EE 461, Microwave Engineering Notes
George W. Hanson
2008
2
Contents
1 Electromagnetic Theory
1.1 Introduction to Microwave Engineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Time-Harmonic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4
1.5
Wave Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Plane Wave Propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
9
9
10
13
14
2 Transmission Line Theory
2.1 The Lumped-Element Circuit Model for a Transmission Line
2.2 Field Analysis of Transmission Lines . . . . . . . . . . . . . .
2.3 The Terminated Lossless Transmission Line . . . . . . . . . .
2.4 The Smith Chart . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 The Quarter-Wave Transformer . . . . . . . . . . . . . . . . .
2.6 Generator and Load Mismatches . . . . . . . . . . . . . . . .
2.7 Lossy Transmission Lines . . . . . . . . . . . . . . . . . . . .
2.8 Transient Transmission Lines . . . . . . . . . . . . . . . . . .
2.8.1 Waveforms and Spectral Analysis . . . . . . . . . . . .
2.8.2 Integrated Circuits and Ground Bounce . . . . . . . .
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17
18
21
22
30
39
40
41
42
51
53
3 Transmission Lines and Waveguides
3.1 General Solutions for TEM, TE, and TM Modes
3.2 Parallel Plate Waveguide . . . . . . . . . . . . . .
3.3 Rectangular Waveguide . . . . . . . . . . . . . .
3.4 Circular Waveguide . . . . . . . . . . . . . . . . .
3.5 Coaxial Line . . . . . . . . . . . . . . . . . . . .
3.6 Surface Waves on a Grounded Dielectric Slab . .
3.7 Stripline . . . . . . . . . . . . . . . . . . . . . . .
3.8 Microstrip . . . . . . . . . . . . . . . . . . . . . .
3.9 The Transverse Resonance Technique . . . . . . .
3.10 Wave Velocities and Dispersion . . . . . . . . . .
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57
57
59
64
65
65
67
70
71
73
73
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.
4 Microwave Network Analysis
79
4.1 Impedance and Equivalent Voltage and Currents . . . . . . . . . . . . . . . . . . . . . . . . . 79
5 Impedance Matching and Tuning
5.1 Matching with Lumped Elements (L Networks)
5.1.1 Lumped Elements . . . . . . . . . . . .
5.2 Single-Stub Tuning . . . . . . . . . . . . . . . .
5.3 Double-Stub Tuning . . . . . . . . . . . . . . .
5.4 The Quarter-Wave Transformer . . . . . . . . .
5.5 The Theory of Small Re‡ections . . . . . . . .
5.6 Binomial Multisection Matching Transformers .
3
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85
. 86
. 88
. 88
. 99
. 105
. 107
. 109
4
CONTENTS
5.7
5.8
5.9
Chebyshev Multisection Matching Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . 115
Tapered Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
Bode-Fano Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
6 Power Dividers and Directional Couplers
131
7 Electromagnetic Compatibility and Interference (EMS/EMI)
141
8 Microwave Filters
147
9 Appendix
149
9.1 Transient transmission line current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
CONTENTS
5
Preliminary Material
“From a long view of the history of mankind - seen from, say, ten thousand years from now
- there can be little doubt that the most signi…cant event of the 19th century will be judged as
Maxwell’s discovery of the laws of electrodynamics”
Richard P. Feynman
6
CONTENTS
What is “Di¤erent”About High Frequencies?
The short, and perhaps obvious, answer is wavelength. For any given object, one usually considers a
frequency to be a “high” frequency if the wavelength ( = c=f ) is small compared to the size of the object.
This is complicated by the fact that an object doesn’t have one “size” (a sphere, for instance, may be an
exception). So if one considers electromagnetic scattering from a raindrop, and scattering from a large
building, you would characterize di¤erent frequencies as being high frequencies. In a broad sense, through,
many people consider high frequencies to be something like f > 100 MHz ( = 3 m in air). This is because
we live in a world of characteristic sizes commiserate with this wavelength. For example, people are typically
5-6 feet tall (1.66-2 meters), cars are typically 8-12 feet long (2.66-4 meters), suburban buildings are typically
10-30 feet tall (3.33-10 meters), etc.
An important point to remember is that electromagnetics is the physics of electrical engineering, and
that the governing equations that describe all classical macroscopic electromagnetic phenomena are Maxwell’s
equations. These are very complicated equations, but they can be simpli…ed in special cases. One such case
is in the event of “low” frequencies— one can derive the classical circuit equations (Ohm’s law, Kirchho¤’s
laws, etc.) as special, simple cases. These simpli…ed equations are often applicable in circuit analysis since
typical circuit dimensions are very small compared to typical wavelengths in common usage.
For example, consider the following discrete (not integrated) circuit)
L= 5 cm
R1
+
+
-
Vout
Vs
R2
+
-
With Vs = V0 cos !t, is Vout =
Case 1 f = 20 kHz !
=
c
f
-
R2
R1 +R2 Vs ?
=
3 1010 cm=s
20 103 =s
= 1; 500; 000 cm.
L
= 3:3
10
6
1:
The circuit length is very short compared to a wavelength— low frequency approximations are applicable,
2
and Vout = R1R+R
Vs :
2
Case 2 f = 6 GHz !
=
c
f
=
3 1010 cm=s
6 109 =s
= 5 cm.
L
= 1:
The circuit length is equal to a wavelength— low frequency approximations are not applicable.
CONTENTS
7
Case 3 f = 300 GHz !
=
c
f
=
3 1010 cm=s
300 109 =s
= 0:1 cm.
L
= 50
1:
The circuit length is long compared to a wavelength— low frequency approximations are not applicable.
In particular, for Cases 2 and 3,
The “voltage” on the line will be a function of position along the line.
The circuit “seen”by the source presents a length-dependent input impedance which must be carefully
matched for e¢ cient power transfer.
Practically, the circuit will not work since it will radiate energy into space. Furthermore, the electrically long distances the signal must travel will result insigni…cant dissipation to the use of imperfect
conductors.
Dispersion will degrade the signal before it reaches the output.
To make it clear that the important thing is wavelength, consider a circuit of length L = 16:66x10
Case 4 f = 6 GHz !
=
c
f
=
3 1010 cm=s
6 109 =s
6
cm.
= 5 cm.
L
= 3:3
10
6
1:
The circuit length is very short compared to a wavelength (same as Case 1)— low frequency approximations
2
are applicable, and Vout = R1R+R
Vs .
2
8
CONTENTS
Brief Chronology of Early Wireless Communications Developments
Magnetic e¤ects known for much of recorded history.
1755: Benjamin Franklin began investigations which would lead to qualitative and quantitative
ideas about electrostatics.
1769: Dr. John Robison began experiments which would lead to the inverse-square law (Coulomb’s
law) governing electrostatics. Similarly by Henry Cavendish in 1773. Neither investigator widely
publicized their work.
1785: Charles Augustin de Coulomb demonstrated the law of electric force, which is now called
Coulomb’s law.
1820: Hans Christian Oersted found that a current carrying wire (electricity) could produce
magnetism.
1821: Jean-Baptiste Biot and Félix Savart quantify the e¤ect discovered by Oersted.
1827: Georg Simon Ohm formulates what is now called Ohm’s law
1831: Michael Faraday demonstrated that a time changing magnetic …eld could produce an
electric current.
1845 (or 1854): Gustav Robert Kirchho¤ formulated Kirchho¤’s laws.
1873: James Clerk Maxwell published the …rst uni…ed theory of electricity and magnetism.
1886: Heinrich Hertz assembled a “radio” system–a spark applied to a transmitting antenna
caused a spark to be produced at a receiving antenna, posited near the antenna.
1901: Guglielmo Marconi developed a system to send signals from England to Newfoundland ('
3000 km) using electromagnetic waves.
1903: Marconi began regular transatlantic message service between England and stations in Nova
Scotia and Cape Cod.
Chapter 1
Electromagnetic Theory
1.1
Introduction to Microwave Engineering
Read
1.2
Maxwell’s Equations
Maxwell’s Equations— Di¤erential Form
Classical macroscopic electromagnetic phenomena are governed by a set of vector equations known collectively as Maxwell’s equations. Maxwell’s equations in di¤erential form are
r D(r; t) =
r B(r; t) =
r
E(r; t) =
r
H(r; t) =
e (r; t);
m (r; t);
(1.1)
@
B(r; t)
@t
Jm (r; t);
@
D(r; t) + Je (r; t);
@t
E is the electric …eld intensity (V=m)
2
D is the electric ‡ux density (C=m )
2
B is the magnetic ‡ux density (Wb=m )
H is the magnetic …eld intensity (A=m)
3
e
is the electric charge density (C=m )
2
Je (= J in text) is the electric current density (A=m )
3
m
is the magnetic charge density (Wb=m )
2
Jm (= M in text) is the magnetic current density (V=m )
V stands for volts, C for coulombs, Wb for webers, A for amperes, and m for meters.
The equations are known, respectively, as Gauss’ law, the magnetic-source law or magnetic Gauss’ law,
Faraday’s law, and Ampère’s law.
9
10
CHAPTER 1. ELECTROMAGNETIC THEORY
1.3
Time-Harmonic Fields
Often the …elds of interest vary harmonically (sinusoidally) with time. Because Maxwell’s equations are
linear, and assuming linear constitutive equations (typical below optical range), then
time-harmonic sources ; J will maintain time-harmonic …elds E; D; B; H.
Assume
E(r; t) = E0 (r) cos(!t +
E );
(1.2)
then, for instance,
n
E(r; t) = E0 (r) Re ej(!t+
j!t
= Re E(r)e
where E(r)
E0 (r)ej
E
E)
;
o
= Re E0 (r)ej!t ej
E
(1.3)
is a complex phasor.
Repeating for the other …eld quantities of interest,
B(r; t)
D(r; t)
H(r; t)
J(r; t)
(r; t)
=
=
=
=
=
B0 (r) cos(!t + B )
D0 (r) cos(!t + D )
H0 (r) cos(!t + H )
J0 (r) cos(!t + J )
)
0 (r) cos(!t +
(1.4)
we obtain the complex phasors
B(r)
D(r)
H(r)
J(r)
(r)
B0 (r)ej B
D0 (r)ej D
H0 (r)ej H
J0 (r)ej J
j
:
0 (r)e
(1.5)
To obtain time-harmonic Maxwell’s equations, consider, for instance, Faraday’s law,
r
E(r; t) =
@
B(r; t)
@t
Jm (r; t):
(1.6)
This can be rewritten as
Re [r
E(r) + j!B(r) + Jm (r)] ej!t = 0;
(1.7)
which must be true for all t.
– When !t = 0; Re fr
– When !t = =2; Im fr
E(r) + j!B(r) + Jm (r)g = 0:
E(r) + j!B(r) + Jm (r)g = 0:
– If both the real part and the imaginary part of a complex number are zero, then the number itself is zero,
and thus
r E(r) = j!B(r) Jm (r):
(1.8)
1.3. TIME-HARMONIC FIELDS
11
Repeating for all Maxwell’s equations and the continuity equations, we get the time-harmonic forms
r D(r) = e (r);
r B(r) = m (r);
r E(r) = j!B(r) Jm (r);
r H(r) = j!D(r) + Je (r);
r Je(m) (r) = j! e(m) (r);
(1.9)
where all quantities are time-harmonic phasors.
Note that in the time-harmonic case we obtain the convenient correspondence
Z
@=@t $ j!
( ) dt $ (j!)
(1.10)
1
:
Time-domain quantities can be recovered from the phasor quantities as, for instance,
E(r; t) = Re E(r)ej!t :
(1.11)
Constitutive Equations
For linear isotropic media
D(r; !) = e(r;!) E(r; !);
B(r; !) = e(r; !) H(r; !);
(1.12)
where e, e are the permittivity and permeability, respectively, of the medium.
e = er "0 , e = er
0
0
is the permittivity of free space (' 8:85
10
0
is the permeability of free space (' 4
10
12
7
F=m)
H=m)
F stands for farads and H for henrys.
For dimensional analysis, C = A s = F V and Wb = V s = H A, where s stands for seconds.
Note that c0 =
p 1
0 "0
= speed of light in free space (vacuum)
Both e and e may be complex. The real parts of e, e are associated with polarization (electric and
magnetic).
The imaginary parts of e, e are due to polarization (molecular) loss, i.e., dipole friction and associated
time-lag,
Ohm’s Law
e = e0
e = e0
je00
e0 :
(1.13)
Another relationship that is often useful for lossy media is the point form of Ohm’s law,
Je (r; !) =
Jm (r; !) =
e (r; !)E(r; !);
m (r; !)H(r; !);
(1.14)
12
CHAPTER 1. ELECTROMAGNETIC THEORY
e
m
(ohms
1
=m ) is the electrical conductivity of the medium
(ohms=m ) is the magnetic conductivity of the medium
V = A ohms:
Complex Constitutive Parameters
In (either) transform domain it becomes particularly easy to separate applied quantities from induced
e¤ects.
In (??) the …eld quantities represent the total …elds at a point in space. Assume that an impressed
current density Jie (r) 6= 0 maintains E; D; H; B 6= 0, which, in turn, results in
Jce (r) =
e
E(r) 6= 0;
(1.15)
where Jce is an induced conduction current density. The total electric current is
i
Je (r) = Je (r) + Jce ;
(1.16)
and Faraday’s law becomes
r
H(r) = j!e
" E(r) + Jie (r) +
= j! e
"
De…ning a new complex permittivity as
j
!
e
"
"
leads to
(1.18)
e
H(r) = j!" E(r) + Jie (r);
r
(1.17)
E(r) + Jie (r):
e
j
!
e E(r)
(1.19)
where we have separated the induced e¤ ects from the applied source.
i
Repeating for Jm (r) = Jm (r)+
m H(r)
i
= Jm (r) + Jcm and noting that
r Jie(m) + j!
leads to
e
j
!
Assuming e
", e are real, the imaginary parts of ",
i
e(m)
m
= 0;
(1.20)
:
(1.21)
account for conduction loss.
In general, one often writes
" = ("0
=( 0
j"00 ) ;
j 00 ) ;
(1.22)
j"00r ) "0 ;
j 00r ) 0 :
(1.23)
or, in terms of the relative permittivity,
" = ("0r
= ( 0r
where the imaginary parts account for all loss mechanisms (conductive and molecular).The following
table lists some typical material parameters for dielectric media.
1.4. WAVE EQUATIONS
13
Material
air
glass
wood
gypsum board
dry brick
dry concrete
fresh water
sea water
snow
ice
moist ground
dry ground
copper
"0r
1.0006
3.8–8
1.5–2.1
2.8
4
4–6
81
81
1.2–1.5
3.2
20–30
3–6
1
"00r
(1/ohm-m)
<0.003 @ 3 GHz
<0.07 @ 3 GHz
0.046 @ 60 GHz
0.05–0.1 @ 4,3 GHz
0.1–0.3 @ 3,60 GHz
<0.006 @ 3 GHz
0.0029 @ 3GHz
0.01–0.001
1–6
0.000001
0.03–0.003
0.005–0.00001
5:7 107
Maxwell’s equations become
r (" E(r)) =
r ( H(r)) =
r
r
E(r) =
i
e (r);
i
m (r);
j! H(r)
H(r) = j!" E(r) +
(1.24)
Jim (r);
Jie (r);
Often the superscript i is omitted in (1.24)–the interpretation of J depends on ; ". For example, a
fairly general form is
r
r
r
r
D(r) = e (r);
B(r) = m (r);
E(r) = j!B(r) Jm (r);
B(r) = j! D(r) + Je (r);
(1.25)
where B = H and D = "E.
– If ; " account only for polarization e¤ects, i.e., if ; " are real-valued, or if ; " are complex-valued
where the imaginary parts are associated with polarization loss (dipole friction), then Jm ; Je are
total currents.
– If ; " contain the conductivities, then Jm ; Je are impressed currents.
1.4
Wave Equations
When ! 6= 0 electric and magnetic quantities are coupled, allowing for wave phenomena.
Vector Wave and Vector Helmholtz Equations for Electric and Magnetic Fields
The independent Maxwell’s equations
r
r
E(r) = j! H(r) Jm (r);
H(r) = j!" E(r) + Je (r);
(1.26)
14
CHAPTER 1. ELECTROMAGNETIC THEORY
represent six scalar equations in six unknowns. However, they are coupled vector partial di¤erential equations. We want to uncouple the equations and obtain one equation which may be solved, assuming a
homogenous medium.
Taking the curl of r
E(r) and of r
r
r
r
r
E(r)
H(r)
H(r) leads to
! 2 " E(r) =
j! Je (r)
2
! " H(r) =
r
j!"Jm (r) + r
Jm (r);
(1.27)
Je (r):
These are the vector wave equations for the …elds.
Noting that r
r
A = r (r A)
r2 A, we also have
r2 E(r) + ! 2 " E(r) = j! Je (r) + r
r2 H(r) + ! 2 " H(r) = j!"Jm (r)
r
r e
;
"
r m
Je (r)+
:
Jm (r)+
(1.28)
These are known as vector Helmholtz equations.
1.5
Plane Wave Propagation
We want to consider the simplest solution of source-free wave equations— these will represent travelling plane
waves which, under typical conditions, model realistic electromagnetic waves.
Time-Harmonic Plane Waves in Free Space
We …rst consider what kind of waves can exist in source-free homogeneous space characterized by "; .
At any point in space where sources are absent the electric …eld satis…es
r2 + k 2 E (r) = 0:
(1.29)
Assume we want to …nd a wave travelling along z, independent of x; y, and polarized in the x-coordinate.
Assume
bE (z)
E=x
The wave equation (1.29) becomes
@2
+ k 2 E (z) = 0
@z 2
which has solution
E (z) = E0+ e
jkz
+ E0 e+jkz :
Plugging into Faraday’s law leads to the magnetic …eld as
where
=
p
=", and so the pair
E0+
e
jkz
b E0+ e
E = x
jkz
b
H=y
b
H = y
E0+
e
E0
e+jkz ;
+ E0 e+jkz ;
jkz
E0
e+jkz ;
1.5. PLANE WAVE PROPAGATION
15
form a wave travelling along z. E ? H ?b
z and the wave is called a TEM (Transverse ElectroMagnetic) wave.
Phase and Attenuation Constants of a Uniform Plane Wave
Assume
k=(
j ):
(1.30)
Then
b E0+ e
E = x
is called the phase constant.
b E0+ e
= x
j(
j )z
j z
e
z
+ E0 e+j(
j )z
+ E0 e+j z e
z
(1.31)
:
is called the attenuation constant.
2
is real-valued and " = "0
Setting k 2 = (
j ) = ! 2 ", where
imaginary parts, leads to
v
u
u1
(!) = !
"0 (!)t
2
v
u
u1
p
(!) = !
"0 (!)t
2
p
s
"00 (!)
s
"00 (!)
2
1+
2
"0 (!)
!
+1
2
1+
j"00 , and equating real and
2
"0 (!)
(1.32)
!
1 :
Time-Domain Waves, Phase Velocity, and Wavelength for Uniform Plane Waves
In the time-domain,
E (r; t)
=
b E0+ e
Re x
b
= x
E0+
z
e
j z
e
z
+ E0 e+j z e
cos (!t
z) + E0 e
z
z
ej!t
(1.33)
cos (!t + z)
For the …rst term, as t increases, the argument of the cosine remains unchanged if z increases
correspondingly— thus the wave travels along the +z-direction. Correspondingly, the second term
travels along the z-direction.
The phase velocity (actually speed since it is a scalar) of the wave is found from
d
(!t
dt
!
z)
=
d
Const = 0
dt
dz
dt
=
0
(1.34)
leading to
vp =
dz
!
= :
dt
(1.35)
16
CHAPTER 1. ELECTROMAGNETIC THEORY
The wavelength of the wave is the distance between adjacent wavefronts that produce the same value
of the cosine function. If z1 and z2 are points on adjacent wavefronts,
z1 =
z2
2
or
= z1
z2 =
2
(1.36)
:
(1.37)
b-direction is propagating in the b
Example 1.1 Assume a plane wave with electric …eld oriented in the x
zdirection through dry soil at 1 MHz. At this frequency typical material properties are " = 3"0 and = 10 5
(ohms 1 =m ). Therefore,
"
=
e
"
3
j
!
=
e
j
10
2 106
3"0
j
10
2 106 "0
5
5
(1.38)
j0:179) "0 = "0
"0 = (3
j"00
such that
(!)
= !
=
(!)
v
u
u1
0
"t
2
0:0363 v
u
p u1
"0 t
= !
2
=
Since k = b
z kz = b
z(
p
!
"002
1 + 02 + 1
"
r
(1.39)
!
r
"002
1 + 02
"
1
0:00108:
j ), the wave has the form
E (r) = E0 e
jk r
and in the time-domain
The phase velocity of the wave is
b E0 e
E (r) = x
vp =
!
=
b E0 e
=x
z
j z
cos (!t
2 106
= 1:73
0:0363
e
z
z) :
108 m/s
(1.40)
(1.41)
(1.42)
and the wavelength is
=
2
=
2
= 173:1m.
0:0363
In travelling 1 km the amplitude will decrease from E0 to E0 e
20 log (0:339) =
z
= 0:339E0 , or by
9:396 dB.
(1.43)
Chapter 2
Transmission Line Theory
Transmission lines are used to transfer electrical signals (information) or electrical power from one point to
another in an electrical system. Transmission lines take a wide variety of forms, from simple wire pairs and
cables to more complicated integrated structures for high-frequency applications. Some common transmission
lines are:
(a)
(c)
(b)
w
conducting strip
ground plane
ε,µ
b
(d)
w
ground planes
ε,µ
b
conducting strip
(e)
17
18
CHAPTER 2. TRANSMISSION LINE THEORY
y
b
ε,µ
x
a
z
(f)
where
(a) parallel wires (two-wire line)
(b) coaxial cable
(c) parallel plates
(d) microstrip
(e) stripline
(f ) rectangular waveguide
The table below lists some characteristics of a few common transmission lines and waveguides.
Structure
Coaxial line
Stripline
microstrip line
rectangular
waveguide
Frequency
Range (GHz)
< 50
< 10
< 100
< 300
Impedance
Range (ohms)
10 100
10 100
10 100
100 500
Power
Rating
medium
medium
low
high
Ease of Device
Mounting
medium
medium
good
good
Low Cost
Production
medium
good
good
poor
In this chapter we will study the analysis of general transmission lines, mostly without considering speci…c
physical structures. In the next chapter we will examine some speci…c geometries of transmission lines and
waveguides.
2.1
The Lumped-Element Circuit Model for a Transmission Line
To illustrate the analysis of transmission lines and transmission-line resonators, consider the generic twoconductor1 TEM transmission line depicted in the …gure below, where Vs and Is represent distributed
sources.
i(z,t)
Vs(z,t)
+
-
+
v(z,t)
Is(z,t)
1 Generally
the term “conductor”is used in transmission-line analysis, and, in fact, the lines are usually conductive. However,
other transmission systems, including those only involving dielectrics (e.g., optical …bers), can be modeled using these techniques.
2.1. THE LUMPED-ELEMENT CIRCUIT MODEL FOR A TRANSMISSION LINE
19
Generic two-conductor TEM transmission line.
We assume that the transmission line has dimensions on the order of a wavelength or larger. In this case,
the traditional circuit equations (v = iz, etc.), which come from Maxwells …eld equations specialized for the
case when the physical dimensions of the structure are small compared to a wavelength (i.e., low frequency),
are no longer valid. The low frequency approximations are valid when applied to a small ( z
) section
of the structure. In this way, we treat transmission lines as many cascaded sections of electrically small
circuits, and apply circuit models to each section.
The lumped-element model for a small segment of the line is shown in the …gure below.
i(z,t)
R ∆z L ∆ z
Vs(z,t) ∆ z
i(z+dz,t)
+
-
+
G ∆z
Is(z,t) ∆z
C ∆z v(z+dz,t)
-
∆z
The circuit elements are
R: series resistance per unit length for both conductors, ohms=m: (R = 0 for perfect conductors)
L: series inductance per unit length for both conductors, H=m
G: shunt conductance per unit length, S=m. (G = 0 for perfect insulators)
C: shunt capacitance per unit length, F=m
is : shunt current source per unit length, A=m
vs : series voltage source per unit length, V=m
The distributed sources may represent, for instance, distributed currents and voltages induced on the
transmission line by an external source. If desired, a localized source can be modeled by vs = v0 (z z 0 ),
and similarly for is . We will assume R; L; G; C 2 R.
Applying Kirchho¤’s voltage law and current law to the circuit of Figure 2.2 yields, respectively,
v(z; t) + vs (z; t) z
L z
R
@i(z; t)
@t
z i(z; t)
v(z +
(2.1)
z; t) = 0
and
i(z; t) + is (z; t) z
C
Dividing these two equations by
G
z v(z +
@v(z + z; t)
z
@t
z and taking the limit as
@v(z; t)
=
@z
@i(z; t)
=
@z
R i(z; t)
G v(z; t)
z; t)
i(z +
(2.2)
z; t) = 0:
z ! 0 lead to
@i(z; t)
+ vs (z; t);
@t
@v(z; t)
C
+ is (z; t);
@t
L
(2.3)
20
CHAPTER 2. TRANSMISSION LINE THEORY
and, upon assuming time-harmonic conditions
dv(z)
=
dz
di(z)
=
dz
R i(z)
j!L i(z) + vs (z);
G v(z)
j!C v(z) + is (z):
(2.4)
These two coupled …rst-order di¤erential equations can be easily decoupled by forming second-order di¤erential equations
d2 v(z)
dz 2
d2 i(z)
dz 2
2
d vs (z)
;
dz
d is (z)
(G + i!C) vs (z) +
;
dz
v(z) =
2
(R + i!L) is (z) +
i(z) =
(2.5)
where
2
= (R + j!L) (G + j!C) ;
(2.6)
and = +j 2 C is called the propagation constant (1=m). The real and imaginary parts of the propagation
constant are known as the attenuation constant ( ) and the phase constant ( ), respectively.
Usually in microwave analysis one …rst considers the homogeneous equations
d2 v(z)
dz 2
d2 i(z)
dz 2
2
v(z) = 0;
2
(2.7)
i(z) = 0;
corresponding to the absence of any source or load. General solutions are found as
v(z) = v0+ e
z
+ v0 e+ z ;
i+
0e
z
+ i0 e+ z ;
i(z) =
(2.8)
which represent voltage and current waves. Exploiting (2.4) leads to the relationship between voltage and
current as
1
i(z) =
v + e z v0 e+ z ;
(2.9)
Z0 0
where
R + j!L
Z0
(2.10)
is called the characteristic impedance (ohms) of the transmission line.
Time-domain form:
Assuming sinusoidal time variation,
v(z; t)
=
=
With
Re v (z) ej!t
nh
Re v0+ e ( +j
)z
+ v0 e+(
+j )z
i
(2.11)
o
ej!t :
v0 = v0 ej
(2.12)
we have
v(z; t) = v0+ cos !t
z+
+
+ v0 cos !t + z +
:
(2.13)
z
The …rst term, corresponding to e
, is a forward (+z-traveling) wave, while the second term, corresponding to e+ z , is a backward ( z-traveling) wave. Also,
=
2
, vp =
!
= f:
(2.14)
2.2. FIELD ANALYSIS OF TRANSMISSION LINES
21
Lossless Lines:
If the line is lossless (R = G = 0), then
=
+j =0+j ;
r
p
2
L
= ! LC =
; Z0 =
:
C
2.2
(2.15)
Field Analysis of Transmission Lines
In these notes we won’t follow the derivations presented in the text, but only summarize some of the results.
Note that the RLGC parameters for a given transmission line are speci…c to that line (depending on the
geometric con…guration of conductors and dielectrics, and on material properties), and are determined by
an electromagnetic …eld analysis of the structure.
The circuit parameters are obtained as
L =
C
=
R
=
G
=
2
Z
H H dS;
jI0 j S
Z
"
(2.16)
E E dS;
2
jV0 j S
Z
Rs
H H dl;
2
jI0 j l
Z
!"00
E E dl;
2
jV0 j l
where E is a linear function of V0 , H is a linear function of I0 , S = cross section surface of the line, l =
conductor boundary, Rs = 1= ( s ) is the surface resistivity of the conductors, and V0 ; I0 are constants that
cancel out with the numerators .
Some representative results are provided in the table below.
a
w
ε,µ
a
D
ε,µ
ε,µ
d
b
L
C
R
G
ln ab
2 "0
b
ln a
Rs 1
1
2
a 00 b
2 !"
b
ln a
2
1
cosh
"0
cosh
1
D
2a
D
( 2a
)
Rs
a
!"00
cosh
1
D
( 2a
)
d
w
"0 w
d
2Rs
w
!"00 w
d
Note that for TEM transmission lines comprised of perfect conductors, the exact EM …eld solution is
in full agreement with the microwave model (with G; L; C appropriately de…ned). For lossy conductors the
microwave model is no longer an exact solution, but usually provides a very good approximation. For non
TEM transmission lines (we’ll discuss later, microstrip is one example), the microwave engineering model is
an approximation of the rigorous EM …eld solution.
22
CHAPTER 2. TRANSMISSION LINE THEORY
2.3
The Terminated Lossless Transmission Line
A terminated transmission line is depicted schematically in the …gure below.
Zg
Vg
Z 0 ,v p
+
-
ZL
Z in
z=-L
z=0
The generator (modeled by Vg and Zg ) can represent, for example, the output of an IC chip, the output of
an antenna, etc., and the load (modeled by ZL ) represents, for example, the input to an IC chip, antenna,
television receiver, etc.
The voltage and current on the line are
v(z) = v0+ e
i(z) =
v0+
Z0
j z
+ v0 e+j
v0 +j
e
Z0
j z
e
z
;
z
(2.17)
:
It is convenient to reform (2.17) as
v(z) = v0+ e
i(z) =
v0+
Z0
j z
+
v0 +j
e
v0+
j z
e
v0 +j
e
v0+
z
;
z
(2.18)
:
De…ning a load re‡ection coe¢ cient
=
v0
v0+
j z
+
L
we have
v(z) = v0+ e
i(z) =
v0+
Z0
e
+j z
Le
j z
;
+j z
Le
(2.19)
:
More generally, de…ne
(z) =
v0 j2
e
v0+
z
;
(2.20)
=
v0
;
v0+
(2.21)
=
v0 j2
e
v0+
where we have the special cases
(0) =
( L)
=
L
in
L
;
such that
(z) =
L
ej2
z
(2.22)
2.3. THE TERMINATED LOSSLESS TRANSMISSION LINE
23
leading to
v(z) = v0+ e
v0+
i(z) =
Z0
e
j z
j z
(1 +
(z)) ;
(1
(z)) :
(2.23)
The total impedance at any point is
Z (z) =
v (z)
1+
= Z0
i (z)
1
(z)
;
(z)
(2.24)
and, conversely,
(z) =
Z (z) Z0
:
Z (z) + Z0
(2.25)
Important special cases:
(0) =
( L)
=
L
in
ZL Z0
;
ZL + Z0
Zin Z0
=
:
Zin + Z0
(2.26)
=
(2.27)
An important formula for Zin can be obtain as follows;
Z (z)
=
v+ e
v (z)
= 0+
v0
i (z)
Z0 (e
+
j z
L
ej
z
L
ej
z)
(2.28)
jZ0 tan ( z)
;
jZL tan ( z)
ZL + jZ0 tan ( L)
= Z ( L) = Z0
:
Z0 + jZL tan ( L)
= Z0
Zin
j z
ZL
Z0
One important use of Zin is that once the input impedance has been determined, the transmission line circuit
can be considered to be
Iin
Zg
+
Vg
+
-
Vin
Z in
-
such that
8
<
1
Zin
1
V
PL = Re fvin iin g = Re
: Zin + Zg g
2
2
Time-average power ‡ow:
Zin
Zin +Zg
Zin
Vg
9
=
;
:
(2.29)
24
CHAPTER 2. TRANSMISSION LINE THEORY
Pav
=
=
=
1
Re fv (z) i (z)g
2
(
1
v0+
Re v0+ e j z + L ej z
e j z
2
Z0
8
9
>
>
+ 2
<
=
1 v0
2
j2 z
j2 z
Re 1 + L e
e
j
j
L
>
>
|
{z L
}
2 Z0
:
;
(2.30)
L
ej
z
)
pure im aginary
=
2
v0+
1
2 Z0
n
1
j
2
Lj
o
= P inc:
P ref: in this case –not generally :
When the line is not matched, all of the available power is not delivered to the load. The return loss is
RL =
20 log j
Lj :
(2.31)
Special cases:
j
j
Lj
=
Lj =
0;
1;
RL = 1;
RL = 0 dB.
(2.32)
Standing waves and standing wave ratio:
In the sinusoidal steady state, for a matched line (
v (z) = v0+ e
For a mismatched line (
L
6= 0),
j z
= 0),
L
jv (z)j = v0+ = constant.
;
v (z) = v0+ e
j z
1+
L
ej2
z
:
(2.33)
(2.34)
Let
L
=j
Lj
ej :
(2.35)
Then,
jv (z)j =
vmax
vmin
=
v0+ 1 + j
j(2 z+ )
Lj e
v0+ (1 + j
L j) ;
v0+
=
(1
j
;
(2.36)
L j) :
The standing wave ratio (SWR), also known as voltage standing wave ratio (VSWR) is
SW R =
vmax
1+j
=
vmin
1 j
Lj
Lj
:
(2.37)
Note that
1
SW R
1;
and that SW R = 1 denotes a matched load.
When will SW R = 1? When j L j = 1. This will occur for
(a) a short circuit, ZL = 0, or
(b) an open circuit, ZL = 1, or
(c) a pure reactive load, ZL = jXL , XL 2 R.
(2.38)
2.3. THE TERMINATED LOSSLESS TRANSMISSION LINE
25
The last case is due to the fact that a pure reactance cannot dissipate power, so everything must be
re‡ected.
Distance between successive minima:
ej(2
zm + )
= ej(2 (zm +dmin )+ ) = ej(2 zm + ) ej2
! 2 dmin = n ; n = 0; 2; 4; :::
so that
dmin =
n
2
=
n
22
=
dmin
n
= :
4
2
(2.40)
To examine the e¤ects of a pure standing wave, consider a short-circuited line (ZL = 0;
v (z)
i (z)
= v0+ e
v0+
=
Z0
j z
e
j z
ej
z
+ ej
z
(2.39)
v0+ 2j sin ( z) ;
=
=
v0+
Z0
L
=
1).
(2.41)
2 cos ( z) :
In the time domain,
v (z; t)
i (z; t)
= v0+ 2 sin ( z) sin (!t) ;
=
v0+
Z0
(2.42)
2 cos ( z) cos (!t) :
1
0.5
-7
-6
-5
-4
bz
-3
-2
0
-1
-0.5
-1
sin ( z) sin (!t) vs.
z for !t = 0;
;
2;
6:
Observations:
1. Voltage is zero at the short and at multiples of =2,
v = 0 at
z=n ;
n = 0; 1; 2; :::
2. Voltage is a maximum at all points such that
z=m ;
2
m = 1; 3; 5; :::
3. Current is maximum at the short circuit and at all points where v = 0. At all points where v = vmax ,
i = 0.
4. The total energy in any length of line a multiple of a quarter wavelength is constant, merely interchanging between energy in the electric …eld (voltage) and energy in the magnetic …eld (current). This
is similar to energy relations in a resonant LC circuit.
26
CHAPTER 2. TRANSMISSION LINE THEORY
For j
Then,
Lj
=
6 1, SW R < 1 and both a standing wave and a travelling wave exist on the line. Let
v (z)
= v0+ e
j z
+ ej(
z+ )
= v0+ e
j z
+ ej(
z+ )
= v0+ (1
+ e
j z
j z
+ v0+ ej(
= v0+ (1
)e
|
{z
j z
+ 2 v0+ ej 2 cos
{z
|
pure travelling wave
= ej .
(2.43)
)e
}
L
j z
e
z+ )
j z
+e
z+
pure standing wave
2
}
1.4
1.2
1
0.8
0.6
0.4
0.2
-7
-6
-5
-4
= 0:5; (1
bz
-3
j z
:5) e
-2
0
-1
+ cos ( z) vs.
z
It is easy to see that no real power ‡ow is associated with standing waves, since
Pav =
1
1
Re fv (z) i (z)g = Re
2
2
2jv0+ sin z
2
v0+
cos ( z)
Z0
= 0:
(2.44)
Example 2.1 A lossless, air-…lled transmission line having characteristic impedance 50 ohms is terminated
with a load ZL . The line is 1:2 m long, and operated at a frequency of 900 MHz. Determine (a) SWR, (b)
Zin , (c) RL for ZL = 100 ohms and ZL = 50 ohms.
Solution:
f = 900 MHz, !
L
=
=
c
3 108
=
= 0:333 m.
f
900 106
1:2
= 3:6;
0:333
L=
2
L = 7:2
(2.45)
(2.46)
ZL = 100 :
(a)
L
=
SW R
=
ZL Z0
100 50
1
=
=
ZL + Z0
100 + 50
3
1 + j Lj
=2
1 j Lj
(2.47)
(b)
Zin = Z0
ZL + jZ0 tan ( L)
= 49:10
Z0 + jZL tan ( L)
j35:03 ohms
(2.48)
2.3. THE TERMINATED LOSSLESS TRANSMISSION LINE
27
(c)
RL =
20 log j
Lj
= 9:54 dB.
(2.49)
ZL = 50 :
(a)
L
=
SW R
=
ZL Z0
50 50
=
=0
ZL + Z0
50 + 50
1 + j Lj
=1
1 j Lj
(2.50)
(b)
Zin = Z0
ZL + jZ0 tan ( L)
= 50 ohms
Z0 + jZL tan ( L)
(2.51)
(c)
RL =
20 log j
Lj
=
1 dB.
(2.52)
Exercise 2.1 (in-class):
Z 0 ,v p
ZL
Z in
z=-L
z=0
If L = =2 and ZL = 0 (short circuit), …nd (a) SW R, and (b) Zin .
Exercise 2.2 (in-class):
1. The input to a television receiver presents an impedance of ZL = 60 + j75 ohms to a Z0 = 75 ohm
coaxial cable. The cable has length 0:2 at the frequency of operation. Determine
(a) SWR
(b)
L
(c) Zin
2. Assume an antenna supplies a signal modeled by an open-circuit voltage of 5 V with source resistance
300 ohms to the transmission line described in (1). Determine the power delivered to the television.
3. A 300 ohm transmission line is short-circuited at the load-end. Determine Zin is the transmission line
has length
(a) l = =4,
(b) l = =2;
(c) l = :
28
CHAPTER 2. TRANSMISSION LINE THEORY
Special cases of lossless terminated lines:
I. ZL = 0 (short circuit)
Zin = jZ0 tan L;
(2.53)
which is pure imaginary for any length line L:
4
2
0
1
2
3
z
4
5
6
7
-2
-4
tan( L) vs.
Zin
Zin
= 0 at L = 0;
= 1 at L =
II. ZL = 1 (open circuit)
Zin =
L
=2;
; 3 =2; :::
=4; 3 =4; 5 =4; :::
jZ0 cot L
(2.54)
(2.55)
which is also pure imaginary for any length L:
4
2
0
1
2
3
z
4
5
6
7
-2
-4
cot( L) vs.
Zin
Zin
L:
= 0 at L =
=4; 3 =4; 5 =4; :::
= 1 at L = 0;
=2;
; 3 =2; :::
(2.56)
III. L = =2,
Zin = ZL
(2.57)
Transmission lines of length =2, or any integer multiple of =2, do not transform the load impedance
–it is seen directly at the input.
2.3. THE TERMINATED LOSSLESS TRANSMISSION LINE
29
IV. L = =4;
Zin =
Z02
:
ZL
(2.58)
This case is know as a quarter-wave transformer, and will be discusses in detail later.
Determining v0+ :
Although we often don’t need to explicitly determine v0+ , it is sometimes useful. To determine v0+ ,
consider the transmission line shown below.
Zg
Vg
Z 0 ,v p
+
-
ZL
Z in
z=-L
z=0
We have
v (z)
v ( L)
= v0+ e
=
j z
v0+ ej L
but
v ( L) = vin =
1+
1+
L
L
ej2
e
z
;
j2 L
;
Zin
Vg
Zin + Zg
(2.59)
(2.60)
Equating (2.59) and (2.60) leads to
v0+ = Vg
Zin
Zin + Zg [ej
1
L
+
Le
j L]
:
(2.61)
Example 2.2 If Vg (t) = 2 cos 108 t + =4 , Zg = 1 ohm, Z0 = 50 ohms, L = 3 m, ZL = 75 ohms, and
vp = 2:8 108 m/s, determine v0+ .
Solution:
=
L
=
Zin
2
!
= 0:357;
L = 1:07;
vp
ZL Z0
75 50
1
=
=
ZL + Z0
75 + 50
5
ZL + jZ0 tan ( L)
Z0 + jZL tan ( L)
75 + j50 tan (1:07)
= 50
50 + j75 tan (1:07)
= 38:23 j13:42 ohms,
= Z0
v0+ = Vg
with Vg = 2 ej 4 leads to
=
Zin
Zin + Zg [ej
v0+ = 2:14
1
L
+
j0:23:
Le
j L]
(2.62)
(2.63)
(2.64)
(2.65)
30
CHAPTER 2. TRANSMISSION LINE THEORY
Example 2.3 You …nd cable of unknown characteristic impedance having length 1:5 m. At a frequency at
which you assume the cable is less than a quarter wavelength, you measure the open- and short-circuit input
impedance to be
oc
Zin
sc
Zin
=
j54:6 ohms,
= j103 ohms.
What is Z0 ?
Solution: Assume the cable is very low loss.
oc sc
Zin
Zin
Notice that you can also …nd
( jZ0 cot L) (jZ0 tan L) = Z02
( j54:6) (j103) = 5623:8
p
) Z0 = 5623:8 = 75 ohms.
=
=
as
oc
Zin
sc
Zin
jZ0 cot L
2
= (cot L)
jZ0 tan L
j54:6
= 0:53
j103
cot L = 0:728
L = 0:941;
= 0:628 1/m.
=
=
)
)
2.4
(2.66)
(2.67)
The Smith Chart
The Smith chart is a graphical tool that serves as an aid in performing transmission line calculations. Although any such calculation can be easily performed on a computer, the Smith chart helps in developing
intuition, and serves as a useful background upon which to plot microwave engineering data (from measurements or simulation)
The Smith chart is really the portion j j 1 of the complex –plane, although the numbers on the Smith
chart are impedance values. The mapping between the Z–plane and the –plane is developed as follows.
Z
Z0
= Z = r + jx
=
=
1+
1
1
=
2
r
(1
(2.68)
1+(
1 (
2
i
2) +
r
+ j i)
r + j i)
+ 2j i
:
2
r
i
Equating real and imaginary components,
r=
2
2
r + i
2) + 2
r
i
1
(1
;
x=
2j
(1
i
2) +
r
2;
i
(2.69)
which can be rearranged to yield
r
(
2
r
1+r
2
r
1) +
i
+
2
i
1
x
2
=
=
1
2;
(1 + r)
1
:
x2
(2.70)
(2.71)
Recall that this is simply
1+
(2.72)
1
separated into real and imaginary components. Both (2.70) and (2.71) are circles in the complex –plane.
Z = Z0
2.4. THE SMITH CHART
31
i. The …rst equation, (2.70), shows that the locus of constant resistance r in the
r
1
circles with centers on the r axis at r = 1+r
; i = 0 with radii 1+r
.
r
i
plane is a family of
Γi
r=0
r=0.5
r=1
1
Γr
ii. The second equation, (2.71), shows that the locus of constant reactance x in the
of circles with centers on the r axis at r = 1; i = x1 with radii x1 :
Γi
r
X=1
r=0
X=2
r=1
|Γ|<=1
. (1,0.5)
Γr
. (1,-0.5)
X=-2
X=-1
Superimposing the previous two plots, and showing only the j j
1 portion, leads to
i
plane is a family
32
CHAPTER 2. TRANSMISSION LINE THEORY
Γi
r=0
X=2
X=0.7
|Γ|=1
Γr
X=-0.7
X=-2
Continuing for the other values
0
r < 1;
1<x<1
(2.73)
generates the Smith chart shown below. Note that on the Smith chart the numbers denote r and x values
instead of r and i values. Re‡ection coe¢ cient values, along with other relevant information, is obtained
from the scales provided at the bottom of the chart.
iii. Since
(z) =
j2 z
Le
and
L
=
ZL Z0
ZL +Z0
= ej ,
= constant, then
(z) = ej ej2 z ;
(2.74)
such that moving along a uniform line toward the load end (i.e., increasing position z) is equivalent
to rotating on a constant j j = circle (constant SW R circle) with increasing angle = 2 z in the
counter-clockwise direction. Moving towards the source results in a clockwise rotation.
Γi
.
.
Γr
Since 2 L +
of L = =2.
= 22 L +
=4
L
+ , a full rotation about the Smith chart corresponds to a distance
2.4. THE SMITH CHART
33
34
CHAPTER 2. TRANSMISSION LINE THEORY
iv. Since
SW R =
then a circle of constant
1+j j
1+
=
1 j j
1
(2.75)
is also a circle of constant SW R.
v. Since
Z=
1+
1
Y
=
=
1 + ej(2
1
ej(2
z+ )
z+ )
;
(2.76)
then
1
=
Z
1+
1
=
1
ej(2 z+
1 + ej(2 z+
ej(2 z+ + )
:
ej(2 z+ + )
)
(2.77)
)
Therefore, for a given point Z on the Smith chart, the corresponding admittance Y is obtained by
rotating radians (one-half rotation) along a constant SW R circle. Thus,
impedance (admittance) and be converted to admittance (impedance) by re‡ecting the impedance
point through the origin.
Γi
Z
.
Γr
Y
Example 2.4 Assume L = 0:35 , ZL = 50
Solution:
1. Start on the Smith chart at Z = 1
.
j200 ohms, Z0 = 50 ohms. Find
L,
Zin ,
in ,
and SW R.
j4.
2. From the SW R legend (or constant SW R circle intersection with the positive-real axis), SW R = 20
( SW R = 17:94 from the equation).
3. From the
equation).
legend and Smith chart angle, j j = 0:9,
6
L
=
27 (
= 0:896
26:57 from the
4. To …nd Zin , rotate 0:35 toward the generator (clockwise) around a constant SW R circle ( 0:287 +
0:35 = 0:637 ). Zin = Z0 (0:13 + j1:17) = 6:5 + j58:5 ( Zin = 6:5206 + j57:67 from the equation).
5.
in
= 0:96 81:7 .
2.4. THE SMITH CHART
35
Note: all possible values of Z (z) are found on the constant SW R circle. Thus, it is very easy to
visualize how Z varies along the line.
Example 2.5 A transmission line is terminated in ZL . Measurements show that the standing wave minima
are 102 cm. apart, and that the last minimum is 35 cm. from the load. The measured SW R is 2:4, Z0 = 250
ohms, and vp = c. Determine (a) frequency and (b) ZL .
Solution:
(a)
2
f
=
=
102 )
= 204cm.,
vp
1010
= 147:06MHz.
204
=
3
(b) vmin is l = 35 cm from the load. l= = 0:172. On Smith chart, draw SW R circle. Start at vmin and
rotate ccw 0:172 . Zin = Z0 (1:16 j0:96) = 290 j240 ohms.
36
CHAPTER 2. TRANSMISSION LINE THEORY
Example 2.6 A lossless 100 ohm transmission line is terminated in 200 + j200. The line is 0:375 long.
Determine (a) L , (b) SW R, (c) Zin , and (d) the shortest length of line for which the impedance is purely
resistive, and the value of the resistance.
Solution:
(a) Z L = 2 + j2.
L
= 0:626 30 = 0:62ej 6 .
(b) SW R = 4:3
(c) Zin = Z0 (0:32 + j0:54) = 32 + j54
(d) l = 0:042 , R = Z0 4:3 = 430 ohms.
2.4. THE SMITH CHART
37
Exercise 2.3 (in-class)
The input to a television receiver presents an impedance of ZL = 60 + j75 ohms to a Z0 = 75 ohm coaxial
cable. The cable has length 0:3 at the frequency of operation. Determine, using the Smith chart,
1. SWR
2.
L
3. Zin
Smith Chart with lumped elements:
38
CHAPTER 2. TRANSMISSION LINE THEORY
C
B
A
XL=30
Z0
Xc=-200
Z0
ZL
Z in
0.06λ
Example 2.7 XL = 30 (Zind = j!L = jXL ), XC =
0.12λ
200 (Zcap =
1
j!C
=
1
j !C
= jXC ). Determine Zin .
Solution:
1. First obtain ZA in the usual way–enter Smith chart at Z L = 2 + j1:5. Rotate 0:12
generator. Z A = 1 j1:3.
toward the
2. Incorporate inductor L1 :
(a) Option a: Z B = ZA +
j30
50
=1
j0:7:
(b) Option b: Note that adding reactance just means moving along a constant resistance contour to
the desired point. The distance moved is j30
50 = j0:6.
3. Convert Z B to Y B . Y B = 0:67 + j0:47.
4. Incorporate capacitor C1 :
(a) Option a: Y C = Y B +
1
j200
50
= 0:67 + j0:72.
(b) Option b: Note that adding susceptance just means moving along a constant conductance contour
1
to the desired point. The distance moved is j200
= j0:25.
50
5. Either rotate 0:06 toward generator to get Yin and convert to Zin , or convert to Z C …rst then rotate
0:06 to get Zin . Zin = Z0 (0:44 j0:39) = 22:4 j18:97 ohms.
2.5. THE QUARTER-WAVE TRANSFORMER
39
Exercise 2.4 (in-class)
Determine, by moving along contours on the Smith chart, the input impedance seen at the beginning of
the line.
20+j10 ohms
50 ohms
0.15
2.5
200+j100 ohms
λ
The Quarter-Wave Transformer
The =4 transformer is used to match two di¤erent real impedance values as shown below.
40
CHAPTER 2. TRANSMISSION LINE THEORY
Z0
RL
Z1
Z in
λ/4
Zin = Z1
RL + jZ1 tan ( L)
Z1 + jRL tan ( L)
(2.78)
but L = =2,
Zin =
Assume we want no re‡ection at the input (
in
Z12
:
RL
(2.79)
= 0, Zin = Z0 ). Choose
Z1 =
p
RL Z0 .
(2.80)
To connect a real-valued load RL with a transmission linephaving Z0 6= RL , use a =4 length section
of transmission line having characteristic impedance Z1 = RL Z0 .
Standing waves do occur on the =4 section of line.
Example 2.8 Design a quarter-wave transformer to match a 200 ohm load to a 50 ohm line.
Z=
p
(50) (200) = 100 ohms.
Quarter-wave transformers are very frequency sensitive. Why? Later we will study methods to broaden
the usable frequency range by using cascaded sections of =4 lines.
The multiple Re‡ection Viewpoint:
Read
2.6
Generator and Load Mismatches
In this section we consider the general case of a source-excited, loaded lossless line, and power delivered to
the load.
Zg
Vg
Z 0 ,v p
+
-
ZL
Z in
z=-L
z=0
2.7. LOSSY TRANSMISSION LINES
PL
=
vin
=
41
1
Re (vin iin ) , since the line is lossless.
2
Zin
Vg
vin
Vg ; iin =
=
;
Zin + Zg
Zin + Zg
Zin
(2.81)
2
noting that ZZ = jZj we get
2
PL
=
=
1
Zin
1
2
Re
jVg j
2
Zin + Zg
Zin
1
Rin
2
jVg j
2
2:
2
(Rin + Rg ) + (Xin + Xg )
(2.82)
1. Special case: load matched to line:
ZL = Z0 ,
L
= 0, SW R = 1, Zin = Z0 :
PL =
1
Z0
2
jVg j
:
2
2
(Z0 + Rg ) + Xg2
(2.83)
1
2 1
jVg j
:
8
Z0
(2.84)
Further, if Zg = Z0 ,
PL =
2. Special case: generator matched to a loaded line:
Zin = Zg ,
in
= 0.
Rg
1
2
jVg j
:
2
4 Rg2 + Xg2
PL =
(2.85)
Depending on conditions, either case may provide the larger power delivered to the load. What determines
maximum power transfer from generator to load? As in circuit theory, conjugate matching.
Zin = Zg
(2.86)
This will result in maximum power transfer to the load for a …xed Zg . The resulting power delivered to the
load is
1
2 1
:
(2.87)
PL = jVg j
2
4Rg
Note that PL is made large by making Rg small.
2.7
Lossy Transmission Lines
Before considering lossless transmission lines we obtained (2.8)
v(z) = v0+ e z + v0 e+ z ;
1
i(z) =
v + e z v0 e+
Z0 0
=
Z0
(2.88)
z
;
p
+ j = (R + j!L) (G + j!C);
s
R + j!L
:
G + j!C
(2.89)
42
CHAPTER 2. TRANSMISSION LINE THEORY
For lossy lines
z
e
=
e| {z z}
j z
e
:
(2.90)
attenuation
In this case, movements along the line no longer are represented by rotation along a constant SW R circle
(for lossy lines j j = 6= constant). For the low-loss case we have
'
1
2
R
+ GZ0 ;
Z0
p
' ! LC;
(2.91)
Z0 '
r
L
.
C
The distortionless line:
The exact equation for
of a lossy transmission line,
p
= (R + j!L) (G + j!C);
(2.92)
indicates that is a complicated function of frequency. If doesn’t have a linear relationship with frequency,
then the phase velocity vp = != will be di¤erent at each frequency (vp = vp (!)). Di¤erent frequency
components of a signal (as in a Fourier decomposition) will travel at di¤erent velocities, reaching the load
at di¤erent times, resulting in dispersion of the signal (a form of distortion). This will be discussed later.
One special case, mostly of academic interest, exists for a lossy line to be dispersion-free. If
R
G
=
L
C
then
p
= ! LC;
(2.93)
=R
r
C
.
L
(2.94)
The resulting line will not distort the signal, although attenuation will still occur.
For a general lossy line power relations are
Pin
PL
PLoss
2
v0+ n
1 j
2Z0
2
v0+ n
=
1 j
2Z0
= Pin PL ;
=
where
v0+ = Vg
Zin
Zin + Zg [e
2
in j
2
Lj
o
o
e2
L
;
;
1
L
+
Le
(2.95)
L]
(2.96)
(compare with (2.61)).
2.8
Transient Transmission Lines
(not in text)
The time-domain transmission line (telegrapher) equations, (2.3), are
@v(z; t)
=
@z
@i(z; t)
=
@z
R i(z; t)
G v(z; t)
@i(z; t)
;
@t
@v(z; t)
C
;
@t
L
(2.97)
(2.98)
2.8. TRANSIENT TRANSMISSION LINES
43
where we have set the distributed sources vs ; is = 0. These …rst-order, coupled partial di¤erential equations
are decoupled by applying @=@z to the …rst equation (2.97) and substituting into the second equation (2.98)
to yield a second-order partial di¤erential equation for v. A similar equation is obtained by applying @=@z
to (2.98) and substituting into (2.97), resulting in
@2
@z 2
LC
@2
@t2
(RC + GL)
@
@t
v (z; t)
i (z; t)
RG
= 0:
(2.99)
For simplicity we will assume a lossless line (R = G = 0), resulting in
@2
@z 2
LC
@2
@t2
v (z; t)
i (z; t)
= 0:
(2.100)
We will solve for v (z; t) and use (2.97) to obtain i (z; t).
In order to solve
@2
@2
LC
v (z; t) = 0
@z 2
@t2
(2.101)
we use Laplace transforms. Recall
L fv (z; t)g = V (z; s) ;
@
v (z; t)
= sV (z; s) v z; t = 0+ ;
L
@t
@2
L
v (z; t)
= s2 V (z; s) sv z; t = 0+
@t2
(2.102)
@
v (z; t)
@t
;
t=0+
where V (z; s) is the Laplace transform of v (z; t). Initial conditions are
1. v (z; t) across C cannot change instantaneously. Assume v (z; t < 0) = 0. Then, v (z; t = 0+ ) = 0.
2. i (z; t) ‡owing in L cannot change instantaneously. Then,
@
v (z; t)
@t
1 @
i (z; t)
C @z
=
t=0+
= 0:
(2.103)
t=0+
Taking the Laplace transform of the wave equation (2.101) results in
@2
@z 2
where
2
V (z; s) = 0
(2.104)
p
= s LC. The solution of (2.104) is
V (z; s)
= V + (s) e
= V
+
(s) e
z
p
(s) e+
+V
s LCz
+V
z
(2.105)
p
+s LCz
(s) e
where V (s) are constants of integration with respect to z. The solution v (z; t) is obtained via inverse
Laplace transform,
n
o
p
p
v (z; t) = L 1 V + (s) e s LCz + V (s) e+s LCz ;
(2.106)
and, using the time-shifting theorem
L ff (t
we obtain
v (z; t) = v + t
t0 )g = F (s) e
p
LCz + v
st0
t+
(2.107)
p
LCz :
(2.108)
44
CHAPTER 2. TRANSMISSION LINE THEORY
p
We can interpret the term LCz by noting that the propagation velocity (the velocity with which an
imaginary observer must move to follow a point of constant amplitude on the wavefront) is obtained from
v
t
p
LCz
p
1
LC
=
constant,
p
) t
LCz = constant,
p
d
t
)
LCz = 0;
dt
dz
dt
dz
dt
=
(2.109)
0;
(2.110)
1
= vp ;
LC
1
.
) vp = p
LC
p
=
Therefore,
v + (t z=vp )
|
{z
}
v (z; t) =
+
forward travelling wave
1
v + (t
Z0
(2.111)
backward travelling wave
The associated current can be obtained as
i (z; t) =
v (t + z=vp ) :
|
{z
}
z=vp )
v (t + z=vp )
(2.112)
as shown in the appendix.
In summary, we have
v (z; t)
=
i (z; t)
=
v + (t z=vp ) + v (t + z=vp ) ;
1
v + (t z=vp ) v (t + z=vp ) ;
Z0
(2.113)
valid for an in…nite line. To analyze a …nite length, source-excited and loaded line we need to consider
conditions at the source-end and load-end of the line.
Terminated transient line:
i(z,t)
i(L,t)
+
+
v(z,t)
Z 0 ,v p
v(L,t)
-
RL
-
With
v (L; t)
=
=
i (L; t) =
=
v + (L; t) + v (L; t)
+
vL
(t) + vL (t) = vL (t)
+
i (L; t) + i (L; t)
i+
L (t) + iL (t) = iL (t)
(2.114)
2.8. TRANSIENT TRANSMISSION LINES
45
and Ohm’s law applied at the load,
vL (t) = iL (t) RL ;
(2.115)
we obtain
+
vL
(t) + vL (t)
=
=
i+
L (t) + iL (t) RL
1
v + (t) vL (t) RL
Z0 L
(2.116)
leading to
vL =
RL Z0 +
v =
RL + Z0 L
+
L vL
(2.117)
(compare with (2.26) for the phasor case).
Launching waves on an in…nite line:
Rg
i(0,t)
i(z,t)
+
Vg
v(0,t)
+
-
Z 0 ,v p
v(z,t)
-
Ohm’s law at z = 0 results in
v (0; t) = vS (t) = Vg (t)
| {z }
vs (t)
for all t. In particular,
vs 0; t = 0+
= Vg t = 0+
= Vg t = 0+
Rg i (0; t)
| {z }
(2.118)
is (t)
Rg is t = 0+
(2.119)
+
Rg
vs (t = 0 )
;
Z0
leading to
vs 0+ =
Z0
Vg 0+ = v + 0; 0+
Z0 + Rg
(2.120)
which gives the initial amplitude of the wave launched on the line. For z; t 6= 0 we obtain
v (z; t)
=
i (z; t)
=
Z0
Vg (t z=vp ) = v + (t
Z0 + Rg
v (z; t)
= i+ (t z=vp ) :
Z0
z=vp ) ;
Complete transient response of a terminated, source-driven transmission line: bounce diagrams:
(2.121)
46
CHAPTER 2. TRANSMISSION LINE THEORY
Rg
Vg
Z 0 ,v p
+
-
RL
z=0
z=L
z
0,0
z=L
z
0,0
v+
t
T
ΓL v+
2T
ΓL Γg v+
3T
i
T
t
- Γ i+
L
ΓL Γg i +
2T
3T
2
ΓL Γg v+
4T
2
-Γ Γ i+
L g
4T
i(z,t)
v(z,t)
T =
v (z; t)
=
i (z; t)
=
L
= one-way transit time for wave transverse line.
vp
Z0
fVg (t
Z0 + Rg
+
z=L
+
2
L g Vg
(t
(t
L Vg
z=vp )
(t
2T + z=vp ) +
2 2
L g Vg
4T + z=vp ) +
1
fVg (t
Z0 + Rg
2
L g Vg
z=vp ) +
L Vg
(t
4T
(t
(t
2T
z=vp )
z=vp ) + ::: ;
2T + z=vp ) +
2 2
L g Vg
4T + z=vp ) +
(t
L g Vg
4T
L g Vg
(t
(2.122)
2T
z=vp ) + :::
z=vp )
(2.123)
Example 2.9 Consider a matched line (RL = Z0 ) excited by an ideal (Rg = 0) source. Determine v (z; t).
Solution:
Vg
Z 0 ,v p
+
-
RL
z=0
z=L
L
(
g
= 0;
g
=
Rg Z0
=
Rg + Z0
1
(2.124)
is not needed since line is matched – no re‡ection comes back towards the source).
v (z; t)
i (z; t)
Z0
Vg (t z=vp ) = Vg (t
Z0 + 0
= Vg (t z=vp ) =Z0
=
z=vp ) ; :
2.8. TRANSIENT TRANSMISSION LINES
47
Vg(t)/Vo
1
t
To
V(L,t)/Vo
leading edge
1
trailing edge
t
T+To
T=L/Vp
Example 2.10 Assume that a step voltage is applied to a line mismatched at both ends. If RL = 3Z0 ,
Rg = 3Z0 , determine v (0; t) and v (L; t).
Solution:
Rg
Vg
Z 0 ,v p
+
-
z=0
Vg (t)
L
v+
RL
z=L
= V0 u (t) ;
RL Z0
3Z0 Z0
1
=
=
= =
RL + Z0
3Z0 + Z0
2
Z0
V
0
=
Vg t = 0+ =
:
Z0 + Rs
4
s;
Then,
v (z; t)
=
V0
1
1
u (t z=vp ) + u (t 2T + z=vp ) + u (t 2T
4
2
4
1
1
+ u (t 4T + z=vp ) + u (t 4T z=vp ) + ::: ;
8
16
v (0; t)
=
=
V0
1
1
u (t) + u (t 2T ) + u (t 2T )
4
2
4
1
1
+ u (t 4T ) + u (t 4T ) + :::
8
16
V0
3
3
u (t) + u (t 2T ) + u (t 4T ) + :::
4
4
16
z=vp )
(2.125)
(2.126)
(2.127)
48
CHAPTER 2. TRANSMISSION LINE THEORY
v (L; t)
=
=
V0
1
1
u (t T ) + u (t 2T + T ) + u (t 2T
4
2
4
1
1
+ u (t 4T + T ) + u (t 4T T ) + :::
8
16
V0 3
3
u (t T ) + u (t 3T ) + :::
4 2
8
T)
(2.128)
(2.129)
V(0,t)/(Vo/4)
3/16
3/4
1
t/T
1
2
3
4
5
6
V(L,t)/(Vo/4)
3/8
1
3/2
t/T
1
2
3
4
5
6
As a check, d.c. analysis yields
v (t) =
and
V0
v (L; t = 1) =
4
(
1+
3Z0
V0
V0 =
3Z0 + 3Z0
2
1
2
1
+
1
2
2
+
(2.130)
1
2
3
)
+ :::
=
V0
2
(2.131)
which agrees with the plot.
Example 2.11 Consider a pulse-excited line mismatched at the load. If RL = 2Z0 , Rg = Z0 , determine the
voltage at the middle of the line.
Solution:
Rg
Vg
Z 0 ,v p
+
-
z=0
RL
z=L
2.8. TRANSIENT TRANSMISSION LINES
49
Vg(t)/Vo
1
t
To
v (z; t)
=
=
v (L=2; t)
=
=
L
=
g
=
2Z0 Z0
1
= ;
2Z0 + Z0
3
0
Z0
fVg (t z=vp ) + L Vg (t 2T + z=vp )g
Z0 + Z0
1
1
Vg (t z=vp ) + Vg (t 2T + z=vp )
2
3
1
1
T
Vg t
+ Vg (t 2T + T =2)
2
2
3
1
1
T
3T
Vg t
+ Vg t
:
2
2
3
2
V(L/2,t)/(1/2)
To<T/2
1
1/3
t/(T/2)
1
(T/2)
I. T0 < T =2
2
3
(T)
(3T/2)
4
5
6
V(L/2,t)/(1/2)
To=T/2
1
1/3
t/(T/2)
1
II. T0 = T =2
(T/2)
2
3
(T)
(3T/2)
4
5
6
(2.132)
(2.133)
(2.134)
(2.135)
(2.136)
50
CHAPTER 2. TRANSMISSION LINE THEORY
V(L/2,t)/(1/2)
To=T
1
1/3
t/(T/2)
1
(T/2)
2
3
4
(T)
(3T/2)
5
6
III. T0 = T
V(L/2,t)/(1/2)
To=2T
1
1/3
t/(T/2)
1
(T/2)
2
3
(T)
(3T/2)
4
5
6
IV. T0 = 2T
Exercise 2.5 (in-class)
Two IC chips are mounted a printed circuit board. Pin 3 on one chip is connected to pin 5 of the other
chip by a printed conducting trace, forming a 20 ohm transmission line. The transmission line has length
l = 1 cm, and the velocity of signal propagation on the line is vp = 2:6 108 m/s. Pin 3 can be modeled
as a voltage source which provides a 1:5 volt pulse starting at t = 0 and lasting for 1 ns, and having source
resistance Rg = 10 ohms. Pin 5 can be modeled as providing a constant 30 ohm resistance. Plot the voltage
waveform at pin 5 versus time.
Example 2.12 Consider the charged-line pulse generator shown below. If Rg = Z0 , RL
the voltage at z = 0.
Solution:
t=0
i(L,t)
i(0,t)
+
=
>>
Z0
+
v(0,t)
Rg
RL
v(L,t)
Z 0 ,v p
+
-
Z0
Z0 determine
-
-
z=0
z=L
V0
Operation: resistor Rg is switched to close the circuit and interrupt the d.c. state. Output voltage v (0; t) is
a pulse of width 2T These types of pulse generators are used to generate very high power, fast pulses.
Initial state (t < 0): v (z; t) = V0 , i (z; t) = 0. For t > 0
v 0; t = 0+
=
g
=
L
=
Rg Z0
= 0;
Rg + Z0
RL Z0
' 1:
RL + Z0
V0 + v + =
+
Rg i 0; t = 0+
v
' v+ .
Z0
) 2v + = V0 (initial pulse amplitude).
=
Rg
(2.137)
(2.138)
2.8. TRANSIENT TRANSMISSION LINES
v (z; t) =
V0
|{z}
51
v + (t z=vp )
|
{z
}
+
d.c. voltage
forw ard w ave (re‡ected on ce at RL )
v (z; t)
= V0
v (0; t)
= V0
=
V0
V0
u (t z=vp )
u (t
2
2
1
1
1
u (t)
u (t 2T )
2
2
V0
fu (t)
2
u (t
+ v (t
|
2T + z=vp )
{z
}
(2.139)
backw ard w ave
2T + z=vp )
(2.140)
2T )g :
v(0,t)/(Vo/2)
1
t
2T
2.8.1
Waveforms and Spectral Analysis
The text (Pozar) deals with time-harmonic signals and associated analysis. However, a time-harmonic
waveform can not be used to transmit information. In the analog domain, modulation is commonly used
to impart information on a high-frequency carrier, which spreads the frequency content out from the single
spectral component of the carrier. Time-harmonic analysis will be valid if the bandwidth of the modulated
signal is su¢ ciently small. In the digital domain, one must consider pulses and pulse trains. In this section
we consider the frequency-domain aspects of some digital signals, and the relation to time-harmonic analysis.
Periodic Signals
Periodic waveforms can be represented Fourier series. Let g (t) be periodic with period T , i.e., g (t) =
g (t + T ). Then,
1
a0 X
n2 t
n2 t
g (t) =
+
+ bn sin
an cos
2
T
T
n=1
where
an
=
bn
=
2
T
2
T
Z
T =2
g (t) cos
n2 t
dt;
T
g (t) sin
n2 t
dt;
T
T =2
Z
T =2
T =2
for n = 0; 1; 2; :::, or in exponential form
g (t) =
1
X
cn ej
n2 t
T
n= 1
where
cn =
1
T
Z
T =2
g (t) e
j n2T
t
dt;
T =2
n = 0; 1; 2; ::: (the equality is in the sense of the Fourier series). For example, for a pulse train consisting
of rectangular pulses having amplitude A, pulse width t0 , and period T , as shown below,
52
CHAPTER 2. TRANSMISSION LINE THEORY
g(t)
t0
A
...
t
T
we have
g (t)
1
X
=
cn ej
n2 t
T
;
n= 1
cn
=
=
Z
1
T
1
T
= A
Z
T =2
j n2T
g (t) e
T =2
t0
j n2T
Ae
t
t
dt
dt
0
t0
e
T
t0
T
jn
sin n Tt0
n t0
T
(for this to be valid t0 < T =2, otherwise the limits of integration need to be adjusted). The quantity f0 = 1=T
is called the fundamental frequency (or pulse repetition rate), in which case
cn = At0 f0 e
jn t0 f0
(sin n t0 f0 )
:
n t0 f0
A plot of jcn j vs. n=T looks like the following.
Cn
c0
1 3
T T
n/T
2
t0
1
t0
Although the spectrum is discrete, the envelope is drawn for convenience. Note that
as T decreases (f0 increases), the density of spectral lines decreases, and the amplitude of the spectral
lines increases,
the shape of the envelope is determined by t0 , the width of a single pulse.
Therefore, the spectral envelope of the pulse train, especially how far out in frequency it extends, is
governed by the pulse width, t0 .
Non-Periodic Signals
If we consider a single pulse, we need to use Fourier transforms,
Z 1
1
g (t) =
G (!) ej!t d!;
2
1
! = 2 f , where
G (!) =
Z
1
1
g (t) e
j!t
dt:
2.8. TRANSIENT TRANSMISSION LINES
53
For a single rectangular pulse,
G (f ) = At0
which has zeros at
f=
sin ( t0 f )
;
t0 f
m
;
t0
m = 1; 2; 3; :::.These are exactly the locations of the zeros of the envelope of cn for a pulse train of period T .
Comparing the spectrum of the pulse train and of the single pulse,
sin ( t0 f )
;
t0 f
(sin n t0 f0 )
jcn j = At0 f0
;
n t0 f0
jG (f )j = At0
one sees that the pulse width t0 e¤ects the amplitude of each spectrum in the same way.
For a pulse train (such as a clock signal), the period T is not too important from a spectral analysis
standpoint. It does a¤ect the density of the discrete spectral components, but not how fast the spectrum
decreases with increasing frequency.
From a spectral analysis standpoint, the pulse width is of paramount importance, for both the pulse
and the pulse train.
For a pulse train, the main importance of the period is logic timing.
For a modulated analog signal, the important spectral region is near the carroer frequency, whereas
for a digital pulse, the important spectral content is near the origin in the frequency domain.
2.8.2
Integrated Circuits and Ground Bounce
Consider a typical integrated circuit (IC) logic device as shown below.
Vcc
A
VA
Q1
Vout
Q2
B
VB
Gnd
When Q1 is on and Q2 is o¤ the output voltage is high (Vout = VA = Vcc ), and when Q1 is o¤ and
Q2 is on the output voltage is low (Vout = VB = 0). However, VA is not actually Vcc volts, and VB is not
actually 0 volts, due to parasitic inductance. Both leads A and B are wires, and, like all wires, they have
inductance. The inductance of a very short (electrically short) straight wire is very small, although perhaps
not negligible depending on the application, leading to the model shown below.
54
CHAPTER 2. TRANSMISSION LINE THEORY
Vcc
Lwire
Q1
Vout
Q2
Lwire
Gnd
Since
vind = Lind
diind (t)
;
dt
(2.141)
then if iind is large enough and/or dt is small enough, vind will be large and Vout (t) = Vcc vind (t) or
Vout (t) = 0 + vind (t). This phenomena is called ground bounce (and Vcc bounce). If the noise margins of
the circuit downstream are too tight, or the timing is too fast (all voltages approach their idealized values
when the currents stop changing) a logic error may follow. To remedy this situation we may 1) slow down
the circuit (not a viable choice), reduce the margins (often not a viable choice), or minimize Lind , which is
usually the preferred method, if possible.
The above described ground bounce occurs internal to a chip. Integrated circuits (ICs) have been traditionally formed on a silicon die, which is glued to a mechanical base (this is referred to as “packaging”
the circuit). Small wires, called wirebonds, connect the die to the package’s external leads. Even through
these wirebonds are typically electrically short, they have, like all wires, inductance (on the order of 2 nH
for a typically wirebond). Therefore, any IC that has large transient currents can induce large, undesirable
voltage transients on the wirebonds. Wirebonds connecting to ground often have large current transients,
resulting in ground bounce as well. For example, if an IC chip has several hundred thousand gates, many of
which are switching at the same time, and all connected to ground through the same wirebond, the switching
current can become quite large, resulting in ground bounce of several volts. To make matters worse, the
inductance of the ground and power planes is added to the inductance of the wirebonds, enhancing this
deleterious e¤ect. Flip-chip technologies (which spread the ground connections over many solder bumps) are
e¤ective at mitigating this e¤ect.
2.8. TRANSIENT TRANSMISSION LINES
55
Vcc
Lwire+Lpower- plane
Q1
Vout
Q2
Lwire+Lground- plane
Gnd
Although ground bounce is not a “transmission line” e¤ect, it is particular to high frequencies, and
relates to the electromagnetic properties of the circuit. Bypass capacitors are often used to remove the
ground and power plane transients from the IC circuits. Since current can change “instantaneously”through
a capacitor, the transient currents on the IC device don’t have to ‡ow through the ground and power planes,
they ‡ow through the capacitor. Wide traces are used if possible (wide traces have lower inductance and
larger capacitance).
56
CHAPTER 2. TRANSMISSION LINE THEORY
Chapter 3
Transmission Lines and Waveguides
In the previous chapter we studied transmission lines in an abstract sense, usually without referring to a
particular physical structure. In this chapter we will study several types of transmission lines and waveguides
commonly used in practice. As is evident from their physical structure, each transmission line and waveguide
has particular advantages and disadvantages depending on the application of interest. Any waveguide or
transmission line that is capable of supporting a TEM mode of propagation can be analyzed using the
transmission line theory discussed in the previous chapter (for that mode). Waveguides and transmission
lines that do not support a TEM mode can be approximately analyzed using the previously developed
analysis methods.
In general, the term “waveguide” is applied to structures consisting of a single conductor. In this case,
the structure cannot support a TEM mode. Two-conductor structures are usually termed “transmission
lines,” and can support a TEM mode (and higher-order modes).
3.1
General Solutions for TEM, TE, and TM Modes
In this section we will study general solutions for propagation modes along structures invariant along the
z axis (and therefore in…nitely long). The conductors will be initially assumed to be perfectly conducting;
later brief mention will be made of attenuation calculations.
It can be show from the theory of Fourier transforms
Z
1
f (x; y; z)e j z dz;
1
Z 1
1
1
f (x; y; )ej z d :
f (x; y; z) = F ff (x; y; )g =
2
1
f (x; y; ) = F ff (x; y; z)g =
(3.1)
(3.2)
that …elds on a z invariant structure can be written as
E (x; y; z) =
H (x; y; z) =
[e (x; y; ) + z ez (x; y; )] e
[h (x; y; ) + z hz (x; y; )] e
j z
(3.3)
j z
where fe; hg represent the (as yet unknown) transverse …elds, fez ; hz g represent the (as yet unknown)
longitudinal …elds, and is the (as yet unknown) propagation constant. The analysis in the text shows how
to obtain these unknown quantities for a given physical structure. Here we will mostly be concerned with
obtaining .
Starting with the source-free curl equations
r
r
E =
j! H;
H = j!"E
57
(3.4)
58
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
and assuming the form (3.3) leads to
@Ez
+ j Ey =
@y
@Ez
j Ex
=
@x
@Ey
@Ex
=
@x
@y
j! Hx ;
(3.5)
j! Hy ;
(3.6)
j! Hz ;
(3.7)
and
@Hz
+ j Hy = j!"Ex ;
@y
@Hz
j Hx
= j!"Ey ;
@x
@Hy
@Hx
= j!"Ez :
@x
@y
(3.8)
(3.9)
(3.10)
The above equations may be solved for the transverse components Hx ; Hy ; Ex ; Ey in terms of the longitudinal components Ez ; Hz as
Hx
=
Hy
=
Ex
=
Ey
=
@Ez
@Hz
j
!"
;
kc2
@y
@x
j
@Hz
@Ez
!"
;
+
kc2
@x
@y
@Hz
@Ez
j
+!
;
kc2
@x
@y
@Ez
@Hz
j
+!
;
kc2
@y
@x
(3.11)
(3.12)
(3.13)
(3.14)
where
kc2
= k2
= !2 "
2
=
Often we write the above as
=
2
(3.15)
2
2
p
k2
2
:
kc2 ;
(3.16)
and once kc is found then is determined (since k is known).
We will study three types of waves that commonly occur on transmission line and waveguiding structures.
1. TEM waves
A transverse electromagnetic (TEM) wave has no longitudinal components, i.e., Ez = Hz = 0. From
(3.11)–(3.14) this would result in E = H = 0, and so no …eld would be present. However, a TEM wave
can exist. The solution of this paradox is that for a TEM wave it must be true that
kc
=
)
0;
(3.17)
=
k=
!
p
"=
2
p
such that TEM waves act like the waves we have studied previously (in Chapter 2 we had = ! LC =
2
). In the text it is shown that TEM …elds are the same as the static …elds (zero frequency) that
exist on the transmission line, and satisfy Laplace’s equation.
3.2. PARALLEL PLATE WAVEGUIDE
59
2. TE waves
It turns out that a transmission line must have two or more conductors for a TEM waves to exist.
However, this is a necessary but not su¢ cient condition. The existence of two or more conductors does
not guarantee that a TEM wave will exist; microstrip is an example of a two-conductor line that does
not support a TEM wave.
Another wave type is a TE (transverse electric) wave, with Ez = 0, Hz 6= 0. In this case all transverse
…elds may be found from (3.11)–(3.14) using the …eld component Hz , which satis…es
@2
@2
@2
+ 2 + 2 + k 2 Hz = 0, or
2
@x
@y
@z
2
2
@
@
2
+ 2
+ k 2 Hz = 0;
@x2
@y
@2
@2
)
+ 2 + kc2 Hz = 0;
2
@x
@y
(3.18)
which is solved subject to the appropriate boundary conditions for a given physical structure, although
will skip this computation. The result of that analysis, through, leads to the determination of kc , and
using
p
= k 2 kc2 ;
(3.19)
to the determination of .
3. TM waves
TM waves have Ez =
6 0, Hz = 0, and the transverse …eld components are found from Ez using (3.11)–
(3.14), where Ez is a solution of
@2
@2
+
+ kc2 Ez = 0:
(3.20)
@x2
@y 2
As with TE waves, kc is determined in the course of solving (3.20). leading to .
In summary, we have
TEM waves
TE waves
TM waves
3.2
Ez = 0
Ez = 0
Ez 6= 0
Parallel Plate Waveguide
Hz = 0
Hz 6= 0
Hz = 0
= k
p
= pk 2 kc2 .
= k 2 kc2
A parallel plate waveguide consists of two conducting plates having width w, separated by spacing d, and
…lled with material characterized by "; , as shown below.
y
ε,µ
z
x
60
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
1. TEM waves
The parallel plate structure supports a TEM mode, which is the dominant mode (i.e., the mode that
exists at the lowest frequencies of operation). The TEM mode is the most important mode from a
practical standpoint. The analysis in the text shows that for the TEM mode,
E (x; y; z)
=
H (x; y; z)
y
= x
V0
e
d
V0
j z
ZT EM d
e
;
( = k) ;
j z
;
and that
d
Z0 =
;
w
=
(3.21)
ZT EM =
r
=
r
"
;
(3.22)
"
independent of frequency.
β
1
k
2. TM modes
Parallel plate waveguides also support TE and TM modes. For TM modes,
Ex
Ey
Ez
Hx
where
r
= Hy = Hz = 0;
j
n
=
y e j nz ;
An cos
kc
d
n
= An sin
y e j nz ;
d
j!"
n
y e j nz ;
=
An cos
kc
d
(3.23)
n 2
n
kc =
.
; n = 0; 1; 2; :::;
d
d
For n = 0 the TM0 mode is really the TEM mode. For n > 0 two situations occur:
n
(a) If k >
n
d
n
d
then
n
=
k2
is real-valued and, therefore, e
j
nz
is a purely propagating mode.
(b) If k <
then n = j n is imaginary-valued and, therefore, e
purely attenuating mode.
Thus, two regimes exist: k > kc =
n
d
(3.24)
(propagation), and k < kc =
n
d
j
nz
=e
j( j
n )z
=e
nz
is a
(attenuation).
Note that this is fundamentally di¤erent that a wave propagating through a lossy medium, where we
obtain
e j ze z;
(3.25)
i.e., propagation and attenuation occurring at the same time. For the TMn mode considered here, the
p
wave either propagates (at higher frequencies, where k = ! " > nd ) or does not (at lower frequencies,
p
where k = ! " < nd ). If the wave is purely attenuating, it is called evanescent, or cuto¤ .
3.2. PARALLEL PLATE WAVEGUIDE
61
The frequency at which a cuto¤ (non-propagating) mode begins to propagate is called the cuto¤
frequency, and is determined by
n
n
kc =
) fc = p .
(3.26)
d
2d "
Real β
k
β
Imag.
β
TEM
TM1
TM2
k1
k
k2
As frequency increases, more modes begin to propagate. This is usually undesirable. Why?
Note that
vp =
but that the speed of light is
What does vp
k, vp
(3.27)
!
:
k
c=
Therefore, since
!
(3.28)
c!
c imply?
Consider the TM1 mode.
1
Ez
=
r
k2
2
d
;
= A1 sin
y e j
8 d
1 < j( y=d
= A
e
{z
2j : |
(3.29)
1z
1 z)
}
plane wave in -y,+z
|e
j( y=d+
{z
1 z)
9
=
} ;:
plane wave in +y,+z
62
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
y
d
+y,+z wave
-y,+z wave
θ
z
With
(
2
1
ky
=
kz
=
d
1
= k sin ;
(3.30)
= k cos
2
+ ( =d) = k 2 ) the phase velocity of each plane wave is
vpP W =
!
= c:
k
(3.31)
The phase velocity of each plane wave in the z direction is
vp =
!
1
=
!
k cos
=
c
:
cos
(3.32)
Since jcos j 1, then vp c. The phase velocity being larger than c is simply due to observing the
wave along one direction (the z direction) when the wave is propagating along another direction (the
direction).
constant phase wavefronts
L
θ
d
With L cos = d, and observer at d sees the wave arrive in T seconds. The observer calculates the wave
velocity to as vp = d=T , but the wave actually travelled a distance L in time T , such that c = L=T .
Since d > L, vp > c. An example of this is a water wave hitting a shoreline at an oblique angle. The
wave “break” travels up the coast faster than the wave is moving. We’ll see later that the energy
velocity is always c.
3. TE waves
For TEn modes,
Ey
Ex
Hy
Hz
= Hx = 0;
n
j!
=
An sin
y e j nz ;
kc
d
j
n
=
An sin
y e j nz ;
kc
d
n
= An cos
y e j nz ;
d
(3.33)
3.2. PARALLEL PLATE WAVEGUIDE
where
n
For n
=
r
63
n
d
k2
2
;
n = 1; 2; 3; :::;
kc =
n
d
.
(3.34)
1 two situations occur, as before:
(a) If k >
n
d
then
n
j
is real-valued and, therefore, e
nz
is a purely propagating mode.
(b) If k < nd then n = j n is imaginary-valued and, therefore, e
purely attenuating mode.
Thus, two regimes exist: k > kc =
n
d
(propagation), and k < kc =
n
d
j
nz
=e
j( j
n )z
=e
nz
is a
(attenuation).
Note that for both the TMn and TEn modes,
n
=
fc
=
r
n
d
k2
2d
n
p
"
2
;
(3.35)
:
In summary, for the parallel plate waveguide, TEM, TMn , and TEn modes exist.
For the TEM mode,
frequency).
=k =!
p
", and no cuto¤ exists (the TEM mode propagates down to zero
q
2
n
For the TMn and TEn modes, n = k 2
(n = 0; 1; 2; ::: for TM modes, and n = 1; 2; 3; ::: for
d
TE modes). Two regimes exist, below cuto¤ (k < nd , pure imaginary), and above cuto¤ (k > nd ,
n
pure real). The dividing point is k = kc = nd , leading to fc = 2dp
".
An in…nite number of modes exist, although at any …nite frequency a …nite number of modes are
propagating.
Are parallel plate waveguides used in practice? Sometimes, but more often an unintentional parallel
plate waveguide is formed by a circuit designer’s attempt to shield a circuit from interference, or to eliminate
undesired radiation (emissions) from a circuit. In this case, on must be aware of the modes of the waveguide,
and especially the cuto¤ frequencies of the allowed modes of propagation.
interference signal
undesired emission
IC chips
coupling via surface wave
ε
initial circuit layout
64
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
undesired emission
IC chips
coupling via surface wave
ε
shielded circuit layout
Or, perhaps, adding a top cover didn’t cause problems at an operating frequency f0 < fc;n . Later,
operating frequency is increased and the circuit doesn’t function properly because f0 > fc;n .(for some n)
and waveguide modes begin to propagate. One solution is to decrease d (to increase fc;n ), or to put in some
lossy dielectric material to absorb the emissions (however, energy loss from the circuit still occurs, perhaps
leading to circuit malfunction).
3.3
Rectangular Waveguide
A rectangular waveguide is a hollow (or dielectric …lled) metal pipe with rectangular cross-section. It can
propagate TE and TM modes, but not TEM modes.
y
b
ε,µ
x
a
z
The table below provides a summary of results for rectangular waveguide.
3.4. CIRCULAR WAVEGUIDE
TEmn mode
p
! "
q
2
2
(m =a) + (n =b)
q
2
k 2 kc;mn
k
kc;mn
mn
c;mn
g;mn
vp;mn
Ez
Hz
Ex
Hx
Hy
Z
TMmn mode
p
! "
q
2
2
(m =a) + (n =b)
q
2
k 2 kc;mn
2 =kc;mn
2 = mn
!= mn
0
Amn cos (m x=a) cos (n y=b) e j mn z
j! n
j mn z
k2
b Amn cos (m x=a) sin (n y=b) e
2 =kc;mn
2 = mn
!= mn
Bmn sin (m x=a) sin (n y=b) e j mn z
0
m
j mn z
a Bmn cos (m x=a) sin (n y=b) e
j
2
kc;mn
j n
2
kc;mn
b Bmn sin (m x=a) cos (n
j!"n
2
kc;mn
b Bmn sin (m x=a) cos (n
j!"m
k2
a Bmn cos (m x=a) sin (n
c;mn
j! m
2
kc;mn
a Amn sin (m
j m
2
kc;mn
a Amn sin (m
j n
k2
b Amn cos (m
Ey
65
c;mn
j
x=a) cos (n y=b) e
mn z
x=a) cos (n y=b) e
j
mn z
x=a) sin (n y=b) e
ZT E = k = mn
j
mn z
c;mn
ZT M =
mn
y=b) e
j
mn z
y=b) e
j
mn z
y=b) e
j
mn z
=k
TE modes:
r
k2
m
a
r
2
2
n
b
m; n = 0; 1; 2; :::;
(3.36)
The …rst (dominant) mode is the TE10 mode, assuming a b.
TM modes:
r
m 2
n 2
k2
; m; n = 1; 2; 3; :::;
mn =
a
b
r
1
m 2
n 2
fc;mn =
+
:
p
2
"
a
b
(3.37)
mn
=
fc;mn
=
2
1
p
"
The …rst mode to propagate is the TM11 mode.
3.4
Circular Waveguide
Omit
3.5
Coaxial Line
TEM modes:
m
a
2
+
;
n
b
2
:
66
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
V=V0
a
ε
V=0
b
Z0 =
E =
H =
b
a
ln
2
.
V0
e j z;
ln (b=a)
r
"
V0
e
ln (b=a)
(3.38)
= k;
j z
(3.39)
:
The common value of 50 ohm for the characteristic impedance of many transmission lines comes from
some attributes of coaxial lines.
– Power handling capacity: The maximum power capacity of a coaxial line is limited by voltage
breakdown, and is given by
a2 Ed2
Pmax =
ln (b=a)
(3.40)
where Ed is the …eld strength at breakdown. Solving for the value of b=a that maximizes Pmax
leads to a characteristic impedance of 30 ohms, as follows: Holding b …xed,
d
Pmax =
da
where
a2
b
a
1
+ 2a ln
a
=0
(3.41)
= Ed2 = . This results in ln (b=a) = 1=2. Then,
1
Z0 =
2
r
"
ln
b
a
' 60
1
2
= 30 ohms
(3.42)
assuming an air-…lled line (long ago most coax was air-…lled, since the available dielectric had too
much loss).
– Attenuation: The attenuation of a coaxial cable due to …nite conductivity of the conductors is
c
=
The equation obtained by minimizing
c
Rs
2 ln
1
a
1
b
.
(3.43)
is
b
ln
a
which has solution
b
a
b
a
=1+
b
= 3:591:
a
b
a
(3.44)
(3.45)
3.6. SURFACE WAVES ON A GROUNDED DIELECTRIC SLAB
67
The corresponding characteristic impedance is
r
1
b
Z0 =
ln
' 60 ln (3:591) = 76:71 ohms.
2
"
a
(3.46)
Therefore, 50 ohms represents a compromise between maximum power handling capability (30
ohms) and minimum attenuation (77 ohms), (30 + 76:71) =2 = 53:35. 50 ohms is also typically
realized using reasonable material dimensions.
Coaxial lines also support higher-order modes (TE and TM modes) (the cuto¤ frequency for the …rst
higher-order modes can be obtained approximately as
fco '
p
c
"r (a + b)
(3.47)
We will not consider them here, but it is important to note that above some frequency these modes will
begin to propagate. Multi-mode propagation is usually to be avoided, since more than one mode carries
power, potentially leading to increased dispersion (as will as input–output coupling e¢ ciency decreases).
3.6
Surface Waves on a Grounded Dielectric Slab
A grounded dielectric slab is a dielectric sheet on a ground plane, and forms the backbone of printed circuit
technology.
x
surface wave
ε0
d
εr
z
This waveguide can support TE and TM surface waves, that are bound to the vicinity of the surface (they
decay exponentially vertically). These waveguide can be used to carry signals in the horizontal direction,
and are, indeed, used for this purpose at very high frequencies (typically optical frequencies). At lower
microwave frequencies, and considering typical slab thickness values used in printed circuit technology, the
surface waves are not strongly bound to the surface, and so are not typically used as intentional waveguides.
However, circuits printed on a ground slab will excite surface waves that may interfere with other circuits
printed on the same substrate.
circuit 1
circuit 2
coupling via surface wave
ε
1. TM modes:
The TM modes have …eld components Ez ; Ex ; and Hy . The propagation constant
the solution of
q
q
q
2
2
2
tan
d = "r
er k02
er k02
k02
is determined by
(3.48)
68
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
and the cuto¤ frequencies from
fc =
nc
p
2d "r
1
;
n = 0; 1; 2; :::
(3.49)
The n = 0 mode (TM0 ) propagates down to zero frequency (although exactly at ! = 0 the …elds
vanish). Thus, this mode is always present, and often leads to undesirable coupling. In general, the
higher the frequency the more energy carried by surface waves, resulting in higher coupling.
2. TE modes:
The TM modes have …eld components Hz ; Hx ; and Ey . The propagation constant
the solution of
q
q
q
2
2
2
cot
d =
er k02
er k02
k02
is determined by
(3.50)
and the cuto¤ frequencies from
fc =
(2n 1) c
p
;
4d "r 1
n = 1; 2; 3; :::
(3.51)
All TE modes have a low-frequency cuto¤.
As discussed previously, surface waves can lead to signi…cant coupling between circuit elements. For
example, consider a grounded slab of thickness d and permittivity "r "0 , and an in…nite line source carrying
current
I = I0 cos (!t) ;
(3.52)
as shown in the …gure below.
x
P
ε0
d
εr
z
3.6. SURFACE WAVES ON A GROUNDED DIELECTRIC SLAB
69
This source produces only TE electromagnetic …elds. For both the source and the observation point P (x; z)
placed on the dielectric interface, the total …eld is given by
Z 1
p2 2
1
k0 x j z
F ( )e
e d
(3.53)
Ey =
2
1
where
p2 2
p2 2
k1 coth
k d
p2 2
p 2 12
2 k2 +
k
coth
k
I0
0
1
1d
p
F( )=
j!"0
2 2 k2
p
with k1 = "r k0 . It can be shown that the …eld can be decomposed into three parts:
p
1+ p
k02
(3.54)
(1)
due to the direct radiation from the source (in the absence of the grounded slab),
(2)
due to interaction of the source and the grounded slab (e.g., multiple partial re‡ections), and
1. Ey
2. Ey
3.
2
(3
Ey
due to surface waves excited in the slab.
The origins of these …eld components are shown in the …gure below.
x
P
E
(1)
E
(2)
ε0
d
εr
E
The three …eld components are, for x = 0; kz
e jkjzj
Ey(1) ' E1 p
;
jkzj
(3)
z
1,
e jkjzj
Ey(2) ' E2 q
;
3
jkzj
Ey(3) ' E3 e
j jzj
(3.55)
where E1;2;3 are constants. Therefore, the surface wave is the dominant contribution to the …eld, and does
not decay with distance from the source (for an in…nite source).
For "r = 2:25 and d = 1 cm, the TE surface waves are given as shown below.
1.5
TE1
1.4
β/ k0
TE3
1.3
TE5
1.2
1.1
1.0
0
10
20
30
40
f (GHz)
50
60
70
70
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
At approximately 6:7 GHz, 20:12 GHz, and 33:5 GHz, the …rst three TE surface waves begin to propagate,
respectively. After those frequencies, a plot of jEy j vs. frequency shows spikes corresponding to surface wave
excitation.
5
E y (a.u.)
4
3
2
1
0
3.7
0
10
20
f (GHz)
30
40
Stripline
Stripline is formed by two ground planes with a conducting strip between the planes.
w
ground planes
ε,µ
b
conducting strip
Stripline supports a TEM mode. The analysis of the TEM (and the higher-order TE and TM modes) is
di¢ cult; some results for the TEM modes are provided below.
=k=!
p
"
since TEM
Z0
'
30
b
;
p
"r we + 0:441b
we
b
=
w
b
0;
0:35
w 2
b
(3.56)
(3.57)
;
w
b
w
b
> 0:35;
:
< 0:35
To obtain strip width w for a given Z0 ,
w
b
=
x =
x;
0:85
30
p
"r Z0
p
0:6
0:441:
p
e Z < 120;
pr 0
;
x;
er Z0 > 120
(3.58)
3.8. MICROSTRIP
3.8
71
Microstrip
Microstrip transmission line is one of the most common types of transmission line extensively used in printedcircuit technology (printed using photolithographic techniques). It consists of a conductor of width w printed
on a grounded dielectric slab.
w
conducting strip
ground plane
ε, µ
b
Microstrip does not support a TEM mode. In fact, microstrip doesn’t support TE or TM modes either,
only hybrid modes (those with all six …eld components). However, the dominant mode is like, at su¢ ciently
low frequencies, a TEM mode, prompting a quasi-TEM analysis. The details are included in the text, with
some results summarized below.
p
p
= k0 "e = ! 0 "0 "e ;
(3.59)
c
!
=p ;
vp =
"e
g
where
"e =
=
p
0
"e
;
" r + 1 "r 1
1
q
+
2
2
1+
If given line dimensions we obtain Z0 as
( 60
8b
w
;
12b
w
1 < "e < "r :
w
4b ;
120
;
p
w
"e [ w
d +1:393+0:667 ln( d +1:444)]
p
Z0 =
"e
ln
+
w
b
w
b
1
1 ;
and for a given characteristic impedance and dielectric constant, the required w=b ratio is
( 8eA
w
w
b < 2;
e2A
h 2;
i
=
"r 1
2
0:61
w
b
B 1 ln (2B 1) + 2"r ln (B 1) + 0:39
; b > 2;
"r
(3.60)
(3.61)
(3.62)
where
A =
B
=
r
Z0 "r + 1 "r 1
+
60
2
"r + 1
377
p :
2Z0 "r
0:23 +
0:11
"r
;
(3.63)
Example 3.1 Design a microstrip line on a "r = 2:25 substrate of thickness b = 0:2 cm to have a characteristic impedance of 50 ohms.
Solution: From the above equations, w=b = 3:037, so w = 0:6074 cm.
Dispersion:
72
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
The previous formulas for Z0 and "e are valid for low frequencies. The e¤ects of frequency can be
approximately accounted for using
"ef f (f )
Z0 (f )
where
F =
p
4bf "r
c
p
p
2
"r
"e p
+
"
;
e
1 + 4F 3=2
r
"e
"ef f (f ) 1
= Z0
;
"e 1
"ef f (f )
=
1 h
w i2
+ 1 + 2 log10 1 +
2
b
1
(3.64)
(3.65)
Plot of "ef f (f ) vs. f . "ef f (0) = "e .
Losses:
Losses on microstrip consist of three components: conductor loss, dielectric loss, and radiation loss. Usually conductor loss dominants over dielectric loss (at least for high-quality microwave substrates). Radiation
loss include energy converted into surface waves, and energy radiated into space. Both e¤ects are most
signi…cant at circuit discontinuities, and are minimized when b= 0
0:01 (e.g., at 2:4 GHz, 0 = 12:5 cm,
and most practical substrates are electrically very thin).
Approximate conductor and dielectric loss formulas as
c
where Rs =
p
!
0 =2
'
Rs
Np/m,
wZ0
(3.66)
is the surface resistivity of the conductor, and
d
=
k0 "r ("e 1) tan
p
2 "e ("r 1)
Np/m,
(3.67)
where tan = Im f"g = Re f"g.
In summary, we have considered several types of transmission lines and waveguiding structures. Some,
like stripline, coaxial line, and parallel plate waveguides, support REM modes, as well as TE and TM modes.
Others, like rectangular waveguide and grounded slab waveguides, only support TE and TM modes. Still
others, like microstrip, only support hybrid modes.
If a structure supports a TEM mode and is lossless, then that mode (there may be others too) can be
modeled using the transmission line theory developed in Chapter 2. For these modes
p
p
= ! " = ! LC;
(3.68)
p
and Z0 = L=C follows from the …eld analysis in Section 2:2.
3.9. THE TRANSVERSE RESONANCE TECHNIQUE
73
If a structure does not support a TEM mode, then no unique de…nition of Z0 exists, although often only
the ratio of impedance values is necessary.
To examine actual high-frequency e¤ects on microstrip transmission lines, consider a line 10 cm long,
1:73 cm wide, on a d = 0:5 cm grounded dielectric slab having "r = 2:2, such that Z0 ' 50 ohms (by
approximate analysis) at 5 GHz. The transmission parameter S21 in dB, as computed by Ansoft Designer
SV, is approximately 0 dB for 1
f
10 GHz (it decreases from 0 dB at 1 GHz to 0:2 dB at 10 GHz.
To compare with a more accurate analysis, the same structure was analyzed by Ansoft Ensemble SV, which
performs an accurate (and time consuming) electromagnetic analysis. The line is fed and terminated by 50
ohm probes. The results is shown below.
10 cm microstrip line, probe fed and terminated.
Full-wave electromagnetic analysis.
w=1.7262 cm, d=0.5 cm,
εr=2.2
0
S 21 (dB)
-10
-20
-30
-40
1
2
3
4
5
6
7
8
9
10
f (GHz)
In large part, the decrease in S21 at high frequency is due to the probes, which can’t be analyzed by the
simple circuit-based analysis. For comparison, the simulation in Designer was virtually instantaneous, while
the more accurate Ensemble simulation took about 20 minutes.
3.9
The Transverse Resonance Technique
Skip
3.10
Wave Velocities and Dispersion
p
For sinusoidal steady-state situations we have c = 1= ", the speed of light in the material characterized
by ; ", and vp = != . For a TEM wave vp = c, whereas for TE and TM modes these velocities di¤er (and
generally vp c). In this case there is another velocity that is useful (and physically signi…cant) to describe
the propagation of signals.
Group Velocity:
The group velocity, vg , is the velocity at which a narrow-band signal propagates. Considering a signal to
be made up of many Fourier components at di¤erent frequencies, one can de…ne a phase velocity for each
wave component. If is not a linear function of !, then vp = != is di¤erent for each Fourier component.
As a result, the wave “breaks up” as it propagates down the line, resulting in dispersion. No single phase
velocity describes the wave. The concept of group velocity overcomes this problem for narrow-band signals.
74
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
Let
f (t)
F (!)
:
:
time-domain signal,
frequency domain (Fourier transform) signal
f (t) $ F (!)
where the Fourier transform pair is
F (!)
f (t)
=
=
Z
1
2
1
f (t) e
1
Z
1
j!t
dt;
(3.69)
F (!) ej!t d!:
1
Modeling the system (in this case, the transmission line) as having transfer function Z (!), then
Fin (ω)
Fo (ω)
Z (ω)
transfer function of transmission line
Fo (!) = Z (!) Fin (!) :
(3.70)
Question: What is the transfer function Z (!)? Consider a lossless, matched line,
Z 0 ,v p
ZL
Z in
z=-L
where ZL = Z0 . In this case v (z) = v0+ e
z=0
j z
, such that
v ( L) = v0+ ej
v (0) = v0+ ;
and so
vout
= Z (!) = e
vin
L
;
j (!)L
(3.71)
:
Therefore, the transfer function of a lossless, matched line of length L is e j (!)L .
Question: What is Fin (!)? Fin (!) is the Fourier components of the signal fin (t).
Example 3.2 If fin (t) is a rectangular pulse of width 2T0 ,
(3.72)
3.10. WAVE VELOCITIES AND DISPERSION
75
f(t)
1
t
2To
Fin (!)
=
=
Z
Z
1
fin (t) e
1
2T0
e
j!t
j!t
dt
(3.73)
j!T0
dt = 2T0 e
0
sin (!T0 )
:
(!T0 )
1
0.8
0.6
0.4
0.2
-10
0
-5
5
x
10
-0.2
Plot of sin( x) = ( x) vs. x.
Therefore, the time-domain output is
fo (t)
=
=
=
=
If
(!) = ! (linear in !;
=
p
Z
1
2
Z
1
2
Z
1
2
Z
1
2
1
1
1
1
1
1
1
Fo (!) ej!t d!
(3.74)
Z (!) Fin (!) ej!t d!
e
j (!)L
Fin (!) ej!t d!
Fin (!) ej(!t
(!)L)
d!:
1
" for TEM waves), then
Z 1
1
fo (t) =
Fin (!) ej!(t
2
1
= fin (t
L) ,
L)
d!
(3.75)
which is an exact replica of fin (t) shifter by L to account for time delay in passing through the network.
p
p
For example, if =
", then L =
"L = L=c = T , the one-way transit time.
76
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
Therefore, lossless TEM lines are dispersionless.
If
(!) is not linear in !, (as for TE and TM waves), then
Z 1
1
Fin (!) ej(!t
fo (t) =
2
1
(!)L)
d! =?
(3.76)
and fo (t) is not, in general, simply a replica of fin (t).
If
(!) is not linear in !, the dispersion will occur.
To consider group velocity, consider a narrow-band signal
s (t) = f (t) cos (! 0 t) = Re f (t) ej!0 t
(3.77)
as in an amplitude modulated waveform with carrier frequency ! 0 . the signal f (t) represents some information that is modulated with a high frequency carrier for transmission. Assume that the highest frequency
component of f (t) is ! m , and that ! m
! 0 . The various Fourier transforms are
Z 1
F (!) =
f (t) e j!t dt;
(3.78)
1
Z 1
Z 1
S (!) =
s (t) e j!t dt =
f (t) ej!0 t e j!t dt
1
1
Z 1
j(! 0 !)t
=
f (t) e
dt
1
= F (!
!0 ) :
F (ω )
ω
S (ω )
−ω 0
ω0
ω
Now, consider the modulated signal to be the input to a transmission line or waveguide having transfer
function Z (!) = e j z .
Sin (!) = F (! ! 0 ) ;
So (!) = F (! ! 0 ) e j z ;
Z 1
1
so (t) =
Re
So (!) ej!t d!
2
1
Z 1
1
=
Re
F (! ! 0 ) e j z ej!t d!
2
1
Z !0 +!m
1
=
Re
F (! ! 0 ) ej(!t z) d!:
2
!0 !m
(3.79)
3.10. WAVE VELOCITIES AND DISPERSION
If ! m
! 0 , expand
77
(!) about ! 0 ,
(!)
=
(! 0 ) +
'
+
0
0
0
d
d!
(!
(!
1 d2
2 d! 2
!0 ) +
!=! 0
(!
2
! 0 ) + :::
(3.80)
!=! 0
!0 ) :
Then,
so (t)
=
=
=
=
1
Re
2
Z
! 0 +! m
! 0 ) ej(!t
F (!
z)
d!
(3.81)
!0 !m
Z !0 +!m
0
1
F (! ! 0 ) ej (!t 0 z 0 (! !0 )z) d!
Re
2
! !
Z !0 m m
0
1
Re
F (y) ej ((y+!0 )t 0 z 0 yz) dy
(c.o.v. y = !
2
!m
Z !m
0
1
F (y) ej (t 0 z)y dy
Re ej(!0 t 0 z)
2
!m
!0 )
which is
so (t)
=
n
Re ej(!0 t
0
0z
= f t
0 z)
f t
cos (! 0 t
0
0z
o
(3.82)
0 z)
which is a time-shifted replica of the input fin (t).
0
The velocity of the envelope f t
0 z is the group velocity,
0
0z
t
d
t
dz
0
0z
= constant,
=
1
vg
=
(3.83)
0;
0
0;
) vg =
1
0
0
=
d
d!
1
:
!=! 0
Waveguide velocities:
p
=
k2
kc2
=
r
!
c
2
vp
=
!
c2
c
;
v
=
=
:
g
c2
!
k0
!
k0 c
=
;
vg
<
c < vp ;
d
d!
=
vp vg =
kc2 ;
k0 c c
= c2 :
k0
(3.84)
78
CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES
Chapter 4
Microwave Network Analysis
4.1
Impedance and Equivalent Voltage and Currents
Voltage and current can be uniquely de…ned for TEM transmission lines
H
+
E
H
-
v=
Z
( )
E dl
(4.1)
(+)
independent of path l from one conductor to another, and
I
I=
H dl;
(4.2)
C+
where C + is any closed path enclosing the (+) conductor. With
Z0 =
v
I
we see that Z0 is uniquely de…ned.
Voltage and current cannot be uniquely de…ned for a non-TEM line.
As an example, consider a rectangular waveguide.
79
(4.3)
80
CHAPTER 4. MICROWAVE NETWORK ANALYSIS
y
b
Ey1;0 = A sin
where ZT1;0
E =k =
1;0 .
x
a
0
x
e
a
j z
Hx1;0 =
;
Ey1;0
ZT1;0
E
;
(4.4)
Then, for the T E10 at z = 0,
v=
Z
( )
E dl =
(+)
Z
b
A sin
0
x
x
dy = Ab sin
a
a
(4.5)
and we see that voltage depends on position x. At x = 0 we obtain v = 0, and at x = a=2 we obtain v = Ab.
Therefore, voltage associated with the T E10 mode is not unique.
What can be done for non-TEM lines? Useful results are often obtained from the following considerations:
1. De…ne voltage and current for a particular mode such that voltage is proportional to the transverse
electric …eld and current is proportional to the transverse magnetic …eld.
2. Voltages and currents should be de…ned so that their product gives the power ‡ow (which is uniquely
de…ned) of the mode.
3. The ratio of voltage to current for a single travelling wave should be chosen equal to the characteristic
impedance of the line. Not that this value is itself arbitrary.
Proceeding with the rectangular waveguide example, let vc be the voltage evaluated along the center of
the guide (at x = a=2); vc = Ab. Let
Iz =
Z
a
0
A
Hx dx = 1;0
ZT E
Z0 =
Z
a
0
sin
x
a A
dx = 2
:
a
ZT1;0
E
vc
Ab
b 1;0
= a A =
Z :
Iz
2 a TE
2 Z 1;0
(4.6)
(4.7)
TE
This is known as the voltage-current de…nition of Z0 . Other de…nitions:
P
;
Iz2
v2
Z0 = c ;
P
Z0 =
power-current,
power-voltage.
(4.8)
4.1. IMPEDANCE AND EQUIVALENT VOLTAGE AND CURRENTS
81
Since
1
2
P =
Z
S
Re fE
H g dS
(4.9)
2
ab jAj
;
4 ZT1;0
E
=
the power-current de…nition yields
2
ab jAj
4 Z 1;0
TE
Z0 =
2
a
A
1;0
ZT
E
2
=
1b
4a
2
2
ZT1;0
E
(4.10)
and the power voltage yields
2
Z0 =
(Ab)
2
ab jAj
4 Z 1;0
TE
b
= 4 ZT1;0
:
a E
(4.11)
Note that each of the three de…nitions results in a di¤erent value for Z0 , but that all de…nitions result in the
form
b 1;0
Z ;
(4.12)
Z0 =
a TE
where is a constant. Thus, methods that only depend on the ratios of impedances will work (and give the
same answer) regardless of how Z0 is de…ned.
If one needs to match a waveguide (or other non-TEM line) to a structure that has a unique impedance
(i.e., a TEM line), then the usual approach is to use the de…nition that gives the best agreement between
experiment and theory.
Often, for non-TEM lines, and for TEM lines as well, one needs to consider the electromagnetic …eld
analysis of the structure (at least to some extent). Consider example 4.2 in the text, which shows the side
view of a waveguide junction.
ε0
y =b
εr ε0
TE10
y =0
z
z=0
a =
"r =
such that
3:485 cm, b = 1:580 cm,
2:56; f = 4:5 GHz,
r
a
=
=
r
k02
2
= 27:5 m
a
1
for the air-…lled guide, TE10 mode, and
d
"r k02
2
a
= 120:89 m
1
for the dielectric …lled guide. Only the TE10 mode propagates in either region.
82
CHAPTER 4. MICROWAVE NETWORK ANALYSIS
Γ
Z0a
Z0b
z=0
Z0a
=
k0
0
= 1292:1 ohms,
a
Z0d
=
=
the same result for
k
d
Z0d
Z0d
= 293:9 ohms,
Z0a
=
+ Z0a
0:629;
would be obtained for any of the three de…nitions of Z0 . So, this is enough analysis.
However, consider the juncture of two waveguides having di¤erent cross-sections but where b=a remains
constant. Assuming the T E10 mode propagates in both sections, Z0 is the same in each section! This would
imply = 0 which doesn’t make sense. So, a one-mode transmission line analysis is not enough here.
Physically, the geometrical change at z = 0 excites an in…nite number of higher-order modes. Those that
propagate (i.e., those that are above cuto¤) will travel away from the discontinuity. Below cuto¤ modes will
decay away from the z = 0, but will store energy in the vicinity of z = 0. This stored energy can be a very
important part of the analysis. In the end, usually a transmission line mode can be obtained, with lumped
reactances modeling the stored energy e¤ects (see text Section 4.6).
4.1. IMPEDANCE AND EQUIVALENT VOLTAGE AND CURRENTS
83
84
CHAPTER 4. MICROWAVE NETWORK ANALYSIS
Chapter 5
Impedance Matching and Tuning
The purpose of a matching network is to provide a good match between a given transmission line and a
given load. In this chapter we will consider ways to design matching networks with consideration to matching
network bandwidth, physical implementation of the network, and the ability to adjust the network. Several
types
Zg
Vg
matching
Z 0 ,v p
+
-
ZL
network
Z in
Transformer Matching
(not in text book)
A transformer can be used to match a real-valued load RL to a line having characteristic impedance Z0
by appropriate choice of the turns ratio.
Zg
n1:n2
i1
i2
+
+
Vg
V1
+
V2
ZL
-
Z1
Z2
-
-
Assume Zg = Z0 and ZL = RL . Assuming an ideal transformer,
V2
n2
=
;
V1
n1
i2
n1
=
:
i1
n2
(5.1)
We want Z1 = Z0 and Z2 = RL , where Z1 = V1 =i1 , Z2 = V2 =i2 . Then
Z1 =
V1
=
i1
n1
n2 V 2
n2
n1 i2
=
n1
n2
85
2
V2
=
i2
n1
n2
2
Z2 :
(5.2)
86
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
Recalling that we want Z1 = Z0 and Z2 = RL we obtain
Z1 = Z0 =
2
n1
n2
Z2 =
and so
Z0 =
n1
n2
2
n1
n2
RL ;
(5.3)
2
RL :
(5.4)
So, given Z0 and RL , a transformer with turns ratio
n1
=
n2
r
Z0
RL
(5.5)
provides a perfect impedance match.
Example 5.1 Match a 300 ohm line to a 75 ohm load.
Solution: Choose
r
r
Z0
300
n1
= 2:
=
=
n2
RL
75
(5.6)
This technique works very well at low frequencies, where a transformer can be constructed to operate in
a somewhat ideal fashion. The common 300 ohm–to–75 ohm adapter used in many television systems
utilizes this technique.
The match is frequency independent over the band of frequencies for which the transformer exhibits
nearly ideal behavior.
5.1
Matching with Lumped Elements (L Networks)
jX
(a)
Z
Z
jB
0
L
A
jX
(b)
jB
Z
L
A
By correct choice of X and B, a perfect match from Z0 to complex-valued ZL can be achieved (at one
frequency).
If ZL =Z0 is inside the unity circle on the Smith chart, choose design (a), else choose design (b).
As in all matching network designs, the goal is to reach the center of the Smith chart.
The procedure is best explained with an example.
5.1. MATCHING WITH LUMPED ELEMENTS (L NETWORKS)
87
Example 5.2 Match a 50 ohm line to a 100 + j100 ohm load using an L network.
Solution: ZL =Z0 = (100 + j100) =50 = 2 + j2; choose design (a).
From the Smith chart,
B
=
0:19 ) inductor,
1
0:19
=
=
!L
50
) !L = 263:16;
Binductor
X
Xcap
=
1:7 ) capacitor,
1
=
= 1:7 (50)
!C
) !C = 0:0118:
(5.7)
(5.8)
Equations are presented in the text to solve this problem as well. For design (a), with ZL = RL + jXL ,
p
p
2 + X2
XL
RL =Z0 RL
Z0 RL
L
B =
;
(5.9)
2
2
RL + X L
1
XL Z0
Z0
X =
+
:
B
RL
BRL
88
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
For the previous example, (5.9) yields
B
X
=
=
0:0038 ) !L = 263:159;
86:6 ) !C = 0:0115:
Example 5.3 Using the calculated values of !L and !C, pick L and C for a perfect match at f = 30 GHz,
and plot the re‡ection coe¢ cient vs. frequency.
1.0
0.8
ρ
0.6
0.4
0.2
0.0
0
5
10
15
20
25
30
35
40
45
f (GHz)
5.1.1
Lumped Elements
Lumped (R; L, and C) elements can be realized at high frequencies if the size of the element is small compared
to . Unfortunately, their performance is usually quite non-ideal. For instance, elements may (and often do)
exhibit stray/parasitic capacitance, inductance, and resistance e¤ects. Elements often have a non-negligible
fringing …eld, and may resonant at certain frequencies (depending on the physical structure of the element).
5.2
Single-Stub Tuning
(single lumped element also)
5.2. SINGLE-STUB TUNING
89
jX
Z
Z
0
1
Z
L
L
Given Z0 and (complex) ZL , the goal is to choose X and (Z1 ; L) such that a perfect match is obtained at
a given frequency. Often Z1 is chosen as Z0 for convenience. The procedure will be illustrated by example.
Example 5.4 Series matching:
jX
Z
Z
0
Z
Z
in
1
Z
L
L
A
Assume Z0 = Z1 = 50 ohms, ZL = 25 + j30 ohms. Then, Z L = ZL =Z0 = 0:5 + j0:6. We want Zin = 50
ohms.
1. Choose L such that Re Z A = 1.
2. Choose X such that Im fZin = ZA + jXg = 0 (at which point we are at the center of the Smith chart.
Proceeding as above,
1. Start at Z L on Smith chart and rotate to either
(a) Point E, Z A = 1 + j1:1 and L = 0:065 , or
(b) Point F, Z A = 1
j1:1 and L = 0:235 .
(a) From Point E, add a series capacitor such that
X Cap =
1:1 )
1
1
=
!Ccap 50
1:1 ) !Ccap = 0:01818;
or,
(b) From Point F, add a series inductor such that
X ind = 1:1 ) !Lind
1
= 1:1 ) !Lind = 55:
50
Therefore, two possible designs are
!Ccap = 0:01818;
L = 0:065
and
!Lind = 55;
L = 0:235 :
The …rst design is better, since L is shorter. Note that 1(a) ! 2(a), and 1(b) ! 2(b).
90
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
If, for example, f = 2 GHz ( = 15 cm),
!Ccap
!Lind
= 0:01818 ) Ccap = 1:45 pF, L = 0:975 cm,
= 55 ) Lind = 4:38 nH, L = 3:53 cm.
1.0
Inductor
0.8
0.6
ρ
Capacitor
0.4
0.2
0.0
1.0
1.5
2.0
2.5
3.0
f (GHz)
3.5
4.0
4.5
5.0
5.2. SINGLE-STUB TUNING
91
92
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
Design using non-lumped elements: short-circuited or open-circuited stubs:
Usually short-circuited stubs are preferable to open-circuited stubs, since they don’t radiate. If using a
stub rather than a lumped element, the design proceeds as described above as far as the determination of X
and L. To generate a certain X, one chooses a stub having the appropriate length Ls using the Smith chart.
For example, to generate X = 1:1 (assuming Z0;stub = Z0 = 50 ohms) for a short-circuited stub, start
at Z = 0 on the Smith chart (corresponding to a short circuit) and rotate toward the generator to X = 1:1
(Point 1). The corresponding length is Ls = 0:368 (Ls = 5:52 cm at 2 GHz).
To generate X = 1:1 (assuming Z0;stub = Z0 = 50 ohms) for a short-circuited stub, start at Z = 0 on the
Smith chart and rotate toward the generator to X = 1:1 (Point 2). The corresponding length is Ls = 0:132
(Ls = 1:98 cm at 2 GHz).
5.2. SINGLE-STUB TUNING
93
There are four possible combinations if we assume Z0;stub = Z0 and if open-circuited stubs are acceptable.
The design resulting in the shortest stub length should generally be chosen, everything else being equal.
Stubs with Z0;stub 6= Z0 :
If we choose stubs having characteristic impedances di¤erent than Z0 of the line, we can often obtain
shorter stubs. For example, if the stub needs to provide X = 1:1, and Z0;stub = 100 ohms, then for a
short-circuited stub
1:1 (50)
= 0:55 ) Ls = 0:08 ( = 1:2 cm at 2 GHz).
100
High characteristic impedance short-circuited stubs are good for implementing inductive reactance.
Try to implement X =
1:1 with Z0;stub = 10 ohms and a shorted line:
1:1 (50)
=
10
5:5 ) Ls = 0:279 ( = 4:128 cm at 2 GHz).
Low characteristic impedance short-circuited stubs are good for implementing capacitive reactance.
94
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
All of the previous steps were for series stubs. Another approach is to use parallel (shunt) stubs, or shunt
lumped elements.
Z
Z
jB
0
Y in
YA
1
Z
L
L
The procedure is similar to that for series stubs, except you begin at Y L rather than Z L .
Note: The procedure using lumped elements and stubs is identical through identifying the required value
of X (or B). After that point, we either determine !Lind or !Ccap , or for a stub design we determine the
required stub length. For all of the designs, it is convenient to use the Smith chart. The Smith chart also
leads to better insight into the problem, compared to using design equations.
Example 5.5 A lossless 200 ohm line is connected to a 100
j150 ohm load. Design
1. a single short-circuited shunt stub matching network (Z0;stub = 200 ohms),
5.2. SINGLE-STUB TUNING
95
2. a single open-circuited shunt stub matching network (Z0;stub = 200 ohms), and
3. a lumped element shunt matching network.
For the stub designs, use the shortest stub lengths possible.
Solution:
Ls
Z
Z
0
Z
0
Y in
Z
0
L
L
1.
ZL
=
YL
=
100
j150
= 0:5
200
0:62 + j0:92:
j0:75;
Enter the Smith chart at Y L and rotate along the constant SW R circle towards the generator until you
intersect the G = 1 circle.
Y A = 1 + j1:3
Y B = 1 j1:3
) LA = 0:036
) LB = 0:194
stub needs to provide j1:3 stub needs to provide +j1:3
) LA
) LB
s = 0:105
s = 0:396
Choose Design A,
LA = 0:036 ;
LA
s = 0:105 .
96
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
2. Same as (1) up through
Y A = 1 + j1:3
) LA = 0:036
stub needs to provide
j1:3
Y B = 1 j1:3
) LB = 0:194
stub needs to provide +j1:3
B
For an open-circuited stub, LA
s = 0:354 and Ls = 0:146 .
Choose Design B,
LB = 0:194 ;
LB
s = 0:146 .
5.2. SINGLE-STUB TUNING
97
Ls
Z
Z
0
Z
0
Y in
ZL = 100
Z
0
L
j150 ohms, Z0 = 200 ohms.
3. Same as (1) up through
Y A = 1 + j1:3
Y B = 1 j1:3
) LA = 0:036
) LB = 0:194
In this case the lumped element must be chosen to supply
B A = j1:3
B B = +j1:3
j
1:3
)choose inductor, !Lind = 200 ; )choose capacitor, !Ccap =
!Lind = 153:85
!Ccap = 0:0065:
1:3
200 ;
L
98
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
To compare designs, assume f0 = 900 MHz ( = 33:33 cm). Then,
1. short-circuited stubs:
LA = 0:036 = 1:20 cm
LA
s = 0:105 = 3:50 cm
LB = 0:194 = 6:466 cm
LB
s = 0:396 = 13:20 cm
2. open-circuit stubs:
LA = 0:036 = 1:20 cm
LA
s = 0:354 = 11:80 cm
LB = 0:194 = 6:466 cm
LB
s = 0:146 = 4:867 cm
3. lumped elements
LA = 0:036 = 1:20 cm
Lind = 2 153:85
900 106 = 27:21 nH
LB = 0:194 = 6:466 cm
Ccap = 2 0:0065
900 106 = 1:15 pF
Short-Circuit Stub
1.0
0.8
0.6
ρ
B
0.4
0.2
0.0
0.6
A
0.7
0.8
0.9
1.0
1.1
1.2
1.1
1.2
f (GHz)
Open-Circuit Stub
1.0
0.8
B
0.6
ρ
A
0.4
0.2
0.0
0.6
0.7
0.8
0.9
f (GHz)
1.0
5.3. DOUBLE-STUB TUNING
99
Lumped Elements
1.0
0.8
ρ
0.6
B
0.4
0.2
A
0.0
0.6
0.7
0.8
0.9
1.0
1.1
1.2
f (GHz)
5.3
Double-Stub Tuning
(double lumped element also)
Single-stub matching can match any load impedance, but (L; Ls ) must be chosen correctly. Since it is
often hard to have a stub that can be moved along a transmission line (if, say, ZL or operating frequency
changed), it would be useful to …x the distance from the load to the stub(s), and simply choose the correct
stub length. The double-stub tuner allows this to be done, and changes in ZL or operating frequency can be
accommodated by changing Ls1 and Ls2 , with L1;2 …xed. Note: the text book places the …rst stub over the
load impedance. The method shown here is more ‡exible, since sometimes the load terminals are not easily
accessible.
L s2
Z
L s1
Z
0
A
Y in
0
Z
L
B
L2
L1
Procedure:
1. Mark Y L on the Smith chart and draw constant SW R circle.
2. Draw G = 1 circle rotated by L2 (Smith chart on the next page shows L2 = =8).
3. Move from Y L a distance L1 on the constant SW R circle (toward generator) to get to Y B + .
4. Add susceptance (moving on constant G circle) to get to intersection with rotated G = 1 circle, Point
Y B . The added susceptance comes form choosing Ls1 correctly.
5. Draw a new constant SW R circle using Point Y B . Rotate on new SW R circle a distance L2 (you
will intersect the non-rotated G = 1 circle at Point Y A+ ).
6. Add susceptance (moving along G = 1 circle to get to the center of the Smith chart, Y A .
100
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
Example 5.6 For the double stub tuner shown below, …nd the shortest lengths of Ls1 and Ls2 for a match
if ZL = 400 + j200 ohms and Z0 = 200 ohms. Assume L1 = 3 =16 and L2 = =8.
Solution:
L s2
Z
L s1
Z
0
A
Y in
1. Enter Smith chart at Y L .
B
L2
ZL
=
YL
=
0
L1
400 + j200
= 2 + j1;
200
1
= 0:4 j0:2:
ZL
Z
L
5.3. DOUBLE-STUB TUNING
101
2. Rotate on constant SW R circle 3 =16.
3. Add susceptance Y =
j0:925 ti intersect the rotated G = 1 circle.
4. Rotate on new SW R circle to intersect non-rotated G = 1 circle (this is the same as de-rotating the
point on the rotated G = 1 circle).
5. Add susceptance Y =
j0:17 to get to the center of the Smith chart.
Now we know we need to add Y = j0:925 at Point B, and Y =
values by choosing Ls1 and Ls2 correctly.
j0:17 at Point A. We obtain these
6. Rotate from Y = 1 Z = 0 to obtain Y =
j0:925: ) Ls1 = 0:131 :
7. Rotate from Y = 1 Z = 0 to obtain Y =
j0:17: ) Ls2 = 0:223 :
Note: We could have used lumped inductors to provide the Y values. Also, at Point 2 on the Smith chart
we could have added susceptance to intersect the rotated G = 1 circle at a di¤ erent point.
102
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
1.0
0.8
ρ
0.6
0.4
0.2
0.0
20
22
24
26
28
30
32
34
36
38
40
f (GHz)
Example 5.7 A 100 + j100 ohm load is to be matched to a 50 ohm line using a double stub tuner as shown
below. Determine the stub lengths to provide a patch, and …nd the forbidden zone of impedance/admittance
values. L1 = 0:4 , L2 = 3 =8.
5.3. DOUBLE-STUB TUNING
103
Solution:
L s2
Z
L s1
Z
0
A
L2
=
YL
=
Z
L
B
Y in
ZL
0
L1
100 + j100
= 2 + j2;
50
0:25 j0:25:
The stub having length Ls1 needs to provide Y = j1:3
Ls2 needs to provide Y = j0:6. From the Smith chart,
Ls1
Ls2
=
=
1:4 + B =
0:1459 ;
0:086 :
:1 ) B = 1:3 . The stub having length
104
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
1.0
0.8
ρ
0.6
0.4
0.2
0.0
20
22
24
26
28
30
f (GHz)
32
34
36
38
40
5.4. THE QUARTER-WAVE TRANSFORMER
5.4
105
The Quarter-Wave Transformer
In this section we will consider the bandwidth performance of a single quarter-wave transformer. In subsequent sections we will determine how to design multi-section quarter-wave transforms that can be used to
achieve a broadband match.
Z0
RL
Z1
Z in
λ/4
p
We know that if Z1 = Z0 RL the above quarter-wave transformer provides a perfect match at the
frequency where L = =4. To determine the behavior of the transformer at other frequencies, consider the
expression
RL + jZ1 tan L
Zin = Z1
:
Z1 + jRL tan L
Let = L = electrical length, such that
2 0
L=
(5.10)
4
such that
2 0
=
(5.11)
L=
2
0 4
at the design frequency f0 (c = 0 f0 ). Then,
=
in
(using Z1 =
p
(5.12)
Z0 RL ). By further manipulations, it is shown that
j
If we assume f ' f0 , then
in j
=h
1
1 + 4Z0 RL = (RL
' =2 and sec2
j
where
Zin Z0
RL Z0
p
=
Zin + Z0
RL + Z0 + j2 tan Z0 RL
in j
=
2
Z0 )
sec2
1, resulting in
jRL Z0 j
p
jcos j =
2 Z0 RL
i1=2 :
jcos j
(5.14)
is a constant. If RL does not vary with frequency,
|Γ|
Γm
θ=β L
θm
π/2
π− θ m
π
(5.13)
106
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
where
2
= L=
0
4
0
=
2
|
f
=
:
2 f0
{z
}
(5.15)
assum ing TEM m o de
If we have a value
bandwidth of
m
for the maximum re‡ection coe¢ cient that can be tolerated, then we obtain a usable
(5.16)
m :
2
The approximation f ' f0 was made to see where the characteristic “vee”shape in the above plot comes
from. We can determine a more useful expression for bandwidth by equating m with the exact expression
for j in j,
=2
j
mj
=
1
j
h
mj
m
m
i1=2 ;
(5.17)
m
2
:
m
m,
cos
Since
sec2
Z0 )
4Z0 RL
1+
=
Solving for cos
2
1 + 4Z0 RL = (RL
2
2 sec
(RL Z0 )
p
2 Z0 RL
1
1+
(RL Z0 ) cos
=
2
1
m
=
m
=
p
1
cos
m
1
p
2 RL Z0
;
2 jRL
Z0 j
m
!
p
2 RL Z0
m
p
:
2 jRL
Z0 j
1
m
(5.18)
= ( =2) (fm =f0 ), the frequency at the lower band edge is
fm =
2
m
f0 :
(5.19)
If we de…ne fractional bandwidth as
f
2 (f0 fm )
=
=2
f0
f0
2fm
=2
f0
we obtain
f
f0
=2
4
m
4
=2
cos
1
p
1
m
4
m
(5.20)
!
p
2 RL Z0
:
2 jRL
Z0 j
m
(5.21)
The above analysis assumes TEM lines, and ignores the e¤ects of reactance associated with discontinuities
of the transmission line when there is a step change in line width. The latter e¤ect can be compensated for
by making a small adjustment in the length of the matching section.
Example 5.8 Design a quarter-wave transformer to match a 75 ohm line to a 30 ohm load at 1 GHz.
Determine the percent bandwidth for which SW R 2.
Solution: Since at f0 = 1 GHz 0 = 30 cm,
L =
Z1
For SW R = 2,
=
30
= 7:5 cm,
4
(75) (30) = 47:43 ohms.
0
4
p
m
=
=
SW R 1
1
=
SW R + 1
3
5.5. THE THEORY OF SMALL REFLECTIONS
107
and so
f
f0
=
2
=
2
=
2
!
p
2 RL Z0
p
cos
2 jRL
Z0 j
1
m
0
1
p
2 (75) (30) A
4
1=3
cos 1 @ q
75j
2 j30
1 (1=3)
4
m
1
4
1
cos
(0:7454) = 2
4
0:7297 = 1:071;
or 107% bandwidth. The means the band extends from
f0
f
= 0:465 GHz
2
f0 +
f
= 1:535 GHz.
2
to
Note: for SW R < 1:1,
as desired SW R #,
For a …xed
m,
m
m
= 0:0476 and
f =f0 = 0:128 (12:8%), or from 0:936 GHz to 1:064 GHz. So,
#, and BW #.
the larger the ratio RL =Z0 (or Z0 =RL ), the smaller the bandwidth, as shown below.
|Γ|
ZL
= 4, 0.25
Z0
ZL
= 2, 0.5
Z0
1
5.5
The Theory of Small Re‡ections
The theory of small re‡ections will facilitate the development of approximate formulas for the design of
multi-section quarter-wave transformers, which will be shown to be relatively broadband.
I. Single Section Transformer:
108
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
T 21
T 12
Z1
Z2
Γ
Γ1
RL
Γ2
Γ3
T 21
1
Γ1
e
e
-j θ
-j θ
Γ3
T 12
Γ2
e
e
-j θ
-j θ
Γ3
T 12
Γ2
where
have
1
=
T21
=
Z2 Z1
;
2 =
1;
Z2 + Z1
1 + 1 ; T12 = 1 + 2 ;
is the overall re‡ection coe¢ cient and
i,
=
3
ZL Z2
;
ZL + Z2
(5.22)
i = 1; 2; 3, are the interfacial re‡ection coe¢ cients. We
=
1
+ T12 T21
3e
j2
=
1
+ T12 T21
3e
j2
+ T12 T21
1
X
2
j4
3 2e
n n
j2n
2 3e
+ :::
(5.23)
:
n=0
Using the geometric series result
1
X
xn =
n=0
leads to
x
for jxj < 1
T12 T21 3 e
1
2 3e
j2
+
e
1
3
:
1 + 1 3 e j2
=
1
=
If the impedance do not di¤er greatly, then
and so
1
1
1
+
and
'
1
3
+
j2
j2
:
(5.25)
j2
are small, and so
3e
(5.24)
1 3
is very small compared to 1
(5.26)
The above expression shows that is dominated by the …rst re‡ection from the initial discontinuity between
Z1 and Z2 , 1 , and the …rst re‡ection from the discontinuity between Z2 and ZL .
II. Multi-Section Transformers:
5.6. BINOMIAL MULTISECTION MATCHING TRANSFORMERS
Now consider the N
109
section transform shown below,
θ
θ
Z1
Z0
θ
...
Z2
ZN
ZL
Γ
Γ0
Γ1
Γ2
ΓN-1
ΓN
where we assume that ZL is real-valued. We have
0
=
Z1 Z0
;
Z1 + Z0
n
=
Zn+1 Zn
;
Zn+1 + Zn
Assume that Zn increases or decreases monotonically. Then
n;
n
n
N
n
=
ZL ZN
:
ZL + ZN
(5.27)
is real-valued and has the same sign for all
> 0 for ZL =Z0 > 1;
< 0 for ZL =Z0 < 1;
(5.28)
and
( )'
0
+
1e
j2
+
2e
j4
+ ::: +
Ne
j2N
(5.29)
from considerations of the single-section transformer.
In some cases it is desirable to simplify further. Assume that the transformer is made symmetrical, in
the sense 0 = N , 1 = N 1 , etc. (this does not imply that the Zn ’s are symmetrical). Then,
n
h
i
h
i
jN
( ) ' e jN
+ e jN + 1 ej(N 2) + e j(N 2) + 2 ej(N 4) + e j(N 4)
0 e
)
N 1
ej + e j ;
N odd
2
;
(5.30)
+ ::: +
N
;
N even
2
leading to
( ) ' 2e
jN
+::: +
f
0
cos N +
N
1
2
1
2
N
2
cos ;
;
1
cos (N
N odd
N even
2) +
)
2
cos (N
4) + ::: +
n
;
cos (N
2n)
(5.31)
which is seen to be a …nite (truncated) Fourier cosine series for ( ). Since a Fourier cosine series can
represent any smooth function , we can synthesize any desired re‡ection coe¢ cient response as a function
of frequency by proper choice of the n values, and by using enough sections (some limitations are imposed
by the Bode-Fano criteria as discussed later). Two choices of desirable functions to implement are presented
next, in each case leading to a method to choose the n values in (5.29) (or (5.30) or (5.31)).
5.6
Binomial Multisection Matching Transformers
A desirable function to implement for matching purposes is given by
g( ) = A 1+e
j2
N
:
(5.32)
110
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
The absolute value is
j
jg ( )j = jAj e
= jAj e
ej + e
jN
ej + e
= jAj ej + e
=
N
j
N
2N jAj jcos j ;
N
j
(5.33)
N
j
N
= jAj j2 cos j
which is plotted below.
(n)
The function (5.32) has the property that at = =2, jf ( )j = 0, n = 1; 2; :::; N 1, where g (n) indicates
the nth derivative of g. Therefore, the response is as ‡at as possible near the center frequency f0 (i.e., at
= 0 = =2). Such a response is known as maximally ‡at.
So far, (5.32) is merely a function with a nice ‡atness property. To obtain a multisection transformer
that behaves like (5.32), we equate the actual (approximate) input re‡ection coe¢ cient of the multisection
transformer, (5.29) (or (5.30) or (5.31)), and (5.32), in order to determine what n values will lead to the
response (5.32).
Since (5.29) is in the form of a series, in order to equate (5.29) (or (5.30) or (5.31)) and (5.32) we use
the binomial expansion
N
X
N
(1 + x) =
CnN xn ;
(5.34)
n=0
where the binomial coe¢ cients are
CnN =
(N
N!
;
n)!n!
(5.35)
to convert (5.32) to series form, leading to
g( )
= A 1+e
= A
N
X
j2
CnN e
N
j2n
(5.36)
:
(5.37)
n=0
Equating (5.37) with (5.29) leads to
g( ) = A
N
X
n=0
CnN e
j2n
=
( )'
0
+
1e
j2
+
2e
j4
+ ::: +
Ne
j2N
:
(5.38)
5.6. BINOMIAL MULTISECTION MATCHING TRANSFORMERS
111
The constant A can be determined by letting f ! 0 ( = L = 0),
ZL Z0
;
ZL + Z0
ZL Z0
) A=2 N
:
ZL + Z0
g (0)
=
A2N =
(0) =
(5.39)
(5.40)
Equating nth terms in (5.38) results in
n
= ACnN = 2
ZL Z0 N
C ;
ZL + Z0 n
N
(5.41)
the sought-after formula for choosing n values to yield a response that is equal to the maximally-‡at
N
response (5.32)1 . Since CnN = CN
n , then n = N n and the transformer turns out to be symmetrical.
The …nal step is determining the desired impedance values. Since
0
=
Z1 Z0
;
Z1 + Z0
n
=
Zn+1 Zn
;
Zn+1 + Zn
N
=
ZL ZN
ZL + ZN
(5.42)
then, for example,
n
=
Zn+1 Zn
=2
Zn+1 + Zn
N
ZL Z0 N
C
ZL + Z0 n
(5.43)
which can be solved for Zn+1 in terms of Zn as
1+2
N ZL Z0 N
ZL +Z0 Cn
1
N ZL Z0 C N
ZL +Z0 n
Zn+1 = Zn
2
:
(5.44)
One can start with n = 0 (Z0 ) to …nd Z1 , etc.
Interestingly, while (5.44) follows exactly from (5.43), making the following approximation to (5.43) leads
to better results. Since
x 1 2
+
x+1 3
x 1
'2
x+1
ln (x) = 2
for x ' 1, then
x 1
x+1
3
+ :::
(5.45)
Zn+1 =Zn 1
1 Zn+1
Zn+1 Zn
=
' ln
Zn+1 + Zn
Zn+1 =Zn + 1
2
Zn
(5.46)
for Zn+1 ' Zn (this assumption was used in developing the theory of small re‡ections, and so the approximation in (5.46) is consistent with the approximations leading to (5.29), leading to a self-consistent formula
that provides accurate results). Therefore, making the same approximations on both sides of (5.43) leads to
n
'
Zn+1
Zn
=
ln Zn+1
=
ln
1 Zn+1
ln
'2
2
Zn
ZL
2 N CnN ln
;
Z0
ln Zn + 2
N
N
ZL Z0 N
C '2
ZL + Z0 n
CnN ln
N
CnN
1 ZL
ln
;
2 Z0
(5.47)
(5.48)
ZL
;
Z0
(5.49)
resulting in
Zn+1 = eln Zn +2
1 Actually,
N
N
Cn
ln
ZL
Z0
since (5.29) is an approximate formula, then the resulting response is approximately that of (5.32).
(5.50)
112
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
which is generally valid for
obtain
1
2 Z0
ZL
2Z0 . Bandwidth can be determined as follows. From (5.32) we
j
when 0
m
mj
= 2N jAj jcos
N
= 2 jAj cos
N
N
mj
(5.51)
m
=2. Therefore,
m
= cos
1
= cos
1
"
"
1
N
j mj
jAj 2N
1
2
j
#
#
1
N
mj
jAj
(5.52)
:
From (5.20) we have
f
f0
2 (f0
=
fm )
4
=2
m
"
4
1 j mj
cos 1
2 jAj
(5.53)
f0
=
2
where
#
(5.54)
1 ZL
ln
2 Z0
N
A=2
1
N
(5.55)
Example 5.9 Design a three section binomial transformer to match a 50 ohm load to a 100 ohm line.
Calculate the bandwidth for j m j = 0:05.
Solution:
= eln Zn +2
Zn+1
ln Zn +2
= e
and
C03 =
3!
= 1;
3!0!
C13 =
N
3
N
Cn
ln
3
Cn
ln
3!
= 3;
2!1!
ZL
Z0
50
100
(5.56)
;
C23 =
(5.57)
3!
= 3.
1!2!
Therefore,
Z1
= eln Z0 +2
3
C03 ln
50
100
Z2
ln Z1 +2
3
C13 ln
50
100
ln Z2 +2
3
C23 ln
50
100
Z3
= e
= e
π/2
100
91.7
= 91:70 ohms,
(5.58)
= 70:71 ohms,
(5.59)
= 54:53 ohms.
(5.60)
π/2
π/2
70.71
54.53
Γ0
Γ1
Γ2
Γ3
-0.043
-0.129
-0.129
-0.043
Z L = 50
5.6. BINOMIAL MULTISECTION MATCHING TRANSFORMERS
113
To compute the bandwidth,
f
4
=2
f0
1
cos
"
j
1
2
mj
1
N
jAj
#
;
(5.61)
and, with
N
A=2
and
m
1 ZL
ln
=2
2 Z0
31
2
ln
50
=
100
0:0433
= 0:05,
f
f0
=2
4
cos
1
"
1
2
A 1+e
1
3
0:05
0:0433
j2 (f ) N
#
= 0:703;
70:3%
vs. f =f0 :
0.30
0.25
ρ
0.20
0.15
0.10
0.05
0.00
10
14
18
22
26
30
34
38
42
46
50
f (GHz)
Example 5.10 Design an N section binomial transformer that needs to operate from 1:2 GHz to 2:8 GHz
with j m j 0:05. Assume ZL = 100 ohms, Z0 = 50 ohms.
Solution: In order to determine the Zn values we need to …rst determine how many sections (N ) are
114
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
required.
f
2 (f0 fm )
2 (2 1:2)
=
=
= 0:8;
f0
f0
2
1 ZL
1 100
A = 2 N ln
= 2 N ln
= 2 N (0:3466)
2 Z0
2
50
"
1 #
f
4
j mj N
1 1
=2
cos
f0
2 jAj
"
1 #
N
4
0:05
1 1
=2
= 0:8:
cos
2 2 N (0:3466
(5.62)
(5.63)
(5.64)
(5.65)
Solving for N we obtain N = 3:64 ) N = 4. With N determined, the design proceeds as in the previous
example.
C04 = 1;
C14 = 4;
C24 = 6;
Z1
= eln Z0 +2
4
C04 ln
100
50
Z2
ln Z1 +2
4
C14 ln
100
50
ln Z2 +2
4
C24 ln
100
50
ln Z3 +2
4
C34 ln
100
50
Z3
Z4
50
= e
= e
= e
π/2
π/2
π/2
52.2
62.1
80.5
C34 = 4;
= 52:214;
= 62:093;
= 80:525;
= 95:76:
π/2
95.8
Z L = 100
0.30
0.25
ρ
0.20
0.15
0.10
0.05
0.00
10
14
18
22
26
30
34
f (GHz)
Note: Pascal’s triangle is also useful for evaluating CnN ,
38
42
46
50
5.7. CHEBYSHEV MULTISECTION MATCHING TRANSFORMERS
n
0
1
2
3
4
5.7
115
CnN
1
1
1
1
2
1
3
3
1
3
4
1
6
4
1
Chebyshev Multisection Matching Transformers
Another function with desirable properties to implement in a multisection transformer is a Chebyshev polynomial. The Chebyshev polynomials are given by
T0 (x) = 1;
T1 (x) = x;
(5.66)
T2 (x) = 2x2
3
T3 (x) = 4x
1;
3x;
4
T4 (x) = 8x
8x2 + 1;
..
.
Tn (x) = 2xTn 1 (x) Tn
2
(x) :
Another, equivalent form can be given where the Chebyshev polynomials look like trigonometric functions
(although they aren’t). Letting x = cos , jxj < 1, then it can be shown that Tn (cos ) = cos (n ), and
Tn (x) = cos n cos
1
Tn (x) = cosh n cosh
x ;
1
x ;
jxj < 1;
(5.67)
jxj > 1.
The two main properties of Chebyshev polynomials are the following:
1. For 1 x 1, jTn (x)j 1 and Tn oscillates between
map this range into the passband of the transformer.
1; Tn (x) has n zeros between
2. For jxj > 1, jTn (x)j > 1. We will map this range to outside the passband.
1
0.5
-1
0
-0.5
0.5
x
-0.5
-1
cos n cos
1
(x) vs. x, n = 1; 2; 3; 4.
1
1. We will
116
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
|Γ|
Γm
θ=β L
θm
π/2
π− θ m
x=+1
π
x=-1
With x = cos , as goes from 0 to , x goes from +1 to 1. Force x = +1 at
=
m . To implement this we replace cos with cos = cos m = sec m cos :
=
=
and x =
1 at
(5.68)
m
cos
cos
Tn
and when
m,
cos
cos m
Tn (cos ) ! Tn
(then when
=
m
= Tn (1) = 1;
(5.69)
m
m,
cos (
cos
Tn
m)
= Tn ( 1) =
1:)
(5.70)
m
Therefore,
T1 (x) = x;
) T1 (sec
2
T2 (x) = 2x
= sec
T3 (x) = 4x3
m
= sec
4
T4 (x) = 8x
m
m
cos ) = 2 (sec
m
cos ;
(1 + cos 2 )
2
1
3
3 (sec
m
cos )
m
cos )
1;
m
cos ) = 4 (sec
(cos 3 + 3 cos )
3 sec
m
m
cos )
m
cos ) + 1;
cos ;
2
8x + 1;
) T4 (sec
= sec4
cos ) = sec
3x;
) T3 (sec
3
m
1;
) T2 (sec
2
(5.71)
m
m
4
cos ) = 8 (sec
m
cos )
8 (sec
4 sec2
(cos 4 + 4 cos 2 + 3)
m
2
(cos 2 + 1) + 1:
Procedure: Equate the actual multisection re‡ection coe¢ cient (5.31) to the N th order Chebyshev polynomial Ae jN TN (sec m cos ) for a given N . That is,
( ) ' 2e
jN
+::: +
f
0
N
1
2
cos N +
cos ;
N ;
1
2
2
1
cos (N
N odd
N even
2) +
)
= Ae
2
cos (N
jN
4) + ::: +
TN (sec
m
n
cos ) :
cos (N
2n)
(5.72)
The constant A is again determined by letting f = 0;
ZL Z0
= ATN (sec
ZL + Z0
ZL Z0
1
)A=
ZL + Z0 TN (sec
(0) =
m) ;
m)
(5.73)
'
1 ZL
1
ln
2 Z0 TN (sec
m)
.
5.7. CHEBYSHEV MULTISECTION MATCHING TRANSFORMERS
117
Since max TN (x) = 1 for jxj < 1, then A is also the maximum allowable re‡ection coe¢ cient magnitude in
the passband,
1 ZL
1
ln
:
(5.74)
m =A=
2 Z0 TN (sec m )
Bandwidth: Using (5.74),
TN (sec
m)
=
1
m
and, since sec
m
= 1= cos
1 ZL
;
ln
2 Z0
(5.75)
1 we use
m
TN (sec
m)
1
= cosh N cosh
) sec
m
= cosh
sec
=
m
1
cosh
N
1
1 ZL
;
ln
Z0
m 2
1 1 ZL
ln
Z0
m 2
1
(5.76)
such that
f
f0
=
=
2 (f0
4
fm )
4
=2
m
f0
4
1
sec 1 cosh
cosh
N
jN
Ae
TN (sec
m
(5.77)
1
1
m
1 ZL
ln
2 Z0
:
cos )
Example 5.11 Design a three section Chebyshev transformer to match a 100 ohm load to a 50 ohm line,
with m = 0:05.
Solution:
A=
sec
m
m
= 0:05;
1
1
cosh 1
N
m
1
1
= cosh
cosh 1
3
0:05
= cosh
= 1:4075;
m
1 ZL
ln
2 Z0
1 100
ln
2
50
= 0:7806:
To determine the Zn values, from (5.72) we have
( )
' 2e j3 f 0 cos 3 + 1 cos g = Ae
) 2 f 0 cos 3 + 1 cos g = AT3 (sec
= A sec3 m (cos 3 + 3 cos ) 3 sec
j3
T3 (sec
m cos )
:
m cos
m
cos )
(5.78)
(5.79)
(5.80)
118
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
Equating similar terms,
cos 3 : 2
= A sec3
0
cos : 2
)
m;
3
= 3A sec
0 = 0:0697;
1
(5.81)
sec m ;
= 0:1036:
m
1
From symmetry,
3
=
0;
π/2
Z0
Z1
Γ0
Γ1
2
=
1:
π/2
π/2
Z2
Z3
Γ2
(5.82)
ZL
Γ3
To determine the Zn values, note the relationship
n
=
Zn+1 Zn
1+
$ Zn+1 = Zn
Zn+1 + Zn
1
n
n
such that
1+
1
1+
= Z1
1
1+
= Z2
1
Z1
= Z0
Z2
Z3
0
= 57:49;
0
1
= 70:77;
1
2
= 86:97:
2
The bandwidth is
f
f0
=
=
2 (f0
2
fm )
4
=2
m
f0
4
0:7806 = 1:006; 100:6%
(compare with 71% for the binomial transformer with N = 3.
(5.83)
5.7. CHEBYSHEV MULTISECTION MATCHING TRANSFORMERS
119
0.30
0.25
ρ
0.20
0.15
0.10
0.05
0.00
10
14
18
22
26
30
34
38
42
46
50
46
50
f (GHz)
0.30
0.25
Chebyshev, N=3
0.20
ρ
Binomial, N=3
0.15
0.10
0.05
0.00
10
14
18
22
26
30
34
38
42
f (GHz)
Note: If given
m
and
f =f0 , determine required N then recalculate
m
using N .
Example 5.12 For the previous example of matching a 100 ohm load to a 50 ohm line with m = 0:05
using Chebyshev polynomials, design an appropriate microstrip implementation. Assume "r = 9:8, f0 = 1
GHz, and that the load is an in…nite length of 100 ohm line.
Solution:
( 8eA
w
;
w
b < 2;
e2Ah 2
i
=
(5.84)
"r 1
2
0:61
w
b
B 1 ln (2B 1) + 2"r ln (B 1) + 0:39
;
"r
b > 2;
where
r
Z0 "r + 1 "r 1
A=
+
60
2
"r + 1
377
B=
p :
2Z0 "r
0:23 +
For
Z1
Z2
Z3
Z0
=
=
=
=
57:49;
70:77;
86:97;
50; ZL = 100
0:11
"r
;
(5.85)
120
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
we obtain
w0
b
w1
b
=
0:978;
=
0:721;
wL
= 0:137;
b
w2
= 0:427;
b
w3
= 0:227
b
Since
c
vp = p = f;
"e
c 1
) =p
:
"e f
(5.86)
With
"e =
we obtain
1
4
"r + 1 "r 1
1
q
+
2
2
1+
= 2:95 cm;
50
57.49
2
4
= 3:01 cm;
70.77
86.97
3
4
(5.87)
12d
w
= 3:06 cm
100
...
...
λ1
λ2
λ3
4
4
4
Note that higher "r leads to smaller and therefore shorter transformers.
The Chebyshev transformer is called an equal ripple transformer. It optimizes bandwidth at the expense
of passband ripple.
Note: Many other transformers are possible, simply by equating the desired function for ( ) with (5.29).
5.8
Tapered Lines
An alternative to using …nite impedance steps such as in the binomial or Chebyshev cases is to use a
continuous taper.
ZL
Z(z)
Z0
z=0
z=L
Considering the continuously tapered line to be made up of a number of incremental section of length
z, with impedance change Z (z) from one section to the next,
5.8. TAPERED LINES
121
∆Γ
Z+∆Z
Z
∆z
and incremental re‡ection coe¢ cient is
=
As
Z ! 0, we can replace
(Z +
(Z +
Z) Z
Z
Z
=
'
.
Z) + Z
2Z + Z
2Z
(5.88)
; Z by d ; dZ such that
dZ
:
2Z
This can be manipulated into a convenient expression as follows.
i
h
d ln Z(z)
Z0
1 dZ (z)
d
=
=2 ;
dz
Z hdz i dz
d =
(5.89)
(5.90)
Z
Form the integral summation of
If Z (z) is known, then
d
1 d ln Z0
=
dz
2
dz
(all partial re‡ections) and add the approximate phase shifts,
Z
Z
1 L
d
e j2 z
ln
=
dz:
2 z=0
dz Z0
(5.91)
can be found.
1. Exponential Taper:
Z (z) = Z0 eaz ;
0 < z < L:
(5.92)
We want
Z (0) = Z0 ;
(5.93)
aL
Z (L) = Z0 e
= ZL :
Therefore,
1 ZL
ZL
) a = ln
:
Z0
L Z0
eaL =
(5.94)
Then,
=
=
=
1
2
1
2
1
2
Z
L
e
j2 z
e
j2 z
e
j2 z
0
Z
L
0
Z
L
0
1 1 ZL
=
ln
2 L Z0
1 ZL
= ln
e
2 Z0
Z
d
Z
ln
dz
dz Z0
d
ln (eaz ) dz
dz
1 ZL
ln
dz
L Z0
L
e
j2 z
dz
0
j L sin
L
L
:
(5.95)
122
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
When L = 0,
performance.
sin L
L
= 1. From the plot, we see that
(2 L >
L >
) L >
2)
for best
2. Triangular Taper:
Another common method is to assume a triangular taper for d ln (Z=Z0 ) =dz,
(
ZL
4z
d ln ZZ0
0 z L=2;
L2 ln Z0 ;
:
=
ZL
4
z
4
ln
;
L=2 z L
dz
L
L2
Z0
This results in
Z (z) =
(
2
Z0 e2(z=L) ln(ZL =Z0 ) ;
2
2
Z0 e(4z=L 2z =L 1) ln(ZL =Z0 ) ;
0
z
L=2
L=2;
:
z L
(5.96)
(5.97)
Integrating the triangular expression leads to
=
1
e
2
j L
ln
ZL sin ( L=2)
Z0
( L=2)
2
:
(5.98)
5.8. TAPERED LINES
123
For L > 2 the peaks of the triangular taper are lower than the corresponding peaks of the exponential
2
taper because of the (sin x=x) factor. However, the …rst null occurs at 2 , whereas the …rst null appears
at for the exponential taper.
3. Klopfenstein Taper:
The Klopfenstein taper has been shown to be optimum in the sense that the re‡ection coe¢ cient is
lowest over the passband, for a given taper length greater than some critical value. Alternatively, upon
speci…cation of m , the Klopfenstein taper yields the shortest length L.
=
0e
j
=
0e
j
cos
L
q
2
( L)
A2
qcosh A
2
cos A2 ( L)
L
cosh A
;
L > A (passband),
;
L < A;
(5.99)
where
0
=
1 ZL
ZL Z0
' ln
;
ZL + Z0
2 Z0
1
A = cosh
0
(5.100)
:
m
The impedance taper must generally be calculated numerically from
ln Z (z) =
1
0
ln (Z0 ZL ) +
A2 (2z=L
2
cosh A
where
(x; A) =
( x; A) =
Z
0
and where I1 (x) is the modi…ed Bessel function.
x
1; A) ;
p
I1 A 1 y 2
p
dy;
A 1 y2
0
z
jx
1j
L;
(5.101)
(5.102)
Example 5.13 Design a triangular taper, an exponential taper, and a Klopfenstein taper (with m = 0:02)
to match a 50 ohm load to a 100 ohm line. Plot the impedance variations and resulting re‡ection coe¢ cient
magnitudes vs. L.
Solution:
(a) Triangular Taper: For the triangular taper
(
2
Z0 e2(z=L) ln(ZL =Z0 ) ;
Z (z) =
2
2
Z0 e(4z=L 2z =L 1) ln(ZL =Z0 ) ;
=
1
e
2
j L
ln
ZL sin ( L=2)
Z0
( L=2)
0 z
L=2
L=2;
;
z L
(5.103)
2
:
(b) Exponential Taper: For the exponential taper
Z (z) = Z0 eaz ;
0<z<L
1 ZL
1
a = ln
= 0:693 ;
L Z0
L
1 ZL j L sin L
= ln
e
:
2 Z0
L
(5.104)
124
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
(c) Klopfenstein Taper: For the Klopfenstein taper
0
'
1 ZL
ln
= 0:346
2 Z0
A = cosh
1
0
(5.105)
= 3:543 = 1:13 ;
m
Z is evaluated numerically and in the passband ( L > A = 1:13 )
=
0e
j
cos
L
q
2
( L)
cosh A
A2
:
(5.106)
Example 5.14 Design an exponentially tapered matching transformer to match a 100 ohm load to a 50
ohm line. Plot j j vs. L, and …nd the length of the matching section (at the center frequency) required to
obtain j j
0:05 over a 100% bandwidth. How many sections would be required if a Chebyshev matching
transformer were used to achieve the same speci…cations?
5.8. TAPERED LINES
125
Solution:
Z (z)
=
a
=
Z0 eaz ;
0<z<L
1 ZL
1
ln
= 0:693 ;
L Z0
L
z
Z (z) = 50e0:693 L
1
ZL j L sin L
ln
e
2
Z0
L
sin L
1
100 sin L
ln
= 0:3466
:
2
50
L
L
)
j j
=
=
For 100% bandwidth,
f
f0
=
4
2
m
) fm =
=
1
f0 ;
2
2 (f0
fm )
f0
=
m
= 1;
4
and so
1
f0
2
3
f0
2
f
represents a 100% bandwidth.
G (x) = 0:3466
Starting at L = 1:72 (where j j
L
sin(x)
x
vs. x= .
0:05)
1:72 )
2
L
1:72
2
1:72 ) L
For
f
:
:
1
3
f0 to f0 ;
2
2
2 0 to 0:666
0:
Therefore,
L
is needed. For example, if f0 = 30 GHz (
0
0:86 (2
= :01 m),
0)
= 1:72
0
= 0:86 :
126
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
0.10
0.08
ρ
0.06
0.04
0.02
0.00
20
22
24
26
28
30
32
34
36
38
40
f (GHz)
For the Chebyshev,
cosh
N=
1
1
cosh
cosh
=
1
2
m
1
cosh
1
ln ZZL0
1
0:05
1
m)
1
100
2 ln 50
(5.107)
(sec
sec
= 2:97;
4
) N = 3:
Since a three section Chebyshev transformer is 3
has length L = 1:72 0 ).
Note that for the Klopfenstein taper,
0
=
A =
0 =4
= 0:75
0
it is smaller then the exponential taper (which
ZL Z0
= 0:333;
ZL + Z0
cosh
1
0
= 2:5846:
m
Since the passband is de…ned as L > A, and the maximum ripple in the passband 0:05, then for our value
of A the ripple will be at most 0:05 for L > A = 2:5846;
L
2
Since
max
=2
> 2:5846;
L >
2:5646 ) L > 0:411 :
0;
L > 0:411 (2
0)
= 0:8227
0
would give a 100% bandwidth. Note that this is quite a bit shorter than the exponential taper, but still longer
than the Chebyshev taper.
5.9
Bode-Fano Criteria
The Bode-Fano criteria provides information about the theoretical limits that constrain the performance of
an impedance matching network.
5.9. BODE-FANO CRITERIA
127
Zg
matching
Z 0 ,v p
+
Vg
-
ZL
network
Z in
Assume that the matching network is lossless, and ZL is generally complex.
1. Can we achieve a perfect match (zero re‡ection) over a speci…ed (nonzero) bandwidth?
2. If not, how good can we do? What is the trade-o¤ between
m
and bandwidth?
The bode-Fano criteria gives the optimum result that can ideally be achieved. As an example, consider
the following.
matching
Γ(ω)
network
The Bode-Fano criteria is, for this case,
Z
1
ln
0
1
d!
j (!)j
RC
:
(5.108)
Assume that we desire the response to be as shown below.
|Γ|
1
Γm
ω
∆ω
Then
Z
0
1
ln
1
d! =
j (!)j
)
Z
ln
!
!
1
j
mj
d! =
RC jln
mj
! ln
:
1
j
mj
RC
;
(5.109)
128
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
Note that as
m
varies from 0 to 1,
RCjln
mj
varies from 0 to 1. So,
: 0 ! 1;
! : 0 ! 1:
(5.110)
m
For a …xed RC, broader bandwidth can only be achieved at the expense of a higher re‡ection coe¢ cient
in the passband.
Also, the passband m cannot be zero unless ! = 0. Thus, a perfect match cannot be maintained
except at possibly a …nite number of frequencies.
As R and/or C increase,
! decreases.
|Γ|
ω
|Γ|
ω
Example 5.15 A parallel RC load having R = 50 ohm and C = 3 pF is to be matched to a 30 ohm line
from 1 to 3 GHz. What is the best m that can be obtained? Assume a square-wave pro…le.
Solution:
|Γ|
1
Γm
2π(1)
Z
0
2π(3)
Z 2 (3 109 )
1
1
d! =
ln
d!
j (!)j
j mj
2 (1 109 )
1
2 (2) ln
20:94 ) m 0:1888:
j mj
1
ln
ω
20:94
109 ;
(5.111)
5.9. BODE-FANO CRITERIA
129
Alternatively, to match from 1 to 10 GHz,
2 (9) ln
1
j
mj
20:94 )
m
0:691:
130
CHAPTER 5. IMPEDANCE MATCHING AND TUNING
Chapter 6
Power Dividers and Directional
Couplers
(Here we will cover Sections 7.4, 7.2, and 7.3)
Directional couplers are passive microwave components that provide power division or power combining.
They are usually represented in the following way:
1 input
2 through
4 isolated
3 coupled
The main idea is that power is input into Port 1. Some fraction of the input power goes through to Port
2 (the “through port”), while the rest goes to Port 3 (the “coupled port”). Ideally, no power is transferred
to Port 4 (the “isolated port”).
C = 10 log10
D = 10 log10
I = 10 log10
P1
, coupling factor in dB,
P3
P3
, directivity in dB,
P4
P1
, isolation in dB.
P4
(6.1)
The coupling factor indicates the fraction of input power coupled to Port 3. Directivity indicates the coupler’s
ability to isolate forward (Port 3) and backward (Port 4) waves. Note that I = C + D. An ideal directional
coupler would have I = D = 1 (P4 = 0).
The device is reciprocal in the sense that power can be applied to Port 2, in which case Port 1 becomes
the through port, Port 4 becomes the coupled port, and Port 3 the isolated port. Also, if power is applied
to both Ports 2 and 3, power is combined and emerges from Port 1.
Bethe Hole Coupler:
131
132
CHAPTER 6. POWER DIVIDERS AND DIRECTIONAL COUPLERS
The principle of operation of this coupler, and of all couplers, is that two separate waves can be combined
in-phase at the desired output port, and can be combined out-of-phase at the desired isolated port. The
details of the single-hole coupler are provided in the text. It can be shown that, for the TE10 mode, the
skewed coupler design leads to
r
2
4k02 r03
C = 20 log10
;
= k02
:
(6.2)
3ab
a
One can thus design a coupler to achieve a certain coupling. This will only hold at one frequency, and is very
frequency sensitive. To decrease sensitivity, multi-hole couplers can be design, analogous to multi-section
quarter-wave transformers.
Two-hole coupler:
Here we use parallel guides. The apertures are small, and placed g =4 apart. Most of the input wave is
transferred to the through port. Aperture 1, referenced at 0 phase, radiates a backward (B1 ) and forward
(F1 ) wave into the upper guide. Aperture 2 does the same, B2 and F2 , respectively. At the reference point,
B1 ej
z
z=0
= B1 ;
F1 e
j z
z=0
= F1
(6.3)
133
due to Aperture 1. At this same reference point, the backward wave from Aperture 2 will be
B2 ej
g =2
= B2 ej =
B2 :
(6.4)
If jB1 j ' jB2 j, these waves will cancel, such that no power comes out the isolated port.
Does any power come out of the coupled port? Yes: at the location of Aperture 2 we have
F1 e
j
g =4
+ F2 e
j
g =4
= (F1 + F2 ) ( j) ;
(6.5)
and so the two forward waves add in-phase.
It is important to note that for exact cancellation of the backward waves, jB1 j must equal jB2 j. This will
not be true (why?), and so we never obtain perfect isolation.
The two-hole coupler is less frequency dependent than the one-hole coupler, even though the spacing will
be g =4 at only one frequency.
Example 6.1 Assume C = 3 dB with in…nite directivity. Determine the power dissipated at ports 2,3, and
4.
Solution:
=
power
At Port 2:
1
2
p
2Z0 Z0
2Z0 + Z0
1
1
=
2
2
at Port 3
2
and at Port 4
with
1
36
watts back out to Port 1.
N + 1 hole couplers:
p
1
:
9
1
4
= watts,
18
9
1
watt,
2
1
2
=
1
1
=
watts,
2
36
134
CHAPTER 6. POWER DIVIDERS AND DIRECTIONAL COUPLERS
All apertures are spaced d = g =4 apart. The amplitude of the wave incident on the input port is A, as
is the amplitude at the output port of the through wave (of course, this is not self-consistent, nevertheless
it yields reasonable results)! Set the phase reference to be at the …rst aperture.
At the coupled port,
N
X
F = Ae j N d
Fn
(6.6)
n=0
(remember we have N + 1 apertures), and at the output of the isolated port,
B=A
N
X
j2 nd
Bn e
:
(6.7)
n=0
Then,
C=
20 log10
D=
20 log10
=
C
F
=
A
20 log10
B
=
F
20 log10
N
X
Fn
dB,
(6.8)
n=0
20 log10
N
X
Bn e
PN
Bn e j2
PN
n=0 Fn
nd
n=0
j2 nd
dB.
n=0
Assume that all aperture are round holes with identical positions, s, relative to the edge of the guide, with
rn being the radius of the nth aperture. It can be shown that
Fn = kf rn3 ;
Bn = kb rn3 ;
(6.9)
where kf;b are dependent on aperture shape, and are slowly varying functions of frequency. Therefore,
C=
D=
20 log10 jkf j
C
20 log10
N
X
rn3
dB,
(6.10)
n=0
20 log10 jkb j
20 log10
|
N
X
rn3 e
j2 nd
n=0
{z
highly frequency sensitive
dB.
}
In the design of multi-section quarter wave transformers we considered binomial and Chebyshev methods
to choose the impedances Zn . Here, the design involves choosing the hole radii using similar methods.
Binomial Response:
Pick rn3 proportional to the binomial coe¢ cients,
rn3 = kCnN
(6.11)
135
(recall
CnN =
N!
):
(N n)!n!
(6.12)
The proportionality constant k can be determined from
C=
=
=
N
X
20 log10 jkf j
20 log10
20 log10 jkf j
20 log10
20 log10 jkf j
20 log10 jkj
rn3
(6.13)
n=0
N
X
kCnN
n=0
20 log10
N
X
CnN :
n=0
From this, and with a desired value of C and a known kf ; N , one can determine k. Note:
"
2k02
jkf j =
sin2
3ab 1;0
"
2k02
jkb j =
sin2
3ab 1;0
2
s
a
2
1;0
k02
2
s 2 1;0
+ 2
a
k0
2
sin
2
sin
s
+
a
s
a
2
2
2
1;0 a
cos
2
s
a
cos
2
s
a
2
2
2
1;0 a
!#
!#
;
(6.14)
:
(6.15)
A Chebyshev response can also be obtained.
Example 6.2 Design a …ve-hole coupler with a binomial response. The center frequency is 10 GHz, and the
required coupling is 18 dB. The physical structure is a rectangular waveguide with round coupling apertures
centered across the broad common wall of the guides.
Solution: For an X band waveguide, a = 0:02286 m, b = 0:01016 m (from the textbook appendix). For
a 5 hole coupler, N = 4. From (6.14)-(6.15),
105 ;
jkf j = 1:113
jkb j = 1:705
106 :
Then,
C
=
)
jkj
=
=
20 log10 jkf j
20 log10 jkj
20 log10
N
X
n=0
20 log10 jkj = C + 20 log10 jkf j + 20 log10
log101
7:071
1
20
10
8
C + 20 log10 jkf j + 20 log10
:
With rn3 = kCn4 ,
r0
r1
r2
r3
r4
=
=
=
=
=
4:135
6:564
7:514
6:564
4:135
mm,
mm,
mm,
mm = r1 ;
mm = r0 :
CnN ;
N
X
CnN ;
n=0
N
X
n=0
CnN
!!
136
CHAPTER 6. POWER DIVIDERS AND DIRECTIONAL COUPLERS
C=
20 log10 jkf j
20 log10 jkj
20 log10
C
20 log10 jkb j
20 log10
CnN
n=0
= 18 dB at f = f0 ;
D=
N
X
N
X
rn3 e
j2 nd
n=0
= 283 dB at f = f0 .
Note that C is very insensitive to frequency (only via kf , which is a slowly-varying function with respect to
frequency; upon taking log10 jkf j we obtain a fairly insensitive function of frequency for C. However, D is
very frequency sensitive due to the summation term.
100
5 Hole Coupler
80
D
60
40
C
20
0
4
6
8
10
12
14
16
14
16
f (GHz)
For a 3 hole coupler (r0 = 6:564; r1 = 8:27; r2 = 6:564 = r0 ),
100
3 Hole Coupler
80
D
60
40
C
20
0
4
6
8
10
f (GHz)
and for a 15 hole coupler
12
137
100
15 Hole Coupler
80
D
60
40
C
20
0
4
6
8
10
12
14
16
f (GHz)
Directional couplers can also be implemented with transmission lines, although the theory will not be
developed here.
Hybrid Couplers:
138
CHAPTER 6. POWER DIVIDERS AND DIRECTIONAL COUPLERS
Hybrid couplers are special cases of directional couplers where C = 3 dB.
1. Quadrature hybrid: There is a 90 phase shift between Ports 2 and 3 when fed at Port 1.
2. Magic-T hybrid: There is a 180 phase shift between Ports 2 and 3 when fed at Port 1.
Power Dividers:
Power dividers are similar to direction couplers, but are usually three-port devices. The idea is still to
split input power into two parts. We will brie‡y study a lossless divider. The resistive divider and Wilkinson
divider are also very common. All of these dividers are T-junction dividers; some physical implementations
are shown below.
139
1
ou
tp
ut
A lossless divider is depicted below.
Z1
input
+
Z0
Vo
jB
-
Yin
Z2
2
ut
tp
ou
The susceptance B accounts for the physical discontinuity in the waveguide or transmission line. We
have
1
1
Yin = jB +
+
;
(6.16)
Z1
Z2
and we want
Yin =
1
:
Z0
(6.17)
If we ignore B then
1
1
1
+
=
:
Z1
Z2
Z0
(6.18)
If Z1 and Z2 satisfy this requirement then we divide the power with no re‡ection. For example, Z1 = Z2 =
2Z0 will work, resulting in a 3 dB divider.
If the divider segments feed lines with Z0 characteristic impedance, quarter-wave transformers can be
used to provide a good impedance match.
140
CHAPTER 6. POWER DIVIDERS AND DIRECTIONAL COUPLERS
Some drawbacks to this type of divider are that there is no isolation between output ports, and looking
into an output port one sees an impedance mismatch.
To determine impedance values for a given power split, the following procedure is used:
Pin =
1 v02
2 Z0
(6.19)
since the divider is match looking into the input, and
P1 =
If I want P1 = xPin and P2 = (1
1 v02
;
2 Z1
P2 =
1 v02
:
2 Z2
(6.20)
x) Pin , then
1 v02
1 v2
Z0
= xPin = x 0 ; ! x =
2 Z1
2 Z0
Z1
1 v02
1 v02
= (1 x) Pin = (1 x)
;! 1
P2 =
2 Z2
2 Z0
P1 =
(6.21)
x=
Z0
:
Z2
Example 6.3 Determine the normalized output impedances for a 3 : 1 power split.
Solution: We want
3
1
Pin ; P2 = Pin ;
4
4
3
Z0
4
)x= =
) Z1 = Z0 ;
4
Z1
3
1
Z0
x= =
! Z2 = 4Z0 :
4
Z2
P1 =
1
(6.22)
Chapter 7
Electromagnetic Compatibility and
Interference (EMS/EMI)
From "Introduction to Electromagnetic Compatibility" by Clayton R. Paul (Wiley: 1992)
Radiated Emissions
The FCC regulates "unintentional radio-frequency devices". It is illegal to sell or advertise for sale
any such products until their radiated and conducted emissions have been measured and found to be in
compliance.
Class A devices are those that are marketed for use in a commercial, industrial or business environments.
Class B devices are those that are marketed for residential use.
Class B limits are more stringent than Class A limits
FCC Emission Limits for Class B Digital Devices, Radiated Emissions (at 3 meters):
Frequency (MHz)
uV/m
dB(uV/m)
30 - 88
100
40
88 - 216
150
44
216 - 960
200
46
> 960
500
54
Hertzian dipole model of a radiating transmission line:
The far …eld (jrj
, Lw ) from a wire of length Lw carrying a constant current I0 along the z axis is
E(r) = b (I0 ) (Lw ) (j! )
E (r) = jI0 Lw 2 10
7
f
e
e jkr
4 r
sin
(7.1)
jkr
r
:
This assumes current is constant (not a bad assumption for Lw
)
Note that the measurement position r = 3 meters may not be in the far …eld
141
142CHAPTER 7. ELECTROMAGNETIC COMPATIBILITY AND INTERFERENCE (EMS/EMI)
For two wires (see …gure),
E (r) = jLw 2 10
7
f
jk(r+ )
e
I1
+ I2
r+
jk(r
e
)
;
r
(7.2)
where
b b
y
r = sin sin
s
= sin sin ;
2
(7.3)
(7.4)
since
cos (90
) = sin
r
z
s/2
∆
y
I2
I1
s
so that
E (r) = jLw 2 10
Di¤erential mode currents (I0 = I1 =
7
f
, sin k 2s sin sin
7
f
7
I0 jLw 2 10
7
f
f
e
jk
e
r
e
:
(7.5)
ejk
(7.6)
jkr
(2j sin (k ))
r
jkr
r
s
sin k sin sin
2
(7.7)
:
(7.8)
' k 2s sin sin ,
2
= 1:32
10
14
1:32
10
14
E dm (r) =
=
jkr
e
E dm (r) = I0 Lw (2 ) 10
Field is maximum for
+ I2 ejk
I2 )
= I0 Lw 4 10
Assuming that s
jk
I1 e
r
E dm (r) = I0 jLw 2 10
=
jkr
e
r
7p
"f 2
I0 Lw f 2 s
e
e
jkr
r
s sin sin
(7.9)
sin sin ;
(7.10)
jkr
r
I0 Lw f 2 s sin jsin j
(7.11)
= 90 ,
E dm;max (r) =
1:32
10
r
14
I0 Lw f 2 s:
(7.12)
143
Maximum radiation occurs in the plane of the wires, broadside to the wire axes.
Radiation can be minimized at a certain frequency and position by
– reducing the current level
– reducing the wire length and/or wire separation (overall, reducing the loop area L
s)
– use ferrite beads (adds series inductance to attenuate high-frequency harmonics)
Common mode currents (I0 = I1 = I2 )
Assuming that s
E cm (r) = I0 jLw 2 10
7
= I0 jLw 2 10
7
= jI0 Lw 4 10
7
, cos k 2s sin sin
f
f
f
e
jkr
e
r
e
+ ejk
(7.13)
jkr
(2 cos (k ))
r
e
jk
jkr
r
(7.14)
s
cos k sin sin
2
;
(7.15)
'1
E cm (r) = 1:26
10
6
1:26
10
6
jE cm (r)j =
r
jI0 Lw f
e
jkr
(7.16)
r
I0 Lw f
(7.17)
Common mode radiation is insensitive to rotating the cable/wires
Radiation can be minimized at a certain frequency and position by
– reducing the current level
– reducing the wire length
– use a common-mode choke (see …gure)
– use ferrite beads (adds series inductance to attenuate high-frequency harmonics)
Example:
Lw = 1 m, I0 = 20 mA, s = 1 mm, f = 50 MHz, r = 3
E dm;max (r) =
1:32
10
14
20 10 3 (1) 50
3
= 220 V/m = 47 dB V/m
106
2
1
10
3
(7.18)
(7.19)
(above the limit)
jE cm (r)j =
=
1:26
10
6
10
6
I0 Lw f
r
1:26
20
3
= 420; 000 V/m
(7.20)
10
(far above the limit)
Q. What I would produce 100 V/m?
A. For di¤erential mode, 9 mA, for common mode, 4:8 A
3
(1) 50
106
(7.21)
(7.22)
144CHAPTER 7. ELECTROMAGNETIC COMPATIBILITY AND INTERFERENCE (EMS/EMI)
Raise frequency to 1 GHz:
E dm;max (r) =
1:32
14
10
20
10
3
(1) 1
3
= 88; 000 V/m = 99 dB V/m
2
109
1
10
3
(7.23)
(7.24)
(way over)
jE cm (r)j =
=
1:26
10
6
10
6
I0 Lw f
r
1:26
(7.25)
20 10
3
= 8; 400; 000 V/m !!!
3
109
(1) 1
(7.26)
(7.27)
Q. What I would produce 500 V/m?
A. For di¤erential mode, 114 A, for common mode, 1:1 A
Susceptibility Models
Circuit Model:
i(z,t)
Vs(z,t)
i(z+dz,t)
+
-
R ∆z L ∆ z
∆z
+
∆z
Is(z,t)
G ∆z
C ∆z v(z+dz,t)
-
∆z
d2 v(z)
dz 2
d2 i(z)
dz 2
2
v(z) =
2
i(z) =
d vs (z)
;
dz
d is (z)
(G + i!C) vs (z) +
;
dz
(R + i!L) is (z) +
(7.28)
where
2
= (R + j!L) (G + j!C) ;
(7.29)
and = +j 2 C is called the propagation constant (1=m). The real and imaginary parts of the propagation
constant are known as the attenuation constant ( ) and the phase constant ( ), respectively.
An incident electromagnetic wave induces voltages and currents on the transmission line, which are
represented by Vs and Is . Faraday’s law can be used to determine Vs . The induced emf around a loop l with
surface S is
I
Z
emf = E(r) dl = j!
B(r) n dS:
(7.30)
l
S
b. The distributed voltage
Assume the y axis is vertical, so that the x axis is into the paper, and n = x
source for two wires separated by distance s is
Z z+ z=2 Z s
Z s
V (z) ' j!
Bxi (y; z) dydz ' j! z
Bxi (y; z) dy
(7.31)
z
V (z)
= Vs (z) =
z
z=2
j!
0
Z
0
0
s
Hxi (y; z) dy:
145
Since the wire separation s is much less than ,
Z s
Hxi (y; z) dy '
Vs (z) = j!
j! sHxi (z)
(7.32)
0
An incident, vertically polarized electric …eld will induce a voltage between the wires, causing a displacement current through the capacitance between the wires (we ignore here conduction current due to a lossy
insulator). Using Ohm’s law,
Z s
I (z)
Eyi (y; z) dy = I (z) Zc =
V (z) =
(7.33)
j!C
0
Z s
Is (z) = j!C
Eyi (y; z) dy ' j!CsEyi (z):
0
Rather than solve (7.28), we can simply things further by assuming that L
one section of line. Replacing z by Lw and setting R = G = 0,
VL =
=
, so that we only need
RL Lw
(Vs + Rs Is )
L LC + j! (Rs RL C + L)
j!RL Lw s
Hxi + Rs CEyi
2
! RL LC + j! (Rs RL C + L)
!2 R
RL + Rs
RL + Rs
Note that the area of the loop formed by the two wires, A = Lw
If frequency is su¢ ciently low,
(7.34)
(7.35)
s is important.
RL Lw
(Vs + Rs Is )
RL + R s
j!ARL
Hxi + Rs CEyi :
=
RL + Rs
VL =
(7.36)
Example:
Assume that the plane wave
jky
E=b
zE z e
;
b
H=x
Ez
e
jky
(7.37)
is incident on a two-wire transmission, with Ez = 1 V/m and ! = 30 MHz. The wires have center-to-center
spacing D = 1:22 cm, radius a = 0:1 cm (this is the standard 300 ohm twin-lead line), and length Lw = 1
m. In this case,
D
"0
; C=
L = 0 cosh 1
;
(7.38)
D
2a
cosh 1 2a
and for this line L = 1 H and C = 11:2 pF. For this incident wave there is only a voltage source,
Vs =
j! sHxi =
such that
Hxi = 4
Assume RL = 1000
, Rs = 10
7
10
j1:22 mV
1
= 3:33
377
10
(7.39)
9
:
(7.40)
. Then,
VL =
j!ARL
RL + Rs
Hxi + 0 =
j1:21 mV.
(7.41)
If, instead, the incident plane wave was
b Ey e
E=y
then
jkx
;
H=b
z
Rs CEyi = 1:12
and
VL =
10
Ey
e
10
j!ARL
0 + Rs CEyi =
RL + Rs
jkx
;
(7.42)
(7.43)
j0:04 mV.
(7.44)
146CHAPTER 7. ELECTROMAGNETIC COMPATIBILITY AND INTERFERENCE (EMS/EMI)
Chapter 8
Microwave Filters
147
148
CHAPTER 8. MICROWAVE FILTERS
Chapter 9
Appendix
9.1
Transient transmission line current
We obtained the transient transmission line voltage as (2.111),
v (z; t) = v + (t
z=vp ) + v (t + z=vp ) :
(9.1)
Using
@v(z; t)
=
@z
L
@i(z; t)
@t
(9.2)
we obtain
@v (z; t)
@z
@v (t z=vp ) @ (t z=vp )
@ (t z=vp )
@z
1
0
v
vp
@i(z; t)
L
;
@t
=
=
=
(chain rule)
(9.3)
leading to
@i(z; t)
@t
1
v +0 (t z=vp )
Lvp
1
) i (z; t) =
v + (t
Lvp
=
v
0
(t + z=vp ) ;
z=vp )
(9.4)
v (t + z=vp ) + c (z)
(9.5)
where c (z) is a constant of integration. To determine c (z), use
@i(z; t)
=
@z
C
@v(z; t)
:
@t
(9.6)
From (9.5) and using (9.3) we have
@
1
i (z; t) =
@z
Lvp2
=
C
v +0 (t
z=vp )
v
0
(t + z=vp ) +
@
c (z)
@z
(9.7)
@v(z; t)
:
@t
With
@v (z; t)
@v (t z=vp ) @ (t z=vp )
=
@t
@ (t z=vp )
@t
= v
0
(1)
149
(9.8)
150
CHAPTER 9. APPENDIX
we have
@
i (z; t) =
@z
=
1
v +0 (t
Lvp2
C
z=vp ) + v
@v(z; t)
=
@t
C v +0 (t
0
(t + z=vp ) +
z=vp ) + v
0
@
c (z)
@z
(9.9)
(t + z=vp ) :
Therefore,
such that
|
1
+ C v +0 (t
Lvp2
{z
}
z=vp ) + v
0
(t + z=vp ) +
@
c (z) = 0;
@z
(9.10)
0
@
c (z) = 0 ) c (z) = constant.
(9.11)
@z
c (z) = c provides at most a d.c. current that provides a d.c. o¤set to the q
time-dependent current. We will
L
1
p
= Z0 , the desired expression
ignore this o¤set (set c = 0) to arrive at, upon noting that Lvp = L LC = C
i (z; t) =
1
v + (t
Z0
z=vp )
v (t + z=vp ) :
(9.12)