Tlines Third Year

7/27/2019 Tlines Third Year

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AN INTRODUCTION TO

T RANSMISSION L INES

Prepared by

Dr ESSAM A. HASHISH

Dept. of Electronics and Communications

Faculty of Engineering, Cairo University

COPYRIGHTS ARE RESERVED FOR THE AUTHOR


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CONTENTS

GENERAL INTRODUCTION

Data use

Data transmission

Data type

Maxwell's Equations

I NTRODUCTION

Statement of Maxwell's equations

Maxwell's Equations in the Integral Form

Maxwell's Equations in Differential Form

Wave Propagation in Source-free Regions

Time Harmonic Fields

Propagation of Uniform Plane Waves

SOLUTION OF THE U NIFORM PLANE WAVE

Propagation in Different Media

Reflection and Refraction of Plane Waves

TRANSMISSION LINES

Introduction

Uses of transmission lines


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Types of transmission lines

Planar Transmission Lines

Analysis of theTwo Parallel-Plate Transmission Line

Voltages and Currents

Line Capacitance

Line Inductance

Circuit model of the transmission line Effect of Dielectric Losses

Effect of Conductor Loss

Complete circuit representation

General Transmission Line Equation

Line parameters of some line configurations

Relation between the two parallel plate line and

other lines

Types of Transmission Lines

Lossless line

Low-loss line

Distortionless line

ATTENUATION CONSTANT FROM POWER RELATION


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Propagation along transmission lines with different

terminations

Propagation of a wave along an infinite line Propagation of a wave along a line terminated with

Z0

Two semi-infinite lines connected together

A line terminated by Z L Special cases Voltages and currents at distance d from the load Spatial Distribution of the Voltage and Current Voltage Standing Wave Ratio (VSWR) Spatial distribution using the unity circle

Graphical Representation of Voltages

THE VOLTAGE AT THE LOAD

The voltage at any point

Properties of the Graphical Representation

1. D IRECTION OF MOTION

2. Location of the voltage maxima and minima

3. Location of open- and short-circuits


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4. Location of different loads

5. Expressing Z in terms of

6. Voltage standing wave ratio

Graphical Representation of Currents

The Smith Chart

Details of the Smith Chart

Applications of the Smith Chart1. Calculation of l from Z L

2. Calculation of Z i and d at a distance d

3. Determination of V max and its first location

4. Determination of V min and its first location

5. Determination of VSWR

6. Transition from impedance to admittance chart

7. Impedance Matching in Transmission Lines

8. Methods of matching

Matching using a /4 section

Matching using a single stub

Double-stub Matching

Transients in Transmission Lines

Scattering matrix


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Two-port Parameters: The Scattering Matrix

Scattering parameters evaluation

General definition of scattering parameters

Important properties of the S matrix

Active source representation

Available power

Case of source and load connection


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GENERAL INTRODUCTION

GOODS DELIVERY

Direct

Car

Train


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Transmission Line

DATA TRANSMISSION

DATA TYPE

Direct Contact

''''''''''''''''''''''''''''''''''''''''

Satellite Link


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Audio

Video

Text

Graph

Picture

Analog

Digital


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MAXWELL'S EQUATIONS

INTRODUCTION

Electrostatic Field Equations

0

4vV V dv

r

E D

E

Where D E

Magnetostatic Field Equations

0

4vdv

r

H J B

JB A A

Where B H

Continuity Equation

0

t cD

J

Maxwell's Equations (Time Varying Case)

These equations establish the relation between the electricand magnetic fields in the general time varying case.

Maxwell's Equations in the Differential Form

0

t t

B DE H J

D B


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Maxwell's Equations in the Integral Form

0

s s

v

d d d d t t

d dv d

B DE l s H l J s

D s B s

Wave Propagation in Source-free Regions

In a source-free region, both and J are equal tozero. Maxwell's equations in this case take the form

(1) (2)

0 (3) 0 (4)t t

H EE H

D B

Taking the curl for equation (1)

0Also

0

2

22

2

22

2

22

t

t

t

t

HH

EE

EEE

HE

The last two equations are called homogeneous wave

equations.


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Time Harmonic Fields

In the case of time harmonic fields, sinusoidal variations are

considered .

t et z y x

t et z y x

m

t j

m

m

t j

m

sinIm,,,or

cosRe,,,

EEE

EEE

Thus, depending whether we consider cos( t ) or

sin( t ) variations, the real or imaginary part of thecomplex exponent is considered.

Taking the above note into consideration, we can simply put

Maxwell's equations in this case reduce to

0 0

j j E H H E

D B

And the wave equation can then have the form

k

k

k

where

0&

022

22

HH

EE

k is called the wave number or the complex phaseconstant in general.

22

2

, , , , ,

, , ,Hence, , , ,

, , ,& , , ,

j t m x y z t x y z e

x y z t x y z t t

x y z t x y z t

t

E E

EE

EE


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Similar to the current and voltage relations, it can beshown that the average of the power density vector P isgiven by

*Re21

HEP av

and the power transmitted over any cross sectional area A is

given by

SP d W A

av . The wave equation is used to get the solution of

propagating waves under any conditions. One of the mostessential types of propagating waves is the uniform planewave.

Propagation of Uniform Plane Waves

In a uniform plane wave, the fields have the samedirection, magnitude and phase in the plane of uniformityat any time t .

If the XY -plane is the plane of uniformity, then

0 y x

Using the div equation of D we get

From which it can be shown the Z -components of E mustequal to zero and similarly that of H is also zero.

This means that the electromagnetic fields are

totally confined to the plane of uniformity.

0D 0 z

z E


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SOLUTION OF THE UNIFORM P LANE W AVE

Consider the XY -plane as the plane of uniformitywhich is called the transverse plane, hence the fieldshave components in general along X and Y -directions.Let the axes be rotated such that the total electric field isalong the X -direction.

The solution is thus given by

x x

t j jkz t j jkz

x

E E

ee E ee E z E 00

which represents two traveling waves. Consider now thefirst term of the above equation, the instantaneousexpression of the field is given by

0, cos

x E z t E t kz

When this expression is sketched with z for progressivelyincreasing values of t , the following remarks can bemade:

0

0

2

2

2

22

x x E k

z E

k EE


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The curves are effectively traveling in the + ve Z -direction.

Consider a point of the same phase on the wave

along the direction of propagation

phvk t z

const z z k t t

const kz t

which is called the phase velocity of the wave, note thatv ph in the free space =3x10 8 m/s.

The distance over which the phase changes by2 radians is called the wavelength

z

E x

t= 0 t= /2 t= /

Direction of Propagation


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f v

k

2

The relation between E and H is given by

1

and

y y

y x

H a

where H E

k

+H

is called the intrinsic impedance or the wave impedanceof the medium. In the case of free space we get,

0 0,120 377

The average power density P av is given by

2*1 Re

2 2av y z H

P E H a

H

aaa

HE

j E

jk

j

x

z y x

0000


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Important notes:

If the uniform plane wave contains an electric

field component E y and is propagating in the + ve Z -direction, then

y x

E H

We note also that in the uniform plane wave, E and H are perpendicular to each other and bothare perpendicular to the direction of propagationsuch that

p H E aaa where a E , a H and a P are vectors in the direction of theelectric field, magnetic field and the wave propagationrespectively.

Propagation in Different MediaIn a general medium, the first Maxwells equation isgiven by

which is called the conduction current density

which is called the displacement current density

EJJJ

DEH

c

d c

where

t

d and j J E


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Examples

z

x

y Hx

E

Direction of propagation

TE WAVE

Hy

x

y

z

Ex H


TM WAVE

Ey

z

x

y H

E


TEM WAVE


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Reflection And Refraction Of Plane Waves

Introduction

Consider an electromagnetic plane wave which is incident from aninfinite homogeneous dielectric medium at an angle i to the interfacewith another infinite homogeneous dielectric medium as shown in thefigure below. Upon this incidence, a part of the incident energy isreflected and another part is transmitted. The fields of the reflected andtransmitted waves are obtained by applying the boundary conditions atthe interface. These boundary conditions imply the equivalence of thetangential electric and magnetic fields at both sides of the interface. In thenext section, the case of normally incident plane wave is only considered.

Normal Incidence

Consider a normally incident plane wave with its electric field E parallel to the interface and directed to the x-axis as shown in the figure. It is

assumed that the interface between the two media is lying in the y-x plane. Therelative prmittivities of these media are assumed as r1 and r2 respectively. The

i t

y

zW t

Wr

W i r1 r2

i x E

t x E r

x E

Reflection of a plane wave at the interface between twoinhomogeneous media (incidence from left to right).


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time dependence is e jt, where is the angular frequency. The incident,reflected and transmitted fields inside these media can be expressed as follows

Incident fields (z 0)The electric field of the incident wave, i x E can be expressed as

1 jk z i x o E E e

and the magnetic field component i H is obtained by the relation

1

1 1

i jk z i

o x y

E E H e

where

1 1 1k and 111

,

Reflected fields (z 0)The electric field of the reflected wave, r x E can be expressed as

1 jk z r x o E R E e

and the magnetic field component r y H is obtained by the relation1

1 1

r jk z r o x

y

E E H R e

Transmitted fields (z 0)The electric field of the transmitted wave, t x E can be expressed as

2 jk z t x o E T E e

and the magnetic field component t y H is obtained by the relation

2

2 2

t

jk z t o x y E E H T e

y

zW t

W r

W i

r1 r2

R

i x E

t x E

r x E

Normal Incidence

T


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where

2 2 2k and2

22

,

and R and T are the reflection and transmission coefficients at the interface between the two media.The boundary conditions at the interface z=0 imply the following

tangential tangential0 0

tangential tangential0 0

z z

z z

E E

and H H

This leads to

0 0

0 0

i r t x x x z z

i r t y y y z z

E E E

and H H H

Applying the above boundary conditions at the interface z=0, the reflectioncoefficient R is then given by

2 1

2 1

R

and the transmission coefficient T is given by2

2 1

21T R

Field Expressions in the First Medium (z 0)i r

i r E E E H H H

or

1 1

1 1

jk z jk z o o x

jk z jk z o o y

E E e RE e a

H H e RH e a

Field Expressions in the Second Medium (z 0)t

t

E E

H H

or 2

2

jk z o x

jk z o y

E T E e a

H T H e a

Surface ImpedanceThe boundary conditions at the interface imply the continuity of the

tangential electric and magnetic fields at the interface. Hence, the ratio betweenthese fields is also continuous at any interface. The surface impedance Z s at anylocation is imposed for this reason as the ratio between the total tangentialelectric and magnetic fields at this location as follows


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1

1

1

i r x x

z z s i r

z y y z

E E R Z

R H H in the region (z 0), and

2

t x z

s t

y z

E Z

H

in the region (z 0)

It follows that the suface impedance in the first medium at a point locatedat distance d from the interface (z=-d) is given as

1

1

1

i r x x z d z d

s i r z d y y

z d

E E R Z

R H H

2

1 2

1 1 21

1 2 1

11

tan( )

tan( )

jkd

jkd

R e R e

j k d

j k d

Case of Dielectric Slab of thickness dThis configuration is shown in the figure below

Special Cases

i) d= 1/2 which implies that k 1d = and hence Z s is given as

2 s Z

and this means that the dielelectric slab is totally transparent in this case

r1 r2

i x E

t x E

r x E

Case of dielectric slab of thickness d

Z s

ro

d


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ii) d= 1 /4which implies that k 1d = /2 and hence Z s is given as

21

2 s Z

and this means that the dielelectric slab is acting as a transformer in thiscase


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TRANSMISSION LINES AND THEIR USES

Transmission lines are used to guide the

electromagnetic waves from one location to another in

the form of

Point-to-point communication:

From source to load.

From one subscriber to another.

Component-to-component connection within thesystem:

From transmitter to antenna.

From antenna to receiver.

From one network to another.


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TYPES OF TRANSMISSION LINES

Two wire line Coaxial cable

Parallel plate line Microstrip line

Rectangular Waveguide Circular Waveguide

Optical Fiber


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P LANAR T RANSMISSION L INES USED IN M ICROWAVEINTEGRATED C IRCUITS

r r

INVERED MICROSTRIP SUSPENDED MICROSTRIP

r1

r2

MICROSTRIP WITH OVERLAY

r1

r2

STRIP DIELECTRIC INVERTED STRIPDIELECTTRIC WAVEGUIDE

r1

r2

r r

r r

MICROSTRIPSLOTLINE

COPLANAR WAVEGUIDE COPLANAR STRIPS


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Analysis of theTwo Parallel-Plate Transmission Line

Electromagnetic Fields

Phasor expression Instantaneous expression

jkz x

jkz y

e E

H

e E E

0

0

t z E

t z H

kz t E t z E

y

x

y

,

,

cos, 0

,k where

y

x z x

y

E

y

x

y

z

Transverse plane Longitudinal direction

H


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Notes

The fields change sign every 2 .

The difference between the field distribution in thetransverse plane and along the direction of propagation.

y

z

Longitudinal direction

E

z

/2 E

Spatial distribution of the electric field E


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Voltages and Currents

d E V

eV V

ed E V

d V

jkz ab

jkz ab

b

aab

00

0

0

where

or

. lE

The ratio between the voltages across the two plates and thecurrents entering them is thus given by

The voltage between the two plates is given by

Also the current entering the two plates is given by

0

0

0

0

where

or

.

E w I

e I I

e E

w H w I

d I

jkz

jkz

x

lH

0

0

0 Z wd

I V

I V

y

x I H Vab

+

_

Plate a

Plate b E

W

d


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Where Z 0 is called the characteristic impedance of the line and itis constant for any line configuration.

Line Capacitance

The line capacitance is defined as the ratio between the chargecarried by the line segment and the potential difference betweenits two conductors. It is calculated as follows

`

z w E Q

z w AQ

E D

d E V

e E E

ya

sa saa

y y sa

yab

jkz

y

0

d

wC

z d w

V Q

C

lengthunit

ab

a

/

x

y

z

E Vab

+

_

Plate a

Plate b

W

d


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Line Inductance

The line inductance is defined as the ratio between the linked

magnetic flux carried by the line segment and the total electriccurrent carried by the conductors. It is calculated as follows

w

d L

z wd

I L

asobtained thusisinductanceThe

z d H

z d B A B

by givenis fluxmagneticThe

H w I e H H H

ascalculated iscurrent The

lengthunit

a

x

x x

xa

jkz

x x

/

0

~

~~

~

~

y

x

H

I

W

z

d


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Circuit model of the transmission line

To this end, the line can be considered from the circuit point of view as distributed sections each of which is an inductance L =L Z and a capacitance C = C Z where L and C are the lineinductance and capacitance per unit length. This circuit model isshown as follows

z

z

L z

C z

Transmission Line

z

Distributed circuit prameters

C z

L z


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Effect of Dielectric Losses

In the case where the dielectric filling material is lossy, then, aleakage current flows across the two conductors. This current isrelated to the voltage between the two plates as

d w

G

z d

w

d

AG

whereGV I

d lengthunit

d d

ableak

/

/

G z

x

y

z

E Vab

+

_

Plate a

Plate b

C z

L z

W


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Effect of Conductor Loss

When the two conductors which guide the wave along thetransmission line are not perfect conductors, an ohmic lossoccurs in the circuit. This is equivalent to the presence of a highfrequency resistance with a depth of one skin depth through themetal surface. The expression of this resistance is as follows.

w R

f depth skin

w z

Al

R

clengthunit

c

cc

1

1where

/

x

y

z

E Vab

+

_

Plate a

Plate b

W

R z


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COMPLETE CIRCUIT REPRESENTATION

The parameters of the transmission line per unit length arethus represented by ( R, L, G and C ).

It is worth noting that the above parameters are related to eachother by the following relations

C G LC ,

L R

C G


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Transmission Line Parameters Examplef = 10 6 Hz Z O=50 d =1 m d= 1000m No of sections = 1000

R= 0.1726 /m L=1.667E-07 H/m G=0.069 m /m C=66.7 pf/m

0 200 400 600 800 1000

0

100

200

300

400

500

I N P U T I M P E D A N C E

NUMBER OF SECTIONS

ZL=ZO

0 2 0 0 4 0 0 6 0 0 8 0 0 1 0 0 0

0

5 0

1 0 0

1 5 0

2 0 0

2 5 0

3 0 0

3 5 0

4 0 0

4 5 0

5 0 0


N U M B E R O F S E C T I O N S

ZL

= I N F I N I T Y


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0 200 400 600 800 1000

0

50

100

150

200

250

Y A x

i s T i t l e

X Axis Title

ZL=0

0 200 400 600 800 1000

0

50

100

150

200

250

300

350

400

450

500


NUMBER OF SECTIONS

ZL=0

ZL=INFINITY

ZL=ZO


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General Transmission Line Equation

To obtain the general transmission line equation, let us examinethe relations between the voltages and currents at the input and

output of the shown segment line section

1

0totendsaslimittheTaking

L j R z I dz

z dV

z

L j R z I z

z V z z V

z L j R z I z V z z V

2C jG z V dz

z dI

C jG z z V z

z I z z I z C jG z z V z I z z I

V z V z+ z

( z ) I z+ z Lz R zCz Gz


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Note that is called the propagation constant and it can beshown that it equals to jk, where k is the wave number that isdefined before.

The general solution of equation (3) is

z z

z z

e I e I z I and

eV eV z V

00

00

which represents two waves propagating in the positive and

negative z -directions. In the above equation, V+

and I+

representa wave propagating along the positive z direction while V - and I - represent a wave propagating along the negative z direction

Using equation (1), it can be shown that

C jG L j R

Z

Z I

V

I

V

0

00

0

0

0

where

In fact, any transmission line is characterized by two parameters, its propagation constant and characteristicimpedance Z 0 which are function of its circuit parameters R, L,C and G.

C jG L j R

I dz

I d V

dz V d

where

3,

(2)and(1)From

2

2

22

2

2


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In general, and Z 0 are complex and have the following forms = + j where and are called the attenuationconstant and the phase constant per unit length.

While Z 0 = R o + j X o where R o and j X o are respectively calledthe characteristic resistance and reactance of the line.

Line parameters of some line configurations

Parameters Coaxial Line T WO -W IREL INE

Planar Line

R( /m)

1 1 12

,c a b

a c b

1ca

a

2cw

t

L(H/m) ln2

ba

1cosh2d a

d

w

G(mho/m)2

ln ba

1cosh2d a

wd

C(F/m)2

lnba

1cosh2d a

( )

wd

w d

where

1

c c f skin depth of the conductor

;cosh ln .

LNM

OQP

12

2 21

d a

d a

if d a


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Relation between the two parallel plate line and other lines

Curved plates

Coaxial line

Two arallel lates


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Two parallel plates

Curved plates

Two wire line


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Types of Transmission Lines (according to their

parameters)

Lossless Line ( R = G = 0)

LC v

LC

LC j j

C L

Z

ph

1

0

real pure0

It is important to note that the phase velocity is independent of frequency. Hence, the signal is not distorted upon propagating insuch a line. This line is an ideal case since Z 0 is real, is zeroand is linearly proportional to .


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Low-loss Line

1,1C

G L

R

The characteristic impedance of the low loss line is given by

Which means that Z 0 is almost real like the lossless line.

The propagation constant is also given by

LC v

LC

GR R R

C jG

L j R

LC j

C jG L j R

ph

1

,

21

221

0

0

Which means that the low loss line is almost distortionless.

0,

221

11

00

0

2121

0

X C L

R

L jG

L j R

C L

Z

C jG

L j R

C L

C jG L j R

Z


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Distortionless Line

C G

L R

The characteristic impedance of the distortionless line Z 0 isgiven by

This means that Z 0 is actually real like the case of the losslessline.

The propagation constant is also given by

LC vand

LC R

R

L j R

LC j

ph

1

,

1

0

0, 0

0

0

X

C L RC

L

C jG L j R

Z


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General expression of the attenuation constant fromthe power relation

For a wave traveling in the positive z -direction

The average power flow W at any distance z is given by

The power loss per unit length is obtained as

*2

0

2

0 0 0

1R e

2

12

z

W V I

W e

w h e re W I R

0

0

0

00

00

where Z I

V

ee I e I I

eeV eV V z z j z

z z j z

power flowlengthunit per loss power

W

P

W

dz dW

W dz

dW

L

2

22

)/(

2

GV R I P L22

2

1

2

1


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Thus, the total attenuation constant is given by

Propagation along Transmission Lines with DifferentTerminations

Case 1 : Propagation of a wave along an infinite line

In the case of an infinite line extension, the wave propagatesalong the line without reflection and the relation between thevoltages and the currents is as follows

This means that the ratio between the voltage and the current at

any point along the line is constant.

2 22

0

20

0

1 11 12 2

12 22

I R V G R G Z

R I R

0

0

0

00 ,

Z

I

V

e I I eV V z z

Z


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Case 2 : Propagation of a wave along a line terminatedwith Z 0

The same in the above case applies for a line terminated by aload equals to the characteristic impedance Z 0.

The currents and voltages have the same relations as

Case 3 : Two semi-infinite lines connected together

In the case of two semi-infinite lines connected together, theincidence of a propagating wave on the junction results in areflected wave in the first line propagating along the negative zdirection and a transmitted one along the second line in the

positive z direction.The voltages and currents in each region are given as follows

Z 0

0

0

0

00 ,

Z I V

e I I eV V z z

Z 01 Z 02

V 1+

V 1

V 2

Z


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1

1

1

1

1

1

1

1

1

1

1

1

1

0

0

0

0

0

001

001

I

V

I

V Z

e I e I I

eV eV V z z

z z

2

2

2

2

2

2

2

2 0

2 0

00

0

z

z

V V e

I I e

V Z I

In terms of V 01 and I 01, the values of 1 1,o oV I and 2 2,o oV I

can be known by

applying the boundary conditions at the junction between the two lines.These boundary conditions at the junction, where Z=0, imply thatV1 = V 2 and I 1 = I 2

V 1+

V 1

V 2

10102

0101

V T V and

V V let Now

02

01

02

02

010101

020101

020101

2121

1

,

1

,:0at

Z

Z T

V Z Z

V V

I I I

T V V V

I I V V z


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Note that is called the reflection coefficient at the junction of the twolines while T is called the transmission coefficient at that junction.

Case 4 : A line terminated by Z L

When the line is terminated by an impedance Z L, the reflectioncoefficient is given by

z L z

z

L

z

ee I I

eeV V

by givenarecurrentsand voltagesThe

0

0

Z 01 Z 02

V 01+

V 01+

TV 01+

z

Z L V +

V

z= 0

z= d

L

0

0

L L

L

Z Z Z Z

0102

0

0102

0102

22 Z Z

Z T and

Z Z Z Z


52/122

51

Special cases Case of an open circuit load Z L = , hence L = 1

Case of an open circuit load Z L = 0 , hence L = -1 Note that along the direction of propagation of the wave, theelectromagnetic fields undergo phase delay and also suffer fromattenuation if the line is lossy.

Voltages and currents at a distance d from the load

For a point at a distance d from the load, i.e. z = - d as shown

Z L V +

V

z= 0

z= d

d d

d d

ee I I

eeV V

0

0

,

d Z Z d Z Z

Z Z

get we , for ng Substituti

L

Lin

tanhtanh

0

00

d d

d d

d d

d d

in

eeee

Z ee I

eeV

I

V Z

0

0

0


53/122

52

2

2

0

1

:loadtheindissipated power The

..,..0For

..0,..For

2:4at

,2atgeneral,In

:2at

i

r i L

in L

in L

Lin

Lin

Lin

W

W W W

co Z c s Z

c s Z co Z

Z Z

Z

d d

Z Z nd Z Z

d d

:thatNote

d jZ Z d jZ Z Z Z

j

L

Lin

tantan

0i.e.for and

0

00


54/122

53

Spatial Distribution of the Voltage and Current

Case (a) : Open-circuit ( Z L= )

d

Z V

jd I

d V eeV d V

L

Ld jd j

L

sin2&

cos2

1

0

V

I

L

d

O.C.


55/122

54

Case (b) : Short-circuit ( Z L= 0)

Note that

d Z V

d I

d jV d V

L

L

cos2&

sin2

1

0

2

0

0

0

cot,

tan

Z Z Z

d jZ Z

d jZ Z

oc sc

oc

sc

V

I

SC L

d


56/122

55

Case (c) : Termination by any load impedance Z L

0 02

0 0

z z

l L

V V e V eV V e

20

20

L L

1

1But = e L

l z z L

z d L

j

V V e e

V e e

20 1 L j d z LV V e e

V

I

Z L L

d

Z


57/122

56

Voltage Standing Wave Ratio (VSWR)

S Z Z

S Z

Z

Z Z Since

V V VSWRS

0max

0min

0

min

max

,

11

1

1


58/122

57

EXAMPLES

i) Resistive load with R L > Z o

In this case, L = L

2/10max nd at occurswhichV V L and

2/)5.0(10min nd at occurswhichV V L

where n=0,1,2,.

0 0 . 2 0 . 4 0 . 6 0 . 8 1 1 . 2 1 . 4 1 . 6 1 . 8 20 . 2

0 . 3

0 . 4

0 . 5

0 . 6

0 . 7

0 . 8

0 . 9

1

1 . 1

1 . 2

N o r m a l i z e d D i s t a n c e

d j L eV V 20 1

V

I

Normalized distance

R L>Z o L

d Z


59/122

58

ii) Resistive load with R L < Z o

In this case, L = -L

2/)5.0(10max nd at occurswhichV V L and

2/10min nd at occurswhichV V L

where n=0,1,2,.

0 0 . 2 0 . 4 0 . 6 0 . 8 1 1 . 2 1 . 4 1 . 6 1 . 8 20 . 2

0 . 3

0 . 4

0 . 5

0 . 6

0 . 7

0 . 8

0 . 9

1

1 . 1

1 . 2

N o r m a l i z e d D i s t a n c e

d j L eV V 20 1

V

I

Normalized distance

L

d Z

R L


60/122

59

iii) Inductive load with Z L = R + jX

In this case, L = Lexp(j L) , 0< L


61/122


62/122

61

Expressing The Wave Voltages in Terms of Voltagesand Currents

At any location, the voltage and current are given by

1 ( )o

o

V V V

and I V V Z

or Z I V V

from which it can be shown that1

( )2

1 ( )2

o

o

V V Z I

and V V Z I

Thus, at the load side, we get

1( )

21

( )2

L L o L

L L o L

V V Z I

and V V Z I

And at the Input side, we get1

( )21

( )2

i i o i

i i o i

V V Z I

and V V Z I

Z L V +

V

L


63/122

62

Relation Between Input and Output Voltages and Currents

1. Input Side Voltages in Terms of The Output Side Voltages

At the load port, the incident and the reflected voltages are given by

1( )

21

( )2

L L o L

L L o L

V V Z I

and V V Z I

The voltages and currents at the input port are given in terms of LV and LV as

1( )

L L

L L

L L

i

io

o i

V V e V e

and I V e V e Z

or Z I V e V e

Hence, we get1 1

( ) ( )2 21 1

( ) ( ) )2 2

i L o L L o L

i L o L L o L

V V Z I e V Z I e

I V Z I e V Z I e

which can be put in the form

V +

V

V i Ii

VL IL


64/122

63

cosh( ) sinh( )

1sinh( ) cosh( )

i L o L

i L L

o

V V Z I

I V I Z

The last two equations can be put in the matrix form

cosh( ) sinh( )

1sinh( ) cosh( )

oi L

i Lo

Z V V

I I Z

i Li L

or

V V T

I I

where T is the transmission matrix of the line, note that1T .In the case of cascaded lines as shown below, the

relation between the input and output voltages andcurrents are given by

1 2 3 4i Li L

V V T T T T

I I

V i

V L

T 1 T 3

T 2

T 4

I i

I L


65/122

64

2. Output Side Voltages in Terms of The Input Side Voltages

In terms of the input voltage and current, the output voltage andcurrent expressions can be expressed as

cosh( ) sinh( )

1sinh( ) cosh( )

oi L

L io

Z V V

I I Z

' L i

L i

or

V V T

I I

where 'T is given by

cosh( ) sinh( )

1 sinh( ) cosh( )

o

o

Z

Z

We note also that' 1T .

The Voltage at Any Point Due to a Connected Source at

the Input Port of the LineCase 1: Matched Source (Z g =Z o )

Consider a transmission line whose length is and characteristicimpedance is Z o. The line is connected at its input to a source whose

voltage is V g and its internal impedance is Z o.The line output is connected

to a load impedance Z L.


66/122

65

The incident voltage oV at the input of the line is calculated on the basis

of dividing the source voltage between the line impedance Z o and the

source impedance which is also Z o in this case. Thus, the incident

voltage oV is given by

12

oo g g

o o

Z V V V

Z Z

The reflected voltage at the input of the line is calculated on the basis of

incident wave traveling to the load and then reflection an return back to

the input port. In this case, the reflected voltage wave can be expressed as

2o o LV V e

With oV and oV known, the voltage at any point can then expressed as

( 2 )1 ( )2

o o z z

z z g L

V V e V e

V e e

It follows that the voltage at the load is given by1

( )2 L g L

V V e e and the voltage at the input of the line is given by

21 (1 )

2o g LV V e

V +

V

V o+

V o

z

Z L

Z o

V g


67/122

66

Another method to get the voltage at any point of the line is to obtain

first the input impedance of the line by the expression

tanhtanh

o Li o

o L

Z Z Z Z Z Z

the input voltage at the line is thus given by

ii g

i o

Z V V

Z Z

while the input current is given by

1i g

i o

V Z Z

With the knowledge of i iV and I then the voltage and the currents can be

obtained by any prescribed method.

Case 2: Unmatched Source (Z g Z o )

The line in this case is connected at its input to a source whose voltage

is V g and its internal impedance is Z g. The line output is connected to a

load impedance Z L.

The input voltage at the line is thus given by

ii g

i g

Z V V

Z Z

while the input current is given by

1i g

i g

V Z Z

With the knowledge of i iV and I , then the voltage and the currents can beobtained.

V +

V o+

V Z L

Z

V


68/122

67

The other method using the concept of the incident and reflected waves

is explained as follows.

The incident voltage 1oV at the input of the line is calculated on the

basis of dividing the source voltage between the line impedance Z o and

the source impedance Z g. Thus, the incident voltage 1oV is given by

1o

o g o g

Z V V

Z Z

The reflected voltage 1oV at the input of the line is calculated on the basis

of incident wave traveling to the load and then reflection a return back to

the input port. In this case, the reflected voltage wave can be expressed as

21 1o o LV V e

Upon the arrival of 1oV at the input side of the line it reflects back towards the load with a reflection coefficient g since the sourceimpedance is not Z o. the back reflected voltage 2oV is expressed as

2 1o g oV V and it travels along the load side till it reaches the load then it

reflects with a reflection coefficient L. The result is an infinite terms of incident and reflected signals. The overall incident and reflected waves at

the input side oV and oV are given by

1 2 3 4 ...o o o o oV V V V V

1 2 3 4 ...o o o o oand V V V V V


69/122

68

21

1

1 2

or

1 ( )

1

1

no o g L

n

o g L

V V e

V e

21

1

1 2

1 ( )

1 1

no o g L

n

o g L

and V V e

V e

The voltage at any location is thus given by

1 1 2

1( )

1

o o

o o

z z

z z

g L

V V e V e

V e V ee

The multiple reflections in this case are shown in the following figure

V +

V o

+

V Z L

Z

V

1oV e L

g

e e

e

e

e

L

L

g

g


70/122

69

Graphical Representation of Voltages

T HE VOLTAGE AT THE LOAD

Consider a transmission line terminated by a loadimpedance Z L. The reflection coefficient at the load, L is given

by

0

0

Z Z

Z Z

L

L

L

The total voltage at the load is

L L

L L

L L

vV V v

V V V V

1Let

1

Z L V +

V

V +

LV +

V L L

1

v L

+

VL

__


71/122

70

The voltage at any point

At any point at a distance d from the load, an expressionfor the normalized voltage can be derived as follows

d j Ld

d j

L

d j

d jd j

d

eV

eeV

eV eV V

2

2

1

1

d

d j

L

d d d

e

V V v

1

1 2

Z L v +

v

vd +

vd

d

1

L

d

2 d

vd

vd


72/122

71

Properties of the Graphical Representation

1. D IRECTION OF MOTION

Constant radius Motion towards the generator is in clockwise

sense ( 2 d ).

Motion towards the load is in anti-clockwise

sense (2 d ).

One circle rotation = /2.

1

d vd to load

to generator

Z L

d

to load

to generator


73/122

72

2. Location of the voltage maxima and minima

The voltage maximum occurs when the reflected voltage

and the incident one are along the same direction.The voltage minimum occurs when the reflected voltage

and the incident one are in opposite directions. The

VSWR is defined in terms of the voltage maximum and

the voltage minimum as

min

max

1

1

V V

VSWR

vmin

vmax


74/122

73

3. Location of open- and short-circuits

In the case of an open circuit, we have . . 1o c

In the case of an open circuit, we have . . 1 s c

. o.c.

Unity circle

s.c.

Unity circle


75/122

74

4. Location of different loads

Z L is pure resistive, i.e. Z L = R L

1,where

11

0

0

0

R Rr

r r

R R R R

L L

L

L

L

L L

Z L is pure reactive i.e. Z L = jX

s.c o.c.

Unity circle

Inductive

Capacitive

Resistive

0

0

1

11

where /

1, 2tan

For 0 : 0

For 0: 0

L

o

L

jX R jx jX R jx

x X R

x

x

x


76/122

75

Spatial distribution using the unity circle

Z L=

Open Ct.

Z L= 0Short Ct.

Z L= R L+jX L Inductive

Z L= R L > R0 Resistive

Z L= R L jX L Capacitive

Z L= R L < R0 Resistive


77/122

76

5. Expressing Z in terms of

The input impedance at any point can be expressed in terms of the

reflection coefficient at that point as

2222

22

0

0

1211

11

1

1

Let

11

11

vu jvvu

jxr

vu

jvu jvu

jvu

jvu z

jxr z and jvu

Z Z

z

Z Z

in

inin

in

22

22

1

1

vu

vur

2212

&vu

v x


78/122

77

Constant resistance curves

011

12

11

22

22

22

r r

r r

uvu

vuvur

which represents the equation of a circle

r= 0

r= 0.5

r= 1

r= 2

r= 4

Unity circle

r =


79/122

78

Constant reactance curves

22

2

22

111

1

2

x xvu

vu

v x

which represents the equation of a circle

Unity circle

x= 0

x= 0.5

x= 1

x= 2

x= 4

x =

x=- 4

x=- 2 x=- 1

x=- 0.5

X is positive

X is negative


80/122

79

r= 0 r= 0.5 r= 1

r= 2 r= 4

r =

Unity circle

x= 0

x= 0.5

x= 1

x= 2

x= 4

x =

x=- 4

x=- 2

x=- 1

x=- 0.5

X is positive

X is negative

Collection of constant r and constant x curves

or THE SMITH CHART


81/122

80

Voltage standing wave ratio

The voltage standing wave ratio is given by

When is positive real value, then S is given by

Where r represents the real normalized impedance, which isgreater than unity.Hence, the SWR can be obtained from the graph as the value of r at the intersection with the positive real axis.

Unity circle

r= S r=1/ S

11

S

1

1S r


82/122

81

Graphical Representation of Currents

The current at any point is given by

Where I+ = V +/Z0 and I- = -V -/Z0

Normalizing I w.r.t. I+

we get

The admittance chart which is based on the graphicalrepresentaion of the current is shown below

Note that the location of the open circuit and the short circuit on

the graph are interchanged in the impedance and admittancecharts.

i

o.c. s.c.

Admittance Chart

imin imax

I I 1 I I

1 I

i


83/122

82


84/122

83


85/122

84

Applications of the Smith Chart

1. Calculation of l from Z L.

2. Calculation of d .3. Calculation of Z in.

4. Calculation of Y in.

5. Calculation of vmax and its location.

6. Calculation of vmin and its location.

7. Calculation of VSWR.

8. Impedance matching


86/122

85

1. Calculation of l from Z L

From Z L and R0, get z L as

jxr R Z z L L0

Locate ( r, x ) Measure | l | and L

E XAMPLE

Find l for Z L = 60 + j 40 , R0 = 50

r | |

L

P

1.2 0.8

0.352 56 L

l

z j


87/122

86

2. Calculation of Z i and d at a distance d

Locate z L at P 1.

Move d towards generator, i.e. rotate d/ in the

clockwise direction along a circle with a constant

radius | l |

Locate P 2.

Get z i= r i + j x i

Get d as :

d d d

l

d

P 1

P 2

d /

0i i i i Z z R R jX


88/122

87

3. Determination of V max and its first location

Locate z L.

Rotate till + ve horizontal axis.

Get d

Get d

1max V V

P 1

P 2

d

V

V max

d


89/122

88

4. Determination of V min and its first location

Locate z L.

Rotate till ve horizontal axis.

Get d

Get d

1min V V

P 1

P 2

d

V

V min

d


90/122

89

5. Determination of VSWR

0when1

1 z

vv

V V

VSWR

Locate z L.

Locate v

Read z = r + j0

SWR = r

P 1

P 2

d


91/122

90

6. Transition from impedance to admittance chart

If we have a point on the line represented by an impedance z on theSmith chart, the chart is called an impedance chart.

1

1111

1

111

Since 0

y

z y

z

R Z

which equals to the value on the Smith chart at the point( ), which is located at 180 rotation.

Hence, the same point on the line can be represented as anadmittance Y on the chart by rotating 180 and the chart in this case istreated as an admittance chart.

P 1

P 2

z

y


92/122

91

Impedance and the admittance Smith charts

o.c. s.c.

z

Impedance Chart

s.c. o.c.

y

Admittance Chart

b is positive

b is negative X is negative

X is positive


93/122

92

7) Calculation of Y from Z

The steps of calculating Y from Z are as follows

Locate z at P 1.

Rotate 180 and locate P 2.

Get y = g + jb

Get Y = y .Y 0 = y/ R0

If P 1 represents z at any point on the line, then P 2

represents Y at the same point on the line


94/122

93

8. Impedance Matching in Transmission Lines

Why is matching required?

To optimize the load power.

To decrease the line losses.

Why is matching preferred near the load?

For matching stability against frequency drift of the

source.

Methods of matching

Using a /4 matching section.

Using a single parallel stub.

Using a double parallel stub.


95/122

94

1. Matching using a /4 sectiona)- Resistive load R L

The condition of the matching is

For a transmission line section of length /4:

Li R Z Z 2

0

0

20

0 0

i

i L

L

Z Z

Z Z R

Z R Z

R L Z' 0 Z 0 Z i

/4


96/122

95

b)- Complex Load

Z i is real at locations of V max and V min.

Matching is possible using a /4 section at theselocations.

At location of voltage maximum, S Z Z 00

At location of voltage minimum, S Z Z 00

Z L

V

Z i=SZ 0 Z i=Z 0/S Z i=SZ 0 Z i=Z 0/S


97/122

96

Matching using a single stub

The stub is a short-circuited line with adjustable length.

The input admittance of the short-circuited stub is pureimaginary and is given by.

s sis

sis jbl j yl j

Y Y

cot

tan0

If the stub is connected to the line at any point having

Z L yi

y s

ytot

L s

Y is Short circuit stub

jb g y i

jbs jb g y tot


98/122

97

To achieve matching, the following relation should be

satisfied

Which means that

1 g bb s

The last condition means that the stub should be connected to

a point along the line where g=1. Therefore, we move along

the line (towards generator) by a distance d such that the real

part of y i equals 1.

These steps of matching are summerized as follows

Locate z L Locate y L by invertion

Rotate with constant radius | | clockwise

Intersect the circle g = 1

Get d /

Get the susceptance of the stub as b s = - b

Get the length of the short circuit stub as l s

01 j jbs jb g ytot

s s

i

l bband

d jb y

get

get1


99/122

98

Note that, when rotating with constant radius as we move along the lineto get the proper point for connecting the single short circuit stub, there

are two points of intersection with the unity circle ( g = 1). The previous

steps are shown in figure below

z L

y L

b

bs= b

d /

l s/


100/122

99

2. Double-stub Matching

Using a single stub, matching cannot be achieved unless we place the stub at a specific distance from the load.

Double stub is easier because the first stub is placed atarbitrary distance.

Case (1): d = 0

Z L

Stub 2 Stub 1

d d s

z L

y L

b

bs2= b

l s/ b1

b b2

bs1=b 2-b1

P 1

P 2

P 3 P 4

P 5


101/122

100

The steps of matching are summarized as follows

Get z L.

Get y L.

Rotate with constant g .

Intersect with the unity g circle rotated by (d s/ )counter clockwise (e.g. = + 90 for d s/ =1/8) .

Rotate the point of intersection 90 clockwise, till it intersects

with the unity g circle.

Case (2): d 0

Move d / .

Continue as case (1) from the new point.

z L

y L

b

bs2= b

d /

l s/

b1

b b2

bs1=b 2-b1

P 1

P 2

P 3

P 4 P 5

P 6


102/122

101

EXAMPLE 1

A transmission line has a characteristic impedance R 0=50 andload impedance Z L =60+40j .

1. Calculate L.

2. Calculate Y L.

3. Calculate the VSWR.

4. Calculate Z in and L at a distance 50 m from the load and

frequency = 2 MHZ.

5. Calculate vmax and its location.

6. Calculate vmin and its location.


103/122

102

P 1

Q 1

P 3

P 1

P 4P 5

P 2

Q 4 Q 5


104/122

103

Calculation of l from Z L.

Since Z L =60+40j and R 0=50 Hence z L = 1.2+j0.8 , Locate P 1 and Q 1 L = 0.35 ang(56 0)

Calculation of Y L.

Invert Z to Y Locate P 2yL =0.57 j 0.38YL =0.0114 j 0.0076 mhos

Calculation of the VSWR.

Locate P 4 and read the corresponding value of r VSWR = 2.08

At a distance 50 m from the load = 150 md/ = 1/3 m

Locate P 3 Zi = 0.47 +j 0.035Zi = 23.5 + j 1.75

d = 0.352 ang(176 0)

Calculate vmax and its location.

Locate P 4 and Q 4 vmax = 1.35

d max / = Q 4 Q 1 = 0.25-0.171 = 0.099

d max = 14.85 m

Calculate vmin and its location.

Locate P 4 and Q 5 vmin = 0.65

d min / = Q 5 Q 1 = 0.099 +0.25 = 0.349

d min = 52.35 m


105/122

104

EXAMPLE 2

A transmission line has a characteristic impedance R 0=50 and

load impedance Z L =100+100j . Find the length and locationof a single shunt stub required to perform matching to the load.

The steps of matching

Location of z L

z L = 2 + 2j point P 1

Locate y L by invertion y L = 0.25 0.25j point P 2

Rotate with constant radius | | clockwise direction and

intersect the unity g circle.

point P 3

Get d /

d / = P6 P5= 0.179+0.041 =0.22

Get the susceptance of the stub as b s = - b

b s = - 1.62

Get the length of the short circuit stub as d s /

d s / = 0.25-0.16=0.09

Another solution

The other point of intersection with the g unity circle leads to another

solution resulting in the following

d / = 0.362

d s / = 0.411


106/122

105

P 1

Q 1

P 1

P 4P 5

P 2

Q 4

P 1

P 2

`P 4

b

b s2= - b

d s/

P 9

P 8

P 5

P 7

P 3

P 6

d/


107/122

106

EXAMPLE 3

A transmission line has a characteristic impedance R 0=50 and

load impedance Z L =100+100j . Find the lengths of two shuntstubs required to perform matching to the load if the first stub is

connected at the load and the other one is separated by /8 .

The steps of matching (one solution only is shown)

Location of z L

z L = 2 + 2j point P 1

Locate y L by invertion y L = 0.25 0.25j point P 2

Move with constant g circle clockwise direction and intersect

the displaced unity g circle, point P 3.

Find b 1 at point P6 and b2 at point P7. Get the susceptance of

the first stub b s1 as

b s1 = b 2 b 1 b s1 = 0.35 (-0.25) = 0.6

Get d s1/ as

d s1/ = 0.25 + 0.086 = 0.336

Rotate with constant radius | | clockwise direction and

Intersect the unity g circle, point P 4 , (1.0 + j 1.92) .

Move with constant g circle clockwise direction and intersectthe origin, point P 5.

Locate b at point P6. Get the susceptance of the second stub as

bs2 = - b

bs2 = - j 1.92

Get the length of the short circuit stub d s2 / as

d s2 /

= 0.334 - 0.25 = 0.084


108/122

107

P 1

P 2

P 3

P 5

b

b s2= - b

d s2/

P 10

P 9

P 6

P 8

b2-b1

P 4

P 7

b1

b2


109/122

108

Transients in Transmission Lines

Case 1: A unit step voltage applied to a lossless line of

infinite length or terminated by Z 0

t

V

LC u

1

E0

E0


110/122

109

Case 2: A unit step voltage applied to a line terminated

in a load= 3Z0.

E0 3Z0

z

V

E0 t < T u

z

V

E0 T< t < 2T

z

V

E0 t > 2T

u


111/122

110

`

V L

t

E 0 1.5 E 0

z

t

T

3T

2T

4T

5T

6T

E 0

E 0/2

E 0/2

E 0/4 E 0/4

E 0/8

E 0/8

E 0/16


112/122

111

Reflection of a pulse

At time t 1 :

V

t 1 Z

E(t) 3Z0

Z

V

t Z


113/122


114/122

113

The Scattering Matrix

In the shown network, if the characteristic impedance isthe same at both ports, the relation between the incidentand reflected voltages can be put in the form

2

1

2

1

2

1

2221

1211

2

1

i

i

o f

r i

r

r

V

V s s

s s

V

V

or V

V

s s

s s

V

V

where S 11 , S 12, S 21 and S 22 are called the scattering parameters of thegiven two-port network.

These parameters are the most suitable representation for multi-portnetworks at high frequencies.

Two-port Network

1 2

V 1+ 1

V 1-

V 2+

V 2-

Z0 Z0


115/122

114

Scattering parameters evaluation

The evaluation of the scattering parameters for the two-port network can be performed as follows

The input and forward S parameters S i and S f can be evaluated as

01

2

2

&iV i

r f V

V s

which means that port 2 should be matched in order that V i2, (or thereflected voltage at port 2) be zero. This is shown in the following figure.

Similarly S 12 and S 22 can be evaluated in the same manner.

01

1

2iV i

r i V

V s

Two-port Network

1 2

V 1+

V 1- V 2- Z0 Z0 Z0


116/122

115

Why the S parameters are the most suitable ones inrepresenting networks at high frequencies?

Matched loads are used in the determination of the S parameters.These matched loads are relatively easier to realize than the openor short circuit which may caus4e instability specially, if activesources are included. When only the magnitudes of the S parameters are required, the

position of the reference planes at the input or the output ports arenot important.

General definition of scattering parameters

The scattering parameters can be evaluated also from the currentrelations; however, these parameters are evaluated using moregeneral variables which are directly related to the power. Thisrepresentaion is put in the form

2

1

2221

1211

2

1

a

a

s s

s s

b

b

or the form

2

1

2

1

a

a s s

s s

b

b

o f

r i

where a 1, a2, b1, and b 2 are defined as follows

*1 11 12

a a averageinputpowerflowingtowardsport

*2 2

12

2a a averageinputpowerflowingtowardsport

*1 1

11

2b b averageinputpowerflowingawayfromport

*2 2

12

2b b averageinputpowerflowingawayfromport


117/122

116

In terms of the voltages and currents at the input and output, thea and b variables can be put in the form

01

22

01

2

2

01

11

01

11

Z V

b

Z V

a

Z V

b

Z V

a

or in the form

02

20222

02

20222

01

10111

01

10111

2

2

2

2

Z I Z V

b

Z I Z V

a

Z I Z V

b

Z I Z V

a


118/122

117

Note that it can be shown that the net power flow inside port 1 is given by

and that the net power flow inside port 2 is given by

Important prperties of the S matrix

Lossless Network

In the case of lossless network, the net power flow inside the network

should be zero. This statement can be written in the form

b b a a i i i

N

j j j

N * *

1 1

0

In matrix notation, we have

b b a a

or

S S I

t t

t

* *

*

0

Reciprocal Network

In the reciprocal network, since Z ij = Z ji, hence it can be proven

that S ij = S ji or that the S matrix is symmetrical

)(21 2

1

2

11 baW f

)(21 2

2

2

22 baW f


119/122

118

Active source representation

s

s s s

s

s s

s s

Z Z

Z V e

Z

ea

baa

I Z V V

0

0

0

,where

Available Power

22 2 2

2

*

2

2

&

then1

1

1

For , with 0

1

s s

l

s

s l

l load s

s l

s l

sav

s

a a b

b a

aa

P a b a

l

a P


120/122

119

Case of source and load connection

so

s f r

oout

l o

l f r

iin

s

s s

s

s

s s s

1,

1

Example 1

Find the S parameters of the shown network

Solution

To get S 11 and S 21, consider a1 only

S

S e j 11

211 2

0.

with a2 = 0,

S S e j

22

121 20 .

L

NM

O

QPS

e

e

j

j

0

0

1 2

1 2

.

.

0.6

Z0


121/122

120

Example 2

Find the S parameters of the shown network

Solution

To get S 11 and S 21, consider a 1 only

S e Z Z Z Z e

e

S e e

e

L j

j

j

j j

j

11 2 6 8

0 0 0 02 6 8

1 5

21

34

12

54

0 5 0 5

13

2

323

( / )

( / )

.

( . ) / ( . )

with a 2 = 0 then S 22 and S 12 are given by,

S e

Z Z Z Z e

S S

L j

j

222 2 4

0 0 0 0

12 21

0 5 0 5

13

( / )

( . ) / ( . )

3 /8

Z0

/4

Z0


122/122

L

N

MMM

O

Q

PPP

S e e

e

j j

j

13

23

23

13

1 5 1 25

1 25

. .

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