Mr. W.D Govender
© 2020 University of South Africa
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University of South Africa
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EHF3701/1/2020
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CONTENTS
Page
PREFACEv
LEARNING UNIT 1 INTRODUCTION1
1.1
1.2
1.3
1.4
Learning outcomes
Introduction to high frequency electronics
Electromagnetic theory
Conclusion
1
1
3
13
LEARNING UNIT 2: HF ELECTRONIC TRANSMISSION LINES AND WAVEGUIDES
14
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
14
14
15
18
20
23
26
26
27
28
28
29
30
Learning outcomes
Introduction
General solutions for TEM, TE and TM waves
Parallel plate wave guide
Rectangular waveguides
Circular waveguide
Coaxial line
Surface waves on a grounded dielectric sheet
Stripline
Microstrip line
The Transverse Resonance Technique
Wave velocities and dispersion
Conclusion
LEARNING UNIT 3 HF NETWORK ANALYSIS
31
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
3.11
31
31
32
35
37
39
40
41
42
43
44
Learning outcomes
Introduction
Impedance and equivalent voltages and currents
Impedance and admittance matrices
The scattering matrix
The Transmission (ABCD) Matrix
Signal flow graphs
Discontuinities and modal analysis
Excitation of waveguides – electric and magnetic currents
Excitation of waveguides – aperture coupling
Conclusion
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LEARNING UNIT 4 HF IMPEDANCE MATCHING AND TUNING
45
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
45
45
46
48
50
52
53
54
55
57
59
60
Learning outcomes
Introduction
Matching with lumped elements (l networks)
Single-stub tuning
Double-stub tuning
The quarter-wave transformer
The theory of small reflections
Binomial multisection matching transformers
Chebyshev multisection matching transformers
Tapered lines
The bode-fano criterion
Conclusion
LEARNING UNIT 5 HF RESONATORS, FILTERS, OSCILLATORS AND MIXERS
5.1
5.2
5.3
5.4
5.5
5.6
Learning outcomes
Introduction
Resonators
Filters
Oscillators and mixers
Conclusion
61
61
61
62
73
78
84
LEARNING UNIT 6 DESIGN OF HF POWER AMPLIFIERS
85
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
85
85
86
88
89
94
95
98
Learning outcomes
Introduction
Two-port power gains
Stability
Single-stage transistor amplifier
Broadband transistor Power amplifiers
Conclusion
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iv
PREFACE
1.
GETTING STARTED
Welcome to High Frequency Electronics (EHF 3701), a module offered by
Unisa’s Department of Electrical and Mining Engineering.
Your study material for this module consists of:
•
•
•
•
•
•
This study guide
Your prescribed textbook
Tutorial Letter 101
Any other tutorial letters you may receive during the year
Any additional information provided on the module website on myUnisa
Any additional electronic communications you may receive, such as
announcements from your lecturer
I will give you the details of your prescribed book later in this preface, and
also in Tutorial Letter 101.
Tutorial Letter 101 will either be part of your study pack or will be posted to
you. However, you can also access it on myUnisa. You can do this by clicking
on Official Study Material in the menu on the left.
Tutorial Letter 101 is just one of the tutorial letters you will be receiving during
the year. Please read it carefully. You will also receive Tutorial Letter 201 during the course of the semester shortly after the assignment due dates. Tutorial
Letter 201 is closely linked to Tutorial Letter 101, and will provide you with
suggested solutions to the assignments.
In this preface, I will give you an overview of and some general information
about this module. I will also tell you more about how to go about studying
this module, how to use myUnisa, and about the assessment for the module.
Your study guide, textbook and tutorial letters contain everything you need
in order to complete this module. However, you may benefit from also using
the module website on myUnisa. Using the site will enable you to:
• submit assignments online (I would really encourage you to submit your
assignment online, as this will ensure that you receive rapid feedback
and comments),
• access your official study material,
• make use of the Unisa library functions,
• communicate online with your lecturer and fellow students and participate
in online discussion forums, and
• access a variety of learning resources.
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Check the site regularly for updates, posted announcements and additional
resources uploaded throughout the semester.
2.
CONTACT DETAILS AND STUDENT SUPPORT SERVICES
My contact details as well as those of the academic department are provided
in Tutorial Letter 101. Should you need to contact the university about matters not related to the content of this module, consult the publication Study @
Unisa, which you received with your study material. This brochure contains
information on how to contact the university (e.g. whom to contact, depending on your query; important telephone and fax numbers and addresses; and
details of the opening and closing times of particular facilities).
You can also make use of the following contact routes:
Unisa website
http://www.unisa.ac.za or http://mobi.
unisa.ac.za
E-mail (general enquiries)
info@unisa.ac.za
Queries related to application and
registration
study-info@unisa.ac.za
Assignment enquiries
assign@unisa.ac.za
Examination enquiries
exams@unisa.ac.za
Study material enquiries
despatch@unisa.ac.za
Student account enquiries
finan@unisa.ac.za
Assistance with myUnisa
myUnisaHelp@unisa.ac.za
Assistance with myLife e-mail
accounts
myLifeHelp@unisa.ac.za
SMS (South Africa only)
32695 (You will receive an auto response
SMS with the various SMS options)
General fax number
012 429 4150
For information about the various student support systems and services available
at Unisa (e.g. student counselling, tutorial classes, language support), again,
consult Study @ Unisa. Note the following support systems that you could use:
• Fellow students
It is always a good idea to have contact with fellow students. You can do this
by means of the Discussions menu option on myUnisa. You can also use the
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Pr e f a ce
Discussions option to find out whether there are students in your area who
would like to form study groups.
• Library
Study @ Unisa lists all the services offered by the Unisa library.
To log in to the library website, you will be asked to provide your login details,
that is, your student number and your myUnisa password, in order to access
the library’s online resources and services. This will enable you to:
• request library material
• view and renew your library material
• use the library’s e-resources
• Unisa Directorate for Counselling and Career Development (DCCD)
DCCD supports prospective and registered students before, during and after
their Unisa studies. There are resources on their website (http://www.unisa.
ac.za/sites/corporate/default/About/Service-departments/Counselling-andCareer-Development), and also printed booklets available to assist you with:
•
•
•
•
•
career advice and how to develop your employability skills
study skills
academic literacy (reading, writing and numeracy skills)
assignment submission
exam preparation
The DCCD can also assist you with regard to your personal wellness: see their website at http://www.unisa.ac.za/sites/corporate/default/
About /Service-departments/Counselling-and-Career-Development /
Personal-development.
• Student health and wellness
Your physical health is an important factor in your learning success. Obtaining
an educational qualification is challenging and may at times involve stress, and
it is therefore vital that you should try to maintain a healthy lifestyle to ensure
that you will cope physically with the demands of your studies.
If you suspect that you may suffer from a chronic condition, or if you know
that you suffer from such a condition but are unsure about medical options
and treatment, you could approach Unisa for further information and support.
See Unisa's Student Health and Wellness website, which you can access from
Unisa's main website: click on About, Service Departments, Student Affairs
and then on Student Health and Wellness. Here you will find details of Unisa's
health and wellness clinics, and also some health and wellness resources.
If you have a health-related condition such as HIV/AIDS, or have a close family member with this or another health condition, then you need to take this
into account in planning your studies. Try not to leave study tasks to the last
minute, as this creates enormous stress, which has a negative effect on both
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EH F3701/1
PR EFACE
your academic performance and your health. Planning your studies is essential
so that you work consistently and make progress.
It would be wise to know your health status (HIV/AIDS, blood pressure,
diabetes, cholesterol, etc.). If you undergo medical tests and obtain reliable
information about your health, with the necessary medical and supportive
interventions you can prolong and improve the quality of your life and your
success in your studies.
If you would like to obtain basic information about the prevention of, testing
for and treatment of HIV/AIDS, you are welcome to follow the web links listed
below:
• http://www.aids.org/topics/aids-factsheets/
• https://www.westerncape.gov.za/documents/public_info/L (click on Living
with HIV/AIDS)
You could also approach the DCCD about counselling in this regard.
• The Advocacy and Resource Centre for Students with Disabilities (ARCSWiD)
You will find more information about this Centre on their web page at
http://www.unisa.ac.za/sites/corporate/default/About/Service-departments/
Student-Affairs/ARCSWiD.
You can also contact the Centre at 012 441 5470/1.
3.
PURPOSE AND OUTCOMES OF THIS MODULE
The purpose of this module is to equip you with knowledge and skills relating
to high frequency RF (radio frequency) and microwave systems consisting of
passive and active microwave components that are applied in modern microwave communication systems and microwave radar systems. On completing
the module, you should have the ability to design high frequency RF and microwave circuits and systems using amplifiers, oscillators, mixers, resonators
and filters. You will also have an understanding of modern computer aided
design (CAD) and electromagnetic (EM) field solvers.
More specifically, the outcomes of this module are that, after completing the
module, you should be able to:
• Apply fundamental underlying models of electromagnetic fields to solve
problems related to high/microwave frequency electronics.
• Apply various wave solutions in HF transmission lines and waveguides.
• Analyse HF electronic networks, and apply basic circuit and network
concepts to solve related design problems of practical interest.
• Design impedance matching and tuning of HF electronic networks.
• Design HF electronic systems such as resonators, filters, oscillators and
mixers.
• Apply modern computer aided design (CAD) and electromagnetic (EM) field
solvers to design HF power amplifiers with maximum gains or for specified
gains and with low noise figures.
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Pr e f a ce
The next section will give you a better idea about how the content of the
module is structured and how the various ideas expressed in the learning
outcomes are related.
4.
HOW THE CONTENT OF THIS MODULE IS ORGANISED
The learning material consists of six learning units, as follows:
Learning unit 1: Introduction. This learning unit is a general introduction
that provides an overview of the fundamental underlying principles of high
frequency electronics operating up to microwave frequencies.
Learning unit 2: HF electronic transmission lines and wave guides. This
learning unit examines the properties of several types of transmission lines
and waveguides that are in common use.
Learning unit 3: HF network analysis. Microwave network analysis is explored
to show how basic circuit and network concepts can be extended to solve
many HF or microwave analysis and design problems.
Learning unit 4: HF impedance matching and tuning. This learning unit explains
impedance matching or tuning, which is done to ensure that maximum power
is delivered when the load is matched to the line, and that power loss in the
feed line is minimised; to improve the signal-to-noise ratio of the system; and
for a number of other reasons.
Learning unit 5: HF resonators, filters, oscillators and mixers. In this learning
unit you will learn how to design high frequency RF and microwave circuits
and systems using amplifiers, oscillators, mixers, resonators and filters.
Learning unit 6: Design of HF power amplifiers. In this unit you will apply
modern CAD and EM field solvers to design two-port power gains that are
useful for HF power amplifier designs and designs for maximum gains or
specified gains, and with low noise figures.
Now that you have a better idea of how the module is structured, let’s look
at what your studies will involve.
5.
LEARNING RESOURCES
Your main learning resources for this module will be your prescribed textbook
and this study guide. These resources will be supported by tutorial letters.
The prescribed textbook to be used in conjunction with the online material is:
David M Pozar, Microwave Engineering, 4th edition, 2012, John Wiley and
Sons, ISBN 978-0-470-63155-3.
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An additional textbook for reference is:
Jeremy Everard, Fundamentals of RF Circuit Design with Low Noise Oscillators,
2001, John Wiley and Sons, ISBN 0 471 49793 2.
The prescribed textbook is a comprehensive guide to the subject field. You will
not be required to study the whole textbook, as I will guide you with regard
to what is needed as you work through these learning units. You will need to
study the chapters that are mentioned at the beginning of each learning unit,
and any recommended reading sections.
In the text of the study guide, the textbook will be referred to as the prescribed
textbook.
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x
1
Learning unit 1
INTRODUCTION
1.1
LEARNING OUTCOMES
By the end of this learning unit, you should be able to apply fundamental underlying
models of electromagnetic fields to solve problems related to high/microwave frequency
electronics.
In particular, you should be able to do the following:
•
•
•
•
•
•
Explain Maxwell’s equations.
Describe fields in media and boundary conditions.
Derive the wave equation and solve basic problems relating to plane waves.
Solve general problems relating to plane waves.
Give equations relating to energy and power.
Solve problems relating to plane wave reflection from a medium.
1.2
INTRODUCTION TO HIGH FREQUENCY ELECTRONICS
Reading: To complete this unit, study the chapter “Electromagnetic Theory” in your
prescribed textbook.
This module deals in general terms with the field of radio frequency (RF) and microwave
engineering, in which the electronic circuits and devices operate alternating current
signals at frequencies in the range of 100 MHz to 1000 GHz. Figure 1.1 shows the
electromagnetic spectrum. Radio frequencies range from very high frequency (VHF) (30–
300 MHz) to ultra-high frequency (UHF) (300–3000 MHz), while the term “microwave”
is typically used for frequencies between 3 and 300 GHz.
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https://upload.wikimedia.org/wikipedia/commons/3/30/EM_spectrumrevised.png
Figure 1.1
The electromagnetic spectrum
We learn electromagnetic fields theory because standard circuit theory offers only
approximations to solve short wavelength or high frequency/microwave network
problems. High frequency electronic problems involve distributed circuit elements or
components (not lumped elements) describable by Maxwell’s equations, since the phase
of the voltage or current changes significantly over the physical extent of the device or
component dimensions. As you can see in Figure 1.1, when frequencies are very low,
the wavelength becomes large as a result of a non-significant phase variation across the
dimensions of the electronic component. The extremely high frequencies are
characterised by a much shorter wavelength. These regions represent optical frequencies,
and in these regions Maxwell’s equations are transformed into geometric optical theories
that are applied to solve optical system problems.
Let’s have a look at some practical applications of high frequency electronics:
•
The gain of an antenna depends on the electrical size of the antenna, with more
antenna gains realised at higher frequencies due to shorter wavelengths.
At higher frequencies, more bandwidth and thus high data rate is achievable.
Waves travelling with high frequencies are promoted by the line of sight. As a result,
these waves are robust with regard to ionospheric distortions.
The effective reflection area of a radar target is usually proportional to the electrical
size of the target.
Various molecular, atomic and nuclear resonances occur at high/microwave
frequencies, creating a variety of unique applications in the areas of basic science,
remote sensing, medical diagnostics and treatment, and heating methods.
•
•
•
•
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Please refer to the section entitled “Introduction to microwave engineering” in the
prescribed textbook for further information about modern applications of HF electronic
engineering and a brief history of microwave engineering. In the next section, you will
be introduced to principles of electromagnetic fields theory that underpin the study of
high frequency electronics.
1.3
ELECTROMAGNETIC THEORY
Please refer to the section entitled “Electromagnetic theory” in the prescribed textbook
for a detailed discussion. This study guide will merely direct you to aspects of the relevant
principles that you need to study in depth.
1.3.1
Maxwell’s equations
Maxwell’s equations are essential to an introduction to electromagnetic theory from the
inductive, or axiomatic, perspective. The time-varying form of Maxwell’s equations can
be written in point, or differential, form as
−∂Β
−Μ,
∂t
∂D
∇ ×=
Η
+J ,
∂t
∇ ⋅ D =ρ ,
∇ ⋅ Β =0 .
∇=
×ε
(1.1)
(1.2)
(1.3)
(1.4)
In this module we follow the MKS system of units. The script quantities represent timevarying vector fields and are real functions of spatial coordinates x, y, z and the time
variable t . The quantities are defined as follows:
ε is the electric field, in volts per meter (V/m).
Η is the magnetic field, in amperes per meter (A/m).
D is the electric flux density, in coulombs per meter squared (Coul/m2).
Β is the magnetic flux density, in webers per meter squared (Wb/m2).
Μ is the (fictitious) magnetic current density, in volts per meter (V/m2).
J is the electric current density, in amperes per meter squared (A/m2).
ρ is the electric charge density, in coulombs per meter cubed (Coul/m3).
The sources of the electromagnetic fields are the currents Μ and J as well as the electric
� being a fictitious magnetic electric current source. We note
charge density ρ , with M
that the flow of electric charge produces an electric current, and so the electric charge
density ρ is the ultimate source of the electromagnetic field. In free-space wave
propagation, the electric and magnetic field intensities and flux densities are related as
follows:
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�
𝐵𝐵� = 𝜇𝜇0 𝐻𝐻
� = 𝜖𝜖0 𝜀𝜀̅
𝐷𝐷
(1.5)
(1.6)
where 𝜇𝜇0 = 4𝜋𝜋 × 10−7 henry/m is the permeability of free-space, and 𝜖𝜖0 = 8.854 ×10−12
farad/m is the permittivity of free-space.
If you have internet access, please watch the following video clip for another explanation
of Maxwell's equations:
Physics – E&M: Maxwell's equations (Differential form) by Michel van Biezen at
https://www.youtube.com/watch?v=v3rBaIK4HQU.
For detailed information about the integral form of Maxwell’s equations, see equations
(1.4) to (1.13) in the prescribed textbook.
ACTIVITY 1.1 REVISE MAXWELL’S EQUATIONS
Give Maxwell's equations and explain each of the quantities in equations (1.1) to (1.4).
1.3.2
Fields in media and boundary conditions
Refer to the section “Fields in media and boundary conditions” in the prescribed textbook
or corresponding sections in any of the recommended reference materials.
We have looked at the electric and magnetic fields existing in free-space, with no material
bodies present. We will now look at some practical situations where material bodies are
often present. When electromagnetic fields exist in material media, the field vectors are
related to each other by the constitutive relations. For a dielectric material, an applied
electric field 𝐸𝐸� causes the polarisation of the atoms or molecules of the material to create
� . This additional
electric dipole moments that augment the total displacement flux, 𝐷𝐷
��, the electric polarisation, where
polarisation vector is called ��
𝑃𝑃𝑃𝑃
� =𝜖𝜖0 𝐸𝐸� + 𝑃𝑃� 𝑒𝑒
𝐷𝐷
(1.7)
𝑃𝑃�𝑒𝑒 = 𝜖𝜖0 𝑋𝑋𝑒𝑒 𝐸𝐸�
(1.8)
� = 𝜖𝜖0 𝐸𝐸� + 𝑃𝑃�e = (1 + 𝑋𝑋𝑒𝑒 )𝐸𝐸� = 𝜖𝜖𝐸𝐸�
𝐷𝐷
(1.9)
Here, Pe , the electric polarisation, is linearly related to the applied electric field in a linear
medium as
where X e , is called an electric susceptibility.
where 𝜖𝜖 = 𝜖𝜖 ′ − 𝑗𝑗𝑗𝑗 ′′ = 𝜖𝜖0 (1 + 𝑋𝑋𝑒𝑒 ).
Equation (1.9) represents the complex permittivity of the medium. The imaginary part of
𝜖𝜖 accounts for loss in the medium (heat) due to the damping of vibrating dipole moments.
Due to the energy conservation, the imaginary part of 𝜖𝜖 must be negative (ϵ’’ positive).
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1.3.2.1 Fields at a general material interface
Consider a plane interface between two media, as shown in Figure 1.5 in your prescribed
textbook. Maxwell’s equations in integral form can be used to deduce conditions
involving the normal and tangential fields at this interface.
Please work through equations (1.31), (1.32), (1.36) and (1.37) in the prescribed textbook
for the most general expressions for the boundary conditions at an arbitrary interface of
materials and/or surface currents.
1.3.2.2 Fields at a dielectric interface
At an interface between two lossless dielectric materials, no charge or surface current
densities will ordinarily exist, such that the following equations hold:
nˆ.D1 = nˆ.D2
(1.10a)
nˆ.B1 = nˆ.B2
(1.10b)
nˆ × E1 =nˆ × E2
(1.10c)
nˆ × H1 =×
nˆ H 2
(1.10d)
In other words, these equations state that the normal components of D and B are
continuous across the interface, and the tangential components of E and H are
continuous across the interface. Thus, if (1.10c) and (1.10d) are enforced for the four
tangential field components, the equations will automatically satisfy for the continuity of
the normal components.
1.3.2.3 Fields at the interface with a perfect conductor (electric wall)
High frequency electronics involve boundaries with good conductors (e.g., metals),
which can often be assumed to be lossless (σ → ∞ ) . In the case of a perfect conductor,
all field components must be zero inside the conducting region. This result can be seen
by considering a conductor with finite conductivity (σ → ∞ ) and noting that the skin
depth (the depth to which most of the high frequency signals penetrate) goes to zero as
(σ → ∞ ) . If we also assume here that M s = 0, which would be the case if the perfect
conductor filled all the space on one side of the boundary, then from (1.10a to 1.10d)
we have the following:
nˆ ⋅ D =
ρs ,
(1.11a)
nˆ ⋅ B =
0,
(1.11b)
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EHF3701/1
nˆ × E =
0,
(1.11c)
nˆ × H =
Js,
(1.11d)
where ρ s and J s are the electric surface charge density and current density respectively
on the interface, and n̂ is the normal unit vector pointing out of the perfect conductor.
This type of boundary is also known as an electric wall because the tangential
components of E are “shorted out,” as seen from (1.11c), and must vanish at the surface
of the conductor.
If you have internet access, please watch the following video clip for an explanation of
electromagnetic fields in media and boundary conditions.
https://www.youtube.com/watch?v=ylDqGlFmXb8
ACTIVITY 1.2 REVISE FIELDS IN MEDIA AND BOUNDARY CONDITIONS
1. Briefly explain the equations used for fields at the different types of interfaces.
2. Describe field equations that hold for an interface between two lossless dielectric
materials.
1.3.3
The wave equation and basic plane wave solutions
Refer to the section “The wave equation and basic plane wave solutions” in the prescribed
textbook or corresponding sections in any of the recommended study materials.
1.3.3.1 The Helmholtz equation
In a source-free, linear, isotropic, homogeneous region, Maxwell’s curl equations in
phasor form are
∇ × E = − jωµ H ,
(1.12a)
∇ × H = jωε E ,
(1.12b)
and they constitute two equations for the two unknowns, E and H . As such, they can
be solved for either E or H . Taking the curl of (1.12a) and using (1.12b) gives
∇ × ∇ × E = − jωµ∇ × H = ω 2 µε E ,
(1.13)
which is an equation for E . This result can be simplified through the use of vector
identity, ∇ × ∇ × A = ∇ ( ∇ ⋅ A ) − ∇ 2 A, which is valid for the rectangular components of an
arbitrary vector A . Then,
∇ 2 E + ω 2 µε E =
0,
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(1.14)
because ∇ ⋅ E = 0 in a source-free region. Equation (1.14) is the wave equation, or
Helmholtz equation, for E . An identical equation for H can be derived in the same
manner:
∇ 2 H + ω 2 µε H =
0,
(1.15)
If you have any difficulties with activities, remember that you are welcome to discuss
them with fellow students in the Discussions space for the module on myUnisa, and
that you are also welcome to direct queries to the lecturer via the Discussions option.
As a way of introducing wave behaviour, we will next study the solutions to the above
wave equations in their simplest forms, first for a lossless medium and then for a lossy
(conducting) medium.
1.3.3.2 Plane waves in a lossless medium
In a lossless medium, ε and μ are real numbers, and so k is real. A basic plane wave
solution to the above wave equations can be found by considering an electric field with
only an x̂ component and that is uniform (no variation) in the x and y directions. Then,
∂/∂x = ∂/∂y = 0,
and the Helmholtz equation of (1.14) reduces to:
∂ 2 Ex
+ k 2 Ex =
0.
2
∂z
(1.16)
Here is an activity that will help you better understand the modelling of wave equations
and the solution of basic plane wave problems.
ACTIVITY 1.3 BASIC PLANE WAVE PARAMETERS
A plane wave propagating in a lossless dielectric medium has an electric field given as
=
Ξ x E0 cos (ωt − β z ) with a frequency of 5.0 GHz and a wavelength in the material of
3.0 cm. Determine the propagation constant, the phase velocity, the relative permittivity
of the medium, and the wave impedance.
Do these calculations on your own first before checking your answer against the solution
below.
SOLUTION
The propagation constant is =
k
2π
=
λ
2π
= 209.4m −1 and the phase velocity is
0.03
ω 2π f
λf =
vp = =
=
( 0.03) ( 5 ×109 ) =1.5 ×108 m / sec .
k
k
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The relative permittivity of the medium can be calculated as follows:
2
2
 c   3.0 × 108 
=
= 4.0
ε r =
 v   1.5 ×108 
 p
The wave impedance is calculated as follows:
=
η η0 / =
εr
377
= 188.5 Ω
4.0
ACTIVITY 1.4 SKIN DEPTH AT HIGH FREQUENCIES
Compute the skin depth of aluminium, copper, gold, and silver at a frequency of 10 GHz.
Do these calculations on your own first before checking your answer against the solution
below.
SOLUTION
Given the conductivities for these metals in the relevant appendix of the prescribed
textbook, we can calculate the skin depth as follows:
2
=
=
δs
ωµσ
1
1
−7
π (10 )( 4π ×10 ) σ
10
1
= 5.03 ×10−3
For aluminium:
For copper:
For gold:
For silver:
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σ
.
1
8.14 ×10−7 m
=
7
3.816 ×10
5.03 ×10−3
δs =
1
6.60 ×10−7 m
=
7
5.813 ×10
5.03 ×10−3
δs =
1
7.86 ×10−7 m
=
7
4.098 ×10
5.03 ×10−3
δs =
1
6.40 ×10−7 m
=
6.173 ×107
5.03 ×10−3
δs =
These results show that most of the current flow in a good conductor occurs in an
extremely thin region near the surface of the conductor.
1.3.4
General plane wave solutions
In this section, we will look at plane waves from a more general point of view and solve
the wave equation by the method of separation of variables. Let us consider a free-space,
and describe the Helmholtz equation for E as follows:
∇ 2 E + k02 E =
∂2E ∂2E ∂2E
+
+
+ k02 E = 0
∂x 2 ∂y 2 ∂z 2
(1.17)
This vector wave equation holds for each rectangular component of E :
∂ 2 Ei ∂ 2 Ei ∂ 2 Ei
+ 2 + 2 + k02 Ei =
0
∂x 2
∂y
∂z
(1.18)
Where the index i = x, y, or z . This equation can be solved by the method of separation
of variables, a standard technique for treating such partial differential equations. Please
see the steps in the solution from equation (1.64) to equation (1.73) in the prescribed
textbook. You will notice that the x, y, and z dependences of the three components of E
in equations (1.71) to (1.73) must be the same because of the divergence condition:
∇.E=
∂Ex ∂E y ∂Ez
+
+
= 0
∂x
∂y
∂z
(1.19)
ACTIVITY 1.5 CURRENT SHEETS AS SOURCES OF PLANE WAVES
An infinite sheet of surface current can be considered as a source for plane waves. If an
electric surface current density J s = J 0 xˆ exists on the z = 0 plane in free-space, find the
resulting fields by assuming plane waves on either side of the current sheet and enforcing
boundary conditions.
SOLUTION
Attempt this activity on your own first, following the discussions in section of 1.5 of the
prescribed textbook, without referring to the full solution. You can then check your
answer against the solution provided in worked example 1.3 in the prescribed textbook.
1.3.5
Energy and power
Refer to the section “Energy and power” in the prescribed textbook or corresponding
sections in any of the recommended study materials.
Any electromagnetic energy source either sets up fields that are stored as electric and
magnetic energy or carries power that is transmitted or dissipated as loss. The time-varying
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stored electric energy in a volume V is given as follows for a sinusoidal steady-state
scenario:
We
=
1
Re ∫ E ⋅ D∗dv
V
4
(1.20)
which, in the case of simple lossless isotropic, homogeneous, linear media, where ε is a
real scalar constant, reduces to
We
=
ε
4∫
V
E ⋅ E ∗dv
(1.21)
Similarly, the time-average magnetic energy stored in the volume V is
Wm
=
1
Re ∫ H ⋅ B ∗dv
V
4
(1.22)
which becomes
Wm
=
µ
4∫
V
H ⋅ H ∗dv
(1.23)
for a real, constant, scalar µ .
1.3.6
Plane wave reflection from a media interface
Refer to the section “Plane wave reflection from a media interface” in the prescribed
textbook or corresponding sections in any of the recommended study materials.
It is interesting to learn the behaviour of electromagnetic fields at the interface of various
types of media, including lossless media, lossy media, a good conductor, or a perfect
conductor. But first we need to study the reflection of a plane wave normally incident
from free-space onto a half-space of an arbitrary material. The geometry is shown in Figure
1.12 in the prescribed textbook, where the material half-space z > 0 is characterised by
the parameters ε , μ, and σ.
1.3.6.1 General medium
If we assume that the incident plane wave has an electric field vector that is oriented
along the x-axis and is propagating along the positive z-axis, then the incident fields can
be written, for z < 0, as
ˆ 0e − jk0 z
Ei = xE
H i = yˆ
10
1
η0
E0e − jk0 z
EHF3701/1
(1.24a)
(1.24b)
where η0 is the impedance of free-space and E0 is an arbitrary amplitude. Also in the
region z < 0, a reflected wave may exist with the form
Er = xˆΓE0e + jk0 z
H r = − yˆ
Γ
η0
(1.25a)
E0e + jk0 z
(1.25b)
where Γ is the unknown reflection coefficient of the reflected electric field. Note that in
(1.25), the sign in the exponential terms has been chosen as positive, to represent waves
traveling in the − ẑ direction of propagation.
If you have internet access, please watch the following video clip for an illustration of
plane wave reflection from a media interface:
https://www.youtube.com/watch?v=_GuLwk-k4KQ
Now, remind yourself of what you know about the theory of plane wave reflection from
a conductor by solving the following problem. Please do the calculations on your own
before looking at the solution given after the problem.
ACTIVITY 1.6 PLANE WAVE REFLECTION FROM A CONDUCTOR
Consider a plane wave normally incident on a half-space of copper. If f = 1 GHz,
compute the propagation constant, intrinsic impedance, and skin depth for the conductor.
Also compute the reflection and transmission coefficients.
SOLUTION
For copper,
=
σ 5.813 ×107 S / m, so the skin depth
2
= 2.088 ×10−6 m
δs
=
ωµσ
and the propagation constant is
1+ j
γ= =
( 4.789 + j 4.789 ) ×105 m−1
δs
The intrinsic impedance is
η=
1+ j
σδ s
=
(8.239 + j8.239 ) ×10−3 Ω
η0 377 Ω ) . The reflection
which is quite small relative to the impedance of free-space (=
coefficient then becomes
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Γ=
η − η0
= 1.0∠179.990
η + η0
The transmission coefficient is
T=
2η
= 6.181×10−5 ∠450
η + η0
ACTIVITY 1.7 OBLIQUE REFLECTION FROM A DIELECTRIC INTERFACE
Plot the reflection coefficients versus incidence angle for parallel and perpendicular
polarised plane waves incident from free-space onto a dielectric region with ε r = 2.55 .
Do these calculations on your own before looking at the solution below.
SOLUTION
The impedances for the two regions are
=
η1 377 Ω,
η2
=
η0
=
εr
377
= 236 Ω .
2.55
We then consider equations for parallel polarisation, that is, equations (1.132a) to (1.138)
and equations for perpendicular polarisation, that is, equations (1.139a) to (1.143b) from
η cos θt − η1 cos θi
the prescribed textbook. In particular, we evaluate Γ = 2
from equation
η2 cos θt + η1 cos θi
2η2 cos θi
from equation (1.143a) versus the incidence angle;
(1.137a) and T =
η2 cos θt + η1 cos θi
we then plot the results as shown in Figure 1.14 in your prescribed textbook.
ACTIVITY 1.8 ADDITIONAL PRACTICE AND GENERAL REVIEW QUESTIONS
1. For additional practice, do the reflective questions in the following online resources:
https://www.sanfoundry.com/electromagnetic-theory-questions-answers-boundaryconditions/
https://shankarkncet.blogspot.com/2018/03/ec-6403-electromagnetic-fieldsquestion_28.html
https://www.utdallas.edu/~dlm/ee4301/old%20exam%20questions.htm
https://pdfs.semanticscholar.org/8bd0/55db2b397061cc59f73c7497590b4374ff24.pdf
(problems at the end)
2. Refer to problems 1.1 to 1.18 in the prescribed textbook and practise the concepts
covered in this unit.
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3. How would you rate your own competence in the mathematical skills required to
achieve the outcomes of this learning unit? Do you feel confident that you have the
necessary knowledge and skills to proceed? If not, be sure to revise the required skills
or to ask for assistance if necessary.
1.4
CONCLUSION
In this learning unit we revised certain fundamental concepts and principles that you need
to know for a more detailed understanding of high frequency electronics. Now that you
have mastered these concepts, we can continue our exploration of high frequency
electronics in the next learning unit.
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Learning unit 1
2
HF ELECTRONIC TRANSMISSION LINES AND
WAVEGUIDES
2.1
LEARNING OUTCOMES
By the end of this learning unit, you should be able to apply various wave solutions to
analyse HF transmission lines and waveguides.
In particular, you should be able to do the following:
•
•
Derive general solutions for TEM, TE and TM waves.
Apply these general solutions for TEM, TE and TM waves to analyse the:
o parallel plate waveguide.
o rectangular waveguide.
o circular waveguide.
Differentiate between coaxial lines, surface waves on a grounded dielectric sheet,
striplines and microstrip lines.
List the steps in the transverse resonance technique.
Apply wave velocities and dispersion.
•
•
•
2.2
INTRODUCTION
Reading: To complete this unit, study the chapter entitled “Transmission Lines and
Waveguides” in your prescribed textbook, or information on this topic in any of the
recommended study materials.
This unit assumes that you already know that a transmission line is characterised by a
propagation constant, an attenuation constant, and characteristic impedance. In this unit
we will use these quantities to analyse the various HF lines and waveguides from
electromagnetic field theory.
If you have internet access, please view the following online video for background and
further explanation of this topic:
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Waveguides – Weekly Whiteboard by Duotech Services Inc at:
https://www.youtube.com/watch?v=YQ_zKHNYn8Q
2.3
GENERAL SOLUTIONS FOR TEM, TE AND TM WAVES
Please refer to the section “General solutions for TEM, TE and TM waves” in the
prescribed textbook for a detailed discussion. This module guide will merely direct you
to aspects of the relevant concepts that you need to study in detail.
This section will introduce you to the general solutions to Maxwell’s equations for the
specific cases of TEM, TE, and TM wave propagation in cylindrical transmission lines or
waveguides. The geometry of an arbitrary transmission line or waveguide with assumed
infinite length and conductor boundaries that are parallel to the z-axis is shown in Figure
3.1 in the prescribed textbook.
For a time-harmonic field with an e jωt dependence and wave propagation along the z axis, the electric and magnetic fields can be written as follows:
ˆ z ( x, y )  e − jβ z
E (=
x, y, z )  e ( x, y ) + ze
(2.1)
ˆ z ( x, y )  e − jβ z
H (=
x, y, z )  h ( x, y ) + zh
(2.2)
where e ( x, y ) and h ( x, y ) represent the transverse
( xˆ, yˆ ) electric
and magnetic field
components, and ez and hz are the longitudinal electric and magnetic field components.
When a transmission line or a waveguide is considered to have no electromagnetic wave
generator, Maxwell’s equations for the above electromagnetic fields will reduce to the
following field component expressions (refer to the section “General solutions for TEM,
TE and TM waves” for detailed steps):
=
Hx
∂H z 
j  ∂Ez
−β
ωε

2 
∂y
∂x 
kc 
(2.3)
=
Hy
−j
∂E
∂H z 
ωε z + β

2 
∂x
∂y 
kc 
(2.4)
=
Ex
− j  ∂Ez
∂H z 
+ ωµ
β

2 
∂y 
kc  ∂x
(2.5)
∂E
∂H z 
j 
E y = 2  − β z + ωµ

∂y
∂x 
kc 
(2.6)
where
2
k=
k2 − β 2
c
(2.7)
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is defined as the cut-off wave number. Equations (2.3) to (2.6) above are the general results
that can be applied to a variety of wave guiding systems provided later.
If you have internet access, please watch the following online video presentation by Prof
Niraj Kumar for a further explanation of how to derive the general solutions for TEM, TE
and TM waves. The video is available at the following link:
https://www.youtube.com/watch?v=45ywQyXNlak
After revising the section “General solutions for TEM, TE and TM waves” in the prescribed
textbook and watching the video presentation recommended above, answer the
following review question.
ACTIVITY 2.1 REVIEW QUESTION
Derive the general solutions for TEM, TE and TM waves.
2.3.1
TEM waves
Let’s begin by analysing transverse electromagnetic (TEM) waveguides. You will notice
that TEM waves are characterised by E
=
H=
0 ; this implies that the transverse fields
z
z
are also all zero. By working through the section “General solutions for TEM, TE and TM
waves” in the prescribed textbook in detail, you will realise that TEM waves can exist
when two or more conductors are present. For the TEM line analysis, follow the
procedure below:
1. Solve Laplace’s equation for Φ ( x, y ) , from ∇t2Φ ( x, y ) =0 . The solution will yield
numerous unknown constants.
2. Apply the boundary conditions for the known voltages on the conductors in order to
find values for these unknown constants.
3. As illustrated in the relevant equations in the prescribed textbook, compute the
transverse electric fields E and electric field component e . Also compute the
transverse magnetic fields H and magnetic field component h .
4. Compute the voltage build-up between two conductors V and the current flow on a
given conductor I using the appropriate equations given in the prescribed textbook.
5. Compute the propagation constant and the characteristic impedance given by
Z0 = V / I .
ACTIVITY 2.2 REVIEW QUESTION
Demonstrate the step-wise analysis of TEM waves propagating through a wave guiding
system, and at each step provide justification as to why that step is necessary. Post your
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answer in the relevant Discussion space of the module website, and respond to other
students' posts where appropriate.
2.3.2
TE waves
Refer to the section “General solutions for TEM, TE and TM waves” in the prescribed
textbook for detailed reference to the concepts summarised here.
Transverse electric (TE) waves are characterised by Ez = 0 and H z ≠ 0 ; thus the general
solutions with regard to the TE waves will reduce to
Hx =
− j β ∂H z
kc2 ∂x
(2.8)
Hy =
− j β ∂H z
kc2 ∂y
(2.9)
Ex =
− jωµ ∂H z
kc2 ∂y
(2.10)
Ey =
jωµ ∂H z
kc2 ∂x
(2.11)
β
where kc ≠ 0 and the propagation constant=
k 2 − kc2 is a function of frequency and
the geometry of the line or waveguide.
2.3.3
TM waves
Refer to the section “General solutions for TEM, TE and TM waves” in the prescribed
textbook for detailed coverage of this concept.
In a nutshell, transverse magnetic (TM) waves are characterised by Ez ≠ 0 and H z = 0
such that the general solutions will reduce to
Hx =
jωε ∂Ez
kc2 ∂y
(2.12)
Hy =
− jωε ∂Ez
kc2 ∂x
(2.13)
Ex =
− j β ∂Ez
kc2 ∂x
(2.14)
Ey =
− j β ∂Ez
kc2 ∂y
(2.15)
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Based on the above solutions, a summarised procedure for analysing TE and TM
waveguides consists of the following steps:
1. Solve the reduced Helmholtz equation illustrated in this section for either magnetic
hz or electric ez field components. The solution will produce several unknown
constants.
2. Apply relevant equations in this section to find the transverse fields from hz or ez .
3. Use the boundary conditions to approximate field components such that the unknown
constants in step (1) as well as kc can be found.
4. The propagation constant and the wave impedance are given by their respective
relevant expressions in this section (refer to the section “General solutions for TEM,
TE and TM waves” for detailed relevant expressions).
After that you will complete a review question, but first learn a little more about TE and
TM waves by watching the following online video lecture by Scott Tyo at the link:
https://www.youtube.com/watch?v=LpAc2vs0zr8
ACTIVITY 2.3 REVIEW QUESTION
Demonstrate the step-wise analysis of TE and TM waves propagating through a wave
guiding system, and explain why each step is necessary. Post your answer in the relevant
Discussion space of the module website, and respond to other students' posts where
appropriate.
2.3.4
Attenuation due to dielectric loss
When we consider transmission lines or waveguides, we encounter an attenuation of
propagating waves due to dielectric and conductor losses. The total attenuation constant
is determined as α= α c + α d , where c and d are the attenuation in a transmission line or
a waveguide due to the conductor and dielectric loss, respectively.
2.4
PARALLEL PLATE WAVE GUIDE
Refer to the section “Parallel plate waveguide” in the prescribed textbook for further
details.
It is important to note that because parallel plate waveguides have two conducting plate
or strips properties, they are capable of supporting TEM, TE and TM modes, as shown in
Figure 3.2 in the prescribed textbook. This structure may also be used to model the
propagation of higher order modes in striplines.
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The solutions for TEM, TM and TE waves are given below. (Refer to the section “Parallel
plate waveguide” in the prescribed textbook for a detailed treatment of this subject.)
2.4.1
TEM modes
The total electric field is described by the following equation:
E ( x, y, z ) = e ( x, y ) e − jkz = − yˆ
V0 − jkz
e
d
(2.16)
where k = ω µε is the propagation constant of the TEM wave and the magnetic field is
described by the following expression:
1
V
H ( x, y , z ) =
h ( x, y ) e − jkz =
zˆ × E ( x, y, z ) =
− xˆ 0 e − jkz
η
ηd
where η =
2.4.2
(2.17)
µ
is the intrinsic impedance of the medium between the parallel plates.
ε
TM modes
For TM waves, the H z = 0 and Ez fields are nonzero; hence it follows that the transverse
field components are given as:
Hx =
jωε
nπ y − jβ z
An cos
e
kc
d
(2.18)
Ey =
nπ y − jβ z
− jβ
An cos
e
kc
d
(2.19)
E
=
H=
0
x
y
(2.20)
where the cut-off frequency of the TMn mode is provided as follows:
fc
=
kc
n
=
2π µε 2d µε
2.4.3
(2.21)
TE modes
TE modes are characterised by Ez = 0 and a nonzero H z . The transverse fields will then
be computed as follows:
Ex =
jωµ
nπ y − jβ z
Bn sin
e
kc
d
(2.22)
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EHF3701/1
Hy =
jβ
nπ y − jβ z
Bn sin
e
kc
d
E=
H=
0
y
x
(2.23)
(2.24)
where the propagation constant of the TEn mode is given as
β
=
2
n
 nπ 
k2 −
 and f c =
2d µε
 d 
(2.25)
If you have internet access, please watch the following video clips for further information
about how to apply the general wave solutions to analyse the parallel plate waveguides:
Electromagnetics – parallel plane wave guide by Tutorials Point (India) at:
https://www.youtube.com/watch?v=t7yMMqaqP1E
Parallel plate waveguide TE and TM modes by Pat Kohl at:
https://www.youtube.com/watch?v=Loyeuji3Q0A ,
and https://www.youtube.com/watch?v=_3qV1-3nHD8
After revising the above notes and watching the video presentation clips, you should be
able to do the following activity.
ACTIVITY 2.4 REVIEW QUESTION
Demonstrate how the general solutions are used to analyse the TEM, TE and TM wave
modes.
2.5
RECTANGULAR WAVEGUIDES
Rectangular waveguides have been in existence for a number of years, and are still used
for many modern applications to transport high frequency waves. There are a variety of
high frequency electronic components, such as couplers, detectors, isolators, attenuators
and slotted lines. These components take the shape of rectangular waveguides, which
can allow signals with frequency bands from 1 to 220 GHz to propagate through them.
Hollow rectangular waveguides are able to propagate TM and TE modes but not TEM
waves, since a waveguide of this type consists of a single conductor.
Refer to the section “Rectangular waveguides” in the prescribed textbook or information
about this topic in any of the recommended study materials for in-depth coverage of this
concept. All step-wise derivations have been set out clearly in the prescribed textbook.
Here we provide you with final results only.
2.5.1
TE modes
The transverse field components of the TEmn mode are expressed as:
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EHF3701/1
Ex =
jωµ nπ
mπ x
nπ y − jβ z
Amn cos
e
sin
2
kc b
a
b
(2.26)
Ey =
− jωµ nπ
mπ x
nπ y − jβ z
Amn sin
e
cos
2
kc a
a
b
(2.27)
Hx =
j β mπ
mπ x
nπ y − jβ z
Amn sin
e
cos
2
kc a
a
b
(2.28)
Hy =
j β nπ
mπ x
nπ y − jβ z
Amn cos
e
sin
2
kc b
a
b
(2.29)
2.5.2
TM modes
TM modes are characterised by fields with H z = 0 and a nonzero Ez . The transverse field
components for the TMmn mode can be computed as follows:
Ex =
− j β mπ
mπ x
nπ y − jβ z
Bmn cos
e
sin
2
akc
a
b
(2.30)
Ey =
mπ x
nπ y − jβ z
− j β nπ
Bmn sin
e
cos
2
bkc
a
b
(2.31)
Hx =
jωε nπ
mπ x
nπ y − jβ z
Bmn sin
e
cos
2
bkc
a
b
(2.32)
Hy =
mπ x
nπ y − jβ z
− jωε mπ
Bmn cos
e
sin
2
akc
a
b
(2.33)
If you have internet access, please watch the following video clips for further explanation
as to how to apply the general wave solutions to analyse rectangular waveguides:
Rectangular waveguides by Pat Kohl at:
https://www.youtube.com/watch?v=IqbUCObqGxI
Transverse magnetic mode rectangular wave guide by Tutorials Point (India) at:
https://www.youtube.com/watch?v=rA_ikvtFLls
Transverse electric mode rectangular wave guide by Tutorials Point (India) at:
https://www.youtube.com/watch?v=4ooo3UqRpyM
You can now analyse the characteristics of a rectangular waveguide using the general TE
and TM wave mode solutions discussed above by doing the following worked examples:
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WORKED EXAMPLE 2.1 CHARACTERISTICS OF A RECTANGULAR WAVEGUIDE
A radio frequency industry in Alberton, South Africa is looking for a high frequency
electronic engineer to design a rectangular waveguide. Imagine that you are asked to find
the cut-off frequencies of the first five propagating modes for a waveguide whose length
is a Teflon-filled, copper K-band having dimensions a = 1.07 cm and b = 0.43 cm. If the
operating frequency is 15 GHz, find the attenuation due to dielectric and conductor
losses.
Attempt these calculations on your own before looking at the solution below.
SOLUTION
For Teflon, ε r = 2.08 and tan δ = 0.0004 . The cut-off frequencies are calculated as follows:
fcmn
=
c
2π ε r
2
 mπ   nπ 

 +

 a   b 
2
Computing fc for the first few values of m and n gives the results:
Thus the TE10, TE20, TE01, TE11 and TM11 modes will be the first five modes to propagate.
At 15 GHz, k = 453.1 m-1 and the propagation constant for the TE10 mode is
2
 2π f ε r  π 2 
π 
345.1m −1
−    =k 2 −   =

c
a 
a

β =

The attenuation due to the dielectric loss is
αd
=
k 2 tan δ
= 0.119 Np/m=1.03dB/m
2β
The surface resistivity of the copper walls is (σ
= 5.8 ×107 S/m )
Rs
=
ωµ0
= 0.032 Ω
2σ
and the attenuation due to conductor loss is
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=
αc
2.6
Rs
3 2
k ) 0.050 Np/m
= 0.434dB/m
2bπ 2 + a=
(
a bβ kη
3
CIRCULAR WAVEGUIDE
Refer to the section “Circular waveguide” in the prescribed textbook or to information on
this topic in any of the recommended study materials for details.
A hollow, round metal pipe with an arbitrary inner radius as shown in Figure 3.11 in the
prescribed textbook is able to support the TE and TM waveguide modes. The transverse
fields in cylindrical coordinates can be derived from Ez or H z field components for the
TM and TE modes, respectively.
The cylindrical components of the transverse fields obtained from the longitudinal
components of Figure 3.11 in the prescribed textbook are given as follows:
=
Eρ
− j  ∂Ez ωµ ∂H z 
+
β
kc2  ∂ρ
ρ ∂φ 
(2.34)
=
Eφ
− j  β ∂Ez
∂H z 
− ωµ

2 
kc  ρ ∂φ
∂ρ 
(2.35)
=
Hρ
∂H z 
j  ωε ∂Ez
−β

2 
∂ρ 
kc  ρ ∂φ
(2.36)
=
Hφ
−j
∂E β ∂H z 
ωε z +

2 
∂ρ ρ ∂φ 
kc 
(2.37)
2
where k=
k 2 − β 2 , and e − jβ z propagation has been assumed. For e + jβ z propagation,
c
replace β with − β in all expressions.
2.6.1
TE modes
For TE modes Ez = 0 and H z is nonzero, and the transverse field components are given
by the following:
− jωµ n
( A cos nφ − B sin nφ ) J n ( kc ρ ) e− jβ z
2
kc ρ
(2.38)
Eφ
=
jωµ
( A sin nφ − B cos nφ ) J n' ( kc ρ ) e− jβ z
kc
(2.39)
=
Hρ
− jβ
( A sin nφ + B cos nφ ) J n' ( kc ρ ) e− jβ z
kc
(2.40)
=
Eρ
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=
Hφ
− jβ n
( A cos nφ − B sin nφ ) J n ( kc ρ ) e− jβ z
2
kc ρ
(2.41)
Here A and B are arbitrary amplitude constants.
The wave impedance is
Z=
TE
Eρ − Eφ η k
= =
Hφ H ρ
β
2.6.2
TM modes
(2.42)
In a similar manner, the transverse fields for TM modes are such that H z = 0 and Ez is
nonzero and must be solved using partial differential equations of the wave equation in
cylindrical coordinates.
− jβ
( A sin nφ + B cos nφ ) J n' ( kc ρ ) e− jβ z
kc
(2.43)
=
Eφ
− jβ n
( A cos nφ − B sin nφ ) J n ( kc ρ ) e− jβ z
kc2 ρ
(2.44)
Hρ
=
jωµ n
( A cos nφ − B sin nφ ) J n ( kc ρ ) e− jβ z
2
kc ρ
(2.45)
=
Hφ
− jωε
( A sin nφ + B cos nφ ) J n' ( kc ρ ) e− jβ z
kc
(2.46)
Eρ
=
The wave impedance is
Z=
TM
Eρ − Eφ ηβ
= =
Hφ H ρ
k
(2.47)
Before you do the following worked example and the unsolved related activity, please
watch the following video clip for further explanation of how to apply general solutions
to analyse the TE and TM waveguides.
Circular waveguide by Basic Building Blocks of Microwave Engineering at:
https://www.youtube.com/watch?v=6Z_ezp4bufE
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WORKED EXAMPLE 2.2 CHARACTERISTICS OF A CIRCULAR WAVEGUIDE
Use the following figure to find the cut-off frequencies of the first two propagating modes
of a Teflon-filled circular waveguide with a = 0.5 cm. If the interior of the guide is gold
plated, calculate the overall loss in dB for a 30 cm length operating at 14 GHz.
SOLUTION
In the figure above, the first two propagating modes of a circular waveguide are TE11 and
TM01 modes. The cut-off frequencies can be obtained as follows:
1.841( 3 ×108 )
p11'
TE
11: fc
=
=
= 12.19GHz
2π a ε r 2π ( 0.005 ) 2.08
2.405 ( 3 ×108 )
p01c
01: fc
TM
=
=
= 15.92GHz
2π a ε r 2π ( 0.005 ) 2.08
So only the TE11 mode is propagating at 14 GHz. The wave number is
9
2π f ε r 2π (14 ×10 ) 2.08
k =
422.9m -1
=
=
8
c
3 ×10
and the propagation constant of the TE11 mode is
2
2
 p' 
2
 1.841 
208.0 m -1
β =k −  11  =( 422.9 ) − 
 =
 0.005 
 a 
2
The attenuation due to dielectric loss is calculated from the following:
k 2 tan δ
αd =
=
2β
( 422.9 ) ( 0.0004
)
=
2 ( 208.0 )
2
0.172 Np/m=1.49dB/m
The conductivity of gold is σ
= 4.1×107 S/m , so the surface resistance is
=
Rs
ωµ
0
=
0.0367 Ω
2σ
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2.7
COAXIAL LINE
Refer to the section “Coaxial line” in the prescribed textbook or discussions of this topic
in any of the recommended study material for more information about this concept. First
watch the following video clip for background information about the coaxial line, and
then complete the following related activity.
The coaxial transmission line by Real Physics at:
https://www.youtube.com/watch?v=-AOL4g5CzgE
ACTIVITY 2.5 HIGHER ORDER MODE OF A COAXIAL LINE
Cable Solutions Ltd in Johannesburg wants you to analyse an RG-401U semi-rigid coaxial
cable, with inner and outer conductor diameters of 1.64 mm and 5.5 mm and a Teflon
dielectric with ε r = 2.2 . What is the highest usable frequency before the TE11 waveguide
mode starts to propagate?
SOLUTION
Your answer should be f c = 17.7 GHz . In practice a 5% safety margin is usually
recommended, so f max =
(0.95) × (17.7 GHz) =
16.8 GHz . The lecturer offering this module
will provide detailed feedback regarding the solution to this activity on myUnisa.
2.8
SURFACE WAVES ON A GROUNDED DIELECTRIC SHEET
Let’s consider the TM and TE surface waves that can be excited along a grounded
dielectric sheet. Surface waves are characterised by a field that decays exponentially away
from the dielectric surface, with most of the field contained in or near the dielectric
material sheet. At higher frequencies the field generally becomes more tightly bound to
the dielectric material sheets, making these waveguides practical.
WORKED EXAMPLE 2.3 SURFACE WAVE PROPAGATION CONSTANTS
Calculate and plot the propagation constants of the first three propagating surface wave
modes of a grounded dielectric sheet with ε r = 2.55 for d/λ 0 = 0 to 1.2.
SOLUTION
The first three propagating surface wave modes are TM0, TE1 and TM1. The cut-off
frequencies for these modes can be found using the following formulae:
TM0: fc =0 ⇒
TE=
1: fc
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c
4d ε r − 1
d
λ0
d
⇒
=
λ0
=0
(4
1
ε r −1
)
TM=
1: fc
c
2d ε r − 1
d
⇒
=
λ0
(2
1
ε r −1
)
The propagation constants can be found from the numerical solutions for TM and TE
modes. This can be done with relatively simple root-finding algorithms.
As a point of interest, you can study in detail the root-finding algorithms that are useful in
numerically finding the root of a transcendental equation, so it may be useful to review
relatively simple but effective algorithms for doing this. The first method is known as the
interval-halving method, where the root of f ( x ) = 0 is first bracketed between the values
x1 and x2 . These values can often be estimated from the problem under consideration. If
a single root lies between x1 and x2 , then f ( x1 ) f ( x2 ) < 0 . An estimate, x3 , of the root is
made by halving the interval between x1 and x2 . The other root-finding algorithm is
known as the Newton-Raphson method, which begins with an estimate, x1 , of the root
of f ( x ) = 0 . Then a new estimate, x2 , is obtained from the formula
x2= x1 −
f ( x1 )
f ' ( x1 )
where f ' ( x1 ) is the derivative of f ( x ) at x1 . You will find a discussion of these two
simple root-finding algorithms in section 3.6 of the prescribed textbook.
2.9
STRIPLINE
A stripline is a planner type of transmission line that lends itself well to HF integrated
electronics, miniaturisation, and photolithographic fabrication. Refer to the section
“Stripline” in the prescribed textbook or to discussions of this topic in any of the other
recommended study materials for detailed coverage of the field components, the
calculation of the cut-off frequencies, attenuation due to dielectric materials or due to
conductors as well as the wave impedance and wavelength on the stripline. You should
then be able to complete the following activity.
If you have access to the internet, please watch the following video lecture clip for further
explanation of different types of transmission lines:
https://www.youtube.com/watch?v=ap9739K_OJo
ACTIVITY 2.6 STRIPLINE
Design the width for a 50 ohm copper stripline conductor with b = 0.32 cm and Er =
2.20. If the dielectric loss tangent is 0.001 and the operating frequency is 10 GHz,
calculate the attenuation in dB/λ . Assume a conductor thickness of t = 0.01mm .
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FEEDBACK
The lecturer offering this module will provide detailed feedback regarding the solution to
this activity on myUnisa.
2.10 MICROSTRIP LINE
Microstrip lines are one of the most popular types of planner transmission lines, primarily
because they can be fabricated by photolithographic processes and are easily
miniaturised and integrated with both passive and active microwave devices. Refer to the
section “Microstrip line” in the prescribed textbook or to discussions of this topic in any
of the other recommended study materials for a detailed analysis of this concept. You will
then be able to design a microstrip line in the following activity.
If you have access to the internet, please watch the following video lecture clip for further
explanation on types of transmission lines:
https://www.youtube.com/watch?v=ap9739K_OJo
ACTIVITY 2.7 MICROSTRIP LINE DESIGN
Design a microstrip line on a 0.5 mm aluminium substrate
( ε r 9.9,
tan δ 0.001 ) for a 50
=
=
ohm characteristic impedance. Find the length of this line required to produce a phase
delay of 2700 at 10 GHz, and compute the total loss on this line, assuming copper
conductors. Compare the results obtained from the approximate formulae with those of
an HF signal CAD package.
FEEDBACK
The lecturer offering this module will provide detailed feedback regarding the solution to
this activity on myUnisa.
2.11 THE TRANSVERSE RESONANCE TECHNIQUE
Refer to the section “The transverse resonance technique” in the prescribed textbook or
discussion of this topic in any of the recommended study materials for a detailed
discussion of this concept.
This technique employs a transmission line model of the transverse cross-section of the
waveguide, and gives a much simpler and more direct solution for the cut-off frequency.
The technique provides a simplified alternative to a field theory solution. The technique
is based on the argument that in a waveguide at cut-off, the fields form standing plane
waves in the transverse plane of the guide, as can be inferred from the bouncing plane
wave interpretation of waveguide modes.
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2.12 WAVE VELOCITIES AND DISPERSION
Refer to the section “Wave velocities and dispersion” in the prescribed textbook or to
discussions of the topic in any of the recommended study materials for more details.
Related to the propagation of electromagnetic waves are the speed of light in a medium
1/ µε , the phase velocity ( v p = ω/β ) and the group velocity. The speed of light in a
(
)
medium is the velocity at which a plane wave would propagate in that medium, while
the phase velocity is the speed at which a constant phase point travels. When the phase
velocity is different for different frequencies, then the individual frequency components
will not maintain their original phase relationships as they propagate down the
transmission line or waveguide, and signal distortion will occur (an effect called
dispersion), since different phase velocities allow the “faster” waves to lead in phase
relative to the “slower” waves, and the original phase relationships will gradually be
dispersed as the signal propagates down the line. However, where the bandwidth of the
signal is relatively small or if the dispersion is not too severe, a group velocity can be
defined to describe the speed at which the signal propagates.
For more detail, watch the following video clip on group and phase velocity by Pat Kohl
at the following link:
https://www.youtube.com/watch?v=uii9clp_DSg
WORKED EXAMPLE 2.4 WAVEGUIDE WAVE VELOCITIES
Calculate the group velocity for a waveguide mode propagating in an air-filled guide.
Compare this velocity with the phase velocity and speed of light.
SOLUTION
The propagation constant for a mode in an air-filled waveguide is
β=
k02 − kc2 =
(ω / c )
2
− kc2 .
Taking the derivative with regard to frequency gives
dβ
=
dω
ω / c2
k
= 0 ,
2
(ω / c ) − kc2 cβ
and the group velocity is
−1
cβ
 dβ 
.
vg =
=

k0
 dω 
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The phase velocity is given as=
v p ω=
/ β ck0 / β . Since β < k0 , vg < c < v p , which
indicates that the phase velocity of a waveguide mode may be greater than the speed of
light, but the group velocity (the velocity of a narrow-band signal) will be less than the
speed of light.
ACTIVITY 2.8 GENERAL PRACTICE AND REVIEW QUESTIONS
1. For practice and revision, please complete problems 3.2, 3.3, 3.5, 3.9, 3.13, 3.19–
3.22 and 3.29 in the prescribed textbook.
Please do these on your own first, after which the lecturer will post the solutions on
myUnisa.
2. Which concepts or calculations did you find fairly easy to understand in this unit, and
which did you struggle with? Remember that you are welcome to discuss difficult
questions with your fellow students on myUnisa, or to contact the lecturer for
clarification.
2.13 CONCLUSION
After studying this unit, you should be able to apply various wave solutions in HF
transmission lines and waveguides. Now that you have derived and applied HF wave
propagation quantities in HF electronics, you should be able to show how basic circuit
and network concepts can be extended to deal with many HF or microwave analysis and
design problems of practical interest, as we will be discussing in the next learning unit.
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3
Learning unit 1
HF NETWORK ANALYSIS
3.1
LEARNING OUTCOMES
By the end of this learning unit, you should be able to analyse HF electronic networks
and apply basic circuit and network concepts to solve related design problems of practical
interest.
In particular, you should be able to do the following:
•
•
•
•
•
•
•
Analyse impedance and equivalence voltages and currents.
Analyse impedance and admittance matrices.
Solve problems relating to scattering matrices.
Solve problems relating to the transmission (ABCD) matrix.
Analyse signal flow graphs.
Solve problems relating to HF electronic discontinuities and modal analysis.
Explain excitation of waveguides with reference to electric and magnetic currents, and
aperture coupling.
3.2
INTRODUCTION
Reading: To complete this unit, study the chapter entitled “HF Microwave Network
Analysis” in your prescribed textbook, or the same topic in any recommended study
materials.
This unit assumes that you are able to apply various wave solutions in HF transmission
lines and waveguides. You can now analyse HF networks and show how basic circuit
and network concepts can be applied in solving a number of HF microwave analysis and
design problems of practical interest.
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3.3
IMPEDANCE AND EQUIVALENT VOLTAGES AND CURRENTS
Please refer to the section “Impedance and equivalent voltages and currents” in the
prescribed textbook for detailed information. This unit will provide you with summary of
the relevant concepts that you need to study in depth.
At high microwave frequencies, voltage or current measurement is reasonably difficult,
since there is no terminal pair. With TEM-type lines such as coaxial cable, microstrip or
stripline, there is a terminal pair, but this is not the case with the non-TEM lines such as
rectangular, circular or surface waveguides. The voltage V across the positive conductor
relative to the negative conductor is determined by
V =∫
− ve
+ ve
E.d 
(3.1)
where the integration path begins on the positive (+ve) conductor and ends on the
negative (-ve) conductor. Likewise, the total current I flowing on the positive conductor
can be determined from Ampere’s law as
I =
∫ + ve H .d 
C
(3.2)
where the integration contour is any closed path enclosing the positive conductor (but
not the negative conductor). Using results in equations (3.1) and (3.2), we have a
characteristic impedance Z 0 defined for the travelling waves as follows:
Z0 =
V
I
(3.3)
Refer to Figure 4.1 in the prescribed textbook for comprehensive illustrations of the
electric and magnetic field lines for any arbitrary two-conductor TEM transmission line.
There are many ways to define equivalent voltage, current and impedance for
waveguides, since these quantities are not unique for non-TEM lines. Let’s look at the
following possible ways of getting the most useful results:
•
For each waveguide mode, voltage and current can be defined such that voltage is
proportional to the transverse electric field and current is proportional to the
transverse magnetic field.
•
For meaningful equivalent voltages and currents, they should be defined such that
their product yields the power flow of the waveguide mode.
•
The ratio of the equivalent voltage to current should give the characteristic impedance
of the line.
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Thus, for any arbitrary waveguide mode with both positively and negatively travelling
waves, the transverse fields are written as follows:
− jβ z
=
Et ( =
x , y , z ) e ( x , y ) ( A+ e − j β z + A
e )
e ( x, y ) + − j β z
(V e + V −e+ jβ z )
C1
(3.4)
− jβ z
Ht ( =
x, y, z ) h ( x, y ) ( A+ e − jβ z + A=
e )
h ( x, y ) + − j β z
( I e − V −e + jβ z )
C2
(3.5)
where e and h are the transverse field variations of the mode, and A+ and A− are the
field amplitudes of the travelling waves. Equations (3.4) and (3.5) give us the equivalent
voltage and current wave equations,
=
V ( z ) V + e − jβ z + V −e jβ z
(3.6)
I ( z ) I + e − jβ z − I −e jβ z
=
(3.7)
+
−
where V=
/ I + V=
/ I − Z0 .
Relating Et and H t from equations (3.4) and (3.5) with wave impedance Z w , we have
h ( x, y ) =
zˆ × e ( x, y )
.
Zw
(3.8)
For more details, watch the following video clip by Prof Amitabha Bhattacharya dealing
with problem-solving relating to equivalent voltages and currents in a waveguide:
https://www.youtube.com/watch?v=wVk0VDY5aD0
You can now attempt some general questions as well as an application question relating
to the concepts discussed above.
ACTIVITY 3.1 EQUIVALENT VOLTAGE AND CURRENT FOR A RECTANGULAR
WAVEGUIDE
1. Why is finding the voltage and current measurement for non-TEM waveguides fairly
difficult at high microwave frequencies?
2. What would one consider when defining equivalent voltage, current and impedance
for waveguides for non-TEM transmission lines?
3. Calculate the equivalent voltages and currents for a TE10 mode in a rectangular
waveguide. First, solve the problem without referring to the solution, and then check
your answer against the solution.
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SOLUTION
For mode TE10 in a rectangular waveguide, the transverse field components and power
flow are given as follows:
Waveguide fields
(A e
+ − jβ z
Ey
=
=
Hx
=
P+
+ A−e jβ z ) sin
Transmission line model
πx
a
−1 + − jβ z
πx
− A−e jβ z ) sin
Ae
(
ZTE
a
=
V ( z ) V + e − jβ z + V −e jβ z
=
I ( z ) I + e − jβ z − I −e jβ z
=
2
−1
ab
=
E y H x∗dxdy
A+
∫
2 S
4 ZTE
1
V + e− jβ z − V −e jβ z )
(
Z0
1
P + = V + I +∗
2
+
+
We obtain the constants
=
C1 V=
/ A+ V − / A− and
=
C2 I=
/ A+ I − / A− such that the
incident powers can be obtained. First, we have
2
ab A+
1 + +∗ 1 + 2
= =
V I
A C1C2∗
4 ZTE
2
2
(3.9)
For Z 0 = ZTE by the designer choice, the following equation holds:
V + C1
= = ZTE
I + C2
(3.10)
So that we have
C1 =
ab
2
C2 =
1
ZTE
(3.11)
ab
2
(3.12)
Expressions (3.9) to (3.12) present us with the desired transmission line equivalence for
the TE10 mode.
If you have access to the internet, watch the following video clip for a further explanation
about how to calculate the impedance, equivalent voltages and currents in a waveguide.
Waveguide: wavelength, impedance and power calculations by Electromagnetic Theory
at:
https://www.youtube.com/watch?v=wGqRGubvcF8
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Now apply the concepts you learnt in the section above by completing the following
activity. Do it on your own before referring to the solution provided.
ACTIVITY 3.2 APPLICATION OF WAVEGUIDE IMPEDANCE
Apply the equivalent transmission model to analyse the reflection coefficient of a TE10
wave incident on the interface from z < 0. Assume a rectangular waveguide with
a = 2.286 cm and b =1.016 cm , air filled for z < 0 , Rexolite filled ( ε r = 2.54 ) for z > 0 . The
operating frequency is 10 GHz.
SOLUTION
The propagation constants within a waveguide filled with air and the dielectric regions
are given as follows:
2
π 
β air = k02 −   =
158.0 m −1
a


2
π 
β dielectric = ε r k02 −   = 304.1 m −1
a


where k0 = 209.4 m −1 . The equivalent characteristic impedances for two lines are given as
Z=
0 air
Z 0 dielectric
=
( 209.4 )( 377
)
=
k0η0
=
β air
kη
=
β dielectric
158.0
k0η0
=
β dielectric
500.0 Ω
377 )
( 209.4 )( =
304.1
259.6 Ω
The reflection coefficient seen looking into the dielectric filled region is then
Γ=
3.4
Z 0 dielectric − Z 0 air
= −0.316 .
Z 0 dielectric + Z 0 air
IMPEDANCE AND ADMITTANCE MATRICES
Refer to the section “Impedance and admittance matrices” in the prescribed textbook or
information on this topic in any of the recommended study material for detailed
derivations.
Please watch the following online video presentation by Chastity Mary Nash for an
introductory explanation of microwave network analysis. You can turn the sound off and
just watch the slides:
https://slideplayer.com/slide/9850453/
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As you will have learnt from this video and the previous section, following the definition
of equivalent voltages and currents at various points in an HF microwave network, the
impedance or admittance matrices of the circuit theory can be used to relate the terminal
or port quantities to each other in order to arrive at a matrix description of the network.
For instance, given an arbitrary N-port HF microwave network, the general impedance
matrix [Z] of the network may relate these voltages and currents as follows:
 V1 
V 
 2=
 ⋅ 
 
VN 
 Z11 Z12 ⋅⋅⋅ Z1N 
Z
  I1 
 21
I 
⋅
 2 
⋅⋅⋅ ⋅⋅⋅ Z NN   ⋅ 

Z N1
 I 

  N 
(3.13)
Using a similar matrix representation form, the admittance matrix [Y] can be defined as
follows:
 I1 
I 
 2=
 ⋅ 
 
IN 
 Y11 Y12 ⋅⋅⋅ Y1N 
Y
  V1 
21

 V 
⋅
 2 
⋅⋅⋅ ⋅⋅⋅ YNN   ⋅ 

YN 1
 V 

  N 
(3.14)
You can now apply these general forms of representing impedance and admittance
matrices by solving the following activity. Review the section “Impedance and admittance
matrices” in the prescribed textbook, and then solve the following problem on your own
before checking your answer against the solution provided.
ACTIVITY 3.3 EVALUATION OF IMPEDANCE PARAMETERS
You work for a high frequency electronics transformer equipment manufacturing
company in the province of Gauteng in South Africa. Your workplace supervisor has
given you the task of carrying out an impedance matching of the overall electronic power
distribution system. However, you decide that before you start, you will determine the Z
parameters of this transformer. Assume that the transformer is treated as a two-port Tnetwork as shown in the figure below. Determine its Z parameters, and make
recommendations to your workplace supervisor.
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SOLUTION
For an open-circulated port 2, we find Z11 as input impedance of port 1 as follows:
Z=
11
V1
Z A + ZC
|I =
=0
I1 2
The transfer impedance Z12 is found by measuring the open-circuit voltage at port 1 when
the current I2 is applied at port 2, such that we have
V1
V
ZC
=
= ZC
|I1 =0 2
I2
I 2 Z B + ZC
Z12
=
Finally, Z22 is
Z=
22
V2
Z B + ZC
|I=
=0
I2 1
Now we can show that the circuit is reciprocal, hence Z21 = Z12, and if the values of ZA,
ZB, and ZC are provided, we can substitute these values into the above expressions to
yield the Z parameters.
3.5
THE SCATTERING MATRIX
Refer to the section entitled “The scattering matrix” in the prescribed textbook and the
online video you watched in the previous section for more details.
Similar to the way that the impedance or admittance matrix were defined, the scattering
matrix entails a complete description of the network as seen at its N ports. While the
impedance and admittance matrices relate to the total voltages and currents as seen at the
ports, the scattering matrix relates the voltage waves incident on the ports to those
reflected from the ports. For an N-port network, the scattering matrix [S] is defined in
relation to these incident and reflected voltage waves as
 S11 S12
V1−  
⋅⋅⋅
 −   S 21
V
 2  = S
 ⋅   N1
⋅⋅⋅
 −   S N +1,1
VN  
⋅⋅⋅
 ⋅
⋅⋅⋅ S1N  +
V 
⋅⋅⋅ ⋅⋅⋅   1+ 
V
 2 
⋅⋅⋅ S NN   ⋅ 
 +
VN 
⋅⋅⋅ ⋅⋅⋅ 
(3.15)
If you have internet access, as practice, answer the multiple-choice questions on
scattering matrices in HF microwave networks by Sanfoundry Technology Education
Blog, available online at the following link:
https://www.sanfoundry.com/microwave-engineering-questions-answers-scatteringmatrices/
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Then, apply the generic form of the scattering matrix to evaluate its parameters in the
activity below.
ACTIVITY 3.4 EVALUATION OF SCATTERING PARAMETERS
What are the scattering parameters of the 3 dB attenuator circuit shown in the following
diagram?
SOLUTION
S11 can be found as the reflection coefficient seen at port 1 when port 2 is terminated in
a matched load (Z0 = 50 ohms):
V1−
Z in(1) − Z 0
(1)
S11 =
=
=
Γ
=
|
|
|Z on port 2
+
+
V2 0
V1+ V2 0=
Z in(1) + Z 0 0
8.56 + [141.8(8.56 + 50)] / (141.8 + 8.56 + 50) =50 Ω , and so S11=0=S22.
But Z in( ) =
1
V2−
S 21 = + |V + =0
V1 2
By applying a voltage V1 at port 1 and using voltage division twice we find V2− = V2 as the
voltage across the 50 Ω load resistor at port 2:
41.44  50 

−
V2=
V=
V1 
=
2

 0.707 V1
 41.44 + 8.56  50 + 8.56 
Thus, S=
S=
0.707
12
21
ACTIVITY 3.5 APPLICATION OF SCATTERING PARAMETERS
Refer to the section entitled “Reciprocal networks and lossless networks” in the prescribed
textbook or information about this topic in any of the recommended study material before
doing the following activity.
Given the following scattering matrix
 0.15∠00
[S ] = 
0
0.85∠45
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0.85∠ − 450 

0.2∠00 
Find out whether the network is reciprocal and lossless. If port 2 is terminated with a
matched load, what is the return loss seen at port 1? If port 2 is terminated with a short
circuit, what is the return loss seen at port 1?
Post your answer in the relevant Discussion space of the module website and respond to
other students' posts where appropriate.
3.6
THE TRANSMISSION (ABCD) MATRIX
The Z, Y and S parameter representations can be used to characterise an HF microwave
network with an arbitrary number of ports, but in practice many microwave networks
consist of a cascade connection of two or more two-port networks. In that case it is
convenient to define a 2 × 2 transmission or ABCD matrix for each two-port network.
Refer to the section entitled “The transmission matrix” in the prescribed textbook for
detailed illustrations of a two-port network and a cascade connection of two-port
networks.
If you have internet access, please watch the online video presentation on microwave
network analysis by Chastity Mary Nash again at the link:
https://slideplayer.com/slide/9850453/.
The ABCD matrix expressed in terms of the total voltages and currents is
=
V1 AV2 + BI 2
(3.16)
=
I1 CV2 + DI 2
(3.17)
Equations (3.16) and (3.17) are written in matrix form as:
V1   A B  V2 
 I  = C D   I 
 2
 1 
(3.18)
Apply equation (3.18) in the following activity, but first do the review questions on the
Sanfoundry Technology Education Blog at:
https://www.sanfoundry.com/microwave-engineering-questions-answers-transmissionmatrix/
ACTIVITY 3.6 EVALUATION OF ABCD PARAMETERS
Determine the ABCD parameters of a two-port network consisting of a series impedance
Z between ports 1 and 2 in Table 4.1, captioned “ABCD parameters of some useful twoport circuits,” in the prescribed textbook.
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SOLUTION
The ABCD parameters are related as follows:
A=
V1
|I =0
V2 2
which indicates that A is found by applying a voltage V1 at port 1 and measuring the
open-circuit voltage V2 at port 2. Therefore, A=1. In a similar manner, we have
B
=
3.7
V1
V1
Z,
=
|V2 =0 =
I2
V / Z1
C
=
I1
|I =0 0 ,
=
V2 2
D
=
I1
I1
= 1.
|V2 =
=0
I2
I1
SIGNAL FLOW GRAPHS
Refer to the section entitled “Signal flow graphs” in the prescribed textbook, or
information about this topic in any of the recommended study material for detailed
coverage of this concept. To help you follow the concepts discussed in this section, please
watch the following online video clip about signal flow graphs by Tutorial Point (India)
at:
https://www.youtube.com/watch?v=aAPi01gajI8
Also watch the following video clip on signal flow graphs from block diagram to equation
by Shrenik Jain at:
https://www.youtube.com/watch?v=uG_RNaXUv78
Signal flow graphs are another useful technique for analysing HF microwave networks in
terms of transmitted and reflected waves. Signal flow graphs consist of nodes and
branches.
Nodes: Each port i of an HF microwave network has two nodes, ai and bi . Node ai
is identified with a wave entering port i , while node bi is identified with a wave
reflected from port i . The voltage at a node is equal to the sum of all signals entering
that node.
Branches: These form a directed path between two nodes representing signal flow
from one node to another. At each branch, there is a scattering parameter or reflection
coefficient. Please go through the flow graph of an arbitrary two-port network as
shown in Figures 4.14 to 4.17 in the prescribed textbook and solve the following
signal flow graph problem.
•
•
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ACTIVITY 3.7 APPLICATION OF SIGNAL FLOW GRAPH
1. Describe four basic decomposition rules of a complex signal flow graph into a single
branch between two nodes such that any desired wave amplitude ratio can be
obtained.
2. Apply signal flow graphs to derive expressions for Γ in and Γ out for the HF
microwave network in Figure 4.17 in the prescribed textbook.
SOLUTION
First convert the network into a signal flow graph network as shown in Figure 4.18. Then
write in terms of node voltages: Γ in is given by the ratio b1 / a1 . The first two steps of the
required decomposition of the flow graph are shown in Figures 4.19 (a) and (b).
By inspection:
Γin =
b1
S S Γ
= S11 + 12 21 
a1
1 − S 22Γ 
Next, find Γ out using the ratio b2 / a2 . The first two steps of this decomposition are shown
in Figures 4.19 (c) and (d), and the results are
Γ out =
3.8
b2
S S Γ
= S 22 + 12 21 s
a2
1 − S11Γ s
DISCONTUINITIES AND MODAL ANALYSIS
Refer to the section “Discontinuities and modal analysis” in the prescribed textbook for a
detailed discussion of this concept.
HF electronic circuits and networks often consist of transmission lines with various types
of discontinuities created either by necessity or by design choice, or which are the
unavoidable result of mechanical or electrical transitions from one medium to another,
such as a junction between two waveguides or a coaxial cable and microstrip transition.
A transmission line discontinuity can be presented as an equivalent circuit at some point
on the transmission line. Depending on the type of discontinuity, the equivalent circuit
may be a simple shunt or series element across the line or, as is more usually the case, a
T - or π -equivalent circuit may be required. Modal analysis is a rigorous and robust
technique that is applied to a number of waveguide and coaxial cable discontinuity
problems, and is easily implemented using computers.
If you have internet access, for more information about discontinuities and modal
analysis, please watch the online video clips at the following links:
Discontinuities of microstrip line by Byron Ellis at:
https://slideplayer.com/slide/7530596/
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Microwave communications by Jaida Drews at:
http://slideplayer.com/slide/4180417/release/woothee
ACTIVITY 3.8 EASY TO SOLVE PROBLEMS RELATED TO DISCONTINUITIES
1. If you have internet access, practise by solving the discontinuity related problems
designed
by
Paul
Dawkins
that
you
will
find
at:
http://tutorial.math.lamar.edu/Problems/CalcI/Continuity.aspx
You can also solve the following discontinuity related problems, designed by Khan
Academy, which you will find at:
https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-10/e/analyzingdiscontinuities-graphical
2. Practise problems relating to HF electronic discontinuities and modal analysis
published online by the Sanfoundry Technology Education Blog at:
https://www.sanfoundry.com/microwave-engineering-questions-answers-reflections/
First do these problems on your own after studying the section entitled “Discontinuities
and modal analysis” in the prescribed textbook, and then check your answers against the
solutions provided online.
3.9 EXCITATION OF WAVEGUIDES – ELECTRIC AND MAGNETIC
CURRENTS
We can consider guided waves with reference to propagation, reflection and transmission
with sources. In waveguides, several propagating modes may be excited, along with
evanescent modes, which store energy. In the section “Excitation of waveguides – electric
and magnetic currents” in the prescribed textbook, you are introduced to formalism,
which is required to determine the excitation of a given waveguide mode as the result of
an arbitrary electric or magnetic current source. Two cases will be introduced: (1) the
current sheets that excite only one waveguide mode, where an infinitely long rectangular
waveguide possesses a transverse sheet of electric surface current density, and (2) mode
excitation arising from an arbitrary electric or magnetic current source, where the electric
current source is located between two transverse planes.
After working through the detailed analysis in the prescribed textbook and watching the
following video clips for a demonstration of how waveguides can be excited by electric
and magnetic currents, please work out the solution to the activity below.
Line excitation of a parallel-plate waveguide by nageljr at:
https://www.youtube.com/watch?v=UDTulutNbcM, and
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Excitation of waveguides by Niklas Manz at: https://www.youtube.com/watch?v=0wPp01wXpg
ACTIVITY 3.9 PROBE-FED RECTANGULAR WAVEGUIDE
1. Review the related theory presented in the section “Excitation of waveguides” in the
prescribed textbook, which is useful in determining the excitation and input
impedance of probe and loop feeds.
2. For a probe-fed rectangular waveguide, determine the amplitudes of the forward- and
backward-travelling TE10 modes, and the input resistance as seen by the probe.
Assume that the TE10 mode is the only propagating mode and that the probe is fed
halfway through the wall of the larger dimension of the waveguide.
3.10 EXCITATION OF WAVEGUIDES – APERTURE COUPLING
Small apertures are able to couple waveguides and other transmission lines just as probe
and loop feeds do. In some cases, small aperture coupling has found its application in the
design of directional couplers and power dividers, in which power from one guide is
coupled to that of another through small apertures. There are two techniques of aperture
coupling, namely coupling through an aperture in a transverse waveguide wall, and
coupling through an aperture in the broad wall of a waveguide.
If you have internet access, please watch the following video clip for further explanation
of the excitation of waveguides – electric and magnetic currents as well as through
aperture coupling – and then solve the following problems so that you can practise the
concepts you have learnt in this unit.
https://slideplayer.com/slide/6050548/
ACTIVITY 3.10 HF NETWORK ANALYSIS AND OVERALL REVIEW QUESTIONS
1. Solve the following unsolved problems in the chapter entitled “HF Network Analysis”
in the prescribed textbook: 4.1–4.10, 4.11, 4.12, 4.14, 4.18, 4.20, 4.21, 4.23, 4.30,
4.32, 4.35, 4.36 and 4.37. Your module lecturer will post the correct answers without
including the detailed steps on the module website for your reference. Remember that
you are welcome to contact your lecturer via the module's online Discussions forum
should you have specific questions.
2. What are some of the practical engineering applications of the knowledge and skills
you have acquired by studying this unit? How might you be able to apply this
knowledge in a practical work situation?
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3.11 CONCLUSION
After studying this unit, you should be able to analyse HF networks and apply basic circuit
and network concepts to solve related design problems of practical interest. In the next
unit, you will learn to analyse and demonstrate impedance matching and tuning of HF
electronic networks.
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4
Learning unit 1
HF IMPEDANCE MATCHING AND TUNING
4.1
LEARNING OUTCOMES
By the end of this learning unit, you should be able to produce impedance matching and
tuning designs for HF electronic networks. A good design should ensure that the power
delivered when the load is matched to the line (assuming that the generator is matched)
is maximum, and power loss in the feed line is minimised; the signal-to-noise ratio of the
system is improved; and amplitude and phase errors of a power distribution network are
reduced.
In particular, you should be able to do the following:
•
•
•
•
•
•
•
•
Match the load with the line impedance using lumped elements (L networks).
Describe how networks are tuned with single and double stubs.
Describe how the load impedances are matched with the line impedance using a
single-section and multisection quarter-wave transformer.
Analyse the theory of small reflections.
Design binomial multisection matching transformers.
Design Chebyshev multisection matching transformers.
Design tapered lines.
Discuss the Bode-Fano criterion.
4.2
INTRODUCTION
Reading: To complete this unit, study the chapter entitled “HF Impedance Matching and
Tuning” in your prescribed textbook or the same topic in any of the recommended study
materials.
This unit assumes that you are able to analyse HF networks and to show how basic circuit
and network concepts can be extended to solve HF microwave analysis and design
problems of practical interest. You will now go on to apply the theory and techniques
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relating to impedance matching, where an impedance matching network is placed
between a load impedance and a transmission line such that the impedance seen when
looking into the matching network is same as that of the characteristic impedance ( Z 0 ).
The process of impedance matching (also known as the tuning process) aims to achieve
the following:
•
Minimise power loss in the feed line and match the line when the generator is
matched, which consequently maximises the power delivered to the load.
Enhance/Improve the signal-to-noise ratio (SNR) of the system when the impedance
of the sensitive receiver components (such as antennas and low-noise amplifiers) are
matched to the characteristic impedance of the transmission line.
Minimise amplitude and phase errors through impedance matching in power
distribution networks (such as an antenna array feed network).
•
•
If you have access to the internet, please watch the following video clip illustrating the
concept of impedance matching in high frequency electronics by Tutorials Point (India),
as this will give you some useful insights into the rest of the concepts discussed in this
unit.
https://www.youtube.com/watch?v=7Xe2xmEAH7M
As you were watching the video, you may have noticed that there are a number of factors
to consider when selecting a particular matching network. These are:
•
Complexity – the simplest design that satisfies the required specifications; for instance,
cheaper, smaller, more reliable and less lossy is generally preferable.
Bandwidth – producing a perfect match at a single frequency or over a band of
frequencies.
Implementation – this depends on the type of transmission line or waveguide being
used. For instance, using tuning stubs for matching waveguides to the load is much
easier than using multisection quarter-wave transformers.
Adjustability – in some applications, the matching network should be adjustable to
match a variable load impedance to the transmission line.
•
•
•
4.3
MATCHING WITH LUMPED ELEMENTS (L NETWORKS)
Please refer to the section “Matching with lumped elements (L networks)” in the
prescribed textbook for a detailed discussion. This learning unit will merely provide you
with a summary of the relevant concepts that you need to study in detail.
If you have internet access, watch the following video presentation by Niraj Chennai for
more information on LC matching or L-section matching network problem-solving using
a Smith chart at:
https://www.youtube.com/watch?v=00oIX5U8zTo
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Note that the matching network of the L-section is the simplest to implement, and that it
uses two reactive elements (either inductors or capacitors, depending on the load
impedance) to match an arbitrary load impedance to a transmission line. As shown in
Figure 5.2 in the prescribed textbook, two possible configurations for an L-section
network exist. The basic idea is to look inside the normalised circle 1 + jx on the Smith
chart to see whether or not it contains the normalised load impedance z L = Z L / Z 0 . The
circuits from Figures 5.2a and Figure 5.2b in the prescribed textbook are then used. The
1 + jx circle is the resistance circle on the impedance Smith chart of which the radius,
r = 1.
You can use either analytic solutions or simple graphical solutions using a Smith chart to
express L-section matching components. The mathematical expressions may in turn be
amenable to a computer-aided design program for L-section matching, or when it is
necessary to have more accurate values than are possible when using the Smith chart.
Instead of applying the tedious mathematical formulas involved, let’s do an activity that
illustrates the process of using a Smith chart to quickly and accurately design L-section
matching networks.
ACTIVITY 4.1 L-SECTION IMPEDANCE MATCHING
Imagine that you are an employee of a transmission line manufacturing company in the
province of Gauteng in South Africa. Your workplace supervisor has asked you to carry
out an L-section impedance matching activity. Your task is to design an L-section
matching network to match a series RC load with an impedance Z L =200 − j100 Ω to a
100 Ω line at a frequency of 500 MHz. Use the Smith chart approach in supplying your
answer.
SOLUTION
The normalised load impedance is z L= 2 − j1 , which is plotted on the Smith chart in
Figure 5.3a in the prescribed textbook. This point is inside the 1 + jx circle, so the
matching circuit in Figure 5.2a in the prescribed textbook is used. The first element from
the load is a shunt susceptance, so we convert to admittance by drawing the SWR circle
through the load, and a straight line from the load through the centre of the chart, as
shown in Figure 5.3a. After we add the shunt susceptance and convert back to
impedance, we want to be on the 1 + jx circle so that we can add a series reactance to
cancel jx and match the load. This means that the shunt susceptance must move us from
yL to the 1 + jx circle on the admittance Smith chart. We therefore construct the rotated
1 + jx circle as shown in Figure 5.3a (centre at r = 0.333 ). Then we see that adding a
susceptance of jb = j 0.3 will move us along a constant-conductance circle to
=
y 0.4 + j.05 (this choice is the shortest distance from yL to the shifted 1 + jx circle).
Converting back to impedance leaves us at z = 1 − j1.2 , indicating that a series reactance
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of x = j1.2 will bring us to the centre of the chart. For comparison purposes the formulas
in analytical solutions provide us with b = 0.29 , x = 1.22 .
4.4
SINGLE-STUB TUNING
Refer to the section “Single-stub tuning” in the prescribed textbook or information on this
topic in any of the recommended study materials for a detailed discussion.
Start by watching the following online video presentation for an introductory explanation
of single-stub tuning using a Smith chart by Prof Shevgaonkar, uploaded by Rushibmehta
at the link:
https://www.youtube.com/watch?v=Wn3A5f24Y8w
Then read up further in the prescribed textbook, from which you will realise that matching
the impedance of a load to the characteristic impedance of the line can be achieved by
other methods. It is possible to use either a single open-circuited or short-circuited length
of transmission line (a stub) connected either in parallel or in series with the transmission
feed line at a certain distance from the load, as shown in Figure 5.4 in the prescribed
textbook. A single-stub tuning circuit such as this is often very convenient, because the
stub can be fabricated as part of the transmission line media of the circuit, and lumped
elements are avoided. Shunt stubs are preferred for microstrip line or striplines, while
series stubs are preferred for slot lines or co-planar waveguides.
The learning activity below will give you more insight into the single-stub shunt tuning
circuit as shown in Figure 5.4a in the prescribed textbook.
ACTIVITY 4.2 SINGLE-STUB SHUNT TUNING
For the company in activity 4.1, design two single-stub (short circuit) shunt-tuning
networks to match a load impedance Z L =60 − j80 Ω to a 50Ω line. Suppose the load is
matched at 2 GHz and that the load consists of a resistor and capacitor in series. What
would the reflection magnitude look like when plotted from 1 to 3 GHz for each solution?
SOLUTION
The first step is to plot the normalised load impedance z=
1.2 − j1.6 , construct the
L
appropriate SWR circle, and convert to the load admittance, yL , as shown on the Smith
chart in Figure 5.5a in the prescribed textbook. For the remaining steps we consider the
Smith chart as an admittance chart. Notice that the SWR circle intersects the 1 + jb circle
at two points, denoted as y1 and y2 in Figure 5.5a. Thus, the distance d from the load
to the stub is given by either of these two intersections. Reading the WTG scale, we
obtain:
d1 = 0.176 − 0.065 = 0.110λ
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d 2 = 0.325 − 0.065 = 0.260λ .
In fact, the distance d around the SWR circle that intersects the 1 + jb circle is infinite. It
is often desirable to keep the matching stub as close as possible to the load in order to
improve the bandwidth of the match and to reduce losses caused by what may be a large
standing wave ratio on the line between the stub and the load.
At the two intersection points, the normalised admittances are:
=
y1 1.00 + j1.47 and
=
y2 1.00 − j1.47 .
The first tuning solution will require a tuning stub with a susceptance of − j1.47 . The
length of a short-circuited stub that gives this susceptance is found on the Smith chart by
starting at y = ∞ (the short circuit) and moving along the outer edge of the chart ( g = 0 )
towards the generator to the − j1.47 point. The stub length is then
1 = 0.095λ .
Similarly, the required short-circuit stub length for the second solution is
1 = 0.405λ .
This completes the two tuner designs.
In order to analyse the frequency dependence of these two designs, we need to know the
load impedance as a function of frequency. The series-RC load impedance is
Z L =60 − j80 Ω at 2 GHz, so =
R 60 Ω and C = 0.995 pF . The two tuning circuits are
shown in Figure 5.5b in the prescribed textbook. Figure 5.5c in the prescribed textbook
shows the calculated reflection coefficient magnitudes for these two solutions. Note that
solution 1 has a significantly better bandwidth than solution 2, since both d and  are
shorter for solution 1, which reduces the frequency variation of the match.
ACTIVITY 4.3 SINGLE-STUB SERIES TUNING
You can use the same problem-solving technique as in activities 4.1 and 4.2, but now
apply it to the single-stub series-tuning problem. Design a matching network of a load
impedance of Z
=
100 + j80 to a 50 Ω line using a single-series open-circuit stub.
L
Assuming that the load is matched at 2 GHz and that the load consists of a resistor and
inductor in series, plot the reflection coefficient magnitude from 1 to 3 GHz.
SOLUTION
Use a similar procedure as for single-stub shunt tuning. You plot the normalised
impedance, z L= 2 + j1.6 , and draw the SWR circle. For the series-stub design the chart is
an impedance chart. The SWR circle intersects the 1 + jx circle at two points, denoted as
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z1 and z2 in Figure 5.6a in the prescribed textbook. The shortest distance, d1 , from the
load to the stub from the WTG scale is
d1 = 0.328 − 0.208 = 0.120λ .
The second distance is
d 2 =( 0.5 − 0.208 ) + 0.172 =0.463λ .
The normalised impedances at the two intersection points are
z1 = 1 − j1.33 ,
z2 = 1 + j1.33 .
The first solution requires a stub with a reactance of j1.33 . The length of an opencircuited stub that gives this reactance can be found on the Smith chart by starting at
z = ∞ (open circuit), moving along the outer edge of the chart ( r = 0 ) toward the generator
to the j1.33 point. This gives a stub length of
 = 0.397λ .
Similarly, the required open-circuited stub length for the second solution is
 = 0.103λ .
This completes the tuner designs.
If the load is a series resistor and inductor with Z L =100 + j80 Ω at 2 GHz, then =
R 100 Ω
and L = 6.37 nH. The two matching circuits are shown in Figure 5.6b in the prescribed
textbook. Figure 5.6c shows the calculated reflection coefficient magnitudes versus
frequency for the two solutions.
4.5
DOUBLE-STUB TUNING
Refer to the section “Double-stub tuning” in the prescribed textbook and the online video
you watched in the previous section for more detail.
The single-stub tuner can match any load impedance (having a positive real part) to a
transmission line, but suffers from the shortfalls of requiring a variable length of line
between the load and the stub. Whenever an adjustable tuner is required, the single-stub
tuner feasible for a fixed matching circuit may not work – hence the need for double-stub
tuners, which consist of two tuning stubs in fixed positions. These tuners are often
fabricated in coaxial line with adjustable stubs connected in shunt to the main coaxial
line.
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If you have internet access, please watch the following online video link by Electrofun
for a further explanation of the process of solving a double-stub tuning problem using a
Smith chart:
https://www.youtube.com/watch?v=gooRTnO0i2I
Now assess your understanding of the concept of a double-stub tuner using a Smith chart
by doing the following learning activity.
ACTIVITY 4.4 DOUBLE-STUB TUNING
Design a double-stub shunt tuner to match a load impedance Z L =60 − j80 Ω to a 50Ω
line. The stubs are to be open-circuited, spaced λ / 8 apart. Assuming that this load
consists of a series resistor and capacitor and that the match frequency is 2 GHz, plot the
reflection coefficient magnitude versus frequency from 1 to 3 GHz.
SOLUTION
The normalised admittance is y=
0.3 + j 0.4 and is plotted on the Smith chart (Figure 5.9a
L
in the prescribed textbook). Next, construct the rotated 1 + jb conductance circle by
moving every point on the g = 1 circle λ / 8 toward the load. Then, find the susceptance
of the first stub, which will be one of two possible values:
b1 = 1.314 or b1' = −0.114
You can now transform through the λ / 8 section of line by rotating along a constant
radius (SWR) circle λ / 8 toward the generator. This brings the two solutions to the
following points:
y2 = 1 − j 3.38 or y2' = 1 + j1.38 .
Then, the susceptance of the second stub should be
b2 = 3.38 or b2' = −1.38 .
The lengths of the open-circuited stubs are then found as
1 = 0.146λ ,  2 = 0.204λ or 1' = 0.482λ ,  '2 = 0.350λ .
This completes both solutions for the double-stub tuner design.
At f = 2 GHz the resistor-capacitor load of Z L =60 − j80 Ω implies that =
R 60 Ω and
C = 0.995 pF . The two tuning circuits are then as shown in Figure 5.9b in the prescribed
textbook, and the reflection coefficient magnitudes are plotted against frequency as
shown in Figure 5.9c. It can be observed that the first solution has a much narrower
bandwidth than the second (primed) solution due to the fact that both stubs for the first
solution are somewhat longer (and closer to λ / 2 ) than the stubs of the second solution.
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4.6
THE QUARTER-WAVE TRANSFORMER
Refer to the section entitled “The quarter-wave transformer” in the prescribed textbook or
to information about this topic in any recommended study materials for a detailed
discussion of this concept.
The quarter-wave transformer is a simple circuit for matching a real load impedance to a
transmission line. Quarter-wave transformers can be extended to multisection designs to
provide broader bandwidth. One drawback of quarter-wave transformers, however, is
that they can only match a real load impedance. For a complex load impedance, a
transformation into a real impedance is required, which is achieved through the use of
an appropriate length of transmission line between the load and the transformer, or an
appropriate series or shunt reactive element. These techniques will usually alter the
frequency dependence of the load, and this often has the effect of reducing the bandwidth
of the match.
If you have internet access, please watch the following online video presentation for a
clear explanation by Prof Niraj Chennai on how to use the quarter-wave transformer to
perform impedance matching in high frequency electronics:
https://www.youtube.com/watch?v=3JOtWxpUtbI
Here is an activity that will give you some practice with regard to quarter-wave
transformers:
ACTIVITY 4.5 QUARTER-WAVE TRANSFOMER BANDWIDTH
Design a single-section quarter-wave matching transformer to match a 10 Ω load to a 50
Ω transmission line at f 0 = 3 GHz . Determine the percentage bandwidth for which the
SWR ≤1.5 .
SOLUTION
The characteristic impedance of the matching section is
=
Z1
Z=
0Z L
10 )
( 50 )(=
22.36 Ω ,
and the length of the matching section is λ / 4 at 3 GHz (the physical length depends on
the dielectric constant of the line). An SWR of 1.5 corresponds to a reflection coefficient
magnitude of
=
Γm
SWR − 1 1.5 − 1
=
= 0.2 .
SWR + 1 1.5 + 1
The fractional bandwidth is computed as follows:
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 Γ
2 Z0Z L 
∇f
4
m
=
=
2 − cos −1 
0.29 or 29%.
f0
π
 1 − Γ m2 Z L − Z 0 
4.7
THE THEORY OF SMALL REFLECTIONS
Refer to the section “The theory of small reflections” in the prescribed textbook or
information on this topic in any of the recommended study materials for a detailed
discussion of this concept.
A quarter-wave transformer provides a simple means of matching any load with a real
impedance with the impedance of any transmission line. However, when a load is being
matched with a transmission line and more bandwidth is required than a single quarterwave section can provide, multisection quarter-wave transformers may be used.
If you have internet access, have a look at the online courses presented by infocobuild at
the following link:
http://www.infocobuild.com/education/audio-videocourses/electronics/MicrowaveIntegratedCircuits-IIT-Bombay/lecture-08.html
After that, study the theory of small reflections classified as single-section and multisection
transformers as presented in the prescribed textbook.
Single-section transformer: This involves a derivation of an approximate expression for
the overall reflection coefficient, Γ , for the single-section matching transformer shown in
Figure 5.13 in the prescribed textbook. The total reflection as an infinite sum of partial
reflections and transmissions is shown to be
Γ + Γ3e −2 jθ
Γ= 1
1 + Γ1Γ3e −2 jθ
(4.1)
If the discontinuities between the impedance Z1 , Z 2 and Z 2 , Z L are small, then
Γ1Γ3 =1 , so the above expression can be approximated as
Γ = Γ1 + Γ3 𝑒𝑒 −2𝑗𝑗𝑗𝑗
(4.2)
This result expresses the intuitive idea that the total reflection is dominated by the
reflection from the initial discontinuity between Z1 and Z 2 ( Γ1 ) and the first reflection
from the discontinuity between Z 2 and Z L ( Γ3e −2 jθ ) .
Multisection transformer: The results from the derivation of an approximate expression
for the total reflection coefficient Γ are given as:
Γ (θ ) = Γ 0 + Γ1e −2 jθ + Γ 2e −4 jθ + ... + Γ N e −2 jNθ
(4.3)
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The importance of this result lies in the fact that we can synthesise any desired reflection
coefficient response as a function of frequency (θ ) by properly choosing the Γ n and using
enough sections ( N ) . This should be clear from the realisation that a Fourier series can
approximate an arbitrary smooth function if enough terms are used.
4.8
BINOMIAL MULTISECTION MATCHING TRANSFORMERS
Refer to the section entitled “Binomial multisection matching transformers” in the
prescribed textbook or information about this topic in any of the recommended study
materials for a detailed discussion of this concept.
Watch the online video courses by infocobuild again to reinforce your understanding of
the concept of binomial multisection matching transformers:
http://www.infocobuild.com/education/audio-videocourses/electronics/MicrowaveIntegratedCircuits-IIT-Bombay/lecture-08.html
Note that the passband response of a binomial matching transformer is optimum in the
sense that the response is as flat as possible near the design frequency for a given number
of sections. This is the maximally flat response, and it is determined for an N-section
transformer by setting the first N-1 derivatives of Γ (θ ) to zero at the centre frequency,
f 0 . This type of response is generated using a reflection coefficient, as you can see below:
Γ (θ ) = A (1 + e −2 jθ )
N
(4.4)
This means that the reflection coefficient magnitude is
=
Γ (θ ) A e − jθ
= 2 N A cos θ
N
e jθ + e − jθ
N
N
(4.5)
To put your knowledge into action, do the following learning activity:
ACTIVITY 4.6 BINOMIAL TRANSFORMER DESIGN
You are required to design a three-section binomial transformer to match a 50 Ω load to
a 100 Ω line and calculate the bandwidth for Γ m =
0.05 . Plot the reflection coefficient
magnitude versus normalised frequency for the exact designs using 1, 2, 3, 4, and 5
sections.
SOLUTION
For N=3, Z=
50 Ω and =
Z 0 100 Ω we have
L
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𝑍𝑍 −𝑍𝑍
1
𝑍𝑍
𝐴𝐴 = 2−𝑁𝑁 𝑍𝑍𝐿𝐿 +𝑍𝑍0 = 2𝑁𝑁+1 𝑙𝑙𝑙𝑙 𝑍𝑍𝐿𝐿 = −0.0433.
𝐿𝐿
0
The bandwidth is computed as follows:
0
1/ N
 
 
∇f
4
−1 1 Γ m
=
2 − cos  
0.70 or 70% .
 =
f0
π
 2  A  


The necessary binomial coefficients are:
=
C03
3!
= 1
3!0!
=
C13
3!
= 3,
2!1!
=
C23
3!
= 3.
1!2!
We use the expression 𝑙𝑙𝑙𝑙
𝑍𝑍𝑛𝑛+1
𝑍𝑍𝑛𝑛
𝑍𝑍
= 2−𝑁𝑁 𝐶𝐶𝑛𝑛𝑁𝑁 𝑙𝑙𝑙𝑙 𝑍𝑍𝐿𝐿, which can be used to find Z n+1 , starting with
0
n = 0 . Therefore, substituting values in this expression, we are able to find the respective
required characteristic impedances as
=
Z1 91.7 Ω ,=
Z 2 70.7 Ω , and =
Z 3 54.5 Ω with n = 0 , n = 1 and n = 2 .
4.9
CHEBYSHEV MULTISECTION MATCHING TRANSFORMERS
Refer to the section entitled “Chebyshev multisection matching transformers” in the
prescribed textbook for a detailed discussion of this concept.
The Chebyshev matching transformer optimises bandwidth at the expense of passband
ripple in a different way from what the binomial transformer does. If the flatness of the
passband response is compensated for by the matching transformer, the bandwidth
produced is better than that of the binomial transformer for a given number of sections.
The design of the Chebyshev transformer entails equating Γ (θ ) to a Chebyshev
polynomial, which has the optimum characteristics needed for this type of transformer.
The Chebyshev polynomials for an nth order is a polynomial of degree n, denoted by
Tn ( x ) The first four Chebyshev polynomials are
T1 ( x ) = x
(4.6)
T2 (=
x ) 2x2 −1
(4.7)
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T3 (=
x ) 4 x3 − 3x
(4.8)
T4 ( x ) = 8 x 4 − 8 x 2 + 1
(4.9)
Higher order polynomials can be found using the following recurrence formula:
=
Tn ( x ) 2 xTn−1 ( x ) − Tn−2 ( x )
(4.10)
ACTIVITY 4.7 CHEBYSHEV TRANSFORMER DESIGN
You are required to design a three-section Chebyshev transformer to match a 100 Ω load
to a 50 Ω line with Γ m =
0.05 , using the above theory. Plot the reflection coefficient
magnitude versus normalised frequency for exact designs using 1, 2, 3 and 4 sections.
SOLUTION
The reflection for N=3 is
=
Γ (θ ) 2e − j 3θ ( Γ 0 cos3θ + Γ1 cos
=
θ ) Ae − j 3θ T3 ( secθ m cos θ )
1

0
−1  ln Z L / Z 0 
Where secθ m cosh
=
=
 cosh 
  1.408 . Thus, θ m = 44.7 . From symmetry we
Γ
N
2
m



also have that
Γ1 =0.1037 ,
Γ3 =Γ 0 =0.0698 ,
Γ 2 =Γ1 =0.1037 .
Then the characteristic impedances are:
=
n 0,=
Z1 57.5 Ω ,
=
n 1,=
Z 2 70.7 Ω ,
=
n 2,=
Z 3 87.0 Ω .
The bandwidth is calculated as follows:
 44.7 0 
∇f
4θ
=
=
2− m =
2 − 4
1.01
0 
π
f0
 180 
or 101%. This is significantly greater than the bandwidth of the binomial transformer of
70% as we calculated it in the previous learning activity, which involved the same
impedance mismatch.
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4.10 TAPERED LINES
Refer to the section “Tapered lines” in the prescribed textbook and/or information about
this topic in any of the recommended study materials.
Please start by watching the following video clip by infocobuild to learn more about
tapered lines and how to match their impedances with the impedance of load, and then
consult the prescribed textbook for more information:
http://www.infocobuild.com/education/audio-videocourses/electronics/MicrowaveIntegratedCircuits-IIT-Bombay/lecture-10.html
By now you are familiar with the process for matching an arbitrary real load impedance
to a line over a desired bandwidth by using multisection matching transformers. You
should have noted that as the number N of discrete transformer sections increases, the
step changes in characteristics impedance between the sections becomes smaller, and
the transformer geometry approaches a continuously tapered line. In practice, any
matching transformer should be of finite length – often not more than a few sections. This
provides the reason for having a continuously tapered transformer instead of considering
discrete sections. Now consider the continuously tapered line in Figure 5.18a in the
prescribed textbook, which is assumed to be made up of a number of incremental sections
of length ∆z , with an impedance change ∆Z ( z ) from one section to the next, as shown
in Figure 5.18b in the prescribed textbook. Thus, the incremental reflection coefficient
from the impedance step at z is given by
(𝑍𝑍+Δ𝑍𝑍)−𝑍𝑍
ΔΓ = (𝑍𝑍+Δ𝑍𝑍)+𝑍𝑍 =
Δ𝑍𝑍
(4.11)
2𝑍𝑍
For an exponential taper, we find the reflection coefficient to be
Γ=
ln Z L / Z 0 − jβ L  sin β L 
e


2
 βL 
(4.12)
This result assumes that β , the propagation constant of the tapered line, is not a
function of z – an assumption generally acceptable only for TEM lines.
For a triangular taper, we find the reflection coefficient given as follows:
1 − jβ L  Z L   sin ( β L / 2 ) 
ln 
e
Γ (θ ) =


2
 Z0   β L / 2 
2
(4.13)
The magnitude of this result is illustrated in Figure 5.20b in the prescribed textbook.
Notice that for β L > 2π , the peaks of the triangular taper are lower than the
corresponding peaks of the exponential taper. However, the first null for the triangular
taper occurs at β L = 2π , whereas for the exponential taper it occurs at β L = π .
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In the case of the Klopfenstein taper, one looks for the best design, considering that there
is an infinite number of possibilities for choosing an impedance matching taper. The
Klopfenstein impedance taper provides the optimum solution such that the reflection
coefficient is minimum over the passband. Therefore, if the passband is defined as
β L ≥ A , then the maximum ripple in the passband is as follows:
Γ
Γm = 0
cosh A
(4.14)
where Γ (θ ) oscillates between ±Γ 0 / cosh A for β L > A .
ACTIVITY 4.8 DESIGN OF TAPERED MATCHING SECTIONS
Suppose that you are undertaking your work-integrated learning (WIL) at a well-known
high frequency electronics consulting firm, which is widely represented across a number
of provinces in South Africa. One of the firm’s clients has asked the firm to design a
triangular taper, an exponential taper and a Klopfenstein taper (with Γ m =
0.02 ) to match
a 50 Ω load to a 100 Ω line. Demonstrate how you would solve this design problem,
since you are part of the technical team. Plot the impedance variations and resulting
reflection coefficient magnitudes versus β L .
SOLUTION
Triangular taper: The impedance variation is
2 z / L ln Z / Z

e( ) L 0

Z ( z ) = Z0 
4 z / L − 2 z 2 / L2 −1) ln Z L / Z 0
e(
2
for 0≤ z ≤ L /2
for L /2≤ z ≤ L
With Z
=
100 Ω and Z=
50 Ω . The resulting reflection coefficient response can be
0
L
obtained using the equation (4.13) above.
Exponential taper: The impedance variation is
=
Z ( z ) Z 0e az for 0 < z < L
=
=
L ) ln Z L / Z 0 0.693 / L . The resulting reflection coefficient response can be
With a (1/
obtained using the equation (4.12) above.
Klopfenstein taper: Γ 0 is given as follows:
=
Γ0
and the value of A is
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1  ZL 
ln
=
  0.346
2  Z0 
−1  Γ 0 
−1  0.346 
=
=
A cosh

 cosh=

 3.543
 0.02 
 Γm 
Thus, the reflection coefficient magnitude is obtained using the equation (4.14) above.
4.11 THE BODE-FANO CRITERION
Refer to the section entitled “The Bode-Fano criterion” in the prescribed textbook for
more insights into this concept.
This is a qualitative concept that offers theoretical limits constraining the performance of
an impedance matching network. The Bode-Fano criterion provides a theoretical limit to
the minimum reflection coefficient magnitude that is obtainable with an arbitrarily
matching network for certain canonical types of load impedance. This criterion represents
an optimum result that is ideally achievable, even though it may only be approximated
in practice. In the case of the lossless network shown in Figure 5.22a in the prescribed
textbook, we see how a network of this type is used to match a parallel RC load
impedance. The Bode-Fano criterion states that
∫
∞
0
ln
1
Γ (ω )
dω ≤
π
RC
(4.15)
where Γ (ω ) is the reflection coefficient determined as a function of angular frequency ω
of the signal, seen looking into the arbitrary lossless matching network. R and C
represent the load resistance and capacitance, respectively. The implications of this result
are discussed in the corresponding sections of the prescribed textbook. Please read these
carefully.
ACTIVITY 4.9 HF IMPEDANCE MATCHING AND TUNING
1. Refer to the chapter “Impedance Matching and Tuning” in the prescribed textbook
and do the following selected unsolved problems: 5.1–5.11, 4.16–5.25. Your module
lecturer will post the correct answers without including the full and detailed solutions
on the module website for your reference.
2. How would you rate your own competence in terms of the applied engineering
problem-solving skills required to achieve the outcomes of this learning unit? Do you
feel confident that you have the necessary knowledge and skills to proceed? If not, be
sure to revise the required skills or to discuss difficult concepts and questions with
your fellow students on myUnisa, or contact the lecturer for clarification.
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4.12 CONCLUSION
After studying this unit, you should be able to design impedance matching and tuning of
HF electronic networks. In the next unit, we will cover the design of HF electronic systems
such as resonators, filters, oscillators and mixers.
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5
Learning unit 1
HF RESONATORS, FILTERS, OSCILLATORS AND MIXERS
5.1
LEARNING OUTCOMES
By the end of this learning unit, you should be able to design HF electronic systems such
as resonators, filters, oscillators and mixers.
In particular, you should be able to do the following:
•
•
•
•
Provide the similarities between the lumped-elements resonators of circuit theory and
microwave resonators.
Derive from first principles the quality factor, Q, for RLC series and parallel resonant
circuits.
Design:
o rectangular cavity resonators.
o gap-coupled microstrip resonators.
o low-pass composite filters.
o bandpass filters.
o Colpitts oscillators.
o transistor oscillators.
Analyse the significance of image frequency in mixer circuits.
5.2
INTRODUCTION
Reading: To complete this unit, study the chapter entitled “Resonators, Filters, Oscillators
and Mixers” in your prescribed textbook or the same topic in any of the recommended
study materials.
Before you study this unit, make sure you are able to design an impedance matching
network placed between a load impedance and a transmission line such that the
impedance seen looking into the matching network is the same as that of the characteristic
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impedance, Z 0 . You should have learnt how to do this in the preceding unit, which
discussed the design of impedance matching and tuning in high frequency electronics. In
this unit, you will analyse and design high frequency resonators, filters, oscillators and
mixers.
Please begin by watching the following video clips, which will introduce you to important
concepts relating to high frequency resonators, filters, oscillators and mixers.
For basic concepts relating to resonators, follow the link below:
https://slideplayer.com/slide/7467873/
For the typical application of resonators, watch the clip by Electronic Specifier available
at:
https://wireless.electronicspecifier.com/5g/dielectric-resonator-antennas-for-5gapplications.
For online lectures about filters, watch the following online videos by Coursera, available
at:
https://www.coursera.org/lecture/linear-circuits-ac-analysis/3-2-lowpass-highpass-filterdesign-soetc.
https://www.coursera.org/lecture/linear-circuits-ac-analysis/3-1-lowpass-and-highpassfilters-OCuSG.
You will find more information about high pass filters in the following online video by
All About Electronics:
https://www.youtube.com/watch?v=9Dx0b0ukNAM
To learn more about oscillators and mixers, watch the following videos:
•
RC High Pass Filter Explained, by All About Electronics, at:
https://www.youtube.com/watch?v=aJAZHPqEUKU
High Frequency Oscillators, by jdflyback, at:
https://www.youtube.com/watch?v=U4co0cC_oVc
•
I hope that watching these video clips will have inspired you to read further about the
relevant concepts in the prescribed textbook.
5.3
RESONATORS
For detailed information about this topic, please refer to the section entitled “Resonators”
in the prescribed textbook. This unit will serve merely as a guideline to highlight certain
aspects, and you need to undertake your own deeper and active learning about the
concepts referred to.
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High frequency (HF) resonators are often used in designing filters, oscillators, frequency
meters and tuned amplifiers. Operations of HF resonators are comparable to those of the
lumped-element resonators in circuit theory, that is, series and parallel RLC resonant
circuits. In this section we will talk about the implementation of resonators at high
frequencies using distributed elements such as transmission lines, rectangular and circular
waveguides and dielectric cavities.
5.3.1
Series and parallel resonant circuits
When frequencies are close to the resonant frequency, the resonator is modelled using a
series or parallel RLC lumped-element equivalent circuit, as you will see from the
following analysis.
For a series RLC resonant circuit
The input impedance is given as
Z in =+
R jω L − j
1
ωC
(5.1)
From the above expression, the complex power delivered to the resonant circuit is
P=
in
1 2
1 
I  R + jω L − j
2 
ωC 
(5.2)
while the power dissipated by the resistor R is
Ploss =
1 2
I R
2
(5.3)
On the other hand, the average magnetic energy stored in the inductor L is
Wm =
1 2
I L
4
(5.4)
The average electric energy stored in the capacitor C is
=
We
1 2
1 2 1
=
Vc C
I
4
4
ω 2C
(5.5)
where Vc is the average voltage across the capacitor. The complex power can be
rewritten as
Pin =+
Ploss 2 jω (Wm − We )
(5.6)
and the input impedance can be rewritten as
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2 Pin Ploss + 2 jω (Wm − We )
=
2
1 2
I
I
2
Z in
=
(5.7)
Resonance occurs when the averaged stored magnetic and electric energies are equal,
or Wm = We . Then the resonant frequency, ω0 , is defined as
ω0 =
1
LC
(5.8)
The quality (Q) factor of the resonant circuit is defined as
Q =ω
=ω
average energy stored
energy loss / sec ond
Wm + We
Ploss
(5.9)
The Q factor measures the loss of a resonant circuit, with lower loss implying a higher Q.
Resonator losses may be due to conductor loss, dielectric loss or radiation loss, and are
represented by the resistance R of the equivalent circuit.
For a parallel resonant circuit
The complex power can be defined as
Pin =+
Ploss 2 jω (Wm − We )
(5.10)
The input impedance can also be expressed as
Z in
=
2 Pin Ploss + 2 jω (Wm − We )
=
2
1 2
I
I
2
(5.11)
Resonance occurs when the average stored magnetic and electric energies are equal, or
Wm = We . Then the resonant frequency, ω0 , is defined as
ω0 =
1
LC
(5.12)
If you have access to the internet, please watch the videos by Eugene Khutoryansky at the
following link for information about the foundational principles of resonance:
https://www.youtube.com/watch?v=Mq-PF1vo9QA
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ACTIVITY 5.1 SERIES AND PARALLEL RESONANT CIRCUITS
Revise the information about RLC series and parallel resonant circuits in the prescribed
textbook and derive from first principles the quality factor (Q) expression. Show all the
steps of your derivation.
5.3.2
Transmission line resonators
Ideal lumped circuit elements are often non-attainable at microwave frequencies, and so
distributed elements are frequently used. This section discusses the use of transmission
line sections of various lengths and terminations (usually open- or short-circuited) to form
resonators.
For a short-circuited λ / 2 line where a length of lossy transmission line is short circuited
at one end, we can represent the line as having a characteristics impedance, Z 0 ,
propagation constant, β , and attenuation constant, α . At the resonant frequency ω = ω0
, the length of the line is  = λ / 2 . The input impedance is then computed as follows:
=
Z in Z 0 tanh (α + j β ) 
(5.13)
This is expanded as follows:
Z in = Z 0
tanh α  + j tan β 
1 + j tan β  tanh α 
(5.14)
Doing the activity below will reinforce your understanding of the concept of the Q factor
of half-wave coaxial line resonators:
ACTIVITY 5.2 Q OF HALF-WAVE COAXIAL LINE RESONATORS
Consider a λ / 2 resonator which is made from a piece of copper coaxial line having an
inner conductor radius of 1 mm and an outer conductor radius of 4 mm. If the resonant
frequency is 5 GHz, compare the unloaded Q of an air-filled coaxial line resonator with
that of a Teflon-filled coaxial line resonator.
SOLUTION
The surface resistivity at 5 GHz using the conductivity of copper,
=
σ 5.813 ×107 S/m ,
Rs =
ωµ0
= 1.84 ×10−2 Ω
2σ
The attenuation due to conductor loss for the air-filled line is
=
αc
Rs
1 1
 + 
2η ln a / b  a b 
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1.84 ×10−2
1 
 1
=
=
+

 0.022 Np/m
2 ( 377 ) ln ( 0.004 / 0.001)  0.001 0.004 
For Teflon, ε r = 2.08 and tan δ = 0.0004 , so the attenuation due to conductor loss for the
Teflon-filled line is
αc
1.84 ×10−2 2.08
1 
 1
=
+

 0.032 Np/m
2 ( 377 ) ln ( 0.004 / 0.001)  0.001 0.004 
The dielectric loss of the air-filled line is zero, but the dielectric loss of the Teflon-filled
line is
α d = k0
εr
2
tan δ
2.08 ( 0.0004 )
= 0.030 Np/m
2
(104.7 )
Finally, the unloaded Qs can be computed as
Q=
air
QTeflon
=
β
104.7
=
= 2380
2α 2 ( 0.022 )
β
104.7 2.08
=
= 1218
2α 2 ( 0.032 + 0.030 )
Thus, the Q of the air-filled line is much higher than that of the Teflon-filled line. The Q
can further be increased by using silver-plated conductors.
For a short-circuited λ / 4 line, the input impedance of a shorted line of length  is
Z in = Z 0
1 − j tanh α  cot β 
tanh α  − j cot β 
(5.15)
where this result was obtained by multiplying both numerator and denominator by
− j cot β  . The resistance of the equivalent circuit is
R=
Z0
α
(5.16)
and the capacitance of the equivalent circuit is
C=
66
π
4ω0 Z 0
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(5.17)
The inductance of the equivalent circuit can be defined as
L=
1
ω02C
(5.18)
The unloaded Q of the resonator is
=
Q0 ω0=
RC
π
β
=
4α  2α
(5.19)
Since  = π / 2 β at resonance.
For an open-circuited λ / 2 line, in microstrip circuits, the resonator often used consists
of an open-circuited length of transmission line. This resonator will behave as a parallel
resonant circuit when the length is λ / 2 or multiples of λ / 2 . Thus, the input
impedance of an open-circuited lossy transmission line of length  is
Z in = Z 0
1 + j tan β  tanh α 
tanh α  + j tan β 
(5.20)
In comparison with the input impedance of a parallel resonant circuit, the resistance of
the equivalent RLC circuit is
R=
Z0
α
(5.21)
and the capacitance of the equivalent circuit is
C=
π
2ω0 Z 0
(5.22)
The inductance of the equivalent circuit is
L=
1
ω02C
(5.23)
The unloaded Q is
=
Q0 ω0=
RC
π
β
=
2α  2α
(5.24)
since  = π / β at resonance.
Use the information given above and the additional detail given in the prescribed
textbook and design the half-wave microstrip resonator in the following activity.
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ACTIVITY 5.3 HALF-WAVE MICROSTRIP RESONATOR
Consider a microstrip resonator constructed from a λ / 2 length of 50 Ω open-circuited
=
=
tan δ 0.0004 ) , with a thickness of 0.159
microstrip line. The substrate is Teflon
(ε r 2.08,
cm, and the conductors are copper. Compute the required length of the line for resonance
at 5 GHz, and the unloaded Q of the resonator. Ignore fringing fields at the end of the
line.
SOLUTION
If we take the width of a 50 Ω microstrip line on this substrate to be w = 0.508 cm and
the effective permittivity is ε e = 1.80 , the resonant length can then be calculated as
=
υp
c
3 × 108
=
=
=
= 2.24 cm
2 2 f 2 f ε e 2 ( 5 ×109 ) 1.80
λ
The propagation constant is
9
2π f 2π ( 5 ×10 ) 1.80
β =
=
= 151.0 rad/m
3 ×108
υp
Thus the attenuation due to conductor loss is
=
αc
Rs
1.84 ×10−2
=
= 0.0724 Np/m
Z 0 w 50 ( 0.00508 )
where the attenuation due to a dielectric loss is
=
αd
k0ε r ( ε e − 1) tan δ
= 0.024 Np/m .
2 ε e ( ε r − 1)
Then the unloaded Q is
Q
=
0
5.3.3
β
= 783
2α
Waveguide cavity resonators
High frequency or microwave resonators can also be constructed from closed sections of
waveguides. The radiation loss from an open-ended waveguide can be significant.
Waveguide resonators are usually short circuited at both ends to form cavities. Electric
and magnetic energy is stored within the cavity enclosure, and power is dissipated in the
metallic walls of the cavity as well as in the dielectric material that may fill the cavity.
The cavity resonator can be coupled by using a small aperture, or a small probe or a loop.
In this section you will learn how to derive an expression for the unloaded Q of the TE10 
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mode. Furthermore, the same procedure can be followed to compute the unloaded Q for
arbitrary TE and TM modes.
Resonant frequencies
The resonant frequencies of a rectangular cavity, under the assumption that the cavity is
lossless, for the TEmn or TM mn mode is given by
f mn=

ckmn
c
=
2π µrε r 2π µrε r
2
2
 mπ   nπ   π 

 +
 +

 a   b   d 
2
(5.25)
If b < a < d , the dominant resonant mode (lowest resonant frequency) will be the TE101
mode, corresponding to the TE10 dominant waveguide mode in a shorted guide of length
λg / 2 and similar to the short-circuited λ / 2 transmission line resonator. The dominant
TM resonant mode is the TM 110 mode.
Unloaded Q of the TE10  mode
The unloaded Q of this mode is computed by finding the stored electric and magnetic
energies, and the power lost in the conducting walls and the dielectric filling. The stored
electric energy is
We =
ε abd
16
E02
(5.26)
while the stored magnetic energy is
=
Wm
 1
π2 
E  2 + 2 2 2
16
 ZTE k η a 
µ abd
2
0
(5.27)
The power lost in conducting walls is given by
Pc =
Rs
2
∫
2
H t ds
(5.28)
walls
where Rs = ωµ0 / 2σ is the surface resistivity of the metallic walls, and H t is the
tangential magnitude field at the surface of the walls. Thus,
Rs E02λ 2  ab bd  2 a d 
=
+
+
+ 
Pc

8η 2  d 2 a 2 2d 2a 
(5.29)
where the symmetry of the cavity is used in doubling the contributions from the walls at
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x = 0 , y = 0 , and z = 0 to account for the contributions from the walls at x = a , y = b ,
and z = d respectively. When both wall losses and dielectric losses are present, the
total power loss is Pc + Pd , where
abdωε '' E0
Pd =
8
2
(5.30)
The total unloaded Q is then found to be
 1
1 
=
Q0  +

 Qc Qd 
−1
(5.31)
Please watch the video presentation by Simarjeet Saini for a further explanation of
rectangular cavity resonator analysis and design, available at this link:
https://www.youtube.com/watch?v=YCKI31Ra1Bo
After that, you will design a rectangular cavity resonator by doing the following learning
activity.
ACTIVITY 5.4 DESIGN OF A RECTANGULAR CAVITY RESONATOR
A rectangular waveguide cavity is made from a piece of copper WR-187 H-band
waveguide with a = 4.755 cm and b = 2.215 cm. The cavity is filled with polyethylene
=
=
tan δ 0.0004 ) . If resonance is to occur at f = 5 GHz , find the required length,
(ε r 2.25,
d, and the resulting unloaded Q for the  = 1 and  = 2 resonant modes.
SOLUTION
The wave number k is
k
=
2π f ε r
= 157.08 m -1
c
m 1,=
n 0)
The required length for resonance can be found as (=
d=
For  = 1 , d = 2.20 cm
For  = 2 , d = 4.40 cm
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π
k 2 − (π / a )
2
The intrinsic impedance is
=
η
377
= 251.3 Ω
εr
The Q due to conductor loss only is
For  = 1 , Qc = 8, 403
For  = 2 , Qc = 11,898
The Q due to the dielectric loss only is, for both  = 1 and  = 2 ,
Qd
=
1
1
=
= 2500
tan δ 0.0004
Then, the total unloaded Qs are
−1
1 
 1
For  = 1 , Q0 =
1927
+

 =
 8403 2500 
−1
1 
 1
+
For  = 2 , Q0 = 
 = 2065
 11,898 2500 
You will notice that the dielectric loss has a dominant effect on the Q; higher Q could be
obtained using an airfield cavity.
5.3.4
Dielectric resonators
Dielectric resonators typically use materials with low loss and a high dielectric constant,
ensuring that most of the fields are contained within the dielectric resonators. Unlike the
metallic cavities, there is some field fringing or leakage from the sides and ends of a
dielectric resonator (which are not metallised), leading to a small radiation loss and
consequent lowering of Q. A dielectric resonator is generally smaller in size and weight
and is less expensive than an equivalent metallic cavity.
Resonant frequencies of TE01δ mode
The unloaded Q of the resonator can be calculated by determining the stored energy
(inside and outside the dielectric cylinder), and the power dissipated in the dielectric
resonator and possibly lost to radiation. If the latter is small, the unloaded Q can be
approximated as 1/ tan δ , as in the case of the metallic cavity resonators.
Basing yourself on the above discussions, do the related learning activity below to further
understand the concepts and applications presented in this subsection.
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ACTIVITY 5.5 RESONANT FREQUENCY AND Q OF A DIELECTRIC RESONATOR
Determine the resonant frequency and approximate unloaded Q for the TE01δ mode of a
dielectric resonator made of titanium, with ε r = 95 and tan δ = 0.001 . The resonator
dimensions are a = 0.413 cm and L = 0.8255 cm.
SOLUTION
2
2
2π f
=
β
ε r k02 − ( 2.405 / a ) , because α and
Given that k=
,α
( 2.405 / a ) − k02 and
0 =
c
β must both be real, the possible frequency range is from f1 to f 2 , where
f1
=
ck0 c ( 2.405 )
=
= 2.853 GHz
2π
2π ε r a
f2
=
ck0 c ( 2.405 )
=
= 27.804 GHz
2π
2π a
Using the interval-halving method to find the root of the above equation gives a resonant
frequency of about 3.152 GHz. This compares with a measured value of about 3.4 GHz,
indicating a 10% error. The approximate unloaded Q, due to dielectric loss, is
=
Qd
5.3.5
1
= 1000 .
tan δ
Excitation of resonators
Resonators are not useful unless they are coupled to external circuitry. Resonators can be
coupled to transmission lines and waveguides. Doing the following learning activity will
help you understand the concept of excitation of resonators.
ACTIVITY 5.6 DESIGN OF A GAP-COUPLED MICROSTRIP RESONATOR
Consider a resonator made from an open-circuited 50 ohm microstrip line gap coupled
to a 50 ohm feedline, as in Figure 6.13a in the prescribed textbook. The resonator has a
length of 2.175 cm, an effective dielectric constant of 1.9 and an attenuation of 0.01
dB/cm near its resonance. Find the value of the coupling capacitor required for critical
coupling, and the resulting resonant frequency.
SOLUTION
The first resonant frequency will occur when the resonator is about  = λg / 2 in length.
Ignoring fringing fields, we find the approximate resonant frequency to be
f 0 υ=
=
p / λg
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c
= 5.00 GHz
2 ε e
This result does not include the effect of the coupling capacitor. The unloaded Q of this
resonator is determined by
=
Q
0
β
π
π
= =
= 628
2α λgα 2α
The normalised coupling capacitor susceptance is
=
bc
π
= 0.05
2Q0
Consequently, the coupling capacitor has a value given by
C
=
0.05
bc
=
= 0.032 pF
ω Z 0 2π ( 5 ×109 ) ( 50 )
which should provide critical coupling of the resonator to the 50 ohm feedline.
ACTIVITY 5.7 UNSOLVED RESONATOR-RELATED PROBLEMS
Do the following problems from the prescribed textbook:
Problems 6.1–6.10 and all the other odd-numbered problems in chapter 6. Your module
lecturer will post the sample solutions on the module website for your reference.
Remember that you can also use the Discussions space on the module website to direct
queries to your lecturer.
5.4
FILTERS
Refer to the section “Filters” in the prescribed textbook or information on the same topic
in any of the recommended study materials for detailed discussions.
Filters are defined as two-port network circuits for controlling the frequency response at
a certain point in an RF or microwave system. Filters provide transmission at frequencies
within the passband of the filter and attenuation in the stopband of the filter. Frequency
responses may be classified as low-pass, high-pass, bandpass and band-reject
characteristics. Filters find many applications, such as in the design of RF or microwave
communication, radar, or test and measurement systems.
5.4.1
Periodic structures
These are infinite transmission lines or waveguides periodically loaded with reactive
elements. The loading elements form discontinuities in the lines, but are modelled as
lumped reactance in shunt or series on transmission lines. When studying the passband
and stopband characteristics of a periodic structure, it is useful to plot the propagation
constant β , versus the propagation constant of the unloaded line, k ( or ω ) . Such a graph
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is called a k − β diagram, or Brillouin diagram (refer to Figure 8.4 in the prescribed
textbook). This diagram can be used to study the dispersion characteristics of many types
of microwave components and transmissions lines.
The dispersion relation for a waveguide mode is written as
=
β
k 2 − kc2 (5.32)
where kc is the cut-off wave number of the mode, k is the free-space wave number and
β is the propagation constant of the mode. The k − β diagram is also useful for
interpreting the various wave velocities associated with a dispersive structure. The phase
velocity is
υ=
p
ω
k
= c (5.33)
β
β
which is seen to be equal to the speed of light times the slope of the line from the origin
to the operating point on the k − β diagram. The group velocity is
υg
=
dω
dk
= c
(5.34)
dβ
dβ
which is the slope of the k − β curve at the operating point. Do the following activity for
an analysis of a periodic structure.
ACTIVITY 5.8 ANALYSIS OF A PERIODIC STRUCTURE
Suppose you have a periodic capacitively loaded line as shown below.
50 Ω , d = 1.0 cm, and C0 = 2.666 pF, sketch the k − β diagram and compute the
If Z=
0
propagation constant, phase velocity and Bloch impedance at f = 3.0 GHz. Assume
k = k0 .
SOLUTION
We can rewrite the dispersion relation as
C Z c
=
cos
β d cos k0 d −  0 0  k0 d sin k0 d
 2d 
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Then
C0 Z 0c
=
2d
2.666 ×10 ) ( 50 ) ( 3 ×10 )
(=
−12
8
2 ( 0.01)
2.0
so we have
cos
β d cos k0 d − 2k0 d sin k0 d
=
At 3.0 GHz, we have
=
k0 d
2π ( 3 ×109 )
8
3 ×10
0.01)
(=
0.6283
= 360
So β d = 1.5 and the propagation constant is β = 150 rad / m . The phase velocity is
υ p = 0.42c
The Bloch impedance is given as follows:
b ωC0 Z 0
= =
1.256
2
2
=
θ k=
360
0d
b
A =cos θ − sin θ =0.0707
2
b
b

=
B j  sin θ + cos θ − =
 j 0.3479
2
2

Then
=
ZB
5.4.2
BZ 0
=
A2 − 1
( j 0.3479 )( 50=
)
2
j 1 − ( 0.0707 )
17.4 Ω
Filter design by the image perimeter method
This method involves the specification of passband and stopband characteristics for a
cascade or simple two-port network. This design approach is simple to apply, and how it
works can be best discussed on the basis of the following learning activity. However, first
watch the following video clip presentation by Greg Lewin on the basics of low-pass filter
design.
https://www.youtube.com/watch?v=WZN42JO_kFo
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ACTIVITY 5.9 LOW-PASS COMPOSITE FILTER DESIGN
At your workplace, you are required to design a low-pass composite filter with a cut-off
frequency of 2 MHz and impedance of 75 ohms. Place the infinite attenuation pole at
2.05 MHz and plot the frequency response from 0 to 4 MHz.
SOLUTION
Use Table 8.2 in the prescribed textbook for all component values and then compute the
inductance and capacitance as follows:
L
=
2 R0
,C
= 11.94 µ H=
ωc
2
= 2.122 nF
R0ωc
For the m-derived sharp cut-off section
2
 f 
1 − m2
mL
L = 12.94 µ H
,
,
m=
1−  c  =
0.2195 ,
= 1.310 µ H mC = 465.8 pF
4
m
f
2
 ∞
For the m = 0.6 matching sections
1 − m2
mL
mC
L = 6.368 µ H
= 636.5 pF ,
= 3.582 µ H ,
2m
2
2
The completed figure design is shown below:
5.4.3
Filter design by the insertion loss method
A perfect filter should have zero insertion loss in the passband, infinite attenuation in the
stopband, and a linear phase response (to avoid signal distortion) in the passband. The
image parameter method may yield a usable filter response for some applications, but
there is no methodical way of improving the design. The insertion loss method allows a
high degree of control over the passband and stopband amplitude and phase
characteristics, with a systematic way to synthesise a desired response. In the insertion
loss method a filter response is defined by its insertion loss or power loss ratio, defined
as follows:
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PLR
=
Power available from source Pinc
1
= =
Power delivered toload
Pload 1 − Γ (ω ) 2
(5.35)
2
This quantity is the reciprocal of S12 if both load and source are matched. The insertion
loss (IL) in dB is
IL = 10log PLR
(5.36)
Please refer to the section “Insertion loss method” for further analysis of maximally flat,
equal ripple and linear phase low-pass filter prototypes.
5.4.4
Filter transformations
In filter transformations, procedures are presented for scaling filter designs such as lowpass filter prototypes in terms of impedance and frequency, and converting them to give
high-pass, bandpass, or bandstop characteristics. For instance, impedance scaling first
assumes that in the prototype design, the source and load impedances are unity. A
source resistance of R0 can be obtained by multiplying all the impedances of the
prototype design by R0 . This would yield the new filter component values as
L' = R0 L , C ' =
C
, Rs' = R0 , RL' = R0 RL
R0
(5.37)
where L , C and RL are the component values for the original prototype. To reinforce
what you have learnt, do the following activity.
ACTIVITY 5.10 BANDPASS FILTER DESIGN
Design a bandpass filter of 0.5 dB equal-ripple response, with N=3. The centre frequency
is 1 GHz, the bandwidth is 10%, and the impedance is 50 ohms.
SOLUTION
Refer to Table 8.4 in the prescribed textbook for element values for a low-pass filter
prototype circuit. You will find that these values are given as:
=
g1 1.5963
= L=
1.5963
= L3=
1.0967
= C=
= RL
, g 4 1.000
1 , g2
2 , g3
The impedance-scaled and frequency-transformed element values are then computed as
follows:
∆
∆R0
L1R0
C2
=
C1' = 0.199 pF,=
L'2 =
0.726 nH,
=
C2'
=
34.91
= 127.0 nH,
ω0 L1R0
ω0 ∆R0
ω0 ∆
ω0 C 2
∆
L3 R0
pF,=
=
C3' = 0.199 pF
L'3 =
127.0 nH and
ω0 L3 R0
ω0 ∆
L1'
=
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ACTIVITY 5.11 FILTERS
Do unsolved problems 8.1–8.10 in the prescribed textbook and check whether you have
the correct answers by checking against the sample solutions posted on the module
website by the module lecturer. Note that the correct answers posted do not show full
and detailed derivation steps, so you will have to show these in your own answers to
provide evidence of your own active and meaningful learning.
5.5
OSCILLATORS AND MIXERS
Refer to the section “Oscillators and mixers” in the prescribed textbook for detailed
information.
Most modern wireless communications, radar, and remote sensing systems are built with
RF and microwave oscillators to provide signal sources for frequency conversion and
carrier generation. For solid-state oscillators, active nonlinear devices such as diodes or
transistors in conjunction with passive circuits are used to convert DC to sinusoidal
steady-state RF signal. Basic transistor oscillator circuits can generally be used at low
frequencies, often with crystal resonators to provide improved frequency stability and low
noise performance. At higher frequencies, diodes or transistors biased to a genitive
resistance operating point can be used with cavity, transmission line, or dielectric
resonators to produce fundamental frequency oscillations up to 100 GHz. Alternatively,
frequency multipliers, in conjunction with a lower frequency source, can be used to
produce power at millimetre wave frequencies.
5.5.1
RF oscillators
In a general sense, oscillators are nonlinear circuits that convert DC power to AC
waveform. RF oscillators provide sinusoidal outputs, which minimise undesired
harmonics and noise sidebands. There are a number of RF oscillator circuits that use
bipolar or field effect transistors in common emitter/source, base/gate or collector/drain
configurations. Various types of feedback networks lead to the well-known Hartley,
Colpitts, Clapp, and Pierce oscillator circuits.
If you have internet access, please watch the following video presentation by All About
Electronics for further explanation of the Colpitts oscillator design. You will then be able
to solve the related design problem in the next learning activity.
The video is at this link:
https://www.youtube.com/watch?v=1fgw-ONlAcc
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ACTIVITY 5.12 COLPITTS OSCILLATOR DESIGN
Design a 50 MHz Colpitts oscillator using a bipolar junction transistor in a common
=
β g=
30 and a transistor input resistance of
emitter configuration with
m / Gi
=
Ri 1/=
Gi 1200 ohms. Use an inductor with L3 = 0.10 µ H and an unloaded Q of 100.
What is the minimum Q of the inductor for which oscillation will be sustained?
SOLUTION
Compute the series combination of C1' and C2 as
C1'C2
1
= =
'
C1 + C2 ω02 L3
1
( 2π )
2
= 100 pF
−6
6 2
50
10
0.1
10
×
×
(
)(
)
'
This value can be obtained in several ways, but here we will choose C=
C=
1
2 200 pF .
However, the unloaded Q of an inductor is related to its series resistance by Q0 = ω L / R
, so the series resistance of the 0.1 µ H inductor is
R
=
ω0 L3
=
Q0
( 2π ) ( 50 ×106 )( 0.1×10−6 )
100
= 0.31 Ω
Thus,
0.31 

C1 =C1' (1 + RGi ) =( 200 pF ) 1 +
 =200 pF
 1200 
which is essentially unchanged from the value found by neglecting the inductor loss.
Thus, the maximum value of series resistance R is
Rmax =
L  7352
1  1+ β
− 3=
= 6.13 Ω
 2
Ri  ω0 C1C2 C1  1200
So the minimum unloaded Q is
=
Q
min
ω0 L3
=
Rmax
( 2π ) ( 50 ×106 )( 0.1×10−6 )
6.13
= 5.1
The above analysis shows that the resonant frequency of an oscillator is determined from
the condition that a 1800 phase shift occurs between the input and output of the transistor.
If the resonant feedback circuit has a high Q, so that there is a very rapid change in the
phase shift with frequency, the oscillator will have good frequency stability.
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5.5.2
Microwave oscillators
These oscillators operate at microwave frequencies, primarily employing negative
resistance diodes or transistors. Refer to Figure 13.6 in the prescribed textbook for a
Rin + jX in is
canonical RF circuit for a one-port negative resistance oscillator, where Z=
in
the input impedance of the active device (e.g., a biased diode or transistor). In general,
this impedance is current (or voltage) dependent, as well as frequency dependent, which
Rin ( I , jω ) + jX in ( I , jω ) . The device is terminated with
we indicate by writing Z=
in ( I , jω )
a passive load impedance,
Z=
RL + jX L
L
(5.38)
ACTIVITY 5.13 NEGATIVE RESISTANCE OSCILLATOR DESIGN
50 Ω )
A one-port oscillator uses a negative resistance diode having Γin= 1.25∠400 ( Z=
0
at its desired operating point, for f = 6 GHz . Design a load-matching network for a 50 Ω
load impedance.
SOLUTION
Either from the Smith chart or by direct calculation, we find the input impedance of the
diode as
Z in =
Z0
1 + Γin
=
−44 + j123 Ω
1 − Γin
Then, the load impedance must be
ZL =
− Z in =
44 − j123 Ω
A shunt stub and series section of line can be used to convert 50 ohms to Z L as shown in
the circuit of Figure 13.17 in the prescribed textbook.
ACTIVITY 5.14 TRANSISTOR OSCILLATOR DESIGN
Design a transistor oscillator at 4 GHz using a GaAs MESFET in a common gate
configuration, with a 5 nH inductor in series with the gate to increase the instability.
Choose a load network to match to a 50 ohm load, and an appropriate terminating
network at the input to the transistor. The scattering parameters of the transistor in a
50 Ω ) S=
common source configuration are ( Z=
0.72∠ − 1160 , =
S12 0.03∠57 0 ,
0
11
0.73∠ − 540 .
=
S 21 2.60∠760 and S=
22
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SOLUTION
The first step is to convert the common source scattering parameters to the scattering
parameters that apply to the transistor in a common gate configuration with a series
inductor. The new scattering parameters are
'
S=
2.18∠ − 350
11
'
S=
1.26∠180
12
=
S 21' 2.75∠960
'
S=
0.52∠1550
22
The output stability circle ( Γ L plane) parameters are calculated using
(S
− ∆ ' S11'∗ )
=
1.08∠330
' 2
' 2
S 22 − ∆
=
C
L
=
RL
'
22
'
S12' S 21
=
0.665
' 2
' 2
S 22 − ∆
For a given value of Γ L , we calculate Γin as
S12' S 21' Γ L
Γ=
S +
= 3.96∠ − 2.40
in
'
1 − S 22Γ L
'
11
or
Z in =
−84 − j1.9 Ω
We can then find Z S as
−R
Z S = in − jX in =
28 + j1.9 Ω
3
Using Rin / 3 creates instability for the start-up of oscillation. One can implement Z S by
using a 90 ohm load with a short length of line. It is likely that the steady-state oscillation
frequency will differ from 4 GHz due to the nonlinearity of the transistor parameters.
5.5.3
Oscillator phase noise
The noise produced by an oscillator or other signal source is important in practice
because it may severely degrade the performance of a communications or radar receiver
system. Phase noise refers to the short-term random fluctuation in the frequency (or phase)
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of the oscillator signal. Phase noise also introduces uncertainty during the detection of
digitally modulated signals. Please do the following activity, which relates to the
application of oscillator phase noise.
ACTIVITY 5:15 GSM RECEIVER PHASE NOISE REQUIREMENTS
The GSM cellular telephone standard requires a minimum of 9 dB rejection of interfering
signal levels of -23 dBm at 3 MHz from the carrier, -33 dBm at 1.6 MHz from the carrier,
and -43 dBm at 0.6 MHz from the carrier, for a carrier level of -99 dBm. Determine the
required local oscillator phase noise at these carrier frequency offsets. The channel
bandwidth is 200 kHz.
SOLUTION
We have
ϕ ( f m )= C ( dBm ) − S ( dB ) − I ( dBm ) − 10log ( B )
where C is the desired signal level in dBm, I is the undesired (interference) signal level
in dBm and B is the bandwidth of the IF filter in Hz. Thus,
ϕ ( f m ) =−99 − 9 − I ( dBm ) − 10log ( 2 ×105 )
You can compute the required LO phase noise as computed from the above expression:
Frequency offset, f m (MHz) ( 3, 1.6 and 0.6 MHz) and interfering signal level (dBm) (i.e.,
-23, -33 and -43 dBm). You can calculate the required local oscillator phase noise at these
carrier frequency offsets: -138 dBc/Hz, -128 dBc/Hz and -118 dBc/Hz. Note that the level
of phase noise requires a phase-locked synthesiser. Bit errors in GSM systems are usually
dominated by the reciprocal mixing effect, while errors due to thermal antenna and
receiver noise are generally negligible.
5.5.4
Frequency multipliers
As the frequency increases into the millimetre wave range, it becomes increasingly
difficult to build fundamental oscillators with good power, stability and noise
characteristics. Thus, there is a need to produce a harmonic of a lower frequency
oscillator using a frequency multiplier. Designing a good-quality frequency multiplier is
a difficult task that generally requires nonlinear analysis, matching at multiple
frequencies, stability analysis, and thermal considerations.
5.5.5
Mixers
Mixers are three-port devices that use a nonlinear or time-varying element to achieve
frequency conversion. An ideal mixer produces an output consisting of the sum and
difference frequencies of its two input signals. The operation of practical RF and
microwave mixers is usually based on the non-linearity provided by either a diode or a
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transistor. A nonlinear component is able to generate a wide variety of harmonics and
other producers of input frequencies, so filtering must be used to select the desired
frequency components. Modern microwave systems typically use several mixers and
filters to perform the functions of frequency up-conversion and down-conversion between
baseband signal frequencies and RF carrier frequencies.
Mixers are best understood by viewing the schematic explanation of the Superheterodyne receivers as presented in the following video link by Engineering Funda:
https://www.youtube.com/watch?v=woltZ-WQWf0
Now put the understanding you have gained from watching this video and information
in the prescribed textbook about mixers into practice to do the following activity:
ACTIVITY 5.16 IMAGE FREQUENCY
The IS-54 digital cellular telephone system uses a receive frequency band of 869–894
MHz, with a first IF frequency of 87 MHz and a channel bandwidth of 30 kHz. What are
the two possible ranges for the LO frequency? If the upper LO frequency range is used,
determine the image frequency range. Does the image frequency fall within the receive
passband?
SOLUTION
The two possible LO frequency ranges are
 956 to 981 MHz
f LO = f RF ± f IF = ( 869 to 894 ) ± 87 = 
782 to 807 MHz
Using the 956–981 MHz frequency range, we find the IF frequency as
f IF =
f RF − f LO =
−87 MHz
(869 to 894 ) − ( 956 to 981) =
Thus, the RF image frequency range is
f IM = f LO − f IF = ( 956 to 981) + 87 = 1043 to 1068 MHz .
This value is well outside the receiver passband.
ACTIVITY 5.17 HF RESONATORS, FILTERS, OSCILLATORS AND MIXERS
1. Refer to the chapter entitled “Resonators, Filters, Oscillators and Mixers” in the
prescribed textbook and solve the following selected unsolved problems: 13.3,
13.5, 13.7, 13.9, 13.15–13.17 and 13.22. Your module lecturer will post the correct
answers on the module website without including the full solutions.
2. When you reflect back on the concepts and designs of the resonators, filters,
oscillators and mixers covered in this learning unit, would you say that you have
acquired sufficient competence in designing high frequency electronics systems? If
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not, be sure to revise the required skills or to discuss difficult concepts and questions
with your fellow students on myUnisa or contact the lecturer for clarification.
5.6
CONCLUSION
After studying this unit, you should be able to design HF electronic systems such as
resonators, filters, oscillators and mixers. In the next unit, you will gain an understanding
of how modern CAD and EM field solvers are used to simulate HF power electronic
amplifiers and to investigate the performance of designed HF power amplifiers in terms
of maximum gains, specified gains and low noise figures for different HF networks
systems.
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6
Learning unit 1
DESIGN OF HF POWER AMPLIFIERS
6.1
LEARNING OUTCOMES
By the end of this learning unit, you should understand how modern computer aided
design (CAD) and electromagnetic (EM) field solvers are used to design HF power
amplifiers with maximum gains or for specified gains and with low noise figures.
In particular, you should be able to do the following:
•
•
•
•
•
Compute power gains for two-port HF electronic networks.
Derive the necessary stability conditions for transistor amplifiers.
Design single-stage transistor amplifiers.
Design broadband transistor amplifiers.
Design power amplifiers.
6.2
INTRODUCTION
Reading: To complete this unit, study the chapter entitled “Microwave Amplifier Design”
in your prescribed textbook or the same topic in any of the recommended study materials.
The dramatic improvements and innovations in solid-state technology that have occurred
since the 1970s have seen most RF and microwave amplifiers today using transistor
devices such as Si BJTs, GaAs or SiGe HBTs, Si MOSFETs, GaAs MESFETs, and GaAs. The
high frequency microwave transistor amplifiers are rugged, low-cost, and reliable and can
be easily integrated in both hybrid and monolithic integrated circuitry. Transistor
amplifiers can be used at frequencies in excess of 100 GHz in a wide range of applications
requiring small size, low noise, broad bandwidth, and medium to high power capacity.
In this unit, you will learn more about transistor amplifier design, which relies primarily
on the terminal characteristics of the transistor, as represented by either scattering
parameters or one of the equivalent circuit models introduced in the previous units.
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If you have internet access, please watch the video clip at the link below, which deals
with power amplifiers:
https://www.youtube.com/watch?v=gRcE2t_28co
6.3
TWO-PORT POWER GAINS
Refer to the section “Two-port power gains” in the prescribed textbook for a detailed
discussion. This unit will serve merely as a guideline to highlight certain aspects, and you
need to engage in deeper active learning of the relevant concepts.
In this unit, you will see how two-port power gains are defined. You will first consider an
arbitrary two-port network, characterised by its scattering matrix [S] connected to source
and load impedance ZS and ZL, respectively as shown in Figure 12.1 in the prescribed
textbook. Thus, the following expressions define three types of power gain in terms of
scattering parameters of the two-port network and the reflection coefficients, Γ S and Γ L ,
of the source and load.
Power gain = G = PL / Pin is the ratio of power dissipated in the load Z L to the power
•
delivered to the input of the two-port network. This gain is independent of Z S ,
although the characteristics of some active devices may be depended on Z S .
•
Available power gain = GA = Pavn / Pavs is the ratio of the power available from the
two-port network to the power available from the source. This assumes conjugate
matching of both the source and the load, and depends on Z S , but not on Z L .
•
Transducer power gain = GT = PL / Pavs is the ratio of the power delivered to the load
to the power available from the source. This depends on both Z S and Z L .
You will find a detailed analysis of these gains in the section entitled “Two-port power
gains” in the prescribed textbook. Alternatively, read the information about this topic in
any reference material. Please revise the concepts learnt in this section by going through
the following online video demonstration of two-port power gains presented by the IIT
Madras, before you go on to solve the learning activity that follows:
https://nptel.ac.in/courses/108106084/3
The video introduced you to fundamentals of two-port power gains. This information in
essence allows you to solve the following analysis problem, which involves finding
power gains.
ACTIVITY 6.1 COMPARISION OF POWER GAIN DEFINITIONS
A silicon bipolar junction transistor has the following scattering parameters at 1.0 GHz,
with a 50 ohm reference impedance:
S=
0.38∠ − 1580
11
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S=
0.11∠540
12
=
S 21 3.50∠800
S=
0.40∠ − 430
22
25 Ω and the load impedance is Z=
40 Ω . Compute the
The source impedance is Z=
S
L
power gain, the available power gain, and the transducer power gain.
SOLUTION
We compute the reflection coefficients at the source and load using the following:
Z − Z 0 25 − 50
ΓS = S
=
=−0.333
Z S + Z 0 25 + 50
Z − Z 0 40 − 50
ΓL = L
=
=−0.111
Z L + Z 0 40 + 50
Also, the reflection coefficients seen by looking at the input and output of the terminated
network are:
Γ=
S11 +
in
S12 S 21Γ L
= 0.365∠ − 1520
1 − S 22Γ L
Γ out
= S 22 +
S12 S 21Γ S
= 0.545∠ − 430
1 − S11Γ S
We can then calculate the power gain as
2
=
G
(
2
)
S 21 1 − Γ L
=
13.1
2
2
1 − Γin 1 − S 22Γ L
(
)
The available power gain is
2
=
GA
(
2
)
S 21 1 − Γ S
19.8
=
2
2
1 − Γ out 1 − S11Γ S
(
)
The transducer power gain is calculated as
2
(
2
)(
2
)
S 21 1 − Γ S 1 − Γ L
=
GT =
12.6
2
2
1 − Γ S Γin 1 − S 22Γ L
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6.4
STABILITY
Refer to the section “Stability” in the prescribed textbook or information on the same
topic covered in any of the recommended study material for detailed discussions.
Begin studying the concept of stability in power amplifiers by watching the following
video clip at the link below:
https://www.youtube.com/watch?v=oiMUn3zjFYA
The above video provides a systematic approach to finding the necessary stability
conditions of high frequency electronic power amplifiers. Watching the video should
help you understand the difficult concepts relating to power amplifier stability as
discussed in the prescribed textbook.
The necessary conditions for stability of a transistor amplifier are as discussed in the video.
Oscillation is possible if either the input or output port impedance has a negative real
part; this would then imply that Γin > 1 or Γ out > 1 . Because Γin and Γ out depend on the
source and load matching networks, the stability of the amplifier depends on Γ S and Γ L
as presented by the matching networks. In this unit, two types of stability are defined as
follows:
Unconditional stability: The network is unconditionally stable if Γ S < 1 and Γ out < 1
•
for all passive source and load impedances (i.e., Γ S < 1 and Γ L < 1 ).
Conditional stability: The network is conditionally stable if Γin < 1 and Γ out < 1 only
•
for a certain range of passive source and load impedances. This case is also referred
to as potentially unstable.
The stability condition of an amplifier circuit is usually frequency dependent, since the
input- and output-matching networks generally depend on frequency. This means that an
amplifier can be stable at its design frequency, but unstable at other frequencies. Please
expand the information in this section of the unit by studying the prescribed textbook and
information on this topic in any of the other recommended materials. Then do the
following learning activity relating to transistor stability.
ACTIVITY 6.2 TRANSISTOR STABILITY
The Triquint T1G6000528 GaN HEMT has the following scattering parameters at 1.9
50 Ω ):
GHz ( Z=
0
=
S11 0.869∠ − 1590
=
S12 0.031∠ − 90
=
S 21 4.250∠610
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=
S 22 0.507∠ − 117 0
Determine the stability of this transistor by using the K − ∆ test and the µ − test, and plot
the stability circles on a Smith chart.
SOLUTION
Let’s work the value of K and ∆ as
=
∆ S11S22 − S12=
S 21 0.336
2
2
2
1 − S11 − S 22 + ∆
=
K =
0.383
2 S12 S 21
Thus, we have ∆ < 1 but not K > 1 , so the unconditional stability criteria are satisfied,
and the device is potentially unstable. The stability of this device can also be evaluated
using the µ − test, for which µ = 0.678 such that a potential instability is observed. The
centres and radii of the stability circles are given by:
C
=
L
(S
− ∆S11∗ )
=
1.59∠1320
2
2
S 22 − ∆
S12 S 21
=
0.915
2
2
S 22 − ∆
=
RL
C
=
S
22
(S
=
RS
∗
− ∆S 22
) 1.09∠1620
=
2
2
S11 − ∆
11
S12 S 21
=
0.205
2
2
S11 − ∆
These data can be used to plot the input and output stability circles, as shown in Figure
12.6 in the prescribed textbook. The unstable regions have been shaded.
6.5
SINGLE-STAGE TRANSISTOR AMPLIFIER
Refer to the section “Single-stage transistor amplifier design” in the prescribed textbook
or information on this topic in any of the recommended study materials for a detailed
discussion.
Please watch the video clip on single-stage transistor amplifier design presented by Learn
and Grow, which is available at the following link:
https://www.youtube.com/watch?v=i2wbaUMwhdg
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You are also welcome to watch the following clip, presented by the Tech Maker, on how
to design a transistor amplifier:
https://www.youtube.com/watch?v=uT0V9gFAVg4
You will notice that the first step in single-stage transistor amplifier design is to determine
the stability of the transistor and to locate on the Smith chart the stable regions for Γ S and
Γ L . As you can see from Figure 12.2 in the prescribed textbook, the maximum power
transfer from the input matching network to the transistor will occur when
Γin =Γ∗S
and maximum power transfer from the transistor to the output matching network will
occur when
Γ out =
Γ∗L
If we assume lossless matching sections, these conditions will maximise the overall
transducer gain. The maximum gain is computed as
GTmax =
1
1 − ΓS
2
S 21
2
1− ΓL
2
1 − S 22Γ L
2
In addition, with conjugate matching and lossless matching sections, the input and output
ports of the amplifier will be matched to Z 0 . The maximum transducer power gain above
occurs when the source and load are conjugately matched to the transistor, as given by
the previous conditions. If the transistor is unconditionally stable, so that K > 1 , the
maximum transducer power gain of the above expression can simply be re-written as
follows:
GTmax=
(
S 21
K − K 2 −1
S12
)
This result can be obtained by substituting (expression 12.40) and (expression 12.41) in
the prescribed textbook for Γ S and Γ L into the conditional expressions for the maximum
transducer gain and then simplying. The following learning activity will provide you with
further ideas on how to design a conjugately matched amplifier.
ACTIVITY 6.3 CONJUGATELY MATCHED AMPLIFIER DESIGN
Design an amplifier for maximum gain at 4 GHz using single-stub matching sections.
Calculate and plot the input return loss and the gain from 3 to 5 GHz. The transistor is a
50 Ω ) :
GaAs MESFET with the following scattering parameters ( Z=
0
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SOLUTION
In many practical applications, the manufacturer furnishes scattering parameters for a
wide range of frequencies, and so the stability over the entire range should be checked.
Calculating the value of K and ∆ from scattering parameters at each frequency in the
above table gives the following results:
We see that K > 1 and ∆ < 1 at 4 and 5 GHz, so the transistor is unconditionally stable
at these frequencies, but it is conditionally stable only at 3 GHz. We can proceed with
the design at 4 GHz, but we should check stability at 3 GHz after we find the matching
networks (which determine Γ S and Γ L ). For maximum gain, we should design the
matching sections for a conjugate match to the transistor. Thus, Γ S =Γ∗in and Γ L =Γ∗out
and Γ S , Γ L can be determined as follows:
2
B1 ± B12 − 4 C1
Γ=
= 0.872∠1230
S
2C1
2
B2 ± B22 − 4 C2
Γ=
= 0.876∠610
L
2C2
The effective gain factors can be calculated as
=
GS
1
4.17
=
= 6.20 dB
2
1 − ΓS
2
=
G0 S=
6.76
= 8.30 dB
21
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2
1 − ΓL
1.67
=
GL
=
= 2.22 dB
2
1 − S 22Γ L
Then the overall transducer gain is
GTmax = 6.20 + 8.30 + 2.22 = 16.7 dB
The matching networks can be easily determined using the Smith chart (refer to the
prescribed textbook for a detailed Smith chart plotting procedure, shown in Figure 12.7).
After working through the learning activity on conjugately matched amplifiers, please
also complete the following learning activity on low-noise amplifier design:
ACTIVITY 6.4 LOW-NOISE AMPLIFIER DESIGN
A GaAs MESFET is biased for minimum noise figure, with the following scattering
50 Ω ): S11= 0.6∠ − 600 , =
S12 0.05∠260 ,
parameters and noise parameters at 4 GHz ( Z=
0
20 Ω . For design
S 21
= 1.9∠810 , S 22= 0.5∠ − 600 , Fmin = 1.6 dB , Γ opt = 0.62∠1000 and R=
N
purposes, assume that the device is unilateral, and calculate the maximum error in GT
resulting from this assumption. Then design an amplifier having a 2.0 dB noise figure with
the maximum gain that is compatible with this noise figure.
SOLUTION
First, calculate that K = 2.78 and ∆ =0.37 , so the device is unconditionally stable even
without the approximation of a unilateral device. Next, compute the unilateral figure of
merit from
=
U
S12 S 21S11S 22
=
0.059
2
2
1 − S11 1 − S 22
(
)(
)
Checking the boundedness of the ratio GT / GTU as follows:
1
(1 + U )
2
<
GT
1
<
GTU (1 − U )2
or
0.891 <
GT
< 1.130
GTU
In dB, this is
−0.50 < GT − GTU < 0.53 dB Where GT and GTU are now in dB. Thus, we should expect
less than about ±0.5 dB error in gain.
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Next, we compute the centre and radius of the 2 dB noise figure circle:
=
N
2
2
F − Fmin
1.58 − 1.445
=
1 + Γ opt
1 + 0.62=
∠1000
0.0986
4 RN / Z 0
4 ( 20 / 50 )
C
=
F
=
RF
Γ opt
= 0.56∠1000
N +1
(
2
)
N N + 1 − Γ opt
= 0.24
N +1
The noise figure circle is plotted in Figure 12.9 in the prescribed textbook. The minimum
0
noise ( Fmin = 1.6 dB ) occurs for Γ S =Γ opt =0.62∠100 . Several input section constant-gain
circles are obtainable as follows:
From Figure 12.9 in the prescribed textbook, the optimum solution is Γ =
0.53∠750 ,
S
yielding GS = 1.7 dB and F = 2.0 dB .
∗
For the output section we choose Γ L = S 22
= 0.5∠600 for a maximum GL of
=
GL
1
1.33
=
= 1.25 dB
2
1 − S 22
The transistor gain is
2
=
G0 S=
3.61
= 5.58 dB
21
so the overall transducer gain will be
GTU = GS + G0 + GL = 8.53 dB
A complete AC circuit for the amplifier, using open-circuited shunt stubs in the matching
sections, is shown in Figure 12.9b in the prescribed textbook. A computer analysis of the
circuit gives a gain of 8.36 dB.
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6.6
BROADBAND TRANSISTOR
Refer to the section “Broadband transistor amplifier design” in the prescribed textbook
for detailed information.
First, click on the following link to watch a video that offers a further explanation of
broadband transistor amplifier design:
https://www.youtube.com/watch?v=6ySLlJ7C0ZQ
The presentation should have indicated that a good amplifier needs to have constant gain
and good input matching over the desired frequency bandwidth. As we have seen in the
previous sections, conjugate matching will give maximum gain only over a relatively
narrow bandwidth, while designing for less than maximum gain will improve the gain
bandwidth, but the input and output ports of the amplifier will be poorly matched. These
problems occur because microwave transistors typically are not well matched to 50
ohms, and large impedance mismatches are governed by the Bode-Fano gain-bandwidth
criterion. The previous section has also shown that S 21 decreases with frequency at the
rate of 6 dB/octave. Thus, special consideration must be given to the problem of designing
broadband amplifiers. Possible solutions to this problem are listed below:
•
Compensated matching networks: Input and output matching can be designed to
compensate for the gain roll-off in S 21 , but generally at the expense of the input and
output matching.
Resistive matching networks: Good input and output matching can be obtained by
using resistive matching networks, with a corresponding loss in gain and increase in
noise figure.
Negative feedback: Negative feedback can be used to flatten the gain response of the
transistor, improve the input and output match, and improve the stability of the
device. Amplifier bandwidths in excess of a decade are possible with this method, at
the expense of gain and noise figure.
Balanced amplifiers: Two amplifiers having 900 couplers at their input and output can
provide good matching over an octave bandwidth or more. The gain is equal to that
of a single amplifier, however, and the design requires two transistors and twice the
DC power.
Distributed amplifiers: Several transistors are cascaded together along a transmission
line, giving good gain, matching, and noise figure over a wide bandwidth. The circuit
is large, and does not give as much gain as a cascade amplifier with the same number
of stages.
Differential amplifiers: Driving two devices in a differential mode, with input signals
of opposite polarity, results in an effective series connection of device capacitance,
•
•
•
•
•
thus roughly doubling fT . Differential amplifiers can also provide a larger output
voltage swing than a single device and common mode noise rejection.
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For more details about the performance and optimisation of a balanced amplifier, please
consult the prescribed textbook and the following websites for insights regarding the
computer aided design (CAD) software for high frequency electronics design:
https://wwwhome.ewi.utwente.nl/~ptdeboer/ham/puff/
http://www.gunthard-kraus.de/PUFF/index_eng.html
Also watch the following online video presentation by Leo Fung on high frequency
electronics CAD software, which will provide you with additional insight concerning how
to use CAD to design high frequency electronics:
https://www.youtube.com/watch?v=Cij4A-efJUE
Now complete the following learning activity:
ACTIVITY 6.5 PERFORMANCE AND OPTIMISATION OF A BALANCED AMPLIFIER
Use the amplifier designed in example 12.4 in the prescribed textbook, that is, an
amplifier with a gain of 11 dB at 4.0 GHz in a balanced configuration operating from 3
to 5 GHz. Use quadrature hybrids, and plot the gain and return loss over this frequency
range. Using microwave CAD software, optimise the amplifier matching networks to give
10 dB gain over this band.
SOLUTION
The amplifier in example 12.4 was designed for a gain of 11 dB at 4 GHz. As seen from
Figure 12.8c in the prescribed textbook, the gain varies by a few dB from 3 to 5 GHz,
and the return loss is no better than 5 dB. We can design a quadrature hybrid, as discussed
in chapter 7 in the prescribed textbook, to have a centre frequency of 4 GHz. Then the
balanced amplifier configuration in Figure 12.11 in the prescribed textbook can be
modelled using a microwave CAD package, with the results shown in Figure 12.12. You
should be able to notice a dramatic improvement in return loss over the band as
compared with the results for the original amplifier in Figure 12.8c. The input matching
is best at 4 GHz, since this was the design frequency of the coupler; a coupler with better
bandwidth will give improved results at the band edges. Also, observe that the gain at 4
GHz is still 11 dB and that it drops by a few dB at the band edges. The results of this
optimisation are shown in Figure 12.12, where it can be seen that the gain response is
much flatter over the operating band. The input match is still very good in the vicinity of
the centre frequency, with a slightly worse result at the low-frequency end.
6.7
POWER AMPLIFIERS
In radar and radio transmitters, power amplifiers are used in the final stages in order to
increase the radiated power level. The output powers for radar or fixed-point radio
systems may be in the range of 1 to 100 W, while that for mobile voice or data
communication systems may be in the order of 100 to 500 mW. Design considerations
for RF and microwave power amplifiers are efficiency, gain, intermodulation distortion,
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and thermal effects. Single transistors can provide output powers of 10 to 100 W at UHF
frequencies, while devices at higher frequencies are generally limited to output power
less than 10 W. Various power-combining techniques can be used in conjunction with
multiple transistors if higher output powers are required.
If you have internet access, watch the following online video clips for more information
on the design and implementation of power amplifiers as presented by Matthew Ozalas
of KeySight EEs of EDA, available at the following links:
https://www.youtube.com/watch?v=WAingaHfBMs
https://www.youtube.com/watch?v=GhPqPVlDRPY
Now that you have learnt about the design methodologies of a power amplifier from the
videos above and the prescribed textbook, the activity below gives you the opportunity
to design a typical power amplifier.
ACTIVITY 6.6 DESIGN OF A CLASS A POWER AMPLIFIER
Design a power amplifier at 2.3 GHz using a Nitronex NPT25100 GaN HEMT transistor,
with an output power of 10 W. The scattering parameters of the transistor for
and
VDS = 28 V
0
I D = 600 mA are as follows:=
S11 0.593 ∠1780 ,=
S12 0.009∠ − 1270 , S=
21 1.77∠ − 106
0
S22 0.958∠175 , and the optimum large-signal source and load impedance are
and =
Z SP =10 − j3 Ω and Z LP =2.5 − j 2.3 Ω . For an output power of 10 W, the power gain is
16.4 dB and the drain efficiency is 26%. Design input and output impedance matching
sections for the transistor, and find the required input power, the required DC drain
current, and the power added efficiency.
SOLUTION
Start by establishing the stability of the device. Using the small-signal scattering
parameters shown as follows,
=
∆ S11S22 − S12=
S21 0.579 < 1
2
=
K
2
2
1 − S11 − S 22 + ∆
= 2.08 > 1 ,
2 S12 S 21
showing that the device is unconditionally stable.
Converting the large-signal source and load impedance to reflection coefficients gives
Γ SP= 0.668∠1870
=
Γ LP 0.905∠ − 1750
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For comparison, using the small-signal scattering parameters in Figure 12.40 of the
prescribed textbook to find the source and load reflection coefficients for conjugate
matching gives
2
B1 ± B12 − 4 C1
Γ=
= 0.508∠1660
S
2C1
2
B2 ± B22 − 4 C2
=
ΓL
= 0.954∠ − 1760
2C2
Note that these values are approximately equal to the large-signal values Γ SP and Γ LP ,
but not exactly, because the scattering parameters used to calculate the source and load
reflection coefficients ( Γ S , Γ L ) do not apply in the case of large power levels. We should
use the large-signal reflection coefficients to design the input and output matching
networks. The AC amplifier circuit is shown in Figure 12.23 of the prescribed textbook.
For an output power of 10 W, the required input drive power is
=
Pin Pout ( dBm ) − G ( =
dB ) 10log (10 000 ) − 16.4
= 23.6 dBm
= 229 mW
PDC
The DC input power can be found from the drain efficiency as=
P=
out / η 38.5 W , so
=
=
/ VDS 1.37 A . The power added efficiency of the amplifier
the DC drain current
is I D PDC
can be found from the following expression:
=
η PAE
Pout − Pin 10.0 − 0.229
=
= 25%
38.5
PDC
At this stage, you should be able to compute power gains for two-port HF electronic
networks, derive the necessary stability conditions for transistor amplifiers, and design
single-stage transistor amplifiers, broadband transistor amplifiers and power amplifiers.
You can now go on to the next activity, which will give you the opportunity to review all
the design concepts you have learnt in this unit.
ACTIVITY 6.7 HIGH FREQUENCY ELECTRONIC SYSTEMS POWER AMPLIFIERS
1. Refer to the chapter entitled “Power Amplifiers” in the prescribed textbook and solve
the following selected unsolved problems: 12.1–12.10 and all the odd-numbered
problems from 12.11–12.21. Please attempt these problems on your own first, and
then check your answers against the correct answers that your module lecturer will
post on myUnisa. The posted answers will not contain the full solutions, however.
Remember that you can also direct queries to the lecturer in the Discussion space for
the module.
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2. When you reflect on the concepts and designs of the HF power amplifiers you have
learnt in this learning unit, how would you rate your own competency in designing
high frequency electronic power amplifiers? Do you feel yourself to be capable of
successfully solving problems related to HF power amplifiers? If not, be sure to revise
the required skills or to discuss difficult concepts and questions with your fellow
students on myUnisa, or contact the lecturer for clarification.
3. Think back on what you have learnt in this module. On the basis of this, do you think
that you will succeed in a career in electronics design? What, would you say, are the
two most useful skills you have learnt? How will you be able to apply them in your
career? Overall, how has this module contributed to the development of your workrelated skills and your career path?
6.8
CONCLUSION
After studying this unit, you should be able to apply modern CAD and EM field solvers
to design HF power amplifiers with maximum gains or for specified gains and with low
noise figures.
Now that you have completed this module, you should be able to analyse HF
transmission lines, waveguides and electronic networks. In addition, you should be able
to design HF impedance matching and tuning and design RF and microwave circuits and
systems using power amplifiers, oscillators, mixers, resonators and filters. This should all
prove very valuable for your career and any further studies you may undertake in
electrical engineering.
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