Transmission Line(KABADI)

7/30/2019 Transmission Line(KABADI)

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DAR ES SALAAM INSTITUTE OF TECHNOLOGY

DEPARTMENT OF ELECTRONICS AND

TELECOMMUNICATIONS ENGINEERING

TRANSMISSION LINES LECTURES


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BY M.D.KABADI (2007)

2


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2.0 STUDY COMPONENTS

2.1 OBJECTIVES

General objectives:

The purpose of this course is to familiarize the students with the basics of transmission lines

principles. Other objectives are:

List the types of transmission lines

Calculate the impedance of transmission lines

Calculate velocity of propagation and delay factor

Analyze wave propagation and reflection for various line configuration

Describe how standing waves are produced

Use the Smith chart to find input impedance

Use the Smith chart to match loads to lines

Critical Sub enabling learning outcomes:

The following sub enabling outcomes are addressed in this module, i.e. at the conclusion of

this module the student will be able to:

Ability to choose group materials Ability to assist in communications developments

Ability to determine load

Ability to identify technological aspects of products Ability to analyse technological aspects of processes

Ability to advise on technological aspects of particular materials

Ability to advise on technological aspects of particular products


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2.2 MODULE STRUCTURE

Study themes

1. Traditional Transmission lines

S/No Topic Lecture Time Hours Week #Lecture

No.

2.1 Definition and types of Transmission

lines6 1-3 1

2.2 Low-loss ,loading and reflection on

transmission line3 4 2

2.2 Traveling and Standing wave concepts,

quarter and half wave section1

52

2.3 Assignment 1 1 52.4 Graphical method( smith chart 4 6-8 3

2.5 Test 1 2 7

2.6 Impedance Matching 8 8-9 4

2. Modern transmission lines

2.1 Micro strip and strip lines 2 10 5

2.2 Modern microwave integrated circuits 2 10 5

Assignment 2 11

Test 2 12

3. Wave guides

2.1 Parameters of waveguides 2 13 6

2.2 Propagation Modes in rectangular Vs

Circular.3 13 6

2.2 Practical waveguides and components 3 14 6

2.3 Application power and attenuation

measurements1 15 6

2.4 Final; Examination 16-17 Final Exam

2.5 END OF SEMESTER - -

TABLE OF CONTENTS

Transmission line Lectures. Prepared by M.D.Kabadi 4


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PAGE

LECTURE 1: INTRODUCTION TO TRANSMISSION LINES .................................................................. 5Such transmission lines include ............................................................................................................. 6

ELECTRICAL CHARACTERISTICS OF TRANSMISSION LINES ................................................... 14

ATTENUATION PER UNIT LENGTH ............................................................................................. 14VELOCITY FACTOR ........................................................................................................................15

ELECTRICAL LENGTH ................................................................................................................... 16

LECTURE 2. TRANSMISSION LINE THEORY ..................................................................................... 182.1 Lumped Element Circuit Model ......................................................................................................... 18

j Leq = j Z0 tan(l) .....................................................................................................................................40

1/(j Ceq) = j Z0 tan(l) ...............................................................................................................................40Time Domain approach ........................................................................................................................ 45

LECTURE 3. SMITH CHART OR GRAPHICAL METHOD .................................................................... 47LECTURE 4: IMPEDANCE MATCHING ................................................................................................. 58

Impedance matching methods.................................................................................................................. 58

LECTURE5: PRINTED CIRCUIT TRANSMISSION LINES .................................................................. 71Microstripline ........................................................................................................................................72

Advantages and disadvantages of stripline ........................................................................................... 79

Stripline equations ................................................................................................................................ 80LECTURE 6: WAVEGUIDES .....................................................................................................................81

Mode ..................................................................................................................................................... 89

REFERENCES ............................................................................................................................................. 95

LECTURE 1: INTRODUCTION TO TRANSMISSION LINESGeneral Considerations



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In its simplest form, a transmission line is a pair of conductors linking together two electrical

systems, components, or devices.

Otherwise stated, transmission line conducts electronic signals from a source component(Sinusoidal signal, Digital pulse or any signal waveform) to a load component.

Such transmission lines include

parallel wires ortwo-wire line: This transmission line consists of a pair of parallel conducting wires

separated by a uniform distance. (telephone wires, power line ) coaxial transmission line : This consists of an inner conductor and a coaxial outer conductor sheath(cover) separated by a dielectric medium. It carries audio and video information to TV sets, or digital

data to computer monitors

parallel plate line: It consists of two parallel plates separated by a dielectric slab of a uniformthickness (Striplines,)

optical fiber ( carrying light waves)

waveguides (for microwave signals f> 1 GHz):it is usually metallic tubes with rectangular, circular,In short,A transmission line is a two port network connecting a generator circuit at the sending end to a

load at the receiving end

TERMINOLOGY



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All transmission lines have two ends (The end of a two-wire transmission line connected to a source is

ordinarily called the INPUT END or the GENERATOR END. Other names given to this end are

TRANSMITTER END, SENDING END, and SOURCE. The other end of the line is called the OUTPUTEND or RECEIVING END. Other names given to the output end are LOAD END and SINK.

Figure 1. - Basic transmission line.

You can describe a transmission line in terms of its impedance. The ratio of voltage to current (E in/Iin) atthe input end is known as the INPUT IMPEDANCE (Z in). This is the impedance presented to the

transmitter by the transmission line and its load, the antenna. The ratio of voltage to current at the output

(E out/Iout) end is known as the OUTPUT IMPEDANCE (Zout). This is the impedance presented to the loadby the transmission line and its source. If an infinitely long transmission line could be used, the ratio of

voltage to current at any point on that transmission line would be some particular value of impedance.

This impedance is known as the CHARACTERISTIC IMPEDANCE.

TYPES OF TRANSMISSION MEDIUMS

Types of TRANSMISSION MEDIUMS in its electronic applications. Each medium (line or waveguide)

has a certain characteristic impedance value, current-carrying capacity, and physical shape and is designedto meet a particular requirement.

The five types of transmission mediums that we will discuss in this chapter include PARALLEL-LINE,TWISTED PAIR, SHIELDED PAIR, COAXIAL LINE, and WAVEGUIDES. The use of a particular line

depends, among other things, on the applied frequency, the power-handling capabilities, and the type of

installation.

Two-Wire Open Line

One type of parallel line is the TWO-WIRE OPEN LINE illustrated in figure 2. This line consists of two

wires that are generally spaced from 2 to 6 inches apart by insulating spacers. This type of line is mostoften used for power lines, rural telephone lines, and telegraph lines. It is sometimes used as a

transmission line between a transmitter and an antenna or between an antenna and a receiver. Anadvantage of this type of line is its simple construction. The principal disadvantages of this type of line are

the high radiation losses and electrical noise pickup because of the lack of shielding. Radiation losses are

produced by the changing fields created by the changing current in each conductor.

Figure 2. - Parallel two-wire line.



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Another type of parallel line is the TWO-WIRE RIBBON (TWIN LEAD) illustrated in figure 3. This typeof transmission line is commonly used to connect a television receiving antenna to a home television set.

This line is essentially the same as the two-wire open line except that uniform spacing is assured by

embedding the two wires in a low-loss dielectric, usually polyethylene. Since the wires are embedded inthe thin ribbon of polyethylene, the dielectric space is partly air and partly polyethylene.

Figure 3. - Two-wire ribbon type line.

Twisted Pair

The TWISTED PAIR transmission line is illustrated in figure 4. As the name implies, the line consists oftwo insulated wires twisted together to form a flexible line without the use of spacers. It is not used for

transmitting high frequency because of the high dielectric losses that occur in the rubber insulation. When

the line is wet, the losses increase greatly.

Figure 4. - Twisted pair.



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Shielded Pair

The SHIELDED PAIR, shown in figure 5, consists of parallel conductors separated from each other and

surrounded by a solid dielectric. The conductors are contained within a braided copper tubing that acts asan electrical shield. The assembly is covered with a rubber or flexible composition coating that protects

the line from moisture and mechanical damage. Outwardly, it looks much like the power cord of a

washing machine or refrigerator.

Figure 5. - Shielded pair.

The principal advantage of the shielded pair is that the conductors are balanced to ground; that is, the

capacitance between the wires is uniform throughout the length of the line. This balance is due to the

uniform spacing of the grounded shield that surrounds the wires along their entire length. The braided

copper shield isolates the conductors from stray magnetic fields.

Coaxial Lines

There are two types of COAXIAL LINES, RIGID (AIR) COAXIAL LINE and FLEXIBLE (SOLID)

COAXIAL LINE. The physical construction of both types is basically the same; that is, each contains two

concentric conductors.

The rigid coaxial line consists of a central, insulated wire (inner conductor) mounted inside a tubular outerconductor. This line is shown in figure 6. In some applications, the inner conductor is also tubular. The

inner conductor is insulated from the outer conductor by insulating spacers or beads at regular intervals.

The spacers are made of pyrex, polystyrene, or some other material that has good insulating characteristicsand low dielectric losses at high frequencies.

Figure 6. - Air coaxial line.



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The chief advantage of the rigid line is its ability to minimize radiation losses. The electric and magneticfields in a two-wire parallel line extend into space for relatively great distances and radiation losses occur.

However, in a coaxial line no electric or magnetic fields extend outside of the outer conductor. The fields

are confined to the space between the two conductors, resulting in a perfectly shielded coaxial line.Another advantage is that interference from other lines is reduced.

The rigid line has the following disadvantages: (1) it is expensive to construct; (2) it must be kept dry to

prevent excessive leakage between the two conductors; and (3) although high-frequency losses are

somewhat less than in previously mentioned lines, they are still excessive enough to limit the practicallength of the line.

Leakage caused by the condensation of moisture is prevented in some rigid line applications by the use of

an inert gas, such as nitrogen, helium, or argon. It is pumped into the dielectric space of the line at a

pressure that can vary from 3 to 35 pounds per square inch. The inert gas is used to dry the line when it isfirst installed and pressure is maintained to ensure that no moisture enters the line.

Flexible coaxial lines (figure 7) are made with an inner conductor that consists of flexible wire insulatedfrom the outer conductor by a solid, continuous insulating material. The outer conductor is made of metal

braid, which gives the line flexibility. Early attempts at gaining flexibility involved using rubber insulatorsbetween the two conductors. However, the rubber insulators caused excessive losses at high frequencies.

Figure 7. - Flexible coaxial line.

Because of the high-frequency losses associated with rubber insulators, polyethylene plastic was

developed to replace rubber and eliminate these losses. Polyethylene plastic is a solid substance thatremains flexible over a wide range of temperatures. It is unaffected by seawater, gasoline, oil, and most



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other liquids that may be found aboard ship. The use of polyethylene as an insulator results in greater

high-frequency losses than the use of air as an insulator. However, these losses are still lower than the

losses associated with most other solid dielectric materials.

WAVEGUIDES

The WAVEGUIDE is classified as a transmission line. However, the method by which it transmits

energy down its length differs from the conventional methods. Waveguides are cylindrical, elliptical, orrectangular (cylindrical and rectangular shapes are shown in figure 8). The rectangular waveguide is used

more frequently than the cylindrical waveguide.

Figure 8. - Waveguides.

The term waveguide can be applied to all types of transmission lines in the sense that they are all used to

guide energy from one point to another. However, usage has generally limited the term to mean a hollowmetal tube or a dielectric transmission line. In this chapter, we use the term waveguide only to mean

"hollow metal tube." It is interesting to note that the transmission of electromagnetic energy along awaveguide travels at a velocity somewhat slower than electromagnetic energy traveling through free

space.

A waveguide may be classified according to its cross section (rectangular, elliptical, or circular), or

according to the material used in its construction (metallic or dielectric). Dielectric waveguides are seldomused because the dielectric losses for all known dielectric materials are too great to transfer the electric

and magnetic fields efficiently.

Waveguide Advantages

1. The large surface area of waveguides greatly reduces COPPER (I2R) LOSSES. Two-wire transmission

lines have large copper losses because they have a relatively small surface area. The surface area of the

outer conductor of a coaxial cable is large, but the surface area of the inner conductor is relatively small.At microwave frequencies, the current-carrying area of the inner conductor is restricted to a very small



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layer at the surface of the conductor by an action called SKIN EFFECT. Skin effect tends to increase the

effective resistance of the conductor.

2. DIELECTRIC LOSSES are also lower in waveguides than in two-wire and coaxial transmission lines.Dielectric losses in two-wire and coaxial lines are caused by the heating of the insulation between the

conductors. The insulation behaves as the dielectric of a capacitor formed by the two wires of thetransmission line. A voltage potential across the two wires causes heating of the dielectric and results in a

power loss

3. The dielectric in waveguides is air, which has a much lower dielectric loss than conventional

insulating materials. However, waveguides are also subject to dielectric breakdown caused by standing

waves. Standing waves in waveguides cause arcing which decreases the efficiency of energy transfer andcan severely damage the waveguide. Also since the electromagnetic fields are completely contained

within the waveguide, radiation losses are kept very low.

4. Power-handling capability is another advantage of waveguides. Waveguides can handle more power

than coaxial lines of the same size because power-handling capability is directly related to the distancebetween conductors. Figure 4 illustrates the greater distance between conductors in a waveguide.

Figure 4. - Comparison of spacing in coaxial cable and a circular waveguide.

In view of the advantages of waveguides, you would think that waveguides should be the only type of

transmission lines used. However, waveguides have certain disadvantages that make them practical foruse only at microwave frequencies

.

Waveguide Disadvantages

1. Physical size is the primary lower-frequency limitation of waveguides. The width of a waveguide must

be approximately a half wavelength at the frequency of the wave to be transported. For example, awaveguide for use at 1 megahertz would be about 500 feet wide. This makes the use of waveguides at

frequencies below 1000 megahertz increasingly impractical. The lower frequency range of any system

using waveguides is limited by the physical dimensions of the waveguides.



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2. Waveguides are difficult to install because of their rigid, hollow-pipe shape. Special couplings at the

joints are required to assure proper operation. Also, the inside surfaces of waveguides are often plated

with silver or gold to reduce skin effect losses. These requirements increase the costs and decrease thepracticality of waveguide systems at any other than microwave frequencies.

LOSSES IN TRANSMISSION LINES

The discussion of transmission lines so far has not directly addressed LINE LOSSES; actually some line

losses occur in all lines. Line losses may be any of three types - COPPER, DIELECTRIC, andRADIATION or INDUCTION LOSSES.

NOTE: Transmission lines are sometimes referred to as rf lines. In this text the terms are used

interchangeably.

Copper Losses

One type of copper loss is I2

R LOSS. In rf lines the resistance of the conductors is never equal to zero.Whenever current flows through one of these conductors, some energy is dissipated in the form of heat.

This heat loss is a POWER LOSS. With copper braid, which has a resistance higher than solid tubing, this

power loss is higher.

Another type of copper loss is due to SKIN EFFECT. When dc flows through a conductor, the movementof electrons through the conductor's cross section is uniform. The situation is somewhat different when ac

is applied. The expanding and collapsing fields about each electron encircle other electrons. This

phenomenon, called SELF INDUCTION, retards the movement of the encircled electrons. The fluxdensity at the center is so great that electron movement at this point is reduced. As frequency is increased,

the opposition to the flow of current in the center of the wire increases. Current in the center of the wire

becomes smaller and most of the electron flow is on the wire surface. When the frequency applied is 100megahertz or higher, the electron movement in the center is so small that the center of the wire could beremoved without any noticeable effect on current. You should be able to see that the effective cross-

sectional area decreases as the frequency increases. Since resistance is inversely proportional to the cross-

sectional area, the resistance will increase as the frequency is increased. Also, since power loss increasesas resistance increases, power losses increase with an increase in frequency because of skin effect.

Copper losses can be minimized and conductivity increased in an rf line by plating the line with silver.

Since silver is a better conductor than copper, most of the current will flow through the silver layer. The

tubing then serves primarily as a mechanical support.

Dielectric Losses

DIELECTRIC LOSSES result from the heating effect on the dielectric material between the conductors.

Power from the source is used in heating the dielectric. The heat produced is dissipated into the

surrounding medium. When there is no potential difference between two conductors, the atoms in thedielectric material between them are normal and the orbits of the electrons are circular. When there is a

potential difference between two conductors, the orbits of the electrons change. The excessive negative

charge on one conductor repels electrons on the dielectric toward the positive conductor and thus distorts



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the orbits of the electrons. A change in the path of electrons requires more energy, introducing a power

loss.

The atomic structure of rubber is more difficult to distort than the structure of some other dielectricmaterials. The atoms of materials, such as polyethylene, distort easily. Therefore, polyethylene is often

used as a dielectric because less power is consumed when its electron orbits are distorted.

Radiation and Induction Losses

RADIATION and INDUCTION LOSSES are similar in that both are caused by the fields surrounding the

conductors. Induction losses occur when the electromagnetic field about a conductor cuts through anynearby metallic object and a current is induced in that object. As a result, power is dissipated in the object

and is lost.

Radiation losses occur because some magnetic lines of force about a conductor do not return to the

conductor when the cycle alternates. These lines of force are projected into space as radiation and these

results in power losses. That is, power is supplied by the source, but is not available to the load.

Table: Some typical media loss values

ELECTRICAL CHARACTERISTICS OF TRANSMISSION LINES

Transmission lines are generally characterized by the following properties:

Attenuation per unit length

Velocity factorElectrical length

Characteristic impedance

ATTENUATION PER UNIT LENGTH

Attenuation per unit length measures how much of the RF signal is lost per unit length of transmission

line. Typically, the attenuation per unit length has units of dB/100ft. Losses in transmission lines arisefrom 3 sources:

Radiation (leakage)

Dielectric losses



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Skin effect losses

Radiation loss occurs in two-wire lines because the fields from one line do not completely cancel outthose from the other line. A small amount of RF is radiated, which is dependent on the separation of the

wires and the frequency of the RF. Radiation occurs in braided coaxial lines because the braid does not

provide 100% shielding. Special types of coax with multiple braids, or a solid outer conductor have nomeasurable radiation losses.

The conductors of a transmission line are separated from one another by an insulating material known as adielectric. This dielectric could be air or a plastic or any insulating material. All dielectrics exhibit losses

that increase as the voltage on the conductors increases. Dielectric losses also increase with increasing

frequency.

Skin effect is a phenomenon that occurs in conductors carrying an AC current. As the frequency increases,

the current tends to be concentrated near the surface of the conductor. At RF, almost no current flows

down the center of wire. It is all on the surface. A copper rod and a copper tube of equal diameter will

have the same resistance above a few MHz, even though the DC resistance of the solid rod is much lower.As frequency increases, the skin effect becomes more pronounced and the loss in conductors increases

dramatically.

The total losses in a transmission line are roughly proportional to the square root of the frequency. If the

attenuation per unit length is known for a particular frequency f1, the loss at any other frequency f2 can beestimated from the following equation:

f2 = attenuation at frequency f2

f1 = attenuation at frequency f1

VELOCITY FACTOR

The radio frequency (RF) current flowing along a transmission line creates a radio wave that is guided bythe transmission line. This guided wave propagates along a transmission line with a velocity given by the

following equation:

Where:

v = the wave velocity

LS is the series inductance per unit lengthCP is the parallel capacitance per unit length

The wave propagation velocity of the guided wave will always be less than the speed of light in a vacuum,which is approximately 300,000,000 m/sec.



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Because the wave velocity is a very large number, manufacturers of transmission lines generally specify

the velocity factor of a transmission line. The velocity factor is simply the wave velocity on thetransmission line divided by the speed of light in a vacuum:

Where:

vf = the velocity factor

LS is the series inductance per unit lengthCP is the parallel capacitance per unit length

c is the speed of light in a vacuum (3.0 * 108 m/sec)

Velocity factors for commercially available transmission lines range from approximately 0.6 to 0.9,

depending on the construction of the line.

ELECTRICAL LENGTH

The electrical length of a cable is its length measured in wavelengths ( ) and is related to the frequency ofthe wave and the velocity with which it propagates along the transmission line. The electrical length of a

transmission line can be computed from the following formula:

l = length of the line in feet

f = frequency in MHz

VF = the velocity factor of the line

The velocity factor is the ratio of the wave velocity to the speed of light. Typical values range from 0.66 to

0.97.

Lets look at an example:

What is the electrical length of 117 feet of RG-8/U coaxial cable at 57 MHz? The velocity factor for this

cable is 0.66.

Solution:

Here is a second example:

A two-wire line has a velocity factor of 0.95 and a length of 3406 ft. What is its electrical length at 2.82

MHz?



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Solution:

Notice that these two transmission lines of very different design and physical length have the same

electrical length.

The concept of electrical length is important because the properties of a resonant transmission line are

periodic with respect to electrical length



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LECTURE 2. TRANSMISSION LINE THEORY

2.1 Lumped Element Circuit Model

When we draw a schematic of an electronic circuit, we use symbols to represent resistors, capacitors,

inductors, diodes, etc. In each case the symbol represents the functionality of the device, rather than its

shape or other attributes. We shall do the same with regard to the transmission lines

A transmission line is represented by a parallel wire configuration, regardless of the specific shape of the

line under consideration.

Same approach will be applied to transmission line

1- Orient the line along the z-direction

2- Subdividing the line into differential sections each of length


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This representation is called the lumped-element circuit model and it is applicable to all TEM

transmission lines. This model consists of four basic elements: two series elements, R and L, and two

shunt elements, G and C.

R: The combined resistance of both conductors per unit length, in /m,

L: The combined inductance of both conductors per unit length, in H/m,

G: The conductance of the insulation medium per unit length, in S/m,

C: The capacitance of both conductors per unit length, in F/m.

R, L, G, and C are called the transmission line parameters.

Expressions for the line parameters R, L, G, and C are given in Table 2-1 for the three types of TEM

transmission lines. They depend on the geometry and characteristics of the materials.

Parameter Coaxial Two wire Unit



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R

)b

1

a

1(

2

Rs +a

Rs

/m

L )a

b(ln2

]1)2(

a2d[ln 2 +

ad H/m

G

ln(b/a)

2]1)2/()2/ln[( 2 + adad

S/m

C

ln(b/a)

2

]1)2/()2/ln[( 2 + adad

F/m

For illustration purpose, lets consider a small section of a coaxial line. It consists of an inner conductor of

radius a separated from an outer conducting cylinder of radius b by a material with permittivity ,

permeability , and conductivity . The two metal conductors are made of a material with conductivity c

and permeability c.

The expression for R will be derived later and it is given by

R= )b

1

a

1(

2

Rs + (/m)



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where Rs represents the surface resistance of the conductor , it is called the intrinsic resistance, and is

given by

Rs =

c

cf

()

Rs depends on the materials properties of the conductor (c and c) and the frequencyf.

For a perfect conductor, c= 0 Rs = 0 and R=0,

Application of Amperes law in Chapter 5 to the definition of inductance leads to the following expression

L = )a

b(ln

2(H/m)

The shunt conductance per unit length G accounts for the current flow between the outer and inner

conductors, made possible by the material conductivity of the insulator. It is given by

G=

ln(b/a)

2(S/m)

If the material separating the inner and outer conductor is a perfect dielectric, = 0, then G=0.

The last line parameter on our list is the capacitance per unit length C, it is due to the presence of

opposite charges on two noncontacting conductors. It is given by

C=

ln(b/a)

2(F/m)

All TEM transmission lines share the following useful relations:

And


L C =

G/C =/


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2.2 Transmission-Line Equations

Equivalent circuit of a differential length zof two-conductor transmission line

Where:

R= series resistance per unit length ( )m/ of the transmission line conductors.

L=series inductance per unit length ( )mH/ of the transmission line conductors (internal

plus external inductance).

G=shut conductance per unit length ( )mS/ of the media between the transmission line

conductors (insulator leakage current).

C= shunt capacitance per unit length ( )mF/ of the transmission line conductors.

We may relate the values of voltage and current at z and +z z by writing KVL and KCL equations for

the equivalent circuit.

KVL

( ) ( ) ( ) ( )tzzVttzIzLtzzIRtzV ,,,, +=

KCL ( ) ( )( )

( )tzzIt

tzzVzCtzzzVGtzI ,

,,, +=

+

+



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Grouping the voltage and current terms of terms and dividing by z gives

( )( ) ( ) ( )

z

tzVtzzV

t

tzILtzzIR

+

=

,,,,

( ) ( ) ( ) ( )z

tzItzzIt

tzzVCtzzzVG

+=

++ ,,,,

Taking the limit as ,0 z the terms on the right hand side of the equations above become partialderivatives with respect to z which gives

( )( )

( )

t

tzILtzRI

z

tzV

=

,,

,

( )( )

( )

t

tzVCtzGV

z

tzI

=

,,

,

For harmonically varying voltage and current, we have

jt

=

The derivatives of the voltage and current with respect to time, it gives

( ) [ ] ( ) ZIzILjRdzzdV =+= and ( ) [ ] ( ) YVzVCjG

dzzdI =+= .

These are called phasor equations.

Where Z= LjR + , the series impedance per unit length of line.CjGY += , the shunt admittance per unit length of line

Even though these equations were derived without any consideration of the electromagnetic fields

associated with the transmission liner ember that circuit theory is based on Maxwells equations.

Just as we manipulated the two Maxwell curl equations to derive the wave equations describing E and H

associated with an unguided wave (plane wave), we can do the same for a guided(transmission line TEM)

wave.

Beginning with the Phasor transmission line equations, we take derivatives of both sides with respect to z.



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Wave propagation on transmission line

Two phasor equations can be solved simultaneously to give wave equation for ( ) ( )zandIzV . That is:

( )[ ]

( )

dz

zdILjR

dz

zVd s

+=2

2

and( )

[ ]( )

dz

zdVCjG

dz

zId+=

2

2

We then insert the first derivates of the voltage and current found in the original Phasor transmission lineequations.

( )[ ][ ] ( ) ( )zVZYzVCjGLjR

dz

zVds ++= 2

2

2

( )[ ][ ] ( ) ( )zYZIzILjRCjG

dz

zIds

s ++= 2

2

If 22

2

=dz

d

The Phasor voltage and current wave equations may be written as

( ) 0)( 2 = zVZY and ( ) 0)( 2 = zIYZ Voltage and current wave equations respectively

This differential equation can be written

.

( ) ( )CjGLjR ++=

Where is the complex propagation constant of the wave on the transmission line given by.

( ) ( )CjGLjRj ++=+=

The real part of the propagation constant ( ) is the attenuation constant while the imaginary part ( ) is

the phase constant. The general equations for and in terms of the per-unit length transmission line

parameters

The general solutions to the voltage and current wave equations are

( ) zoz

o eVeVzV + += and ( ) zo

z

o eIeIzI + +=



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Where the ze term represents wave propagation in +z-direction and ze represents wavepropagation in -z-direction.

Similar from current equation

( ) )( zoz

oo eVeVYzI + =

The characteristic impedance of the line is defined as:

CjG

LjR

Y

Z

YZ

o

o

++===

1

The transmission line characteristic impedance is, in general, complex and can be defined by

ooo jXRZ +=

oR Resistive component of oZ

0X Reactive component of oZ

The voltage and current wave equation can be written in terms of the voltage coefficients and the

characteristic impedance (rather than the voltage and current coefficients using the relationship

0Z

VI oo

+

+ = ,o

o

o

Z

VI

= ando

o

o

o

o

I

VZI

V

+

+

==

( ) zo

z

o eVeVzV + += ( ) )(

1 zo

z

o

o

eVeVZ

zI + =

Wavelength

2= and Phase velocity

fVp ==

Where = angular frequency and phase constant or wave number



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Special Case # 1 Lossless Transmission Line

A lossless transmission line is defined by perfect conductors and a perfect insulator between the

conductors. Thus, the ideal transmission line conductors have zero resistance ( )0,0, === GR while

the ideal transmission line insulating medium has infinite resistance ( ).0,0 == G

The propagation constant on the lossless transmission line reduces to

( ) ( ) LCjCjGLjRj =++=+=

0= LC =

Given the purely imaginary propagation constant, the transmission line equations for the lossless line are

( ) zjzj eVeVzV 00 + += ( ) [ ]zjzj eVeV

ZzI 00

0

1 + =

The characteristic impedance of the lossless transmission line is purely real and given by

C

L

CjG

LjRZ =

++

=

0

The velocity of propagation and wavelength on the lossless line are

LCVp

1==

and

LCfLC

122 ===

mrad/ = (rad/m)

where and are , respectively, the magnetic permeability and electrical permittivity of the insulatingmaterial separating the conductors.

The phase velocity for the lossless transmission line is independent of frequency. For such case, the

medium is called nondispersive.


Vp = 1

=r

c

(m/s)


27/101

Transmission lines are designed with conductors of high conductivity and insulators of low conductivity

in order to minimize losses. The lossless transmission line model is an accurate representation of an actual

transmission line under most conditions.

Also ,At extremely low frequency.

CGLR > >> > ,

G

R

CjG

LjRZ =

++

=

0

And, at extremely high frequency. CGLR > >> > ,

Then.

C

L

CjG

LjR

Z =+

+

=

0

Example

The parameters of a transmission line are:

pFpermeterandCnHpermeterLmSpermeterGerohmspermetR 23.0,8,5/0,2 ===

If the signal frequency is 1GHZ. Calculate its characteristic impedance Z and propagation constant

CjGLjRZo

++= = ohmsxxxjx

xxxj 1293

99

1023.0102105.01081022 =

++

= ohmxjx

j33 104451.1105.0

2655.502 +

+=

radx

rad

2377.11029.15

531.131.504

= 04.839.181 ohm = ohmj 51.2644.179 +

And ZY= = .)2377.11029.15()531.131.50( 4 radXxrad

= jmjm +=+= 110 2726.00514.031.792774.0

Therefore, ,/0514.0 mNp= and mrad/2726.0=

Class Exercise



28/101

At a frequency of 100MHz, the following values are appropriate for a certain transmission line:./8,/15.0,/80,/25.0 mSandgmRmpFCmHL ==== Calculate (a) the propagation constant,

, j+=(b) The signal wavelength,(c) The phase velocity, and

(d) The characteristic impedance.

Special Case#2. For low loss line- Lossy Transmission LineWhen the loss is small, some approximations can be made that simplify the expressions for the general

transmission line parameters of , j+= and 0z . The general expression for propagation constants is

( ) ( )

( ) ( )

+

+=

++=

Cj

G

Lj

RCjLj

CjGLjR

11

LC

RG

C

G

L

RjLCj

21

+=

For low loss: ,LR and .CG Then LCRG 2

+=

C

G

L

RjLCj

1

,11 x

C

G

L

Rj

+=

+ this term is expressed in Taylor series expansive as ......2/1 ++ x

+

C

G

L

RLCj

2

11

So that:-



29/101

=

=

=

=++

+=

+

CL

CjG

LjR

Lc

Gzz

R

C

LG

L

cR

Z

0

2

1

2

10

0

This is also the same for the high frequency.

Special Case#3 Distortion less Transmission Line

On a lossless transmission line, the propagation constant is purely imaginary and given by

LCjj ==

The phase velocity on the lossless line is

LCv 1==

Note that the phase velocity is a constant (independent of frequency) so that all frequencies propagate

along the lossless transmission line at the same velocity. Many applications involving transmission lines

require that a band of frequencies be transmitted (modulation, digital signals, etc) as opposed to a single

frequency. From Fourier theory, we know that any time-domain signal may be represented as a weighted

sum of sinusoids. A Single rectangular pulse contains energy over a band of frequencies. For the pulse to

be transmitted down the transmission line without distortion, all of the frequency components must

propagate at the same velocity. This is the case on a lossless transmission line since the velocity of

propagation is a constant. The velocity of propagation on the typical non-ideal transmission line is a

function of frequency so that signals are distorted as different components of the signal arrive at the load

at different times. This effect is called dispersion. Dispersion is also encountered when an unguided wave



30/101

propagates in a non-ideal medium. A plane wave pulse propagating in a dispersive medium will suffer

distortion. A dispersive medium is characterized by a phase velocity which is a function of frequency.

For a low-loss transmission line, on which the velocity of propagation is near constant, dispersion may or

may not be a problem, depending on the length of the line. The small variations in the velocity of

propagation on a low-loss line may produce significant distortion if the line is very long. There is a special

case of lossy line with the linear phase constant that produces a distortion less line.

A transmission line can be made distortion less (linear phase constant) by designing the line such that the

per unit-length parameters satisfy

C

G

L

R = (Distortion less line)

Inserting the per-unit length parameter relationship into the general equation for the propagation constant

on a lossy line gives.

( ) ( )CjGLjR ++=

+

+=

GCjG

RLjR 11

2

1

+=

R

LjRG

jR

LjRG +=

+= 1

RG= LCLL

CL

R

G

R

LRG ====

Although the shape of the signal is not distorted, the signal will suffer attention as the wave propagates

along the line since the distortion less line is a lossy transmission line.



31/101

Note that the attenuation constant for a distortion less transmission line is independent of frequency. If this

were not true, the signal would suffer distortion due to different frequencies being attenuated by different

amounts.

In the previous derivation, we have assumed that the per-unit-length parameters of the transmission line

are independent of frequency. This is also an approximation that depends on the spectral content of the

propagating signal. For very wideband signals, the attenuation and phase constants will, in general, both

be functions of frequency.

For most practical transmission lines, we find that RC .GL> In order to satisfy the distortion less line

requirement, series loading coils are typically placed periodically along the line to increase L

Example.

A signal propagating through a 50- distortion less transmission line attenuates at the rate of 0.01 dB per

meter. If this line has a capacitance of 100 pF per meter, find (a) R, (b) L, (c) G, and (d) Vp

Solution

Since the line is distortionless,

500 ==C

LZ

And,

mNpxmNpmdB /1015.1/69.8/01.0/01.0

3

==

Hence

(a) R = mxxC

L/057.0501015.1 3 ==

(b) L = mHMHXCZ /25.0/50102102

0 =

(c) G = mSMsZ

R

L

RC/8.22/

50

0057.022

0

===

(d) smxLC

vP /1021 8==



32/101

A dispersive transmission line is one on which the wave velocity is not constant as a function of the

frequencyf.

Propagation Modes

Transmission line may be classified into two basic types:

Transverse electromagnetic (TEM) transmission lines

Higher-order transmission lines

Transverse electromagnetic (TEM) transmission lines: The electric field and the magnetic field are

perpendicular (transverse) to the direction of propagation. This is called a TEM mode. Example:

coaxial line, two-wire line, and parallel-plate line.

A good example is the coaxial line. The electric field lines are in the radial direction between the inner

and outer conductors. The magnetic field form circles around the inner conductor and it is

perpendicular to E. Both are perpendicular to the direction propagation.



33/101

Higher-order transmission lines: waves propagating along these lines have at least one significant field

component in the direction of propagation. Example: optical fibers, hollow conducting waveguides,



34/101

TERMINATED LOSSLESS TRANSMISSION LINE

For fields having sinusoidal time dependence and steady-state conditions, a field analysis of a terminatedlossless transmission line results in the following relations:

Figure Diagram of lossless transmission line with load Zl

If an incident wave of the form , where is the phase constant or wave number , is incident

from the -zdirection then the total voltage on the line can be written as a sum of incident and reflectedwaves:

( ) [ ]zjzj eeVzV += +0

The total current on the line is

( ) [ ]zjzj eeZ

VzI =

+

0

0

Where is the characteristic impedance of the transmission line .The total voltage and current at the load

are related by the load impedance, at z=0.

0

00

00

0

0 XZVV

VV

I

VZL +

+

+

==

Solving for

0V gives:

( )

( )

+

+

== 00

0

0

0

0 XVZZ

ZZ

I

VV

L

L

The amplitude of the reflected voltage wave normalized to the amplitude of the incident voltage wave is

known as the voltage reflection coefficient,

0

0

0

0

ZZ

ZZ

V

V

L

L

+

== +

where is the load impedance.



35/101

The total voltage and current waves on the line can then be written in terms of the reflection coefficient as

( ) [ ]zjzj eeVzV += +0 And ( ) [ ]zjzj

eeZ

VzI

= +

0

0

From the previous equations we see that the voltage and current on the line are a superposition of anincident and reflected wave. If the system is static, i.e. if and are not changing in time, the

superposition of waves will also be static. This static superposition of waves on the line is called a

standing wave. Note. When 0= , no reflection ,For 0= , the load impedance 0ZZL = of thetransmission line and the load is said to be matched to the line,

STANDING-WAVE RATIO

The measurement of standing waves on a transmission line yields information about equipment operating

conditions. Maximum power is absorbed by the load when ZL = Z0. If a line has no standing waves, the

termination for that line is correct and maximum power transfer takes place.

Because of the complicated shape of this standing wave, the voltage will vary with position along the line,

from some minimum value to some maximum value. The ratio of to is one way to quantify the

mismatch of the line. This mismatch is called the standing wave ratio (SWR) or voltage standing waveratio (VSWR) and can be expressed as

+

==1

1

min

max

V

VSWR



36/101

The SWR is a real number such that 1


37/101

INSERTION LOSSInsertion loss of a device is defined as a ratio of transmitted power ( power available at the output port) tothat of power incident at its input. Since transmitted power is equal to the difference of incident and

reflected powers for lossless device, the insertion loss can be expressed as follows.

Insertion loss of a lossless device= ( )dB210 1log10

TRANSMISSION COEFFICIENTS

It is sometimes useful to define a transmission coefficient may be defined as the ration of the voltage on

the load to the amplitude of the incident voltage. Since

zj

ov

zj eVeVzV += 0)(

The voltage at the load is V(z=0), and it is given by

( ) ( )vVV += 10 0

Since the amplitude of the incident voltage is 0V we have

( )

oL

Lv

o

oZZ

Z

V

V

+=+=

21

0

INPUT IMPEDANCE OF THE LOSSLESS TRANSMISSION LINE.

The variation in impedance along the transmission line caused by the line/load mismatch can be written.

Where is the distance from the load.

If we substitute the expression for in terms of the impedances, the generalized input impedance of theload plus transmission line simplifies to:

ljZZ

ljZZZZ

L

L

in

tan

tan

0

0

0 ++

=

With this equation the impedance anywhere along the line can be calculated if the load impedance andcharacteristic impedance are known.



38/101

In the most basic sense, then, if the load impedance equals the line impedance, the reflection coefficient is

zero and the load is said to be matched to the line. All of the microwave impedance matching techniques

can be reduced to this simple idea: minimize the reflection of the incident wave to as nearly zero aspossible.

Example

A load impendence of + 10050 j terminates a loss less, quarter wavelength-long transmission line. Ifcharacteristic impendence of the line is ,50 the find the impedance at (a) its input end, (b) the loadreflection coefficient (c) the VSWR on this transmission line.

Solution

090

2

1

4

2===

x

a) oZZin = ( ) ( )

( ) ( )

90tan1005050

90tan501005050

tan

tan

0

0

jj

jj

jZZ

jZZ

L

L

++++

=++

( )( )

20101005010050

25005050 j

jjj

jZin =

++=

=

b)

90210090100

100100100

50100505010050

0

0

=+=++ +=+= xjjj

jj

ZZZZ

L

L

= 0457071.0 there is a maximum at the load, and if ZoRL < there is a minimum. The location of voltagemaxima and minima must be at those points when Zin is a pure resistance. Purely resistive input

impedance must occur 0=x (the axis) of the smith chart voltage maximum or current minimum occurwhen 1>r or at wtg 25.0= and voltage minima or current maxima occur when 1r ,In the example the intersection is marked point C, and 2.4= thus .2.4=SWR

Normalized admittance

The smith chart may also be used for normalized admittance, this is

oo jBGZ

y +==0

0

1and jBG

zy +== 1

Then the normalized admittance is

jbgzz

Z

y

yy o

o

+==== 1

Use the r circles as g circles and x circles as b circles. The two differences then are that ,1>g 0=b corresponds to a voltage minimum and 1800 must be added to the angle of as read from the perimeter ofthe chart.

Figure 5 shows a smith chart which superimpose constant resistance, and constant reactance circles

The characteristics of the smith chart are summarized as follows.

1. The constant r and x loci from two families orthogonal.

2. The constant r and x circles all pass through the point 0= i3. The upper half of the diagram represents jx+4. The lower half of the diagram represents jx



54/101

5. For admittance the constant r circles becomes constant g circles, and constant x circles

becomes constant suspectance circles.

6. The distance around the smith chart once is one-half wavelength (

7. At a point of Z min =SWR

1,there is a Vmin out line

8. At a point of Zmix = ,SWR there is a Vmax on the line.9. The horizontal radius to the right of the chart corresponds to Zmax and SWR.

10. The horizontal radius to the left of the chart centre corresponds to Zmix andSWR

1

11. Since the normalized admittance y is a reciprocal of the normalized impendence, Z, the

corresponding quantities in the admittance chart are 180 out phase with the in the impedance.

12. The normalized impedance or admittance is repeated for every half wavelength of distance.13. The distance is given in wavelengths toward the generator and away toward the load.

Example 2

A load impendence of + 10050 j terminates a loss less, quarter wavelength-long transmission line. Ifcharacteristic impendence of the line is ,50 the find the impedance at (a) its input end, (b) the loadreflection coefficient (c) the VSWR on this transmission line.

Solution

0902

1

4

2===

x

a) oZZin = ( ) ( )

( ) ( )

90tan1005050

90tan501005050

tan

tan

0 jj

jj

jZZ

jZZ

L

oL

++++

=++

( )( )

201010050

5050 j

jj

jZin =

+=

b)

902100

90100

100100

100

5010050

5010050

0

0

=+=+++

=+

= xjjj

j

j

ZZ

ZZ

L

L

= 0457071.0



55/101

c) 8283.52929.0

7071.1

7071.01

7071.01

1

1==

+

=+

=VSWR

8283.5=VSWR

Solving the problem using the smith chart (graphical method)

1. Find the normalized impedance i.e50

10050

0

j

Z

Zz L

+==

21 jz +=

2. Locate this point on the smith chart as shown in figure 7.

3. Draw a circle that passes through 21 j+ with point jo+1 as its centre.

4. The radius of this circle is equal to the magnitude of the reflection coefficient

5. The input port of the line that is a quarter wavelengths away from the load i.e /== d islocated clockwise movement on this circle to move away from the lead. This point on circle is

located after moving by to a point at 438.0 on wavelength toward generator scale.4.02.0 jZin =

( )4.02.050 jZin =

= 2010 j

6. Thus from the smith chart

4571.0,8.5 == LandVSWR



56/101

Figure 7. Solution for example 2

Example 3.

Given the normalized load impendence 11 jz += and operating wavelength .5cm= .Determine the firstmax,V first Vmin, from the load.

Solution:



57/101

1. Enter 11 jZ += on the chart as shown in figure 8.

Figure 8. solution for example 3.

2. Read 662.0 on the distance scale by drawing a dashed straight line from the centre of the chartthrough the load point and intersection the distance scale.

3. More a distance from the point at 162.0 towards the generator and first the voltage maximum onthe right hand real axis at .25.0

( ) ( ) 088.0165.025.0max1 == xVdBut ,5cm= i.e ( ) 40.05088.0max1 == xVd

4. Similarly, more a distance from the point of 0.162 toward the generator first stops at the voltage

minimum on the left-hand real axis at 0.5 ;

Then

( ) ( ) ( ) ( ) 690.15338.0162.05.0min2 === Vd cm



58/101

LECTURE 4: IMPEDANCE MATCHING

Impedance matching is the practice of attempting to make the output impedance of a source equal to the

input impedance of the load to which it is ultimately connected, usually in order to maximise the power

transferand minimise reflections from the load. This only applies when both are lineardevices.

Sometimes, in electric circuits, it is not the power transfer that is to be maximised, but the voltage transfer.

In this case the term used is impedance bridging where the load impedance is much larger than the source

impedance.

The concept of impedance matching was originally developed forelectrical power, but has since been

generalised to other fields ofengineering where any form of energy (not just electrical) is transferred

between a source and a load.

WHY IMPEDANCE MATCHNING?

Reflections lead to variations in the input impedance of the line. The input impedance changes

with length and frequency.

Power is wasted. An impedance match provides maximum power taransfer to the load.

VSWR >1 means there will be voltage maxima on the line. These can lead to voltage breakdown at

high power levels.

ADVANTAGES OF MATCHING.

Maximum power is delivered when the load is matched to the line (

The Input Impedance remains constant at the value Zo. Therefore, the input impedance is

independent of length, and frequency( Over the bandwidth of the matching network)

VSWR =1. Therefore there is no voltage peaks on the line.

Impedance matching methods.

Most commonly, matching is performed in one or more of the following techniques:

Matching with lumped elements

http://en.wikipedia.org/wiki/Impedancehttp://en.wikipedia.org/wiki/Maximum_powerhttp://en.wikipedia.org/wiki/Maximum_powerhttp://en.wikipedia.org/wiki/Linearityhttp://en.wikipedia.org/wiki/Impedance_bridginghttp://en.wikipedia.org/wiki/Electrical_powerhttp://en.wikipedia.org/wiki/Engineeringhttp://en.wikipedia.org/wiki/Impedancehttp://en.wikipedia.org/wiki/Maximum_powerhttp://en.wikipedia.org/wiki/Maximum_powerhttp://en.wikipedia.org/wiki/Linearityhttp://en.wikipedia.org/wiki/Impedance_bridginghttp://en.wikipedia.org/wiki/Electrical_powerhttp://en.wikipedia.org/wiki/Engineering


59/101

Matching with tuning stub with short sections of line.

Matching with lumped elements

Matching with lumped elements is done with one or more reactive L-section( a series and shunt

reactance) A lumped element means inductor or capacitors, as opposed to a section of line.At

microwave frequencies, it is not easy to realize a pure inductive or capacitance, because what is an

inducatance at lower frequencies will have a significant capacitive part at higher frequencies, and vice

versa. Hower, at lower frequencies, such as those used for cellurar radio, lumped element matching is

common and ineexpensive.

Examples of matching with lumped elements arrangements.

The T network provides a larger transformation than an L network.



60/101

A Pi network provides independent variation of phase shift and transformation ratio through the circuit.

Matching with tuning stub.

In this technique, the following methods are used:

Single stub matching.

Double stub matching.

Quarter-wave section matchnig.

We will look at examples of single stub and quarter wave section matching.

As long as the load impedance, Zl, has some nonzero real part, a matching network can always be

found. Many choices are available, however, and we will discuss the design and performance of several

types of practical matching networks. Factors that may be important in the selection of a particular

matching network include the following:

Complexity-As with most engineering solutions, the simplest design that satisfies the required

specifications is generally the most preferable. A simpler matching network is usually cheaper,

more reliable, and lossy than a more complex design.



61/101

Bandwidth-Any type of matching network can ideally give a perfect match (zero reflection) at

single frequencies. There are several ways of doing this with ,of course, a corresponding increase

in complexity

Implementation depend on the type of transmission line or waveguide being used, one type of

matching network may be preferable compared to another for example, tuning tubs are much

easier to implement in waveguide than are multi- section quarter-wave transformer

Adjustabilityin some application the matching network may require adjustment to match variable

load impedance. Some types of matching networks are more amenable than others in this regard.

Why stubs? Stubs are shorted or open circuit lengths of transmission line which produce a pure reactance

at the attachment point. Any value of reactance can be made, as the stub length is varied from zero to half

a wavelength.

SINGLE STUB MATCHING

A matching technique that uses a single open circuited or short circuited length of transmission line (a

stub), connected either in parallel or in series with the transmission line (a stub), connected either in

parallel or in series with the transmission feed line at a certain distance from a microwave fabrication

aspect, since lumped elements are not required. The shunt tuning stub is especially easy to fabricate in

micro strip-line or stripline form.

In single stub tuning, the two adjustable parameters are the distance, d, from the load to the stub

position, and the value of susceptance or reactance provided by the shunt or series stub. For the shunt-stub

case, the basic idea is to select d so that the admittance, y, seen looking into the line at distance d from the

load is of the form Yo+jB.Then

.the stub susceptance is chosen as-jb, resulting in a matched condition. For the series stub case, the

distance d is of the form zo+jx. Then the stub reactance is chosen as-jx, resulting in a matched condition.



62/101

The proper length of open or shorted transmission line can provide any desired value of reactance or

susceptance. For a given susceptance or reactance, the difference in lengths of an open or short circuited

stub is /4. For transmission line media such as micro strip or strip line, open circuited stubs are easier to

fabricate since a via hole through the substrate to ground plane is not needed. For lines like coax or

waveguide, however, short-circuited stubs are usually preferred, because the cross-sectional area of such

open-circuited line may be large enough (electrically) to radiate, in which case the stub is no longer purely

reactive

FIGURE1. Single- stub tuning circuits. (a) Shunt stub (b). Series stub

Graphical method (Smith chart method)

These networks can also be graphically designed using a smith chart. The procedure is similar

for both series as well as shunt-connected element; except that the former is based on the

normalized impedance while the letter works with normalized admittance. It can be

summarized in the following steps.



63/101

1. Determined the normalized impedance of the load and locate that point on the smith chart

2. Draw the constant VSWR circle. If the stub needs to be connected in parallel, move quarter-

wavelength away from the load impedance point. This point is located at the other end of the

diameter that connects the load point with the center of the circle. For a series-stub, at the

normalized impedance point

From the point found in step 2, move toward the generator (clockwise) on the VSWR circle until it

intersects the unity resistance (or conductance) circle. Distance traveled to at this intersection point from

the load is equal to d s. There will be at least two such point within one-half wavelength from the load. A

matching element can be placed at either one of these points.

4. If the admittance in the previous step is 1+jb, then a susceptance of jb in shunt is needed for

matching. This can be a discrete reactive element (inductor or capacitor, depending upon a

negative or positive susceptance value) or a transmission line stub.

5. In the case of a stub, the required length is determined as follows. Since its other end will have an

open or a short, VSWR on it will be infinite. It is represented by the outermost circle of the smith

chart. Locate the desired susceptance point (i.e,0-jb)on this circle and then move toward load

(counterclockwise)until an open circuit (i.e. ,a zero susceptance)or a short circuit (an infinite

susceptance)is found .This distance is equal to the stub length ls.

For a series reactive element or stub, steps 4 and 5 will be same except that the normalized reactance

replaces the normalized susceptance.

Single Shunt Stub Tuner Design Procedure

1. Locate normalized load impedance and draw VSWR circle (normalized load admittance point is 0180

from the normalized impedance point).

2. From the normalized load admittance point, rotate CW (toward generator) on the VSWR circle until

it intersects the 1=r circle. This rotation distance is the length d of the terminated section of t-tline. The

normalized admittance at this point is .1 jb+

3. Beginning at the stub end (rightmost smith chart point is the admittance of a short-circuit, leftmost

smith chart point is the admittance of a short-circuit, leftmost smith chart point is the admittance of an



64/101

open-circuit),rotate CW(toward generator) until the point at jb0 is reached. This rotation distance is the

stub length .l

Single Series Stub Tuner Design Procedure

1. Locate normalized load impedance and draw VSWR circle.

2. From the normalized load impedance point, rotate CW (toward generator) on the VSWR circle until it

intersects the 1=r circle. This rotation distance is the length dof the terminated section of t-tline . The

normalized impedance at this point is .1 jx+

3. Beginning at the stub end (leftmost Smith chart point is the impedance of a short-

circuit, rightmost smith chart point is the impedance of an open circuit), rotate CW (toward generator)

until the point at jx0 is reached. This rotation distance is the stub length .l

Example 1. . Single-stub shunt Tuning

A uniform lossless 100 ohm line is connected to a load of 50 j75 ohm, as illustrated in figure below.

A single stub of 100-ohm characteristic impedance is connected in parallel at distance ds from the

load. Find the shortest values of ds and stub length ls for a match.



65/101

The following steps are needed for solving this example graphically with the smith chart.

1. Determine the normalized load admittance. Zl=50-j75/100=

0.5-j0.75.

2. Locate the normalized load impedance point on the smith chart. Draw the VSWR circle as shown

in figure below.

3. From the load impedance point, move to the diametrically opposite point and locate thecorresponding normalized load admittance. It is point 0.62+j0.91 on the chart.

4. Locate the point on the VSWR circle where the real part of the admittance is unity. There are two

such points with normalized admittance values 1+j1.3(say, point A)and 1-j1.3(say, point

B)respectively.

5. Distance ds of 1+j1.3 (point A) from the load admittance can be determined as 0.036 (i.e.,0.17

-0.134 )and for point B (1-j1.3)as 0.195 (i.e.,0.329-0.134 ).

6. If a susceptance of j1.3 is added at point A or j1.3 at point B the load will be matched.

7. Locate the point j1.3 along the lower circumference of the chart and from there move toward the

load (counterclockwise) until the short circuit the short circuit (infinity on the chart) is reached.

Separation between these two points is as 0.25 -0.146 =0.104 . Hence a 0.104- long

transmission line with a short circuit at its real end will have the desired susceptance for point A.

8. For a matching stub at point B, locate the point j1.3 on the upper circumference of the chart and

then move toward the load up to the short circuit (i.e., the right hand end of the chart ).hence the

stub length ls for this case is determined as 0.025+0.146 =0.396 .

Therefore, a 0,104 -long stub at 0.036 from the load (point A) or a 0.396 long stub at 0.195

(point B) from the load will match the load. Point A is preferred over point B in matching network

design because it is closer to the load and also the stub length in this case is shorter.



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Figure 2. Graphical design of matching circuit for example 1.



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Example 2. Single-stub series tuning

For a load impedance ZL=100+j80 , design two single-stub series tuning networks to match this

load to a 50 line.

Solution

1. Determine the normalized load admittance. Zl=100+j80/50=

2+j1.6.

2. Plot the normalized load impedance ZL=2+j1.6, Construct the appropriate SWR circle.

For series- stub design, the chart is impedance chart.

3. SWR circle intersection the 1+ jx circle at two points, denoted as z1 and z2 as shown

in figure 4

4. Thus the shortest distance d1, from the load to the stub is from the WTG scale

d1=0.328-0.208 =0.120 while d2= (0.5-0.208) +0.172=0.463 .

1. As in the shunt stub case, additional rotations around the SWR circle lead to additional

solutions, but these are usually not of practical interest. The normalized impedance s at the

two intersection points are z1=1-j1.33, z2 = 1 +j1.33

2. The first solution requires a stub with a reactance of j1.33. The length of an open circuited

stub that gives this reactance can be found on the smith chart starting at z = (open circuit),

and moving toward the generator to the j1.33 point. This gives a stub length of l1=0.397.

3. Similarly, the required open-circuited stub length for the second solution is l2= 0.103

This completes the tuner designs.



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FIGURE 3 Solution to Example 2 (a) smith chart for the series- stub tuners.



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Example 3. Single-stub shunt Tuning

For a load impedance ZL=15+j10 , design two single-stub shunt tuning networks to match this

load to a 50 line.

Solution

1. Determine the normalized load admittance. Zl=15+j10/50=

0.3+j0.2.

3. plot the normalized load impedance ZL=0.3+j0.2, Construct the appropriate SWR circle

4. Convert load impedance to the load admittance, yl, as shown on the smith chart Figure 4. At the

two intersection points, the normalized admittances are y1=1-j1.33 and y2=1+j1.33

5. Thus the distance d, from the load to the stub (y1) and y2 is given by either of these two

intersections. Reading the WTG scale, we obtain d1=0.328-0.284 =0.044 and d2= (0.5-0.284)

+0.171=0.387 .

6. If a susceptance of j1.3 is added at point y1 or j1.3 at point y2 the load will be matched Thus, the

first tuning solution requires a stub with a susceptance of j1.33, The length of an open-circuited

stub that gives this susceptance can be found on the smith chart by starting at y = 0 (the open

circuit) and moving along the outer edge of the chart (g = 0) toward the generator to the j1.33

point. The length is then l1= 0.147.

7. similarly, the required open-circuit stub length for the second solution is l2=0.353.

This completes the tuner designs.



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FIGURE 4 Solution to Example 3 (a) smith chart for the shunt- stub tuners



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LECTURE5: PRINTED CIRCUIT TRANSMISSION LINES

At high frequencies (e.g. microwave) the physical dimensions of printed circuit board (PCB) traces

become significant relative to the wavelength of the signal. At these frequencies, controlled impedance

transmission lines are used to move signals around a printed circuit board. By controlling the impedance

and electrical length we can easily predict its behavior in a circuit .The most commonly used transmission

lines (strip line and microstrip line

ADVANTAGES OF PRINTED TRANSMISSION LINE.

PRINTED transmission lines are widely used, and for good reason.

They are broadband in frequency. They provide circuits that are compact and light in weight.

(small, easy and fast to make, and cheap).

They are generally economical to produce since they are readily adaptable to hybrid and

monolithic integrated circuit (IC) fabrication technologies at RF and microwave frequencies.

DISADVANTAGE

They cannot handle very high power levels and

They are lossier than coax or waveguide. At higher frequencies (above 20 GHz), the dielectric losses

limit the performance

Uses of microstrip

Microstrip finds wide use in microwave circuits.

It used as the "interconnect" between components that have to be mounted in a system that is

matched from one stage to the other. Not only do the components have to be of known input and

output impedance the interconnecting transmission lines need to be matched to the system

impedance, in order to get maximum power transfer between stages.

microstrip is in the design of impedance matching transformers. These are sections of microstrip

line used as series and shunt stubs to effect.

These are important in the design of microwave amplifiers



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Microstrip has the very signicant advantage that it is a planar structure and can be

used as a "circuit board" for microwave circuits.

It is readily mass-produced making low cost microwave circuits feasable. The satellite TV

down-converter is a good example of this, as are the new generation of digital cellular phones

Microstripline

A microstrip is a thin, flat electrical conductor separated from a ground plane by a layer ofinsulation or

an air gap. Microstrips are used in printed circuit designs where high frequency signals need to be routed

from one part of the assembly to another with high efficiency and minimal signal loss due to radiation.

They are of a class of electrical conductors called transmission lines, having specific electrical properties

that are determined by conductor width and resistivity, spacing from the ground plane, and dielectric

properties of the insulating layer. A microstrip transmission line is similar to a stripline, except that the

stripline is sandwiched between two ground planes and respective insulating layers.

Therefore, A microstrip circuit uses a flat rod of metal which isparallel to a ground plane, such a circuit

can be made by having a strip of copper on one side of aprinted circuit board while the other side is a

plain ground plane. The insulating material of the PCB forms a dielectric. The width of the strip, the

thickness of the PCB and the relative permittivity of the PCB board determine the characteristicimpedance.

Microstrips can also be designed to launch electromagnetic waves into space, in which case they are

called microstrip antennas.

http://en.wikipedia.org/wiki/Ground_planehttp://en.wikipedia.org/wiki/Insulatorhttp://en.wikipedia.org/wiki/Printed_circuithttp://en.wikipedia.org/wiki/Transmission_lineshttp://en.wikipedia.org/wiki/Dielectrichttp://en.wikipedia.org/w/index.php?title=Stripline&action=edithttp://en.wikipedia.org/wiki/Parallelhttp://en.wikipedia.org/wiki/Ground_planehttp://en.wikipedia.org/wiki/Printed_circuit_boardhttp://en.wikipedia.org/wiki/Relative_permittivityhttp://en.wikipedia.org/wiki/Characteristic_impedancehttp://en.wikipedia.org/wiki/Characteristic_impedancehttp://en.wikipedia.org/wiki/Electromagnetic_waveshttp://en.wikipedia.org/wiki/Microstrip_antennahttp://en.wikipedia.org/wiki/Ground_planehttp://en.wikipedia.org/wiki/Insulatorhttp://en.wikipedia.org/wiki/Printed_circuithttp://en.wikipedia.org/wiki/Transmission_lineshttp://en.wikipedia.org/wiki/Dielectrichttp://en.wikipedia.org/w/index.php?title=Stripline&action=edithttp://en.wikipedia.org/wiki/Parallelhttp://en.wikipedia.org/wiki/Ground_planehttp://en.wikipedia.org/wiki/Printed_circuit_boardhttp://en.wikipedia.org/wiki/Relative_permittivityhttp://en.wikipedia.org/wiki/Characteristic_impedancehttp://en.wikipedia.org/wiki/Characteristic_impedancehttp://en.wikipedia.org/wiki/Electromagnetic_waveshttp://en.wikipedia.org/wiki/Microstrip_antenna


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Figure. Microstrip transmission line printed on a grounded dielectric substrate

.

The equations below provide reasonable approximations for (effective dielectric constant) and Z0.

What do you want for free? Note that there are separate solutions for cases where W/H is greater than or

equal to 1, and W/H is less than 1.

The effective dielectric constant is calculated by:

The line impedance is a function of the ratio of the height to the width of the transmission line, as shown

in the following equation.



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For a give characteristic impedance 0Z and permittivity r , theh

wratio can be found as

Where,

Figure. The cross section of a microstrip line and the characteristic impedanceZ0

as a function of the ratio of strip width to substrate height w/h for different

substrate materials.

Microstrip parameters

Aside from the difficulty of calculating the value of eff, there is another important effect. It is clear that

eff will depend on both W and h. Hence, the phase velocity along the microstrip will depend on these

parameters. Assuming the relative permeability of all materials in the line design is well approximated by

r = 1,



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Phase Velocity will be given by:eff

p

CV

=

Where,

C is the speed of light,

eff = effective relative dielectric constant of substrate. ( Not equal to r ).

Wavelength =eff

p

f

C

f

V

=

Where f= operating frequency.

Propagation constant = eff 00

Figure for design purpose.

Advantages and disadvantages of Microstrip line compared to striplinesAdvantages

The major advantage of microstrip over stripline is that all active components can be mounted on

top of the board. Therefore, the microstrip structure is the "open" line which makes it very easy to

connect components.



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Another advantage is that microstrips can be packed together with fairly high density (multiple

channels) with only minimal "crosstalk" interference, and therefore lends itself well to RF and

microwave IC design

Disadvantages

The disadvantages are that when high isolation is required such as in a filter or switch, some external

shielding may have to be considered. Given the chance, microstrip circuits can radiate, causing unintended

circuit response.

A minor issue with microstrip is that it is dispersive, meaning that signals of different frequencies travel

at slightly different speeds (usually not a big deal, but this property is what causes the asymmetric

frequency of bandpass filters, for example).

Here are some rules of thumb to remember:

The higher the dielectric constant, the thinner the line, keeping the thickness of the dielectric and

the impedance of the line constant.

The thinner the dielectric, the thinner the line is, keeping the dielectric constant and impedance of

the line constant.

The higher the dielectric constant, the smaller the circuit is (why?).

The wider the line, the lower the impedance.

SOURCE OF LOSSES IN MICROSTRIP LINE.

Conductor loss in the strip and ground plane

Dielectric and conduction losses in the substrate.

Radiation loss.

Surface wave loss.

Some Considerations in the Choice of Microstrip Substrate Materials

Important qualities of the dielectric substrate for microstrip design include:

1. The microwave dielectric c

2. onstant



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3. The frequency dependence of this dielectric constant which gives rise to "material dispersion" in

which the wave velocity is frequency-dependent

4. The surface finish and flatness

5. The dielectric loss tangent, or imaginary part of the dielectric constant, which sets the dielectric loss

6. The cost

7. The thermal expansion and conductivity

8. The dimensional stability with time

9. The surface adhesion properties for the conductor coatings

10. The manufacturability (ease of cutting, shaping, and drilling)

11. The porosity (for high vacuum applications we don't want a substrate which continually "outgasses"

when pumped)

Common substrate materials

Duroid, quartz, alumina and silicon

Microstrip design

It is necessary to connect two microwave ICs using a microstrip transmission line. The line needs to be 10

cm long; to be constructed using copper microstrip and backplane conductors with a thickness (t = 0.15

mm) separated by an epoxy fiberglass ( r = 4.9) circuit board with a thickness (h = 0.8 mm). Impedance

matching requires the line to have a characteristic impedance (Zo = 50 Ohms). Determine the following:

1. The actual width (W) of the microstrip

2. The effective dielectric constant ( eff ) for the microstrip design.

Stripline

A stripline circuit uses a flat strip of metal which is sandwiched between two parallel ground planes, The

insulating material of the substrate forms a dielectric. The width of the strip, the thickness of the substrate

and the relative permittivity of the substrate determine the characteristic impedance of the strip which is a

transmission line

http://en.wikipedia.org/wiki/Parallelhttp://en.wikipedia.org/wiki/Ground_planehttp://en.wikipedia.org/wiki/Relative_permittivityhttp://en.wikipedia.org/wiki/Characteristic_impedancehttp://en.wikipedia.org/wiki/Parallelhttp://en.wikipedia.org/wiki/Ground_planehttp://en.wikipedia.org/wiki/Relative_permittivityhttp://en.wikipedia.org/wiki/Characteristic_impedance


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Stripline is a conductor sandwiched by dielectric between a pair of groundplanes, much like a coax cable

would look after you ran it over with your small-manhood indicating SUV (let's not go there...) In

practice, stripline is usually made by etching circuitry on a substrate that has a groundplane on the

opposite face, then adding a second substrate (which is metalized on only one surfac

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Transmission Line(KABADI)