (t)
(15.5)
Obviously, the difference is
d(t) = a(t) − b(t)
= [CMa − CMb ] + [(Aa + Ab ) sin ω t] + [n a (t) − n b (t)]
(15.6)
It is interesting to note that (15.6) implies that if n a (t) = n b (t) = n(t), i.e. both
signals a(t) and b(t) are exposed to the same interference n(t), then the common
noise is canceled. Similarly, if CMa = CMb = CM, then the resulting signal is at
zero common mode level, i.e. CMd = 0. Finally, if the two amplitudes are equal, i.e.
Aa = Ab = A, then this special case of (15.6) reduces to
d(t) = 2 A sin ω t = D sin ω t
(15.7)
where, D = 2 A. From Figs. 15.3 (right) and (15.7) we observe two important
consequences of deriving the differential signal. First, both the common noise and
the common mode DC level are cancelled. Second, the amplitude of the resulting
amplitude of the differential signal is twice the amplitude of either a(t) or b(t) signals, i.e. D = 2 A. Thus, mathematical operation of subtracting these two signals
results in the new signal that is effectively amplified by the factor of two, i.e. there
is inherent 6 dB voltage gain in this operation.
These two properties of a differential signal, i.e. the noise cancelation and the
inherent gain, are attractive enough to justify the additional effort required to actually
implement the subtraction function and to operate with two signals instead of one.
Therefore, it is valid to ask the following question, how realistic is to expect that both
signals a(t) and b(t) are subjected to the same noise, and what is the implementation
cost of the differential signal processing system?
When two wires carrying the two signals, a(t) and b(t), are physically located
close and parallel to each other along their full lengths, it is reasonable approximation that through either inductive or capacitive coupling any external signal (thus
perceived as the “noise”) will be equally injected into both wires. Therefore, we
establish an engineering rule for designing a “differential signal path” by specifying
that the two wires must be close and parallel to each other, as much as possible,
along the full length of the path. For instance, that is one of the reasons why “twisted
pair” cable is used in phone connections.
The cost of benefiting from the noise cancellation and the achieved 6 dB gain is
that two wires and two signals must be used instead of one. In addition, an operational amplifier is needed to perform the subtraction of the two signals. Thus,
112
15 Electrical Noise: Solutions
processing differential signals necessitates the use of more wires, and the use of both
single–ended–to–differential and differential–to–differential operational amplifiers.
That is, the overall hardware complexity is drastically increased.
Then one may also ask, why do not we use only single–ended signal processing
instead, which is simpler do design and implement? To answer this question, we
should keep in mind the main purpose of communication electronics, which is to
transmit a signal as far as possible while using as little as possible amount of energy.
One possible analogy could be with the need to drive a car as far as possible with
the minimum amount of fuel. After travelling a long distance, the signal inevitably
becomes attenuated and distorted by the noise. At this moment, the removal of
noise and the 6 dB gain make the whole difference in the world. Therefore, to make
efficient and elegant systems, whenever possible we must eventually take advantage
of differential signal processing.
Of course, in order to achieve full benefits of differential design, we must make
sure that the two paths and all associated components are perfectly matched so that
the desired cancelations are achieved.
As further exercise, plot more graphs similar to Fig. 15.3, where for example the
noise is injected only into one of the two signals, or where the two CM levels are not
identical, or any other mismatch combination that you could think of. In all cases,
try to estimate the resulting SNR.
3.4. By noting that the RLC network in Fig. 3.1 consists of a resistor R in series with
the parallel impedance Z LC = Z L ||Z C , we write
1
ZC =
Z L = jωL
(15.8)
jωC
∴
ZC Z L
jωL
(15.9)
Z LC =
=
ZC + Z L
1 − ω2 LC
Then, we simply apply the voltage divider rule for R and Z LC , Fig. 3.1, and directly
write expression for the voltage transfer function H ( jω) as2
H ( jω) =
vout
Z LC
=
=
vin
R + Z LC
1
R
Z LC
+1
=
jωL
R(1 − ω2 LC) +
jωL
∴
|H ( jω)| = H (ω)
=
ωL
R 2 (1 − ω 2 LC)2 + ω 2 L 2
=
R
1
1
ω 2 L2
− 2 CL + ω 2 C 2 + 1
(15.10)
2
|z| =
√(z)2 + →(z)2
15 Electrical Noise: Solutions
113
Fig. 15.4 Solution 3.4: frequency domain plots of a resonator transfer function H (s)
After choosing component values R = 1 π, 10 π, 100 π, L = 1 H , and C = 1 F, it
is now trivial to plot either of the two forms of function (15.10) and create Fig. 15.4
for the three resistor values.
However, let us stop here for a moment and have the following discussion. First,
let us take a closer look at Fig. 3.1. From strictly electrical point of view we observe
that the node aiis shorted to the ground if either L = 0 H or C = ∞ F, i.e. if in
accordance to (15.8) if either Z L = 0 or Z C = 0. That is to say, in either of these two
cases the output voltage becomes Vout = 0 V , which leads into |H (ω )| = 0. But,
we also know that Z L (ω ) = 0 if ω = 0 (i.e. at DC), and Z C (ω ) = 0 if ω = ∞ (as
shown by (15.8)) even if the two components (L , C) have non–zero finite values.
Same conclusion may be reached from strictly mathematical perspective, by
finding the following limits of function (15.10) as
lim H (ω ) =
ω ≤0
R
lim H (ω ) =
ω ≤∞
R
1
1
0
− 2 CL + 0 + 1
=
1
1
∞
− 2 CL + ∞ + 1
1
=0
∞
=
1
=0
∞
As expected, our conclusion based on engineering reasoning is consistent with
the mathematical behaviour of function (15.10), i.e. gain of its equivalent network
in Fig. 3.1, for the two extreme frequencies. Similarly, we make additional observation that the maximum value of H (ω ) = 1, simply because that is the maximum
gain of a passive RLC network. In this particular case the maximum gain happens
when Z LC becomes an open connection. Obviously, open connection implies infinite impedance, which from (15.9) leads into conclusion that Z LC = ∞ when
≥
ω = 1/ LC . Note that the same conclusion is reached by using strictly calculus
methods and by finding maximum of the first derivative of (15.10). It is really matter
of convenience which method is used. For instance, from the first form of |H (ω )| in
(15.10) we easily find that the maximum is achieved when
114
15 Electrical Noise: Solutions
lim
ω ≤ ≥1
LC
H (ω ) = R 2 (1 −
L
C
1
LC
LC)2 +
L
C
=
L
C
L
C
=1
(15.11)
These observations reveal to us that gain of the RLC network, as described by H (ω ),
has very interesting property that, as the frequency ω changes from zero to infinity,
its value starts at zero, rises to one, and falls again to zero. Intuitively, we can say
that having a function whose gain changes with frequency must be useful for some
engineering purpose.
In the case of this particular function, we note that if H (ω ) is multiplied by
another function G(ω ), then
G(ω ) H (ω ) = G(ω ) if, and only if ω = ω 0 = ≥
1
LC
(15.12)
where, ω 0 is referred to as the resonant frequency and is arguably the most important
variable in wireless communications. Here is an example of how we interpret and
use this seemingly trivial property of the RLC network. If instead of being a single tone signal, i.e. G(ω 1 ), the function G(ω ) consist of several harmonics, i.e.
G(ω 1 , ω 2 , . . . , ω 0 , . . . , ω n , . . .), then by applying the RLC network function we
can “extract” the signal whose frequency is equal to the resonant frequency (because
that tone is multiplied by one), and “remove” all the others (because they those tones
are multiplied either by less then one or by zero). For instance, if we set ω 0 = ω 2
then
G(ω 1 , ω 2 , . . . , ω 0 , . . . , ω n , . . .) H (ω 2 ) = G(ω 2 ) if ω 2 = ω 0 = ≥
1
LC
(15.13)
In this scenario, G(ω 1 , ω 2 , . . . , ω 0 , . . . , ω n , . . .) may represent the mix of all
various frequencies of the signals present around us due to transmission of all radio
and TV stations, cell phones, satellites…. However, if the RLC network is incorporated into our receiver, we effectively create the “entry door” into the system that
rejects all other frequencies except the ω 0 , which means that the receiver is tuned
into, i.e. it hears only the transmitter working on the ω 0 frequency and completely
rejects all the other transmitters working at different frequencies. What is more, by
changing values of either L or C we can change value of ω 0 and tune the receiver to
some other transmitting frequency.
On the other hand, value of the real resistance R is directly responsible for the
width of the frequency bandwidth BW. We will come back to this topic again, but
for the moment we keep in mind that design of narrowband RLC filters assumes that
we use an ideal inductor made of ideal wire with zero resistance, i.e. there are no
internal thermal losses in the system and the corresponding energy loss.
A brick–wall type bandpass (BP) filter transfer function is defined as being
centered around frequency ω 0 , to have gain A = 1 within the frequency range
BW , and gain A = 0 for any other frequency (Fig. 15.4, right). The bandwidth BW
15 Electrical Noise: Solutions
(a)
115
(b)
(c)
Fig. 15.5 Solution 3.5: frequency domain noise plots
itself is defined by two points on the transfer curve that are 3 dB below the function’s
maximum value. Therefore, the consequence of applying BP filter to any multi–
tone signal (i.e. using multiplication operation) is that all frequencies outside of the
BW frequency range are completely suppressed (because amplitudes of those tones
are multiplied by zero gain), while the power level inside the BW frequency range
stays unchanged (because amplitudes of those tones are multiplied by the gain of
one). Obviously, the “brick–wall” filter is only a crude approximation of the real BPF
transfer function, nevertheless it is very practical and drastically simplifies our system
analysis. This approximation is even more justified if we keep in mind that −3 dB
is equivalent to 1/2 of the original power level. From the engineering perspective,
when the power level drops by half or more, we justify the BPF approximation and
neglect those low power levels (while keeping in mind that the calculated SNR will
be overly optimistic).
3.5. To the first approximation, power spectral density (PSD) function of a white
noise type signal is assumed flat across all frequencies in the (DC to f ≤ ∞) range,
that is the PSD function is at the constant power level P. In other words, when
comparing any two 1 Hz wide frequency windows, as illustrated by red boxes in
Fig. 15.5a, the measured power levels are always the same. For example, the same
power is contained in (10–20 Hz) as is in (1,000–1,010 Hz) frequency window. A
brick–wall type bandpass (BP) filter transfer function is defined as being centered
around frequency ω 0 , to have gain A = 1 within the frequency range BW , and gain
A = 0 for any other frequency (Fig. 15.5b). Therefore, the consequence of applying
BP filter to the white noise (i.e. using multiplication operation) is that all frequencies
outside of the BW frequency range are completely suppressed (because amplitudes
of those tones are multiplied by zero gain), while the power level inside the BW
frequency range stays unchanged (because amplitudes of those tones are multiplied
by the gain of one) (Fig. 15.5c).
We dully note that the total amount of energy contained within white noise is, by
definition, infinite. Direct consequence is that if the complete white noise is allowed
to enter an amplifier, then the amplifier would need to draw an infinite amount of
energy from the energy source (e.g. the battery) in order to process the noise signal.
And that is even before the wanted signal is introduced into the system.
Naturally, this is not realistic engineering solution. Therefore, we must apply BP
filtering operation before the the amplification, so that BW of the BP filter is wide
116
15 Electrical Noise: Solutions
enough to let through only the signal itself, and at the same time to allow only limited
amount of white noise to “sneak by” and enter the system. With this approach, both
the signal and noise are amplified by using finite energy source (even though we
never wanted to amplify the noise in the first place, but that is unavoidable). By
using BP filter, we now calculate SNR by comparing complete signal energy with
the amount of noise energy that was allowed to enter (which is now finite).
3.6. By definition, for the known P S D, the total power P contained within a given
frequency ( f 1 , f 2 ) window is calculated by solving integral
P=
f2
PSD df
(15.14)
f1
which, for the trivial case of P S D = const and ( f 1 , f 2 ) = (0, ∞), reduces to
P = PSD
∞
d f = P S D × ∞ = ∞ [W]
(15.15)
0
for any positive value of (P S D > 0). Or, in geometrical sense, the integral (15.14) is
just another way of calculating rectangular area under the P S D function, i.e. greyed
areas in Fig. 15.6. Obviously, if at least one side of a rectangle is infinitely long then
its area must be infinite. Similarly, we calculate noise power within bandwidth of the
human voice (i.e. BW = 20 Hz, 20 kHz) while holding the same P S D as
Pn = P S D
20×103
d f = 1 µW/Hz × 19, 980 Hz = 19, 980 µW
(15.16)
20
within the given BW. Thus, the total noise power is, simply, the area under P S D
function in Fig. 15.6 (right). This example clearly illustrates the need for using BP
filtering function in front of the amplification function. For instance, for a given
Ps = 100 mW voice signal, we calculate the two SNR levels as
SNR =
100 mW
Ps
= −∞ dB
=
Pn
∞
and
SNR =
Ps
100 mW
7 dB
=
Pn
19, 980 µW
In other words, without the application of a BPF to help us limit power of the
surrounding noise, the wanted signal is completely swamped by the noise and it
would be impossible to recover it again. In that case, the noise power is infinitely
greater than the signal power. Thus, we always try to design BPF whose BW is wide
just enough to let the wanted signal through, which results in the maximum possible
SNR for the given design.
15 Electrical Noise: Solutions
117
Fig. 15.6 Solution 3.6: power spectrum density (PSD) functions
As a side note, we keep in mind that any realistic electronic system (including
amplifiers, of course) has its own inherent BW due to the fact that the realistic
components used to manufacture the electronic system are frequency limited on
their own. We keep this point in mind throughout the rest of the book.
3.7. By definition, instantaneous amplitude of a random noise source is in accordance
with Gaussian probability distribution function, thus, variance φ 2 is equivalent to the
average mean–square variation of the amplitude around its average value. In other
words, the average mean-square current or voltage variation around its average value,
i 2 or e2 , is equivalent to the variance φ 2 , which leads into conclusion that the rms
current/voltage value equals to the standard deviation φ .
For instance, in the case of two random voltage sources acting in series, this
statistical property of random signals results in the following conclusion for the total
average mean–square voltage e2
2
2
2
2
2
2
2
2
2
2
ettr
= e1rms
+ 2 e1rms
e2rms
+ e2rms
= e1rms
+ e2rms
ms = e1rms + e2rms
(15.17)
because, the average value of product of two independent random values is zero, i.e.
2
2
e1rms
e2rms
= 0. Thus, we generalize (15.17) for n thermal voltage sources
2
2
2
+ e2rms
+ · · · + enrms
e2 = e1rms
(15.18)
Therefore, we keep in mind that the noise powers of random signals add up, not the
noise voltages. For simplicity, in the rest of this book we assume the average and
rms values and simply write that
e=
e12 + e22 + · · · + en2
(15.19)
For the specific numbers used in this example we write,
e=
e12 + e22 = 12 + 102 V = 10.05 V
(15.20)
118
15 Electrical Noise: Solutions
The above result clearly illustrate that within a network that contains more than one
source, main contributor to the noise voltage is the source with the highest amplitude.
In this particular case, by simply ignoring the ten times smaller source the overall
error is only 5 % relative to the exact value. By recognizing this fact, we can quickly
estimate the total voltage contribution without using exact numerical calculations.
3.8. Thermal noise generated by a resistor is modelled as a simple ideal voltage
source en whose voltage amplitude is function of the internal resistance R [π], the
environment temperature T [K], and frequency bandwidth B = B [Hz] as
en (R) =
≥
4kT R B
V
(15.21)
where, R represents equivalent resistance of the resistive network, and k is Boltzmann
constant.
Following up on discussion in Problem 3.7, and while keeping in mind that the
noise powers add up, not the noise voltages, first, we find the equivalent network
resistance, and then finding the equivalent thermal noise generator is straightforward
by using (15.21).
When two resistors R1 and R2 are connected in series, then the equivalent
resistance is R S = R1 + R2 , and their equivalent mean–square noise voltage of
the Thévenin voltage source
2
2
2
= en1
+ en2
= 4kT B(R1 + R2 ) = 4kT B R S
en12
V2
(15.22)
Similarly, when two resistances are connected in parallel, 1/R P = 1/R1 + 1/R2 ,
we write expression for their equivalent mean–square noise current of the Norton
current source by using equivalent noise conductances G = 1/R as
2
2
2
= i n1
+ i n2
= 4kT B(G 1 + G 2 ) = 4kT B(1/R1 + 1/R2 )
i n12
∴
2
= 4kT B(R1 ||R2 ) = 4kT B R P
en12
V2
(15.23)
Therefore, we find,
(a) Each of the two stand–alone separate resistors generates thermal noise as follows:
en2 (R1 ) = 4 × 20 kπ × 1.3806488 × 10−23 J/K × 290 K × 20 kHz
= 6.406 × 10−12 V2
en2 (R2 ) = 4 × 50 kπ × 1.3806488 × 10−23 J/K × 290 K × 20 kHz
= 16.016 × 10−12 V2
∴
en (R1 ) = 2.531 µV
en (R2 ) = 4.002 µV
15 Electrical Noise: Solutions
119
Fig. 15.7 Problem 3.8:
model of a realistic resistor
(left), and illustration for
available noise power (right)
(b) Equivalent resistance Rs of the two resistors in series is
R S = (20 kπ + 50 kπ) = 70 kπ ∴ en (R S ) = 4.735µV.
(c) Equivalent resistance R p of the two resistors in parallel is
R P = (20 kπ||50 kπ) = 14.286 kπ ∴ en (R P ) = 2.139 µV.
(d) With the given square value of the noise voltage en2 and its associated resistance
R, it is straightforward to calculate available noise power. Available, i.e. maximum possible power, is delivered to the load if the load impedance equals
the source impedance. Under the matched impedances condition Z S = Z ∗L ,
Fig. 15.7 (right), only one half of the source voltage is available to the load,
therefore for a single resistor R S we write the expression for available noise
power as
Pn =
(en/2)2
e2
4R S kT B
= n =
= kT B
RS
4 RS
4 RS
(15.24)
For two resistors R1 and R2 in series, the equivalent resistance is R S = R1 + R2 ,
which after using (15.22) gives expression for the noise power as
Pn (R1 + R2 ) =
(en/2)2
4kT B(R1 + R2 )
= kT B
=
R1 + R2
4(R1 + R2 )
(15.25)
Finally, when two resistances are in parallel, we write expression for their available noise power as
Pn (R1 ||R2 ) =
(en/2)2
4kT B(R1 ||R2 )
= kT B
=
R1 ||R2
4(R1 ||R2 )
(15.26)
Therefore, the resulting available noise power is always the same, Pn = kT B =
8.008 ×10−17 W. We dully note that passive networks do not have power amplification.
Finally, we note that in order to reduce the thermal noise, everything else being
equal, the equivalent resistance needs to be reduced. Thus, we have one guideline for
design of low–noise circuits: the internal circuit nodes should have low–resistance
as much as possible, which is often contradictory to other requirements that we try
to achieve with the same circuit. Thus, in the circuit design process we always have
to make some engineering compromises.
120
15 Electrical Noise: Solutions
3.9. For the series resistor combination in Fig. 3.2 (middle), for the two resistors at
two different temperatures, we write
en2 = e12 + e22 = 4k∂f R1 T1 + 4k∂f R2 T2
= 4k∂f (R1 T1 + R2 T2 + R1 T2 − R1T 2)
= 4k∂f [T2 (R1 + R2 ) + R1 (T1 − T2 )]
= 1.680 × 10−11 V2
(15.27)
where all of the above expressions are identical and they result in the same numerical
answer. Expression for the available noise power Pn is found by definition, where
the noise power is calculated across the series connection of the two resistors, as
(en/2)2
4k∂f [T2 (R1 + R2 ) + R1 (T1 − T2 )]
=
R1 + R2
4(R1 + R2 )
R1
= kT2 ∂f + k∂f (T1 − T2 )
R1 + R2
= 6 × 10−17 W
Pn =
(15.28)
We observe that if (T1 = T2 = T ) the second term in (15.28) cancels itself and
the noise power expression collapses to the already known expression Pn = kT ∂f .
However, when two resistors are at different temperatures then, through the voltage
“gain” factor R1 /(R1 + R2 ) in the second term, the resistors create voltage divider
effect for the noise voltage sources.
3.10. By definition, noise factor F is calculated as the ratio of the output noise power
Po and the input noise power Pin . Therefore, by using the same reasoning and results
from problems 3.8 and 3.9, we write
(a) Expression for the output noise power of serial resistor network is already found
as in (15.28). The “input” noise power is, however, only due to thermal noise of
resistor R1 distributed over the series resistance R S = R1 + R2 . Thus, we write
Pin =
(en (R1 )/2)2
4kT1 B(R1 )
R1
= kT1 B
=
R1 + R2
4(R1 + R2 )
R1 + R2
(15.29)
We can now easily find expression for the noise figure NF by substituting (15.28)
and (15.29) as
15 Electrical Noise: Solutions
F=
Po
=
Pin
121
kT2 B + k B(T1 − T2 )
R1
R1 + R2
R1
R1 + R2
T2 R1 + R2
=1+
−1
T1
R1
∴
T2 R1 + R2
−1
NF = 10 log F = 10 log 1 +
T1
R1
kT1 B
dB
(15.30)
Obviously, if R1 ∓ R2 , then [(R1 + R2 )/R1 ] ≤ 1, which leads into F ≤ 1 and
NF ≤ 0 dB. In all other cases NF becomes greater than 0 dB. Equation (15.30)
implies that in the case when the load is connected in series to the source, in
order to reduce NF the source impedance should be as high as possible, i.e. much
higher than the load impedance.
(b) In the case of parallel resistive network, but this time we use G = 1/R terms, the
output noise power is found as
Po = kT2 B + k B(T2 − T1 )
R2
R1 + R2
(15.31)
while the input available noise power is found as
Pin = kT1 B
R2
R1 + R2
(15.32)
which leads directly into
Po
F=
=
Pin
kT2 B + k B(T2 − T1 )
R2
R1 + R2
R2
R1 + R2
T2 R1 + R2
=1+
−1
T1
R2
∴
T2 R1 + R2
−1
NF = 10 log F = 10 log 1 +
T1
R2
kT1 B
dB
(15.33)
Obviously, this time, if R2 ∓ R1 , then [(R1 + R2 )/R2 ] ≤ 1, which leads
into F ≤ 1. In all other cases NF becomes greater than 0 dB. Equation (15.33)
implies that in the case when the load is connected in parallel to the source, in
order to reduce NF the load impedance should be as high as possible, i.e. much
higher than the source impedance.
122
15 Electrical Noise: Solutions
3.11. In (15.24) we already confirmed that power and temperature are directly
proportional, thus the equivalent temperature is yet another unit of expressing power
of the incoming signal within the given frequency bandwidth B.
For the case of a signal that is present at the antenna being calibrated against the
local thermal etalon, we can derive the following relationship.
PR = kTR B W
⇒
PR = 10 log(kTR B) dB
∴
PS = PR + 3.01 dB
= 10 log(kTR B) + 3.01 dB
= 10 log(kTR B) + 10 log(2)
= 10 log(2kTR B) dB ⇒ PS = 2kTR B W
(15.34)
where, PS is the measured power of the antenna signal. To find the equivalent antenna
temperature Ta , from (15.24) we write
Ta =
2kTR B
PS
=
= 2TR = 5 K
kB
kB
(15.35)
Similarly, after setting PS = 6.02 dB = 10 log(4) W, PS = 9.03 dB = 10 log(8) W,
etc., by repeating as same calculation as above, we respectively calculate Ta =
5 K, 10 K, 20 K, . . ., which illustrates the relationship between noise power and temperature.
Additionally, we note that the noise temperature is also used in the radio astronomy
to determine physical temperatures of the celestial bodies (once some other variables
are also known).
3.12. Although noise analysis of amplifiers may be involved, it follows already known
principles: (a) each noisy component is replaced with its equivalent noiseless component model plus the ideal voltage/current source that models the associated noise
voltage/current itself, and (b) networks containing multiple sources are solved by
applying the superposition principle.
Thus, the noisy operational amplifier is replaced with its equivalent noiseless
version plus the thermal input–referred voltage noise source en at the positive input
node, Fig. 15.8. Doted line represents boundaries of the original noisy operational
amplifier. We note that polarity of ideal thermal noise sources is meaningless because
the signal is random. In addition, in the noise analysis we use the squared voltage
values for the statistical addition of multiple noise signals (which simplifies the
analysis because the signal power is proportional to the squared voltage).
Contribution of voltage noise source en : is found by shorting the input signal source
v S , thus for the non–inverting configuration input–output relationship we write
eout (en ) = i R1 (R1 + R2 ) =
en
(R1 + R2 )
R1
(15.36)
15 Electrical Noise: Solutions
123
Fig. 15.8 Solution 3.12:
schematic diagram of a noisy
operational amplifier
Fig. 15.9 Solution 3.13: noise
calculation schematic diagram
for operational amplifier
Therefore, for the given numerical data we write
≥
≥
10 kπ
5 nV/ Hz = 55 nV/ Hz
eout = 1 +
1 kπ
(15.37)
It would be now trivial to calculate
≥ the total noise voltage within a certain BW, simply
by multiplying the last result by BW . For example, if the amplifier bandwidth were
specified as BW = 400 Hz–19 kHz, the total measured noise voltage at the output
terminal would have been
≥
(15.38)
eout (total) = 55 nV/ Hz × (19 k − 400)Hz = 7.5 µV
3.13. The noisy operational amplifier is replaced with its equivalent noiseless version
plus the thermal voltage noise source en at the positive input node, and the thermal
current noise sources i n− and i n+ associated with its input terminals (Fig. 15.9).
Doted line represents boundaries of the original noisy operational amplifier.
Now, there is more than one noise source, and we have to apply the superposition
principle. Each of the noise sources is considered on its own as follows.
Contribution of voltage noise source en is already found in (15.36), therefore for the
given numerical data
124
15 Electrical Noise: Solutions
Fig. 15.10 Solution 3.13:
schematic diagram for noise
calculations
≥
≥
10 k
5.082 nV/ Hz = 55.902 nV/ Hz
eout (en ) = 1 +
1k
∴
(eout (en ))2 = 3.125 × 10−15
V2/Hz
(15.39)
Contribution of current noise source i n− : is found after shorting the noise voltage
source, and after the current noise source i n+ is left open (Fig. 15.10). Under these
circumstances, we can reason that there is no noise current flowing through R1
because both of its nodes are at the same potential (i.e. gnd). Therefore, the complete
noise current i n− is forced through R2 , which generates noise voltage at the output
node (keep in mind that one node of R2 is at the gnd potential)
ein− = i n− R2
∴
2
ein−
= (i n− R2 )2
(15.40)
Therefore, for the given numerical data we write
≥
≥
ein− = i n− R2 = 5pA/ Hz × 10 kπ = 50 nV/ Hz
∴
2
ein−
= 2.5 × 10−15 V2 /Hz
(15.41)
Contribution of current noise source i n+ : is found after shorting the noise voltage
source, and after the current noise source i n− is left open, Fig. 15.11. We reason that
there is no resistor for i n+ to develop voltage, i.e. (i n+ × 0 π) = 0, because the
current source is shorted by ideal wire (of course, in this case we are not discussing
the theoretical dissipated power by the ideal source).
Therefore, total noise contribution of the two combined sources is found by
statistical addition (i.e. by adding the noise powers) of (15.39) and (15.41) as
15 Electrical Noise: Solutions
125
Fig. 15.11 Solution 3.13:
schematic diagram for noise
calculaitons
Fig. 15.12 Solution 3.14:
schematic diagram for noise
calculations
2
eout
=
2
een
2
+ ein−
2
R2
en + (i n− R2 )2
= 1+
R1
= (3.125 + 2.5) × 10−15 V2 /Hz = 5.625 × 10−15 V2 /Hz
∴
≥
eout = 75 nV/ Hz
(15.42)
3.14. Now, both the noisy operational amplifier and noisy resistors are replaced
with their equivalent noiseless versions plus the respective thermal voltage noise
sources (Fig. 15.12). Important point to keep in mind is that the thermal noise voltage
appears across the noisy resistor terminal. That is, in our model that consists of
both noiseless resistor and the associated thermal voltage generator, the thermal
noise appears between the same circuit nodes where the noisy resistor resistor was
connected. For example, in Fig. 15.12 the noise voltage e R2 is assumed to be between
the negative input terminal and the output terminal of the operational amplifier. That
is exactly where the noisy resistor R2 was connected, and therefore in this noise
model we do not account for the voltage drop through the corresponding noiseless
R2 element.
Thus, relative to the already found solution (15.42) in Problem 3.13, we need to
add contribution of the two noisy resistors.
Contribution due to noise source e R1 : is found from the equivalent circuit network
(Fig. 15.13, left) as follows. The analysis can be done in a number of ways, here we
use engineering reasoning to say that voltage e R1 appears at the negative input of the
126
15 Electrical Noise: Solutions
operational amplifier (remember: e R1 voltage is generated across the noisy version
of R1 , which was connected between the ground node and the negative input node).
Therefore, the noise current through R1 must be i R1 = e R1 /R1 . The same current
must flow through R2 , thus we write
eout (R1 ) = i R1 R2 =
∴
2
eour
(R1 ) = e2R1
R2
R1
e R1
R2
R1
2
= 4kT R1 ∂f
= 4 × 1.3806488 × 10
−23
R2
R1
2
V2 = 4kT R1
× 276.9 × 1 k ×
10 k
1k
R2
R1
2 2
V2 /Hz
V2 /Hz
= 1.529 × 10−15 V2 /Hz
(15.43)
where, in order to simplify the rest of calculations, for the moment we prefer to use
units.
Contribution due to noise source e R2 : is found from the equivalent circuit network
(Fig. 15.13, right) as follows. The noise voltage is is due to the noisy R2 resistor that
was connected between the amplifier output node and virtual ground, therefore we
simply write
V2/Hz
2
(R2 ) = 4kT R2 ∂f V2 = 4kT R2 V2 /Hz
eout (R2 ) = e R2 ∴ eour
2
eour
(R2 ) = 4 × 1.3806488 × 10−23 × 276.9 × 10 k V2 /Hz
= 0.153 × 10−15 V2 /Hz
(15.44)
Therefore, after combining (15.42) to (15.44), the complete solution is
2
eout
=
R2
1+
R1
2
en
+ (i n− R2 ) + 4kT R1
2
R2
R1
2
+ 4kT R2
V2 /Hz
∴
≥
(3.025 + 2.5 + 1.529 + 0.153) × 10−15 V/ Hz
≥
= 84.895 nV/ Hz
eout =
(15.45)
In the above analysis it is useful to keep the same order of magnitude for the
intermediate results, so that it becomes obvious which component is the main contributor to the noise. Also, by keeping the B information away, simplifies the calculations. Therefore, it is now convenient to calculate the total noise rms voltage by
performing the B calculation only once as
15 Electrical Noise: Solutions
127
Fig. 15.13 Solution 3.14: schematic diagram for noise calculaions
Fig. 15.14 Solution 3.15: schematic diagram for noise calculations
eout (total) = eout ×
= 12µV
≥
≥
B = 84.895 nV/ Hz × (20 kHz − 20Hz)
(15.46)
The above analysis is valid under assumption that the noise spectrum density function
is constant within the given bandwidth. Otherwise, the exact definition for a resistor’s
thermal noise that includes the frequency dependent integral must be used.
3.15. Main purpose of calculating noise figure NF of a device is to quantify how much
noise power is actually created internally within the device itself, which makes NF
one of the device’s very useful quality measures.
We keep in mind that square voltage is equivalent to normalized power (because
V2 /1 π ∼ W), and also that division of a square voltage by the unit of frequency
(i.e. V2 /Hz) is equivalent to power spectral density en2 .
Complete schematic diagram suitable for amplifier noise analysis includes all
thermal noise sources, Fig. 15.14, along with their respective noiseless components.
By definition, noise factor F is a ratio of SNRi at the input of a device, and Signal–
to–Noise–Ratio at its output SNRo . Thus, in order to calculate F we have to find
ratio of the noise power PI that was delivered to the amplifier by the source (i.e.
presented at the input node 1i), and the total noise power eout at the output of the
128
15 Electrical Noise: Solutions
amplifier. The fact that eout is measured at the amplifier’s output node complicates the
analysis because the source noise power is measured at the input node. Additionally,
the output noise power eout is the sum of the amplified input noise power plus the
internally generated noise power. It is important to realize, therefore, that in order
to compare the input noise power PI , as delivered by the source to the input node,
with the internally generated noise power, the amplifier noise contribution needs to
be referenced relative to the same node. Thus, we introduce the term input–referred
voltage/current noise power PA . In accordance with this definition, PA is equivalent
to the complete internally generated noise power that is integrated at the input node,
i.e. before the amplification.
Naturally, if an amplifier is ideal (i.e. noiseless) the input–referred noise power
PA = 0, which is equivalent to F = 1. In that case, the output noise power is
due to only the source noise power amplified by the amplifier. We see how any noise
internally generated by the amplifier is easily quantified by the noise factor. Assuming
the amplifier gain is A, the total output noise power is then A(PI + PA ), the input
signal power is PS and, with this in mind, it is convenient to rewrite expression for
noise figure F as
PS
SNRi
PA
PI
=
(15.47)
F=
=1+
A PS
SNRo
PI
A (PI + PA )
which clearly illustrates the point. We now derive input–referred noise and the input
noise powers as follows.
Contribution due to noise source e R S : thermal noise voltage generated by the source
resistance R S is delivered to the input node through the resistive voltage divider
(R S , R3 ), thus at node 1iwe write
2
R3
R S + R3
(15.48)
When the source resistance is matched to the amplifier input impedance, i.e. R S = R3
from (15.48) we have
eI =
R3
eRS
R3 = 4kT R S
R S + R3
R S + R3
∴
PI = e2I = kT R3
V2
Hz
e2I = 4kT R S
∴
PI = 5.083 × 10−18 V2 /Hz
(15.49)
Contribution due to noise source e R3 : thermal noise voltage generated by the source
resistance R3 is delivered to the input node through the resistive voltage divider
(R3 , R S ), thus at node 1iwe write
15 Electrical Noise: Solutions
129
2
RS
∴
= 4kT R3
R S + R3
(15.50)
When the source resistance is matched to the amplifier input impedance, i.e. R S = R3
from (15.51) we have
RS
e R3
e3 =
R S = 4kT R3
R S + R3
R S + R3
e32 = kT R3
e32
V2
Hz
(15.51)
Contribution due to noise source e R1 : thermal noise voltage generated by the source
resistance R1 is delivered to the input node through the resistive voltage divider
(R1 , R2 ), thus at node 1iwe write
2
R2
R1 + R2
(15.52)
(Noise voltage source e R2 is shorted (its internal resistance is zero), and we remember
that an operational amplifier is also a voltage source, thus its output impedance is
zero. That is, for purposes of this analysis R2 is effectively connected to the gnd node
through the operational amplifier’s output.)
Contribution due to noise source e R2 : thermal noise voltage generated by the source
resistance R2 is delivered to the input node through the resistive voltage divider
(R2 , R1 ), thus at node 1iwe write
e1 =
R2
e R1
R2 = 4kT R1
R1 + R2
R1 + R2
∴
e12 = 4kT R1
2
R1
R1 + R2
(15.53)
Contribution due to noise voltage source en : the internal thermal noise voltage generated by operational amplifier is directly connected to the input node, thus at node
1iwe write
2
= en2
(15.54)
en1 = en ∴ en1
e1 =
R1
e R2
R1 = 4kT R2
R1 + R2
R1 + R2
∴
e12 = 4kT R2
Contribution due to noise current source i n+ : this current source forces noise current
through parallel connection R3 ||R S , which generates voltage directly at the input
node,
ein+ = i n+ R3 ||R S
∴
2 R3 R S
R3 2
2
ein+
= i n+
= i n+
R3 + R S
2
(15.55)
Contribution due to noise current source i n− : this current source forces noise current
through parallel connection R1 ||R2 , which generates voltage directly at the negative
input node of operational amplifier, which appears also at the positive input node,
thus
2
R1 R2
2
(15.56)
ein− = i n− R1 ||R2 ∴ ein− = i n−
R1 + R2
130
15 Electrical Noise: Solutions
Therefore, after combining the six amplifier thermal noise power terms (15.51)
to (15.56), the complete solution for the internal noise power spectrum density is
PA =
2
eout
2
2
R2
R1
+ 4kT R2
+
R1 + R2
R1 + R2
R3 R S 2
R1 R2 2
2
+ i n−
V2 /Hz
R3 + R S
R1 + R2
= kT R3 + 4kT R1
2
en2 + i n+
∴
PA = (5.083 + 16.803 + 1.680 + 25 + 20.661 + 6.25) × 10−18 V2 /Hz
= 75.447 × 10−18 V2 /Hz
(15.57)
It is now straightforward to substitute (15.57) and (15.49) into (15.47) and to calculate
F =1+
75.477 × 10−18 V2 /Hz
= 15.849
5.083 × 10−18 V2 /Hz
∴
NF = 10 log(15.849) = 12 dB
(15.58)
3.16. Using the same methodology from the previous problems, while paying attention to the subtle differences between gains of non–inverting and inverting amplifiers,
with reference to Fig. 15.15 we derive NF as follows.
Contribution due to noise source e R S : thermal noise voltage generated by resistance
R S is delivered to the input node through the resistive voltage divider (R S , R1 ||R4 ),
thus at node 1iwe write
eR S =
R1 ||R4
eRS
(R1 ||R4 ) = 4kT R S
R S + R1 ||R4
R S + R1 ||R4
∴
(eR S )2 = 4kT R S
R1 ||R4
R S + R1 ||R4
2
V2/Hz
(15.59)
In order to match the source resistance to the amplifier input impedance first we
calculated R4 so that R S = R1 ||R4 , i.e. R4 = R1 R S /(R1 − R S ), then from (15.59)
we write
PI = (eR S )2 = kT R S
V2
Hz
∴
PI = 3.909 × 10−18 V2 /Hz
(15.60)
15 Electrical Noise: Solutions
131
Fig. 15.15 Solution 3.16: schematic diagram for noise calculations
Fig. 15.16 Solution 3.16: schematic diagrams for noise calculations
Contribution due to noise source e R3 : thermal noise voltage due to resistance R3
is directly
to the positive input node of the operational amplifier. Thus,
≥ connected
≥
e R3 = 4kT R3 V/ Hz, is also measured at the negative input node of the operational
amplifier (Fig. 15.16, left).
Our reasoning goes as following. We know that thermal voltage e R3 is amplified
and is therefore measured as eout (e R3 ). This output voltage due to e R3 is then easily
found by following current i 0 through the (R2 + R1 + R S ||R4 ) resistive chain. The
feedback current i 0 is controlled by e R3 potential at the negative input terminal of
the operational amplifier that forces i 0 down through (R1 + R S ||R4 ) resistor chain,
i.e.
eout (e R3 ) = i 0 (R2 + R1 + R S ||R4 ) =
e R3
(R2 + R1 + R S ||R4 ) (15.61)
R1 + R S ||R4
Nevertheless, our goal is to find an equivalent voltage source eR3 that, if connected
to 1i, would also generate eout (e R3 ). We note that under conditions of this particular
setup where eR3 is the only signal source there is no current flow through R3 , which
implies that the positive node of the operational amplifier is at ground potential,
Fig. 15.16 (right). Thereafter, we write
132
15 Electrical Noise: Solutions
eout (e R3 ) =
−i 0
e
R2 = − R3 R2
R1
∴
(eR3 )2
=
R1
eout (e R3 )
R2
2
(15.62)
From (15.61) and (15.62) it follows that
(eR3 )2
2
R1 R2 + R1 + R S ||R4
=
e R3
R2
R1 + R S ||R4
R1 R2 + R1 + R S ||R4 2
=
4kT R3
R2
R1 + R S ||R4
V2/Hz
(15.63)
which, the given numerical data results in R4 = 4 kπ, and
(eR3 )2 =
1 k 10 k + 1 k + 666.667 2
−23
× 4 × 1.3806488 × 10
× 353.95 K × 1 k V2/Hz
10 k
1 k + 666.667
= 9.578 × 10−18 V2/Hz
(15.64)
Contribution due to noise source e R1 : in order to derive thermal noise voltage
generated by resistance R1 , as seen from the input node 1ithrough the resistive
voltage divider (R1 , R4 ||R S ), we use similar reasoning as for eR3 above, thus we
write
eR1 = i R1 R1 =
∴
(eR1 )2 = 4kT R1
R1
e R1
R1 = 4kT R1
R1 + R4 ||R S
R1 + R4 ||R S
R1
R1 + R4 ||R S
2
(15.65)
which, the given numerical data results in
(eR1 )2 = 4 × 1.3806488 × 10−23 × 353.95 K × 1 k
= 7.037 × 10−18
V2/Hz
1k
1 k + 666.667
2
V2/Hz
(15.66)
Contribution due to noise source e R2 : thermal noise voltage generated by resistance
R2 is delivered to the input node through the resistive network (R2 , R1 ). Current
through R1 and R2 is set by e R2 , i.e. i R1 = i R2 , thus at node 1iwe write
15 Electrical Noise: Solutions
133
Fig. 15.17 Solution 3.16:
schematic diagrams for noise
calcualtions
R1
e R2
R1 = 4kT R2
R2
R2
eR2 = i R1 R1 =
∴
(eR2 )2
= 4kT R2
R1
R2
2
V2/Hz
(15.67)
which, the given numerical data results in
(eR2 )2 = 4 × 1.3806488 × 10−23 × 353.95 K × 10 k
= 1.955 × 10−18
1k
10 k
2
V2/Hz
V2/Hz
(15.68)
Contribution due to noise source e R4 : thermal noise voltage generated by resistance
R4 is delivered to the input node through the resistive divider (R4 + R1 ||R S ). Thus,
with reference to Fig. 15.17 at node 1iwe write
eR4 = i R4 R S ||R1 =
∴
(eR4 )2
= 4kT R4
e R4
R S ||R1
R4 + R S ||R1
R S ||R1
R4 + R S ||R1
2
V2/Hz
(15.69)
which, the given numerical data results in
(eR4 )2
= 4 × 1.3806488 × 10
= 0.782 × 10−18
−23
V2/Hz
× 353.95 K × 4 k
444.444
4 k + 444.444
2
V2/Hz
(15.70)
Contribution due to noise voltage source en : the internal thermal noise voltage
generated by operational amplifier is directly connected to the positive input node of
operational amplifier, thus it is at the same position as e R3 in the network (Fig. 15.16,
left). Thus, we reuse (15.63) and write
134
15 Electrical Noise: Solutions
Fig. 15.18 Solution 3.16:
schematic diagram for noise
calculations
(en )2 =
R1 R2 + R1 + R S ||R4
R2
R1 + R S ||R4
en2
V2/Hz
(15.71)
which, the given numerical data results in
(en )2 = 17.5 × 10−18
V2/Hz
(15.72)
Contribution due to noise current source i n+ : this current source forces noise current
through resistance R3 , which generates voltage directly at the positive input node of
the operational amplifier ein+ = i n+ R3 . Again, we reuse (15.63) and write
)2 =
(ein+
R1 R2 + R1 + R S ||R4
R2
R1 + R S ||R4
2
(i n+ R3 )2
V2/Hz
(15.73)
which, the given numerical data results in
)2 = 12.25
(ein+
V2/Hz
(15.74)
Contribution due to noise current source i n− : this noise source forces a current
through R2 , because due to the virtual ground both terminals of (R1 + R S ||R4 )
resistive network is at ground potential, which implies that i n− flows through R2 ,
Fig. 15.18. With that in mind, eout (i n− ) = i n− R2 is seen at the output node. Thus,
referencing the output voltage back to node 1ithrough voltage divider R1 , R2 we
write
ein−
=
eout (i n− )
i n− R2
R1 ==
R1
R2
R2
∴
(ein−
)2 = (i n− R1 )2
(15.75)
which, the given numerical data results in
(ein−
)2 = (5p × 1 k)2
V2/Hz
= 25 × 10−18
V2/Hz
(15.76)
Therefore, after combining the seven amplifier thermal noise power terms (15.63)
to (15.76), the complete solution for the internal noise power spectrum density is
15 Electrical Noise: Solutions
135
2 = (e )2 + (e )2 + (e )2 + (e )2 + (e )2 + (e )2 + (e )2 V2/Hz
PA = eout
n
in+
in−
R1
R2
R3
R4
∴
PA = (7.037 + 1.955 + 9.578 + 0.782 + 17.5 + 12.25 + 25) × 10−18 V2 /Hz
= 74.102 × 10−18 V2 /Hz
(15.77)
It is now straightforward to substitute (15.77) and (15.60) into (15.47) and to calculate
74.102
= 19.955
3.909
NF = 10 log(19.955) = 13 dB
F =1+
∴
(15.78)
3.17. The input noise power is calculated as
N I = kT = 1.3806488 × 10−23 J/K × 288.15K = 3.978 × 10−21 W/Hz
∴
N I = −174 dB m/Hz
(15.79)
We note that the −174 dBm/Hz is is often taken as the minimal noise power reference
at room temperature. Then, this thermal noise power is amplified by gain G, i.e.
No = N I + G
(15.80)
However, the amplifier itself contributes the additional NF, which adds up to the
output noise power. That means the total noise power at the output node is
No = N I + G + NF = −174 dBm + 12 dB + 3 dB = −159 dBm
(15.81)
3.18. Within the given bandwidth B for the signal before amplification (narrow red
arrow), Fig. 15.19, we estimate SNR I as the difference between the signal peak and
the noise level, i.e.
SNR I = −80 dBm − (−100 dBm) = 20 dB
(15.82)
The amplifier provides G = 10 dB of power gain for both noise and the signal.
However, it also provides internally generated noise power N F = 3 dB, which
is added only to the amplified noise level. Therefore, the two amplified power
levels are −80 dBm + 10 dB = −70 dBm for the signal (wide blue arrow), and
−100 dBm + 10 dB + 3 dB = −87 dBm for the nose floor (blue dots). Therefore,
after the amplification we have
SNRo = −70 dBm − (−87 dBm) = 17 dB
which reflects the amplifier’s noise figure NF = 3 dB as expected.
(15.83)
136
15 Electrical Noise: Solutions
Fig. 15.19 Solution 3.25:
signal to noise frequency
spectrum
3.19. It is convenient to express both the total cell phone power and the thermal noise
power radiated by the bird in units of W/Hz, so that
PS =
Pcell
B
W/Hz
PN = kT
and
W/Hz
(15.84)
At the average lunar distance D the total cell phone power is distributed over surface
of a sphere with radius D. Similarly, the total thermal noise power generated by the
bird is distributed over surface of a sphere with radius H . Thus, we find the two
respective power fluxes as
PS f =
PS
4π D 2
PN f =
and
PS
4π H 2
(15.85)
The two signals are received by the whole area A of the antenna dish, which gives
us expressions for the total received signal and noise powers as
S=
PS
A
4π D 2
N=
and
PS
A
4π H 2
(15.86)
We are now ready to actually write explicit expression for SNR as
PS
A
2
S
PS H 2
Pcell H 2
SNR =
= 4π D
=
=
PS
N
PN D 2
kT B D 2
A
2
4π H
(15.87)
which, for the given numerical data works as:
(a) From (15.87) when signal–to–noise–ratio is set to SNR = 0 dB, that is to say
SNR = 1, we write
Pcell H 2
=1
kT B D 2
∴
H=
kT B D 2
= 56.247m
Pcell
(b) After setting H = h from (15.87) we write
(15.88)
15 Electrical Noise: Solutions
137
SNR =
Pcell h 2
1
= −15 dB
=
kT B D 2
31.623
(15.89)
which clearly illustrates how difficult is to receive a week signal coming from
the space.
3.20. We already found expression for a signal power delivered to the matched load,
for example in (15.24). Here, BPF is matched to the signal source both in frequency
bandwidth and resistance, while the source resistance generates the thermal noise.
(a) Straightforward implementation of (15.24) results in
2
vin
2
vin
S
4Rin
SNR =
=
=
= 50.138 = 17 dB
N
kT1 B1
4Rin kT1 B1
(15.90)
(b) After the signal and BP filter bandwidths change to B2 , then from (15.90) we
write
2
vin
= 88 K
(15.91)
T2 =
4Rin k B2 S N R
3.21. In the textbook(s), we already showed expressions for dynamic resistance R D
of an R LC resonator at resonance ω 0 , for thermal noise voltage, and for effective
bandwidth to be
RD =
Q
ω0C
en2 = 4kT R D Beff
and
and
Beff =
π
B-3 dB (15.92)
2
which, after combining the three of them together leads into
en2 = 4 k T
∴
en =
Q f0
kT
Q π
B = 2π k T
=
ω0C 2
ω0C Q
C
1.3806488 × 10−23 J/K × 290.15 K
= 20 µV
10pF
(15.93)
3.22. Thermal noise of a resistor R at temperature T within a bandwidth B that is
controlled by a capacitor C leads into
en2 = 4kT R B
and
en2 =
kT
C
(15.94)
∴
B=
1
= 25 kHz
4RC
(15.95)
138
15 Electrical Noise: Solutions
Now, the total integrated thermal voltage within the bandwidth B is calculated by
using either of the two expressions in (15.94), e.g.
en =
kT
= 65 µV
C
(15.96)
3.23. First, all given gain and noise figures specifications have to be expressed as
−6 dB ≡ 1/4
10 dB ≡ 10
3 dB ≡ 2 8 dB ≡ 6.31
20 dB ≡ 100
Then, straightforward implementation of Friss formula for a five–stages system gives
F2 − 1
F3 − 1
F4 − 1
F5 − 1
+
+
+
A1
A1 A2
A1 A2 A3
A1 A2 A3 A4
6.31 − 1
10 − 1
2−1
10 − 1
+
+
=2+
+
100
100 × 100 100 × 100 × 1/4 100 × 100 × 1/4 + 100
= 2.270
(15.97)
F(tot) = F1 +
NF = 10 log F = 3.561 dB
(15.98)
Tn = (F − 1) T = (2.270 − 1) 299.15 K = 380 K
(15.99)
3.24. By definition (15.47) we find
PI
5 × 10−6
−6
SNR I
N
F=
= I = 1 × 10 −3 = 4
PO
SNR O
50 × 10
NO
40 × 10−3
∴
NF = 10 log(4) = 6 dB
(15.100)
3.25. A simple model developed in solid state physics describes a diode noise manly
as a “shot–noise” i sn , which is function of the diode’s biasing current I D . It is
very practical to remember approximate values of diode parameters at 1 mA biasing
current at room temperature 27◦ over 1 MHz bandwidth, i.e.
2
= 2 q ID B
i sn
∴
i sn =
2 × 1.602176565 × 10−19 C × 1 mA × 1 × 106 Hz = 17.90 nA
At the same time, voltage across a pn junction and dynamic diode resistance r D are
15 Electrical Noise: Solutions
139
VT =
kT
= 25.865 mV ≈ 26 mV
q
∴
rD =
VT
= 25.865 π ≈ 26 π
I DC
Therefore, the shot–noise current i sn flowing through the diode resistance r D generates noise voltage
en = i sn r D =
kT
2 q ID B
= kT
q ID
= 1.3806488 × 10
−23 J
2B
q ID
/K × 300.15 K ×
(15.101)
2 × 1 MHz
1.602176565 × 10−19 C × 1 mA
= 463 nV
In order to reduce a diode noise, (15.101) implies that we need to reduce threshold
voltage VT , thus to lover the internal resistance r D , which is achieved by lowering
temperature and by increasing the biasing current, or by decreasing bandwidth.
For example, by doubling the biasing current to I D = 2 mA, everything else being
equal, the dynamic resistance r D is halved and the diode shot–noise voltage reduced
to en = 327 nV. Or, equivalently, in order to achieve the same noise reduction by
controlling the temperature while keeping the same I D = 1 mA biasing current we
wold have to cool the diode down to −61◦ c. Concurrent biasing current increase
and temperature reduction results in en = 231 nV.
We could go even further and create, for example, a plot (15.101) to show how
the noise voltage changes relative to the biasing current I D = 100µA to 10 mA,
and relative to temperature T = 50–400 K(in steps of 50 K) (Fig. 15.20). (Note that
both axis use logarithmic scale.) Within this temperature range for a given biasing
current, e.g. I D = 1 mA, the noise voltage changes over order of magnitude, where
the temperature dependance is more pronounced at lower temperatures. Similar plot
could be created if we wanted to see how the noise voltage changes with, for instance,
bandwidth and the biasing current.
3.26. We already know that wider amplifier’s bandwidth allows more noise to enter
the system, thus to increase the integrated noise power level. As soon as the noise
power level brings SNR of this amplifier down to zero, we state that the maximum
allowed bandwidth of this amplifier is reached, because beyond that particular value
of bandwidth B the input signal is completely swamped with the noise without any
chance of being recovered latter.
In this problem, we identify three sources of noise at the input terminals of the
amplifier and apply the superposition principle, Fig. 15.21.
(a) The shot–noise current i sn passing through the source resistor R S (while v S is
shorted) causes voltage
140
15 Electrical Noise: Solutions
-6
10
en [V]
400K
10-7
50K
B=1MHz
10-8
-4
10
-3
-2
10
10
ID [A]
Fig. 15.20 Solution 3.25: noise vs. drain current parametric plot
Fig. 15.21 Solution 3.26:
illustration diagram
2
i sn
= 2q I D B
∴
2
esn
(R S ) = (i sn × R S )2 = 2q I D B R S2
(15.102)
This voltage is forced by the current through R S resistor and it is therefore
directly connected to the input node of the amplifier.
(b) the thermal noise voltage generated by the source resistance R S
en (R S ) =
4kT R S B
(15.103)
is spread across (Rin , R S ) divider, therefore at the input terminals
≥
4kT R S B
Rin
en (R Sin ) =
R S + Rin
∴
2
Rin
2
en (R Sin ) = 4kT R S B
R S + Rin
= 4kT B R S
where,
R S = R S
Rin
R S + Rin
2
(15.104)
15 Electrical Noise: Solutions
141
(c) Similarly, thermal noise generated by the amplifier input resistance Rin
en2 (Rin ) = 4kT Rin B
= 4kT B Rin
where,
= Rin
Rin
RS
R S + Rin
2
(15.105)
RS
R S + Rin
2
Therefore, from (15.102) (15.104) and (15.105) the total noise at the input of
the amplifier is
2
(R S ) + en2 (R Sin ) + en2 (Rin )
(15.106)
en2 = esn
while the SNR(in) definition leads into
SNR(in) = 20 log
vS
=0
en
⇒
v S = en
∴
v2S
= en2
2
= esn
(R S ) + en2 (R Sin ) + en2 (Rin )
B
= 2q I D B R S2 + 4kT R S B + 4kT Rin
∴
B=
v2S
)
2q I D R S2 + 4kT (R S + Rin
= 10.208 MHz
(15.107)
3.27. In order to compare and manipulate various units, it is much easier if all data
is converted to the common unit, in this case a plain numbers, therefore
NF = 12 dB
∴
F = 15.85
and
A = 50 dB = 1 × 105
(15.108)
which leads into
TRx = (F − 1) T = (15.85 − 1) × 300 = 4, 455 K
and
4, 455 K
Tsys = 90 K +
≈ 90 K
1 × 105
(15.109)
This example illustrates dominance of the first stage in the overall system level noise
calculations.
142
15 Electrical Noise: Solutions
Fig. 15.22 Solution 3.28:
noise spectrum
3.28. In order to estimate the total input noise current, the input current power spectral
density plot is approximated with piecewise linear sections (Fig. 15.22).
By doing so, the process of integration is reduced to simple additions of regular
surfaces under the PSD curve. We easily recognize the following three geometrical
shapes.
1. Rectangular area (the noise ground level is at 0 A2/Hz, thus :
i t2 = 1 × 10−25 A2/Hz × (100 × 106 − 1 × 103 )Hz
≈ 10 × 10−18 A2
(15.110)
2. LF triangular area:
1 (1 × 10−23 − 1 × 10−25 )A2/Hz × (2 × 104 − 1 × 103 )Hz
2
(15.111)
= 94.05 × 10−21 A2
i L2 F =
3. HF triangular area:
1 (1 × 10−24 − 1 × 10−25 )A2/Hz × (1 × 108 − 3 × 107 )Hz
2
(15.112)
= 31.5 × 10−18 A2
2
iH
F =
Now we estimated the total noise current as
2
i n = i t2 + i L2 F + i H
F = 6.45 nA
(15.113)
Chapter 16
Electronic Devices: Solutions
Complex electronic systems are designed by following the modular, similar to Lego
Bricks, approach. Each of the basic electronic building blocks is first designed and
studied separately, and thereafter it becomes only the designers imagination and
creativity that limits how complicated or how unusual the new circuit design will be.
Nevertheless, thorough understanding the basic building blocks behaviour is the first
step on the road to great designs and inventions.
In this section we review important voltage/current divider based devices,
frequency limitations of fast pulse waveforms, and the elements of non-linear device
behaviour, which is fundamental for the overall wireless circuit functionality.
Solutions:
4.1. Rapid amplitude changes of a current flowing through an inductive element
generate induced voltage at terminals of the inductive element. As per (4.1), amplitude of the induced instantaneous voltage is directly proportional to the slope (i.e. rate
of change) of current that is causing it. Therefore, if for instance the current waveform takes shape of an ideal square pulse (i.e. a pulse whose rising and falling edges
are vertical), by definition (4.1) the induced voltage amplitude is infinite, Fig. 16.1.
It is very important to be aware of existence of these high-voltage spikes in RF
circuits simply because the voltage amplitude may take values much higher than
the corresponding power supply value, which is against the conventional wisdom
assumed in the design of low frequency amplifiers where maximum amplitudes
of the internal voltages are limited by the power supply. Consequently, the other
circuit components may be damaged if they are not rated for high enough operational
voltages. Naturally, in real circuits it is impossible to achieve infinitely short rise/fall
pulse edges, which is to say that amplitude of the induced voltage can not really
take the infinite value. However, with high quality inductors (as quantified by their
respective Q factor) and fast pulses the generated voltage amplitude may become
high enough to cause the damage.
R. Sobot, Wireless Communication Electronics by Example,
143
DOI: 10.1007/978-3-319-02871-2_16, © Springer International Publishing Switzerland 2014
144
16 Electronic Devices: Solutions
Fig. 16.1 Problem 4.1: time
domain current diagram.
Current waveform (blue dotted
line), and voltage waveform
(red solid line)
4.2. Given the initial conditions, at the time t = 0 s when the capacitor is discharged,
i.e. vC (t = 0) = 0 V, and the pulse source changes its output from 0 V to v(max),
the total source voltage is distributed across the resistor, i.e. v R (t = 0) = v(max).
Over the first half of its period T a square function shape holds its high level,
while over the second half period it holds its low level, Fig. 16.2.
In accordance to Kirchhoff’s voltage law, for 0 √ t √ T/2 we write expression
for the voltage inside the RC loop as
v(max) = v R (t) + vC (t)
T/2
1
= i(t) R +
i(x) d x
C 0
∴
di(t)
+ i(t)
0 = RC
dt
(16.1)
(16.2)
where (16.2) is derivative of (16.1). Solution of a first–order differential equation
(16.2) is well known as
t
v(max)
t
i(t) = I0 exp −
exp −
=
RC
R
RC
Fig. 16.2 Problem 4.2: time
domain current diagram
(16.3)
16 Electronic Devices: Solutions
145
where, the initial current at t = 0 is I0 = v(max)/R , and time constant of an RC
circuits is defined as τ = RC. We note that the product of two electrical variables,
i.e. R and C, is measured in units of time, which is arguably the most important
variable in physics and engineering.
After substituting (16.3) into (16.1) we easily write expression for voltage across
the capacitor as
t
vC (t) = v(max) 1 − exp −
τ
t
= 10V 1 − exp −
1ms
(16.4)
Within the first half period time interval, plot of (16.4) shows exponential growth of
voltage across the capacitor and at the same time exponential reduction of the current
flowing inside the circuit. Once the capacitor is fully charged, i.e. vC = v(max) then
voltage across the resistor drops to zero, thus no current flow.
It is important to note that, in accordance to (16.4), it takes infinite amount of
time for the voltage vC to become truly equal to v(max). Thus, saying that the
capacitor is “fully charged” is, strictly speaking, not correct. Indeed, when we say
“fully charged” we mean charged “close enough” to the maximum possible level. This
is another example of differences between perfect mathematics and the engineering
approximations.
It is convenient to express progression of time in units of τ = RC so that, for
example, t = 5 τ expression (16.4) becomes
5ø
= 10V × 0.993 = 9.93V
vC (t) = 10V 1 − exp −
ø
(16.5)
which leads into convenient engineering rule of thumb that after five timing constants
a capacitor is considered fully charged (regardless of the absolute time value). In other
words, any other event in the circuit (for example, falling edge of the pulse) should
wait at least 5RC units long until the maximum capacitor voltage is achieved. An
example of important practical implication of this conclusion is that pulse width of
clocks used in our computers is limited to 5τ to 6τ , which sets the minimum clock
period to T → 10τ to 12τ .
Similarly, if the capacitor is already fully charged to vC = v(max) and the source
voltage becomes zero, e.g. the second half period of a pulse function, then the capacitor releases its charge through the resistor in accordance to Kirchhoff’s current
law,
146
16 Electronic Devices: Solutions
Source pulse
τ=5.0 RC
τ=0.2 RC
τ=1.0 RC
1
amplitude
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
12
time [RC]
Fig. 16.3 Problem 4.2: influence of the timing constant τ relative to the pulse period T = 12τ on
the charge discharge capacitor function
C
dvC (t) vC (t)
+
=0
dt
R
∴
t
vC (t) = v(max) exp −
RC
(16.6)
which is well known formula for exponential decay. Therefore, full charge–discharge
cycle of a capacitor is combination of (16.4) and (16.6), for the first half period and
the second half period respectively.
Figure 16.3 illustrates importance of RC timing constant relative to a clock pulse
(red dotted line). If the timing constant is too long, e.g. τ = 5RC, then the capacitor
voltage does not have enough time to charge up to the “high” voltage level. Instead,
when the falling edge of the pulse comes, the capacitor must start the discharging
process, which results in waveform whose shape is closer to a triangular instead of
square waveform. On the other hand, if the timing constant is short, e.g. τ = 0.2RC,
then the capacitor voltage (black dash–dot line) is very close to the ideal pulse shape.
The blue solid line shows the capacitor voltage when τ = RC. After 6RC time units
the capacitor is practically fully charged.
4.3. In analog signal processing circuits, one of the most common operations is to
derive one voltage from another voltage source. Voltage division, or if you prefer
voltage multiplication with a factor smaller than one, is used to set voltage reference
point that is needed, for example, to set biasing condition of a transistor. Simplest
voltage reference is derived from the power supply by the means of two resistors in
series, Fig. 16.4 However, for high precision stable voltage reference are designed
using sophisticated feedback circuits. For example, a stable and precise voltage reference standard commonly used in modern electronic systems is provided by a circuit
known as a bandgap voltage reference.
16 Electronic Devices: Solutions
147
Fig. 16.4 Problem 4.3: voltage reference based on a resistive divider
In this example we learn how to design the simplest voltage reference, Fig. 16.4, by
calculating the appropriate resistor values. Naturally, since there are two unknown
variables, R1 and R2 , there must be two equations so that the system is solvable.
Otherwise, we would be forced to arbitrary pick value of one resistor and then to
calculate the other.
Based on the given data, the two constrains are derived as follows. The reference
voltage Vref is derived as
Vref = I R2 R2 =
VD D
R2
R1 + R2
∴
R1 + R2 = 3.0 R2
(16.7)
after substitution of Vref = 1 V and VD D = 3.0 V. Second constraint is derived
from the requirement that the equivalent Thévenin resistance of this reference is
Rth √ 100 ω. Output resistance, as seen by looking into Vref node is obviously
parallel resistance R1 ||R2 , thus in the worst case
Rth =
R1 R2
R1 + R2
∴
R1 R2 = 100 (R1 + R2 )
(16.8)
It is straightforward to solve the two equations system (16.7) and (16.8) and calculate
R1 = 300 ω, and R2 = 150 ω. We keep in mind that this simple voltage reference
is directly proportional to the power supply, while by industrial standards the power
supply voltage alone is allowed to vary ±10 %. Therefore, this voltage reference is
not suitable for high precision applications, however it is almost exclusively used to
setup biasing conditions for transistors.
If the power supply voltage is increased by ±10 % to VD D = 3.3 V, then the
reference voltage is found to be
148
16 Electronic Devices: Solutions
Fig. 16.5 Problem 4.5: voltage reference based on a diode
Vref =
R2
VD D = 1.1V
R1 + R2
while for VD D = 2.7 V we find that Vref = 0.9 V. In other words, for a 20 %
variation in the power supply voltage, the reference voltage also varies by 20 %.
4.4. The three ideal linearized elements (R,L,C) are the cornerstone of the traditional
circuit theory, while the memristance M is joining the group as the fourth element.
However, these elements are only idealized mathematical models of the real material
behaviour. We already know that, under various external conditions, even a single
piece of metallic wire behaves as a complicated R LC network. Thus, we use the
ideal linearized element models only within a certain limited range of operation.
Nevertheless, as we will find later in this book, it is actually the property of nonlinearity of real materials that enables us to perform operations such as frequency
shifting, which is essential for operation of a radio.
In this example we take a look at basic active nonlinear element known as a diode.
Voltage–current characteristics of this two terminal device obeys “exponential law”,
hence a diode mathematical model illustrates its “exponential nature” as in (4.2),
while the approximation is valid if (VD ∞ VT ). For the given data straightforward
calculation results in:
1. for VD = 600 mV (i.e. VD /VT = 23.120),
I D = 0.64684449361 A and
I D ≤ 0.64684441671 A,
therefore the approximation error is λ = 11.89 × 10−6 %; and
2. for VD = 50 mV (i.e. VD /VT = 1.927),
I D = 290.40168723 nA and
I D ≤ 213.50168723 nA,
therefore the approximation error is λ = 26.5 %
which illustrates validity of using the approximative model for strongly biased (VD ∞
VT ) diodes.
4.5. In the previous problem 4.3 we used a simple voltage divider to design a voltage
reference. We found that main disadvantage of using that topology is that voltage variations of the power supply directly propagate to the reference voltage node
Fig. 16.5.
16 Electronic Devices: Solutions
149
In this example we demonstrate a bit better voltage reference topology based on
a diode. The operational principle is very simple, we already know that the diode
current I D depends upon the diode voltage VD . However, another way of stating the
same is to say that if current I D is forced through the diode, then the voltage VD
across diode must obey the same (4.2) equation.
If we rewrite (4.2) to solve VD , we have
I D ≤ I S exp
VD
n VT
∴
VD ≤ n VT ln
ID
IS
Where, for I D = 1 A we find that VD = 616 mV. If the current I D changes ±10 %,
i.e. from I D = 1.1 A to I D = 0.9 A, then the diode voltage is within the range of
VD = 620 mV to VD = 612 mV. In other words, for 20 % variation of the I D current
value, the diode voltage VD changes only 1.3 %. This effect is due to “levelling–off”
shape of the ln(x) function, which changes very little for the large argument values.
4.6. In principle, a BJT transistor is often modelled as two “back to back‘” diodes
(not exactly correct, but very useful model), where the forward biased base–emitter
diode is used to control the collector current. Similarly to the simple diode model,
we use the exponential model to relate collector current IC and the corresponding
base–emitter voltage VB E as
VB E
VB E
− 1 ≤ I S exp
IC = I S exp
VT
VT
∴
VB E ≤ VT ln
IC
IS
where the approximative expression is valid under the condition that VB E ∞ VT .
For the given data, I S = 5 × 10−15 A, IC = 1 mA, and VT = 25 mV we easily find
that VB E = 650.540 mV. By definition, gm gain is
gm =
IC
VT
(16.9)
therefore, at this biasing point, gm gain is 40 mS.
In order to develop better sense of how the base–emitter voltage controls the
collector current, we for example recalculate the following values
VB E , V IC , mA
0.50
0.55
0.60
0.65
0.70
0.75
0.80
gm , mS
0.002
0.080
0.017
0.680
0.132
5.280
0.978
39.120
7.231
289.240
53.432 2,137.200
394.814 15,792.560
150
16 Electronic Devices: Solutions
which illustrate the importance of tight control of base–emitter voltage VB E , i.e.
correct biasing point setting, because the collector current IC is exponentially related
and even a small change of VB E causes large variations in the collector current.
However, this relationship also illustrates amplification effect of a transistor, in this
case small change in the input voltage causes large change in the output current, thus
in this mode the transistor is said to behave as a gm amplifier.
4.7. When considering resistive elements, first question to ask is if the element
changes its resistance relative to the frequency of signal at its terminals. Based on
this criteria, naturally there are two distinct groups, resistances are either frequency
independent or not. Beauty of the frequency dependent resistances (commonly refereed to as impedances) is that networks containing such elements simultaneously
provide different gain for single tones at different frequencies.
When we couple that behaviour with the Fourier’s frequency spectrum principle,
we achieve very power mechanism to arbitrary shape frequency spectrum of the given
multi–tone signal, and therefore to modify its time–domain shape. This mechanism
is fundamental principle used in signal processing techniques.
In order to develop better sense of how the resistive network affects the frequency
spectrum of a given signal, let us assume that a multi–tone signal s(t) is defined as
the sum of single tone signals, all of them with the same amplitude normalized to
one,
s(t) = 1 + sin(2π t) + sin(2π × 1 × 103 t) + sin(2π × 1 × 104 t)
+ sin(2π × 1 × 106 t) + sin(2π × 1 × 108 t)
(16.10)
Each of the single–tone components will be affected differently after the signal
s(t) is passed through the resistive networks. By following rules for serial/parallel
equivalent resistances, and by knowing expression for impedance Z C of capacitive
elements, we write the four respective equivalent impedances as:
Ra = R + 10R = 11R ≤ 10R
Rb = R + Z C = R −
Rc = R||Z C =
j
ωC
∴
R2
R 2 ω2 C 2
Rd = R||10R ≤ R
∴
+1
∴
Ra ≥= f (ω)
|Rb | =
∴
R2 +
1
2
ω C2
∴
Rb = f (ω)
Rc = f (ω)
Rd ≥= f (ω)
For the given numerical values, in the following table we summarize the frequency
dependance of these four equivalent impedances which illustrates how impedance
of these four networks changes with the frequency. In the case of resistive networks
only, i.e. Ra and Rd , one is series combination and one is parallel combination of two
resistors. In the case of the series combination, the equivalent resistance is greater
than the greater of the two, i.e. Ra > max(10R, R). However, since one is ten
16 Electronic Devices: Solutions
151
f, Hz Ra , ω Rb , ω Rc , ω Rd , ω
DC
1
1k
10k
1M
100M
∗
10k
∗
10k
1G
10k
1M
10k 100k
10k 1.414k
10k
1k
10k
1k
1k
1k
1k
1k
707
10
0
1k
1k
1k
1k
1k
1k
1k
or more times larger than the other, it is reasonable to approximate the equivalent
resistance with the larger one, i.e. Ra = 10 kω + 1 kω = 11 kω ≤ 10 kω. In the
case of parallel connection of two resistors, however, the equivalent resistance is
smaller than the smaller of the two, i.e. Rd < min(10R, R). Again, if one resistance
is ten or more times greater than the other, and approximative error of 10 % or less
is acceptable, then it is reasonable to write Rd = 10R||R ≤ R.
The situation becomes much more interesting for Rb and Rc equivalent impedances
– they change with the frequency. We already know that impedance of a given capacitance changes from Z C = ∗ to Z C = 0 as the frequency changes form DC to ∗.
Thus, in the case of series RC connection, impedance the sum R + Z C ) changes from
∗ to R, while in the case of parallel connection R||Z C , the equivalent impedance
changes from R to zero.
Considering that we already know Ohm’s law which states that voltage across a
resistor is calculated as V = I R, it should be trivial to conclude that if a multi–tone
signal is applied across a frequency dependent impedance, then each harmonic of the
frequency spectrum is affected by different gain factor. Consequently, in accordance
with Fourier’s theorem, amplitudes of all harmonics is changed relative to the original
frequency spectrum, and therefore the signal also changes its shape in the time
domain. That change of the waveform shape may be intentional and desired, which is
what signal processing is all about. Or, it may be unwanted and beyond our control,
for example caused by parasitic frequency dependent components, which is what
signal distortion is.
4.8. Being able to mentally approximate gain boundaries of a voltage divider is one
of the essential engineering skills. In this example we estimate the voltage gains
by inspection of the circuit topologies and by knowing at least some relationship
between the components.
By now we should be able to just write down voltage gain expression of a general
voltage divider similar to (Z 1 , Z 2 ) as
Vout,B = i Z 1 ,Z 2 Z 2 =
V A,B
Z2
Z1 + Z2
∴
Vout,B
Z2
=
V A,B
Z1 + Z2
which is easily evaluated, first assuming Z 1 = 10 Z 2 and no frequency dependance
(i.e. R1 = 10 R2 ) as
152
16 Electronic Devices: Solutions
AV =
Vout,B
Z2
Z2
=
=
= 0.0909 ≤ −20dB
V A,B
Z1 + Z2
11 Z 2
in other words, the 10 : 1 impedance ratio results in about ten times attenuation (i.e.
around −20dB). In the extreme case of Z 1 ∞ Z 2 we conclude that the voltage gain
A V ∓ 0. Similarly, if Z 2 = 10 Z 1 we write
AV =
Vout,B
Z2
10 Z 1
=
=
= 0.909 ≤ 1
V A,B
Z1 + Z2
11 Z 1
which is to say that the (1 : 10) ratio causes almost no voltage signal loss. In the
extreme case of Z 2 ∞ Z 1 the voltage gain A V ∓ 1.
In the case when the voltage divider includes frequency dependent components,
e.g. a capacitor, we first evaluate the voltage gain at DC and then at f ∓ ∗. At
DC, a capacitor is an open connection (i.e. Z C ∓ ∗) and can be removed from the
parallel combination. Under that condition, it is straightforward to conclude that
Vout,B = i R R =
V A,B
R
R+R
∴
AV =
Vout,B
= 0.5 = −6dB
V A,B
in other words, two equal resistors making voltage divider must split the voltage
in half, which is equivalent of saying that the voltage gain equals A V = −6dB.
Similarly, the other extreme is when f ∓ ∗, which leads into Z C ∓ 0. The
parallel combination R||Z C , therefore, results in the zero equivalent impedance.
Direct consequence is that the output voltage Vout,B develops across a zero resistance,
therefore the output voltage must be zero. In other words, network in Fig. 4.3 (right)
shapes the frequency spectrum by setting A V = 0 for high frequencies, while the
maximum gain A V = 0.5 is for DC and low frequencies.
Design of RLC networks used to shape frequency spectrum of continuous signals
is engineering discipline known as analog filter design. If the same shaping function
of the frequency spectrum is implemented using digital logic and discrete signals,
then we say that we did digital filter design.
4.9. We now continue our discussion about mentally evaluating circuits and determining useful boundaries. In this example we learn to recognize voltage divider inside
a typical amplifier section. We already know that one of the main applications of a
voltage divider is to provide voltage levels required to setup biasing conditions of a
transistor. In addition, for the purpose of setting up the biasing conditions, we keep
in mind the back–to–back diode model of a BJT transistor. To set a BJT transistor in
the forward active mode, we know that base–emitter diode must be forward biased,
while at the same time the base–collector diode must be reverse biased.
1. Two resistors R1 , R2 in Fig. 4.4 (left) create a simple voltage divider used to
scale down the VCC voltage level. So, what voltage is required at the node B?
Resistor R2 is in parallel with the base–emitter diode, thus voltage across R2
resistor equals to VB E . In the case of ideal base–emitter diode, any forward bias
16 Electronic Devices: Solutions
153
voltage VB E > 0 turns the diode on. Therefore, assuming any current (in this
case I R2 = 1 mA) being forced through R2 , any R2 > 0 generates the required
non–zero positive V R2 = VB E > 0 voltage. Setting up the base–emitter diode is
only the first step in setting up the BJT transistor biasing point. Second step is to
keep the base–collector diode reverse biased. That is achieved by setting potential
at the collector VC to the same or higher level relative to the base potential, i.e.
VC → VB > 0. In this approximative analysis the base current is assumed zero.
2. If the base–emitter diode is realistic, that is, if its turn on voltage is greater than
zero, procedure for setting up the biasing conditions is still as same as in the
ideal case. In order to provide VB E = 1 V voltage across R2 must be set to
V R2 → 1 V. If the current I R2 = 1 mA then R2 = 1 V/1 mA = 1 kω. Again, in
order to keep base–collector diode reverse biased, potential at collector must be
VC → VB > 1 V.
More precise numerical analysis is always left to the circuit simulators, however by
doing our “back of the envelope” analysis we are capable to quickly and intuitively
predict the circuit operation before committing the time for simulations.
4.10. Following up on the results of 4.9, the problem is now reduced to design of a
network to provide the required biasing voltage at the gate node. Circuit is Fig. 4.4
(right) contains emitter resistor R E whose role is separate the emitter potential from
the ground level. Current through this resistor forces the emitter node potential to
VE = i R E = 1 V. From this point on, there are two given scenarios.
1. If base–emitter diode of transistor Q 1 is ideal, then base potential must be at
VB > VE = 1 V. For example, we can set VB = 1.1 V. Given the power supply
voltage VCC = 10 V we write
VB =
∴
VB
=
VCC
R2
VCC
R2 = VCC
R1 + R2
R1 + R2
1
R1
R2
+1
∴
VCC
10V
R1
−1
=
−1=
R2
VB
1.1V
∴
R1
≤8
R2
where, the any choice of the actual resistor values is acceptable as long is
R1 : R2 = 8 : 1. In practice, we put the additional constrain to limit the total power consumption. Infinite number of (R1 , R2 ) values satisfy the proportion
requirement, thus we are free to take practical resistor values that allow minimal
current through (R1 + R2 ). Of course, it is not practical to take too large resistor
values and make the current too small, because we keep in mind the noise floor
as well as practical realization of large resistor values in integrated circuit (IC)
technologies.
2. If base–emitter diode is not ideal, i.e. VB E > 0, then for the given data the base
potential must be set to VB > VE + Vth (B E) = 1 V + 1 V = 2 V. For example,
we can set VB = 2.1 V, which leads into
154
16 Electronic Devices: Solutions
R1
VCC
10V
−1
=
−1=
R2
VB
2.1V
∴
R1
≤ 3.8
R2
This example illustrates one of the first design steps of setting up voltage divider
that serves the purpose of providing stable DC voltage level for the base terminal.
4.11. Estimate of a resistance by “looking into a node” is very useful technique
that is not difficult to master, as long as we can visualize serial/parallel combination
of impedances. In this example, three typical cases commonly found in almost all
amplifying stages, are demonstrated. Naturally, this technique is an approximation,
however very good one while more precise results can be achieved using numerical
simulators.
Although it is usually not specifically drawn in circuit schematic diagrams, the
power supply source is assumed to be an ideal voltage source, thus its internal impedance is zero. Therefore, for the purposes of finding an equivalent resistance between a certain node and the ground, the positive power supply rail is shorted to the
ground line.
That being the case, voltage divider R1 , R2 in Fig. 4.5a becomes parallel connection R1 ||R2 from the perspective of Z out node, i.e.
Z out =
R1 R2
R1 + R2
(16.11)
However, if an active component is involved, Fig. 4.5b, then while looking into
base we must take into account influence of the emitter resistor R E . It is shown in
the textbook that due to feedback effect, resistance R E is projected to and perceived
by the base node as if it were multiplied by (β + 1) factor, where β is gain of Q 1 .
In this example it was indicated that i B ≤ 0, which implies that the internal base
resistance r B itself is r B ∓ ∗, therefore
Z in = r B ||(β + 1) R E ≤ (β + 1) R E
(16.12)
which is the upper bound, because for finite r B the equivalent parallel resistance Z in
is lower.
In order to estimate Z out resistance looking into the emitter node Fig. 4.5c we
start with the voltage divider R1 , R2 . Resistance at the base node consists of three
resistance in parallel R1 ||R2 ||r B , which is approximated with R1 ||R2 because the
internal base resistance r B ∓ ∗ (i B ≤ 0). Whatever resistance is associated with
the base node, it is perceived from the emitter side as if it were divided by the (β + 1)
factor, i.e. in this case (R1 ||R2 )/(β + 1). This mental transformation of the base side
resistance to the emitter side is illustrated in Fig 16.6.
When looking into emitter, we should see the emitter resistor R E to the ground.
In addition, the internal BJT emitter resistance is
re =
VT
IC
(16.13)
16 Electronic Devices: Solutions
155
Fig. 16.6 Problem 4.11: schematic of a BJT output impedance network
and for IC = 1mA at room temperature (i.e. VT ≤ 25 mV) it follows that re ≤ 25 ω.
Obviously, resistance value of parallel R1 ||R2 divided by relatively large value of β
is also expected to be small. Typically, emitter resistor R E is much larger than the
others, Fig 16.6 (left). Therefore, depending upon the biasing collector current, it is
expected that resistance looking into BJT emitter is rather small (definitely smaller
than re ), that is the emitter node behaves as a voltage source.
4.12. Value of the BJT thermal voltage VT is very important, because it directly
influences small signal emitter resistance re , which in return controls voltage gain of
the BJT device. Design engineers must be aware of the temperature dependance of VT
and either try to reduce the circuit temperature sensitivity, or try use this sensitivity
to design, for example, temperature sensing circuits.
It is straightforward to calculate thermal voltage by using its definition
VT =
kT
q
(16.14)
which for the given temperature results in the following three values
VT (−55 ⇒ C, 25 ⇒ C, 125 ⇒ C) = (18.8, 25.7, 34.3 mV)
(16.15)
This simple example demonstrates that within the given temperature range, the
thermal voltage changed almost 100 %, which is very significant and must be taken
into account during the design process.
4.13. In this example we practice to design a voltage divider that involves a resistor
and a nonlinear element, i.e. a diode. As we already learned this combinations of
elements is very often used to provide voltage reference to the rest of the circuitry.
However, due to existence of the nonlinear device, the main problem is that in order
to set diode voltage, we need to set diode current, which is set by the diode voltage.
Therefore, the solution is found by iterations.
In this case, we start the iterative process by assuming an ideal diode, i.e. VD = 0,
which enables us to calculate the initial current through resistor as I R = VCC /R =
9.000 mA. The diode and resistor currents must be equal, therefore, diode current is
alsoI D = 9.000 mA. But then, this current forces the diode voltage to be
156
16 Electronic Devices: Solutions
VD = VT ln
ID
IS
=
kT
9.000 mA
ln
= 336.030 mV
q
18.8 nA
which is definitely not zero. Taking this diode voltage as the starting point, in the
second iteration we find resistor current as
IR =
9 V − 336.030 mV
VCC − VD
=
= 8.664 mA
R
1 kω
which is of course different than the initially assumed 9.000 mA, and thus we must
start another iteration by using I D = 8.664 mA as the diode current.
By repeating the last two steps again, the next iteration gives VD = 335.052 mV
which leads into I R = 8.665 mA. The following iteration gives VD = 335.055 mV
and again I R = 8.665 mA. Therefore, being satisfied with three decimal places, we
converge to VD = 335.055 mV.
In this simple example we demonstrated the same process that numerical simulators repeat millions of times per second when used to resolve operating points of
complex electronic circuits.
4.14. In this example we demonstrate temperature dependance of a BJT transistor,
that is as same as foe any other pn junction. Here we already know that
VB E ≤ VT ln
IC
IS
where the temperature dependance is included only through the VT =
Assuming I S = 100 fA and by direct implementation of (16.16) we find
VB E (−55 ⇒ C, 25 ⇒ C, 125 ⇒ C) = (432.8, 591.6, 790 mV)
(16.16)
kT/q
term.
(16.17)
while for I S = 200 fA the base emitter voltage VB E takes the following values
VB E (−55 ⇒ C, 25 ⇒ C, 125 ⇒ C) = (419.8, 573.8, 766.2 mV)
(16.18)
which illustrates importance of the environment temperature where the electronic
equipment is intended to operate.
Additional important point to note is that (16.17) and (16.18) imply that the
base–emitter voltage VB E increases with the temperature, i.e. that it has positive
temperature coefficient. However, that conclusion is not correct because base–emitter
voltage actually has negative temperature coefficient, i.e. its value reduces with the
increase of temperature. The reason for this wrong conclusion in our example is
that we completely ignored temperature dependance of the leakage current (i.e. the
reverse saturation diode current) I S = f (T ). This temperature dependance is much
stronger than the VT = f (T ) dependance, thus the overall VB E = f (T ) temperature
coefficient is negative. Detailed temperature sensitivity analysis of a BJT transistor
16 Electronic Devices: Solutions
157
is beyond scope of this course, thus for now we keep in mind only the information
about the negative temperature coefficient.
4.15. Following up on the results of 4.14, we now derive biasing voltage at the gate,
while considering temperature dependance of VB E and presence of R E resistor. For
large β values, it is reasonable to assume that I E ≤ IC , which simplifies calculation
of potential at the emitter node, VE = I E R E = 100 mV. At 25 ⇒ C the base potential
must be at least VB E = 591.6 mV higher, which sets VB → 691.6 mV.
4.16. This kind of estimate is done simply by inspection of the schematic diagram.
The emitter resistor is “projected” to the base node as if being magnified by factor
(β + 1), thus, Z in = (β + 1) R E , which for the given data results in Z in = 10 kω
and Z in = ∗. In this case, the base current is negligible, i.e. rπ ∓ ∗, thus only the
multiplied value of emitter resistor is seen from the input side.
We note that very high input impedance (i.e. negligible input current) is property
usually associated with CMOS type transistor. However, in the first approximation of
BJT, it is often useful to ignore base current and estimate biasing parameters without
the use of numerical solvers.
4.17. This problem is an example of under–constrained design specification set,
which happens relatively often in engineering. In order to divide 3.3 V that is available
at the positive power supply rail down to VB → 691.6 mV that is required by the base
terminal, there is an infinite number of possible R1 , R2 values that would achieve
that. Here are several possible reasons that could be used to add more constrains to
the design specifications. By building up on the results in Problem 4.16 we develop
the following arguments.
The input signal amplitude is important factor. For the sake of argument, let us say
that in a given application the input signal amplitude is vin = 5 mVpp . That means
VB = 691.6 mV voltage must be increased by 2.5 mV in order to accommodate for
the full swing of the input signal. In other words, the input common–mode voltage
must be set to VB = 694.1 mV. Therefore, in this setup the base voltage fluctuates as
VB = 694.1 ± 2.5 mV, which guarantees that even the minimum input side voltage
is high enough to keep the transistor VB E diode turned on.
We can now proceed and determine the required ratio (R1 /R2 ), which is found
by setting up the proportion
R1
≤ 3.754
R2
(16.19)
At this point, there are at least two possible additional design goals the could help
us to set constrains leading into unique solution for (R1 , R2 ) pair of resistors. One
possible design goal would be to achieve certain input resistance Z in value, which,
by inspection we simply write as
VB
VCC
=
R1 + R2
R2
∴
R1
VCC
3.3V
−1
=
−1=
R2
VB
694.1mV
Z in = R1 ||R2 ||(β + 1) R E
∴
(16.20)
158
16 Electronic Devices: Solutions
Another possible design goal would be to reduce the overall power consumption of
the circuit. Reference current VCC /(R1 + R2 ) is for all practical purposes wasteful,
because the goal is to set VB , which can be achieved by infinite number of resistor
values, as long as the ratio (16.19) is satisfied. Larger (R1 , R2 ) resistor values allow
lower current, which reduces power consumption. We already discussed practical
limits of large resistors.
In this example we illustrated some of the design compromises that must be made
while designing electronic circuits, which makes analog circuit design as much art as
it is engineering. Indeed, elegant design solutions are always admired by colleagues
and users.
4.18. It is straightforward to find from the given data for VB E , T , and I S that IC =
1 mA. By definition, change of the output current i C due to change of the input
voltage v B E at the operating point IC is refereed to as gm , which in the case of BJT
transistor takes form
VB E
di C d
I
exp
−
1
=
gm ∼
S
dv B E iC =IC
dv B E
VT
i =I
C C
VB E IS
i C IC
=
exp
=
=
(16.21)
V
V
V
V
T
T
i C =IC
T i C =IC
T
which is also illustrated in Fig. 16.7. For the given data and (16.21)
gm =
1 mA
IC
q IC
=
= 38.922 mS
=
VT
kT
25.692 mV
(16.22)
Internal emitter diode resistance re is found by definition of Ohm’s law as
vB E
α v B E α
1
=
=
≤
rE ∼
iE
i C iC =IC
gm
gm
which, for the given data gives r E = 25.692 ω.
Fig. 16.7 Problem 4.18:
illustration for gm definition
(16.23)
16 Electronic Devices: Solutions
159
It is now easy to tabulate BJT parameter variations for various biasing currents, and
various temperatures. Good rule of thumb is that at room temperature and IC = 1 mA,
results in r E ≤ 25 ω. By noticing the inverse relation between r E and IC , one can
easily conclude that if the collector current doubles, the emitter resistance drops by
half, and so on.
Chapter 17
Electrical Resonance: Solutions
In the circuit theory an inductor and capacitor are two of the four fundamental
electronic elements, resistor and memristor being the other two. In addition, the
filter theory is very well developed, which enable us to cary out analytical work with
high level of accuracy. In this chapter we use complex analysis to derive all important
parameters of an LC resonator: the resonant frequency f 0 , bandwidth B, Q factor,
and dynamic resistance R D .
Solutions:
5.1. By inspection of the two RLC networks in Fig. 5.1, it is straightforward to write
the two expressions.
1
jω C
1
Yab (ω ) = G + j jω C −
jωL
Z ab (ω ) = R + j
jωL −
5.2. Another way of saying that the impedance Z ab (ω0 ) is real is to say that its
imaginary part is √(Z ab (ω0 )) = 0, i.e.
√(Z ab (ω0 )) = 0
→
ω0 L −
1
ω0 C
=0
→
ω0 L =
1
ω0 C
→
ω0 C =
1
ω0 L
∴
ω0 = ∞
1
∴
f0 =
1
∞
LC
2λ LC
(and, for parallel RLC network)
1
=0
ω0 C −
√(Yab (ω0 )) = 0 →
ω0 L
∴
R. Sobot, Wireless Communication Electronics by Example,
161
DOI: 10.1007/978-3-319-02871-2_17, © Springer International Publishing Switzerland 2014
162
17 Electrical Resonance: Solutions
ω0 = ∞
1
LC
∴
f0 =
2λ
1
∞
LC
(17.1)
In other words, for ideal case of lossless inductor and capacitor, both series and
parallel LC resonator have the same resonant frequency ω0 . In other words, the
“ideal case” means wire resistance of the inductor is R = 0, i.e. Q factor is Q = ≤.
5.3. By inspection of the two RLC networks in Fig. 5.2, it is straightforward to write
the two expressions.
Series configuration:
j
Z ab = Z L + Z C = jωL −
ω tC
1
at resonance (i.e. ω t = ω0 ), |Z L | = |Z C | → ω t L =
ω tC
∴
Z ab = jω0 L − jω0 L = 0
Parallel configuration:
1
1
1
j
=
+
= jω C −
Z ab
ZC
ZL
ωtL
at resonance (i.e. ω = ω0 ), |Z L | = |Z C |
∴
1
1
1
=
−
=0
Z ab
jω0 L
jω0 L
∴
Z ab = ≤
5.4. Model of an ideal LC resonator assumes no thermal losses whatsoever. In reality,
inductor is made of a wire with a finite resistance R, while the capacitor uses realistic
dielectric that allows for a small but not zero leakage current. Thus, there is a minor
thermal energy loss in the resonator, and the important consequence is that the loss
must be compensated for, otherwise the resonance is not sustainable. Quantitative
measure of the internal thermal losses is expressed through the Q factor.
1
+ jω C
R + jωL
R − jωL
= 2
+ jω C
R + (ω L)2
R
ωtL
= 2
+
j
ω
tC
−
R + (ω L)2
R 2 + (ω t L)2
Y (ω t) =
at resonance (i.e. ω t = ω p0 ), |Z L | = |Z C |
→
√(Y ) = 0
(17.2)
17 Electrical Resonance: Solutions
163
ω p0 C =
R2
ω p0 L
+ (ω p0 L)2
R 2 + (ω p0 L)2 =
∴
L
C
ω p0 =
R2
1
− 2
LC
L
(17.3)
In other words, the resonant frequency ω p0 of a parallel LC network that includes
realistic inductance has the additional term (R/L)2 due to the finite wire resistance, which slightly reduces the resonant frequency relative to the case of ideal LC
resonator. When R ≥ 0 Eq. (17.3) reduces to (17.1) for ideal LC resonator, i.e.
ω p0 ≥ ω0 .
5.5. The imaginary part of √(Y ) determined the resonant frequency, while the real
part ∗(Y ), from (17.2), determines the dynamic resistance R D , i.e. real resistance of
the LC resonator at the resonant frequency, as,
R
+ (ω p0 L)2
R
=
1
R2 +
LC −
∗(Y (ω p0 )) =
=
R2
R2
L2
2
L2
R
L
+ C − R2
RC
=
L
∴
L
1
=
RD =
∗(Y (ω p0 ))
RC
R2
(17.4)
For ideal case, i.e. R = 0, the dynamic resistance of LC resonator becomes R D = ≤.
It should be noted that from the perspective of the resonant current, which circulates
inside the RLC loop, the three elements are in series. Hence, reducing the resistance
associated with the inductive branch is desirable in order to increase the dynamic
impedance being perceived by the network external to the RLC resonator.
5.6. By definition, the quality factor Q of a RLC circuit is the ratio between the
energy stored in the imaginary parts (L or C) and the energy dissipated by the real
part, i.e. R, of the network during a complete cycle,
Q ∓ 2λ
maximum energy stored
energy dissipated per cycle
(17.5)
164
17 Electrical Resonance: Solutions
1. In ideal case, energy stored in the magnetic field of the inductor is eventually
converted without losses into energy of the electrostatic filed of the capacitor.
At the resonant frequency, the maximum energy stored in the network keeps
bouncing back and forth between the inductor and capacitor without losses and,
therefore, is calculated either at the moment when the capacitor is fully discharged
(and therefore the inductor holds the full amount of energy W L ), or when the
capacitor is fully charged (and therefore temporarily holds the full amount of the
energy WC ). That is to say,
WL =
T
0
v(t)i(t) dt =
T
i(t)L
0
Ip
di(t)
1
2
dt = L
i di = L I p2 = L Irms
(17.6)
dt
2
0
or, similarly
Vp
T
T
dv(t)
1
2
dt = C
v(t)i(t) dt =
v(t)C
v dv = C V p2 = C Vrms
WC =
dt
2
0
0
0
(17.7)
∞
∞
where, I p = 2Imax is the peak current through the inductor, and V p = 2Vmax
is the peak voltage across the capacitor.
The energy dissipated in the resistor W R during one full resonant cycle
T0 =
1
2λ
=
f0
ω0
(17.8)
is simply, by definition, power times the time, i.e.
2
W R = PR × T0 = R Irms
× T0 =
2λ
2
R Irms
ω0
(17.9)
which means that (17.5) becomes (using either W L or WC ) for series RLC
Q s = 2λ
WL
= 2λ
WR
2
L Irms
2λ
ω0
R
2
Irms
=
ω0 L
R
(17.10)
2. At resonance, the resonant frequency ω0 , inductance L, and capacitance C are
connected as in (17.1), therefore the three equivalent formulations of Q s are
ω0 = ∞
1
LC
∴
1
1
ω0 L
=
=
Qs =
R
ω0 RC
R
L
C
(17.11)
17 Electrical Resonance: Solutions
165
which are the answers (b) and (c) for series RLC network. Expressions (17.10)
and (17.11) for the quality factor Q are very important and are used to quantify
a number of specifications in radio design.
It is important to note that for ideal inductor, i.e. R = 0 the Q factor becomes
Q = ≤. It is desirable to increase the Q factor for many reasons that will be seen
throughout this book. In addition, it should be noted that in series configuration Q
factor is inversely proportional to the resistance R. For parallel RLC configuration
Fig. 5.1 (right), however, it is desirable to have the R as high as possible in order
to reduce the power dissipation (i.e. to reduce the current through the R branch
of the RLC network), which is to say that the three equivalent quality factor Q p
formulations for a parallel RLC network are
C
R
Qp =
= ω0 RC = R
ω0 L
L
(17.12)
To elaborate the point, it is also useful to find out how much is the difference between
resonant frequencies of series and parallel resonant RLC networks. In problem 5.4
we already derived expression (17.3) for resonant frequency ω p0 of a parallel RLC
network, which can be reformulated as
ω p0 =
1
R2
− 2 =
LC
L
2
2
R
R
1
2 −
ωs0
= ωs0 1 − 2 2 = ωs0 1 − 2
L2
Q
ωs0 L
s
∴
ω p0 ⇒ ωs0 for
(Q s > 10)
(17.13)
where, we appropriately introduced series resonant frequency ωs0 through series
Q s factor. It is now obvious that for ideal or high Q networks (i.e. Q > 10) there
is very small error in calculating resonating frequencies ωs0 and ω p0 of the series
and parallel circuits, hence they can be used interchangeably, as long as the high Q
condition applies.
Finally, expression for dynamic resistance (17.4) can also be reformulated in terms
of the Q factor, after using (17.11), as:
RD =
L
Q
= ω0 L Q =
= Q2 R
RC
ω0 C
(17.14)
which is, again, resistance of a realistic RLC tank in resonance as perceived by the
external network. Important distinction to make is that the resistance R is physical
entity and in series RLC network needs to be as small as possible, while in parallel
configuration it needs to be as large as possible. However, at resonance, this small
resistance is perceived by the external network as if being magnified by the Q 2
factor. In ideal case, i.e. when serial the resistance R = 0, the Q factor goes to
infinity. Hence, the last expression (17.14) is only mathematical approximation.
166
17 Electrical Resonance: Solutions
5.7. Often, it is useful to transform a series RLC network into its equivalent parallel
configuration and vice versa. This transformation must be done only at a single
frequency, which does not affect the Q factor of the networks.
Serial and parallel Q factors are
Xs
Rs
Rp
Qp =
Xp
Qs =
(17.15)
(17.16)
so that, assuming Q s = Q p = Q at the given frequency
(17.17)
Z s = Rs + j X s = Rs + j Q s Rs = Rs (1 + j Q s )
1 − jQ
1
1
Q
1
1
=
=
−j
=
Yp =
Zs
Rs (1 + j Q)
Rs (1 + j Q) 1 − j Q
Rs (1 + Q 2 )
Rs (1 + Q 2 )
(17.18)
Q
1
1
1
−jX
−j
(17.19)
=
=
s
2
Rs (1 + Q 2 )
R
X
p
p
(1 + Q )
Q
∴
R p = Rs (1 + Q 2 )
1
X p = Xs 1 + 2
Q
(17.20)
(17.21)
after replacing (17.15) and (17.16) into (17.17) to (17.19). Again, for large Q, i.e.
Q > 10,
R p ⇒ Q 2 Rs
(17.22)
X p ⇒ Xs
(17.23)
5.8. Let us consider series RLC network from the perspective of voltage source with
resistance R driving impedance Z = j (ω L − 1/ω C ), Fig. 17.1. Maximum power
transfer, therefore, happens when the source is matched to the load, i.e. R = |Z |.
Otherwise, at DC the capacitor becomes open while inductor becomes short connection; and at the other side of the frequency spectrum, at very high frequencies, the
Fig. 17.1 Solution 5.8:
schematic diagram of a
series LC network
17 Electrical Resonance: Solutions
167
capacitor becomes short while inductor becomes open connection. In both extreme
cases there is no power transfer because the loop current must drop to zero.
Hence, condition for maximum power transfer R = |Z |, leads into
Vout =
Vin
Vin
Vin
Vin
|Z | =
R=
=∞
|R + Z |
|R ± j R|
|1 ± j|
2
(17.24)
which happens at two frequency points; let us label them (for the time being) as ωU
and ω L (for “upper” and “lower” frequency, respectively), so that R = |Z | is written
as
1
R = ωU L −
(17.25)
ωU C
1
−R = ω L L −
ωL C
(17.26)
which, after using (17.1) to replace C, leads into
R = ωU L −
1
ωU
−R = ω L L −
1
ω02 L
1
ωL
1
ω02 L
∴ R = ωU L −
∴ −R = ω L L −
ω02 L
ωU
ω02 L
ωL
∴
R
ω0 L
=
ωU
ω0
∴ − ωR0 L =
ωL
ω0
−
ω0
ωU
−
ω0
ωL
(17.27)
and Q = ω0 L/R, which is to say
1
ωU
ω0
=
−
Q
ω0
ωU
ωL
1
ω0
− =
−
Q
ω0
ωL
(17.28)
(17.29)
First, by adding (17.28) and (17.29) it follows that
ωU
ωL
ω0
ω0
+
=
+
ω0
ω0
ωU
ωL
∴
ω02 = ωU ω L
(17.30)
and now, using (17.28) and (17.30) we write
ω2 − ω02
ωU
1
ω0
=
−
= U
Q
ω0
ωU
ωU ω0
∴
ωU ω0
ωU ω0
ω0
ω0
Q= 2
= 2
=
=
2
ω
−
ω
∆ω
ωU − ω0
ωU − ωU ω L
U
L
(17.31)
The last expression is very important, because the two frequencies ωU and ω L are
used to define the resonator’s bandwidth, as the two of them are at −3 dB points
168
17 Electrical Resonance: Solutions
0
Fig. 17.2 Solution 5.8: frequency domain plot to illustrate the BP bandwidth definition
relative to the maximum amplitude of the resonator (which is at ω0 ). Also, (17.31)
shows that narrow band is achieved by using high Q components, or in other words,
high Q is needed to achieve narrow band around the resonant frequency.
As a side note, in series RLC configuration high Q also means very low resistance
R and high inductance L, which is to say that it is good for matching with a low
impedance source, such as an antenna for example, whose impedance is usually in
order of 50 π. Otherwise, if the source impedance is very high then the parallel
RLC configuration must be used, where high Q means high resistance and very low
inductance.
5.9. By definition, Q factors of inductive and capacitive branches in the LC network
(after substitution ESR = r2 ) at resonant frequency ω0 are:
XL
ω0 L
=
= tan φ1 ∴ φ1 = arctan Q 1
r1
r1
XC
1
Q2 ∓
=
= tan φ2 ∴ φ2 = arctan Q 2
r2
ω0 C r2
Q1 ∓
(17.32)
(17.33)
where φ1 and φ2 are respective phase angles in inductor and capacitor due to the
thermal losses (resistances r1,2 denote the internal resistances of the coil and the effective series resistance of the capacitor respectively). We also define, after including
(17.11),
ω0 L
Z 1 = r1 + jω0 L =
+ jω0 L
Q1
∴
ω0 L 2
1
2
|Z 1 | =
+ (ω0 L) = ω0 L 1 + 2
(17.34)
Q1
Q1
17 Electrical Resonance: Solutions
169
as well as,
Z 2 = r2 +
∴
1
1
1
=
+
jω0 C
Q 2 ω0 C
jω0 C
|Z 2 | =
1
1
1
+
=
2
2
(Q 2 ω0 C)
(ω0 C)
ω0 C
1+
1
Q 22
(17.35)
From (17.32) and (17.33) in addition to straightforward application of trigonometric identities,1, 2 we write
sin φ1 = sin φ2 = Q1
1
; cos φ1 = 1 + Q 21
1 + Q 21
Q2
1
; cos φ2 = 1 + Q 22
1 + Q 22
(17.36)
(17.37)
Derivation for the resonant frequency ω0
If an AC voltage Vin = V cos φ1 is applied to the resonator, Fig. 17.3, the total
current needed to compensate for the thermal losses i = i 1 + i 2 is split between the
two branches.
The inductive branch current i 1 has two components: one that is in phase with the
source voltage Vin , i.e. (V cos φ1 )/Z 1 , and one that is lagging by 90∼ , (V sin φ1 )/Z 1 .
At the same time, the capacitive branch current i 2 also has two components: one that
is in phase with the source voltage Vin , i.e. (V cos φ2 )/Z 2 , and one that is leading
the source voltage Vin by 90∼ , i.e. (V sin φ2 )/Z 2 .
At resonance, the two quadrature current components must be opposite and equal
(so that the vector sum is zero), which leads into the following expressions (after
using results (17.34) to (17.37))
Fig. 17.3 Notification in the realistic parallel LC network for solution of the problem 5.9
1
2
∞
cos[arctan x] = 1/ 1 + x 2 .
∞
sin[arctan x] = x/ 1 + x 2 .
170
17 Electrical Resonance: Solutions
1
1
= (V sin φ2 )
Z1
Z2
∴
Q1
1
ω0 C
Q2
=
1
1
1 + Q 21 ω0 L 1 + Q 2
1 + Q 22 1 + Q 2
(V sin φ1 )
1
Q1
2
1 + Q1
2
Q2
=
1
2
1 + Q2
1 + Q2 1 +
2
ω0 LC
(17.38)
1
Q 22
1
where both the left and right side of (17.38) contain algebraic term that can be
simplified as follows:
x
(1 + x 2 )(1 + x12 )
x2
x
=
2
x+
x + x1
=
1
x
x2
(1 + x 2 )(1 +
=
1
)
x2
=
x2
x2 + 2 +
1
1+
1
x2
=
(17.39)
1
x2
Using (17.39) it is straightforward to rewrite (17.38) as:
1
1+
1
Q 21
=
1
1+
1
Q 22
ω02 LC
∴
1 1 +
ω0 = ∞
1+
LC
1
Q 22
1
Q 21
⇒∞
1
LC
;
(Q 1,2 1)
(17.40)
which is the solution for the resonant frequency of an LC resonator with non-ideal
inductor and non-ideal capacitor. Naturally, for very good L and C components
the thermal losses are negligible, in other words Q 1,2 1, hence (17.40) can be
approximated with the expression for the resonant frequency ω0 that was already
derived in problem 5.2. for the ideal case of LC resonator. However, note that the
assumption of high Q is not always valid, a very dramatic example is the case of
on-chip inductors manufactured in standard CMOS process that are used in modern
wireless devices—their Q is in order of five. Consequently, additional design and
technological techniques must be employed to improve performance of integrated
LC resonators.
Derivation for the dynamic resistance R D
At resonance, the sum of complex quadrature components of the two branch currents
is zero, which leaves only the two in-phase current components. Similarly to the
17 Electrical Resonance: Solutions
171
previous derivation, we write,
i=V
⎧
cos φ2
cos φ1
+
Z1
Z2
⎨
1
1
= V ⎪
1 + Q 2 ω0 L 1 +
1
∴
=V
1
Q 21
+
Q1
Q 2 ω0 C
+
2
ω0 L(1 + Q 1 )
1 + Q 22
1
1 + Q 22
ω0 C
1+
1
Q 22
⎩
(17.41)
It is now convenient to introduce substitution for the ω0 C term in (17.41) by rewriting
(17.40) as follows,
ω02 LC =
1 + 12
Q2
1 + 12
Q1
∴
Q 22 ω0 C
1 + Q 22
=
Q 21
(1 + Q 21 )ω0 L
∴ ω0 C =
1 + Q 22
Q 21
Q 22
(1 + Q 21 )ω0 L
(17.42)
and after the substitution (17.41) becomes,
i=V
Q1
Q2
+
ω0 L(1 + Q 21 ) 1 + Q 22
1 + Q 22
Q 21
Q 22
(1 + Q 21 )ω0 L
=V
Q1
ω0 L(1 + Q 21 )
1+
Q1
Q2
(17.43)
which now leads straight into the expression for dynamic resistance R D as
RD ∓
Q 1 + Q11
V
= ω0 L
Q1
i
1+ Q
2
(17.44)
for the case of non-ideal inductor and non-ideal capacitor. As is, (17.44) shows
dependance of dynamic resistance versus Q factors of L and C components. In
case of very good (but still not perfect) inductors, i.e. Q 1 1 or in other words
(1/Q 1 ) ⇒ 0, equation (17.44) is than written as very close approximation,
R D = ω0 L
Q1
1+
Q1
Q2
= ω0 L
Q1 Q2
Q1 + Q2
(17.45)
Modern capacitors, however, are made using very good dielectrics, which is to say
that Q 2 is not only large but could be approximated as Q 2 ≥ ≤, in other words
Q 2 Q 1 , in other words Q 1 /Q 2 ⇒ 0. Therefore, in case of lossless capacitor
(17.45) is further approximated as
R D = ω0 L Q 1
(17.46)
172
17 Electrical Resonance: Solutions
which is the case most commonly used in practice because, in comparison with
capacitors, inductors are much harder components to build.
Finally, in the extreme approximation that is good only for fast “back of the
envelope” hand analysis even the inductor is assumed to be perfectly lossless, i.e.
Q 1 ≥ ≤, which means that (17.46) becomes simply
RD ≥ ≤
(17.47)
which is what already was concluded earlier in (17.14).
All four expressions (17.44) to (17.47), in addition to (17.31), for dynamic resistance R D are useful, as long as the applied assumptions are kept in mind.
5.10. In a series RL model, Q factor of an inductor is simply the ratio of its complex
impedance ω0 L and its real resistance R. High quality, i.e. high Q factor, inductors
are made of very low resistance wire. However, in parallel RL model, high Q values
is achieved if the parallel resistance is very high. For the given data, first we note
that because Q 10 series and parallel Q values are approximately same, i.e.
Qs ⇒ Q p = Q
ω0 L
Rs
Rp
Qp =
ω0 L
Qs =
2λ f L
= 628 mπ
Q
∴
Rs =
∴
R p = 2λ f Q L = 25 kπ
Known inductance value L resonates with only one capacitance value C at the given
resonant frequency f 0 ,
f0 =
2λ
1
∞
∴
LC
C=
1
⇒ 126 pF
(2λ f 0 )2 L
Once Q factor and the resonant frequency f 0 are known, expression for bandwidth
B is already found in 5.8 (17.31) as
Q=
f0
B
∴
B = 50 kHz
With the given resonant frequency f 0 , the required bandwidth B = 500 kHz is
achieved if Q factor of the LC resonator is Q 1 = 20. By working backwards, we
first conclude that the total parallel resistance must be
R P = 2λ f Q 1 L = 2.5 kπ
which is to say that the problem is reduced to finding what resistance R in parallel
with R p = 25 kπ gives the total resistance of R P = 2.5 kπ. From rules for the
equivalent parallel resistance we find
17 Electrical Resonance: Solutions
173
Fig. 17.4 Problem 5.11: Low pass RC filter schematic (left), and high pass RC filter (right)
1
1
1
=
+
RP
Rp
R
∴
R = 2.778 kπ
In this example we have exercised numerical relationships among an LC resonator’s
parameters, and also we explored differences between serial and parallel LC resonators. We realize that series and parallel resonators have different impedances,
which is important data point to know for circuit design because we want to maximize power transfer with the following stages inside radio circuits.
5.11. When LP filter is followed by a HP filter and their respective poles overlap, the
outcome is effectively a bandpass filter whose bandwidth is the overlapped frequency
region. In order to illustrate the point, in this example we consider the following
simplified analysis of lowpass (LP) and highpass (HP) RC filters, Fig. 17.4.
Transfer function HL P ( jω ) of an RC LP filter is derived following the same
procedure as for the voltage divider. By taking the output voltage vout across the
capacitor we have
HL P ( jω ) =
vout
ZC
1
=
=
vin
R + ZC
jω RC + 1
∴
|HL P ( jω )| = 1
1 + (ω RC)2
(17.48)
where, we already know that the −3 dB point is defined as the frequency value
1
|H ( jω −3 dB )| = ∞ = −3 dB
2
(17.49)
which is relative to its maximum voltage value that is normalized to one. Therefore,
from (17.48) and (17.49) it follows that pole of LP RC filter is
ω −3 dB L P =
1
RC
(17.50)
174
17 Electrical Resonance: Solutions
Similarly, transfer function H HP ( jω ) of an RC LP filter is derived as
H HP ( jω ) =
vout
R
jω RC
=
=
vin
R + ZC
jω RC + 1
∴
|H HP ( jω )| = ω RC
(17.51)
1 + (ω RC)2
which leads us into expression for HP RC filter pole as
ω −3 dB HP =
1
RC
(17.52)
We realize that expressions (17.48) and (17.51) are scalable and can be used to
set the poles to any frequency. In general, we normalize all filter transfer function
expressions to 1 Hz so that the designers can scale them to any other frequency. Thus,
normalized versions of LP and HP filter transfer functions are found by setting pole
locations at ω −3 dB = 1, that is RC = 1, as
1
HL P (ω ) = ∞
1+ω2
ω
H HP (ω ) = ∞
1+ω2
(17.53)
(17.54)
where both maximum amplitudes are one, which is easy to verify by setting ω = 0
for LP, and ω ≥ ≤ for HP filter.
From the circuit theory, we know that if two transfer functions are chained one
after another, the resulting function is found as their product, thus we write
|HL P−HP ( jω )| = |HL P ( jω )| |H HP ( jω )|
(17.55)
However, from (17.50), (17.52) and (17.55) we conclude that maximum amplitude
∞
∞
of |HL P−HP ( jω )| must equal (1/ 2) (1/ 2) = 1/2, which means that we need to
multiply (17.55) by two so that its amplitude is also normalized to one, Fig. 17.5.
Series RLC network transfer function is derived, for example, with reference to
schematic in Fig. 17.6, as
vout =
vin
R=
R + X L + XC
vin
R + jω L − j
∴
vout = |H R LC (ω )| = vin R
1 2
R2 + ω L −
ωC
1
ωC
R
(17.56)
17 Electrical Resonance: Solutions
175
At the same time, we already know that Q factor for series RLC network is
Q=
1
ω0 L
=
R
ω0 RC
(17.57)
where, ω0 is the resonant frequency. Simple substitution of (17.57) into (17.56) leads
into
1
(17.58)
|H R LC (ω )| = ω0 2 2
ω
−
Q
1+
ω0
ω
Fig. 17.5 Problem 5.11:
normalized transfer functions
of LP, HP, R LC, and 2 ×
(LP × HP) amplitudes
normalized v out
In order to normalize resonant frequency of (17.58) we set ω0 = 1 and made plot,
Fig. 17.5 for Q = 10, which is very modest value of Q factor for RLC resonators
(typically, RLC resonators have Q factors in the range of few hundred).
With a little bit of algebra, it could be shown that RLC resonator whose Q = 0.5
degenerates in RC LPHP bandpass filter. Alternatively, by re-plotting Fig. 17.5 with
linear frequency axis and by experimenting with the Q factor, it is easy to reach the
same conclusion. Thus, it should be obvious why we typically do not use LPHP RC
filter when a narrowband BP filter is needed.
Although a bit simplified, this example illustrates a line of thought in the design process of a BP filter. Parallel RLC resonator is solved by following the same
reasoning as in this example, while keeping in mind the duality of voltage/current series/parallel passive circuits. While the filter specifications are very clear, the choice
and design of required components is constrained by practical issues related to the
achievable component values and material properties. That discussion, however, is
beyond the scope of this book.
RLC, Q=10
LP
HP
LPHP
0dB
-3dB
0.1
1
normalized frequency
Fig. 17.6 Problem 5.11:
series RLC circuit network
10
176
17 Electrical Resonance: Solutions
5.12. If a radio receiver were built by using only fixed discrete components, as it
was at the very beginning when there were only a few radio transmitters in the
whole world, the whole wireless transmission concept would have been completely
unpractical. And, consequently, we would not have our cellphones, WiFi, radio, TV,
satellites all transmitting at the same time, all the time. Indeed, very dark prospect
by today’s standards.
In order to enable the receiving equipment to isolate the wanted carrier frequency
(out of the very noisy and loud EM environment, where everyone is talking at the
same time and all the time) and to receive the message, we use the concept of BP filter.
Effectively, a BP filter serves as a very narrow “door” into the receiver and allows
only one carrier frequency at ω0 to pass in, while all the other signals at different
frequencies are suppressed by the BP transfer function, i.e. they are rejected at the
entrance.
Instead of building one receiver equipment for each possible frequency, which is
practically the impossible task, we developed concept of tuneable components. For
technological reasons, first tuneable component that was developed for the use in RF
equipment was a rotating metal plate capacitor, Fig. 17.7.
Centre frequency ω0 of an ideal LC resonator is set by the two component values
ω0 = ∞
1
LC
(17.59)
where for a given L = 2.533 nH it follows that f 0 (C = 1.3 nF) ⇒ 87.7 MHz, while
f 0 (C = 857.345 pF) ⇒ 108 MHz. Therefore, this tuneable capacitor is suitable for
the use in FM radio receivers designed for North American region.
5.13. We are now ready for drill design problems, in this example we design LC
resonator using only components found on the shelf. Typically, the inductor must be
designed specifically for the intended application, thus by using L = 2.533 nH we
need to connect in parallel C = 100 pF. This example is a short exercise in using the
available resources to design required circuit.
To create the equivalent C = 100 pF we consider series/parallel combinations of
available capacitors,
(a) C = C1 + C2 + C3 = 10 nF + 40 nF + 50 nF = 100 pF, i.e we connect the four
capacitors in parallel.
Fig. 17.7 Problem 5.12:
tuneable capacitor using
rotating metal plates with
air gap in between each pair
of capacitive plates
17 Electrical Resonance: Solutions
177
(b) In this case we find that serial combination of the four capacitors gives C =
100 nF as
1
1
1
1
1
1
1
1
1
=
+
+
+
(17.60)
+
+
+
=
C
C1
C2
C3
C4
200 nF 400 nF 0.5 µF 2 µF
(c) Combination of capacitor C1 in parallel with series connection of C2 and C3
gives the right value of C = 100 nF as
C = C1 + C2 ||C3 = 70 nF +
60 nF × 60 nF
60 nF + 60 nF
(17.61)
and we have one capacitor to spare. It is common practice to “hack” into prototype
boards and make unusual component combinations until the circuit being developed
is completely debugged and ready for volume production.
5.14. This example is a simple drill design problem to demonstrate sensitivity of Q
factor relative to the quality of wire and the resonant frequency. In this example we
design LC resonator using only components found on the shelf.
With the given inductor of L = 2.0 nH and the given wire resistance, it is straightforward to calculate Q factor as
2λ × 10 × 106 Hz × 2 nH
ωL
=
= 40
r
λ × 10−3 π
ωL
2λ × 100 × 106 Hz × 2 nH
Q=
=
= 400
r
λ × 10−3 π
Q=
which illustrates dependance of Q factor, and therefore the bandwidth, versus frequency. This relation is linear through the inductor impedance Z L = jω L, while
Q factor is inverse function of the wire resistance, which is also correlated to the
inductance value, and all together affecting the bandwidth. That is why the inductors
are designed specifically for the given frequency and intended signal bandwidth.
5.15. Given high enough frequency even a single piece of wire starts behaving as
a complicated RLC network. Existence of both inductive and capacitive elements
implies unavoidable condition for resonance. To our surprise, given a stimuli at the
right frequency a single piece of metal becomes LC resonator. This phenomena is
refereed to as self-resonance. Due to this effect, at very high frequencies it is not
possible to used discrete components, instead designers exploit electrical properties
of various metallic shapes to design passive networks, albeit for very narrow band
of operation.
In this example, we explore operation of a realistic inductive device, Fig. 17.8,
and effects of approaching the self-resonating mode. First, for the given data let us
establish the self-resonant frequency f L0 of this coil as
1
f L0 =
= 71.2 MHz
2λ 1 µH × 5 pF
178
17 Electrical Resonance: Solutions
Fig. 17.8 Problem 5.15: RLC
model af a realistic inductor
and, its Q factor is
2λ × 25 MHz × 1 µH
= 31.4
5π
Q=
With the self resonant frequency known, under normal circumstances of operation,
we assume that the operational frequency is at least one decade (i.e. ten times) below
the self resonant frequency. Otherwise, we must use effective values for the realistic
inductance, which are derived from model in Fig. 17.8 as
1
1
RL
∼
+ jω C L =
+ j ω CL −
YL =
R L + jω L
(ω L)2
ωL
2
RL
1 − ω LC L
=
= ∗(Y L ) + j √(Y L )
−j
2
(ω L)
ωL
(17.62)
where, effective resistance Reff and inductance L eff of this inductor are found from
the real and imaginary parts of (17.62) respectively, as
Reff =
(ω L)2
RL
and L eff =
L
L
=
2
1 − ω LC L
1 − (ω/ω0L )2
(17.63)
which we use to calculate
L
L eff =
1−
f
f L0
2 =
1−
1 µH
25 MHz
71.2 MHz
2 = 1.14 µH
At the circuit resonance frequency ω0 the dynamic resistance R D of the LC tank in
Fig. 17.8 is
(17.64)
R D = Q ω0 L
In the same time, dynamic resistance R D of the equivalent circuit is described using
effective values
(17.65)
R D = Q eff ω0 L eff
17 Electrical Resonance: Solutions
179
Due to equivalence of these two circuits it follows that
Q ω0 L = Q eff ω0 L eff
∴
2 ω
2
Q eff = Q 1 − ω LC L = Q 1 −
ω0
(17.66)
(17.67)
where Q = ω0 L/R as defined in (17.4). We now find the numerical value
Q eff = Q 1 −
f
f L0
2 = 31.4 1 −
25 MHz
71.2 MHz
2 = 27.5
This result shows that effective Q factor Q eff of realistic LC tank decreases with
increased frequency and the difference relative to the ideal resonator model.
5.16. In this short drill problem we explore behaviour of a resonant RLC circuit that is
counterintuitive relative to behaviour of passive circuits studied in linear electronics.
At f = 10 kHz we have,
X L = 2λ × 10 kHz × 3 mH = 188.5π
1
= 159.2π
XC =
2λ × 10 kHz × 100 nF
Z = R 2 + (X L − X C )2 = 302 + (188.5 − 159.2)2 π = 41.9π
We note that at 10 kHz the serial RLC circuit looks more inductive.
At f = 5 kHz we have,
X L = 2λ × 5 kHz × 3 mH = 94.2π
1
= 318.3π
XC =
2λ × 5 kHz × 100 nF
Z = R 2 + (X L − X C )2 = 302 + (94.2 − 318.3)2 π = 226.1π
We note that at 5 kHz the serial RLC circuit looks more capacitive. Obviously, the
results are on two opposite slopes of RLC frequency transfer curve, thus the resonant
frequency must be somewhere in between, at f 0 = 9188.8 Hz to be more precise.
5.17. Arguably the most important transfer function in communication electronics,
an LC resonator’s band-pass (BP) transfer function, contains all relevant information
about the resonator and its operation, thus practicing visual inspection of the BP plot
and extracting the information is very important.
Wireless communication is based on the idea that a single tone high-frequency
signal is used as a carrier of the wanted information, i.e. speech, music, video. The
problem is that a single tone by itself does not carry much information, aside from
180
17 Electrical Resonance: Solutions
revealing the very existence of the signal source. Thus, the single tone communication
systems are limited to on/off coding mechanism, Morse code for instance. On the
other side the human speech consists of a number of single-tone signals organized
in a very specific manner so that our hearing system and brain interpret them as an
articulate and meaningful message.
In the case of the human hearing system, typically a healthy person can distinguish
tones from 20 Hz to 20 kHz, i.e. the human hearing system covers frequency bandwidth of approximately B = 20 kHz . The human speech, however, is synthesized
by using frequencies in the range of approximately 300 Hz–3.4 kHz, which explains
why a phone was relatively easy to invent. It becomes more difficult to transmit classic music that stimulates the full hearing range, or digital video signal that requires
about 6 MHz wide transmission channel.
We already learned about role of a noise in signal transmission mechanism and
concluded that in order to increase the communication distance we must keep the
noise level at the minimum. Consequently, we must design bandwidth of the resonator
to allow passage of only desired frequencies, and in addition the bandwidth must be
centred around the carrier frequency f 0 , as in Fig. 5.5.
By inspection of the given data and frequency plot, we write
B ∓ f 2 − f 1 = (460 − 450) kHz = 10 kHz
(17.68)
which is more than needed for a human speech transmission, but not enough for
Hi–Fi reproduction of classical music.
By definition, Q factor is ratio of the centre frequency f 0 and bandwidth B
Q∓
f0
455 kHz
=
= 45.5
B
10 kHz
(17.69)
This value is reasonable and relatively easy achieved by today’s technology. In contrast, for the same bandwidth specification, cell phone communication system that
works at 2.4 GHz would require a resonator with
Q∓
2.4 GHz
f0
=
= 240,000
B
10 kHz
(17.70)
which is very difficult (but not impossible) to manufacture, and it is good to note
that in this case it is not possible to use the traditional LC resonator due to its wire
resistance. For this reason, LC resonators can achieve Q factors in the range of up to
a hundred, while quartz oscillators are manufactured with Q factors in the range of
ten thousand to a million.
If a C = 100 pF capacitor is available then, in order to set the resonant frequency
at 455 kHz,
f0 =
2λ
1
∞
LC
∴
L=
1
= 1.223 µH
(2λ f 0 )2 C
(17.71)
17 Electrical Resonance: Solutions
181
inductor must be used in this design. Obviously, because the B > 0 it means that
this inductor is realistic, thus its internal resistance is found from the Q factor as
ωL
R
ωL
2λ × 455 kHz × 1.223 µH
=
= 76.877 mπ
Q
45.5
(17.72)
Based on these specifications, designer now can choose type of metallic wire, the
number of turns, and other geometrical parameters of this inductor.
Q=
∴
R=
5.18. In the textbook we find that series and parallel RLC network are different in
their effective impedances presented to the rest of the circuit. For the power matching
purposes, one configuration is chosen over the other. In this drill example we practice
to recalculate serial to parallel resonator parameters.
The two networks must have the same Q factor, which is found by definition as
QS =
XS
1
1
=
=
=2
RS
2λ C S R S
2λ × 7.95 pF × 10 π
(17.73)
then it is straight forward to use the conversion formulas derived in the textbook and
to we write
R p = Rs (1 + Q 2 ) = 10 π (1 + 22 ) = 50 π
1
1
1
1 + 2 = 25.024 π
X p = Xs 1 + 2 =
Q
2λ × 1 GHz × 7.95 pF
2
∴
C P = 6.36 pF
Chapter 18
Matching Networks: Solutions
Although there is number of different ways to design matching networks of various
levels of complexity, in this book we study Q-matching design methodology based on
a single-stage passive matching network topology that uses only two components, an
inductor and a capacitor. This most basic matching network topology is also known
as L-network, where one of the two components is connected either in series and the
other one in parallel with the source and load respectively, or vice versa. Aside from
its simplicity, this methodology does not require sophisticated numerical simulators,
instead hand calculations are sufficient. More elaborated two-stage variant of the
topology based on the use of three components is commonly refereed to as either ω
or T -network, depending how the three components are oriented.
Solutions:
6.1. Filter theory and matching network design procedures are very well developed
and they are one of a few examples where the design flow is actually very well
defined and straightforward to implement. In this book we consider only L-shape
passive LC matching networks, and there are therefore only four basic cases of the
network topology to learn. The source impedance Rs is either greater or smaller than
the load impedance R L and, at the same time, connection between the source and
the load terminals must be either DC or AC.
In this example, we consider the case of load resistance being larger then the source
resistance, R L > Rs with the additional requirement to maintain DC connection
between the source and load impedances. Therefore, to reach “the middle ground”
impedance of the source side R S resistance needs to be increased (i.e. an impedance
X S must be added in series to Rs ) while, at the same time, load side resistance R L
must be reduced (i.e. an impedance X p must be added in parallel with R L ).
In accordance to notation in Fig. 18.1 and Q-matching technique (where Z s = Z ∗p )
we design the LC matching network by executing the following steps:
R. Sobot, Wireless Communication Electronics by Example,
183
DOI: 10.1007/978-3-319-02871-2_18, © Springer International Publishing Switzerland 2014
184
18 Matching Networks: Solutions
Fig. 18.1 Problem 6.1: Typical case of non matched source and load resistances, Rs < R L
1. Calculation of Q factor:
RL
−1=
Rs
Qs = Q p =
50 λ
−1=3
5λ
2. Calculation series impedance X s that needs to be added to the source side:
Qs =
Xs
Rs
∴
X s = Q s Rs = 3 × 5 λ = 15 λ
3. Calculation parallel impedance X p that needs to be added to the load side:
Qp =
RL
Xp
∴
Xp =
50 λ
RL
=
= 16.667 λ
Q
3
4. The above calculation of two complex impedances X s and X p still does not
say which component (i.e. either L or C) should be used to implement each of
them. At this point of design procedure, there are two possible solutions and
both are equally valid if there are no additional constrains. In this example, the
given condition is that there must be DC connection between the source and load,
which means that X s component must be able to conduct DC, in other words
an inductor must be used. Once X s is set as an inductor, X p must be set to the
opposite, i.e. it must be a capacitor. With this additional constrain, the solution for
matching network becomes unique and the actual component values at 10 MHz
are calculated as
X s = 15 λ
∴
X p = 16.667 λ
Xs
= 238.732 nH
2π × 10 MHz
1
Cp =
= 954.9 pF
2π × 10 MHz × X p
Ls =
∴
18 Matching Networks: Solutions
185
and the final schematic of the solution is shown in Fig. 18.2. In the following
examples, we illustrate some other possible cases of relationships between the
source and load impedances.
6.2. Matching of two real resistance should be solved first when working on design
of matching networks, i.e. all complex impedances should be added later as shown
in the following examples. If the case of source resistance being greater than the load
resistance, Fig. 6.1 (right), the design procedure is still as same as in problem 6.1,
with the only difference that greater resistance is used as a reference to calculate the
Q factor,
In order to reach “the middle ground” this time impedance of the source side R S
needs to be reduced (i.e. an impedance X p must be added in parallel to Rs ) while,
at the same time, load side resistance R L must be increased (i.e. an impedance X p
must be added in series with R L ). According to notation in Fig. 18.3 and Q-matching
technique we design the LC matching network by executing the following steps:
1. Calculation of Q factor:
Qs = Q p =
Rs
−1=
RL
50 λ
−1=3
5λ
Fig. 18.2 Solution 6.1: Matching network for the case of Rs = 5λ < R L = 50λ at f = 10MHz
Fig. 18.3 Problem 6.2: Typical case of non matched source and load resistances, Rs > R L
186
18 Matching Networks: Solutions
2. Calculation parallel impedance X p that needs to be added to the source side:
Qp =
Rs
Xp
∴
Xp =
50 λ
Rs
=
= 16.667λ
Q
3
3. Calculation series impedance X s that needs to be added to the load side:
Qs =
Xs
RL
∴
X s = Q s R L = 3 × 5 λ = 15 λ
4. The above calculation of two complex impedances X s and X p is valid for two
possible solutions, depending whether AC or DC connection should be maintained through the matching network. In this example, the given condition is that
there must be AC connection between the source and load, which means that
X s component must be able to conduct AC, in other words a capacitor must be
used. Once X s is set as capacitor, X p must be set to the opposite, i.e. it must
be an inductor. With this additional constrain, the solution for matching network
becomes unique and the actual component values at 10 MHz are calculated as
X p = 16.667 λ
X s = 15 λ
∴
Xp
= 265.258 nH
2π × 10 MHz
1
Cs =
= 1.061 nF ≈ 1 nF
2π × 10 MHz × X s
∴
Lp =
and the final schematic of the solution is shown in Fig. 18.4. In the following
examples, we illustrate some other possible cases of relationships between the
source and load impedances.
6.3. In this short drill example we practice conversion from series to parallel matching
network configuration by implementing formulas already derived in the textbook.
Series inductor impedance at 10 MHz is calculated as
X s = π L S = 2π × 10 MHz × 238.732 nH = 15 λ
which is followed by calculation of serial Q as
Fig. 18.4 Solution 6.2:
Matching network for the case
of Rs = 50 λ > R L = 5 λ
at f = 10 MHz
(18.1)
18 Matching Networks: Solutions
187
Qs =
Xs
=3
Rs
(18.2)
Therefore, serial to parallel conversion formulas give:
R p = Rs (1 + Q 2 ) = 50 λ
1
X p = X s (1 + 2 ) = 16.667 λ
Q
∴
L p = 265.258 nH
6.4. One of the consequences of having non-equal source and load impedances is
that part of the incoming signal energy is reflected back at the interface point and
never reaches the load. This situation is very similar to a sun beam entering water in
a lake where part of the incoming light energy is reflected at the air–water surface
while part of the incoming light energy is absorbed in the water. Depending upon
how large is the mismatch between the source and load impedances, the incoming
energy is reflected to the various degrees, form none to all, and therefore the received
signal is weekended accordingly.
In this example we illustrate quantitative measure of signal reflection at the interface of source and load impedances. By definition mismatch coefficient φ is calculated as:
φ =
Zs − Z p
Zs + Z p
One way to compare Z s and Z p impedances is to transform (X p , R L ) parallel
network into its equivalent serial network (which is not rally necessary, the existing
parallel configuration gives the same result), and then to confirm that they are actually
matched at the given frequency.
With reference to Figs. 18.1 and 18.2, by looking into the source side impedance
to the left it is straightforward to write by inspection
Z s = Rs + j X s = (5 + j 15) λ
(18.3)
By looking into the load side impedance to the right, and after converting the parallel
R L , C impedances into their equivalent series impedances (Rs (R p ), X s (X p )) by
using the transformation formulas as
R p = Rs (1 + Q 2 )
∴
1
)
Q2
∴
X p = X s (1 +
Rp
50 λ
=
= 5λ
(1 + Q 2 )
(1 + 32 )
Xp
16.667 λ
=
= 15 λ
X s (X p ) =
1
(1 + Q 2 )
(1 + 312 )
Rs (R p ) =
188
18 Matching Networks: Solutions
we write expression for the equivalent series network at the load side as
X p = (5 − j15) λ
(18.4)
In other words, impedances on both sides are equal |Z s | = |Z p | = 15.811 λ.
Therefore, at matching frequency of 10 MHz, from (18.3) and (18.4),we have,
φ =
Zs − Z p
(15.811 − 15.811)λ
= 0 = ∞ dB
=
Zs + Z p
(15.811 + 15.811)λ
which, again by definition, gives the mismatch loss ML as
ML =
1
= 1 = 0 dB
1−φ2
In other words, mismatch loss ML = 0dB obtained in this case indicates perfect loss–
less matching that is valid only if inductor L s = 238.732 nH and capacitor C p =
954.9 pF components are used at 10 MHz. At all other frequencies, the calculated
series/parallel impedances map into different component values.
6.5. As same as any other passive network that contains RLC components, matching network is a narrow–band circuit whose bandwidth is determined by the RLC
component values. In the first approximation we assume that the frequency response
curve of the matching network is symmetrical, i.e. equivalent to an ideal LC resonator, which in reality is not the case. In order to find the network’s impedance,
first step is to transform the series part of the network (i.e. at the source side) into
its equivalent parallel network, see example 6.3, which gives component values of
R p = 50 λ and L p = 265.258 nH.
After this series to parallel transformation, the equivalent matching network looks
as in Fig. 18.5 (right). A quick check confirms that the equivalent parallel LC resonator indeed resonates at
Fig. 18.5 Solution 6.3: Conversion of the series R–L subnetwork portion of the matching network
into its equivalent parallel R–L subnetwork, for purposes of calculating the overall bandwidth
18 Matching Networks: Solutions
f0 =
2π
1
√
LC
=
2π
189
1
265.258 nH × 954.9 pF
≈ 10 MHz
(18.5)
which means that the LC resonator’s dynamic impedance R D is infinite, hence the
network resistance is determined only with the two real resistors Rs and R L in parallel.
In other words, the equivalent resonator’s resistance R = 50 λ||50 λ = 25 λ at
the resonant frequency. Resonator bandwidth can now be estimated by using either
inductor or capacitor impedance (they are equal at the resonant frequency) versus the
equivalent parallel resistor value to calculate Q factor, and therefore the bandwidth.
At f 0 = 10 MHz,
Q=
25
R
≈ 1.5
=
XC
16.667
∴
BW3dB =
10 MHz
f0
=
= 6.667 MHz
Q
1.5
Once again, in this approximated estimate it was assumed that the matching network
is symmetrical, which is not the case. In reality, in this case the matching network
has BW3dB a bit wider that estimated in this example. For more precise estimate,
however, we must use a numerical simulator.
6.6. Once the components of the matching network are fixed, effectiveness of the
network is tied to the particular frequency that was used in the design process.
As demonstrated in example 6.4, in the ideal case mismatch losses are reduced to
ML = 0 dB. In reality, that is impossible goal to reach because of finite bandwidth
required by realistic signals (as oppose to a single tone signal) and also because
of the component value variations, thus specifications of realistic systems include
allowed mismatch losses levels for the given communication standard. For the sake
of argument, in this example we examine behaviour of matching network in Fig. 18.2
if a signal whose frequency bandwidth B is in 8 to 12 MHz range, which is centred
around the carrier frequency of f 0 = 10 MHz.
On the lower side of the frequency spectrum at f = 8 MHz the impedance values
are:
X s = π L S = 2π × 8 MHz × 238.732 nH = 12 λ
∴
Q=
Xs
12 λ
= 2.4
=
Rs
5λ
As earlier, it is more practical to convert the parallel RC connection on the load side
into its serial connection and derive
Xp =
1
1
=
= 20.833 λ
πC
2π × 954.931 pF
and then convert parallel RC network on the load side into its equivalent series
network
190
18 Matching Networks: Solutions
50 λ
RL
=
= 7.396 λ
2
(1 + Q )
(1 + 2.42 )
Xp
20.833 λ
=
= 17.751 λ
X s (X p ) =
1
(1 + Q 2 )
(1 + 2.41 2 )
Rs (R L ) =
That is, the left side and right side looking impedances are
Z s = (5 + j 12) λ
∴
|Z s | = 13 λ
Z p = (7.936 − j 17.751) λ
∴
|Z p | = 19.231 λ
which clearly illustrates the mismatch between the source and the load side of
the matching network. Consequentially, there is a portion of signal energy being
reflected, which at 8 MHz is quantified as
φ =
Z p − Zs
(19.231 − 13)λ
= −0.193 = −14.275 dB
=
Zs + Z p
(19.231 + 13)λ
And, therefore mismatch loss is
ML =
1
= 1.039 = 0.331 dB
1−φ2
Repetition of the above procedure at 12 MHz gives the following results
∴
|Z s | = 18.682 λ
Z s = (5 + j 18) λ
Z p = (3.582 − j 12.894) λ
∴
|Z p | = 13.382 λ
which at 12 MHz translates into
φ = −0.165 = −15.636 dB
and
ML = 1.028 = 0.241 dB
Moving away from the matching frequency results in the matching network becoming less effective, a portion of the incoming signal power is reflected back and dissipated while creating destructive interference with the incoming signal wave. Direct
consequence is that amplitude of the signal is lowered, thus SNR reduced.
6.7. Realistic sources and loads are never purely resistive except only as an approximation at very low frequencies. Very elegant technique for designing matching networks is to include any pre-existing parasitic either inductance or capacitance into
the calculations and “absorb” them either partially or completely into the required LC
matching components that are calculated for the case of pure resistive source driving
pure resistive load. This technique is based on the fact that dynamic resistance R D
of a pure LC resonator at resonant frequency is resistive and ideally infinite, thus it
does not change value of the total parallel resistance of the matching network.
18 Matching Networks: Solutions
191
In this example, the source is not ideal and it contains the inductive component.
In the first step of the matching network design procedure, the parasitic inductance
is ignored and network in Fig. 6.2 (left) is assumed to be as shown in Fig. 6.1 (left),
(because in this problem we use the same values for source and load resistances).
In problem 6.1 we already found values of the matching network for the case
of a 5 λ resistive source driving a 50 λ resistive load as L s = 238.732 nH and
C p = 954.9 pF. However, the source already contains 138.732 nH parasitic inductance. In this case we are lucky because of the DC connection requirement, thus in
order to achieve L s = 238.732 nH all that is needed is to add additional X s = 100 nH
inductance in series, Fig. 18.6, and therefore completely absorb the parasitic component into the required inductive value.
6.8. In this example we demonstrate two of the possible ways to treat the parasitic
components, such as C L in Fig. 6.2 (left). Both methods are based on the idea that
adding an inductor in parallel with capacitor creates LC resonator whose dynamic
resistance R D is infinite at the resonant frequency, and therefore it does not contribute
to the total equivalent parallel resistance. The two possible ways to “resonate out”
the parasitic capacitance are:
1. to resonate out the full value of the parasitic component; and,
2. to resonate out only the “excess” part of the parasitic value.
Both approaches are equally valid and the final decision is made by considering
additional criteria for instance, the number of used components or other practical
aspects of the design. Let us take a look at these two possible solutions and illustrate
the point.
1. In the first approach, we take the whole value of the parasitic capacitance and
use it to create LC resonator. By “resonating out” the whole value of the C L
parasitic capacitor, the given network is reduced to the problem of matching two
resistances as in problem 6.1. Hence, an LC resonator at f 0 = 10 MHz connected
in parallel with the full value of the parasitic requires inductor whose value is
calculated as
f0 =
2π
1
√
L CL
Fig. 18.6 Solution 6.7:
Inductive source driving
resistive load
∴
L=
1
= 253.303 nH
(2 π 10 MHz)2 1 nF
192
18 Matching Networks: Solutions
Therefore, at 10 MHz the capacitor will “disappear” because the newly build
LC resonator is transparent at this particular frequency (i.e. 10 MHz tone is not
attenuated at all). After addition of the inductor, the current problem is reduced to
matching 5–50 λ resistor, hence we already know the solution, L s = 238.732 nH
and C p = 954.9 pF, Fig. 18.7.
2. The second possibility is, after already solving the case of the 5–50 λ matching case, to resonate out only the excess portion of C L and used only C0 =
1 nF − 954.9 pF = 45.1 pF and inductor L 0 = 5.616 nH to create ideal “virtual”
L 0 C0 resonator at 10 MHz. Thus, we temporarily imagine that the C L = 1 nF
capacitor consists of two capacitors in parallel, i.e. C L = 954.9 pF + 45.1 pF,
where at 10 MHz the 45.1 pF portion of the load capacitor becomes part of the
L 0 C0 resonator alongside with L 0 , Fig. 18.8. The resulting circuit is more elegant
engineering solution because it uses only two inductive components, as oppose
the solution in Fig. 18.7 that needs two inductive plus one capacitive component.
Note that, for the convenience, in the last several examples we keep reusing
the 5–50 λ matching case, it is understood however that R S and R L resistances
can take any values. Still, regardless of the component values, the single–stage
matching network design flow presented here stays the same. What is more, now
it should not be difficult to also work out other cases including R S > R L , AC or
Fig. 18.7 Solution 1 for 6.8: Inductive source driving resistive load
Fig. 18.8 Solution 2 for 6.8: Imaginary splitting of the C L capacitor for the purpose of partial
resonating out
18 Matching Networks: Solutions
193
DC connection, and with various combinations of (L, C) parasitic components
on each side.
6.9. As simple and elegant as they are, single stage L–shaped matching networks
have limited number of degrees of freedom. Once the components of a single stage
matching network are calculated based on Q matching requirements, bandwidth of
the matching network is also set and can not be changed anymore. In order to add
more degrees of freedom, two or more stages network is required.
In the following examples we explore some of simpler cases of two stage matching network design by using a technique that enables us to reduce the problem of
two-stage to the problem of one-stage network design repeated twice. This design
technique is based on concept of the intermediate (i.e. “ghost”) resistance R I N T that
serves as imaginary load resistance for the first single stage and imaginary source
resistance for the second single stage of the two stage matching network. However,
there are two possibilities, shown below, depending how R I N T is related to Rs , R L
resistances, Fig. 18.9.
When one of the design goals is to decrease bandwidth of the matching network
then ghost resistance R I N T should be set outside of the resistance range set by Rs
and R L values, i.e. either R I N T < min(R S , R L ) or R I N T > max(R S , R L ). Both
possibilities are equally valid, and it is convenient to arbitrarily set R I N T value such
that the resulting Q factor is a round number. Then simply proceed with the design
of two single stage LC matching networks, the first one matching Rs to R I N T , and
the second one matching R I N T to R L .
Fig. 18.9 Problem 6.9: Two stage matching network design concept, Z out = R I N T = Z in
194
18 Matching Networks: Solutions
1. Note that the following two calculated Q values are not equal. However, bandwidth
is limited by the higher Q, which should be used in the design. For the given data let
us arbitrary set R I N T to a value greater than R L , i.e. R I N T > max(5 λ, 50 λ) =
250 λ, Fig. 18.9 (bottom), which produces round numbers for Q factors
Q1 =
RI N T
− 1 = 7.0
Rs
Q2 =
and
RI N T
− 1 = 2.0
RL
Obviously, Q 2 = 7 is the one limiting the total bandwidth.
2. First stage: Matching 5–250 λ results in
X s1 = Q 1 Rs = 35 λ
and
X p1 =
RI N T
= 35.714 λ
Q1
3. Second stage: Matching 250–50 λ results in
X s2 = Q 2 R L = 100 λ
and
X p2 =
RI N T
= 125 λ
Q2
4. General solution, i.e. with no specified types of components is in Fig. 18.10. Each
complex value is either “+” or “−” depending whether the component is an inductor or capacitor respectively. Numerical AC simulation would confirm decrease in
the bandwidth. In addition, if possible then the two parallel components should be
merged into one to reduce the number of components, thus this general solution
would become a T–network.
6.10. When one of the design goals is to increase bandwidth√of the matching network
then the intermediate resistance value is set to R I N T = Rs R L , i.e. geometrical
mean of the two termination resistances. Then, again the two-stage design problem is
reduced to design of two single-stage LC matching networks, the first one designed
to match source resistance Rs to R I N T , and the second one to match R I N T to the
load R L . Once again, resistance R I N T is not a real resistor component added in the
Fig. 18.10 Solution 6.9: General solution that does not yet specify exactly the type, i.e. L or C, of
the complex components
18 Matching Networks: Solutions
195
Fig. 18.11 Problem 6.10: Two stage matching network design concept, Z out = R I N T = Z in
network, it is only a fictitious number used to set impedance of the inner node in
between the two matching network stages (Fig. 18.11).
For the given data given in this example, the design flow is as follows.
1. Calculation of the intermediate resistance:
√
R I N T = R S R L = 5 λ × 50 λ = 15.811 λ
(18.6)
2. Calculation of Q factor at R I N T looking left into the first stage, and looking right
into the second stage:
Q=
Q=
RI N T
−1=
RS
RL
−1=
RI N T
15.811 λ
− 1 = 1.470
5λ
50 λ
− 1 = 1.470
15.811 λ
3. First stage (LP) components: matching 5–15.81 λ results in
X S1 = Q Rs = 7.352 λ
∴
L S1 = 117 nH
RI N T
= 10.753 λ
∴
C P1 = 1.480 nF
X P1 =
Q
where, series inductor and parallel capacitor are used to create LP filter.
4. Second stage (HP): matching 15.811–50 λ results in
X S2 = Q R I N T = 23.250 λ
RL
= 10.753 λ
∴
X P2 =
Q
∴
C S2 = 684.533 pF
L P1 = 117.113 nH
AC simulation of this matching network would confirm the increase in the bandwidth.
With different requirement in terms of use of LP–HP sections other valid solutions
can also be derived (Fig. 18.12).
196
18 Matching Networks: Solutions
Fig. 18.12 Solution 6.10: Two stage matching network schematic
6.11. In this classical design problem, the design of a single stage LC matching
network is straightforward, i.e.
1. Q–calculation:
Rin
−1=
RA
Qs = Q p =
2 kλ
− 1 = 6.25
50 λ
2. Source side impedance calculation:
Qs =
Xs
Rs
∴
X s = Q s Rs = 6.25 × 50 λ = 312.5 λ
3. Load side impedance calculation:
Qp =
Rin
Xp
∴
Xp =
2 kλ
Rin
=
= 320 λ
Q
6.25
4. Calculation of the complex impedances X s and X p still does not say which
component (L or C) should be used to implement them, Fig. 18.13. Both possible
solutions are equally valid. Possible component values are (at f 0 = 665 kHz)
depending whether AC or DC connection through the matching network is
required:
Fig. 18.13 Solution 6.11:
General matching network
solution
18 Matching Networks: Solutions
197
X s = 312.5 λ,
∴
L s = 74.79 µH or Cs = 765.9 pF
X p = 320.0 λ
∴
C p = 747.9 pF or L p = 76.56 µH
6.12. One of the first problems to solve in RF receiver design process is to calculate components of the antenna-to-RF amplifier input matching network. Typically,
the antenna presents itself to the amplifier as a voltage source V A with the internal
resistance of R A = 50 λ. On the input side of the amplifier, impedance of infinitely
large capacitance C0 is obviously zero at all frequencies, while input impedance of a
FET device is assumed infinite. Thus, the input side of the amplifier presents itself to
the antenna through node 1ias parallel connection of the voltage divider resistors
Rin = R1 ||R2 , which is also connected in parallel with capacitor C2 , Fig. 18.14.
Therefore, the original problem is reduced to the design of single stage matching
network for the case of a 50 λ source driving (R1 ||R2 ) load with parasitic C2 capacitance.
Equivalent resistance Rin at the amplifier input side, assuming infinite gate
impedance of FET, is simply the voltage divider’s parallel resistnace (R1 , R2 ), i.e.
Rin = R1 ||R2
(18.7)
From the given condition that V (1) = 1/4 VD D and the laws of voltage divider, it
follows that
VD D
1
R2 = V D D
R1 + R2
4
∴
3R2 = R1
∴
R2 =
5.2 kλ
= 1733.333 λ
3
(18.8)
which after substitution into (18.7) gives Rin = 1300 λ. Therefore, the matching
problem is reduced to matching the R A = 50 λ antenna resistance to the C2 ||Rin
load impedance.
It is now easy to calculate the matching network component values by first solving
the case of 50–1300 λ resistance matching.
Fig. 18.14 Problem 6.12:
Equivalent schematic diagram
of RF amplifier’s input stage
connected to antenna through
matching network
198
18 Matching Networks: Solutions
1300 λ
−1=5
50 λ
X s = Q s Rs = 5 × 50 λ = 250 λ
Rin
1300 λ
= 260 λ
=
Xp =
Qp
5
Qs = Q p = Q =
(18.9)
Duo the required DC connection between the input and output terminals of the
matching network, it follows that X s must be inductive and X p capacitive impedance.
Therefore,
X s = π L s = 250 λ
1
Xp =
= 260 λ
π Cp
∴
∴
L s = 3.979 µH
C p = 61.213 pF
Obviously, because there is already C2 = 11.213 pF capacitor in parallel with the
load resistance Rin , thus, an additional C = 50 pF must be added in parallel to make
up for the required C p = 61.213 pF, which is the final solution for this problem,
Fig. 18.15.
6.13. From the given data, we conclude that real source resistance of R S = 5 λ must
be matched to real load resistance R L = 50 λ. In addition, there is series parasitic
inductor at the source side, X S = 10 λ, and parallel parasitic capacitance at the load
side, X L = 30 λ.
At 10 MHz the source inductance L S is calculated as
XS = π LS
∴
LS =
10 λ
XS
=
= 159.155 nH
π
2π × 10 MHz
while the load capacitance C L is calculated as
XL =
1
π CL
∴
CL =
1
= 530.516 pF
2π × 10 MHz × 30 λ
Fig. 18.15 Solution 6.12: Schematic of matching network
18 Matching Networks: Solutions
199
Fig. 18.16 Solution 6.13: Schematic of matching network
From example 6.13 we already know that matching network for the 5–50 λ case
consists of L s = 238.732 nH and C p = 954.9 pF. Therefore, both the input side
inductance and the load side capacitance should be absorbed into the solution for
the matching network, which means only the differences to L s = 238.732 nH and
C p = 954.9 pF values should be added, i.e.
L = L s − L S = (238.732 − 159.155) nH = 79.577 nH
C = C p − C L = (954.9 − 530.516) pF = 424.383 pF
In this example, both parasitic inductance and parasitic capacitance are absorbed into
the matching network components, Fig. 18.16.
Chapter 19
RF and IF Amplifiers: Solutions
By no means complete and self sufficient, the following tutorials should give enough
guidance to the reader, so that the very difficult subject of analogue circuit design and
analysis becomes more understandable and intuitively acceptable. Of course, analogue circuit design work is based on conceptual thinking developed in applied mathematics, thus the reader is encouraged to review mathematical techniques learned
in the previous schooling and to use this subject as a playground to apply all those
abstract concepts in real life engineering problems.
Solution:
7.1. In this example we practice to quickly estimate impedances associated with the
internal nodes of amplifier circuit simply by inspection and by knowing a few basic
techniques.
Circuit network in Fig. 7.1b demonstrates technique for estimating resistance Z out
as seen by looking into (R1 , R2 ) voltage divider. It is simply Z out = R1 ||R2 . We
reached this conclusion because there are only three components present in the network, resistors R1 , R2 and voltage source VCC . The internal resistance of a voltage
source is zero, thus from the perspective of circuit’s internal resistance nodes G N D
and VCC are shorted through the voltage source. Which leave only two resistors
whose respective terminals are shorted together in parallel.
Circuit network in Fig. 7.1c demonstrates technique for estimating the input
impedance of a BJT transistor whose current gain ω is very large (thus, base current
is i √ 0). We keep in mind the “magnification effect” of emitter resistor R E when it
is looked through the base terminal. It is seen as magnified by ω factor, i.e.
Z in √ ω R E
(19.1)
which makes the base terminal look like very high resistance (base resistance r B is
much smaller then ω R E ). We note that resistance ω R E is not real resistance, instead
it is only perceived resistance at the base node, which is caused by a feedback effect
in this circuit (as derived in textbook),
R. Sobot, Wireless Communication Electronics by Example,
201
DOI: 10.1007/978-3-319-02871-2_19, © Springer International Publishing Switzerland 2014
202
19 RF and IF Amplifiers: Solutions
Circuit network in Fig. 7.1d demonstrates technique for estimating the output
impedance as seen at BJT emitter node. First, between the output node and ground
there is R E resistance. In addition, there is projected resistance due to resistors
(R1 , R2 ) at the input side. We already found that the equivalent resistance of a
voltage divider is R1 ||R2 , which is then projected to the emitter node by being scaled
down by ω factor. Subsequently, the output resistance is found as
Z out = R E || [ω (R1 ||R2 )]
(19.2)
which makes the emitter node look like very low resistance for large values of ω.
7.2. In this short drill example we practice the magnifying emitter resistance concept
by solving the following equations
RB
ω
= ω RE
Rout =
Rin
∴
∴
RB
∴ R B = 10 kλ
100
100 kλ = 100 R E ∴ R E = 1 kλ
100 λ =
This approximative technique is very useful to do mentally and quickly acquire a
feeling for the circuit in hands.
7.3. From results found in the previous problems, by inspection we write:
1. Assuming the required Vth (B E) = 0 V, because I R2 > 0 then any positive
value of R2 > 0 results in VB = Vth (B E) → 0 V, which is sufficient to turn
base–emitter diode on. With the first condition for turning on BJT transistor, the
second condition becomes simply that VC > VB → 0 V.
2. Assuming the required Vth (B E) = 1 V, then because VB = Vth (B E) = 1 V, it
follows that
R2 →
VB
= 1 kλ
I R2
(19.3)
which means that potential at collector node must be held at VC → VB → 1 V.
As a reminder, the condition that base–collector diode must be reverse biased is
satisfied even if terminals of the base–collector diode are shorted, i.e. at the same
potential, thus we include the equality sign in the solution.
7.4. In this example we practice to apply and combine the previous results within a
typical case of CE amplifier.
1. assuming ideal base–emitter diode, and by knowing both current and resistance
at the emitter node, it is straightforward to find the emitter node potential as
VE = I R E R E = 1 V. That means potential at base node must be VB → VE = 1 V.
The biasing design problem is now reduce to finding scaling factor that is required
to derive 1 V from the VCC = 10 V voltage source. Obviously, given that BJT
base does not draw any current from the voltage divider, we write
19 RF and IF Amplifiers: Solutions
203
VB =
VCC
VCC
R2 = R
1
R1 + R2
R2 + 1
∴
R1
VCC
=
−1∞9
R2
VB
(19.4)
where, because VB → 1 V, the resistor ratio must be equal or less then nine. The
transistor is turned on as long as VC → VB → 1 V.
2. assuming realistic base–emitter diode, procedure for finding the resistor ration
stays the same, except that the base potential is shifted up to VB → 2 V. Therefore,
VCC
R1
=
−1∞4
R2
VB
(19.5)
which illustrates reasoning behind values calculated for the voltage divider. As
always, VC → VB → 2 V.
7.5. In the following drill examples we use the “back of the envelope” approach to
to estimate bounds of various circuits parameters in isolation from all other aspects
of circuit behaviour. By focusing at one aspect of circuit behaviour at the time, we
learn to recognize them within a content and mentally evaluate their bounds.
By inspection of Fig. 7.2d we recognize that the input AC signal (capacitor C
stops DC component of the input signal) is connected to the gate terminal, while the
output voltage is taken at the collector terminal of BJT transistor, thus the circuit a
classical common–emitter (CE) amplifier. After carefully going through CE amplifier
related derivations in textbook, with the given set of numerical data, and assuming
that the transistor is set in the active gain mode we conclude that it is possible to
estimate upper bound of CE amplifier voltage gain as the ratio of collector and emitter
resistance,
RC
10 kλ
= 100 V/V = 40 dB
=
(19.6)
Av ∞
RE
100 λ
where the equality sign is valid only in the ideal case of infinite base resistance, that
is equal collector and emitter currents, i.e. infinite current gain of BJT transistor.
Thus, voltage in all realistic cases must be lower than 40 dB.
By visual inspection of (19.6) one could make the following argument. Most often
our design goal is achieve as high gain of an amplifier stage as possible, then why
not choose either RC ≤ ≥ or R E ≤ 0 ? In both cases the voltage gain A V ≤ ≥
becomes immensely high.
In short, it is not realistic approach because (19.6) is an overly simplified
approximation, nevertheless exceptionally useful, that helps us to evaluate the upper
gain limit by simply looking at resistances attached to collector and emitter terminals
of the transistor.
204
19 RF and IF Amplifiers: Solutions
Fig. 19.1 Solution 7.5: Illustration of expression (19.6)
To illustrate the point, let us take a closer look what happens with the amplifier’s
gain if we really push for RC ≤ ≥ or R E ≤ 0, which causes circuit in 7.2d to
morph into the one in Fig. 19.1 (right).
By setting RC = ≥ collector terminal becomes effectively disconnected from
the positive power supply line. In other words, the collector current becomes IC = 0,
which inevitably forces the gain to become zero, i.e. the transistor ceases its primary
function. Therefore, everything else being equal, the increase of RC value causes the
collector current to reduce, thus the gain must drop.
The second choice of reducing R E initially appears more promising, until we
eventually find by experiment that even if R E = 0 the voltage gain is still limited to
a value much lower than the infinite gain. Obviously, there are some other aspects
of the circuit that are not accounted for in (19.6) that limit the gain. At this moment
we state only the first two most important reasons. First, inside a linear amplifier the
maximum signal amplitude is limited by the power supply voltage, and second, in
order to more accurately estimate the gain there is an additional internal resistance
inside the transistor itself that must be included in (19.6), as we see in the following
examples.
7.6. With the given set of numerical data we calculate the following circuit parameters
one after the other.
1. From the given temperature it follows that transistor thermal voltage VT is
VT =
kT
= 25.693 mV
q
(19.7)
2. From the given leakage current I S , base emitter voltage VB E , and the calculated
thermal voltage VT , collector current IC is calculated as
IC = I S
VB E
exp
VT
− 1 = 987.728 mA
(19.8)
3. At the same time, from the calculated thermal voltage VT and collector current
IC , small signal emitter resistance re is calculated as
19 RF and IF Amplifiers: Solutions
205
re √
1
VT
=
= 26 λ
gm
IC
(19.9)
4. From the given emitter resistance R E , input signal frequency f , and the parallel
capacitance C, the total impedance Z E in the emitter branch is
Z E = R E ||Z C =
1
ωC
R E + ω1C
RE
(19.10)
where, at f = 10 MHz capacitor impedance is calculated as Z C (1 µF) =
15.915 λ, and Z C (≥) = 0 λ. By substituting these two impedance values into
(19.10), it follows that
Z E (1 µF) = 15.912 λ
Z E (≥) = 0 λ
(19.11)
5. With the calculated impedances Z E in the emitter branch alongside the calculated
small emitter resistance re , it is now possible to calculate more accurate voltage
gain of CE amplifier, again as the ratio of collector and emitter impedances, as
AV =
RC
re + Z E
∴
10 kλ
= 238.595 V/V = 47.553 dB
26 λ + 15.912 λ
10 kλ
= 384.615 V/V = 51.7 dB
A V (≥) =
26 λ
A V (1 µF) =
The above calculations demonstrate very important technique used in amplifier
circuit design to control the small signal gain while maintaining desired biasing point.
Fixed resistances in collector and emitter branches are used to set BJT biasing point,
which also sets voltage to current gain gm as the first derivative of I (C) = f (VB E )
function (19.8) at biasing point (IC , VB E ).
One of the implications of (19.6) is that reduction of R E leads into higher gain
values. But, change of R E value also changes the biasing point, which controls the
gain. Thus, this is classical case of the “chicken and egg” problem. Fortunately,
biasing point is set by DC current/voltage values, while a signal is by definition
time variance of the biasing point, i.e. it is an AC quantity. After recognizing these
fundamental relations, it is now straightforward to come up with the solution that
satisfies both requirements, one to keep the biasing point stable and the other to
enable high gain of the signal. The solution is to provide two signal paths in parallel,
one for DC and one for AC component of the signal. We already have devices that
can be used to implement this idea, a resistor for DC path and a capacitor for AC
path, Fig. 19.2. (Strictly speaking, AC path is provided by Z E ||R E .)
206
19 RF and IF Amplifiers: Solutions
Fig. 19.2 Solution 7.6: Illustration of two–paths, DC and AC, enabled by addition of C E , with
base–emitter diode resistance re explicitly shown
Fig. 19.3 Solution 7.7: Illustration of CE voltage gain dependance versus emitter resistance
This example illustrates how formal decoupling of DC and AC signal components
enables very powerful practical technique for circuit design, as illustrated by the
effects due to presence of emitter capacitor C E , Fig. 19.2.
7.7. To continue discussion from the previous examples, let us take a look what
happens when there is no external R E resistor in the circuit, instead both gain and
biasing points are set by relying on the internal emitter resistance, Fig. 19.3.
In this example, there is only small emitter resistance to set the gain, which is
derived by following the following reasoning.
IC =
VCC − Vout
RC
∴
re =
VT
VT RC
=
IC
VCC − Vout
∴
AV =
RC
VCC − Vout
=
= (100, 200, 392) V/V
re
VT
when Vout = (7.5, 5, 0.2) V, which illustrates how voltage gain keeps changing with
the output signal. If the output voltage swing becomes too high, the biasing conditions for setting mode of transistor operation may start falling apart, which pushes
19 RF and IF Amplifiers: Solutions
207
transistor first into its linear region, i.e. transistor changes its behaviour from the
current source into voltage controlled resistor, and then eventually the transistor turns
off. Recognizing this dynamic behaviour of a transistor is important for determining
correct range of operation of an amplifier.
7.8. By now we realize that amplifier design using active components is very fine
balancing act of looking for the best compromise among large number of competing
parameters. That is why it is very difficult to formalize the analogue circuit design
flow, similarly to what has been achieved with the digital circuit design. Instead,
analogue circuit design is as much art as it is engineering, which sometimes makes
analogue designers look like magicians. Nevertheless, fundamental physical principles always apply, and it is only matter of learning to recognize as many or them
as possible within the content, and then to find the optimal solution. This particular
skill is what clearly separates brilliant analogue from the rest.
In this example we again evaluate relationship between the controlling VB E voltage and the output collector current IC . Two distinct base–emitter voltages force
two distinct collector currents, assuming VB E ∗ VT , the approximative formula for
collector current is valid, thus for two distinct biasing voltages
IC1
IC2
IC2
IC1
VB E
= I S exp
VT
VB E + πVB E
= I S exp
VT
∴
BE
exp VB E +πV
VT
πVB E
=
= exp
VT
exp VB E
VT
which yields the following two numerical results
πVB E = 18 mV
∴
πVB E = 60 mV
∴
IC2
√ 5 = 6.254 dB
IC1
IC2
√ 11 = 21 dB
IC1
which illustrates the need for keeping very tight control over VB E voltage, due to the
exponential realtionship the output current changes too easily by factor two or more,
which is not negligible by any measure.
7.9. Let us now start looking into important frequency dependent phenomena hiding
inside active circuits. For instance, another way of seeing a pn junction of a diode
is that in reality it is a capacitance. There are two electrodes, anode and cathode,
at different potential and there is an isolation layer in between, depletion region,
thus all elements of a capacitor are present. These internal parasitic capacitances are
208
19 RF and IF Amplifiers: Solutions
Fig. 19.4 Solution 7.9:
Illustration of Miller
capacitance
small and unavoidable because they are fundamental to the diode function. In the
case of a BJT transistor, there is parasitic capacitance in between each pair of the
three electrodes that are, due to their size, usually noticeable at higher frequencies.
However, under special circumstances the collector–gate capacitance CC B becomes visible at much lower frequencies. Specifically, if the following three conditions are met
1. Parasitic impedance provides the return path from the output terminal back to
the input terminal;
2. There is relatively large voltage gain (A V ∗ 1) between the input and output
terminals; and
3. The output signal is inverted.
then, due to the feedback action, the bridging impedance Z C is perceived at the gate
node as
(19.12)
C M = (A V + 1) CC E
which, for large voltage gain becomes significant. Similarly to the magnification
effect of emitter resistance, Miller capacitance is not a real capacitor, however, it is
perceived as real large capacitance when looking into the input terminal, Fig. 19.4.
For the given numerical data, first estimated upper bound of voltage gain is
AV =
CM
RC
= 99
RE
∴
= (99 + 1) 1 pF = 100 pF
(19.13)
which illustrates the point. Due to this large capacitance, for realistic signal sources
with finite source resistance the amplifier input side becomes frequency limited due
to the RC product and therefore unavoidable creation of low–pass (LP) filter.
7.10. In this example we combine several of the previously learned concepts and
demonstrate how their combined effect creates yet another limitation of an amplifier,
Fig. 19.5.
19 RF and IF Amplifiers: Solutions
209
Fig. 19.5 Solution 7.10:
Illustration of input side LP
filter due to Miller capacitance
CM
By inspection of Fig. 7.2b , we write in the following order
1. Input impedance:
1
1
1
1
=
+
+
Rin
R1
R2
ω RE
∴
Rin = 950 λ
(19.14)
2. Voltage gain:
AV =
RC
= 99
RE
(19.15)
3. Miller capacitance:
C M = (ω + 1) CC B =
100
pF
φ
(19.16)
4. Bandpass of LP filter crated by Rin ||C M is found by
ω 3 dB =
1
Rin C M
∴
f 3 dB =
1
2φ 950 λ ×
100
φ
pF
= 5.263 MHz
The calculated LP bandwidth is clearly not infinite and, what is most important, definitely not suitable for amplification for RF signals that are routinely above 10 MHz.
For this reason, it is common to use a two stage cascoded amplifier topology instead
of the single stage CE emitter, where the second common–base (CB) stage is used
mostly to cancel the first of three conditions required to by Miller effect, i.e. to break
the direct feedback from the output back to the input node.
7.11. In this example we practice another typical case of Miller capacitance limiting
performance of an RF amplifier. Here, there is existing inductive component connected to the base node. From the previous examples we already learned to recognize
existence of Miller capacitance, which in this case becomes parallel to the existing
inductor L. By definition, an LC resonator is crated whose centre frequency is calculated from the given data as
210
19 RF and IF Amplifiers: Solutions
ω0 = ∓
1
L CM
∴
f0 =
2φ
1
2.533 µH × 100 pF
= 10 MHz
(19.17)
which gives us illustration how BP filter is created at the input side of an RF amplifier
based on CE single stage. From the given data we conclude that this amplifier should
be used for a 10 MHz signal, however unless we have data about the inductor’s
resistance we can not estimate working bandwidth.
7.12. Concept of Miller capacitance is very important for circuit design, and in this
drill example, technically, same question is asked as in the previous example 7.11
where we learned how collector–base capacitance CC B is reflected into base node in
parallel with the existing inductance L so that LC resonator is formed with resonant
LC frequency set as
1
(19.18)
f0 =
∓
2φ L C M
where, the Miller capacitance, which is calculated as
C M = A V C = 100 pF
Therefore, we conclude that L = 1 µH inductance should be used if all no other
capacitance is present in the circuit.
7.13. In this example we illustrate how frequency characteristics of an amplifier becomes complicated as we include more and more details at the same time. In doing
hand analysis we trade the numerical accuracy for simplicity and the quick estimate
of the circuit behaviour, whose numerical accuracy can be improved by using simulators. From the previous examples we already estimated the input side resistance
of the amplifier as Rin = Rin ||R E ||ω R E = 950 λ, and that Miller capacitance is
C M = 100 pF. With the input signal generator being AC coupled through capacitor
C = 1 µF, the input side network is reduced to the network in Fig. 19.6. Considering that C ∗ C M , it is reasonable approximation to account only for C at low
frequencies and to account only for C M at high frequencies.
Fig. 19.6 Solution 7.13:
Illustration of input side
HP–LP filter due to the
coupling capacitor C and
Miller capacitance C M
19 RF and IF Amplifiers: Solutions
211
At low frequencies capacitor C and input resistance Rin form a high pass (HP)
filter whose −3 dB point is approximately (Miller capacitance is neglected)
f 3 dB (HP) √
1
= 167.6 Hz
2φ Rin C
(19.19)
while, at high frequencies Miller capacitance C M and input resistance Rin form a
low pass (LP) filter (impedance of capacitor C is very low) and its −3 dB point is
approximately
1
= 1.675 MHz
(19.20)
f 3 dB (L P) √
2φ Rin C M
In both cases the frequencies are underestimated, for the HP filter the actual −3 dB
point is only slightly higher (about 188 Hz), however, the error at larger for the LP
−3 dB point because Z C is in parallel with Rin , which pushes the bandpass of LP
filter above 1.675 MHz. It is an iterative procedure to calculate the exact numerical
values, for example simulator finds f 3 dB (L P) √ 14 MHz.
Nevertheless, this quick approximative analysis correctly revealed reasons for the
trend of this amplifier that is band–limited between f 3 dB (HP) and f 3 dB (L P), i.e. it
rejects both low frequencies from DC to about a couple of hundred hertz, and high
frequencies above several megahertz.
7.14. A general amplifier circuit needs two input and two output terminals. In addition, the signal being amplified can take the form of either voltage or current,
therefore there are four possible types of ideal amplifier functions: 1) voltage input –
voltage output, referred to as voltage amplifier whose gain is A V = vout/vin ; 2) current
input – current output, referred to as current amplifier whose gain is Ai = iout/iin ;
3) voltage in – current out, referred to as transconductance amplifier whose gain is
G m = i out i in iout/vin ; and 4) current input – voltage output, referred to as transresistance amplifier whose gain is A R = vout/iin i in .
We keep in mind that these four ideal mathematical amplifying functions are
impossible to implement separately from each other. To various extents, all realistic
amplifiers embody each of the four ideal amplifying functions at the same time.
Therefore, classification of realistic amplifiers is based solely on which one of the four
functions is dominant. Conclusion about the dominant role is reached by observing
whether the input/output terminals are better suited for voltage or current signals,
which brings us back to the study of voltage/current dividers and their properties.
And, of course, voltage/current dividers are modelled by ideal voltage/current source
whose main functionality is related to their terminal impedances; a voltage source
presents low (ideally zero) impedance, while a current source presents high (ideally
infinite) impedance at their respective output terminals. A transistor is very versatile
device that can amplify both voltage and current signals, Furthermore, transistor is
modelled as a three terminal active device, which means if transistor is to be used as
an amplifier then one terminal must be shared between the input and output sides of
the single stage amplifier network. Thus, there are three possible configurations that a
212
19 RF and IF Amplifiers: Solutions
Fig. 19.7 Solution 7.14:
Common base amplifier
schematics
single stage amplifier can have, each named after the sharing terminal, i.e. common
emitter (CE), common base (CB), and common collector (CC). Each of the three
configuration has its own set of characteristics in terms of input/output impedance and
voltage/current gain. By combining single stage amplifiers it is possible to implement
any of the four fundamental amplifying functions with high level of fidelity.
In this example we practice how to quickly evaluate common–base (CB) amplifier.
Assumption that ω = ≥ implies approximations I B = 0 and IC = I E , which
simplifies the analysis very much.
By inspection of Fig. 19.7, we note that the base terminal is connected to the
small signal ground (details of biasing setup are not shown in Fig. 19.7), while both
collector and emitter terminals are accessible by the external world. By identifying
the grounded terminal it is straightforward to classify this amplifier as common base.
From the CB amplifier analysis we known that it is suitable to serve as a current buffer,
because the input terminal is taken at the emitter node (which is low impedance node,
thus suitable to accept current signals) while the output terminals is taken at collector
node (which is high impedance node, thus suitable for generating current). With the
ω = ≥ approximation, which leads into IC = I E , i.e. the output current equals the
input current, we easily conclude that the current gain of CB is Ai = 1. The other
characteristics of CB amplifier are as follows.
1. By knowing potential at VCC rail, collector resistance RC , and collector current
IC = I E = I = 1 mA it is straightforward to conclude that collector potential
VC is below VCC as
VC = VCC − RC IC = 5 V − 7.5 kλ × 0.5 mA = 1.25 V
(19.21)
2. By convention capital letter G m is used to denote voltage to current gain of a
multi–device circuit, while small letter gm is used to denote voltage to current
gain of a single transistor. By definition, gm is change (i.e. derivative) of the output
current due to change of input voltage at the biasing point (IC , VB E ), thus for
small signal changes we write
gm =
IC
0.5 mA
= 20 mS
=
VT
25 mV
(19.22)
19 RF and IF Amplifiers: Solutions
213
3. Infinitely large capacitance C implies short connection for AC signal, and DC
signals are blocked. Thus an input AC signal can be injected into emitter node
without disturbing DC biasing point set by the current source. That leaves collector
as the output terminal. By inspection, we recognize that for this BJT transistor
VB E = −vi , which leads into the following signal analysis
i o = gm vi = gm (−VB E )
(19.23)
therefore, by taking output current at the terminal node i o = i C the small signal
voltage gain is found as the ratio of voltage across the load resistor RC and the
input voltage vi , i.e.
vC = RC i C = −RC gm VB E = RC gm vi
∴
AV =
vC
= gm RC = 20 mS × 7.5 kλ = 150 V/V = 43.522 dB
vi
(19.24)
which illustrates that CB amplifier is effectively used to force current through
the resistive load RC , thus to provide high voltage gain, while at the same time
the current gain is unity. These properties make CB amplifier very useful, especially in combination with the other single–stage amplifiers to design multi–stage
amplifiers such as cascoded amplifier.
Chapter 20
Sinusoidal Oscillators: Solutions
The following tutorial exercises are based on simple idea that the study of oscillators
can be done by splitting this new concept into three already existing concepts, i.e.
concept of a signal amplifier, concept of resonant LC circuit, and concept of feedback
systems.
Solutions:
8.1. All oscillations are, by definition, caused by a feedback path from output of an
amplifier back through the feedback network to the amplifier input, Fig. 20.1. If the
total gain around the loop is one or more, even a small noise signal found at the
input terminal of the amplifier is sufficient to start the oscillations by being amplified
and then routed back to the input, where it is amplified more, and the cycle repeats
until the output signal reaches maximum possible amplitude. Depending upon the
actual loop gain, within a few cycles, the closed loop enters a steady state with stable
periodic waveform present at the amplifier’s output terminal.
In this example we review the general feedback loop equation. By following the
input signal through the loop, first along the forward path through the amplifier, and
then back through the feedback network, the loop equations are written by inspection
as,
Vout = A (Vin + V f b )
V f b = ω Vout
∴
Vout = A (Vin + ω Vout )
∴
Vout
A
=
Vin
1 − Aω
(20.1)
which implies that for Aω = 1 the loop gain becomes infinite, i.e. the loop is unstable
and it starts to oscillate.
R. Sobot, Wireless Communication Electronics by Example,
215
DOI: 10.1007/978-3-319-02871-2_20, © Springer International Publishing Switzerland 2014
216
20 Sinusoidal Oscillators: Solutions
Fig. 20.1 Solution 8.1: a
general closed loop block
diagram
Fig. 20.2 Solution 8.2:
simplified schematic diagram of a phase oscillator
There are number of criteria for establishing conditions for oscillations, none of
them truly general and all applicable, for purposes of our discussion we will adopt
Barkhausen stability criterion, which can be summarized as
1. The loop gain is equal unity, |ω A| = 1, where in reality |ω A| is taken just a
little bit more than unity to compensate for the internal losses;
2. The phase shift around the loop must be an integer multiple of 2λ .
8.2. A phase shift oscillator produces sinusoidal waveform and its operational principle is very educational for purposes of studying oscillating closed loops. It consists
of an inverting amplifier in the forward path, and RC feedback passive network.
Common emitter amplifier is of inverting type, therefore, by inspection, the total
loading resistance Ro and output voltage vout are
R o = RC
vout = −Ro i c = −Ro ω i b
From the perspective of the feedback network CE amplifier behavies as a voltage
source with Ro source resistance, while the feedback loop is maintained through the
base branch with current i b . Therefore, the circuit equations are set in accordance
with the equivalent circuit in Fig. 20.2. Voltage loop equations are:
j
i1 − R i2
−Ro ω i b = Ro + R −
ωC
j
i2 − R ib
0 = −R i 1 + R + R −
ωC
j
ib
0 = −R i 2 + R + R −
ωC
(20.2)
(20.3)
(20.4)
20 Sinusoidal Oscillators: Solutions
217
The system (20.2) to (20.4) can be solved in number of ways, one possible approach
is to introduce substitution Z = R − j/ωC and then eliminate the three currents to
arrive at
(Ro + Z )(Z 2 + 2R Z ) − R(R 2 + R Z − ω R Ro ) = 0
(20.5)
which, after simple algebra rearangements, results in the following polynomial
Z 3 + (2R + Ro )Z 2 + (2R Ro − R 2 )Z + ω R 2 Ro − R 3 = 0
(20.6)
A complex number equals zero if both its real and complex part are zero, which is
to say that the real and complex parts of (20.6) are
√:
R3 −
3R
1
2
+
(2R
+
R
)
R
−
o
(ωC)2
(ωC)2
+ (2R Ro − R 2 )R + ω R 2 Ro − R 3 = 0
1
3R 2
2R
1
2
+
− (2R Ro − R )
=0
→: j −
− (2R + Ro )
ωC
(ωC)3
ωC
ωC
(20.7)
(20.8)
Solving the imaginary part (20.8) for ω , after substituting Z , leads into formula for
the resonant frequency ω 0 as
ω0 =
1
Ro
+6
RC 4
R
(20.9)
At the same time, the real part (20.7) is solved for ω as
ω = 23 + 4
Ro
R
+ 29
R
Ro
(20.10)
Equation (20.10) is solvable for ω = f ( Ro/R ) as function of the resistor ratio x =
Ro /R, Fig. 20.3. Minimum of this function is easily found by setting its first derivative
to zero as
ω(x) =
29
+ 4x + 23
x
∴
ω ∞ (x) = −
29
+4
x2
∴
x ≤ ±2.6926
218
20 Sinusoidal Oscillators: Solutions
Fig. 20.3 Solution 8.2: plot of ω(x) and ω ∞ (x) functions
There are two possible solutions for x, in this case the positive value is taken to
calculate the resistors, hence Ro /R ≤ 2.6926 that, after being substituted in (20.10),
produces
ωmin ≤ 44.5
(20.11)
It is very interesting to note that the ω value does not depend upon specific values
of Ro and R, only on their ratio. From (20.11) and Ro = RC = 10 kπ, it follows
that R = Ro /2.6926 = 3.714 kπ. As the last step, at f = 10 MHz from (20.9) it
follows that C ≤ 1 pF.
8.3. Similarly to any other LC resonator circuit, assuming Q > 10, we can easily
estimate the resonant frequency by following the energy flow inside LC loop. From
the energy flow perspective, energy stored in the capacitor must be delivered to the
series connection of the two inductors, and then returned back to the capacitor.
Therefore, circulating resonant current i c in a tapped L, centre grounded network
perceives the two inductors L 1 , L 2 , and C components in series, which means that
L eff = L 1 + L 2 = 2 µH. It is then straightforward to calculate the resonant
frequency as
f0 =
2λ
≥
1
= 10 MHz
L eff C
This example illustrates how resonant frequency of an oscillating closed loop circuit is
still set by values of its internal LC components, which are fundamentally responsible
for the oscillations in any case.
8.4. When using our strategy of analyzing the forward path (i.e. amplifier) and the
feedback path (i.e. LC network) separately, we must keep in mind that output terminal
20 Sinusoidal Oscillators: Solutions
219
of the amplifier is loaded by input impedance of the feedback path, while at the same
time, output node of the feedback network is loaded by the input impedance of the
amplifier.
By inspection, the feedback network input voltage (i.e. voltage generated by the
amplifier) is distributed across L 2 inductor (note the central ground node in between
the two inductors), while the output voltage of the feedback network is distributed
across L 1 inductor. As we already stated in example 8.3, the same resonating current
i c is circulating through both inductive components, thus it is straightforward to write
vin = i c jω L 2
vout = −i c jω L 1
∴
vout
i c jω L 1
L1
0.5 µH
= −0.333
ω=
=−
=−
=−
vin
i c jω L 2
L2
1.5 µH
(20.12)
that is, amplitude of signal passing through the feedback network is multiplied by
factor ω that is smaller than one, which must be later compensated for by the amplifier
gain.
8.5. As it was shown in textbook, theoretical derivation of this relationship is rather
involved, and very approximative. For the purposes of hand analyses, exercises of
this kind are therefore reduced to direct implementation of the derived formulas.
Using formula provided in textbook, and knowing that the R L resistor of the
feedback network, Fig. 8.1 (middle), is in fact the input resistance of the amplifier,
straightforward implementation of the textbook formula yields
L 2 2 Qω 0 L 22
||
L1
L1 + L2
1.5 µH 2 50 × 2λ × 10 MHz × (1.5 µH)2
= 10 kπ
||
0.5 µH
2 µH
= 90 kπ || 3.534 kπ = 3.4 kπ
Reff = R L
For purposes of practicsing, repeat problems 8.3 to 8.5 for the other three types of
feedback network shown in the textbook.
8.6. Although, by now we already know how to estimate the resonant frequency of
this circuit, let us take the opportunity and develop one possible methodology to
solve this kind of circuits in a more general way.
First, let us rearrange the circuit network so that it becomes more obvious how
the network equations are going to be written. In Fig. 20.4 (left) two paths, p1 and
p2 , from the collector node through the feedback network to ground are identified.
By following the components on each of the two paths it is straightforward to redraw
the equivalent circuit diagram to look as Fig. 20.4 (right).
220
20 Sinusoidal Oscillators: Solutions
By doing so it becomes easy to generalize components in each branch of the
circuit and to introduce the equivalent subnetwork that represents the amplifier itself
(grey box that contains BJT and RC ), whose function is to be a voltage controlled
current source, i.e. collector current i C = f (V1 ).
Second, by inspection, equivalent resistance Ro at the collector node A is
Ro = RC || rC || R D
(20.13)
where, RC is the given real resistor (looking up into the RC from the A node), rC
is the output resistance of BJT at its biasing point (looking down into the collector),
and R D (looking left into the LC resonator) is the dynamic resistance at resonant
frequency. From the signal perspective the three resistances are in parallel, i.e. connected between the A node and signal ground. In addition, the condition Q ∗ ∓
implies that R D = f (Q 2 ) ∗ ∓, i.e. it has no influence on Ro and can be ignored
in (20.13).
Third, the BJT amplifier (the grey box in Fig. 20.4), is then replaced with its
equivalent current source whose current is gm v1 and the output resistance is Ro ,
Fig. 20.5 (left). Note that the feedback loop is maintained through the controlling
voltage v1 . Finally, in order to simplify the incoming analytical expressions, the last
step in this network transformation is to substitute the real RLC components with
general Z 1 , Z 2 , and Z 3 impedances, Fig. 20.5 (right), where
Fig. 20.4 Solution 8.6: simplified schematic transformation of LC oscillator
Fig. 20.5 Simplified schematic transformation of LC oscillator, problem 8.6
20 Sinusoidal Oscillators: Solutions
221
1
ω C1
1
Z 2 = Ro || − j
ω C2
Z 3 = jω L
Z1 = − j
With this last substitution and transformation, it becomes straightforward to write
the KCL current equation at the node A, after recognizing that the same current flows
through Z 1 and Z 3 , as,
v2
v2 − v1
+
Z2
Z3
v1
v2
=
Z1
Z1 + Z3
−gm v1 =
(20.14)
(20.15)
which leads into,
v2
v2
v1
+
−
Z2
Z3
Z3
Z1
v1 =
v2
Z1 + Z3
−gm v1 =
(20.16)
(20.17)
by substituting (20.17) into (20.16) it follows that
1
Z1
Z1
1
−gm
−
v2 = v2
+
v2
Z1 + Z3
Z2
Z3
Z1 + Z3
∴
Z1 + Z3
Z1 + Z3
Z1 Z1 + Z3
+
−
−gm Z 1 =
Z2
Z3
Z3 Z1 + Z3
Z1
Z3
Z1
Z1
−gm Z 1 =
+
+
+1−
Z2
Z2
Z3
Z3
−gm Z 1 Z 2 = Z 1 + Z 2 + Z 3
(20.18)
The result (20.18) is rather general in sense that the three impedances Z 1 , Z 2 , and
Z 3 may be any other combination of RLC components, not necessarily the ones
we started with in this problem. Indeed, CE amplifiers that use the other feedback
networks studied in this course could be solved by applying the same methodology
as in this example.
In this particular example, it is easier to switch to admittances for the reactive
components (of course, the final result must be the same). Following up with (20.18)
we write
222
20 Sinusoidal Oscillators: Solutions
−gm =
∴
Z1 + Z2
Z3
1
1
1 1
+
=
+
+
Z 3 = Y1 + Y2 + Y1 Y2 Z 3
Z1 Z2
Z1 Z2
Z1
Z2
Z1 Z2
1
1
+ jωC2 + ( jωC1 )
+ jωC2 jωL
−gm = jωC1 +
Ro
Ro
1
ω 2 LC1
=
+ jω (C1 + C2 ) − ω 2 LC1 C2
−
Ro
Ro
(20.19)
Condition of resonance is that the imaginary part of (20.19) equals zero, which
directly leads into expression for the resonant frequency as,
C1 + C2 = ω 2 LC1 C2
∴
ω 20 =
1
C2
L CC11+C
2
=
1
LCs
(20.20)
where, Cs is equivalent series capacitance of C1 and C2 . Expression (20.20) is what
we already have known by inspection of the resonant loop in Fig. 8.2, and it is
only reconfirmed by this derivation. For the given data, from (20.20) it follows that
f 0 = 10 MHz.
Under condition of oscillation (20.19) is left only with its real part, i.e. after
substituting ω 0 from (20.20), we write
−gm =
ω 2 LC1
1
− 0
Ro
Ro
∴
gm =
1 C1
Ro C 2
(20.21)
Therefore, from (20.21) for the given data
Ro = RC || rc = 5 kπ
∴
1
× 1 = 200 µs
gm =
5 kπ
The case of finite Q = 50 is worked out by using Z 3 = r + jωL all along,
where r = ω L/Q = 2.513 π. Since the expressions become a bit more complicated,
it maybe be beneficial to use some of the analytical software tools, for example
MAPLE.
It is also recommended that the same problem is solved for other types of LC
feedback networks by using the same methodology.
20 Sinusoidal Oscillators: Solutions
223
8.7. To conclude this chapter on oscillators, we revisit a modern version of tuneable
oscillators used virtually in all modern wireless circuits, that is based on a voltage
controlled capacitive element, a varicap. In the previous chapters we already met
mechanical tuneable rotary capacitor that served its role for decades. Modern varicap
element is nothing more than our old friend a semiconductor diode, disguised as a
voltage controlled capacitor. From the fundamental operation of a pn junction we
remember that the width of depletion region between p and n layers varies based on
the diode biasing voltage VD .
At the same time we know that p type layer, depletion region, and n type layer
form a capacitor. From the definition of capacitance C we know that
C =φ
S
d
(20.22)
where, φ is property of the dielectrics, S is area of the capacitor plates, and d is
distance between the two plates forming the capacitor, which in the case of varicap
diode equals the width of of the depletion region. Therefore, by changing diode
biasing voltage VD , width d of the depletion region changes, which changes C,
which means that varicap capacitance is function of diode voltage, C = f (VD ).
Thus, we created voltage controlled capacitance that can be used as part of LC
resonator to design voltage controlled LC resonator, which is to say tuneable radio
receiver.
In this example we practice to evaluate resonators whose capacitance C D is due
to a biased varicap diode,
CD = C0
|VD |
1+
0.5
=
20 pF
| − 0.7|
1+
0.5
= 5.16 pF
By inspection of Fig. 8.2, from the perspective of the resonating current, the three
capacitances, C1 , C2 , and C D are perceived by the resonating loop as being in series,
hence the total loop capacitance C at zero bias is,
1
1
1
1
=
+
+
= 0.0567 1/pF
C
C1
C2
C0
∴
C = 17.65 pF
which means that the zero biasing frequency is
f0 =
1
≤ 3.789 MHz
2λ 17.65 pF × 100 µH
If the biasing voltage is set to VD = −7V the total loop capacitance becomes
1
1
1
1
=
+
+
C
C1
C2
CD
∴
C = 4.998 pF
224
20 Sinusoidal Oscillators: Solutions
which means that the zero biasing frequency is now found as
f0 =
1
≤ 7.126 MHz
≥
2λ 4.998 pF × 100 µH
Therefore, this oscillator is continuously tuneable from 3.789–7.126 MHz by the
means of a varicap diode whose biasing voltage changes by 7 V. This tiny element that
replaced the bulky rotary capacitor enabled virtually all modern miniature wireless
circuits.
Chapter 21
Frequency Shifting: Solutions
In the high-school algebra we already learned that multiplication of two sinusoidal
functions is done by directly applying trigonometry identities
1
[cos(x − y) + cos(x + y)]
2
1
sin x sin y = [cos(x − y) − cos(x + y)]
2
1
sin x cos y = [ sin (x + y) + sin (x − y)]
2
1
cos x sin y = [ sin (x + y) − sin (x − y)]
2
cos x cos y =
which, in the context of radio communication theory, are fundamental for the concept of frequency shifting. Once the two general mathematical variables x and y
take physical meaning of two frequencies, f 1 and f 2 , then it becomes clear that
the operation of multiplication of two sinusoidal function produces the sum of two
other sinusoidal functions, where one of the two new functions has high frequency
argument ( f 1 + f 2 ) while the second new sinusoidal function has low frequency
argument of ( f 1 − f 2 ). That is, if we start, for instance, with 10 and 11 MHz tones at
the input of a multiplying circuit, then output signal’s frequency spectrum consists
of only two new tones widely separated, one is found at 1 MHz and one at 21 MHz.
Surprisingly, the two original tones, 10 and 11 MHz, are not anymore present in the
output spectrum.
This is extremely useful property of mathematical multiplication of two sinusoidal
functions that can be exploited for practical applications. Obviously, by controlling
one of the two input frequencies, we can “search and isolate” any other tone already
existing in the space, thus we can “tune in” and receive message carried by that
particular carrier.
In summary, the problem of shifting a single tone waveform whose frequency is
A, is solved by multiplying it with another single tone waveform whose frequency is
B, so that the output waveform is a dual-tone waveform that is synthesized from two
R. Sobot, Wireless Communication Electronics by Example,
225
DOI: 10.1007/978-3-319-02871-2_21, © Springer International Publishing Switzerland 2014
226
21 Frequency Shifting: Solutions
single tone waveforms at frequencies ( A + B) and (A − B). All that it left to do is
to design the multiplying circuit, and then filter out either the high or low frequency
tone, and -voilà!- you just learned the most basic principle a radio.
Solutions:
9.1. Direct implementation of the trigonometry identities leads into
S(t) = S1 (t) × S2 (t)
= 3V sin (2π × 1 MHz × t) × 4V sin (2π × 20 MHz × t)
= 12V [cos(2π × 19 MHz × t) − cos(2π × 21 MHz × t)]
(21.1)
First point to observe is that the output waveform S(t) is not a single tone signal,
instead there are two tones present, one at 19 MHz (the difference tone) and one
at 21 MHz (the sum tone). Its main frequency is relatively high, here it is 20 MHz,
however, its amplitude varies in time in accordance with the low-frequency signal,
here 1 MHz. This conclusion becomes obvious once the product of the two signals
S(t) is plot together with the low-frequency single-tone signal S1 (t), where we note
that both positive and negative versions of the low-frequency tone S1 (t) are embedded
into amplitude of the high-frequency tone as its envelope, Fig. 21.1.
This particular example illustrates how an amplitude modulated (AM) waveforms are created mathematically. The high frequency AM signal is referred to as
the carrier, while the envelope itself is the actual low frequency message signal.
This example is trivial because the envelope is just another single tone signal, not
for example modulated speech or music. If that were the case, the envelope shape
would be much more complicated, nevertheless, it would still be exact replica of the
transmitted message. Finally, we note that there are two identical envelopes, both
carrying identical signal waveform, thus by recovering either of the two envelopes
results in the same recovered message.
9.2. The trigonometric identity sin x × sin y = 1/2[cos(|x − y|) − cos(x + y)] shows
that frequency spectrum of the product of two single tones contains another two
tones (and not the original tones), one low–frequency tone with frequency calculated
Fig. 21.1 Solution 9.1:
waveform plots
20
S(t)
S1(t)
-S1(t)
S(t), S1(t), S2(t)
15
10
5
0
-5
-10
-15
-20
0
0.5
1
time
1.5
2
21 Frequency Shifting: Solutions
227
as difference f L F = | f 1 − f 2 |, and one with high-frequency calculated as the sum
f H F = f1 + f2 .
If the carrier frequency tone is known, for example f L O = 10 MHz, in order to
find carrier frequencies f C1 and f C2 , which after multiplication with the 10 MHz
tone both produce a 1 kHz signal, we need to look at the right side of trigonometric
identity to realize that there are actually two possible signals, one on each sides of
the carrier frequency at f L O ± 1 kHz, nevertheless at the same distance of 1 kHz
from f 0 , i.e.
f L F = f 2 − f 0 = 10.001 MHz − 10.000 MHz = 1 kHz
f L F = | f 1 − f 0 | = |9.999 MHz − 10.000 MHz| = 1 kHz
Hence, multiplication of a 10 MHz tone with these two tones, i.e. 9.999 and
10.001 MHz, results in two overlapping 1 kHz LF tones in the output frequency
spectrum, Fig. 21.2. At the same time, the summing tones in the output spectrum
are not identical, one is at 20.001 MHz while the other is at 19.999 MHz, still, their
difference is double of the LF 1 kHz tone.
This example illustrates very important (and not desirable) side effect of multiplication of sinusoidal functions. For a given waveform frequency f L O there are two
other HF signals that produce the same LF signals. However, only one of them carries the wanted signal, while the second therefore produces unwanted interference.
Imagine how it would sound if at the same time a radio receivers plays a classical
music concert and live commentary of a hockey game. Depending on which of the
two programs was wanted by the person listening to the radio, one of the two carriers is perceived as unwanted and it is referred to as the ghost or image frequency.
Existence of the ghost frequencies is unavoidable because it comes from the rules of
mathematics not from the physics, and in practice it must be resolved by regulations
and laws. That is why a radio transmitting equipment must have the licence to operate
at a given frequency, so that the assigned frequency is not creating ghost image to
the already existing radio station.
Fig. 21.2 Solution 9.2: frequency domain plot of two image signals
228
21 Frequency Shifting: Solutions
9.3. Function of frequency multiplier, i.e. frequency shifter, is implemented by a
circuit known as mixer. As oppose to a mixer used to combine musical soundtracks,
which is implements by liner amplifiers, the radio mixer is implemented using nonlinear circuits. In the following examples we practice the frequency shifting principle.
1. Received signal and the local oscillator waveforms are mixed (i.e. multiplied)
inside the receiver’s mixer, therefore an ideal output waveform contains only
two tones that the sum and the difference of the two input frequencies, that is:
(sum:)1435 kHz + 980 kHz = 2415 kHz
(difference:) 1435 kHz − 980 kHz = 455 kHz
2. As oppose to a transmitter whose function is to upconvert the message frequency,
a receiver is intended to downconvert frequency of the incoming carrier waveform, thus in receivers the lower frequency tone is used as IF, f I F = 455 kHz.
By declaring standard numbers to the intermediate frequencies simplifies design
of commercial radio equipment.
3. Working backwards from the mixer output, it is straightforward to find two
frequencies presented at the mixer’s input terminals that can produce same f I F =
455 kHz for the given local oscillator f L O = 1435 kHz as:
(sum:)1435 kHz + 455 kHz = 1890 kHz
(difference:) 1435 kHz − 455 kHz = 980 kHz
In other words, aside from the wanted station operating at 980 kHz, another
station operating at 1890 kHz also produces the same IF frequency after its
carrier waveform is multiplied with f L O by the mixer, therefore it would not be
possible to separate the two transmitted programs, which means that the other
frequency is the image of the wanted frequency.
4. Graphical representation, for case of f L O > f R F , is shown in Fig. 21.3.
9.4. Even though frequency spectrum space is regulated, the number of radio transmitting stations is huge, which is to say that it is not practical to assign neighbouring
frequencies too far apart because the number of available frequencies would be very
small. Instead, practical solution is to also control maximum power level of the image
signals that falls close to the frequency band of the wanted signal. This is possible
because the image signal, by definition, is not at resonant frequency of LC resonator
in radio receiver’s front end electronics.
That means, any signal that is not at the resonant frequency f 0 is attenuated to
some extent depending how far its frequency is from f 0 , Fig. 21.4.
In the textbook we already derived formula for relative amplitudes Ar of frequencies that fall inside bandwidth of LC resonator as:
21 Frequency Shifting: Solutions
229
1
Ar = 1+
Q2
f
f0
−
f0
f
2
Therefore, for the given data Q = 20, f / f 0 = 1.1/1 = 1.1 and f 0 / f = 1/1.1 =
0.909, we find relative amplitude of an image tone that is 10 % away from f 0 as
Ar = 0.253 = −11.925 dB. For sake of comparison, if high quality resonator is used,
for example one whose Q = 200, then the image tone amplitude would be attenuated
to Ar = −31.640 dB, and if Q = 500 then the attenuation is Ar = −39.596 dB
relative to the tone at resonant frequency.
9.5. From KVL, for two equal resistors voltage potential at the node A is at half the
sum of the two inputs, i.e.
vD = vA =
1
1
(v1 + v2 ) = [V1 · cos(ω1 t) + V2 · cos(ω2 t)]
2
2
(21.2)
Following the signal path after the node A, the diode voltage v D is converted into
current. Sum of the input voltage signals is assumed to be small, (implying that
v D < Vt so that higher order terms in (21.5) keep approaching zero), which allows
the exponential term
Fig. 21.3 Solution 9.3:
frequency relationship of
wanted waveform, its image,
and the IF signal. (The sum of
frequencies not shown.)
Fig. 21.4 Solution 9.4:
illustration of partial
suppression of image
frequency by LC resonator
230
21 Frequency Shifting: Solutions
vD
−1
i D = I S exp
Vt
(21.3)
to be expanded into Taylor series around the diode’s biasing point. Series expansion
for exponential function is well known as
√
x2
x3
x4
xn
e =
=1+x +
+
+
+ ···
n!
2
6
24
x
(21.4)
n=0
hence, after substitution x = v D /Vt , equation (21.3) becomes
1
vD
+
1+
Vt
2
I D = IS
vD
Vt
2
1
+
6
vD
Vt
3
1
+
24
vD
Vt
4
+ ···
−1
(21.5)
We can now examine each of the terms on the right side of (21.5) separately and find
out about the signal’s total spectrum (note: the “1” and “-1” terms cancel). Obviously,
the exact series include an infinite number of terms, hence, in the first approximation,
since the assumption is that the signal is small, the third and higher orders terms are
neglected (i.e. they are smaller and smaller numbers being divided by larger and
larger numbers). After substituting (21.2) into (21.5), we have the following two
terms:
1. The linear term:
vD
1
=
[V1 · cos(ω1 t) + V2 · cos(ω2 t)] = f (ω1 , ω2 )
Vt
2Vt
(21.6)
which is to say that the linear term of the series expansion contributes as same
frequencies to the output spectrum as the original signal v D , i.e. ω1 and ω2 . We
already have that at the input, hence this term is not much useful.
2. The square term:
1
2
vD
Vt
2
2
1
1
·
cos(ω
t)
+
V
·
cos(ω
t)]
[V
1
1
2
2
2Vt2 2
1 2
V1 cos2 (ω1 t) + 2V1 V2 cos(ω1 t) cos(ω2 t)
=
8Vt2
+V22 cos2 (ω2 t)
1
1
V12 (1 + cos(2ω1 t))+
=
2
2
8Vt
V1 V2 (cos(|ω1 − ω2 |t) + cos((ω1 + ω2 )t)+
1
V22 (1 + cos(2ω2 t))
2
=
21 Frequency Shifting: Solutions
231
= f [(ω1 − ω2 ), 2ω1 , 2ω2 , (ω1 + ω2 )]
which is to say that aside from the wanted sum and difference tones ω1 and ω2 ,
there are the additional unwanted tones, i.e. 2ω1 , 2ω2 , that are not part of the
ideal multiplication operation.
The multiplying term, i.e. the square term, of interest in expression for diode current
I Ds , after replacing the numerical data for this problem, is
V1 V2
[cos(|ω1 − ω2 |t) + cos((ω1 + ω2 )t)]
8Vt2
1V × 1V
= 1μA
[cos(2π × 1 kHz × t) + cos(2π × 20.001 MHz × t)]
8 × (25mV )2
= 200μA [cos(2π × 1 kHz × t) + cos(2π × 20.001 MHz × t)]
I Ds = I S
which is very often used because a diode is the simplest device that can serve as a
mixer, and also works over wide range of frequencies. Moreover, the small signal
condition can be now stated a bit more specifically as V1 V2 < Vt2 . In this particular
case, the diode generates 1 kHz and 20.001 MHz signals that are widely separated
and easily used in the subsequent circuits. The addition, unwanted tones are usually
filtered out with an LC resonator.
In conclusion, using a diode as a nonlinear element for the purpose of multiplying
two single tone signals does produce the desired theoretical tones ((ω1 − ω2 ) and
(ω1 + ω2 )), however it also produces tones that are not part of the ideal solution
(i.e. ω1 , ω2 , 2ω1 , 2ω2 , ...). In addition, when higher order harmonics in (21.5) are
not neglected, many more tones are created in the output frequency spectrum as
the inter-multiplication terms. All tones that are not needed must be filtered out
afterwards, implying that a diode is rather inefficient multiplying element.
9.6. Depending on the frequency range of interest, it is more common that the simple
diode mixer is replaced with the active BJT based circuit. The two variants of BJT
transistor mixers in Fig. 9.2 are very similar, therefore the following two equations
are written by inspection :
VB E (Q 1 ) =
1
(v1 + v2 )
2
VB E (Q 2 ) = v1 − v2
(21.7)
(21.8)
Relationship between collector current IC of a BJT transistor versus the base-emitter
voltage VB E is as same as for a diode,
iC = I S
vB E
exp
Vt
−1
(21.9)
232
21 Frequency Shifting: Solutions
which is to say that expression for the square term of interest in the output frequency
spectrum of the circuit in Fig. 9.2 (left) is similar to the one for a diode, with the only
addition of the β factor,
ICs = β I S
V1 V2
[cos(|ω1 − ω2 |t) + cos((ω1 + ω2 )t)]
8Vt2
where the corresponding expression for the circuit in Fig. 9.2 (right) is only slightly
different. It is important to note that due to the β factor a BJT mixer has much
better efficiency relative to a simple diode mixer, and it is possible even to have a
conversion gain. That means that it is possible to have the output tone (usually the
low-frequency one, |ω1 − ω2 |) with higher power than the input signal power. On
the other hand, BJT transistor mixer still has the same limitation as for the diode
in terms of the input signal amplitude relative the the Vt voltage. Both circuits in
Fig. 9.2 have LC resonators in the collector branch that can be tuned to either of the
two tones of interest, i.e. either to |ω1 − ω2 | or ω1 + ω2 , and immediately filter out
all other unwanted tones in the frequency spectrum.
9.7. Using similar procedure as in the previous problem with main difference that
current-voltage characteristics between drain current I D and the gate-source voltage
vG S of a JFET transistor obeys the following relationship
I D = I DSS
vG S 2
1−
Vp
(21.10)
where, I DSS is the JFET saturation drain current, VG S is gate-source voltage, and
V p is pinch-off voltage. In JFET case there is no exponential term, which makes the
derivation a bit simpler. Therefore, expansion of (21.10) leads into
I D = I DSS
v2
vG S
+ G2S
1−2
Vp
Vp
(21.11)
By focusing only on the non-linear terms in (21.11), the square term is
I D → −I DSS
→ −I DSS
1 [V1 · cos(ω1 t) + V2 · cos(ω2 t)]2
4
V p2
V1 V2
[cos(|ω1 − ω2 |t) + cos((ω1 + ω2 )t)]
2 V p2
(21.12)
where, (21.12) focuses only at the cosine product term from the previous step.
It is interesting to note that because there was only second order term in (21.11)
without the higher order terms, there was no need to apply power series expansion
as in the cases of a diode and BJT, which is to say that there is no strict limitation
to amplitudes of V1 and V2 , as long as the JFET is not cut off or become forward
21 Frequency Shifting: Solutions
233
biased. Again, similar to the BJT circuits from the previous example, the LC resonator
simultaneously filters out all other harmonics except the desired one. JFET transistors
are commonly used in RF mixer applications due to tolerance to high signal levels
and good conversion efficiency.
9.8. Miller effect drastically limits useful frequency range of single stage CE or
CS amplifiers that are intended for RF applications. To overcome that limit, and
remove the output to input node feedback path that is the cause of Miller effect, it
is common to use casoded amplifier topology instead of the single stage amplifier.
Effectively, casoced amplifier is a two stage amplifier, where in a standard cascoded
configuration, transistor M1 is set as the CS amplifier for v1 signal, while M2 serves as
CG current buffer. Therefore, assuming v2 = const = VDC2 , we can write equations
for M1 transistor in saturation (ignoring its nonlinear effects) as
I D = k(VG S − Vth )2 = k(v1 − Vth )2
∴
d ID
= 2k(v1 − Vth ) = 2k[V1 sin (ω1 t) − Vth ]
gm ∞ ≤
d VG S
(21.13)
where, k = (μn Cox W )/(2L) and Vth is MOS threshold voltage, and gm ∞ is the
circuit’s overall gm under condition that the gate of M2 is at its small signal ground.
Drain current I D passes through the current buffer M2 with no loss, (i.e. the two
transistors have the same drain current), which is followed by LC load. Therefore,
voltage across the LC load at resonance is found by its dynamic resistance R D
multiplied by the drain current, which is as same as the output voltage Vout relative
to VD D . In other words,
Vout = I D R D = gm ∞ v1 R D = gm ∞ R D V1 sin (ω1 t)
(21.14)
That is to say, when gate of M2 is at the small signal ground, in respect to signal v1
the circuit works as a CS cascoded amplifier. However, when gate of M2 is used as
the input terminal for another signal, for example in case of dual-gate MOS mixer
signal v2 comes from local oscillator (LO), the common drain current is additionally
controlled by v2 as well. Variation of drain current due to variation of v2 voltage is
manifested as change of the circuit’s overall gm , as
I D = k[(VDC2 + VG S2 ) − Vth ]2
∴
d ID
= 2k[(VDC2 + VG S2 ) − Vth ]
gm ≤
d(VDC2 + VG S2 )
= 2kVDC2 + 2k(V2 sin (ω2 t) − Vth )
→ gm ∞ + gmΔ sin (ω2 t)
(21.15)
234
21 Frequency Shifting: Solutions
where, gm ∞ is part of the circuit’s gm due to v1 (21.13), while gmΔ is variation of the
circuit’s gm due to v2 whose common mode is (i.e. it is centred around) at VDC2 .
It is important to note that VDC2 is not constant anymore and that this arrangement
works because the two transistors are identical with the same drain current.
After replacing gm ∞ in (21.14) with gm from (21.15) it follows
Vout = [gm ∞ + gmΔ sin (ω2 t)] R D V1 sin (ω1 t)
= gm ∞ R D V1 sin (ω1 t) + gmΔ R D V1 sin (ω2 t) sin (ω1 t)
→ gmΔ R D V1 [cos(|ω1 − ω2 |t) + cos((ω1 + ω2 )t)]
(21.16)
where (21.16) again focuses only on the cosine product term, the LC resonator is
then tuned to either of the two desired tones and filters out all the other harmonics.
In conclusion, dual-gate FET mixer is commonly used for design of RF mixers
for purposes of multiplying the incoming RF signal with the local oscillator (LO).
Setting appropriate LO frequency, the RF signal is then precisely shifted in frequency
domain, i.e. either “down-converted” or “up-converted”. In addition, from (21.16) it
becomes obvious that v2 signal amplitude should be as large as possible so that gmΔ
term is maximized as well, which is another advantage of this circuit.
Chapter 22
Modulation: Solutions
Examples given in this chapter serve as an introduction in AM, FM, and phase
modulation techniques and basic principles. It is recommended that other books are
also consulted for more drill type problems.
Solutions:
10.1. AM modulation is relatively simple and easy to implement, while some of its
basic aspects are illustrated in this example.
1. Given 1.5 kHz audio signal is always positive because its common mode voltage
is set to 3 V, while its amplitude is 1.5 V p , which means that its amplitude swing
is between 1.5 V and 4.5 V. At the same time the 50 kHz carrier signal is set at
zero common mode voltage, Fig. 22.1 (left).
2. AM modulation means that the amplitude of the carrier is multiplied by the audio
signal waveform, thus it is straightforward to write
S(t) = Aa (t) × Ac (t)
= 6 [3 + 1.5 sin(2ω × 1500 × t)] × cos(2ω × 50, 000 × t)
where maximum amplitude of modulated signal is 6 × (3 + 1.5) = 27, and
therefore the minimum amplitude is −27, Fig. 22.1 (right).
3. Modulation factor m is also known as percentage modulation or modulation index,
and it is defined the ratio of the baseband (i.e. audio) and the carrier maximum
amplitudes. In this example,
m=
1.5
= 0.25 = 25 %
6
Preferred AM transmission is when the modulation index is most of the time at
100 %, which implies the case when the signal and carrier amplitudes are equal.
In that case of m = 1, AM modulation results in largest possible demodulated
signal amplitude, which is important to maintain the overall signal to noise ratio
(SNR) in the transmission system.
R. Sobot, Wireless Communication Electronics by Example,
235
DOI: 10.1007/978-3-319-02871-2_22, © Springer International Publishing Switzerland 2014
236
22 Modulation: Solutions
10
Audio
Carrier
Amplitude, V
Amplitude, V
5
0
-5
-10
0
20
40
60
time, µs
80
100
35
30
25
20
15
10
5
0
-5
-10
-15
-20
-25
-30
-35
AM
0
20
40
60
80
100
time, µs
Fig. 22.1 Solution 10.1: illustration of a 1.5 kHz and 50 kHz waveforms (left), and AM modulated
waveform (right)
4. By inspection of the given equations, the carrier frequency is f c = 50 kHz and
the signal frequency is f s = 1.5 kHz.
5. After expanding trigonometry equations describing AM modulated waveform
S AM (t) as
S AM (t) = e(t) sin λ c t
= (C + B sin λ b t) sin λ c t
B
= C 1 + sin λ b t sin λ c t
C
= C (1 + m sin λ b t) sin λ c t
= sin λ c t + m sin λ b t sin λ c t
m
= sin λ c t + [cos |λ c − λ b | t − cos (λ c + λ b ) t]
2
where, index c indicates carrier, and index b indicates baseband audio waveform,
C is the baseband common mode, B is the baseband maximum amplitude, and
C is the carrier maximum amplitude.
This analysis shows that spectrum of AM waveform contains three tones,
1. the carrier itself f c = 50 kHz;
2. the lower–sideband (LSB) tone at f c − f s = (50 − 1.5) kHz = 48.5 kHz; and
3. the upper–sideband frequency (USB) f c + f s = (50 + 1.5) kHz = 51.5 kHz.
This result is very important, because it also shows that energy of AM signal is divided
among the carrier itself (not much useful), and two other tones who carry the same
message. This observation revels that AM modulation is not very power efficient, and
it also gives hints about possible techniques for design of AM transmission systems
with better power efficiency.
10.2. By inspection of Fig. 22.2 and the discussion in Problem 10.1 we observe that
AM signal occupies two times signal frequency bandwidth, which is the distance
22 Modulation: Solutions
237
Fig. 22.2 Solution 10.1
frequency domain plot of
two image signals
between the upper–sideband (USB) and lower–sideband (LSB) frequencies, i.e.
10 kHz. Therefore only 10 stations can fit within a 100 kHz frequency space.
10.3. In this drill exercise we simply use formula developed in textbook that relates
the total power to the carrier power, thus we write:
m2
PT = Pc 1 +
2
0.852
1200 W = Pc 1 +
2
∴
Pc = 881.5 W
Sum of the carrier power and power in two sidebands PS B equals the total power,
therefore,
PS B = PT − Pc = 318.5 W
One half of the total sideband power PS B is in the upper sideband (USB) and one
half in the lower sideband (LSB), i.e.
PU S B = PL S B =
PS B
= 159.25 W
2
which leads us to conclude that 72 % of the total energy of an AM modulated
waveform is held in the carrier itself, Fig. 22.2, while each of the tones actually carrying the information uses only approximately 13 % of the total power. This result
reveals that AM modulation, although very simple, is not very efficient in terms of
energy used to actually transmit the message.
10.4. Change of modulation index for an AM waveform causes the change of power
contained in the sidebands, while the carrier power is kept constant. By using the
equation which relates total power and the carrier power, expression for power in
each of the sidebands is
238
22 Modulation: Solutions
1. m = 0.7:
PU S B = PL S B =
m 2 Pc
0.72 × 1500 W
=
= 183.75 W
4
4
2. m = 0.5:
PU S B = PL S B =
0.52 1500 W
m 2 Pc
=
= 93.75 W
4
4
This example illustrates importance and the role of modulation index in AM
waveforms, which should be compared with the role of modulation index in FM
waveforms.
10.5. Frequency modulated waveform has spectrum that is significantly different
than spectrum of AM waveform. The main difference is that the modulation index
in FM waveform can take values outside from the (0, 1) range. In addition, due to
the properties of Bessel functions, change of modulation index of FM waveform
changes the internal distribution of the energy among the harmonics, however the
total energy stays constant.
1. Using Carson’s rule we write:
B F M = 2(m f + 1) f b = 2(1.5 + 1) 10 kHz = 50 kHz
2. Using Bessel’s function table (see the textbook) we write:
PT
= J02 + 2 (J12 + J22 + J32 + J42 + J52 )
Pc
= 0.5122 + 2 (0.5582 + 0.2322 + 0.0612 + 0.0122 + 0.0022 )
= 1.000258
(22.1)
where the PT → 1 for infinite number of terms, or, in other words, the total power
is constant; it is just redistributed between the carrier and sidebands as function
of the modulation index.
3. according to the Bessel’s function table, for m f = 1.5 the first sideband frequency
signal has the highest amplitude, J1 = 0.558 relative to the amplitude of the
unmodulated signal. We note that for certain values of the modulation index, the
tone whose frequency corresponds to the carrier frequency can be completely
suppressed, Fig. 22.3, however the total energy of the FM waveform is still the
same.
10.6. In this case the intermediate frequency is difference between the carrier and
the local oscillator frequencies, i.e.
1. for f L O > f c it is straightforward to write
22 Modulation: Solutions
239
Fig. 22.3 Solution 10.5: illustration of an FM waveform frequency spectrum for two modulation
indexes
2. for f c > f L O
f I F = f L O − fc
∴
f L O = 995 kHz
f I F = fc − f L O
∴
f L O = 85 kHz
This short example illustrates relationship between the local oscillator frequency
and the carrier frequency, where regardless to which of the two frequency is higher,
they are always multiplied in the mixer and an IF waveform is produced. Both of
the possible solutions are used, and the choice which one to implement is left to the
designer and possible practical limitations of the design at hand.
10.7. FM waveforms keep constant amplitude, however, depending on amplitude of
the baseband signal that is used for the modulation, the carrier frequency keeps moving around the centre frequency, where the distance of the instantaneous frequency
from the centre frequency (i.e. the carrier unmodulated frequency) is proportional to
the baseband signal’s amplitude.
1. given frequency deviation of f = 50 kHz, it is another way to say that the carrier
swing is 100 kHz, i.e. it is “deviating” on both sides of the carrier frequency).
2. the highest frequency is one deviation above the carrier, i.e. 107.65 MHz, and the
lowest frequency is one deviation bellow the carrier, i.e. 107.66 MHz,
3. By definition,
f
50 kHz
= 7.143
=
mf =
fm
7 kHz
which illustrates the role and range of FM modulation index, in comparison with
AM modulation index.
10.8. We continue discussion on power of FM waveform by taking a look at powers
of the individual harmonics. We note that for unmodulated FM signal, the total
power equals to the carrier power, PT = Pc , (i.e. for m = 0). Also, the total power
does not change for various modulation indexes, it only becomes redistributed. That
means that the carrier power, which is initially 100 W, is reduced by its respective
J0 coefficient, while the rest of the power will be “assigned” to the sideband signals.
240
22 Modulation: Solutions
Fig. 22.4 Solution 10.1:
equivalent schematic of
reactance modulator used
in parallel with LC resonator
1. Using equation for FM power and values of Bessel function (as tabulated in
textbook) for m f = 2.0 it follows:
PT = Pc J02 + 2 J12 + J22 + J32 + J42 + J52 + J62
and
J0 = 0.224, J1 = 0.577, J2 = 0.353,
J3 = 0.129, J4 = 0.034, J5 = 0.007, J6 = 0.001
In other words,
P0 = 100 W × 0.2242 = 5.0176 W
P1 = 100 W × 2 × 0.5772 = 66, 5858 W
P2 = 100 W × 2 × 0.3532 = 24.9218 W
P3 = 100 W × 2 × 0.1292 = 3.3282 W
P4 = 100 W × 2 × 0.0342 = 0.2312 W
P5 = 100 W × 2 × 0.0072 = 0.0098 W
P6 = 100 W × 2 × 0.0012 = 0.0002 W
which gives, again, the total power of 100 W.
2. Using Carson’s rule, estimated bandwidth is (for m f = 2),
B F M = 2(m f + 1) f m = 6 × 1.0 kHz = 6 kHz
This example illustrates practical use of Bessel function whose coefficients are used
to scale the corresponding harmonics in FM waveform.
10.9. An elegant way of implementing a voltage controlled capacitance is by circuit
known as reactance modulator, where its output capacitance is function of the control
voltage Vctrl . In this example it is attached to the L T C T resonator to create a voltage
controlled resonator, Fig. 22.4.
For a MOS transistor implementation its equivalent capacitance at the f out node
is given as Ceq = gm RC. Therefore, for the given data, resonant frequency of the
22 Modulation: Solutions
241
voltage controlled resonator is set by:
f out =
1
2ω L T (C T + Ceq )
∴
C T + Ceq = 103.4 nF
∴
Ceq = 20 nF
10.10. Another elegant method to create a voltage controlled resonator is to use
varicap diode inside LC resonator. For phase modulator shown in Fig. 10.2 it was
shown in textbook that phase deviation constant is calculated as
K =−
Q
(1 + 2V0 )(1 + n)
where, C = n Cd0 , and Cd0 is the varicap diode capacitance for the biasing voltage
V0 . Straightforward calculation gives K = −0.2 rad/V.
Chapter 23
Signal Demodulation: Solutions
Design of peak detectors is done with help of numerical solver, however, preliminary
calculations can be done using approximative model. In order to design peak detector
whose operation becomes close to the circuit model based on ideal diode, in realistic
peak detector it is common to use active diode circuit instead of a single diode.
Problems:
11.1. Peak detector, or envelope detector, is a very versatile and useful circuit in
analog signal processing. Its main function is to measure instantaneous time–varying
amplitude of a signal waveform. In RF communication systems, information about
the instantaneous amplitude is essential for analog algorithms used to extract the
embedded message from both AM and FM waveforms. Functionality of a peak
detector is based on an ideal diode model, i.e. voltage drop across the diode is
assumed to be zero. Role of the diode is to serve as unidirectional current valve, which
enables charges to flow into capacitor but not in the opposite direction. By itself, the
capacitor serves as storage space for the incoming charge, where the instantaneous
voltage difference across the capacitor indicates the amount stored charges, thus the
voltage is proportional to the signal amplitude.
As long as the amplitude of incoming waveform keeps increasing, the diode is
turned on and the voltage across the capacitor can follow because the influx of
charges is stored in the capacitor and, therefore the capacitor voltage increases too.
However, once amplitude of the incoming waveform starts to drop relative to its
previous value, then the diode is turned off because the capacitor voltage becomes
higher than voltage amplitude of the incoming waveform, thus the diode becomes
reverse biased. At this moment, unless there is another path to ground for the charges
to flow, voltage across the capacitor would have stayed at the same level indefinitely,
because the path back through the diode is closed. That is why resistor R is needed in
parallel with the capacitor. It should now become apparent that with the right balance
between RC time constant and period T of the incoming signal it is possible to adjust
the overall dynamics of the circuit so that the output voltage becomes relatively close
approximation of the signal’s envelope, Fig. 11.1 (right).
R. Sobot, Wireless Communication Electronics by Example,
243
DOI: 10.1007/978-3-319-02871-2_23, © Springer International Publishing Switzerland 2014
244
23 Signal Demodulation: Solutions
In practice, realistic diodes have the minimum turn on voltage in order of a few
hundred mV, which is much higher than amplitude of weak RF signals arriving to the
peak detector. As solution to this problem a circuit known as an “active diode”, which
consists of operational amplifier and a realistic diode, was developed to approximate
ideal diode behaviour by inserting very low voltage drop across its input and output
terminals.
In this example, we practice an approximate method to estimate component values
of a peak detector. At the same time we keep in mind that methodology presented
in the textbook is very approximate, and that the final component values are found
with the help of numerical solvers.
1. It was shown in textbook that input impedance of the envelope detector should
be approximated as Z in = R/2, therefore Z in = 1 kω.
2. Amplitude of unmodulated input signal (i.e. carrier amplitude vc ( pk)) equals to
the average value of its envelope voltage venv . From the given data and inspection
of Fig. 11.1, we write
vi (avg) =
1.5 V + 0.5 V
venv (max) + venv (min)
=
= 1.0 V = vc ( pk)
2
2
By definition, RMS power of the carrier is
Pc =
vc2 ( pk)
= 0.5 mW
2Z in
From the same data we can find the modulation index as
m=
1.5 − 0.5
= 0.5
1.5 + 0.5
Therefore,
0.52
Pt = 1 +
0.5 mW = 562.5 µW
2
3. Output voltage of the envelope detector should be as same as the envelope of the
input waveform, except for the shift due to diode voltage drop. From the given
data (right side graph) by inspection we conclude that the carrier envelope is
shifted by 0.2 V because of the diode voltage drop.
Therefore, relative to the already calculated minimum and maximum values of
the input signal, maximum value of the envelope is v0 (max) = 1.3V, minimum
value of the envelope is v0 (min)0.3 V, and the average (DC) output is v0 (DC) =
(1.3 V + 0.3 V)/2 = 0.8 V, i.e. also shifted by 200 mV relative to the peak
detector’s input side level.
4. With known average (DC) current and output resistance (R), the output current
must be I0 (DC) = 0.8 V /2 kω = 400 µA.
23 Signal Demodulation: Solutions
245
5. At the end of peak detector analyses, in textbook it was shown that good balance
for choosing the capacitor value, with respect to the clipping conditions at 5 kHz
(1/m a )2 − 1
C=
= 7.7 nF
2λ R f m (max)
Again, design of a peak detector circuits starts with very approximate hand analysis
to find the initial solution, however correct solution requires iterative use of numerical
solvers.
11.2. As we found in the previous chapters, AM carrier is bound by two envelopes
(positive and negative) that are embodiments of the same message. It is in the envelop
detector where it is decided which one of the two envelopes is going to be used. In
Fig. 11.1 diode is forward biased if the input signal amplitude is greater than the
capacitor voltage, thus the positive envelope is stored in the capacitor. In contrast,
diode in Fig. 11.3 is forward biased when the input signal amplitude is lower than
the capacitor voltage. That is to say that in this case negative envelope is stored in the
capacitor. Another addition to note in Fig. 11.3 is presence of LP filter that consists
of (R1 , C2 ) that removes the ragged edges (i.e. the high frequency components) and
smoothens shape of the recovered envelope.
1. Output signal spectrum, at R L , is due to the nonlinear characteristics of the diode.
Both the carrier and the signal are being processed by the diode and, therefore,
they produce sidetones. With the stated assumption of the problem, the incoming
IF signal has carrier at 665 kHz which was modulated by a 5 kHz signal. As a
result, the incoming IF signal contains 665 kHz, as well as two sidetones at 660
and 670 kHz with the respective relative amplitudes, Fig. 23.1. These three tones
are processed by the diode nonlinear characteristics, i.e. multiplied again, and
Fig. 23.1 Solution 11.2: frequency spectrum of diode output signal
246
23 Signal Demodulation: Solutions
produce the following tones:
Sum frequencies:
Difference frequencies:
(660 + 665) kHz = 1325 kHz
(660 + 670) kHz = 1330 kHz
(665 + 670) kHz = 1335 kHz
(670 − 665) kHz = 5 kHz
(665 − 660) kHz = 5 kHz
(670 − 660) kHz = 10 kHz
Note that in the low frequency range the 5 kHz tone is easily isolated by a LP
fillter.
2. Fast changing signal is the carrier, slow changing signal is the signal envelope.
Note that due to the diode orientation the negative amplitude signal envelope is
recovered. The last stage serves purpose of removing the DC offset in the signal
envelope by using the blocking capacitor C, Fig. 23.2.
3. At 5 kHz: impedance of capacitors and diode resistances are as follows:
1
= 144.68 kω ≈ 145 kω
2λ × 5 kHz × 220 pF
1
Z C2 =
= 1.4468 Mω ≈ 1.45 Mω
2λ × 5 kHz × 22 pF
V
0.7V
RD =
=
= 100 ω
I
7 mA
Z C1 =
where diode resistance R D is found from the diode transfer characteristics graph,
Fig. 11.2, and the transformer is modelled as an ideal voltage source element.
Hence, the peak detector is modelled with the equivalent voltage divider, where
first resistance consists of R D = 100 ω, and the second resistance R is
R = Z C1 ||(R1 + (Z C2 ||R2 ||R L )) = 4.6 kω
Fig. 23.2 Solution 11.2: envelope detector internal waveforms
(23.1)
23 Signal Demodulation: Solutions
247
That means that the voltage amplitude at node 3 relative to the voltage amplitude
of the input signal, is as same ratio as
A=
R
V (3)
=
= 0.978
Vin
RD + R
(23.2)
i.e. the 5 kHz signal is attenuated only approximately 2.2 % relative to its input
side amplitude.
4. At 665 kHz: impedance of capacitors are as follows:
1
= 1087.86 ω ≈ 1.1 kω
2λ × 665 kHz × 220 pF
1
= 10.878 kω ≈ 11 kω
=
2λ × 665 kHz × 22 pF
Z C1 =
Z C2
while the diode resistance is not function of frequency, i.e. diode resistance is
still R D = 100 ω. The equivalent circuit network is as same as at 5 kHz, however
this time R = 840 ω. Subsequently, A = 0.894, i.e. the 665 kHz carrier tone is
attenuated approximately 10.6 % relative to its input side amplitude.
5. By choosing component values, designer has control over how much the carrier
tone is attenuated relative to the envelope signal, as well as control over the
internal timing constants.
11.3. In this drill example we practice to the use of dB and dBm power level units at
the system level.
1. Input power is calculated by definition
Pin =
∴
(8 µV)2
V2
=
= 1.28 pW
R
50 ω
Pin ≡ 10 log
1.28 pW
= −88.9 dBm = −118.9 dBW
1 mW
where dBm is power relative to 1 mW reference, while dBW is relative to 1 W
reference level.
2. One of the advantages of using dB unit is that the total system power is found
by simple addition of either dB or dBm units along the system chain, i.e.:
Pout = − 88.9 dBm + 8 dB + 3 dB + 24 dB + 26 dB + 26 dB − 2 dB + 34 dB
= 30.1 dBm
∴
Pout ≡ 1W
248
23 Signal Demodulation: Solutions
11.4. Regardless of the modulation signal shape, AM operation of mixer produces
modulated waveform, Fig. 23.4. By inspection, we observe that the carrier amplitude
swings up to maximum possible value, thus the modulation index is m = 1, or
m=
4−0
Vmax − Vmin
=1
=
Vmax + Vmin
4+0
11.5. Assuming constant antenna impedance, for AM waveform the total power PT
of modulated waveform is related to the unmodulated carrier power PC and the
modulation index m as
m2
m2
2
2
∴ I T R = IC R 1 +
PT = PC 1 +
2
2
∴
2
I
T
m = 2
−1
IC2
thus, for the given data,
1.1A 2
m= 2
− 1 = 0.648 = 64.8 %
1A
11.6. Distribution of harmonic powers within an FM waveform is calculated by using
Bessel function coefficients, where each coefficient is use to scale the respective
harmonic tone. For the given example
S F M (t) = 2000 sin 2λ × 108 + 2 × λ × 104 cos (λ × 104 t) t
(23.3)
it follows that:
1. By inspection of the given FM signal equation:
fc =
(2λ × 108 ) rad/s
πc
=
= 100 MHz
2λ
2λ
2. Given FM peak voltage of 2000 V, by definition
P=
√
(2000 V/ 2)2
= 40 kW
50 ω
3. Modulation index of an FM waveform is found from definition of the waveform
frequency π (t), which is described as
23 Signal Demodulation: Solutions
249
π (t) = π c + k B0 cos π b t
(23.4)
where, π c is the carrier frequency, k is frequency deviation constant, B0 is maximum amplitude of the modulating signal, and π b is frequency of the modulating
signal. With this annotation, modulation index m f is defined as
mf =
k B0
πb
(23.5)
which, by inspection of the FM signal Eq. (23.3) with reference to (23.4) results
in:
mf =
2 × λ × 104
=2
λ × 104
4
4
3
3
2
2
amplitude
amplitude
Fig. 23.3 Solution 11.2: equivalent envelope detector circuit
1
0
-1
-2
-3
-4
1
0
-1
-2
-3
0
0.5
1
time
1.5
2
-4
0
0.5
1
time
Fig. 23.4 Solution 11.4: envelope of a carrier modulated by triangular waveform
1.5
2
250
23 Signal Demodulation: Solutions
4. By definition,
fb =
(λ × 104 )
πb
=
= 5 kHz
2λ
2λ
5. By reading the table with Bessel functions, for m f = 2, significant bands exist
to J4 , which means that on one side of the unmodulated carrier frequency 4 ×
f s = 4 × 5 kHz = 20 kHz bandwidth is needed. Thus, due to the symmetry of
bandwidth, Fig. 23.3, relative to the carrier frequency whose scaling factor is J0 ,
B = 2 × 20 kHz = 40 kHz.
6. From table with Bessel functions, J1 is the largest sideband at 0.58 times the
unmodulated carrier amplitude. The smallest sideband is J4 = 0.03. Hence
√
(0.58 × 2000 V/ 2)2
= 13.5 kW (i.e. 27 kW for the two–sided band)
50 ω √
(0.03 × 2000 V/ 2)2
= 36 W (i.e. 72 W for the two–sided band)
P4 =
50 ω
P1 =
In this example we reviewed basic definitions related to FM waveforms.
Chapter 24
RF Receivers: Solutions
System level analysis concepts presented in the following tutorials are taken one at
the time, so that the reader can recognize them within the content. Although we use
mostly AM radio receiver working at relatively low frequencies by today’s standard
to demonstrate the principles, by now the reader should be comfortable with applying
all relevant concepts and techniques, with the understanding that all the other wireless
communication systems follow the same principles and similar design philosophy.
Solutions:
12.1. The traditional heterodyne radio receiver architecture, Fig. 24.1, was invented
by Edwin Armstrong in 1918 and has been used in virtually all radio equipment. Usually, if the radio receiver contains one VCO circuit, i.e. one IF frequency, it is referred
to as heterodyne receiver. As the radio frequencies moved into higher ranges, due to
problems related to image frequencies, it was necessary to introduce second IF frequency and performs down–converting operation in two steps. When two VCOs and
two IF frequencies are used for down–conversion, the radio architecture is referred
to as super heterodyne radio.
The space around us is filled with RF EM waves v R F from various sources emitting
all at the same time, and it is in a radio antenna that the EM waves are converted
into the induced voltage signals. At the first radio stage, i.e. in the matching network
and RF amplifier, the RF amplifier’s internal LC resonator allows only the desired
waveform whose frequency is ω c to enter the system. The filtered v AM waveform
is mixed (i.e. multiplied) with the local voltage controlled oscillator’s tone vV C O
to produce down–converted waveform v AM− . It is role of IF amplifier to filter and
amplify the lower-side band waveform v I F , which is then passed through envelope
detector to extract the transmitted message v pk . The envelope waveform v pk filtered
with low-pass filter to remove the ragged edges (i.e. HF components) and to deliver
high fidelity version of the transmitted message v pk L P, which is then amplified
with audio amplifier to deliver the recovered vb .
12.2. In this example we practice calculations related to tuneability of an AM receiver
and variations in bandwidth due to the component values.
R. Sobot, Wireless Communication Electronics by Example,
251
DOI: 10.1007/978-3-319-02871-2_24, © Springer International Publishing Switzerland 2014
252
24 RF Receivers: Solutions
Fig. 24.1 Solution 12.1: heterodyne radio architecture and time domain plots of the internal waveforms
1. By definition, Q factor of an LC resonator is calculated as
Q=
1050 kHz
f0
=
= 105
B
10 kHz
Assuming constant Q factor at f 0 = 1600 kHz the achieved bandwidth is
f max
1600 kHz
=
= 15.238 kHz
Q
105
B f max =
which requires capacitance
C=
1
(2 π f max )2 L
= 9.895 nF
(24.1)
2. Similarly, at f min = 500 kHz for the given data we fand
B f min =
f min
500 kHz
=
= 4.762 kHz
Q
105
which requires that the tuneable capacitor is set to
C=
1
(2 π f min )2 L
= 101.321 nF
(24.2)
In this example we illustrate that with the fixed inductance, the tuneablity comes
with the price in increased overall bandwidth. Although at the centre of the tuning
range the designed bandwidth is B = 10 kHz, at the low end of the tuning range a
24 RF Receivers: Solutions
253
signal whose bandwidth is B = 4.762 kHz can be processed. At the same time, at
the high end of the tuning range, a signal whose bandwidth is up to 15.238 kHz can
be processed. In summary, this receiver should be used only to process signals whose
bandwidth is B √ 4.762 kHz, however if the channel spacing Δf between various
AM radio stations is constant, then it must be set to Δf → B f max = 15.238 kHz to
avoid the inter-channel interference.
12.3. Design of tuneable components is an important engineering issue because there
are practical limits to the possible tuneable range that can be actually manufactured
given the state of the art technology. In this example we take a look at design of
tuneable dual capacitor that is intended for simultaneous use in the RF amplifier and
LO oscillator sections.
1. Starting from the input RF signal frequency range it follows that, for a given
inductor, frequency tuneability ratio R f R F and capacitor tuneability ratio Rc R F
are
f R F (max)
1600 kHz
=
= 3.2
f R F (min)
500 kHz
∴
1
∞
Cmax
f R F (max)
2π LCmin
=
=
1
f R F (min)
Cmin
∞
2π LCmax
∴
C R F (max)
f R F (max) 2
=
= 10.24
=
C R F (min)
f R F (min)
R f RF =
Rc R F
That is, in order to enable frequency tuning ratio of 3.2 times a tuneable capacitor
whose tuning ratio is at least 10.24 is needed.
2. Frequency of a signal generated by the mixer, in this example the intermediate
frequency IF, is calculated as absolute difference Δ between the local oscillator
oscillator and RF frequencies, Δ = f I F = | f R F − f L O |. That is to say, there
are two possible cases for relationships between the RF and LO frequencies that
generate the same intermediate frequency.
a. Case 1: f L O > f R F
The graph in Fig. 24.2 shows how an RF signal is shifted down to IF frequency.
Although in this example a range of frequencies is considered, for a band
of RF frequencies it is sufficient to follow only what happens to the two
frequencies that define the band edges, i.e. f min and f max (all frequencies in
between are bounded by these two extremes). Hence, by simple inspection of
the graph, if RF frequency shifts the LO frequency must follow at the same
distance Δ, so that the same IF frequency is generated, i.e.
254
24 RF Receivers: Solutions
Fig. 24.2 Problem 12.3:
relationship among
frequencies of the incoming
RF signal f R F , local oscillator
f L O , and the intermediate
frequency f I F for the case of
fLO > fRF
f L O (min) = f R F (min) + f I F = 500 kHz + 465 kHz = 965 kHz
f L O (max) = f R F (max) + f I F = 1600 kHz + 465 kHz = 2065 kHz
Hence, the local oscillator’s frequency tuneability ratio R f L O and capacitor
tuneabilility ratio RcL O are
R f LO =
f L O (max)
2065 kHz
=
= 2.1399
f L O (min)
965 kHz
∴
RcL O
C L O (max)
=
=
C L O (min)
f L O (max)
f L O (min)
2
= 4.5792
(24.3)
b. Case 2: f L O < f R F
Similarly to the Case 1, we calculate
f L O (min) = f R F (min) − f I F = 500 kHz − 465 kHz = 35 kHz
f L O (max) = f R F (max) − f I F = 1600 kHz − 465 kHz = 1135 kHz
Hence, the local oscillator’s frequency tuneability ratio R f L O and capacitor
tuneabilility ratio RcL O are
R f LO =
f L O (max)
1135 kHz
=
= 32.429
f L O (min)
35 kHz
∴
RcL O =
C L O (max)
=
C L O (min)
f L O (max)
f L O (min)
2
= 1051.6
(24.4)
3. Results (24.3) and (24.4) give us a hint about reasonable choice for the LO frequency. Although both cases are mathematically equally valid, manufacturing
tuneable capacitor whose capacitance tuneability range is about 4.5 times is not
much difficult, however manufacturing a capacitor that is required to change its
capacitance for more then 1,000 times is not easily done. At the same time, design
of a VCO whose tuning range is just over 2 is much easier than design of a VCO
24 RF Receivers: Solutions
255
whose tuning range is more than 32 as required in the second case.
Therefore, choosing Case 1: f L O > f R F , i.e. f L O = 965 − 2065 kHz is more
practical engineering solution.
12.4. Continuing discussion from the previous examples, in this case intermediate
frequency IF is generated as f I F = f L O − f R F = 1 MHz. However, the same IF is
generated by another transmitter whose working frequency is fimage = f L O + f I F =
12 MHz, which must be declared forbidden frequency, i.e. the image frequency to
the already existing transmitter’s frequency.
12.5. An ideal amplifier is intended to implement a simple functions of multiplying amplitude of the input signal and providing the multiplied version at its output
terminal, i.e. ideal amplifying function is
y=Gx
(24.5)
where, y is the output signal amplitude, x is the input signal amplitude, and G is the
multiplication factor. In the ideal case, the multiplication factor can take any value for
instance (G = −≤, ...−1, ...0, ...+1, ...+≤), where absolute value |G| represents
the amplitude multiplication factor while the ± sign indicates the phase of output
signal (i.e. negative gain implies inverted amplifier). In addition, if the multiplication
factor |G| > 1 it is common to use term “gain”, if |G| < 1 it is common to use term
“attenuation”, and if |G| = 1 it is common to use term “buffer” to indicate the
outcome of the amplitude multiplication operation. We dully note that, as defined,
the amplification function is linear.
Realistic amplifiers, however, can not provide an arbitrary gain. Instead, only
within a limited range of the input signal amplitudes an amplifier can be reasonably
well approximated by the linear function. In order to design a system level function,
which assumes linear amplification of the used amplifiers, is important to define
boundaries of the linear operation for the realistic amplifier being used in the system.
If the output signal level, as predicted by (24.5), becomes too high the amplifier
starts to deliver only as high as it realistically can. In other words, realistic output
becomes less then (24.5) calls for. Commonly accepted practice is to assume that
an RF amplifier is linear as long as the difference between its realistic output and
predicted linear output levels is less than 1 dB. The input level for which the difference
becomes equal to 1 dB is appropriately named the 1 dB compression point. Alternate
measure of amplifier nonlinearity is based on existence of higher harmonics, which
always indicates non–linear signal processing. Extrapolated intercept point where
powers of the first and third harmonics in the output signal spectrum become equal
is referred to as the third order intercept point (IIP3).
In this example we practice to read an amplifier specifications from measured
relationship input–output signal levels.
By inspection of the graph in Fig. 12.3 we can conclude the following:
1. Gain: in the linear part of the transfer characteristics for the input of −50 dBm
the output power is −30 dBm, hence the gain is 20 dB. The same gain is expected
256
24 RF Receivers: Solutions
Fig. 24.3 Problem 12.3:
relationship among
frequencies of the incoming
RF signal f R F , local oscillator
f L O , and the intermediate
frequency f I F for the case of
fLO < fRF
for the input level of −20 dB because the predicted linear output level is 0 dB,
i.e. 20 dB above the input level, which is by definition the amplifier gain.
2. 1 dB compression point: the linear part of the characteristics extends to approximately −20 dBm of the input power, when the output power becomes −1 dBm instead of the expected 0dBm. Therefore the 1 dBcompression point is at −20 dBm
of the input power.
3. The third order intercept point IIP3: harmonics power of the third harmonic is
extrapolated until its intersection with the extrapolated linear gain, and the crossing point is found at the output power of approximately +9.6 dBm, which is only
extrapolated point, not the real measurement point. Keep in mind that the amplifier output never reaches that level of output power, it has already saturated close
to the 1 dB compression point level. It should be apparent that 1 dB compression
IIP3 points are closely related and can be used interchangeably to quantify the
amplifier nonlinearity.
12.6. As already implied by the existence of image frequencies, creating RF transmission scheme that ic free of interference among the transmitting stations is rather
complicated. We already learned that the image frequency entering the receiving
mixer is shifted and it also falls into the desired IF band. In addition, in case when
transmitted waveform contains (either intentionally or unintentionally) both the carrier and its second harmonic frequency it opens possibility that the second harmonic
falls over the desired RF frequency band. In this example we explore one such possibility.
Receiver’s RF front end serves as an entrance door that allows only tones whose
frequencies are within the LC resonator’s bandwidth to enter the mixer stage. We
also keep in mind that the local oscillator frequency f L O is not visible to the outside
world, i.e. only frequencies that pass through the LC bandpass filter are multiplied
by the oscillator’s frequency. We also note that, while frequency shifting merely
translates bandwidth along the frequency axis without changing its width (because
the shifting is result of addition/subtraction of two frequencies), plain frequency
multiplication (i.e. multiplication by a constant) affects the bandwidth width as well.
Depending upon relationship between f R F and f T x , (because f R F ≥ = f T x−I ) there
are two possibilities we should consider.
1. Case f T x > f R F : If the local oscillator’s frequency f L O is lower than the interfering Tx signal, then both f T x−I and it’s second harmonic f T x−I I = 2 × f T x−I are
24 RF Receivers: Solutions
257
Fig. 24.4 Problem 2.6: relationship among frequencies of a receiver and the interfering transmitter
for f T x > f R F case
far away from the receiver’s input RF frequency band f R F , Fig. 24.4. Therefore,
both frequencies are rejected by the input LC bandpass filter.
2. Case f T x < f R F : if the interfering signals are to reach the mixer their frequencies must be aligned with the LC bandpass filter’s frequency range f R F =
950 ± 200 kHz. The first harmonic is already prohibited from that range, thus we
take a look what happens if the second harmonic is aligned with the f T x−I I = f R F
band, Fig. 24.5.
In that case, the first harmonic must be further down the frequency axis, i.e.
f T x−I = 1/2 f T x−I = 475 ± 5 kHz. This frequency alignment results in the
transmitter’s second harmonic f T x−I I entering the receiver along with the intended RF signal f R F and reaches the mixer’s input terminals. Output terminals
of the mixer then deliver I F waveform that is modulated by both the intended RF
signal as well as the unintended T x signal, thus the received message is ruined
and can not be recovered. Therefore, in this case, we conclude that transmitting
stations should not be allowed to operate in the f T x = 475 ± 5 kHz range.
12.7. A bundle of frequencies close to each other can be visualized as a multi–wire
cable, where each wire is reserved for one broadcasting channel. In this example,
each channel occupies 10 kHz wide frequency space (similar to wire diameter), thus
within the medium wave AM band it is possible to have the total of n channels, i.e.
n=
(1610 − 540) kHz
= 1070
10 kHz
258
24 RF Receivers: Solutions
Fig. 24.5 Problem 12.6: relationship among frequencies of a receiver and the interfering transmitter
for f T x < f R F case
if the assumptions are that there is no need for “guard bands” (i.e. frequency spacing
in between the neighbouring channels to account for manufacturing tolerances) and
that all channels are available for communication (i.e. for the moment we ignore
the image frequencies inside the AM band). In order to shift all channels down to
455 kHz the local VCO must be able to generate frequencies
f L Omin = f I F + f R Fmin = 455 kHz + 540 kHz = 995 kHz
f L Omax = f I F + f R Fmax = 455 kHz + 1610 kHz = 1965 kHz
∴
f L Omax
= 1.975
f L Omin
which is considered an easy tuning ratio to design. Let us now estimate bandpass
filter’s Q factor that can provide B = 10 kHz for each channel.
f R Fmax
1610 kHz
=
= 161
B
10 kHz
f R Fmin
540 kHz
=
=
= 54
B
10 kHz
Q max =
Q min
Obviously, holding a constant B = 10 kHz over the whole AM band is not trivial
requirement because Q factor is set by properties of RLC components in the RF
resonator (see also the previous examples in this chapter). One way to deal with the
problem would be to provide the fixed 10 kHz bandwidth filtering at the IF stage,
24 RF Receivers: Solutions
259
Fig. 24.6 Problem 12.7: illustration of using IF bandpass filter to set fixed bandwidth for each RF
channel
while the RF stage resonator is designed to provide minimal bandwidth that allows
all individual channels to pass (i.e Q = 54), and then further trim the bandwidth at
the IF frequency (which is constant), Fig. 24.6. If a bandpass filter is designed at IF
frequency then
455 kHz
fI F
=
= 45.5
Q=
B
10 kHz
which is fixed and easily achieved value. This example illustrates one advantage of
using standard IF frequency for a given band, which simplifies the overall receiver
design because it is easier to design tuneable VCO than tuneable fixed bandwidth
bandpass filter.
12.8. When it is not practical to do the frequency downconversion in one step, it is
necessary to use super–heterodyne receiver architecture where two IF frequencies
are used. In this example we practice relationships among two LO frequencies and
the internally generated frequencies.
Following the previously reached conclusion that it is more practical to set the
LO frequency higher than the second frequency at the mixer input terminals, in order
to shift the carrier frequency f R F = 20 MHz down to f I F1 = 10.7 MHz obviously
the first LO frequency must be f L O1 = f R F + f I F1 = 30.7 MHz.This relationship
between f L O1 and f R F must also produce the sum frequency at the I F1 node,
i.e. f R F + f L O1 = 50.7 MHz, Fig. 24.7. It is now more logical and practical to
Fig. 24.7 Solution 12.8: super–heterodyne radio architecture
260
24 RF Receivers: Solutions
choose frequency f L O2 = 11.155 MHz for the second V C O2 , which then produces
f L O2 − 10.7 MHz = 455 kHz, instead of choosing f L O2 = 51.155 MHz to produce
(51.155MHz − 50.7 MHz) = 455 kHz.
Sums and differences between the pair (50.7, 10.7 MHz) and 11.155 MHz lead
into the second IF frequencies taking values of (61.855, 39.545, 21.855 MHz,
455 kHz). We note that in this approach the wanted 455 kHz waveform is widely
separated from the other three higher frequency waveforms, thus it is very easy to
design IF amplifier whose LC bandwidth is centred at 455 kHz and narrow enough
to attenuate the other three waveforms, Fig. 24.7.
We recognize that the other available multiple choices would still result in
mathematically correct solutions, however in the design process it is important to
take into account practicality of the chosen component values, as well as relationship
with the applicable industrial standards.
12.9. In this example we practice to use frequency multiplier, as oppose to frequency
shifter (which requires a mixer). Plain frequency multiplication does not produce the
sums and difference frequencies, and its only input terminal is to accept the waveform
whose frequency needs to by multiplier by a constant. Therefore, it is straightforward
to calculate:
a) The output carrier frequency is f c = 27 × f 0 = 94.5 MHz,
b) The output carrier deviation is Δ f c = 27×Δ f 0 = ±43.2 kHz, which is equivalnet
to finding a difference between
Δf cout = f cmax − f cmin = 94.5432 MHz − 94.4568 MHz = 86.4 kHz
c) Frequency modulation is
mf =
Δf c
43.2 kHz
× 100 % =
× 100 % = 57.6 %
B/2
75 kHz
where we keep in mind that frequency modulation occupies half bandwidth for
the positive deviation and half bandwidth for the negative frequency deviation.
d) By using proportion
Vmax
100
=
57.6
3.6 V pp
∴
Vmax = 6.25 V pp
12.10. Signal processing is, in reality, just implementation of system level mathematical functions. A hypothetical system in Fig. 12.5 takes a range of tones and applies
mathematical operations to produce the output tone whose power is measured.
In this example we practice to recognize difference between frequency shifting
and frequency multiplication, and compare to linear addition of two functions.
1. The input FM signal occupies frequency range of 200 kHz ± 200 Hz, that is from
199.8 to 200.2 kHz range. Therefore,
24 RF Receivers: Solutions
261
64 × 200.2 kHz = 12.8128 MHz
64 × 199.8 kHz = 12.7872 MHz
∴
f 1 = 12.8 MHz ± 12.8 kHz
2. Similarly,
54 × 200 kHz = 10.8 MHz
∴
f 2 = 10.8 MHz ± 10.8 kHz
3. It is important to note that linear addition of two waveforms does not introduce
new tones in the output waveform (frequency shifting is done by multiplication of
two waveforms, which is non–linear operation). Thus, the output of the summing
block contains only two frequency bands, f 1 ±Δ f 1 , and f 2 ±Δ f 2 , which is to say
that LSB is centred around 10.8MHz with the bandwidth of B = Δf 2 = 21.6 kHz.
Thus the LSB bandwidth occupies from 10.7892 to 10.8108 MHz and it requires
Q=
10.8 MHz
f2
=
= 500
B
21.6 kHz
4. Waveform bandwidth B is defined at one half of the maximum power level at
the centre frequency, which is by definition −3.010 dB, and in this example it
is centred around LSB band at f 2 = 10.8 MHz. In other words, if the measured
power at the edge of LSB waveform bandwidth is 1 mW, which is by definition
equivalent to 0 dBm, than the maximum power is by definition P f 0 = 2 mW,
which is equivalent to 3.010 dBm.
The upper side band (USB) in this example occupies frequency space from
f 1 (max) = 12.8128 MHz to f 1 (min) = 12.7872 MHz. In order to find attenuation of these two tones, we apply the attenuation formula for tones that are not
centred at the resonant frequency f 0 and for the given Q, as
f 1 (max)
f2
12.8128 MHz
10.8 MHz
−
=
−
= 0.343
f2
f 1 (max)
10.8 MHz
12.8128 MHz
f 1 (min)
f2
12.7872 MHz
10.8 MHz
y2 =
−
=
−
= 0.339
f2
f 1 (min)
10.8 MHz
12.7872 MHz
y1 =
which we use to calculate the attenuation relative to the maximum level as
262
24 RF Receivers: Solutions
1
= −22.349 dB
171.735
1 + (Q × y1 )2
1
1
=
= −22.297 dB
Ar (12.7872MHz) = 2
169.706
1 + (Q × y2 )
Ar (12.8128MHz) = 1
=
These two equations gave us relative ratio of the maximum power (+3 dBm) and
each of the tones, thus we write directly
Ar (12.8128 MHz) = 3.010 dBm − 22.349 dB = −19.338 dBm
Ar (12.7872 MHz) = 3.010 dBm − 22.297 dB = −19.287 dBm
These results are summarized in Fig. 24.8.
12.11. By inspection of the equivalent block diagram of a three stage RF amplifier,
Fig. 24.9, we write,
1. Power gain of the second stage A2 is found as difference between power levels
at node 2iand node 1i(i.e. P2 − P1 ) and by following the signal power from
the input node, thus we simply write
Fig. 24.8 Solution 12.10: frequency spectrum of output signal (not to scale)
Fig. 24.9 Solution 12.11: simplified block diagram a three stage RF amplifier
24 RF Receivers: Solutions
Pin = −24 dBm and
263
A1 = 10 dB
∴
P1 = −24 dBm + 10 dB = −14 dBm
P2 = 10 mW = 10 dBm
∴
A2 = P2 − P1 = 10 dBm − (−14 dBm) = 24 dB
2. Gain of the last stage in amplifier, is found by looking at difference between the
output power and power level at the node 2ias
2 R = 1 W = 30 dBm ∴
Pout = i out
L
A3 = Pout − P2 = 30 dBm − 10 dBm = 20 dB
which now gives us the total gain Atot of the three stage amplifier as
Atot = A1 + A2 + A3 = 10 dB + 24 dB + 20 dB = 54 dB = 251, 188 V/V
It is now possible to follow the input noise power as
Pn (in) = k T, Beff = k T
π
B = 1.951 × 10−15 W = −117.0976 dBm
2
By knowing the input signal power Pin and the input noise power Pn (in) by
definition we find the input side SNRin as
SNRin =
Pin
= −24 dBm − (−117.0976 dBm) = 93.0976 dB
Pn (in)
(24.6)
while signal to noise ratio at the output of the first stage is given as SNRo =
90.09655 dB. From these two numbers we find noise figure of the first stage by
definition as
NF1 = SNRo − SNRin = 3.010 dB
Now we can summarize noise figure of the three stages and apply Friis’s formula
for multistage amplifiers, as
NF1 = 3.010 dB
NF2 = 6.020 dB
∴
F1 = 2
∴
∴
F1 = 4
F1 = 8
NF3 = 9.030 dB
∴
F2 − 1
F3 − 1
3
7
F = F1 +
+
=2+
+
= 2.303
A1
A1 A2
10 2511
which means that the total noise figure is NF = 10 log 2.303 = 3.622 dB.
Therefore, the total output noise power is found as the sum of amplified input
noise power plus the internally generated noise, i.e.
264
24 RF Receivers: Solutions
Fig. 24.10 Solution 12.11: diagram of relationship among RF amplifier specifications and the total
dynamic range
Pn (out) = Pn (in) + A(tot) + NF(tot)
= −59.4756 dBm ∴ Pn (out) = 1.128 nV
vn (out) = Pn (out) R L ∗ 336 μV
∴
3. We have now all information to establish dynamic range of this amplifier. It may
be easier if the solution is presented graphically, Fig. 24.10. The last piece of
information that is calculated is the total signal to noise ratio at the output node,
SNRtot =
Ps (out)
= 30 dBm − (−54.4756 dBm) = 89.4756 dBm
Pn (out)
where we conclude that this amplifier needs quite a bit of improvement, because
as it stands it can not process any useful signal, its DR = 0 dB.
12.12. Reception of wanted but weak signal from a faraway transmitter in the presence
of unwanted but strong signal from a nearby transmitter, for example in a crowded bus
when two persons standing next to each other use their cell phones at the same time,
may be associated with a couple of quite unexpected side effects. In this example we
explore some of the non–linear effects in an RF receiver.
1. Quantitative measure of an amplifier’s non–linearity is its 1dB compression point,
which can be calculated directly from the non–linear transfer function as
A1 dB =
|a1 |
0.145
=
|a3 |
0.145
2
= 1.042
0.267
∴
A1 dB = 10 log 1.042 ∓
= 0.179 dB
2. Similarly, the third order intercept point IIP3 is also calculated directly from the
non–linear transfer function
24 RF Receivers: Solutions
265
IIP3 =
4 |a1 |
=
3 |a3 |
4 2
= 3.16
3 0.267
∴
IIP3[dB] = 10 log 3.16 ∓
= 5 dB
3. Sensitivity of a receiver is calculated relative to the thermal noise floor. At the
room temperature (290K), for the given data the receiver sensitivity is
S = −174 dBm + NF + 10 log B + SNR
= −174 dBm + 20 dB + 10 log 106 + 0 dB
= −94 dBm
4. Dynamic range DR is calculated as difference between the sensitivity level (i.e.
the noise floor) and either the IIP3[dB] or A1 dB [dB], hence
2
99 dB = 66 dB
3
2
DR = A1 dB − S = 0.179 dB − (−94 dBm) ∓
= 94 dB ∴ DReff = 94 dB ∗ 62.7 dB
3
DR = IIP3 − S = 5 dB − (−94 dBm) = 99 dB
∴
DReff =
5. The output signal level y(x) drops down to zero, and therefore the intended
receiving signal is blocked, when
y(x) = 0
∴
A2 =
∴
3
2
a 1 + a3 A 2 = 0
2
2a1
=
3(−a3 )
2×2
= 2.234 V
3 × 0.267
Appendix A
Physical Constants and Engineering Prefixes
1. Basic physical constants
Physical constant
Symbol
Value
Speed of light in vacuum
Magnetic constant (vacuum permeability)
Electric constant (vacuum permittivity)
Characteristic impedance of vacuum
Coulomb’s constant
Elementary charge
Bohr’s radius
Boltzmann constant
Plank’s constant
Atomic mass unit
Electron mass
Proton mass
Avogadro’s number
c
μ0
λ0 = 1/(μ0 c2 )
Z 0 = μ0 c
ke = 1/4ω λ0
e, q
a0
k
h
u
me
mp
NA
2.99792458 × 108 m/s
4ω × 10−7 N/A2
8.854187817 × 10−12 F/m
376.730313461
8.987551787 × 109 Nm2 /C2
1.602176565 × 10−19 C
5.2917721092 × 10−11 m
1.3806488 × 10−23 J/K
6.62606957 × 1034 Js
1.660538921 × 10−27 kg
9.10938291 × 10−31 kg
1.672621777 × 10−27 kg
6.0221415 × 1023 1/mol
2. Basic engineering prefix system
Tera Giga Mega Kilo Hecto Deca Deci Centi Milli Micro Nano Pico
T
G
1012 109
M
106
k
103
h
102
da
101
d
c
m
µ
10−1 10−2 10−3 10−6
Femto Atto
n
p
f
a
10−9 10−12 10−15 10−18
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014
267
268
Appendix A: Physical Constants and Engineering Prefixes
3. Some useful physical quantities
Name
Symbol
Quantity
Unit
Astronomical unit
The Earth radius
Mass of the Earth
Mass of the Sun
Gravitational acceleration
Gravitational constant
AU
r
mE
mS
g
G
149.5978707 × 109
6.3568 × 106
5.9736 × 1024
1.9891 × 1030
9.806650
6.67384 × 10−11
m
m
kg
kg
ms−2
Nm2 kg−2
4. SI system of fundamental units
Name
Unit
Quantity
Symbol
Metre
Kilogram
Second
Ampere
Kelvin
Candela
Mole
m
kg
s
A
K
cd
mol
Length
Mass
Time
Electric current
Thermodynamic temperature (−273.16√ C)
Luminous intensity
Amount of substance
l
m
t
I
T
Iv
n
5. Derived units
Name
Unit
Quantity
Dimension
Hertz
Rad
Newton
Joule
Watt
Coulomb
Volt
Farad
Ohm
Siemens
Weber
Tesla
Henry
Hz
rad
N
J
W
C
V
F
S
Wb
T
H
Frequency
Angle
Force, weight
Energy, work, heat
Power
Electric charge
Voltage
Electric capacitance
Electric resistance
Electric conductance
Magnetic flux
Magnetic field strength
Inductance
s−1
kg m s−2
N m = C V = W s = kg m2 s−2
V A = J s−1 = kg m2 s−3
As
W A−1 = J C−1 = kg m2 s−3 A−1
C V−1 = kg−1 m−2 s4 A−2
V A−1 = kg m2 s−3 A−2
π −1 = kg−1 m−2 s3 A2
J A−1 = kg m2 s−2 A−1
V s m−2 = Wb m−2 N A−1 m−1 = kg s −2 A−1
V s A−1 = kg m2 s −2 A−2
Appendix B
Maxwell’s Equations
Complete set of Maxwell’s equation is listed here for the reference.
1. Gauss’s law for electric fields
D · ds = q free, enc
S
∇ · D = φ free
integral form
(B.1)
differential form
(B.2)
2. Gauss’s law for magnetic fields
S
B · ds = 0
integral form
(B.3)
∇·B = 0
differential form
(B.4)
3. Faraday’s law
E · dl = −
L
∇×E
d
dt
=−
B · ds
integral form
(B.5)
S
∂B
∂t
differential form
(B.6)
4. Ampere–Maxwell law
H · d l = I free, enc +
L
∇×H
d
dt
= J free +
D · ds integral form
(B.7)
S
∂D
∂t
differential form
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014
(B.8)
269
Appendix C
Second Order Differential Equation
The three basic elements have voltages at their respective terminals as:
v R = i R vL = L
di
dt
vC =
q
C
(C.1)
and, if they are put together in a series circuit that includes a voltage source v(t),
after applying KVL the circuit equation is
v(t) = v L + v R + vC
∴
v(t) = L
q
di
+i R+
dt
C
(C.2)
However, we know that a current is derivative of charge in respect to time, hence we
have the second order differential equation
v(t) = L
d 2q
1
dq
+ q
+R
dt 2
dt
C
∴
v(t) =
d 2q
1
R dq
+
q
+
2
dt
L dt
LC
(C.3)
which is now solved, starting with its auxiliary quadratic equation
0 = x2 +
1
R
x+
L
LC
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014
(C.4)
271
272
Appendix C: Second Order Differential Equation
and its general solution with complex roots
r1,2
⎛
⎞
2
R
1⎝ R
4 ⎠
=
−
− ±
2
L
L
LC
(C.5)
Appendix D
Complex Numbers
A complex number is a neat way of presenting a point in (mathematical) space with
two co-ordinates or, equivalently, it is a neat way to write two equations in the form
of one. A general complex number is Z = a + jb, where a, b are real numbers
referred to as real and imaginary parts, i.e. →(Z ) = a, and ∞(Z ) = b. Here is a
reminder for basic operations with complex numbers, keep in mind that j 2 = −1.
(a + jb) + (c + jd) = (a + c) + j (b + d)
(D.1)
(a + jb) − (c + jd) = (a − c) + j (b − d)
(a + jb) (c + jd) = (ac − bd) + j (bc + ad)
(a + jb) (c − jd)
ac + bd
(a + jb)
bc − ad
=
= 2
+j 2
2
(c + jd)
(c + jd) (c − jd)
c +d
c + d2
≤
(a + jb) = (a − jb)
|(a + jb)| = (a + jb)(a − jb) = (a 2 + b2 )
(D.2)
(D.3)
(D.4)
(D.5)
(D.6)
It is much easier to visualize complex numbers and operations if we use vectors and
trigonometry of right triangle, i.e. Pythagoras’ theorem. Imaginary part is always
taking value at the y–axis and the real part is always on the x–axis.
Therefore, alternative view of complex numbers is based on geometry, i.e.
(a + jb) ≥ (|Z |, ε )
(D.7)
where, of course, the absolute value of Z is just the length of the hypotenuse, and
real and imaginary parts are just the two legs of the right–angled triangle, i.e.
|Z | =
∗
Z
Z≤
= (a 2 + b2 ),
b
ε = arctan
a
(D.8)
where, ε is the phase angle, which, after using Euler’s formula becomes
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014
273
274
Appendix D: Complex Numbers
Fig. D.1 Illustration of complex numbers in [→(Z ), ∞(Z )] space, its equivalence to Pythagoras’
theorem and the vector arithmetics
e j x ≥ cos x + j cos x
(D.9)
enables us to write really compact form of complex numbers
Z = a + jb = |Z | e jε
(D.10)
which leads into another simple way of doing complex arithmetic, simply by using
the absolute values and the arguments in combination with the algebra rules of
exponential numbers, for example
A e jε A
B e jε B = AB e j (ε A +ε B )
(D.11)
and we have the final link,
A e jε ≥ A (cos ε + j sin ε )
(D.12)
where,
→ A e jε = A cos ε
∞ A e jε = A sin ε
(D.13)
Appendix E
Basic Trigonometric Identities
sin(α + ω/2) = + cos α
cos(α + ω/2) = − sin α
(E.1)
(E.2)
sin(α + ω ) = − sin α
cos(α + ω ) = − cos α
(E.3)
(E.4)
sin(α ± ν) = sin α cos ν ± cos α sin ν
cos(α ± ν) = cos α cos ν ∓ sin α sin ν
(E.5)
(E.6)
sin2 α = 1/2 (1 − cos 2α)
(E.7)
cos α = 1/2 (1 + cos 2α)
(E.8)
sin α = 1/4 (3 sin α − sin 3α)
(E.9)
cos α = 1/4 (3 cos α + cos 3α)
(E.10)
2
3
3
sin2 α cos2 α = 1/8 (1 − cos 4α)
sin α cos α =
3
3
1/32 (3 sin 2α
− sin 6α)
(E.11)
(E.12)
cos α cos ν = 1/2 (cos(α − ν) + cos(α + ν))
sin α sin ν = 1/2 (cos(α − ν) − cos(α + ν))
(E.13)
(E.14)
sin α cos ν = 1/2 (sin(α + ν) + sin(α − ν))
cos α sin ν = 1/2 (sin(α + ν) − sin(α − ν))
(E.15)
(E.16)
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014
275
276
Appendix E: Basic Trigonometric Identities
α∓ν
α±ν
cos
2
2
α−ν
α+ν
cos
cos α + cos ν = 2 cos
2
2
α−ν
α+ν
sin
cos α − cos ν = −2 sin
2
2
sin α ± sin ν = 2 sin
(E.17)
(E.18)
(E.19)
Appendix F
Useful Algebra Equations
1. Binomial formula
(x ± y)2 = x 2 ± 2x y + y 2
(F.1)
(F.2)
(x ± y) = x ± 3x y + 3x y ± y
n(n − 1) n−2 2 n(n − 1)(n − 2) n−3 3
x
x
(x ± y)n = x n +nx n−1 +
y +
y · · · + yn
2!
3!
(F.3)
3
3
2
2
3
where, n! = 1 · 2 · 3 · · · n and 0! ≥ 1.
2. Special cases
x 2 − y 2 = (x − y)(x + y)
(F.4)
x 3 − y 3 = (x − y)(x 2 + x y + y 2 )
(F.5)
x + y = (x + y)(x − x y + y )
3
3
2
2
(F.6)
x − y = (x − y )(x + y ) = (x − y)(x + y)(x + y )
4
4
2
2
2
2
2
2
(F.7)
3. Useful Taylor series
f (x) =
⇒
f (n) (a)
(x − a)n (general expression around x = a point)
n!
n=0
⇒
x2
x3
xn
=1+x +
+
+ ···
ex =
n!
2!
3!
(F.8)
n=0
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014
277
278
Appendix F: Useful Algebra Equations
sin x =
cos x =
tan x =
⇒
(−1)n 2n+1
x3
x5
=x−
x
+
− ···
(2n + 1)!
3!
5!
n=0
⇒
n=0
⇒
n=1
x4
(−1)n 2n
x2
x =1−
+
− ···
(2n)!
2!
4!
for all x
for all x
2x 5
B2n (−4)n (1 − 4n ) 2n−1
x3
x
+
+ ···
=x+
(2n)!
3
15
(F.9)
(F.10)
for |x| <
ω
2
(F.11)
Appendix G
Bessel Polynomials
1. Bessel differential equation
x2
d2 y
dy
+x
+ (x 2 − α 2 )y = 0
dx2
dx
(G.1)
2. Relation with trigonometric functions
cos(x sin α) =J0 (x) + 2 [J2 (x) cos 2α + J4 (x) cos 4α + · · · ]
sin(x sin α) =2 [J1 (x) sin α + J3 (x) sin 3α + J5 (x) sin 5α + · · · ]
cos(x cos α) =J0 (x) − 2 [J2 (x) cos 2α − J4 (x) cos 4α
+J6 (x) cos 6 − J8 (x) cos 8α · · · ]
(G.2)
sin(x cos α) =2 [J1 (x) cos α − J3 (x) sin 3α + J5 (x) sin 5α + · · · ]
(G.5)
(G.3)
(G.4)
3. Bessel series
x2
x4
x6
J0 (x) = 1 − 2 + 2 2 − 2 2 2 + · · ·
2
2 ·4
2 ·4 ·6
x4
x2
x
+
+
·
·
·
1− 2
J1 (x) =
2
2 · 2 2 · 24 · 2 · 3
x4
x2
xn
Jn (x) = n
+
+
1− 2
4
2 n!
2 · (n + 1) 2 · 2 · (n + 1) · (n + 2)
(−1) p x 2 p
+ ···
p! 22 p (n + 1)(n + 2) · · · (n + p)
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014
(G.6)
(G.7)
(G.8)
279
280
Appendix G: Bessel Polynomials
4. Bessel approximations
For very large x Bessel function reduces to
Jn (x) =
nω
ω
2
cos x −
−
ωx
2
4
(G.9)
Glossary
The following glossary of technical terms is provided for reference only. The reader
is advised to further study the terms in appropriate books, for example a technical
dictionary.
Absolute zero is the theoretical temperature at which entropy would reach its minimum value. By international agreement, absolute zero is defined as 0 K on the Kelvin
scale and as −273.15 √ C on the Celsius scale.
Active device is an electronic component that has signal gain larger then one, for
example a transistor. Compare to passive device.
Active mode is a condition for a BJT transistor where emitter–base junction is
forward biased, while the collector–base junction is reverse biased.
Admittance reciprocal of impedance, is the measure of how easy AC current flows
in a circuit, measured in Siemens [S].
Ampere [A] is the unit of electric current defined as the flow of one coulomb of
charge per second.
Ampère’s Law states that a current flowing into a wire generates a magnetic flux that
encircles the wire following the “right hand rule”, where the thumb points the direction of the current flow and the other curled fingers show direction of the magnetic
field. Study Maxwell’s equations for more details.
Amplifier is a liner device that implements mathematical equation y = A x, where
y is the amplified output signal, A is the gain coefficient, and x is the input signal.
Analogue is the general class of devices and circuits meant to process a continuous
signal. Compare with digital and sampled signals.
Attenuation is gain lower then one.
Attenuator is a device that reduces a gain, without introducing phase of frequency
distortion.
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014
281
282
Glossary
Automatic Gain Control a closed loop feedback system designed to hold the overall
gain as constant as possible.
Average power is the power averaged over one period.
Bandwidth is the difference between upper and lower frequencies at which the
amplitude response is 3 dB bellow the maximum level. It is equivalent to half–power
bandwidth.
Base is the region of BJT transistor between the emitter and collector.
Bel [B] is a dimensionless unit used to express ratio of two powers. More practical
unit is decibel [dB].
Beta β is the current gain of a BJT transistor. It is the ratio of the change of collector
current relative to change of the base current, ν = d IC /d I B .
Bias is a steady current or voltage used to set operating conditions of the device.
Breakdown voltage is the voltage at which the reverse current of a reverse biased
pn junction suddenly rises. If the current is not limited the device is destroyed.
Capacitance is the ratio between the electric charge two conductors and voltage
between the two.
Capacitor is a device made of two conductors separated by an insulating material
for purposes of storing electric charge, i.e, energy.
Celsius [◦ C] is a unit increment of temperature unit defined as 1/100 between the
freezing (0 √ C) and boiling (100 √ C) point of water. Compare to Kelvin and Fahrenheit.
Characteristic Curve a family of I–V plots shown for several parameter values.
Characteristic impedance is the entry point impedance of an infinitely long transmission line.
Charge a basic property of elementary particles of matter (electrons, protons, etc.)
responsible for creating a force field.
Circuit the interconnection of devices, both passive and active, for the purpose of
synthesizing a mathematical function.
Common mode average value of a sinusoidal waveform.
Coulomb [C] the unit of electric charge defined as the charge transported through
a unity area in one second by an electric current of one ampere. An electron has a
charge of 1.602 × 10−19 C.
Coulomb’s Law defines force between two electric charges in space.
Common–base configuration a single BJT amplifier configuration where the base
potential is fixed, emitter serves as the input and collector as the output terminal.
Glossary
283
Also known as current buffer. Equivalent to common–gate configuration for MOS
amplifiers.
Common–collector combination a single BJT amplifier transistor configuration
where the collector potential is fixed, the base serves as the input and emitter as
the output terminal. Also known as voltage buffer or voltage follower. Equivalent to
common–drain configuration for MOS amplifiers.
Common–emitter a single BJT amplifier transistor configuration where the emitter
potential is fixed, while the base serves as the input and collector as the output
terminal. Also known as gm stage. Equivalent to common–source configuration for
MOS amplifiers.
Conductivity represents the ability of a matter to conduct electricity.
Conductor a material that easily conducts electricity.
Current transfer of electrical charge through a unity size area per unit of time.
Current gain the ratio of current at output and current at the input terminals of a
device or circuit.
Current source a device capable of provided constant current value regardless of
the voltage at its terminals.
DC direct current, i.e. current that flows in one direction only.
DC biasing a process of setting stable operating point of a device.
DC load line a straight line across a family of I–V curves that shows movement of
the operating point as the output voltage changes for a given load.
DC analysis mathematical procedure to calculate stable operating point.
Decibel [dB] is a dimensionless unit used to express ratio of two powers. Ten times
smaller then bel [B].
Device a single discrete device, for instance resistor, transistor, capacitor, etc.
Dielectric a material that is not good in conducting electricity, i.e. opposite of a
conductor. Characterized by dielectric constant.
Differential amplifier an amplifier that operates on differential signals.
Differential signal a difference between two sinusoidal signals of same frequency,
same amplitude, same common mode, and with phase difference of 180√ C.
Digital is the general class of devices and circuits meant to process a sampled signal.
Compare with analogue and continuous signals.
Diode a nonlinear two terminal device which obeys exponential transfer function.
Used as unidirectional switch.
284
Glossary
Discrete device an individual electrical component that exhibits behaviour associated with resistor, transistor, capacitor, inductor, etc. Compare with distributed
components.
Dynamic range the difference of the maximum acceptable signal level and the
minimum acceptable signal level.
Electric field a field generated by an electric charge, detected by existence of the
electric force within space surrounding the charge.
Electrical noise any unwanted electrical signal.
Electromagnetic (EM) wave a phenomena exhibited by flow of electromagnetic
energy through the space. In special case of standing wave this definition may need
more explanation.
Electron a fundamental particle that carries negative charge.
Electronics is the branch of science and technology which makes use of the controlled
motion of electrons through different media and vacuum.
Electrostatics is the branch of science that deals with the phenomena arising from
stationary or slow–moving electric charges.
Emitter a region of a BJT transistor from which charges are injected into the base.
One of the three terminal points at BJT device.
Energy a concept that can be loosely defined as the ability of a body to perform
work.
Equivalent circuit a simplified version of the original circuit that still performs the
same function.
Equivalent noise temperature the absolute temperature at which a perfect resistor
would generate the same noise as its equivalent real component at the room temperature.
Fall time the time during a pulse decreases from 90 to 10 % of its maximum value
(sometimes defined between the 80 and 20 % points).
Farad [F] the unit of capacitance of a capacitor. One Farad is very large, capacitance
of the Earth’s ionosphere with respect to the ground is around 50 mF.
Faraday’s Law the law of electromagnetic induction. See also Faraday’s cage.
Faraday cage an enclosure that blocks out external static electric fields.
Feedback the process of coupling output and input terminals through an external
path. Negative feedback increases stability of an amplifier for the price of reduced
gain, positive feedback boosts gain and is needed for creating oscillating circuits.
Field a concept that describes a flow of energy through the space.
Glossary
285
Field–Effect Transistor (FET) a transistor controlled by two perpendicular electrical
fields used to change resistivity of semiconductor material underneath the gate terminal and force current between the source and drain terminals.
Flicker noise or 1/f noise. A random noise in semiconductors whose power spectral
density is, to the first approximation, inverse to frequency.
Frequency the number of complete cycles per second.
Frequency response a curves showing gain and phase change of a device as a
function of frequency.
Gain the ratio of signal values measured at output and input.
Gauss’s Law is a law relating the distribution of electric charge to the resulting
electric field.
Ground an arbitrary potential reference point that all other potentials in a circuit are
compared against. Difference between the ground potential and the node potential
is expressed as voltage at that node. The ground node may or may not be the lowest
potential in the circuit.
Henry [H] the unit for self and mutual inductance.
Hertz [Hz] the unit for frequency, equal to one cycle per second.
Impedance resistance of a two terminal device at any frequency.
Inductance is the property in an electrical circuit where a change in the electric
current through that circuit induces an electromotive force (EMF) that opposes the
change in current.
Inductor is a passive electrical component that can store energy in a magnetic field
created by the electric current passing through it.
Input current, voltage, power, or other driving force applied to a circuit or device.
Insertion loss the attenuation resulting from inserting a circuit between source and
load.
Insulator a material with very low conductivity.
Intermediate Frequency (IF) is a frequency to which a carrier frequency is shifted
as an intermediate step in transmission or reception.
Intermodulation Products additional harmonics created due to non–linear device
processing two or more single tone signals.
Junction a joining of two semiconductor materials.
Junction capacitance capacitance associated with pn junction region.
Kelvin [K] the unit increment of temperature on the absolute temperature scale.
Kirchhoff’s Current Law (KCL) is the law of conservation of charge.
286
Glossary
Kirchhoff’s Voltage Law (KVL) is based on the conservation of “energy given/taken
by potential field” (not including energy taken by dissipation).
Large signal a signal with large enough amplitude to move the operating point of
a device far away from its original biasing point. Hence, non–linear model of the
device must be used.
Law of conservation of energy is the fundamental law of nature. It states that energy
can neither be created nor destroyed, it can only be transformed from one state to
another.
Large signal analysis a method used to describe behaviour of devices stimulated
by large signals. Hence, the nonlinear devices in terms of the underlying nonlinear
equations.
Linear network A network in which the parameters of resistance, inductance, and
capacitance are constant with respect to current or voltage, and in which the voltage
or current of sources is independent of or directly proportional to other voltages and
currents, or their derivatives, in the network.
Load a device that absorbs energy and converts it into another form.
Local Oscillator (LO) an oscillator used to generate single tone signal that is needed
for upconversion and downconversion operations.
Lossless a theoretical device that does not dissipate energy.
Low Noise Amplifier (LNA) is an electronic amplifier used to amplify very weak
signals captured by an antenna.
Lumped element a self contained and localized element that offers one particular
property, for instance, resistance over a range of frequencies.
Magnetic field a field generated by magnetic energy, detected by existence of the
magnetic force within space surrounding the magnet.
Matching circuit a passive circuit designed to interface two networks for purpose
of enabling maximum energy transfer between the two networks.
Matching a concept of connecting two networks for purpose of enabling maximum
energy transfer between them.
Maxwell’s equations are a set of four partial differential equations that relate the
electric and magnetic fields to their sources, charge density and current density.
These equations can be combined to show that light is an electromagnetic wave.
Individually, the equations are known as Gauss’s law, Gauss’s law for magnetism,
Faraday’s law of induction, and Ampère’s law with Maxwell’s correction. These
four equations, together with the Lorentz force law are the complete set of laws of
classical electromagnetism.
Metal Oxide Semiconductor Field Effect Transistor (MOSFET) Originally, a
sandwich of aluminum–Silicone Dioxide–Silicon was used to manufacture FET transistors. Although, a metal is not used anymore for creating gates for FET transistors,
the name has stuck.
Glossary
287
Microwaves waves in the frequency range of 1–300 GHz, i.e. with a wavelength of
three hundreds to one millimetre.
Mixer a nonlinear three port device used for frequency shifting operation.
Negative resistance a resistance of a device or circuit where an increase in the
current entering a port results in a decreased voltage across the same port.
Noise any unwanted signal that interferes with the wanted signal.
Noise Figure (NF) is a measure of degradation of the signal–to–noise ratio (SNR),
caused by components in a radio frequency (RF) signal chain.
Nonlinear a system which does not satisfy the superposition principle, or whose
output is not directly proportional to its input.
Norton’s Theorem is the dual of Thévenin’s theorem, states that any collection
of voltage sources, current sources, and resistors with two terminals is electrically
equivalent to an ideal current source in parallel with a single resistor.
NPN transistor a transistor with p–type base and n–type collector and emitter.
Octave the interval between any two frequencies having a ration of 2:1
Ohm [] unit of resistance, as defined by Ohm’a law.
Ohm’s Law states that the change of current through a conductor between two points
is directly proportional to the change of voltage across the two points, and inversely
proportional to the resistance between them.
One–dB gain compression point the point at which the power gain at the output
of a nonlinear device or circuits is reduced by 1 dB relative to its small signal linear
model predicted value.
Open loop gain the ratio of the output signal and the input signals of an amplifier
with now feedback path present.
Oscillator an electronic device the generates a single tone (or some other regular
shape) signal at predetermined frequency.
Output current, voltage, power, or driving force delivered at the output terminals.
Pasive a component that does not have gain larger then one.
Phase the angular property of a wave.
Phase shifter a two port network which provides a controllable phase shift of the
RF signals.
Phasor a mathematical representation of a sine wave by a rotating vector.
Power the rate at which work is performed.
Quality factor (Q factor) is a dimensionless parameter that characterizes a resonator’s bandwidth relative to its centre frequency.
288
Glossary
Radio frequency (RF) any frequency at which coherent electromagnetic radiation
of energy is possible.
Reactance is the opposition of a circuit element to a change of current, caused by
the build–up of electric or magnetic fields in the element.
Reactive element an inductor and capacitor.
Reflected waves the waves reflected from a discontinuity in the medium they are
traveling in.
Resistance of an object is a measure of its opposition to the passage of a steady
electric current.
Resistor a lumped element designed to have a certain resistance.
Resonant frequency the frequency at which a given system or circuit responds with
maximum amplitude when driven by an external single tone.
Root Mean Square (RMS) is the square root of the arithmetic mean (average) of
the squares of the original values.
Saturation a circuits condition whereby an increase of the input signal does not
produce expected change at the output.
Self–resonant frequency the frequency at which all real devices or circuits start to
oscillate due to the internal parasitic inductances and capacitances.
Signal an electrical quantity containing an information that is carried by voltage or
current.
Single ended circuit a circuit operating on single ended signals, as oppose to differential signals.
Skin effect is the tendency of an alternating electric current (AC) to distribute itself
within a conductor so that the current density near the surface of the conductor is
greater than that at its core. That is, the electric current tends to flow at the “skin” of
the conductor, at an average depth called the skin depth.
Small signal a low amplitude signal the occupies very narrow region that is centred
at the biasing point. Hence, linear model alway applies.
Small signal amplifier an amplifier that operates only in the linear region.
Space is the boundless, three–dimensional extent in which objects and events occur
and have relative position and direction.
Stability ability of a circuit to stay away from the self–resonating frequency.
Standing wave is a wave that remains in a constant position. It can arise in a stationary
medium as a result of interference between two waves traveling in opposite directions.
For waves of equal amplitude traveling in opposing directions, there is on average
no net propagation of energy.
Glossary
289
Standing Wave Ratio (SWR) the ratio of the maximum to the minimum value of
current or voltage in a standing wave.
Thévenin’s theorem is the dual of Norton’s theorem, states that any combination
of voltage sources, current sources and resistors with two terminals is electrically
equivalent to a single voltage source and a single series resistor.
Third order intercept point (IP3) is a measure for weakly nonlinear systems and
devices, for example receivers, linear amplifiers and mixers.
Time a concept used to order the sequence of events.
Transmission line any system of conductors capable of efficiently conducting electromagnetic energy.
Tuned circuit a circuit consisting of inductance and capacitance that can be adjusted
for resonance at the desired frequency.
Tuning process of adjusting resonant frequency of the tuned circuit.
Varactor a two terminal pn junction used as a voltage controlled capacitor.
Volt [V] a unit for potential difference.
Voltage controlled oscillator (VCO) an oscillator whose output frequency is controlled by a voltage.
Voltage divider is a simple linear circuit that produces an output voltage that is a
fraction of its input voltage.
Voltage follower amplifier also known as voltage buffer amplifier provides electrical
impedance transformation from one circuit to another.
Voltage source a device capable of provided constant voltage value regardless of the
current at its terminals.
Wave a disturbance that progresses from point in space to another.
Wavefront a surface having constant phase.
Wavelength space distance between two consecutive points having the same phase.
Wave propagation journey of the wave through space.
White noise random signal that consists of all possible frequencies from zero to
infinity.
Work the advancement in space of a point under application of force.
Bibliographies
1. S.W. Amos, Principles of Transistor Circuits. Number 0–408-04851-4 (Butterworths, 1990).
2. [BG03a] L. Besser, R. Gilmore, Practical RF Circuit Design for Modern Wireless Systems I.
Number 1–58053-521-6 (Artech House, 2003).
3. [BG03b] L. Besser, R. Gilmore, Practical RF Circuit Design for Modern Wireless Systems II.
Number 1–58053-522-4 (Artech House, 2003).
4. [BMV05] J.S. Beasley, G.M. Miller, J.K. Vasek, Modern Electronic Communication. Number
0–13-113037-4 (Pearson, Prentice Hall, 2005).
5. J.J. Brophy, Basic Electronics for Scientist. Number 0–07-008147-6 (McGraw-Hill Inc., 1990).
6. P. Bubb, Understanding Radio Waves. Number 0–7188-2581-0 (Lutterworth Press, 1984).
7. [CC03a] D. Comer, D. Comer, Advanced Electronic Circuit Design. Number 0–471-22828-1
(John Wiley & Sons Inc., 2003).
8. [CC03b] D. Comer, D. Comer, Fundamentals of Electronic Circuit Design. Number 0–47141016-0 (John Wiley & Sons Inc., 2003).
9. [CL62] D.R. Corson, P. Lorrain, Introduction to Electromagnetic Fields and Waves. Number
62–14193 (Freeman Co., 1962).
10. [DA01] W.A. Davis, K.K. Agarwal, Radio Frequency Circuit Design. Number 0–471-35052-4
(Wiley Interscience, 2001).
11. [DA07] W.A. Davis, K.K. Agarwal, Analysis of Bipolar and CMOS Amplifiers. Number 1–
4200-4644-6 (CRC Press, 2007).
12. R.S. Elliott, Electromagnetics. Number 66–14804 (McGraw Hill, 1966).
13. D. Fleisch, A Student’s Guide to Maxwell’s Equations. Number 978–0-521-87761-9 (Cambridge University Press, 2008).
14. [FLS05] R.P. Feynman, R.B. Leighton, M. Sands, The Feynman Lectures on Physics. Number
0–8053-9047-2 (Pearson Addison Wesley, 2005).
15. [GM93] P.G. Gray, Robert G. Meyer, Analysis and Design of Analog Integrated Circtuits.
Number 0–471-57495-3 (John Wiley & Sons Inc., 1993).
16. S. Goldman, Frequency Analysis, Modulation and Noise. Number TK6553.G58 1948
(McGraw-Hill, 1948).
17. B. Green, The Fabric of Cosmos. Number 0–375-72720-5 (Vintage Books, 2004).
18. J. Gribbin, In Search of Schrödinger’s Cat, Quantum Physics and Reality. Number 0–55334253-3 (Bantam Books, 1984).
19. [HH89a] T.C. Hayes, P. Horowitz, Student Manual for The Art of Electronics. Number 0–52137709-9 (Cambridge University Press, 1989).
20. [HH89b] P. Horowitz, W. Hill, The Art of Electronics. Number 0–521-37095-7 (Cambridge
University Press, 1989).
R. Sobot, Wireless Communication Electronics by Example,
DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014
291
292
Bibliographies
21. P.G. Huray, Maxwell’s Equations. Number 978–0-470-54276-7 (Wiley, 2010).
22. [II99] U.S. Inan, A.S. Inan, Electromagnetic Waves. Number 0–201-36179-5 (Prentice Hall,
1999).
23. [JK93] W.H. Hayt Jr., J.E. Kemmerly, Engineering Circuit Analysis. Number 0–07-027410-X
(McGraw Hill, 1993).
24. [JN71] R.H.Good Jr., T.H. Nelson, Classical Theory of Electric and Magnetic Fields. Number
78–137-628 (Academic Press, 1971).
25. [Jr.89] W.H. Hayt Jr, Engineering Electromagnetics. Number 0–07-024706-1 (McGraw Hill,
1989).
26. [KB80] H.L. Krauss, C.W. Bostian, Solid State Radio Engineering. Number 0–471-03018-X.
(Wiley, 1980).
27. G.C. King, Vibrations and Waves. Number 978–0-470-01189-8 (Wiley, 2009).
28. J.A. Kong, Theory of Electromagnetic Waves. Number 0–471-50190-5 (Wiley, 1975).
29. [LB00] R. Ludwig, P. Bretchko, RF Circuit Design, Theory and Applications. Number 0–13095323-7 (Prentice Hall, 2000).
30. T.H. Lee, The Design of CMOS Radio-Frequency Integrated Circtuis. Number 0–521-63922-0
(Cambridge University Press, 2005).
31. W.F. Lovering, Radio Communication. Number TK6550.L546 1966 (Longmans, 1966).
32. [PP99] Z. Popovic, D. Popovic, Electromagnetic Waves. Number 0–201-36179-5 (Prentice
Hall, 1999).
33. E.M. Purcell, Electricity and Magnetism. Number 0–07-004908-4 (McGraw Hill, 1985).
34. M.M. Radmanesh, Radio Frequency and Microwave Electronics. Number 0–13-027958-7
(Prentice Hall, 2001).
35. B. Razavi, RF Microelectronics. Number 0–13-887571-5 (Prentice Hall, 1998).
36. [RC84] D. Roddy, J. Coolen, Electronic Communications. Number 0–8359-1598-0 (Reston
Publishing Company, 1984).
37. [RR67] J.H. Reyner, P.J. Reyner, Radio Communication (Sir Isaac Pitman & Son Ltd, 1967).
38. D.B. Rutledge, The Electronics of Radio. Number 0–521-64136-5 (Cambridge University
Press, 1999).
39. [SB00] B. Streetman, Sanjay Banerjee, Solid State Electronic Devices. Number 0–13-025538-6
(Prentice Hall, 2000).
40. R.J. Schoenbeck, Electronic Communications Modulation and Transmission. Number 0–67521311-8 (Prentice Hill, 1992).
41. M.G. Scroggie. Foundations of Wireless and Electronics, 10 edn. Number 0–408-01202-1
(Newnes Technical Books, 1984).
42. S. Seely, Radio Electronics. Number 55–5696 (McGraw Hill, 1956).
43. R.E. Simpson, Introductory Electronics for Scientist and Engineers. Number 0–205-08377-3
(Allyn and Bacon Inc., 1987).
44. R. Sobot, Wireless Communication Electronics, Introduction to RF Circuits and Design Techniques. Number 978–1-4614-1117-8 (Springer, 2012).
45. S.M. Sze, Physics of Semiconductor Devices. Number 0–471-05661-8 (John Wiley and Sons,
1981).
46. D. Terrell, Electronics for Computer Technology. Number 0–7668-3872-2 (Thompson Delmar,
Learning, 2003).
47. M.T. Thompson, Intuitive Analog Circuit Design. Number 0–7506-7786-4 (Newnes, 2006).
48. Wikipedia.org. Electromagnetic wave equation. http://en.wikipedia.org/wiki/electromagnetic_
wave_equation, September 2010
49. Wikipedia.org. Waves, wavelength. http://en.wikipedia.org/wiki/wave, July 2010
50. D. H. Wolaver. Phase-Locked Loop Circuit Design. Number 0–13-662743-9. Prentice Hall,
1991.
51. P.H. Young, Electronic Communication Techniques. Number 0–13-048285-4 (Pearson, Prentice Hill, 2004).
Download
Related flashcards
Create Flashcards