EELE 3332 – Electromagnetic II

Chapter 11

Transmission Lines

Islamic University of Gaza

Electrical Engineering Department

Dr. Talal Skaik

2012 1

2

11.1 Introduction

Wave propagation in unbounded media is used in radio or TV

broadcasting, where the information being transmitted is meant for

everyone who may be interested.

Another means of transmitting power or information is by guided

structures. Guided structures serve to guide (or direct) the

propagation of energy from the source to the load.

Typical examples of such structures are transmission lines and

waveguides.

Waveguides are discussed in the next chapter; transmission lines

are considered in this chapter.

3

Transmission lines are commonly used in power distribution (at

low frequencies) and in communications (at high frequencies).

A transmission line basically consists of two or more parallel

conductors used to connect a source to a load.

Typical transmission lines include coaxial cable, a two-wire line, a

parallel-plate line, and a microstrip line.

Coaxial cables are used in connecting TV sets to TV antennas.

Microstrip lines are particularly important in integrated circuits.

Transmission line problems are usually solved using EM field

theory and electric circuit theory, the two major theories on which

electrical engineering is based.

Introduction

4

Figure 11.1 Typical transmission lines in cross-sectional view: (a) coaxial line, (b) two-wire line, (c) planar line, (d) wire above conducting plane, (e) microstrip line.

Introduction

Transmission line parameters are:

5

11.2 Transmission Line Parameters

R: Resistance per unit length. (Ω/m)

L: Inductance per unit length. (H/m)

G: Conductance per unit length. (S/m)

C: Capacitance per unit length. (F/m)

Distributed parameters of a two-conductor

transmission line

6

Transmission Line Parameters

7

The line parameters R, L, G, and C are uniformly distributed

along the entire length of the line.

For each line, the conductors are characterized by σc,µc,εc,=ε0,

and the homogeneous dielectric separating the conductors is

characterized by σ,µ,ε.

G ≠1/R; R is the ac resistance per unit length of the conductors

comprising the line and G is the conductance per unit length due to

the dielectric medium separating the conductors.

For each line:

Transmission Line Parameters

and G

LCC

8

Fields inside transmission line

•Transmission lines transmit TEM waves.

•V proportional to E,

•I proportional to H

9

11.3 Transmission Line Equations

Two-conductor transmission lines support a TEM wave; E and H

are perpendicular to each other and transverse to the direction of

propagation.

E and H are related to V and I:

Using V and I in solving the transmission line problem is simpler

than solving E and H (requires Maxwell’s equations).

Examine an incremental portion of length Δz of a two-conductor

transmission line.

E. , I= H.V dl dl

10

Transmission Line Representation

11

Transmission Line Equations ( , )

Using KVL:- ( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( , )

Taking the limit as z 0 leads to:

( , ) ( , ) ( , )

I z tV z t R zI z t L z V z z t

t

V z z t V z t I z tor RI z t L

z t

V z t I z tRI z t L

z t

equivalent circuit model

of a two-conductor T.L.

of differential length z.

12

Using KCL:- ( , ) ( , )

( , ) ( , )= ( , ) ( , )

( , ) ( , ) ( , ) ( , )

Taking the limit as z 0 leads to:

(

I z t I z z t I

V z z tI z t I z z t G zV z z t C z

t

I z z t I z t V z z tor GV z z t C

z t

I z

, ) ( , )( , )

t V z tGV z t C

z t

The time domain form of the transmission line equations:

( , ) ( , )( , )

( , ) ( , )

( , )

If we assume harmonic time dependence so that:

V( , )

V z t I z tRI z t L

z t

I z t V z tGV z t C

z t

z t

j t

s

j t

s

s s

=Re[ (z) ]

I( , )=Re[ (z) ]

where and are the phasor forms of ( , ) and ( , ),

( )

( )

ss

ss

V e

z t I e

V I V z t I z t

dVR j L I

dz

dIG j C V

dz

13

Transmission Line Equations

s

2

2

s

2

2

( ) , ( )

To solve the previous equations, take second derivative of gives

( )( )

Now take second derivative of gives

( )(

s ss s

ss

s

dV dIR j L I G j C V

dz dz

V

d VR j L G j C V

dz

I

d IG j C R j

dz

22

2

22

2

)

, the for voltage and current become

0

, where ( )( )

0

s

ss

ss

L I

Hence

d VV

dzj R j L G j C

d II

dz

wave equations

14

Transmission Line Equations

( )( )

: is the propagation constant

: attenuation constant (Np/m or dB/m)

: phase constant ( rad/m)

2wavelength is: = , wave velocity is:

The soluti

j R j L G j C

u f

0 0 0 0

0 0 0 0

ons to the wave equations are:

,

where , , , are wave amplitudes.

wave traveling along +z direction.

- sign wave traveling along -z direction

z z z z

s sV V e V e I I e I e

V V I I

sign

. 15

+z -z +z -z

16

0 0

0 0

0 0

In time domain:

( , ) Re[ ( ) ]

( , ) cos( ) cos(

z z

s

z z

s

j t

s

z z

V V e V e

I I e I e

V z t V z e

V z t V e t z V e

0 0

)

Similarly for current:

( , ) cos( ) cos( )z z

t z

I z t I e t z I e t z

Transmission Line Equations

0 0

0 0

0 0

0 0

0

( ) , ( )

( )since ( ) ( ),

( ) ( )

Characterestic 1 ,

Admittance

z z

z z

o o o

o o o

V z V e I z I e

dV zR j L I z

dz

V e R j L I e

V R j L R j L VZ R jX

I G j C I

R j LZ R jX Y

G j C Z

017

Characteristic Impedance, Z0

The Characteristic Impedance Z0 of the line is the ratio of the

positively travelling voltage wave to the current wave at any point on

the line.

For lossless line, R=G=0

Since = ( )( )

0, ,

1 2 ,

0o o o

j R j L G j C

j LC

u fLC

LX Z R

C

18

Lossless Line (R=G=0) A transmission line is said to be lossless if the conductors of the line

are perfect (σc ≈ ∞) and the dielectric medium separating them is

lossless (σ ≈ 0)

19

Any signal that carries significant information must has some

non-zero bandwidth. In other words, the signal energy (as well as

the information it carries) is spread across many frequencies.

If the different frequencies that comprise a signal travel at

different velocities, that signal will arrive at the end of a

transmission line distorted. We call this phenomenon signal

dispersion.

Recall for lossless lines, however, the phase velocity is

independent of frequency—no dispersion will occur!

Of course, a perfectly lossless line is impossible, but we find

phase velocity is approximately constant if the line is low-loss.

Distortionless Line

1/u LC

A distorionless line results if the line parameters are such that

, = ( )( )= 1 1

= 1

R G

L C

j L j CThus R j L G j C RG

R G

j CRG j

G

or RG

,

/ 1/ (frequency independent)

LC

u LC

20

Distortionless Line (R/L=G/C)

A distortionless line is one in which the attenuation constant α is

frequency independent while the phase constant β is linearly

dependent on frequency.

α does not depend on

frequency, whereas β is a linear

function of frequency.

Notes:

Shape distortion of signals happen if α and u are frequency

dependent.

u and Z0 for distortionless line are the same as lossless line.

A lossless line is also a distortionless line, but a distortionless line

is not necessarily lossless.

Lossless lines are desirable in power transmission, and telephone

lines are required to be distortionless. 21

Distortionless Line (R/L=G/C)

0

1 / (Real)

1 /

1u=

R j L RR j L R LZ

G j C G j C G G C

LC

To achieve the required condition of R/L=G/C for a transmission line,

L may be increased by loading the cable with a metal with high

magnetic permeability (μ).

A common practice is to replace repeaters in long lines to maintain the

desired shape and duration of pulses for long distance transmission. 22

Distortionless Line – Practical use

Zo

General

Lossless

Distortionless

CjG

LjRZo

o o

LZ R

C

))(( CjGLjR

C

LRZ oo

Summary

0 j LC

RG j LC

24

Example 11.1 An air line has characteristic impedance of 70 Ω and a phase

constant of 3 rad/m at 100 MHz. Calculate the inductance per meter

and the capacitance per meter of the line.

0 0

0

6

0

An air line can be regarded as lossless line because 0

. 0 and =0

= =

1Deviding the two equations yields:

3 6

2 100 10 (70)

cand Hence R G

LZ R

C

LC

R

C

or CR

2 2 12

0

8.2 pF

= (70) (68.2 10 ) 334.2 nH/mL R C

0

0

3

0

2 6

0

1A distortionless line has or ,

, =

(20 10 )(60) 1.2 /m

400 10 = 333 S/m

1.2

1 by gi

RCRC GL G u

L LC

L C RZ RG R

C L Z

R Z

Since RG GR

LDividing Z u

C LC

0

8

ves

60= 333 nH/m

0.6(3 10 )

ZL

u

25

Example 11.2

A distortionless line has Z0=60 Ω, α=20 mNp/m, u=0.6c, where c is

the speed of light. Find R,L,G , C and λ at 100 MHz.

26

0

0 8

0

8

8

1 by gives

1 1 1= 92.59 pF/m

0.6(3 10 )60

0.6(3 10 )= 1.8 m

10

LMultiplying Z u

C LC

uZ CC uZ

u

f

Example 11.2 – solution continued

27

11.4 Input impedance, standing wave ratio, power

Consider a transmission line of length l, characterised by and Z0,

connected to a load ZL. Generator sees the line with the load as an

input impedance Zin.

0 0

0 0 0 00

0 0 0 0

( )

At generator terminals (sen

( ) , =

ding e

n

Let

d):

z z

s

z z

s

V z V e V e

V V V VI z e e Z

Z Z I I

0 0

0 0 0 0 0 0 0

0 00

0 0 0 00 0

in

0 0

( 0), ( 0), Substitute in prev. equs.:

1 2

... (1) 1

2

If the input impedance at the terminals is Z ,

then , gin

g

in g

V V z I I z

V V V V V Z I

V VI

V V Z IZ Z

VZV V I

Z Z Z

in gZ 28

Input impedance

0 00 0

0 0

0 0 0 0

0 0

00 0

( ) , ( )

Let ( ), ( ), Substitute in prev.

At the load:

equs.

1 2

1

2

z z z z

s s

L L

l l lL L L

l l

L

V VV z V e V e I z e e

Z Z

V V z l I I z l

V V e V e V V Z I e

V VI e e

V VZ Z

0

in

... (2)

Now determine the input impedance Z = ( ) / ( ) at any

point on the line.

l

L L

s s

Z I e

V z I z

29

Input impedance

0 00 0 0 0

0 0

0 0 00in

0 0 0

At the generator, recall , ,

( )Z =

( )

Substituting eq. 2 and utilizing

the fact that:

cosh , sinh , 2 2

or t

s

s

l l l l

V VV V V I then

Z Z

Z V VV z V

I z I V V

e e e el l

00

0

sinhanh

cosh

tanhwe get (General - Lossy Lin

nh)

te

a

l l

l l

Lin

L

l e el

l e e

Z Z lZ Z

Z Z l

30

Input impedance

00

0

00

0

tanh (General - Lossy Line)

tanh

For a lossless line, = , tanh tan , then

tan

tan

Lin

L

Lin

L

Z Z lZ Z

Z Z l

j j l j l

Z jZ lZ Z

Z jZ l

( )

Note: To find at a distance

Loss

less L

' from load,

e

rep

in

inZ

l

00

0

lace   by ' :-

tan '

tan '

Lin

L

l l

Z jZ lZ Z

Z jZ l

31

Input impedance (Lossless Line)

βl is known as electrical length, in degrees or radians

0

0

0 0

0 0

Define as the voltage reflection coefficient (at the load), as the

ratio of the voltage reflection wave to the incident wave at the load,

1

2 , an

1

2

L

l

L l

l

L L

l

L L

V e

V e

V V Z I e

Since

V V Z I e

0

0

Volt

d

age Reflection coefficient at l

oa d ( )

L L L

LL

L

V Z I

Z Z

Z Z

32

Reflection Coefficient, (at load)

0

0

0 00 0

0 0

0 0 0 0

0 0

Define (at 0) as the voltage reflection coefficient at the source,

as the ratio of the voltage reflection wave to the incident wave at source,

1

2

1

2

z

V e V

V e V

V V Z I

Since

V V Z

0 0

0 0

00

0

V

, and

oltage Reflection coefficient at

( )source

in

in

in

V Z I

I

Z Z

Z Z

33

Reflection Coefficient, (at generator)

20 0

0 0

0 0

That is: ( )

Thus the current reflection coefficient at the load is

/

zz

z

l l

L

V e Vz e

V e V

I e I e

34

Reflection Coefficient

The current reflection coefficient at any point on the line is the

negative of the voltage reflection coefficient at that point.

The voltage reflection coefficient at any point on the line is the ratio

of the reflected voltage wave to that of the incident wave.

max max 0

min min 0

0

The standing wave ratio s is defined as: (as we did for plane waves)

1=

1

When load is ( ) Total Transmission

L LL

L L

L

V I Z Zs

V I Z Z

perfectly matched Z Z

0 1

When load is a :

Total Reflection

1 1

When load is an :

Total Reflection

1 1

L

L L

L L

s

short circuit

s

open circuit

s

35

Whenever there is a reflected wave, a standing wave will form

out of the combination of incident and reflected waves.

Standing Wave Ratio

36

*

2

20

0

The time-average power flow along the line at the point is:

1Re[ ( ) ( )]. For a line, this can be reduced to:

2

1 , 2

The average powe

ave s s

ave

z

P V z I z lossless

VP

Z

2 2 2

0 0 0 0

r flow is constant at any point on the line.

The total power delivered to the load ( ) is equal to the incident

power ( / 2 ) the reflected power ( / 2 )

If 0, maximum p

minus

av

l

P

V Z V

le s

Z

oss s

ower is delivered to the load, while no power

is delivered for 1.

The above discussion assumes that the generator is matched.

Power

Incident

Power (Pi)

Reflected

Power (Pr)

2 22

0 0 0 0/ 2 / 2aveP V Z V Z

Short Circuited Line

(ZL=0)

Open Circuited Line

(ZL=∞)

Matched Line

(ZL=Z0)

37

Special Cases , ZL=0, ZL=∞, ZL=Z0

38

Shorted Line (ZL=0)

00 0

0

0

0

tantan

tan

1, (Total Reflection)

Lin

L

LL

L

Z jZ lZ Z jZ l

Z jZ l

Z Zs

Z Z

39

Open-Circuited Line (ZL=∞)

00

0

0

00 0

0

0

0

tan, ( )

tan

1 tan

cot tan

tan

1, (Total Reflection)

Lin L

L

L

in

L

LL

L

Z jZ lZ Z Z

Z jZ l

Zj l

Z ZZ Z jZ l

j lZj l

Z

Z Zs

Z Z

40

Matched Line (ZL=Z0)

0 0

0

0

Most desired case from practical point of view.

0, 1

The whole wave is transmitted, and there is no reflection.

The incident power is fully absorbed by the load.

(

in in

LL

L

Since Z Z Z Z

Z Zs

Z Z

Maximum power transfer)

( ) Since 1 Np=8.686 dB

8 = 0.921 /

8.686

= + 0.921 1

=2(0.921 1)

Solution

a

Np m

j j

l j

1.84 2j

41

Example 11.3

A certain transmission line 2 m long operating at ω=106 rad/s has

α=8 dB/m, β=1 rad/m, and Z0= 60+j40 Ω. If the line is connected

to a source of 10∟00 V, Zg=40 Ω and transmitted by a load of

20+j50 Ω, determine

(a) The input impedance

(b) The sending end current

(c) The current at the middle of the line.

00

0

tanh tanh 1.84 2 1.033 0.03929

tanh

tanh

20 50 (60 40)(1.033 0.03929)(60 40)

60 40 (20 50)(1.033 0.03929)

60.25 38.79

( ) The sending end cu

Lin

L

in

in

l j j

Z Z lZ Z

Z Z l

j j jZ j

j j j

Z j

b

0

0

rrent is ( 0)

1093.03 21.15

60.25 38.79 40

g

in g

I z I

VI mA

Z Z j

42

Example 11.3 – Solution continued

0 0

0

0 0 0

0 0

0 0 0 0

( ) To find the current at any point, we need V and V .

93.03 21.15

(71.66 32.77 )(0.09303 21.15 ) 6.667 11.62

1 1V 6.667 11.62 (60 40)(0.09303 21.15)

2 2

6.687 12.

in

c But

I mA

V Z I

V Z I j

0

0 0 0 0

0

08

1 1V 6.667 11.62 (60 40)(0.09303 21.15)

2 2

0.0518 260

V Z I j

43

Example 11.3 – Solution continued

0 0

/2 /20 0

0 0

12.08 0.921 1 260 0.921 1

At the middle of the line, / 2, z= / 2 0.921+j1, Hence the

current at this point is:

V V( / 2)

(6.687 ) (0.0518 ) =

60 40 60 40

No

l l

s

j j j j

z l l

I z l e eZ Z

e e e e

j j

0 0 0 0

0 0

0

0

12.08 0.921 57.3 260 0.921 57.3

33.69 33.69

78.91

te that 1 is in radians and is equivalent to j57.3 ,( 1 180/ ):-

(6.687 ) (0.0518 )( / 2)=

72.1 72.1

0.0369 0.001805

j j j j

s j j

j

j j

e e e e e eI z l

e e

e

0283.61

0

6.673 34.456 mA

=35.10 281 mA

je

j

44

Example 11.3 – Solution continued