Microwave Engineering

Microwave Engineering Notes

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Contents

1 Electromagnetic Theory 91.1 Introduction to Microwave Engineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2 Maxwell�s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Time-Harmonic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4 Wave Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.5 Plane Wave Propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Transmission Line Theory 172.1 The Lumped-Element Circuit Model for a Transmission Line . . . . . . . . . . . . . . . . . . 182.2 Field Analysis of Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3 The Terminated Lossless Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.4 The Smith Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.5 The Quarter-Wave Transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.6 Generator and Load Mismatches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.7 Lossy Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.8 Transient Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2.8.1 Waveforms and Spectral Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.8.2 Integrated Circuits and Ground Bounce . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3 Transmission Lines and Waveguides 573.1 General Solutions for TEM, TE, and TM Modes . . . . . . . . . . . . . . . . . . . . . . . . . 573.2 Parallel Plate Waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.3 Rectangular Waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643.4 Circular Waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.5 Coaxial Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.6 Surface Waves on a Grounded Dielectric Slab . . . . . . . . . . . . . . . . . . . . . . . . . . . 673.7 Stripline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703.8 Microstrip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713.9 The Transverse Resonance Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.10 Wave Velocities and Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

4 Microwave Network Analysis 794.1 Impedance and Equivalent Voltage and Currents . . . . . . . . . . . . . . . . . . . . . . . . . 79

5 Impedance Matching and Tuning 855.1 Matching with Lumped Elements (L Networks) . . . . . . . . . . . . . . . . . . . . . . . . . . 86

5.1.1 Lumped Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885.2 Single-Stub Tuning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885.3 Double-Stub Tuning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 995.4 The Quarter-Wave Transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1055.5 The Theory of Small Re�ections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1075.6 Binomial Multisection Matching Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

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4 CONTENTS

5.7 Chebyshev Multisection Matching Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . 1155.8 Tapered Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1205.9 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 1499.1 Transient transmission line current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

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CONTENTS 5

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6 CONTENTS

What is �Di¤erent�About High Frequencies?

The short, and perhaps obvious, answer is wavelength. For any given object, one usually considers afrequency 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 anexception). So if one considers electromagnetic scattering from a raindrop, and scattering from a largebuilding, 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 becausewe live in a world of characteristic sizes commiserate with this wavelength. For example, people are typically5-6 feet tall (1.66-2 meters), cars are typically 8-12 feet long (2.66-4 meters), suburban buildings are typically10-30 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�sequations. These are very complicated equations, but they can be simpli�ed in special cases. One such caseis in the event of �low� frequencies� one can derive the classical circuit equations (Ohm�s law, Kirchho¤�slaws, etc.) as special, simple cases. These simpli�ed equations are often applicable in circuit analysis sincetypical circuit dimensions are very small compared to typical wavelengths in common usage.For example, consider the following discrete (not integrated) circuit)

+

Vs

R1

R2

Vout

L= 5 cm

+ +

With Vs = V0 cos!t, is Vout = R2

R1+R2Vs?

Case 1 f = 20 kHz ! � = cf =

3�1010cm=s20�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,and Vout = R2

R1+R2Vs:

Case 2 f = 6 GHz ! � = cf =

3�1010cm=s6�109=s = 5 cm.

L

�= 1:

The circuit length is equal to a wavelength� low frequency approximations are not applicable.

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CONTENTS 7

Case 3 f = 300 GHz ! � = cf =

3�1010cm=s300�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 carefullymatched for e¢ cient power transfer.

� Practically, the circuit will not work since it will radiate energy into space. Furthermore, the electri-cally long distances the signal must travel will result insigni�cant dissipation to the use of imperfectconductors.

� 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�6 cm.

Case 4 f = 6 GHz ! � = cf =

3�1010cm=s6�109=s = 5 cm.

L

�= 3:3� 10�6 � 1:

The circuit length is very short compared to a wavelength (same as Case 1)� low frequency approximationsare applicable, and Vout = R2

R1+R2Vs.

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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 quantitativeideas about electrostatics.

1769: Dr. John Robison began experiments which would lead to the inverse-square law (Coulomb�slaw) governing electrostatics. Similarly by Henry Cavendish in 1773. Neither investigator widelypublicized their work.

1785: Charles Augustin de Coulomb demonstrated the law of electric force, which is now calledCoulomb�s law.

1820: Hans Christian Oersted found that a current carrying wire (electricity) could producemagnetism.

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 anelectric 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 antennacaused 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 NovaScotia and Cape Cod.

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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 col-lectively as Maxwell�s equations. Maxwell�s equations in di¤erential form are

r �D(r; t) = �e(r; t);r �B(r; t) = �m(r; t); (1.1)

r�E(r; t) = � @@tB(r; t)� Jm(r; t);

r�H(r; t) = @

@tD(r; t) + Je(r; t);

� E is the electric �eld intensity (V=m)

� D is the electric �ux density (C=m2)

� B is the magnetic �ux density (Wb=m2)

� H is the magnetic �eld intensity (A=m)

� �e is the electric charge density (C=m3)

� Je (= J in text) is the electric current density (A=m2)

� �m is the magnetic charge density (Wb=m3)

� Jm (=M in text) is the magnetic current density (V=m2)

� 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.

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10 CHAPTER 1. ELECTROMAGNETIC THEORY

1.3 Time-Harmonic Fields

Often the �elds of interest vary harmonically (sinusoidally) with time. Because Maxwell�s equations arelinear, and assuming linear constitutive equations (typical below optical range), then

time-harmonic sources �;J will maintain time-harmonic �elds E;D;B;H.

AssumeE(r; t) = E0(r) cos(!t+ �E); (1.2)

then, for instance,

E(r; t) = E0(r)Renej(!t+�E)

o= Re

�E0(r)e

j!tej�E

(1.3)

= Re�E(r)ej!t

;

where E(r) � E0(r)ej�E is a complex phasor.

Repeating for the other �eld quantities of interest,

B(r; t) = B0(r) cos(!t+ �B) (1.4)

D(r; t) = D0(r) cos(!t+ �D)

H(r; t) = H0(r) cos(!t+ �H)

J(r; t) = J0(r) cos(!t+ �J)

�(r; t) = �0(r) cos(!t+ ��)

we obtain the complex phasors

B(r) � B0(r)ej�B (1.5)

D(r) � D0(r)ej�D

H(r) � H0(r)ej�H

J(r) � J0(r)ej�J

�(r) � �0(r)ej�� :

To obtain time-harmonic Maxwell�s equations, consider, for instance, Faraday�s law,

r�E(r; t) = � @@tB(r; t)� Jm(r; t): (1.6)

This can be rewritten asRe�[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�E(r) + j!B(r) + Jm(r)g = 0:

�When !t = �=2; Im fr�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)

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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); (1.9)

r�H(r) = j!D(r) + Je(r);

r � Je(m)(r) = �j! �e(m)(r);

where all quantities are time-harmonic phasors.

� Note that in the time-harmonic case we obtain the convenient correspondence

@=@t $ j! (1.10)Z(�) dt $ (j!)

�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; !); (1.12)

B(r; !) = e�(r; !)H(r; !);where e�, e� are the permittivity and permeability, respectively, of the medium.� e� = e�r"0, e� = e�r�0� �0 is the permittivity of free space (' 8:85� 10�12 F=m)

� �0 is the permeability of free space (' 4� � 10�7 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 = 1p�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 andmagnetic).

� The imaginary parts of e�, e� are due to polarization (molecular) loss, i.e., dipole friction and associatedtime-lag,

e� = e�0 � je�00 (1.13)e� = e�0 � e�0:Ohm�s Law

Another relationship that is often useful for lossy media is the point form of Ohm�s law,

Je(r; !) = �e(r; !)E(r; !); (1.14)

Jm(r; !) = �m(r; !)H(r; !);

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� �e (ohms�1=m ) is the electrical conductivity of the medium

� �m (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 inducede¤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) = �eE(r) 6= 0; (1.15)

where Jce is an induced conduction current density. The total electric current is

Je(r) = Jie(r) + J

ce; (1.16)

and Faraday�s law becomes

r�H(r) = j!e"E(r) + Jie(r) + �eE(r) (1.17)

= j!

�e"� j

!�e

�E(r) + Jie(r):

De�ning a new complex permittivity as

" ��e"� j

!�e

�(1.18)

leads to

r�H(r) = j!"E(r) + Jie(r); (1.19)

where we have separated the induced e¤ects from the applied source.

Repeating for Jm(r) = Jim(r)+�mH(r) = J

im(r) + J

cm and noting that

r � Jie(m) + j!�ie(m) = 0; (1.20)

leads to

� ��e�� j

!�m

�: (1.21)

� Assuming e", e� are real, the imaginary parts of ", � account for conduction loss.� In general, one often writes

" = ("0 � j"00) ; (1.22)

� = (�0 � j�00) ;

or, in terms of the relative permittivity,

" = ("0r � j"00r ) "0; (1.23)

� = (�0r � j�00r )�0:

where the imaginary parts account for all loss mechanisms (conductive and molecular).The followingtable lists some typical material parameters for dielectric media.

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1.4. WAVE EQUATIONS 13

Material "0r "00r � (1/ohm-m)

air 1.0006glass 3.8�8 <0.003 @ 3 GHzwood 1.5�2.1 <0.07 @ 3 GHzgypsum board 2.8 0.046 @ 60 GHzdry brick 4 0.05�0.1 @ 4,3 GHzdry concrete 4�6 0.1�0.3 @ 3,60 GHzfresh water 81 0.01�0.001sea water 81 1�6snow 1.2�1.5 <0.006 @ 3 GHz 0.000001ice 3.2 0.0029 @ 3GHzmoist ground 20�30 0.03�0.003dry ground 3�6 0.005�0.00001copper 1 5:7� 107

Maxwell�s equations become

r � ("E(r)) = �ie(r);r � (�H(r)) = �im(r); (1.24)

r�E(r) = �j!�H(r)� Jim(r);r�H(r) = j!"E(r) + Jie(r);

� Often the superscript i is omitted in (1.24)�the interpretation of J depends on �; ". For example, afairly general form is

r �D(r) = �e(r);r �B(r) = �m(r); (1.25)

r�E(r) = �j!B(r)� Jm(r);r�B(r) = j!�D(r) + �Je(r);

where B =�H and D = "E.

� If �; " account only for polarization e¤ects, i.e., if �; " are real-valued, or if �; " are complex-valuedwhere the imaginary parts are associated with polarization loss (dipole friction), then Jm;Je aretotal 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�E(r) = �j!�H(r)� Jm(r); (1.26)

r�H(r) = j!"E(r) + Je(r);

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represent six scalar equations in six unknowns. However, they are coupled vector partial di¤erential equa-tions. We want to uncouple the equations and obtain one equation which may be solved, assuming ahomogenous medium.

Taking the curl of r�E(r) and of r�H(r) leads to

r�r�E(r)� !2"�E(r) = �j!�Je(r)�r� Jm(r); (1.27)

r�r�H(r)� !2"�H(r) = �j!"Jm(r) +r� Je(r):

These are the vector wave equations for the �elds.

Noting that r�r�A =r (r �A)�r2A, we also have

r2E(r) + !2"�E(r) = j!�Je(r) +r� Jm(r)+r�e"; (1.28)

r2H(r) + !2"�H(r) = j!"Jm(r)�r� Je(r)+r�m�

:

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 planewaves 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 + k2

�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

E = bxE (z)The wave equation (1.29) becomes �

@2

@z2+ k2

�E (z) = 0

which has solutionE (z) =

�E+0 e

�jkz + E�0 e+jkz

�:

Plugging into Faraday�s law leads to the magnetic �eld as

H = by�E+0�e�jkz � E

�0

�e+jkz

�;

where � =p�=", and so the pair

E = bx �E+0 e�jkz + E�0 e+jkz� ;H = by�E+0

�e�jkz � E

�0

�e+jkz

�;

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1.5. PLANE WAVE PROPAGATION 15

form a wave travelling along z. E ? H ?bz and the wave is called a TEM (Transverse ElectroMagnetic) wave.

Phase and Attenuation Constants of a Uniform Plane Wave

Assumek = (� � j�) : (1.30)

Then

E = bx�E+0 e�j(��j�)z + E�0 e+j(��j�)z� (1.31)

= bx �E+0 e�j�ze��z + E�0 e+j�ze�z� :� � is called the phase constant.

� � is called the attenuation constant.

� Setting k2 = (� � j�)2 = !2�", where � is real-valued and " = "0 � j"00, and equating real andimaginary parts, leads to

� (!) = !p�"0 (!)

vuut1

2

s1 +

"00 (!)2

"0 (!)2 + 1

!(1.32)

� (!) = !p�"0 (!)

vuut1

2

s1 +

"00 (!)2

"0 (!)2 � 1

!:

Time-Domain Waves, Phase Velocity, and Wavelength for Uniform Plane Waves

In the time-domain,

E (r; t) = Re�bx �E+0 e�j�ze��z + E�0 e+j�ze�z� ej!t (1.33)

= bx �E+0 e��z cos (!t� �z) + E�0 e�z cos (!t+ �z)�

� For the �rst term, as t increases, the argument of the cosine remains unchanged if �z increasescorrespondingly� thus the wave travels along the +z-direction. Correspondingly, the second termtravels along the �z-direction.

� The phase velocity (actually speed since it is a scalar) of the wave is found from

d

dt(!t� � z) =

d

dtConst = 0 (1.34)

! � � dzdt

= 0

leading to

vp =dz

dt=!

�: (1.35)

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� The wavelength of the wave is the distance between adjacent wavefronts that produce the same valueof the cosine function. If z1 and z2 are points on adjacent wavefronts,

� z1 = � z2 � 2� (1.36)

or� = z1 � z2 =

2�

�: (1.37)

Example 1.1 Assume a plane wave with electric �eld oriented in the bx-direction is propagating in the bz-direction through dry soil at 1 MHz. At this frequency typical material properties are " = 3"0 and � = 10�5

(ohms�1=m ). Therefore,

" ��e"� j

!�e

�=

�3"0 �

j

2�10610�5

�(1.38)

=

�3� j

2�106"010�5

�"0 = (3� j0:179) "0 = "0 � j"00

such that

� (!) = !p�"0

vuut1

2

r1 +

"002

"02+ 1

!(1.39)

= 0:0363

� (!) = !p�"0

vuut1

2

r1 +

"002

"02� 1!

= 0:00108:

Since k = bz kz = bz (� � j�), the wave has the formE (r) = E0e

�jk�r = bxE0e�j� ze��z (1.40)

and in the time-domain

E (r) = bxE0e�� z cos (!t� � z) : (1.41)

The phase velocity of the wave is

vp =!

�=2�106

0:0363= 1:73� 108m/s (1.42)

and the wavelength is

� =2�

�=

2�

0:0363= 173:1m. (1.43)

In travelling 1 km the amplitude will decrease from E0 to E0e�� z = 0:339E0, or by

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Chapter 2

Transmission Line Theory

Transmission lines are used to transfer electrical signals (information) or electrical power from one point toanother in an electrical system. Transmission lines take a wide variety of forms, from simple wire pairs andcables to more complicated integrated structures for high-frequency applications. Some common transmissionlines are:

(c)(a)

(b)

bε,µ

conducting strip

w

ground plane

(d)

conducting strip

w

bε,µ

ground planes

(e)

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18 CHAPTER 2. TRANSMISSION LINE THEORY

y

x

z

b

a

ε,µ

(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 Frequency Impedance Power Ease of Device Low CostRange (GHz) Range (ohms) Rating Mounting Production

Coaxial line < 50 10� 100 medium medium mediumStripline < 10 10� 100 medium medium goodmicrostrip line < 100 10� 100 low good goodrectangular < 300 100� 500 high good poorwaveguide

In this chapter we will study the analysis of general transmission lines, mostly without considering speci�cphysical structures. In the next chapter we will examine some speci�c geometries of transmission lines andwaveguides.

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 two-conductor1 TEM transmission line depicted in the �gure below, where Vs and Is represent distributedsources.

v(z,t)

i(z,t) Vs(z,t)

Is(z,t)

+ +

1Generally 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.

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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 thecase 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 � �) sectionof the structure. In this way, we treat transmission lines as many cascaded sections of electrically smallcircuits, and apply circuit models to each section.The lumped-element model for a small segment of the line is shown in the �gure below.

∆

z∆

zVs(z,t)

Is(z,t)

i(z,t) R L

G C

z +

z

z

z z

∆ ∆

∆ ∆

∆

+

v(z+dz,t)

i(z+dz,t)

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 thetransmission line by an external source. If desired, a localized source can be modeled by vs = v0�(z � z0),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 �R�z i(z; t) (2.1)

�L�z @i(z; t)@t

� v(z +�z; t) = 0

and

i(z; t)+ is(z; t)�z �G�z v(z +�z; t) (2.2)

�C�z @v(z +�z; t)@t

� i(z +�z; t) = 0:

Dividing these two equations by �z and taking the limit as �z ! 0 lead to

@v(z; t)

@z= �R i(z; t)� L@i(z; t)

@t+ vs(z; t); (2.3)

@i(z; t)

@z= �Gv(z; t)� C @v(z; t)

@t+ is(z; t);

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and, upon assuming time-harmonic conditions

dv(z)

dz= �R i(z)� j!L i(z) + vs(z); (2.4)

di(z)

dz= �Gv(z)� j!C v(z) + is(z):

These two coupled �rst-order di¤erential equations can be easily decoupled by forming second-order di¤er-ential equations

d2v(z)

dz2� 2v(z) = � (R+ i!L) is(z) +

d vs(z)

dz; (2.5)

d2i(z)

dz2� 2i(z) = � (G+ i!C) vs(z) +

d is(z)

dz;

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 propagationconstant are known as the attenuation constant (�) and the phase constant (�), respectively.Usually in microwave analysis one �rst considers the homogeneous equations

d2v(z)

dz2� 2v(z) = 0; (2.7)

d2i(z)

dz2� 2i(z) = 0;

corresponding to the absence of any source or load. General solutions are found as

v(z) = v+0 e� z + v�0 e

+ z; (2.8)

i(z) = i+0 e� z + i�0 e

+ z;

which represent voltage and current waves. Exploiting (2.4) leads to the relationship between voltage andcurrent as

i(z) =1

Z0

�v+0 e

� z � v�0 e+ z�; (2.9)

where

Z0 �R+ j!L

(2.10)

is called the characteristic impedance (ohms) of the transmission line.

Time-domain form:

Assuming sinusoidal time variation,

v(z; t) = Re�v (z) ej!t

(2.11)

= Renhv+0 e

�(�+j�)z + v�0 e+(�+j�)z

iej!t

o:

Withv�0 =

��v�0 �� ej�� (2.12)

we havev(z; t) =

��v+0 �� cos �!t� �z + �+�+ ��v�0 �� cos �!t+ �z + �� : (2.13)

The �rst term, corresponding to e� z, is a forward (+z-traveling) wave, while the second term, corre-sponding to e+ z, is a backward (�z-traveling) wave. Also,

� =2�

�, vp =

!

�= �f: (2.14)

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2.2. FIELD ANALYSIS OF TRANSMISSION LINES 21

Lossless Lines:

If the line is lossless (R = G = 0), then

= �+ j� = 0 + j�; (2.15)

� = !pLC =

2�

�; Z0 =

rL

C:

2.2 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 byan electromagnetic �eld analysis of the structure.The circuit parameters are obtained as

L =�

jI0j2ZS

H �H�dS; (2.16)

C ="

jV0j2ZS

E �E�dS;

R =Rs

jI0j2Zl

H �H�dl;

G =!"00

jV0j2Zl

E �E�dl;

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 thatcancel out with the numerators .Some representative results are provided in the table below.

ε,µb

a ε,µ

a

D

w

dε,µ

L �2� ln

ba

�� cosh

�1 � D2a

��dw

C 2�"0

ln ba

�"0

cosh�1( D2a )"0wd

R Rs

2�

�1a �

1b

�Rs

�a2Rs

w

G 2�!"00

ln ba

�!"00

cosh�1( D2a )!"00wd

Note that for TEM transmission lines comprised of perfect conductors, the exact EM �eld solution isin full agreement with the microwave model (with G;L;C appropriately de�ned). For lossy conductors themicrowave model is no longer an exact solution, but usually provides a very good approximation. For nonTEM transmission lines (we�ll discuss later, microstrip is one example), the microwave engineering model isan approximation of the rigorous EM �eld solution.

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2.3 The Terminated Lossless Transmission Line

A terminated transmission line is depicted schematically in the �gure below.

Z g

Z 0 ,v p

z=0z=L

in

Z L+

gVZ

The generator (modeled by Vg and Zg) can represent, for example, the output of an IC chip, the output ofan 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) = v+0 e�j� z + v�0 e

+j� z; (2.17)

i(z) =v+0Z0e�j� z � v

�0

Z0e+j� z:

It is convenient to reform (2.17) as

v(z) = v+0

�e�j� z +

v�0v+0e+j� z

�; (2.18)

i(z) =v+0Z0

�e�j� z � v

�0

v+0e+j� z

�:

De�ning a load re�ection coe¢ cient

�L =v�0v+0

we have

v(z) = v+0�e�j� z + �Le

+j� z�; (2.19)

i(z) =v+0Z0

�e�j� z � �Le+j� z

�:

More generally, de�ne

� (z) =v�0v+0ej2� z; (2.20)

where we have the special cases

� (0) = �L =v�0v+0; (2.21)

� (�L) = �in =v�0v+0ej2� L;

such that� (z) = �L e

j2� z (2.22)

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2.3. THE TERMINATED LOSSLESS TRANSMISSION LINE 23

leading to

v(z) = v+0 e�j� z (1 + � (z)) ; (2.23)

i(z) =v+0Z0e�j� z (1� � (z)) :

The total impedance at any point is

Z (z) =v (z)

i (z)= Z0

1 + � (z)

1� � (z) ; (2.24)

and, conversely,

� (z) =Z (z)� Z0Z (z) + Z0

: (2.25)

Important special cases:

� (0) = �L =ZL � Z0ZL + Z0

; (2.26)

� (�L) = �in =Zin � Z0Zin + Z0

: (2.27)

An important formula for Zin can be obtain as follows;

Z (z) =v (z)

i (z)=v+0�e�j� z + �L e

j� z�

v+0Z0(e�j� z � �L ej� z)

(2.28)

= Z0ZL � jZ0 tan (�z)Z0 � jZL tan (�z)

;

Zin = Z (�L) = Z0ZL + jZ0 tan (�L)

Z0 + jZL tan (�L):

One important use of Zin is that once the input impedance has been determined, the transmission line circuitcan be considered to be

gZ

+gV inZ

Iin

+

Vin

such that

PL =1

2Re fvini�ing =

1

2Re

8<:�

ZinZin + Zg

Vg

� � Z�inZ�in+Z

�gV �g

�Z�in

9=; : (2.29)

Time-average power �ow:

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Pav =1

2Re fv (z) i� (z)g (2.30)

=1

2Re

(�v+0�e�j� z + �L e

j� z��v+0

Z0

�e�j� z � �L ej� z

��)

=1

2

��v+0 ��2Z0

Re

8><>:1 + �Lej2�z � ��Le�j2�z| {z }pure imaginary

� j�Lj2

9>=>;=

1

2

��v+0 ��2Z0

n1� j�Lj2

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�Lj = 0; RL =1; (2.32)

j�Lj = 1; RL = 0 dB.

Standing waves and standing wave ratio:

In the sinusoidal steady state, for a matched line (�L = 0),

v (z) = v+0 e�j�z; jv (z)j =

��v+0 �� = constant. (2.33)

For a mismatched line (�L 6= 0),v (z) = v+0 e

�j� z �1 + �L ej2� z� : (2.34)

Let�L = j�Lj ej�: (2.35)

Then,

jv (z)j =��v+0 �� 1 + j�Lj ej(2�z+�)�� ; (2.36)

vmax =��v+0 �� (1 + j�Lj) ;

vmin =��v+0 �� (1� j�Lj) :

The standing wave ratio (SWR), also known as voltage standing wave ratio (VSWR) is

SWR =vmaxvmin

=1 + j�Lj1� j�Lj

: (2.37)

Note that1 � SWR � 1; (2.38)

and that SWR = 1 denotes a matched load.When will SWR =1? When j�Lj = 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.

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The last case is due to the fact that a pure reactance cannot dissipate power, so everything must bere�ected.

Distance between successive minima:

ej(2�zm+�) = ej(2�(zm+dmin)+�) = ej(2�zm+�)ej2�dmin (2.39)

! 2�dmin = n�; n = 0; 2; 4; :::

so that

dmin =n�

2�=n��

22�=n�

4=�

2: (2.40)

To examine the e¤ects of a pure standing wave, consider a short-circuited line (ZL = 0;�L = �1).

v (z) = v+0�e�j� z � ej� z

�= �v+0 2j sin (�z) ; (2.41)

i (z) =v+0Z0

�e�j� z + ej� z

�=v+0Z02 cos (�z) :

In the time domain,

v (z; t) = v+0 2 sin (�z) sin (!t) ; (2.42)

i (z; t) =v+0Z02 cos (�z) cos (!t) :

1

0.5

0

0.5

1

7 6 5 4 3 2 1bz

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 inter-changing between energy in the electric �eld (voltage) and energy in the magnetic �eld (current). Thisis similar to energy relations in a resonant LC circuit.

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For j�Lj 6= 1, SWR <1 and both a standing wave and a travelling wave exist on the line. Let �L = �ej�.Then,

v (z) = v+0

�e�j� z + � ej(� z+�)

�(2.43)

= v+0

�e�j� z + � ej(� z+�) + �e�j� z � �e�j� z

�= v+0 (1� �) e�j� z + �v+0

�ej(� z+�) + e�j� z

�= v+0 (1� �) e�j� z| {z }

pure travelling wave

+ 2�v+0 ej �2 cos

��z +

�

2

�| {z }

pure standing wave

0

0.2

0.4

0.6

0.8

1

1.2

1.4

7 6 5 4 3 2 1bz

� = 0:5;��(1� :5) e�j�z + cos (�z)��vs. �z

It is easy to see that no real power �ow is associated with standing waves, since

Pav =1

2Re fv (z) i� (z)g = 1

2Re

��2jv+0 sin�z

��2v+0Z0cos (�z)

��= 0: (2.44)

Example 2.1 A lossless, air-�lled transmission line having characteristic impedance 50 ohms is terminatedwith 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, ! � =c

f=

3� 108900� 106 = 0:333 m. (2.45)

L

�=

1:2

0:333= 3:6; �L =

2�

�L = 7:2� (2.46)

ZL = 100 :

(a)

�L =ZL � Z0ZL + Z0

=100� 50100 + 50

=1

3(2.47)

SWR =1 + j�Lj1� j�Lj

= 2

(b)

Zin = Z0ZL + jZ0 tan (�L)

Z0 + jZL tan (�L)= 49:10� j35:03 ohms (2.48)

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(c)RL = �20 log j�Lj = 9:54 dB. (2.49)

ZL = 50 :

(a)

�L =ZL � Z0ZL + Z0

=50� 5050 + 50

= 0 (2.50)

SWR =1 + j�Lj1� j�Lj

= 1

(b)


Z0 + jZL tan (�L)= 50 ohms (2.51)

(c)RL = �20 log j�Lj = �1 dB. (2.52)

Exercise 2.1 (in-class):

Z 0 ,v p

z=0z=L

in

Z LZ

If L = �=2 and ZL = 0 (short circuit), �nd (a) SWR, 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 ohmcoaxial 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 resistance300 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 linehas length

(a) l = �=4,

(b) l = �=2;

(c) l = �:

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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

2

4

1 2 3 4 5 6 7z

tan(�L) vs. �L

Zin = 0 at L = 0;��=2;��;�3�=2; ::: (2.54)

Zin = 1 at L = ��=4;�3�=4;�5�=4; :::

II. ZL =1 (open circuit)Zin = �jZ0 cot�L (2.55)

which is also pure imaginary for any length L:

4

2

0

2

4

1 2 3 4 5 6 7z

cot(�L) vs. �L:

Zin = 0 at L = ��=4;�3�=4;�5�=4; ::: (2.56)

Zin = 1 at L = 0;��=2;��;�3�=2; :::

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.

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IV. L = �=4;

Zin =Z20ZL: (2.58)

This case is know as a quarter-wave transformer, and will be discusses in detail later.

Determining v+0 :

Although we often don�t need to explicitly determine v+0 , it is sometimes useful. To determine v+0 ,consider the transmission line shown below.

Z g

Z 0 ,v p

z=0z=L

in

Z L+

gVZ

We have

v (z) = v+0 e�j� z �1 + �L ej2� z� ;

v (�L) = v+0 ej� L

�1 + �L e

�j2� L� ; (2.59)

but

v (�L) = vin =Zin

Zin + ZgVg (2.60)

Equating (2.59) and (2.60) leads to

v+0 = VgZin

Zin + Zg

1

[ej�L + �Le�j�L]: (2.61)

Example 2.2 If Vg (t) = 2 cos�108t+ �=4

�, Zg = 1 ohm, Z0 = 50 ohms, L = 3 m, ZL = 75 ohms, and

vp = 2:8� 108 m/s, determine v+0 .Solution:

� =2�

�=!

vp= 0:357; �L = 1:07; (2.62)

�L =ZL � Z0ZL + Z0

=75� 5075 + 50

=1

5


Z0 + jZL tan (�L)(2.63)

= 5075 + j50 tan (1:07)

50 + j75 tan (1:07)

= 38:23� j13:42 ohms,

v+0 = VgZin

Zin + Zg

1

[ej�L + �Le�j�L](2.64)

with Vg = 2 ej�4 leads to

v+0 = 2:14� j0:23: (2.65)

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Example 2.3 You �nd cable of unknown characteristic impedance having length 1:5 m. At a frequency atwhich you assume the cable is less than a quarter wavelength, you measure the open- and short-circuit inputimpedance to be

Zocin = �j54:6 ohms,Zscin = j103 ohms.

What is Z0?Solution: Assume the cable is very low loss.

ZocinZscin = (�jZ0 cot�L) (jZ0 tan�L) = Z20 (2.66)

= (�j54:6) (j103) = 5623:8) Z0 =

p5623:8 = 75 ohms.

Notice that you can also �nd � as

ZocinZscin

=�jZ0 cot�LjZ0 tan�L

= � (cot�L)2 (2.67)

=�j54:6j103

= �0:53

) cot�L = 0:728

) �L = 0:941; � = 0:628 1/m.

2.4 The Smith Chart

The Smith chart is a graphical tool that serves as an aid in performing transmission line calculations. Al-though any such calculation can be easily performed on a computer, the Smith chart helps in developingintuition, and serves as a useful background upon which to plot microwave engineering data (from measure-ments 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 (2.68)

=1 + �

1� � =1 + (�r + j�i)

1� (�r + j�i)

=

�1� �2r � �2i

�+ 2j�i

(1� �2r) + �2i:

Equating real and imaginary components,

r =1�

��2r + �

2i

�(1� �2r) + �2i

; x =2j�i

(1� �2r) + �2i; (2.69)

which can be rearranged to yield ��r �

r

1 + r

�2+ �2i =

1

(1 + r)2 ; (2.70)

(�r � 1)2 +��i �

1

x

�2=

1

x2: (2.71)

Recall that this is simply

Z = Z01 + �

1� � (2.72)

separated into real and imaginary components. Both (2.70) and (2.71) are circles in the complex ��plane.

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2.4. THE SMITH CHART 31

i. The �rst equation, (2.70), shows that the locus of constant resistance r in the �r ��i plane is a family ofcircles with centers on the �r axis at �r = r

1+r ;�i = 0 with radii11+r .

rΓ

Γi

r=1

r=0.5

r=0

1

ii. The second equation, (2.71), shows that the locus of constant reactance x in the �r��i plane is a familyof circles with centers on the �r axis at �r = 1;�i = 1

x with radii1x :

r

Γi

r=1

Γ

r=0

|Γ|<=1

X=1

X=1

X=2

X=2

. (1,0.5)

. (1,0.5)

Superimposing the previous two plots, and showing only the j�j � 1 portion, leads to

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r

Γ

Γ

i

X=2

r=0

|Γ|=1

X=0.7

X=0.7

X=2

Continuing for the other values0 � 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 valuesinstead of �r and �i values. Re�ection coe¢ cient values, along with other relevant information, is obtainedfrom the scales provided at the bottom of the chart.

iii. Since � (z) = �Lej2�z and �L = ZL�Z0ZL+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 equivalentto rotating on a constant j�j = � circle (constant SWR circle) with increasing angle � = 2�z in thecounter-clockwise direction. Moving towards the source results in a clockwise rotation.

rΓ

.

Γi

.

� Since 2�L+� = 2 2�� L+� = 4�L� +�, a full rotation about the Smith chart corresponds to a distance

of L = �=2.

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iv. Since

SWR =1 + j�j1� j�j =

1 + �

1� � (2.75)

then a circle of constant � is also a circle of constant SWR.

v. Since

Z =1 + �

1� � =1 + �ej(2�z+�)

1� �ej(2�z+�) ; (2.76)

then

Y =1

Z=1� �ej(2�z+�)1 + �ej(2�z+�)

(2.77)

=1 + �ej(2�z+�+�)

1� �ej(2�z+�+�) :

Therefore, for a given point Z on the Smith chart, the corresponding admittance Y is obtained byrotating � radians (one-half rotation) along a constant SWR circle. Thus,

� impedance (admittance) and be converted to admittance (impedance) by re�ecting the impedancepoint through the origin.

r

Z

Γ

Y

Γi

.

.

Example 2.4 Assume L = 0:35�, ZL = 50� j200 ohms, Z0 = 50 ohms. Find �L, Zin, �in, and SWR.Solution:

1. Start on the Smith chart at Z = 1� j4.

2. From the SWR legend (or constant SWR circle intersection with the positive-real axis), SWR = 20(SWR = 17:94 from the equation).

3. From the � legend and Smith chart angle, j�j = 0:9, 6 �L = �27� (� = 0:896 � 26:57� from theequation).

4. To �nd Zin, rotate 0:35� toward the generator (clockwise) around a constant SWR 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�.

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� Note: all possible values of Z (z) are found on the constant SWR circle. Thus, it is very easy tovisualize how Z varies along the line.

Example 2.5 A transmission line is terminated in ZL. Measurements show that the standing wave minimaare 102 cm. apart, and that the last minimum is 35 cm. from the load. The measured SWR is 2:4, Z0 = 250ohms, and vp = c. Determine (a) frequency and (b) ZL.Solution:

(a)

�

2= 102) � = 204cm.,

f =vp�=3� 1010204

= 147:06MHz.

(b) vmin is l = 35 cm from the load. l=� = 0:172. On Smith chart, draw SWR circle. Start at vmin androtate ccw 0:172�. Zin = Z0 (1:16� j0:96) = 290� j240 ohms.

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Example 2.6 A lossless 100 ohm transmission line is terminated in 200 + j200. The line is 0:375� long.Determine (a) �L, (b) SWR, (c) Zin, and (d) the shortest length of line for which the impedance is purelyresistive, and the value of the resistance.Solution:

(a) ZL = 2 + j2. �L = 0:626 30� = 0:62ej�6 .

(b) SWR = 4:3

(c) Zin = Z0 (0:32 + j0:54) = 32 + j54

(d) l = 0:042�, R = Z04:3 = 430 ohms.JNTU W

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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:

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Z 0 Z 0

inZZ L

0.06λ 0.12λ

C A

XL=30

B

Xc=200

Example 2.7 XL = 30 (Zind = j!L = jXL), XC = �200 (Zcap = 1j!C = �j

1!C = jXC). Determine Zin.

Solution:

1. First obtain ZA in the usual way�enter Smith chart at ZL = 2 + j1:5. Rotate 0:12� toward thegenerator. ZA = 1� j1:3.

2. Incorporate inductor L1:

(a) Option a: ZB = ZA +j3050 = 1� j0:7:

(b) Option b: Note that adding reactance just means moving along a constant resistance contour tothe desired point. The distance moved is j30

50 = j0:6.

3. Convert ZB to Y B. Y B = 0:67 + j0:47.

4. Incorporate capacitor C1:

(a) Option a: Y C = Y B + 1�j20050

= 0:67 + j0:72.

(b) Option b: Note that adding susceptance just means moving along a constant conductance contourto the desired point. The distance moved is 1

�j20050

= j0:25.

5. Either rotate 0:06� toward generator to get Yin and convert to Zin, or convert to ZC �rst then rotate0:06� to get Zin. Zin = Z0 (0:44� j0:39) = 22:4� j18:97 ohms.

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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

λ0.15

200+j100 ohms

2.5 The Quarter-Wave Transformer

The �=4 transformer is used to match two di¤erent real impedance values as shown below.

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inZZ 0 R LZ 1

λ/4

Zin = Z1RL + jZ1 tan (�L)

Z1 + jRL tan (�L)(2.78)

but �L = �=2,

Zin =Z21RL: (2.79)

Assume we want no re�ection at the input (�in = 0, Zin = Z0). Choose

Z1 =pRLZ0. (2.80)

� To connect a real-valued load RL with a transmission line having Z0 6= RL, use a �=4 length sectionof transmission line having characteristic impedance Z1 =

pRLZ0.

� 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 broadenthe 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 tothe load.

Z g

Z 0 ,v p

z=0z=L

in

Z L+

gVZ

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2.7. LOSSY TRANSMISSION LINES 41

PL =1

2Re (vini

�in) , since the line is lossless. (2.81)

vin =Zin

Zin + ZgVg; iin =

VgZin + Zg

=vinZin

;

noting that ZZ� = jZj2 we get

PL =1

2jVgj2

�� ZinZin + Zg

��2Re� 1

Z�in

�(2.82)

=1

2jVgj2

Rin

(Rin +Rg)2+ (Xin +Xg)

2 :

1. Special case: load matched to line:

ZL = Z0, �L = 0, SWR = 1, Zin = Z0:

PL =1

2jVgj2

Z0

(Z0 +Rg)2+X2

g

: (2.83)

Further, if Zg = Z0,

PL =1

8jVgj2

1

Z0: (2.84)

2. Special case: generator matched to a loaded line:

Zin = Zg, �in = 0.

PL =1

2jVgj2

Rg

4�R2g +X

2g

� : (2.85)

Depending on conditions, either case may provide the larger power delivered to the load. What determinesmaximum power transfer from generator to load? As in circuit theory, conjugate matching.

Zin = Z�g (2.86)

This will result in maximum power transfer to the load for a �xed Zg. The resulting power delivered to theload is

PL =1

2jVgj2

1

4Rg: (2.87)

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) = v+0 e� z + v�0 e

+ z; (2.88)

i(z) =1

Z0

�v+0 e

� z � v�0 e+ z�;

= �+ j� =p(R+ j!L) (G+ j!C); (2.89)

Z0 �

sR+ j!L

G+ j!C:

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For lossy linese� z = e��z| {z }

attenuation

e�j�z: (2.90)

In this case, movements along the line no longer are represented by rotation along a constant SWR circle(for lossy lines j�j = � 6= constant). For the low-loss case we have

� ' 1

2

�R

Z0+GZ0

�; (2.91)

� ' !pLC; Z0 '

rL

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 frequencycomponents of a signal (as in a Fourier decomposition) will travel at di¤erent velocities, reaching the loadat 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

L=G

C(2.93)

then

� = !pLC; � = R

rC

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 =

��v+0 ��22Z0

n1� j�inj2

oe2�L; (2.95)

PL =

��v+0 ��22Z0

n1� j�Lj2

o;

PLoss = Pin � PL;

where

v+0 = VgZin

Zin + Zg

1

[e L + �Le� 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= �R i(z; t)� L@i(z; t)

@t; (2.97)

@i(z; t)

@z= �Gv(z; t)� C @v(z; t)

@t; (2.98)

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2.8. TRANSIENT TRANSMISSION LINES 43

where we have set the distributed sources vs; is = 0. These �rst-order, coupled partial di¤erential equationsare 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 @=@zto (2.98) and substituting into (2.97), resulting in�

@2

@z2� LC @

2

@t2� (RC +GL) @

@t�RG

��v (z; t)i (z; t)

�= 0: (2.99)

For simplicity we will assume a lossless line (R = G = 0), resulting in�@2

@z2� 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

@z2� LC @

2

@t2

�v (z; t) = 0 (2.101)

we use Laplace transforms. Recall

Lfv (z; t)g = V (z; s) ; (2.102)

L�@

@tv (z; t)

�= sV (z; s)� v

�z; t = 0+

�;

L�@2

@t2v (z; t)

�= s2V (z; s)� sv

�z; t = 0+

�� @

@tv (z; 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,

@

@tv (z; t)

��t=0+

= � 1C

@

@zi (z; t)

��t=0+

= 0: (2.103)

Taking the Laplace transform of the wave equation (2.101) results in�@2

@z2� 2

�V (z; s) = 0 (2.104)

where = spLC. The solution of (2.104) is

V (z; s) = V + (s) e� z + V � (s) e+ z (2.105)

= V + (s) e�spLCz + V � (s) e+s

pLCz

where V � (s) are constants of integration with respect to z. The solution v (z; t) is obtained via inverseLaplace transform,

v (z; t) = L�1nV + (s) e�s

pLCz + V � (s) e+s

pLCz

o; (2.106)

and, using the time-shifting theorem

Lff (t� t0)g = F (s) e�st0 (2.107)

we obtainv (z; t) = v+

�t�

pLCz

�+ v�

�t+

pLCz

�: (2.108)

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We can interpret the termpLCz 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�

pLCz

�= constant, (2.109)

) t�pLCz = constant,

) d

dt

�t�

pLCz

�= 0;

1�pLC

dz

dt= 0; (2.110)

dz

dt= � 1p

LC= �vp;

) vp =1pLC

.

Therefore,v (z; t) = v+ (t� z=vp)| {z }

forward travelling wave

+ v� (t+ z=vp)| {z } :backward travelling wave

(2.111)

The associated current can be obtained as

i (z; t) =1

Z0

�v+ (t� z=vp)� v� (t+ z=vp)

�(2.112)

as shown in the appendix.In summary, we have

v (z; t) =�v+ (t� z=vp) + v� (t+ z=vp)

�; (2.113)

i (z; t) =1

Z0

�v+ (t� z=vp)� v� (t+ z=vp)

�;

valid for an in�nite line. To analyze a �nite length, source-excited and loaded line we need to considerconditions at the source-end and load-end of the line.

Terminated transient line:

+

+

Z 0 ,v p v(L,t)LRv(z,t)

i(z,t) i(L,t)

With

v (L; t) = v+ (L; t) + v� (L; t) (2.114)

= v+L (t) + v�L (t) = vL (t)

i (L; t) = i+ (L; t) + i� (L; t)

= i+L (t) + i�L (t) = iL (t)

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and Ohm�s law applied at the load,vL (t) = iL (t)RL; (2.115)

we obtain

v+L (t) + v�L (t) =

�i+L (t) + i

�L (t)

�RL (2.116)

=1

Z0

�v+L (t)� v

�L (t)

�RL

leading to

v�L =RL � Z0RL + Z0

v+L = �Lv+L (2.117)

(compare with (2.26) for the phasor case).

Launching waves on an in�nite line:

g

+

i(0,t)

Z 0 ,v p

i(z,t)

+gV

R

v(0,t) v(z,t)

Ohm�s law at z = 0 results in

v (0; t)| {z }vs(t)

= vS (t) = Vg (t)�Rgi (0; t)| {z }is(t)

(2.118)

for all t. In particular,

vs�0; t = 0+

�= Vg

�t = 0+

��Rgis

�t = 0+

�(2.119)

= Vg�t = 0+

��Rg

vs (t = 0+)

Z0;

leading to

vs�0+�=

Z0Z0 +Rg

Vg�0+�= v+

�0; 0+

�(2.120)

which gives the initial amplitude of the wave launched on the line. For z; t 6= 0 we obtain

v (z; t) =Z0

Z0 +RgVg (t� z=vp) = v+ (t� z=vp) ; (2.121)

i (z; t) =v (z; t)

Z0= i+ (t� z=vp) :

Complete transient response of a terminated, source-driven transmission line: bounce diagrams:

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g

v+

v+ΓL

ΓL g v+

ΓL g v+Γ2

Γ

0,0

t

z

T

2T

3T

4T

z=L

ΓL g

Z 0 ,v p

i(z,t)

L+

gV

z=0 z=L

R

R

Γ

0,0

t

z

T

2T

3T

4T

z=L+

+ΓL

+

Γ2

+ΓL g

i

i

i

i

v(z,t)

T =L

vp= one-way transit time for wave transverse line.

v (z; t) =Z0

Z0 +RgfVg (t� z=vp) + �LVg (t� 2T + z=vp) + �L�gVg (t� 2T � z=vp)

+�2L�gVg (t� 4T + z=vp) + �2L�2gVg (t� 4T � z=vp) + :::; (2.122)

i (z; t) =1

Z0 +RgfVg (t� z=vp)� �LVg (t� 2T + z=vp) + �L�gVg (t� 2T � z=vp)

��2L�gVg (t� 4T + z=vp) + �2L�2gVg (t� 4T � 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:

Z 0 ,v p R L+

gV

z=0 z=L

�L = 0;�g =Rg � Z0Rg + Z0

= �1 (2.124)

(�g is not needed since line is matched �no re�ection comes back towards the source).

v (z; t) =Z0

Z0 + 0Vg (t� z=vp) = Vg (t� z=vp) ; :

i (z; t) = Vg (t� z=vp) =Z0

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t

Vg(t)/Vo

To

1

leading edge

t

V(L,t)/Vo

1

T+ToT=L/Vp

trailing edge

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:

g

Z 0 ,v p

z=L

L+

gV

R

R

z=0

Vg (t) = V0u (t) ;

�L =RL � Z0RL + Z0

=3Z0 � Z03Z0 + Z0

=1

2= �s;

v+ =Z0

Z0 +RsVg�t = 0+

�=V04:

Then,

v (z; t) =V04

�u (t� z=vp) +

1

2u (t� 2T + z=vp) +

1

4u (t� 2T � z=vp)

+1

8u (t� 4T + z=vp) +

1

16u (t� 4T � z=vp) + :::

�; (2.125)

v (0; t) =V04

�u (t) +

1

2u (t� 2T ) + 1

4u (t� 2T )

+1

8u (t� 4T ) + 1

16u (t� 4T ) + :::

�(2.126)

=V04

�u (t) +

3

4u (t� 2T ) + 3

16u (t� 4T ) + :::

�(2.127)

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v (L; t) =V04

�u (t� T ) + 1

2u (t� 2T + T ) + 1

4u (t� 2T � T )

+1

8u (t� 4T + T ) + 1

16u (t� 4T � T ) + :::

�(2.128)

=V04

�3

2u (t� T ) + 3

8u (t� 3T ) + :::

�(2.129)

V(0,t)/(Vo/4)

1

t/T

1 2 3 4 5 6

3/4

3/16

3/8

1

t/T

1 2 3 4 5 6

V(L,t)/(Vo/4)

3/2

As a check, d.c. analysis yields

v (t) =3Z0

3Z0 + 3Z0V0 =

V02

(2.130)

and

v (L; t =1) = V04

(1 +

�1

2

�1+

�1

2

�2+

�1

2

�3+ :::

)=V02

(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 thevoltage at the middle of the line.Solution:

g

Z 0 ,v p

z=L

L+

gV

R

R

z=0

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t

Vg(t)/Vo

To

1

�L =2Z0 � Z02Z0 + Z0

=1

3; (2.132)

�g = 0

v (z; t) =Z0

Z0 + Z0fVg (t� z=vp) + �LVg (t� 2T + z=vp)g (2.133)

=1

2

�Vg (t� z=vp) +

1

3Vg (t� 2T + z=vp)

�(2.134)

v (L=2; t) =1

2

�Vg

�t� T

2

�+1

3Vg (t� 2T + T=2)

�(2.135)

=1

2

�Vg

�t� T

2

�+1

3Vg

�t� 3T

2

��: (2.136)

I. T0 < T=2

1

To<T/2

1 2 3 4 5 6

t/(T/2)

V(L/2,t)/(1/2)

1/3

(T) (3T/2)(T/2)

II. T0 = T=2

1

To=T/2

1 2 3 4 5 6

t/(T/2)

V(L/2,t)/(1/2)

1/3

(T) (3T/2)(T/2)

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III. T0 = T

1

1 2 3 4 5 6

t/(T/2)

V(L/2,t)/(1/2)

1/3

(T) (3T/2)(T/2)

To=T

IV. T0 = 2T

1

To=2T

1 2 3 4 5 6

t/(T/2)

V(L/2,t)/(1/2)

1/3

(T) (3T/2)(T/2)

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 lengthl = 1 cm, and the velocity of signal propagation on the line is vp = 2:6 � 108 m/s. Pin 3 can be modeledas a voltage source which provides a 1:5 volt pulse starting at t = 0 and lasting for 1 ns, and having sourceresistance Rg = 10 ohms. Pin 5 can be modeled as providing a constant 30 ohm resistance. Plot the voltagewaveform at pin 5 versus time.

Example 2.12 Consider the charged-line pulse generator shown below. If Rg = Z0, RL � Z0 determinethe voltage at z = 0.Solution:

LR

Z 0

Z 0

Z 0=

i(L,t)

z=0 z=L

V0R g

t=0

+

v(0,t)

i(0,t)

+

+

v(L,t)

>>

,v p

Operation: resistor Rg is switched to close the circuit and interrupt the d.c. state. Output voltage v (0; t) isa 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

�g =Rg � Z0Rg + Z0

= 0; (2.137)

�L =RL � Z0RL + Z0

' 1:

v�0; t = 0+

�= V0 + v

+ = �Rgi�0; t = 0+

�(2.138)

= �Rgv+

Z0' �v+.

) 2v+ = �V0 (initial pulse amplitude).

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v (z; t) = V0|{z}d.c. voltage

+ v+ (t� z=vp)| {z }forward wave (re�ected once at RL)

+ v� (t� 2T + z=vp)| {z }backward wave

(2.139)

v (z; t) = V0 �V02u (t� z=vp)�

V02u (t� 2T + z=vp) (2.140)

v (0; t) = V0

�1� 1

2u (t)� 1

2u (t� 2T )

�=

V02fu (t)� u (t� 2T )g :

t

1

2T

v(0,t)/(Vo/2)

2.8.1 Waveforms and Spectral Analysis

The text (Pozar) deals with time-harmonic signals and associated analysis. However, a time-harmonicwaveform can not be used to transmit information. In the analog domain, modulation is commonly usedto impart information on a high-frequency carrier, which spreads the frequency content out from the singlespectral component of the carrier. Time-harmonic analysis will be valid if the bandwidth of the modulatedsignal is su¢ ciently small. In the digital domain, one must consider pulses and pulse trains. In this sectionwe 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,

g (t) =a02+

1Xn=1

�an cos

n2�t

T+ bn sin

n2�t

T

�where

an =2

T

Z T=2

�T=2g (t) cos

n2�t

Tdt;

bn =2

T

Z T=2

�T=2g (t) sin

n2�t

Tdt;

for n = 0; 1; 2; :::, or in exponential form

g (t) =1X

n=�1cne

j n2�tT

where

cn =1

T

Z T=2

�T=2g (t) e�j

n2�tT dt;

n = 0;�1;�2; ::: (the equality is in the sense of the Fourier series). For example, for a pulse train consistingof rectangular pulses having amplitude A, pulse width t0, and period T , as shown below,

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A

g(t)

...0

t

T

t

we have

g (t) =1X

n=�1cne

j n2�tT ;

cn =1

T

Z T=2

�T=2g (t) e�j

n2�tT dt

=1

T

Z t0

0

Ae�jn2�tT dt

= At0Te�jn�

t0T

�sin n�t0T

�n�t0T

(for this to be valid t0 < T=2, otherwise the limits of integration need to be adjusted). The quantity f0 = 1=Tis called the fundamental frequency (or pulse repetition rate), in which case

cn = At0f0e�jn�t0f0 (sinn�t0f0)

n�t0f0:

A plot of jcnj vs. n=T looks like the following.

0t2T

3

nC

n/T

c 0

1T 1

t 0

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 spectrallines 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, isgoverned by the pulse width, t0.

Non-Periodic Signals

If we consider a single pulse, we need to use Fourier transforms,

g (t) =1

2�

Z 1

�1G (!) ej!td!;

! = 2�f , where

G (!) =

Z 1

�1g (t) e�j!tdt:

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For a single rectangular pulse,

G (f) = At0sin (�t0f)

�t0f;

which has zeros atf =

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,

jG (f)j = At0

�� sin (�t0f)�t0f

�� ;jcnj = At0f0

�� (sinn�t0f0)n�t0f0

�� ;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 analysisstandpoint. It does a¤ect the density of the discrete spectral components, but not how fast the spectrumdecreases with increasing frequency.

� From a spectral analysis standpoint, the pulse width is of paramount importance, for both the pulseand 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, whereasfor 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.

Q1

Q2

Vcc

Gnd

Vout

VA

VB

A

B

When Q1 is on and Q2 is o¤ the output voltage is high (Vout = VA = Vcc), and when Q1 is o¤ andQ2 is on the output voltage is low (Vout = VB = 0). However, VA is not actually Vcc volts, and VB is notactually 0 volts, due to parasitic inductance. Both leads A and B are wires, and, like all wires, they haveinductance. The inductance of a very short (electrically short) straight wire is very small, although perhapsnot negligible depending on the application, leading to the model shown below.

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Q1

Lwire

Q2

Lwire

Vcc

Vout

Gnd

Since

vind = Linddiind (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) orVout (t) = 0 + vind (t). This phenomena is called ground bounce (and Vcc bounce). If the noise margins ofthe circuit downstream are too tight, or the timing is too fast (all voltages approach their idealized valueswhen the currents stop changing) a logic error may follow. To remedy this situation we may 1) slow downthe circuit (not a viable choice), reduce the margins (often not a viable choice), or minimize Lind, which isusually the preferred method, if possible.

The above described ground bounce occurs internal to a chip. Integrated circuits (ICs) have been tra-ditionally 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 throughthese wirebonds are typically electrically short, they have, like all wires, inductance (on the order of 2 nHfor a typically wirebond). Therefore, any IC that has large transient currents can induce large, undesirablevoltage 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 ofwhich are switching at the same time, and all connected to ground through the same wirebond, the switchingcurrent can become quite large, resulting in ground bounce of several volts. To make matters worse, theinductance of the ground and power planes is added to the inductance of the wirebonds, enhancing thisdeleterious e¤ect. Flip-chip technologies (which spread the ground connections over many solder bumps) aree¤ective at mitigating this e¤ect.

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Q1

Q2

Vcc

Vout

Gnd

Lwire+Lpower plane

Lwire+Lground plane

Although ground bounce is not a �transmission line� e¤ect, it is particular to high frequencies, andrelates to the electromagnetic properties of the circuit. Bypass capacitors are often used to remove theground and power plane transients from the IC circuits. Since current can change �instantaneously�througha 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 andlarger capacitance).

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Chapter 3

Transmission Lines and Waveguides

In the previous chapter we studied transmission lines in an abstract sense, usually without referring to aparticular physical structure. In this chapter we will study several types of transmission lines and waveguidescommonly used in practice. As is evident from their physical structure, each transmission line and waveguidehas particular advantages and disadvantages depending on the application of interest. Any waveguide ortransmission line that is capable of supporting a TEM mode of propagation can be analyzed using thetransmission line theory discussed in the previous chapter (for that mode). Waveguides and transmissionlines that do not support a TEM mode can be approximately analyzed using the previously developedanalysis 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 �transmissionlines,�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 thez�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

f(x; y; �) = F ff(x; y; z)g =Z 1

�1f(x; y; z)e�j�zdz; (3.1)

f(x; y; z) = F�1 ff(x; y; �)g = 1

2�

Z 1

�1f(x; y; �)ej�zd�: (3.2)

that �elds on a z�invariant structure can be written as

E (x; y; z) = [e (x; y; �) + z ez (x; y; �)] e�j� z (3.3)

H (x; y; z) = [h (x; y; �) + zhz (x; y; �)] e�j� z

where fe;hg represent the (as yet unknown) transverse �elds, fez; hzg represent the (as yet unknown)longitudinal �elds, and � is the (as yet unknown) propagation constant. The analysis in the text shows howto obtain these unknown quantities for a given physical structure. Here we will mostly be concerned withobtaining �.Starting with the source-free curl equations

r�E = �j!�H; (3.4)

r�H = j!"E

57

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58 CHAPTER 3. TRANSMISSION LINES AND WAVEGUIDES

and assuming the form (3.3) leads to

@Ez@y

+ j�Ey = �j!�Hx; (3.5)

�j�Ex �@Ez@x

= �j!�Hy; (3.6)

@Ey@x

� @Ex@y

= �j!�Hz; (3.7)

and

@Hz@y

+ j�Hy = j!"Ex; (3.8)

�j�Hx �@Hz@x

= j!"Ey; (3.9)

@Hy@x

� @Hx@y

= j!"Ez: (3.10)

The above equations may be solved for the transverse components Hx;Hy; Ex; Ey in terms of the longi-tudinal components Ez;Hz as

Hx =j

k2c

�!"@Ez@y

� � @Hz@x

�; (3.11)

Hy =�jk2c

�!"@Ez@x

+ �@Hz@y

�; (3.12)

Ex =�jk2c

��@Ez@x

+ !�@Hz@y

�; (3.13)

Ey =j

k2c

�� @Ez

@y+ !�

@Hz@x

�; (3.14)

where

k2c = k2 � �2 (3.15)

= !2�"� �2

=

�2�

�

�2� �2:

Often we write the above as� =

pk2 � k2c ; (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 wavecan exist. The solution of this paradox is that for a TEM wave it must be true that

kc = 0; (3.17)

) � = � k = �!p�" = �2��

such that TEM waves act like the waves we have studied previously (in Chapter 2 we had � = !pLC =

2�� ). In the text it is shown that TEM �elds are the same as the static �elds (zero frequency) thatexist on the transmission line, and satisfy Laplace�s equation.

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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 doesnot guarantee that a TEM wave will exist; microstrip is an example of a two-conductor line that doesnot 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

@x2+@2

@y2+@2

@z2+ k2

�Hz = 0, or (3.18)�

@2

@x2+@2

@y2� �2 + k2

�Hz = 0;

)�@2

@x2+@2

@y2+ k2c

�Hz = 0;

which is solved subject to the appropriate boundary conditions for a given physical structure, althoughwill skip this computation. The result of that analysis, through, leads to the determination of kc, andusing

� =pk2 � k2c ; (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

@x2+@2

@y2+ k2c

�Ez = 0: (3.20)

As with TE waves, kc is determined in the course of solving (3.20). leading to �.

In summary, we have

TEM waves Ez = 0 Hz = 0 � = � kTE waves Ez = 0 Hz 6= 0 � =

pk2 � k2c

TM waves Ez 6= 0 Hz = 0 � =pk2 � k2c

.

3.2 Parallel Plate Waveguide

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.

z

ε,µ

y

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1. TEM waves

The parallel plate structure supports a TEM mode, which is the dominant mode (i.e., the mode thatexists at the lowest frequencies of operation). The TEM mode is the most important mode from apractical standpoint. The analysis in the text shows that for the TEM mode,

E (x; y; z) = �yV0de�j�z; (� = k) ; (3.21)

H (x; y; z) = xV0

ZTEM de�j�z; ZTEM = � =

r�

";

and that

Z0 =�d

w; � =

r�

"(3.22)

independent of frequency.

1β

k

2. TM modes

Parallel plate waveguides also support TE and TM modes. For TM modes,

Ex = Hy = Hz = 0; (3.23)

Ey =�j�kc

An cos�n�dy�e�j�nz;

Ez = An sin�n�dy�e�j�nz;

Hx =j!"

kcAn cos

�n�dy�e�j�nz;

where

�n =

rk2 �

�n�d

�2; n = 0; 1; 2; :::;

�kc =

n�

d

�. (3.24)

For n = 0 the TM0 mode is really the TEM mode. For n > 0 two situations occur:

(a) If k > n�d then �n is real-valued and, therefore, e

�j�nz is a purely propagating mode.

(b) If k < n�d then �n = �j�n is imaginary-valued and, therefore, e�j�nz = e�j(�j�n)z = e��nz is a

purely attenuating mode.

Thus, two regimes exist: k > kc = n�d (propagation), and k < kc = n�

d (attenuation).

Note that this is fundamentally di¤erent that a wave propagating through a lossy medium, where weobtain

e�j�ze��z; (3.25)

i.e., propagation and attenuation occurring at the same time. For the TMn mode considered here, thewave either propagates (at higher frequencies, where k = !

p�" > n�

d ) or does not (at lower frequencies,where k = !

p�" < n�

d ). If the wave is purely attenuating, it is called evanescent, or cuto¤ .

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The frequency at which a cuto¤ (non-propagating) mode begins to propagate is called the cuto¤frequency, and is determined by

kc =n�

d) fc =

n

2dp�". (3.26)

βReal

βImag.

k

TEM

TM1 TM2

k k k

β

1 2

As frequency increases, more modes begin to propagate. This is usually undesirable. Why?

Note thatvp =

!

�(3.27)

but that the speed of light isc =

!

k: (3.28)

Therefore, since � � k, vp � c!What does vp � c imply?

Consider the TM1 mode.

�1 =

rk2 �

��d

�2; (3.29)

Ez = A1 sin��dy�e�j�1z

= A1

2j

8<: ej(�y=d��1z)| {z }plane wave in -y,+z

� e�j(�y=d+�1z)| {z }plane wave in +y,+z

9=; :JN

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y,+z wave

y

zθ

d+y,+z wave

With

ky =�

d= k sin �; (3.30)

kz = �1 = k cos �

(�21 + (�=d)2= k2) the phase velocity of each plane wave is

vPWp =!

k= c: (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 thewave along one direction (the z� direction) when the wave is propagating along another direction (the�� direction).

θ

constant phase wavefronts

d

L

With L cos � = d, and observer at d sees the wave arrive in T seconds. The observer calculates the wavevelocity 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. Thewave �break� travels up the coast faster than the wave is moving. We�ll see later that the energyvelocity is always � c.

3. TE waves

For TEn modes,

Ey = Hx = 0; (3.33)

Ex =j!�

kcAn sin


Hy =j�

kcAn sin


Hz = An cos�n�dy�e�j�nz;

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where

�n =

rk2 �

�n�d

�2; n = 1; 2; 3; :::;

�kc =

n�

d

�. (3.34)

For n � 1 two situations occur, as before:

(a) If k > n�d then �n is real-valued and, therefore, e

�j�nz is a purely propagating mode.

(b) If k < n�d then �n = �j�n is imaginary-valued and, therefore, e�j�nz = e�j(�j�n)z = e��nz is a

purely attenuating mode.

Thus, two regimes exist: k > kc = n�d (propagation), and k < kc = n�

d (attenuation).

Note that for both the TMn and TEn modes,

�n =

rk2 �

�n�d

�2; (3.35)

fc =n

2dp�":

� In summary, for the parallel plate waveguide, TEM, TMn, and TEn modes exist.

� For the TEM mode, � = k = !p�", and no cuto¤ exists (the TEM mode propagates down to zero

frequency).

� For the TMn and TEn modes, �n =qk2 �

�n�d

�2(n = 0; 1; 2; ::: for TM modes, and n = 1; 2; 3; ::: for

TE modes). Two regimes exist, below cuto¤ (k < n�d , � pure imaginary), and above cuto¤ (k >

n�d ,

� pure real). The dividing point is k = kc = n�d , leading to fc =

n2dp�" .

� An in�nite number of modes exist, although at any �nite frequency a �nite number of modes arepropagating.

Are parallel plate waveguides used in practice? Sometimes, but more often an unintentional parallelplate waveguide is formed by a circuit designer�s attempt to shield a circuit from interference, or to eliminateundesired 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.

ε

coupling via surface wave

interference signal undesired emission

IC chips

initial circuit layout

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shielded circuit layout

ε

IC chips

undesired emission

coupling via surface wave

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 somelossy dielectric material to absorb the emissions (however, energy loss from the circuit still occurs, perhapsleading to circuit malfunction).

3.3 Rectangular Waveguide

A rectangular waveguide is a hollow (or dielectric �lled) metal pipe with rectangular cross-section. It canpropagate TE and TM modes, but not TEM modes.

y

x

z

b

a

ε,µ

The table below provides a summary of results for rectangular waveguide.

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3.4. CIRCULAR WAVEGUIDE 65

TEmn mode TMmn mode

k !p�" !

p�"

kc;mn

q(m�=a)

2+ (n�=b)

2q(m�=a)

2+ (n�=b)

2

�mn

qk2 � k2c;mn

qk2 � k2c;mn

�c;mn 2�=kc;mn 2�=kc;mn�g;mn 2�=�mn 2�=�mnvp;mn !=�mn !=�mnEz 0 Bmn sin (m�x=a) sin (n�y=b) e

�j�mnz

Hz Amn cos (m�x=a) cos (n�y=b) e�j�mnz 0

Exj!�n�k2c;mnb

Amn cos (m�x=a) sin (n�y=b) e�j�mnz �j�m�

k2c;mnaBmn cos (m�x=a) sin (n�y=b) e

�j�mnz

Ey�j!�m�k2c;mna

Amn sin (m�x=a) cos (n�y=b) e�j�mnz �j�n�

k2c;mnbBmn sin (m�x=a) cos (n�y=b) e

�j�mnz

Hxj�m�k2c;mna

Amn sin (m�x=a) cos (n�y=b) e�j�mnz j!"n�

k2c;mnbBmn sin (m�x=a) cos (n�y=b) e

�j�mnz

Hyj�n�k2c;mnb

Amn cos (m�x=a) sin (n�y=b) e�j�mnz �j!"m�

k2c;mnaBmn cos (m�x=a) sin (n�y=b) e

�j�mnz

Z ZTE = k�=�mn ZTM = �mn�=k

TE modes:

�mn =

rk2 �

�m�a

�2��n�b

�2; m; n = 0; 1; 2; :::; (3.36)

fc;mn =1

2�p�"

r�m�a

�2+�n�b

�2:

The �rst (dominant) mode is the TE10 mode, assuming a � b.TM modes:

�mn =

rk2 �

�m�a

�2��n�b

�2; m; n = 1; 2; 3; :::; (3.37)

fc;mn =1

2�p�"

r�m�a

�2+�n�b

�2:

The �rst mode to propagate is the TM11 mode.

3.4 Circular Waveguide

Omit

3.5 Coaxial Line

TEM modes:

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V=V0

a

bε

V=0

Z0 =�

2�ln

�b

a

�. (3.38)

E = �V0

� ln (b=a)e�j�z; � = k; (3.39)

H = �

r"

�

V0� ln (b=a)

e�j�z:

� The common value of 50 ohm for the characteristic impedance of many transmission lines comes fromsome attributes of coaxial lines.

�Power handling capacity: The maximum power capacity of a coaxial line is limited by voltagebreakdown, and is given by

Pmax =�a2E2d�

ln (b=a) (3.40)

where Ed is the �eld strength at breakdown. Solving for the value of b=a that maximizes Pmaxleads to a characteristic impedance of 30 ohms, as follows: Holding b �xed,

d

daPmax = �

��a2 1

a+ 2a ln

�b

a

��= 0 (3.41)

where � = �E2d=�. This results in ln (b=a) = 1=2. Then,

Z0 =1

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 toomuch loss).

�Attenuation: The attenuation of a coaxial cable due to �nite conductivity of the conductors is

�c =Rs

2� ln�ba

� �1a� 1b

�. (3.43)

The equation obtained by minimizing �c is

b

aln

�b

a

�= 1 +

b

a(3.44)

which has solutionb

a= 3:591: (3.45)

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3.6. SURFACE WAVES ON A GROUNDED DIELECTRIC SLAB 67

The corresponding characteristic impedance is

Z0 =1

2�

r�

"ln

�b

a

�' 60 ln (3:591) = 76:71 ohms. (3.46)

Therefore, 50 ohms represents a compromise between maximum power handling capability (30ohms) and minimum attenuation (77 ohms), (30 + 76:71) =2 = 53:35. 50 ohms is also typicallyrealized using reasonable material dimensions.

Coaxial lines also support higher-order modes (TE and TM modes) (the cuto¤ frequency for the �rsthigher-order modes can be obtained approximately as

fco 'c

�p"r (a+ b)

(3.47)

We will not consider them here, but it is important to note that above some frequency these modes willbegin to propagate. Multi-mode propagation is usually to be avoided, since more than one mode carriespower, 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 circuittechnology.

rε

x

zd

ε 0

surface wave

This waveguide can support TE and TM surface waves, that are bound to the vicinity of the surface (theydecay 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 lowermicrowave frequencies, and considering typical slab thickness values used in printed circuit technology, thesurface 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 circuitsprinted on the same substrate.

ε

coupling via surface wavecircuit 1 circuit 2

1. TM modes:

The TM modes have �eld components Ez; Ex; and Hy. The propagation constant � is determined bythe solution of q

erk20 � �2 tan

�qerk20 � �

2d

�= "r

q�2 � k20 (3.48)

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and the cuto¤ frequencies from

fc =nc

2dp"r � 1

; n = 0; 1; 2; ::: (3.49)

The n = 0 mode (TM0) propagates down to zero frequency (although exactly at ! = 0 the �eldsvanish). Thus, this mode is always present, and often leads to undesirable coupling. In general, thehigher 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 � is determined bythe solution of q

erk20 � �2 cot

�qerk20 � �

2d

�= �

q�2 � k20 (3.50)

and the cuto¤ frequencies from

fc =(2n� 1) c4dp"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. Forexample, consider a grounded slab of thickness d and permittivity "r"0, and an in�nite line source carryingcurrent

I = I0 cos (!t) ; (3.52)

as shown in the �gure below.

P

zd

ε 0

x

rεJN

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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

Ey =1

2�

Z 1

�1F (�) e�

p�2�k20xej�zd� (3.53)

where

F (�) =I0j!"0

1 +

p�2�k20�

p�2�k21 coth

�p�2�k21d

�p�2�k20+

p�2�k21 coth

�p�2�k21d

�2p�2 � k2

(3.54)

with k1 =p"rk0. It can be shown that the �eld can be decomposed into three parts:

1. E(1)y due to the direct radiation from the source (in the absence of the grounded slab),

2. E(2)y due to interaction of the source and the grounded slab (e.g., multiple partial re�ections), and

3. E(3y due to surface waves excited in the slab.

The origins of these �eld components are shown in the �gure below.

(2)

(1)

(3)E

E

PE

zd

x

0

r

ε

ε

The three �eld components are, for x = 0; kz � 1,

E(1)y ' E1e�jkjzjpjkzj

; E(2)y ' E2e�jkjzjqjkzj3

; E(3)y ' E3e�j�jzj (3.55)

where E1;2;3 are constants. Therefore, the surface wave is the dominant contribution to the �eld, and doesnot 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.

0 10 20 30 40 50 60 70

f (GHz)

1.0

1.1

1.2

1.3

1.4

1.5

β/k0

TE1

TE3TE5JN

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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 jEyj vs. frequency shows spikes corresponding to surface waveexcitation.

0 10 20 30 40f (GHz)

0

1

2

3

4

5

E y (a.

u.)

3.7 Stripline

Stripline is formed by two ground planes with a conducting strip between the planes.

conducting strip

w

bε,µ

ground planes

Stripline supports a TEM mode. The analysis of the TEM (and the higher-order TE and TM modes) isdi¢ cult; some results for the TEM modes are provided below.

� = k = !p�" since TEM (3.56)

Z0 ' 30�p"r

b

we + 0:441b; (3.57)

web

=w

b��0; w

b > 0:35;�0:35� w

b

�2; w

b < 0:35:

To obtain strip width w for a given Z0,

w

b=

�x;

perZ0 < 120;

0:85�p0:6� x; p

erZ0 > 120; (3.58)

x =30�p"rZ0

� 0:441:

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3.8. MICROSTRIP 71

3.8 Microstrip

Microstrip transmission line is one of the most common types of transmission line extensively used in printed-circuit technology (printed using photolithographic techniques). It consists of a conductor of width w printedon a grounded dielectric slab.

bε,µ

conducting strip

w

ground plane

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¢ cientlylow frequencies, a TEM mode, prompting a quasi-TEM analysis. The details are included in the text, withsome results summarized below.

� = k0p"e = !

p�0"0"e; (3.59)

vp =!

�=

cp"e;

�g =�0p"e;

where"e =

"r + 1

2+"r � 12

1q1 + 12b

w

; 1 < "e < "r: (3.60)

If given line dimensions we obtain Z0 as

Z0 =

(60p"eln�8bw +

w4b

�; w

b � 1120�p

"e[wd +1:393+0:667 ln(wd +1:444)]

; wb � 1

; (3.61)

and for a given characteristic impedance and dielectric constant, the required w=b ratio is

w

b=

(8eA

e2A�2 ;wb < 2;

2�

hB � 1� ln (2B � 1) + "r�1

2"r

�ln (B � 1) + 0:39� 0:61

"r

�i; w

b > 2;(3.62)

where

A =Z060

r"r + 1

2+"r � 1"r + 1

�0:23 +

0:11

"r

�; (3.63)

B =377�

2Z0p"r:

Example 3.1 Design a microstrip line on a "r = 2:25 substrate of thickness b = 0:2 cm to have a charac-teristic impedance of 50 ohms.Solution: From the above equations, w=b = 3:037, so w = 0:6074 cm.

Dispersion:

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The previous formulas for Z0 and "e are valid for low frequencies. The e¤ects of frequency can beapproximately accounted for using

"eff (f) =

�p"r �

p"e

1 + 4F�3=2+p"e

�2; (3.64)

Z0 (f) = Z0"eff (f)� 1"e � 1

r"e

"eff (f);

where

F =4bf

p"r � 1c

�1

2+h1 + 2 log10

�1 +

w

b

�i2�(3.65)

Plot of "eff (f) vs. f . "eff (0) = "e.

Losses:

Losses on microstrip consist of three components: conductor loss, dielectric loss, and radiation loss. Usu-ally conductor loss dominants over dielectric loss (at least for high-quality microwave substrates). Radiationloss include energy converted into surface waves, and energy radiated into space. Both e¤ects are mostsigni�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 'RswZ0

Np/m, (3.66)

where Rs =p!�0=2� is the surface resistivity of the conductor, and

�d =k0"r ("e � 1) tan �2p"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. Stillothers, 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�" = !

pLC; (3.68)

and Z0 =pL=C follows from the �eld analysis in Section 2:2.

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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 onlythe 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 (byapproximate analysis) at 5 GHz. The transmission parameter S21 in dB, as computed by Ansoft DesignerSV, 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, whichperforms an accurate (and time consuming) electromagnetic analysis. The line is fed and terminated by 50ohm probes. The results is shown below.

1 2 3 4 5 6 7 8 9 10

f (GHz)

40

30

20

10

0

S21

(dB)

10 cm microstrip line, probe fed and terminated.Fullwave electromagnetic analysis.w=1.7262 cm, d=0.5 cm, εr=2.2

In large part, the decrease in S21 at high frequency is due to the probes, which can�t be analyzed by thesimple circuit-based analysis. For comparison, the simulation in Designer was virtually instantaneous, whilethe more accurate Ensemble simulation took about 20 minutes.

3.9 The Transverse Resonance Technique

Skip

3.10 Wave Velocities and Dispersion

For sinusoidal steady-state situations we have c = 1=p�", 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 (andgenerally vp � c). In this case there is another velocity that is useful (and physically signi�cant) to describethe propagation of signals.

Group Velocity:

The group velocity, vg, is the velocity at which a narrow-band signal propagates. Considering a signal tobe made up of many Fourier components at di¤erent frequencies, one can de�ne a phase velocity for eachwave 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 phasevelocity describes the wave. The concept of group velocity overcomes this problem for narrow-band signals.

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Let

f (t) : time-domain signal,

F (!) : frequency domain (Fourier transform) signal

f (t)$ F (!)

where the Fourier transform pair is

F (!) =

Z 1

�1f (t) e�j!tdt; (3.69)

f (t) =1

2�

Z 1

�1F (!) ej!td!:

Modeling the system (in this case, the transmission line) as having transfer function Z (!), then

Fin Fo(ω)

transfer function of transmission line

Z(ω) (ω)

Fo (!) = Z (!)Fin (!) : (3.70)

Question: What is the transfer function Z (!)? Consider a lossless, matched line,

Z 0 ,v p

z=0z=L

in

Z LZ

where ZL = Z0. In this case v (z) = v+0 e

�j�z, such that

v (�L) = v+0 ej�L; (3.71)

v (0) = v+0 ;

and sovoutvin

= Z (!) = e�j�(!)L: (3.72)

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,

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3.10. WAVE VELOCITIES AND DISPERSION 75

f(t)

2To

1

t

Fin (!) =

Z 1

�1fin (t) e

�j!tdt (3.73)

=

Z 2T0

0

e�j!tdt = 2T0e�j!T0 sin (!T0)

(!T0):

0.2

0

0.2

0.4

0.6

0.8

1

10 5 5 10x

Plot of sin(�x) = (�x) vs. x.

Therefore, the time-domain output is

fo (t) =1

2�

Z 1

�1Fo (!) e

j!td! (3.74)

=1

2�

Z 1

�1Z (!)Fin (!) e

j!td!

=1

2�

Z 1

�1e�j�(!)LFin (!) e

j!td!

=1

2�

Z 1

�1Fin (!) e

j(!t��(!)L)d!:

If � (!) = �! (linear in !; � =p�" for TEM waves), then

fo (t) =1

2�

Z 1

�1Fin (!) e

j!(t��L)d! (3.75)

= fin (t� �L) ,

which is an exact replica of fin (t) shifter by �L to account for time delay in passing through the network.For example, if � =

p�", then �L =

p�"L = L=c = T , the one-way transit time.

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� Therefore, lossless TEM lines are dispersionless.

If � (!) is not linear in !, (as for TE and TM waves), then

fo (t) =1

2�

Z 1

�1Fin (!) e

j(!t��(!)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 (!0t) = Re�f (t) ej!0t

(3.77)

as in an amplitude modulated waveform with carrier frequency !0. the signal f (t) represents some informa-tion that is modulated with a high frequency carrier for transmission. Assume that the highest frequencycomponent of f (t) is !m, and that !m � !0. The various Fourier transforms are

F (!) =

Z 1

�1f (t) e�j!tdt; (3.78)

S (!) =

Z 1

�1s (t) e�j!tdt =

Z 1

�1f (t) ej!0te�j!tdt

=

Z 1

�1f (t) ej(!0�!)tdt

= F (! � !0) :

F (ω)

ω

S (ω)

ω−ω ω 0

0

Now, consider the modulated signal to be the input to a transmission line or waveguide having transferfunction Z (!) = e�j�z.

Sin (!) = F (! � !0) ; (3.79)

So (!) = F (! � !0) e�j�z;

so (t) =1

2�Re

Z 1

�1So (!) e

j!td!

=1

2�Re

Z 1

�1F (! � !0) e�j�zej!td!

=1

2�Re

Z !0+!m

!0�!mF (! � !0) ej(!t��z)d!:

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3.10. WAVE VELOCITIES AND DISPERSION 77

If !m � !0, expand � (!) about !0,

� (!) = � (!0) +d�

d!

��!=!0

(! � !0) +1

2

d2�

d!2

��!=!0

(! � !0)2 + ::: (3.80)

' �0 + �00 (! � !0) :

Then,

so (t) =1

2�Re

Z !0+!m

!0�!mF (! � !0) ej(!t��z)d! (3.81)

=1

2�Re

Z !0+!m

!0�!mF (! � !0) ej(!t��0z��

00(!�!0)z)d!

=1

2�Re

Z !m

�!mF (y) ej((y+!0)t��0z��

00yz)dy (c.o.v. y = ! � !0)

=1

2�Re

�ej(!0t��0z)

Z !m

�!mF (y) ej(t��

00z)ydy

�which is

so (t) = Renej(!0t��0z)f

�t� �00z

�o(3.82)

= f�t� �00z

�cos (!0t� �0z)

which is a time-shifted replica of the input fin (t).The velocity of the envelope f

�t� �00z

�is the group velocity,

t� �00z = constant, (3.83)d

dz

�t� �00z

�= 0;

1

vg= �00; ) vg =

1

�00=

�d�

d!

��1��!=!0

:

Waveguide velocities:

� =pk2 � k2c =

r�!c

�2� k2c ; (3.84)

d�

d!=

!

c2�; vg =

c2�

!=c�

k0:

vp =!

�=k0c

�;

vg < c < vp; vpvg =k0c

�

c�

k0= c2:JN

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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

E

H

H

+

v =

Z (�)

(+)

E�dl (4.1)

independent of path l from one conductor to another, and

I =

IC+

H�dl; (4.2)

where C+ is any closed path enclosing the (+) conductor. With

Z0 =v

I(4.3)

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

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80 CHAPTER 4. MICROWAVE NETWORK ANALYSIS

0 a

b

y

x

E1;0y = A sin�x

ae�j�z; H1;0

x =�E1;0yZ1;0TE

; (4.4)

where Z1;0TE = k�=�1;0. Then, for the TE10 at z = 0,

v =

Z (�)

(+)

E�dl =Z b

0

A sin�x

ady = Ab sin

�x

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 TE10 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 transverseelectric �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 uniquelyde�ned) of the mode.

3. The ratio of voltage to current for a single travelling wave should be chosen equal to the characteristicimpedance 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 ofthe guide (at x = a=2); vc = Ab. Let

Iz = �Z a

0

Hxdx =A

Z1;0TE

Z a

0

sin�x

adx = 2

a

�

A

Z1;0TE: (4.6)

Z0 =vcIz=

Ab

2 a�AZ1;0TE

=�

2

b

aZ1;0TE : (4.7)

This is known as the voltage-current de�nition of Z0. Other de�nitions:

Z0 =P

I2z; power-current, (4.8)

Z0 =v2cP; power-voltage.

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4.1. IMPEDANCE AND EQUIVALENT VOLTAGE AND CURRENTS 81

Since

P =1

2

ZS

Re fE�H�g � dS (4.9)

=ab

4

jAj2

Z1;0TE;

the power-current de�nition yields

Z0 =

ab4jAj2

Z1;0TE�2 a�

AZ1;0TE

�2 = 1

4

b

a

��2

�2Z1;0TE (4.10)

and the power voltage yields

Z0 =(Ab)

2

ab4jAj2Z1;0TE

= 4b

aZ1;0TE : (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 theform

Z0 = �b

aZ1;0TE ; (4.12)

where � is a constant. Thus, methods that only depend on the ratios of impedances will work (and give thesame 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 betweenexperiment 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 sideview of a waveguide junction.

ε0y=b

z=0y=0 z

TE10

ε0εr

a = 3:485 cm, b = 1:580 cm,

"r = 2:56; f = 4:5 GHz,

such that

�a =

rk20 �

��a

�2= 27:5 m�1

for the air-�lled guide, TE10 mode, and

�d =

r"rk20 �

��a

�2= 120:89 m�1

for the dielectric �lled guide. Only the TE10 mode propagates in either region.

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z=0

ΓZ0a Z0b

Za0 =k0�0�a

= 1292:1 ohms,

Zd0 =k�

�d= 293:9 ohms,

� =Zd0 � Za0Zd0 + Z

a0

= �0:629;

the same result for � 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 remainsconstant. Assuming the TE10 mode propagates in both sections, Z0 is the same in each section! This wouldimply � = 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 thatpropagate (i.e., those that are above cuto¤) will travel away from the discontinuity. Below cuto¤modes willdecay away from the z = 0, but will store energy in the vicinity of z = 0. This stored energy can be a veryimportant part of the analysis. In the end, usually a transmission line mode can be obtained, with lumpedreactances modeling the stored energy e¤ects (see text Section 4.6).

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4.1. IMPEDANCE AND EQUIVALENT VOLTAGE AND CURRENTS 83

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Chapter 5

Impedance Matching and Tuning

The purpose of a matching network is to provide a good match between a given transmission line and agiven load. In this chapter we will consider ways to design matching networks with consideration to matchingnetwork bandwidth, physical implementation of the network, and the ability to adjust the network. Severaltypes

Z g

Z 0 ,v p Z Lin

+gV

Z

matching

network

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.

Z L

1Z

gZ

+gV

n1:n2

Z2

i1 i2

+

+

V1 V2

Assume Zg = Z0 and ZL = RL. Assuming an ideal transformer,

V2V1=n2n1;

i2i1=n1n2: (5.1)

We want Z1 = Z0 and Z2 = RL, where Z1 = V1=i1, Z2 = V2=i2. Then

Z1 =V1i1=

n1n2V2

n2n1i2=

�n1n2

�2V2i2=

�n1n2

�2Z2: (5.2)

85

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86 CHAPTER 5. IMPEDANCE MATCHING AND TUNING

Recalling that we want Z1 = Z0 and Z2 = RL we obtain

Z1 = Z0 =

�n1n2

�2Z2 =

�n1n2

�2RL; (5.3)

and so

Z0 =

�n1n2

�2RL: (5.4)

So, given Z0 and RL, a transformer with turns ratio

n1n2=

rZ0RL

(5.5)

provides a perfect impedance match.

Example 5.1 Match a 300 ohm line to a 75 ohm load.Solution: Choose

n1n2=

rZ0RL

=

r300

75= 2: (5.6)

� This technique works very well at low frequencies, where a transformer can be constructed to operate ina somewhat ideal fashion. The common 300 ohm�to�75 ohm adapter used in many television systemsutilizes this technique.

� The match is frequency independent over the band of frequencies for which the transformer exhibitsnearly ideal behavior.

5.1 Matching with Lumped Elements (L Networks)

Z L

jX

jB

A

(a)

Z LjB

jX

Z 0

A

(b)

By correct choice of X and B, a perfect match from Z0 to complex-valued ZL can be achieved (at onefrequency).

� 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.

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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, (5.7)

Binductor = � 1

!L= �0:19

50) !L = 263:16;

X = �1:7) capacitor, (5.8)

Xcap = � 1

!C= �1:7 (50)

) !C = 0:0118:

Equations are presented in the text to solve this problem as well. For design (a), with ZL = RL + jXL,

B =XL �

pRL=Z0

pR2L +X

2L � Z0RL

R2L +X2L

; (5.9)

X =1

B+XLZ0RL

� Z0BRL

:

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For the previous example, (5.9) yields

B = 0:0038) !L = 263:159;

X = �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.

0 5 10 15 20 25 30 35 40 45

f (GHz)

0.0

0.2

0.4

0.6

0.8

1.0ρ

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 comparedto �. 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-negligiblefringing �eld, and may resonant at certain frequencies (depending on the physical structure of the element).

5.2 Single-Stub Tuning

(single lumped element also)

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5.2. SINGLE-STUB TUNING 89

LZ 0 Z

jX

Z 1

L

Given Z0 and (complex) ZL, the goal is to choose X and (Z1; L) such that a perfect match is obtained ata given frequency. Often Z1 is chosen as Z0 for convenience. The procedure will be illustrated by example.

Example 5.4 Series matching:

LZ 0 Z

A

Z 1

L

jX

Z Zin

Assume Z0 = Z1 = 50 ohms, ZL = 25 + j30 ohms. Then, ZL = ZL=Z0 = 0:5 + j0:6. We want Zin = 50ohms.

1. Choose L such that Re�ZA= 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 ZL on Smith chart and rotate to either

(a) Point E, ZA = 1 + j1:1 and L = 0:065�, or

(b) Point F, ZA = 1� j1:1 and L = 0:235�.

(a) From Point E, add a series capacitor such that

XCap = �1:1) � 1

!Ccap

1

50= �1:1) !Ccap = 0:01818;

or,

(b) From Point F, add a series inductor such that

Xind = 1:1) !Lind1

50= 1:1) !Lind = 55:

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).

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If, for example, f = 2 GHz (� = 15 cm),

!Ccap = 0:01818) Ccap = 1:45 pF, L = 0:975 cm,

!Lind = 55) Lind = 4:38 nH, L = 3:53 cm.

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

f (GHz)

0.0

0.2

0.4

0.6

0.8

1.0

ρ Capacitor

Inductor

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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 astub rather than a lumped element, the design proceeds as described above as far as the determination of Xand 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).

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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 obtainshorter stubs. For example, if the stub needs to provide X = 1:1, and Z0;stub = 100 ohms, then for ashort-circuited stub

1:1 (50)

100= 0:55) Ls = 0:08� ( = 1:2 cm at 2 GHz).

� 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.JN

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All of the previous steps were for series stubs. Another approach is to use parallel (shunt) stubs, or shuntlumped elements.

LZ 0 Z

Y

Z 1

Lin A

jB

Y

The procedure is similar to that for series stubs, except you begin at Y L rather than ZL.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 therequired stub length. For all of the designs, it is convenient to use the Smith chart. The Smith chart alsoleads 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),

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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:

1.

LZ 0

L s

Z 0

Z 0

Z

LinY

ZL =100� j150

200= 0:5� j0:75;

Y L = 0:62 + j0:92:

Enter the Smith chart at Y L and rotate along the constant SWR circle towards the generator until youintersect 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) LAs = 0:105� ) LBs = 0:396�

Choose Design A,LA = 0:036�; LAs = 0:105�.

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2. Same as (1) up through

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

For an open-circuited stub, LAs = 0:354� and LBs = 0:146�.

Choose Design B,LB = 0:194�; LBs = 0:146�.

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LZ 0

L s

Z 0

Z 0

Z

LinY

ZL = 100� 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

BA = �j1:3 BB = +j1:3

)choose inductor, � j!Lind

= � 1:3200 ; )choose capacitor, !Ccap = 1:3

200 ;

!Lind = 153:85 !Ccap = 0:0065:

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To compare designs, assume f0 = 900 MHz (� = 33:33 cm). Then,

1. short-circuited stubs:

LA = 0:036� = 1:20 cm LB = 0:194� = 6:466 cmLAs = 0:105� = 3:50 cm LBs = 0:396� = 13:20 cm

2. open-circuit stubs:

LA = 0:036� = 1:20 cm LB = 0:194� = 6:466 cmLAs = 0:354� = 11:80 cm LBs = 0:146� = 4:867 cm

3. lumped elements

LA = 0:036� = 1:20 cm LB = 0:194� = 6:466 cmLind =

153:852�900�106 = 27:21 nH Ccap =

0:00652�900�106 = 1:15 pF

0.6 0.7 0.8 0.9 1.0 1.1 1.2

f (GHz)

0.0

0.2

0.4

0.6

0.8

1.0

ρ B

ShortCircuit Stub

A

0.6 0.7 0.8 0.9 1.0 1.1 1.2

f (GHz)

0.0

0.2

0.4

0.6

0.8

1.0

ρ

B

OpenCircuit Stub

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5.3. DOUBLE-STUB TUNING 99

0.6 0.7 0.8 0.9 1.0 1.1 1.2

f (GHz)

0.0

0.2

0.4

0.6

0.8

1.0

ρB

Lumped Elements

A

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 frequencychanged), it would be useful to �x the distance from the load to the stub(s), and simply choose the correctstub length. The double-stub tuner allows this to be done, and changes in ZL or operating frequency can beaccommodated by changing Ls1 and Ls2, with L1;2 �xed. Note: the text book places the �rst stub over theload impedance. The method shown here is more �exible, since sometimes the load terminals are not easilyaccessible.

LZ 0 Z 0

L s1L s2

Z

A

inY L L2 1

B

Procedure:

1. Mark Y L on the Smith chart and draw constant SWR 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 SWR 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, PointY B� . The added susceptance comes form choosing Ls1 correctly.

5. Draw a new constant SWR circle using Point Y B� . Rotate on new SWR circle a distance L2 (youwill 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� .

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Example 5.6 For the double stub tuner shown below, �nd the shortest lengths of Ls1 and Ls2 for a matchif ZL = 400 + j200 ohms and Z0 = 200 ohms. Assume L1 = 3�=16 and L2 = �=8.Solution:

LZ 0 Z 0

L s1L s2

Z

A

inY L L2 1

B

ZL =400 + j200

200= 2 + j1;

Y L =1

ZL= 0:4� j0:2:

1. Enter Smith chart at Y L.

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2. Rotate on constant SWR circle 3�=16.

3. Add susceptance�Y = �j0:925

�ti intersect the rotated G = 1 circle.

4. Rotate on new SWR circle to intersect non-rotated G = 1 circle (this is the same as de-rotating thepoint 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 = �j0:17 at Point A. We obtain thesevalues by choosing Ls1 and Ls2 correctly.

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 chartwe could have added susceptance to intersect the rotated G = 1 circle at a di¤erent point.

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20 22 24 26 28 30 32 34 36 38 40

f (GHz)

0.0

0.2

0.4

0.6

0.8

1.0

ρ

Example 5.7 A 100+ j100 ohm load is to be matched to a 50 ohm line using a double stub tuner as shownbelow. Determine the stub lengths to provide a patch, and �nd the forbidden zone of impedance/admittancevalues. L1 = 0:4�, L2 = 3�=8.

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Solution:

LZ 0 Z 0

L s2

Z

s1

inY L L2 1

BA

L

ZL =100 + j100

50= 2 + j2;

Y L = 0:25� j0:25:The stub having length Ls1 needs to provide Y = j1:3

��1:4 +B = �:1) B = 1:3

�. The stub having length

Ls2 needs to provide Y = j0:6. From the Smith chart,

Ls1 = 0:1459�;

Ls2 = 0:086�:

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20 22 24 26 28 30 32 34 36 38 40

f (GHz)

0.0

0.2

0.4

0.6

0.8

1.0

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5.4. THE QUARTER-WAVE TRANSFORMER 105

5.4 The Quarter-Wave Transformer

In this section we will consider the bandwidth performance of a single quarter-wave transformer. In subse-quent sections we will determine how to design multi-section quarter-wave transforms that can be used toachieve a broadband match.

inZZ 0 R LZ 1

λ/4

We know that if Z1 =pZ0RL 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 theexpression

Zin = Z1RL + jZ1 tan�L

Z1 + jRL tan�L:

Let � = �L = electrical length, such that

�L =2�

�

�04

(5.10)

such that

�L =2�

�0

�04=�

2(5.11)

at the design frequency f0 (c = �0f0). Then,

�in =Zin � Z0Zin + Z0

=RL � Z0

RL + Z0 + j2 tan �pZ0RL

(5.12)

(using Z1 =pZ0RL). By further manipulations, it is shown that

j�inj =1h

1 +�4Z0RL= (RL � Z0)2

�sec2 �

i1=2 : (5.13)

If we assume f ' f0, then � ' �=2 and sec2 � � 1, resulting in

j�inj =jRL � Z0j2pZ0RL

jcos �j = � jcos �j (5.14)

where � is a constant. If RL does not vary with frequency,

m θm

mΓ

θ

L

π− π

|Γ|

π/2

θ=βJNTU W

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where

� = �L =2�

�

�04=

�

2

�0�=�

2

f

f0| {z }assuming TEM mode

: (5.15)

If we have a value �m for the maximum re�ection coe¢ cient that can be tolerated, then we obtain a usablebandwidth of

�� = 2��2� �m

�: (5.16)

The approximation f ' f0 was made to see where the characteristic �vee�shape in the above plot comesfrom. We can determine a more useful expression for bandwidth by equating �m with the exact expressionfor j�inj,

j�mj =1h

1 +�4Z0RL= (RL � Z0)2

�sec2 �m

i1=2 ; (5.17)

1

j�mj2= 1 +

4Z0RL

(RL � Z0)2sec2 �m

= 1 +

�2pZ0RL

(RL � Z0)1

cos �m

�2:

Solving for cos �m,

cos �m =�mp1� �2m

2pRLZ0

jRL � Z0j; (5.18)

�m = cos�1

�mp1� �2m

2pRLZ0

jRL � Z0j

!:

Since �m = (�=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

f0=2 (f0 � fm)

f0= 2� 2fm

f0= 2� 4

��m (5.20)

we obtain�f

f0= 2� 4

��m = 2�

4

�cos�1

�mp1� �2m

2pRLZ0

jRL � Z0j

!: (5.21)

The above analysis assumes TEM lines, and ignores the e¤ects of reactance associated with discontinuitiesof the transmission line when there is a step change in line width. The latter e¤ect can be compensated forby 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 SWR � 2.Solution: Since at f0 = 1 GHz �0 = 30 cm,

L =�04=30

4= 7:5 cm,

Z1 =p(75) (30) = 47:43 ohms.

For SWR = 2,

�m =SWR� 1SWR+ 1

=1

3

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5.5. THE THEORY OF SMALL REFLECTIONS 107

and so

�f

f0= 2� 4

�cos�1

�mp1� �2m

2pRLZ0

jRL � Z0j

!

= 2� 4

�cos�1

0@ 1=3q1� (1=3)2

2p(75) (30)

j30� 75j

1A= 2� 4

�cos�1 (0:7454) = 2� 4

�0:7297 = 1:071;

or 107% bandwidth. The means the band extends from

f0 ��f

2= 0:465 GHz

to

f0 +�f

2= 1:535 GHz.

Note: for SWR < 1:1, �m = 0:0476 and �f=f0 = 0:128 (12:8%), or from 0:936 GHz to 1:064 GHz. So,

� as desired SWR #, �m #, and BW #.

� For a �xed �m, the larger the ratio RL=Z0 (or Z0=RL), the smaller the bandwidth, as shown below.

L

0

Z L

Z

1

|Γ|

0Z= 4, 0.25

Z= 2, 0.5

5.5 The Theory of Small Re�ections

The theory of small re�ections will facilitate the development of approximate formulas for the design ofmulti-section quarter-wave transformers, which will be shown to be relatively broadband.

I. Single Section Transformer:

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2

j θe

Γ3

Γ3

Γ3

Γ1

Γ1

T 21

T 12

T 21

Γ2

Γ2

T 12

T 12

Γ2

Z

1

LRΓZ 1

e j θ

e j θ

e j θ

�1 =Z2 � Z1Z2 + Z1

; �2 = ��1; �3 =ZL � Z2ZL + Z2

; (5.22)

T21 = 1 + �1; T12 = 1 + �2;

where � is the overall re�ection coe¢ cient and �i, i = 1; 2; 3, are the interfacial re�ection coe¢ cients. Wehave

� = �1 + T12T21�3e�j2� + T12T21�

23�2e

�j4� + ::: (5.23)

= �1 + T12T21�3e�j2�

1Xn=0

�n2�n3 e�j2n�:

Using the geometric series result1Xn=0

xn =1

1� x for jxj < 1 (5.24)

leads to

� = �1 +T12T21�3e

�j2�

1� �2�3e�j2�(5.25)

=�1 + �3e

�j2�

1 + �1�3e�j2�:

If the impedance do not di¤er greatly, then �1 and �3 are small, and so �1�3 is very small compared to 1and so

� ' �1 + �3e�j2�: (5.26)

The above expression shows that � is dominated by the �rst re�ection from the initial discontinuity betweenZ1 and Z2, �1, and the �rst re�ection from the discontinuity between Z2 and ZL.

II. Multi-Section Transformers:

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5.6. BINOMIAL MULTISECTION MATCHING TRANSFORMERS 109

Now consider the N� section transform shown below,

0Z Z Z Z1 2 N

θ θ θ

...

Γ0 Γ Γ Γ Γ1 2 N1 N

Z L

Γ

where we assume that ZL is real-valued. We have

�0 =Z1 � Z0Z1 + Z0

; �n =Zn+1 � ZnZn+1 + Zn

; �N =ZL � ZNZL + ZN

: (5.27)

Assume that Zn increases or decreases monotonically. Then �n is real-valued and has the same sign for alln;

�n > 0 for ZL=Z0 > 1; (5.28)

�n < 0 for ZL=Z0 < 1;

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,

� (�) ' e�jN�n�0�ejN� + e�jN�

�+ �1

hej(N�2)� + e�j(N�2)�

i+ �2

hej(N�4)� + e�j(N�4)�

i+ :::+

�N�12

�ej� + e�j�

�; N odd

�N2

; N even

); (5.30)

leading to

� (�) ' 2e�jN� f�0 cosN� + �1 cos (N � 2) � + �2 cos (N � 4) � + :::+ �n cos (N � 2n) �

+:::+�N�1

2cos �; N odd

12�N

2; N even

); (5.31)

which is seen to be a �nite (truncated) Fourier cosine series for � (�). Since a Fourier cosine series canrepresent any smooth function , we can synthesize any desired re�ection coe¢ cient response as a functionof frequency by proper choice of the �n values, and by using enough sections (some limitations are imposedby the Bode-Fano criteria as discussed later). Two choices of desirable functions to implement are presentednext, 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)

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The absolute value is

jg (�)j = jAj��e�j� �ej� + e�j��N �� (5.33)

= jAj��e�jN�� ej� + e�j��N ��

= jAj��ej� + e�j��N = jAj j2 cos �jN

= 2N jAj jcos �jN ;

which is plotted below.

The function (5.32) has the property that at � = �=2, jf (�)j(n) = 0, n = 1; 2; :::; N � 1, where g(n) indicatesthe 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 multisectiontransformer, (5.29) (or (5.30) or (5.31)), and (5.32), in order to determine what �n values will lead to theresponse (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

(1 + x)N=

NXn=0

CNn xn; (5.34)

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�j2�

�N(5.36)

= ANXn=0

CNn e�j2n�: (5.37)

Equating (5.37) with (5.29) leads to

g (�) = ANXn=0

CNn e�j2n� = � (�) ' �0 + �1e�j2� + �2e�j4� + :::+ �Ne�j2N�: (5.38)

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The constant A can be determined by letting f ! 0 (� = �L = 0),

g (0) = A2N = � (0) =ZL � Z0ZL + Z0

; (5.39)

) A = 2�NZL � Z0ZL + Z0

: (5.40)

Equating nth terms in (5.38) results in

�n = ACNn = 2�N

ZL � Z0ZL + Z0

CNn ; (5.41)

the sought-after formula for choosing �n values to yield a response that is equal to the maximally-�atresponse (5.32)1 . Since CNn = CNN�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 � Z0Z1 + Z0

; �n =Zn+1 � ZnZn+1 + Zn

; �N =ZL � ZNZL + ZN

(5.42)

then, for example,

�n =Zn+1 � ZnZn+1 + Zn

= 2�NZL � Z0ZL + Z0

CNn (5.43)

which can be solved for Zn+1 in terms of Zn as

Zn+1 = Zn

�1 + 2�N ZL�Z0

ZL+Z0CNn

��1� 2�N ZL�Z0

ZL+Z0CNn

� : (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

ln (x) = 2x� 1x+ 1

+2

3

�x� 1x+ 1

�3+ ::: (5.45)

' 2x� 1x+ 1

for x ' 1, thenZn+1 � ZnZn+1 + Zn

=Zn+1=Zn � 1Zn+1=Zn + 1

' 1

2lnZn+1Zn

(5.46)

for Zn+1 ' Zn (this assumption was used in developing the theory of small re�ections, and so the approxi-mation in (5.46) is consistent with the approximations leading to (5.29), leading to a self-consistent formulathat provides accurate results). Therefore, making the same approximations on both sides of (5.43) leads to

�n ' 1

2lnZn+1Zn

' 2�N ZL � Z0ZL + Z0

CNn ' 2�NCNn1

2lnZLZ0; (5.47)

lnZn+1Zn

= 2�NCNn lnZLZ0; (5.48)

lnZn+1 = lnZn + 2�NCNn ln

ZLZ0; (5.49)

resulting in

Zn+1 = elnZn+2

�NCNn ln

ZLZ0 (5.50)

1Actually, since (5.29) is an approximate formula, then the resulting response is approximately that of (5.32).

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which is generally valid for 12Z0 � ZL � 2Z0. Bandwidth can be determined as follows. From (5.32) we

obtain

j�mj = 2N jAj jcos �mjN (5.51)

= 2N jAj cosN �m

when 0 � �m � �=2. Therefore,

�m = cos�1

"�j�mjjAj 2N

� 1N

#(5.52)

= cos�1

"1

2

�j�mjjAj

� 1N

#:

From (5.20) we have

�f

f0=

2 (f0 � fm)f0

= 2� 4

��m (5.53)

= 2� 4

�cos�1

"1

2

�j�mjjAj

� 1N

#(5.54)

where

A = 2�N1

2lnZLZ0

(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�mj = 0:05.Solution:

Zn+1 = elnZn+2�NCN

n lnZLZ0 (5.56)

= elnZn+2�3C3

n ln50100 ; (5.57)

and

C30 =3!

3!0!= 1; C31 =

3!

2!1!= 3; C32 =

3!

1!2!= 3.

Therefore,

Z1 = elnZ0+2�3C3

0 ln50100 = 91:70 ohms, (5.58)

Z2 = elnZ1+2�3C3

1 ln50100 = 70:71 ohms, (5.59)

Z3 = elnZ2+2�3C3

2 ln50100 = 54:53 ohms. (5.60)

Γ0 Γ

π/2

Γ1 2

Z L= 50100 91.7 70.71 54.53

0.043 0.129 0.129

Γ3

0.043

π/2 π/2

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To compute the bandwidth,

�f

f0= 2� 4

�cos�1

"1

2

�j�mjjAj

� 1N

#; (5.61)

and, with

A = 2�N1

2lnZLZ0

= 2�31

2ln50

100= �0:0433

and �m = 0:05,

�f

f0= 2� 4

�cos�1

"1

2

�0:05

0:0433

� 13

#= 0:703; 70:3%

A�1 + e�j2�(f)

�Nvs. f=f0:

10 14 18 22 26 30 34 38 42 46 50

f (GHz)

0.00

0.05

0.10

0.15

0.20

0.25

0.30

ρ

Example 5.10 Design an N section binomial transformer that needs to operate from 1:2 GHz to 2:8 GHzwith j�mj � 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

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required.

�f

f0=2 (f0 � fm)

f0=2 (2� 1:2)

2= 0:8; (5.62)

A = 2�N1

2lnZLZ0

= 2�N1

2ln100

50= 2�N (0:3466) (5.63)

�f

f0= 2� 4

�cos�1

"1

2

�j�mjjAj

� 1N

#(5.64)

= 2� 4

�cos�1

"1

2

�0:05

2�N (0:3466

� 1N

#= 0:8: (5.65)

Solving for N we obtain N = 3:64 ) N = 4. With N determined, the design proceeds as in the previousexample.

C40 = 1; C41 = 4; C42 = 6; C43 = 4;

Z1 = elnZ0+2�4C4

0 ln10050 = 52:214;

Z2 = elnZ1+2�4C4

1 ln10050 = 62:093;

Z3 = elnZ2+2�4C4

2 ln10050 = 80:525;

Z4 = elnZ3+2�4C4

3 ln10050 = 95:76:

= 100

π/2 π/2 π/2 π/2

50 52.2 62.1 95.880.5 Z L

10 14 18 22 26 30 34 38 42 46 50

f (GHz)

0.00

0.05

0.10

0.15

0.20

0.25

0.30

ρ

Note: Pascal�s triangle is also useful for evaluating CNn ,

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5.7. CHEBYSHEV MULTISECTION MATCHING TRANSFORMERS 115

n CNn0 11 1 12 1 2 33 1 3 3 14 1 4 6 4 1

5.7 Chebyshev Multisection Matching Transformers

Another function with desirable properties to implement in a multisection transformer is a Chebyshev poly-nomial. The Chebyshev polynomials are given by

T0 (x) = 1; (5.66)

T1 (x) = x;

T2 (x) = 2x2 � 1;

T3 (x) = 4x3 � 3x;

T4 (x) = 8x4 � 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 x

�; jxj < 1; (5.67)

Tn (x) = cosh�n cosh�1 x

�; 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 �1; Tn (x) has n zeros between �1. We willmap this range into the passband of the transformer.

2. For jxj > 1, jTn (x)j > 1. We will map this range to outside the passband.

1

0.5

0

0.5

1

1 0.5 0.5 1x

cos�n cos�1 (x)

�vs. x, n = 1; 2; 3; 4.

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m θm

mΓ

θ

x=1

π− π

|Γ|

π/2

θ=β L

x=+1

With x = cos �, as � goes from 0 to �, x goes from +1 to �1. Force x = +1 at � = �m, and x = �1 at� = � � �m. To implement this we replace cos � with cos �= cos �m = sec �m cos �:

Tn (cos �)! Tn

�cos �

cos �m

�(5.68)

(then when � = �m

Tn

�cos �mcos �m

�= Tn (1) = 1; (5.69)

and when � = � � �m,Tn

�cos (� � �m)cos �m

�= Tn (�1) = �1:) (5.70)

Therefore,

T1 (x) = x; (5.71)

) T1 (sec �m cos �) = sec �m cos �;

T2 (x) = 2x2 � 1;

) T2 (sec �m cos �) = 2 (sec �m cos �)2 � 1

= sec2 �m (1 + cos 2�)� 1;T3 (x) = 4x

3 � 3x;) T3 (sec �m cos �) = 4 (sec �m cos �)

3 � 3 (sec �m cos �)= sec3 �m (cos 3� + 3 cos �)� 3 sec �m cos �;

T4 (x) = 8x4 � 8x2 + 1;

) T4 (sec �m cos �) = 8 (sec �m cos �)4 � 8 (sec �m cos �)2 + 1;

= sec4 �m (cos 4� + 4 cos 2� + 3)� 4 sec2 �m (cos 2� + 1) + 1:

Procedure: Equate the actual multisection re�ection coe¢ cient (5.31) to the Nth order Chebyshev poly-nomial Ae�jN�TN (sec �m cos �) for a given N . That is,

� (�) ' 2e�jN� f�0 cosN� + �1 cos (N � 2) � + �2 cos (N � 4) � + :::+ �n cos (N � 2n) �

+:::+�N�1

2cos �; N odd

12�N

2; N even

)= Ae�jN�TN (sec �m cos �) : (5.72)

The constant A is again determined by letting f = 0;

� (0) =ZL � Z0ZL + Z0

= ATN (sec �m) ; (5.73)

) A =ZL � Z0ZL + Z0

1

TN (sec �m)' 1

2lnZLZ0

1

TN (sec �m).

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Since maxTN (x) = 1 for jxj < 1, then A is also the maximum allowable re�ection coe¢ cient magnitude inthe passband,

�m = A =1

2lnZLZ0

1

TN (sec �m): (5.74)

Bandwidth: Using (5.74),

TN (sec �m) =1

�m

��12 ln ZLZ0�� ; (5.75)

and, since sec �m = 1= cos �m � 1 we use

TN (sec �m) = cosh�N cosh�1 sec �m

�=

1

�m

��12 ln ZLZ0�� ; (5.76)

) sec �m = cosh

�1

Ncosh�1

�1

�m

��12 ln ZLZ0��

such that

�f

f0=

2 (f0 � fm)f0

= 2� 4

��m (5.77)

= 4� 4

�sec�1

�cosh

�1

Ncosh�1

�1

�m

��12 ln ZLZ0�� :

��Ae�jN�TN (sec �m 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 = �m = 0:05;

sec �m = cosh

�1

Ncosh�1

�1

�m

��12 ln ZLZ0��

= cosh

�1

3cosh�1

�1

0:05

��12 ln 10050��

= 1:4075; �m = 0:7806:

To determine the Zn values, from (5.72) we have

� (�) ' 2e�j3� f�0 cos 3� + �1 cos �g = Ae�j3�T3 (sec �m cos �) (5.78)

) 2 f�0 cos 3� + �1 cos �g = AT3 (sec �m cos �) (5.79)

= A�sec3 �m (cos 3� + 3 cos �)� 3 sec �m cos �

�: (5.80)

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Equating similar terms,

cos 3� : 2�0 = A sec3 �m; (5.81)

cos � : 2�1 = 3A�sec3 �m � sec �m

�;

) �0 = 0:0697; �1 = 0:1036:

From symmetry,

�3 = �0; �2 = �1: (5.82)

0Z 3

Γ0 Γ Γ1 2

Z L

Γ3

π/2 π/2 π/2

Z Z Z1 2

To determine the Zn values, note the relationship

�n =Zn+1 � ZnZn+1 + Zn

$ Zn+1 = Zn1 + �n1� �n

(5.83)

such that

Z1 = Z01 + �01� �0

= 57:49;

Z2 = Z11 + �11� �1

= 70:77;

Z3 = Z21 + �21� �2

= 86:97:

The bandwidth is

�f

f0=

2 (f0 � fm)f0

= 2� 4

��m

= 2� 4

�0:7806 = 1:006; 100:6%

(compare with 71% for the binomial transformer with N = 3.

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10 14 18 22 26 30 34 38 42 46 50

f (GHz)

0.00

0.05

0.10

0.15

0.20

0.25

0.30

ρ

10 14 18 22 26 30 34 38 42 46 50

f (GHz)

0.00

0.05

0.10

0.15

0.20

0.25

0.30

ρ

Binomial, N=3

Chebyshev, N=3

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 = 0:05using Chebyshev polynomials, design an appropriate microstrip implementation. Assume "r = 9:8, f0 = 1GHz, and that the load is an in�nite length of 100 ohm line.Solution:

w

b=

(8eA

e2A�2 ;wb < 2;

2�

hB � 1� ln (2B � 1) + "r�1

2"r

�ln (B � 1) + 0:39� 0:61

"r

�i; w

b > 2;(5.84)

where

A =Z060

r"r + 1

2+"r � 1"r + 1

�0:23 +

0:11

"r

�; (5.85)

B =377�

2Z0p"r:

For

Z1 = 57:49;

Z2 = 70:77;

Z3 = 86:97;

Z0 = 50; ZL = 100

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we obtain

w0b

= 0:978;wLb= 0:137;

w1b

= 0:721;w2b= 0:427;

w3b= 0:227

Since

vp =cp"e= �f; (5.86)

) � =cp"e

1

f:

With

"e ="r + 1

2+"r � 12

1q1 + 12d

w

(5.87)

we obtain�14= 2:95 cm;

�24= 3:01 cm;

�34= 3:06 cm

λ 1

4

50 57.49 70.77 86.97 100

λ

4

λ

42 3

......

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 acontinuous taper.

Z0

z=0 z=L

ZL

Z(z)

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,

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5.8. TAPERED LINES 121

Z Z+∆Z

∆Γ

∆ zand incremental re�ection coe¢ cient is

�� =(Z +�Z)� Z(Z +�Z) + Z

=�Z

2Z +�Z' �Z

2Z. (5.88)

As �Z ! 0, we can replace ��;�Z by d�; dZ such that

d� =dZ

2Z: (5.89)

This can be manipulated into a convenient expression as follows.

dhln Z(z)Z0

idz

=1

Z

dZ (z)

dz= 2

d�

dz; (5.90)

d�

dz=1

2

dhln Z

Z0

idz

Form the integral summation of � (all partial re�ections) and add the approximate phase shifts,

� =1

2

Z L

z=0

e�j2�zd

dzlnZ

Z0dz: (5.91)

If Z (z) is known, then � can be found.

1. Exponential Taper:

Z (z) = Z0eaz; 0 < z < L: (5.92)

We want

Z (0) = Z0; (5.93)

Z (L) = Z0eaL = ZL:

Therefore,

eaL =ZLZ0

) a =1

LlnZLZ0: (5.94)

Then,

� =1

2

Z L

0

e�j2�zd

dzlnZ

Z0dz (5.95)

=1

2

Z L

0

e�j2�zd

dzln (eaz) dz

=1

2

Z L

0

e�j2�z1

LlnZLZ0dz

=1

2

1

LlnZLZ0

Z L

0

e�j2�zdz

=1

2lnZLZ0e�j�L

sin�L

�L:

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When �L = 0, sin �L�L = 1. From the plot, we see that �L > � ( 2�� L > � ) L > �

2 ) for bestperformance.

2. Triangular Taper:

Another common method is to assume a triangular taper for d ln (Z=Z0) =dz,

d ln ZZ0

dz=

(4zL2 ln

ZLZ0; 0 � z � L=2;�

4L � 4

zL2

�ln ZLZ0 ; L=2 � z � L : (5.96)

This results in

Z (z) =

(Z0e

2(z=L)2 ln(ZL=Z0); 0 � z � L=2;Z0e

(4z=L�2z2=L2�1) ln(ZL=Z0); L=2 � z � L: (5.97)

Integrating the triangular expression leads to

� =1

2e�j�L ln

ZLZ0

�sin (�L=2)

(�L=2)

�2: (5.98)

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For �L > 2� the peaks of the triangular taper are lower than the corresponding peaks of the exponentialtaper because of the (sinx=x)2 factor. However, the �rst null occurs at 2�, whereas the �rst null appearsat � 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 islowest over the passband, for a given taper length greater than some critical value. Alternatively, uponspeci�cation of �m, the Klopfenstein taper yields the shortest length L.

� = �0e�j�L cos

q(�L)

2 �A2

coshA; �L > A (passband), (5.99)

= �0e�j�L cos

qA2 � (�L)2

coshA; �L < A;

where

�0 =ZL � Z0ZL + Z0

' 1

2lnZLZ0; (5.100)

A = cosh�1�0�m

:

The impedance taper must generally be calculated numerically from

lnZ (z) =1

2ln (Z0ZL) +

�0coshA

A2� (2z=L� 1; A) ; 0 � z � L; (5.101)

where

� (x;A) = �� (�x;A) =Z x

0

I1

�Ap1� y2

�Ap1� y2

dy; jx � 1j (5.102)

and where I1 (x) is the modi�ed Bessel function.

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¢ cientmagnitudes vs. �L.Solution:

(a) Triangular Taper: For the triangular taper

Z (z) =

(Z0e

2(z=L)2 ln(ZL=Z0); 0 � z � L=2;Z0e

(4z=L�2z2=L2�1) ln(ZL=Z0); L=2 � z � L; (5.103)

� =1

2e�j�L ln

ZLZ0

�sin (�L=2)

(�L=2)

�2:

(b) Exponential Taper: For the exponential taper

Z (z) = Z0eaz; 0 < z < L (5.104)

a =1

LlnZLZ0

= 0:6931

L;

� =1

2lnZLZ0e�j�L

sin�L

�L:

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(c) Klopfenstein Taper: For the Klopfenstein taper

�0 '1

2lnZLZ0

= 0:346 (5.105)


= 3:543 = 1:13�;

Z is evaluated numerically and in the passband (�L > A = 1:13�)

� = �0e�j�L cos

q(�L)

2 �A2

coshA: (5.106)

Example 5.14 Design an exponentially tapered matching transformer to match a 100 ohm load to a 50ohm line. Plot j�j vs. �L, and �nd the length of the matching section (at the center frequency) required toobtain j�j � 0:05 over a 100% bandwidth. How many sections would be required if a Chebyshev matchingtransformer were used to achieve the same speci�cations?

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Solution:

Z (z) = Z0eaz; 0 < z < L

a =1

LlnZLZ0

= 0:6931

L;

) Z (z) = 50e0:693zL

j�j =1

2

��ln ZLZ0 e�j�L sin�L�L

��=

1

2

��ln 10050 sin�L�L

�� = 0:3466 �� sin�L�L

�� :For 100% bandwidth,

�f

f0= 2� 4

��m =

2 (f0 � fm)f0

= 1;

) fm =1

2f0; �m =

�

4

and so1

2f0 � f �

3

2f0

represents a 100% bandwidth.

G (x) = 0:3466�� sin(x)x

�� vs. x=�.Starting at �L = 1:72� (where j�j � 0:05)

�L � 1:72� ) 2�

�L � 1:72� ) L � 1:72��

2�= 0:86�:

For

f :1

2f0 to

3

2f0;

� : 2�0 to 0:666�0:

Therefore,

L � 0:86 (2�0) = 1:72�0

is needed. For example, if f0 = 30 GHz (�0 = :01 m),

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20 22 24 26 28 30 32 34 36 38 40

f (GHz)

0.00

0.02

0.04

0.06

0.08

0.10

ρ

For the Chebyshev,

N =cosh�1

�1�m

�� 12 ln ZLZ0 ��cosh�1 (sec �m)

(5.107)

=cosh�1

�10:05

�� 12 ln

10050

��cosh�1

�sec �4

� = 2:97;

) N = 3:

Since a three section Chebyshev transformer is 3�0=4 = 0:75�0 it is smaller then the exponential taper (whichhas length L = 1:72�0).Note that for the Klopfenstein taper,

�0 =ZL � Z0ZL + Z0

= 0:333;


= 2:5846:

Since the passband is de�ned as �L > A, and the maximum ripple in the passband 0:05, then for our valueof A the ripple will be at most 0:05 for �L > A = 2:5846;

�L > 2:5846;2�

�L > 2:5646) L > 0:411�:

Since �max = 2�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 longerthan the Chebyshev taper.

5.9 Bode-Fano Criteria

The Bode-Fano criteria provides information about the theoretical limits that constrain the performance ofan impedance matching network.

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5.9. BODE-FANO CRITERIA 127

Z g

Z 0 ,v p Z Lin

+gV

Z

matching

network

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, considerthe following.

Γ(ω)matching

network

The Bode-Fano criteria is, for this case,Z 1

0

ln1

j� (!)jd! ��

RC: (5.108)

Assume that we desire the response to be as shown below.

ω

∆ω

Γm

1

|Γ|

Then Z 1

0

ln1

j� (!)jd! =Z�!

ln1

j�mjd! = �! ln

1

j�mj� �

RC; (5.109)

) �! � �

RC jln �mj:

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Note that as �m varies from 0 to 1, �RCjln �mj varies from 0 to 1. So,

�m : 0! 1; (5.110)

�! : 0!1:

� For a �xed RC, broader bandwidth can only be achieved at the expense of a higher re�ection coe¢ cientin the passband.

� Also, the passband �m cannot be zero unless �! = 0. Thus, a perfect match cannot be maintainedexcept at possibly a �nite number of frequencies.

� As R and/or C increase, �! decreases.

ω

|Γ|

ω

|Γ|

Example 5.15 A parallel RC load having R = 50 ohm and C = 3 pF is to be matched to a 30 ohm linefrom 1 to 3 GHz. What is the best �m that can be obtained? Assume a square-wave � pro�le.Solution:

ωΓm

1

|Γ|

2π(1) 2π(3)

Z 1

0

ln1

j� (!)jd! =Z 2�(3�109)

2�(1�109)ln

1

j�mjd! � 20:94� 109; (5.111)

2� (2) ln1

j�mj� 20:94) �m � 0:1888:

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5.9. BODE-FANO CRITERIA 129

Alternatively, to match from 1 to 10 GHz,

2� (9) ln1

j�mj� 20:94) �m � 0:691:

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Chapter 6

Power Dividers and DirectionalCouplers

(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:

2 through

3 coupled4 isolated

1 input

The main idea is that power is input into Port 1. Some fraction of the input power goes through to Port2 (the �through port�), while the rest goes to Port 3 (the �coupled port�). Ideally, no power is transferredto Port 4 (the �isolated port�).

C = 10 log10P1P3, coupling factor in dB, (6.1)

D = 10 log10P3P4, directivity in dB,

I = 10 log10P1P4, isolation in dB.

The coupling factor indicates the fraction of input power coupled to Port 3. Directivity indicates the coupler�sability to isolate forward (Port 3) and backward (Port 4) waves. Note that I = C +D. An ideal directionalcoupler 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 appliedto both Ports 2 and 3, power is combined and emerges from Port 1.

Bethe Hole Coupler:

131

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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 combinedin-phase at the desired output port, and can be combined out-of-phase at the desired isolated port. Thedetails of the single-hole coupler are provided in the text. It can be shown that, for the TE10 mode, theskewed coupler design leads to

C = �20 log104k20r

30

3ab�; � =

rk20 �

��a

�2: (6.2)

One can thus design a coupler to achieve a certain coupling. This will only hold at one frequency, and is veryfrequency sensitive. To decrease sensitivity, multi-hole couplers can be design, analogous to multi-sectionquarter-wave transformers.

Two-hole coupler:

Here we use parallel guides. The apertures are small, and placed �g=4 apart. Most of the input wave istransferred 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,

B1ej�z��z=0

= B1; F1e�j�z��

z=0= F1 (6.3)

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133

due to Aperture 1. At this same reference point, the backward wave from Aperture 2 will be

B2ej��g=2 = B2e

j� = �B2: (6.4)

If jB1j ' jB2j, 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

F1e�j��g=4 + F2e

�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, jB1j must equal jB2j. 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, and4.

Solution:

�power =

�2Z0 � Z02Z0 + Z0

�2=1

9:

At Port 2:1

2� �p

1

2=1

2� 1

18=4

9watts,

at Port 31

2watt,

and at Port 41

2�p1

2=1

36watts,

with 136 watts back out to Port 1.

N + 1 hole couplers:

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All apertures are spaced d = �g=4 apart. The amplitude of the wave incident on the input port is A, asis the amplitude at the output port of the through wave (of course, this is not self-consistent, neverthelessit yields reasonable results)! Set the phase reference to be at the �rst aperture.At the coupled port,

F = Ae�j�NdNXn=0

Fn (6.6)

(remember we have N + 1 apertures), and at the output of the isolated port,

B = ANXn=0

Bne�j2�nd: (6.7)

Then,

C = �20 log10��FA�� = �20 log10

��NXn=0

Fn

�� dB, (6.8)

D = �20 log10��BF�� = �20 log10

��PN

n=0Bne�j2�ndPN

n=0 Fn

��= �C � 20 log10

��NXn=0

Bne�j2�nd

�� dB.

Assume that all aperture are round holes with identical positions, s, relative to the edge of the guide, withrn being the radius of the nth aperture. It can be shown that

Fn = kfr3n; Bn = kbr

3n; (6.9)

where kf;b are dependent on aperture shape, and are slowly varying functions of frequency. Therefore,

C = �20 log10 jkf j � 20 log10

��NXn=0

r3n

�� dB, (6.10)

D = �C � 20 log10 jkbj � 20 log10

��NXn=0

r3ne�j2�nd

��| {z }highly frequency sensitive

dB.

In the design of multi-section quarter wave transformers we considered binomial and Chebyshev methodsto choose the impedances Zn. Here, the design involves choosing the hole radii using similar methods.

Binomial Response:

Pick r3n proportional to the binomial coe¢ cients,

r3n = kCNn (6.11)

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135

(recall

CNn =N !

(N � n)!n! ): (6.12)

The proportionality constant k can be determined from

C = �20 log10 jkf j � 20 log10

��NXn=0

r3n

�� (6.13)

= �20 log10 jkf j � 20 log10

��NXn=0

kCNn

��= �20 log10 jkf j � 20 log10 jkj � 20 log10

��NXn=0

CNn

�� :From this, and with a desired value of C and a known kf ; N , one can determine k. Note:

jkf j =�� 2k203ab�1;0

"sin2

�s

a�2�21;0k20

sin2�s

a+

�2

�21;0a2cos2

�s

a

!#�� ; (6.14)

jkbj =�� 2k203ab�1;0

"sin2

�s

a+2�21;0k20

sin2�s

a� �2

�21;0a2cos2

�s

a

!#�� : (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 therequired coupling is 18 dB. The physical structure is a rectangular waveguide with round coupling aperturescentered 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),

jkf j = 1:113� 105; jkbj = 1:705� 106:

Then,

C = �20 log10 jkf j � 20 log10 jkj � 20 log10

��NXn=0

CNn

�� ;) �20 log10 jkj = C + 20 log10 jkf j+ 20 log10

��NXn=0

CNn

�� ;jkj = log�110 � 1

20C + 20 log10 jkf j+ 20 log10

��NXn=0

CNn

��!!

= 7:071� 10�8:

With r3n = kC4n,

r0 = 4:135 mm,

r1 = 6:564 mm,

r2 = 7:514 mm,

r3 = 6:564 mm = r1;

r4 = 4:135 mm = r0:

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C = �20 log10 jkf j � 20 log10 jkj � 20 log10

��NXn=0

CNn

��= 18 dB at f = f0;

D = �C � 20 log10 jkbj � 20 log10

��NXn=0

r3ne�j2�nd

��= 283 dB at f = f0.

Note that C is very insensitive to frequency (only via kf , which is a slowly-varying function with respect tofrequency; upon taking log10 jkf j we obtain a fairly insensitive function of frequency for C. However, D isvery frequency sensitive due to the summation term.

4 6 8 10 12 14 16

f (GHz)

0

20

40

60

80

100

C

D

5 Hole Coupler

For a 3 hole coupler (r0 = 6:564; r1 = 8:27; r2 = 6:564 = r0),

4 6 8 10 12 14 16

f (GHz)

0

20

40

60

80

100

C

D

3 Hole Coupler

and for a 15 hole coupler

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137

4 6 8 10 12 14 16

f (GHz)

0

20

40

60

80

100

C

D

15 Hole Coupler

Directional couplers can also be implemented with transmission lines, although the theory will not bedeveloped here.

Hybrid Couplers:

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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 tosplit input power into two parts. We will brie�y study a lossless divider. The resistive divider and Wilkinsondivider are also very common. All of these dividers are T-junction dividers; some physical implementationsare shown below.

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139

A lossless divider is depicted below.

Z0

1

Z2

jB+

Vo

Yin

Z

input

2 output

1 ou

tput

The susceptance B accounts for the physical discontinuity in the waveguide or transmission line. Wehave

Yin = jB +1

Z1+1

Z2; (6.16)

and we want

Yin =1

Z0: (6.17)

If we ignore B then

1

Z1+1

Z2=1

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 beused to provide a good impedance match.

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Some drawbacks to this type of divider are that there is no isolation between output ports, and lookinginto an output port one sees an impedance mismatch.To determine impedance values for a given power split, the following procedure is used:

Pin =1

2

v20Z0

(6.19)

since the divider is match looking into the input, and

P1 =1

2

v20Z1; P2 =

1

2

v20Z2: (6.20)

If I want P1 = xPin and P2 = (1� x)Pin, then

P1 =1

2

v20Z1

= xPin = x1

2

v20Z0;! x =

Z0Z1

(6.21)

P2 =1

2

v20Z2

= (1� x)Pin = (1� x)1

2

v20Z0;! 1� x = Z0

Z2:

Example 6.3 Determine the normalized output impedances for a 3 : 1 power split.Solution: We want

P1 =3

4Pin; P2 =

1

4Pin; (6.22)

) x =3

4=Z0Z1

) Z1 =4

3Z0;

1� x = 1

4=Z0Z2

! Z2 = 4Z0:JNTU W

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Chapter 7

Electromagnetic Compatibility andInterference (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 saleany such products until their radiated and conducted emissions have been measured and found to be incompliance.

� 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 4088 - 216 150 44216 - 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�jkr4�r

�sin � (7.1)

E� (r) = jI0Lw2�10�7f

e�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

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142CHAPTER 7. ELECTROMAGNETIC COMPATIBILITY AND INTERFERENCE (EMS/EMI)

For two wires (see �gure),

E� (r) = jLw2�10�7f

�I1e�jk(r+�)

r +�+ I2

e�jk(r��)

r ��

�; (7.2)

where

by � br = sin � sin� (7.3)

� =s

2sin � sin�; (7.4)

sincecos (90� �) = sin �

I1

z

y

r

s/2

s

∆

I2

so that

E� (r) = jLw2�10�7f

e�jkr

r

�I1e

�jk� + I2ejk��: (7.5)

Di¤erential mode currents (I0 = I1 = �I2)

Edm� (r) = I0jLw2�10�7f

e�jkr

r

�e�jk� � ejk�

�(7.6)

= �I0jLw2�10�7fe�jkr

r(2j sin (k�)) (7.7)

= I0Lw4�10�7f

e�jkr

rsin�ks

2sin � sin�

�: (7.8)

Assuming that s� �, sin�k s2 sin � sin�

�' k s2 sin � sin�,

Edm� (r) = I0Lw (2�)210�7

p�"f2

e�jkr

rs sin � sin� (7.9)

= 1:32� 10�14I0Lwf2se�jkr

rsin � sin�; (7.10)��Edm� (r)

�� = 1:32� 10�14r

I0Lwf2s sin � jsin�j (7.11)

Field is maximum for � = � = 90�,��Edm;max� (r)�� = 1:32� 10�14

rI0Lwf

2s: (7.12)

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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)

Ecm� (r) = I0jLw2�10�7f

e�jkr

r

�e�jk� + ejk�

�(7.13)

= I0jLw2�10�7f

e�jkr

r(2 cos (k�)) (7.14)

= jI0Lw4�10�7f

e�jkr

rcos�ks

2sin � sin�

�; (7.15)

Assuming that s� �, cos�k s2 sin � sin�

�' 1

Ecm� (r) = 1:26� 10�6jI0Lwfe�jkr

r(7.16)

jEcm� (r)j = 1:26� 10�6r

I0Lwf (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��Edm;max� (r)

�� = 1:32� 10�143

�20� 10�3

�(1)�50� 106

�2 �1� 10�3

�(7.18)

= 220 �V/m = 47 dB�V/m (7.19)

(above the limit)

jEcm� (r)j = 1:26� 10�6r

I0Lwf (7.20)

=1:26� 10�6

3

�20� 10�3

�(1)�50� 106

�(7.21)

= 420; 000 �V/m (7.22)

(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

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Raise frequency to 1 GHz:��Edm;max� (r)�� = 1:32� 10�14

3

�20� 10�3

�(1)�1� 109

�2 �1� 10�3

�(7.23)

= 88; 000 �V/m = 99 dB�V/m (7.24)

(way over)

jEcm� (r)j = 1:26� 10�6r

I0Lwf (7.25)

=1:26� 10�6

3

�20� 10�3

�(1)�1� 109

�(7.26)

= 8; 400; 000 �V/m !!! (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:

∆

z∆

zVs(z,t)

Is(z,t)

i(z,t) R L

G C

z +

z

z

z z

∆ ∆

∆ ∆

∆

+

v(z+dz,t)

i(z+dz,t)

d2v(z)

dz2� 2v(z) = � (R+ i!L) is(z) +

d vs(z)

dz; (7.28)

d2i(z)

dz2� 2i(z) = � (G+ i!C) vs(z) +

d is(z)

dz;

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 propagationconstant are known as the attenuation constant (�) and the phase constant (�), respectively.

An incident electromagnetic wave induces voltages and currents on the transmission line, which arerepresented by Vs and Is. Faraday�s law can be used to determine Vs. The induced emf around a loop l withsurface S is

emf =

Il

E(r) � dl = �j!ZS

B(r) � n dS: (7.30)

Assume the y�axis is vertical, so that the x� axis is into the paper, and n = bx. The distributed voltagesource for two wires separated by distance s is

V (z) ' �j!Z z+�z=2

z��z=2

Z s

0

Bix(y; z) dydz ' �j!�zZ s

0

Bix(y; z) dy (7.31)

V (z)

�z= Vs (z) = �j!�

Z s

0

Hix(y; z) dy:

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145

Since the wire separation s is much less than �,

Vs (z) = �j!�Z s

0

Hix(y; z) dy ' �j!�sHi

x(z) (7.32)

An incident, vertically polarized electric �eld will induce a voltage between the wires, causing a displace-ment current through the capacitance between the wires (we ignore here conduction current due to a lossyinsulator). Using Ohm�s law,

V (z) =

Z s

0

Eiy (y; z) dy = I (z)Zc =I (z)

j!C(7.33)

Is (z) = �j!CZ s

0

Eiy(y; z) dy ' �j!CsEiy(z):

Rather than solve (7.28), we can simply things further by assuming that L � �, so that we only needone section of line. Replacing �z by Lw and setting R = G = 0,

VL =RLLw

RL +Rs � !2RLLC + j! (RsRLC + L)(Vs +RsIs) (7.34)

=�j!RLLws

RL +Rs � !2RLLC + j! (RsRLC + L)��Hi

x +RsCEiy

�(7.35)

Note that the area of the loop formed by the two wires, A = Lw � s is important.If frequency is su¢ ciently low,

VL =RLLwRL +Rs

(Vs +RsIs) (7.36)

=�j!ARLRL +Rs

��Hi

x +RsCEiy

�:

Example:Assume that the plane wave

E = bzEze�jky; H = bxEz�e�jky (7.37)

is incident on a two-wire transmission, with Ez = 1 V/m and ! = 30 MHz. The wires have center-to-centerspacing D = 1:22 cm, radius a = 0:1 cm (this is the standard 300 ohm twin-lead line), and length Lw = 1m. In this case,

L =�0�cosh�1

�D

2a

�; C =

�"0

cosh�1�D2a

� ; (7.38)

and for this line L = 1 �H and C = 11:2 pF. For this incident wave there is only a voltage source,

Vs = �j!�sHix = �j1:22 mV (7.39)

such that�Hi

x = 4� � 10�71

377= 3:33� 10�9: (7.40)

Assume RL = 1000 , Rs = 10 . Then,

VL =�j!ARLRL +Rs

��Hi

x + 0�= �j1:21 mV. (7.41)

If, instead, the incident plane wave was

E = byEye�jkx; H = bzEy�e�jkx (7.42)

thenRsCE

iy = 1:12� 10�10; (7.43)

and

VL =�j!ARLRL +Rs

�0 +RsCE

iy

�= �j0:04 mV. (7.44)

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Chapter 8

Microwave Filters

147

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148 CHAPTER 8. MICROWAVE FILTERS

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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

(chain rule)

=�v��0�� 1

vp

�(9.3)

= �L@i(z; t)@t

;

leading to

@i(z; t)

@t=

1

Lvp

�v+0 (t� z=vp)� v�0 (t+ z=vp)

�; (9.4)

) i (z; t) =1

Lvp

�v+ (t� z=vp)� 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

@

@zi (z; t) =

1

Lv2p

��v+0 (t� z=vp)� v�0 (t+ z=vp)

�+@

@zc (z) (9.7)

= �C @v(z; t)@t

:

With

@v�(z; t)

@t=@v� (t� z=vp)@ (t� z=vp)

@ (t� z=vp)@t

(9.8)

=�v��0(1)

149

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150 CHAPTER 9. APPENDIX

we have

@

@zi (z; t) = � 1

Lv2p

�v+0 (t� z=vp) + v�0 (t+ z=vp)

�+@

@zc (z) (9.9)

= �C @v(z; t)@t

= �C�v+0 (t� z=vp) + v�0 (t+ z=vp)

�:

Therefore, �� 1

Lv2p+ C

�| {z }

0

�v+0 (t� z=vp) + v�0 (t+ z=vp)

�+@

@zc (z) = 0; (9.10)

such that@

@zc (z) = 0) c (z) = constant. (9.11)

c (z) = c provides at most a d.c. current that provides a d.c. o¤set to the time-dependent current. We will

ignore this o¤set (set c = 0) to arrive at, upon noting that Lvp = L 1pLC

=q

LC = Z0, the desired expression

i (z; t) =1

Z0

�v+ (t� z=vp)� v� (t+ z=vp)

�: (9.12)

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Education

Microwave Engineering