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Amanogawa, 2006 Digital Maestro Series 1

Complex Numbers, Phasors and Circuits

Complex numbers are defined by points or vectors in the complexplane, and can be represented in Cartesian coordinates

1 z a jb j =or in polar (exponential) form

exp( ) cos sin

cos

sin

z A j A jA

a A

b A

= = = =

r

imagina

ea

ry

l part

part

where

2 2 1tanb

A a ba

= =

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exp( ) exp( 2 ) z A j A j j n = Note :

z

Re

Im

A

b

a

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Every complex number has a complex conjugate

* * z a jb a jb =

so that

22 2 2

* ( ) ( ) z z a jb a jb

a b z A

=

= = =

In polar form we have

* exp( ) * exp( )

exp 2cos sin

z A j A j

A j j A jA

= =

= =

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The polar form is more useful in some cases. For instance, when

raising a complex number to a power, the Cartesian form

( ) ( ) ( )n z a jb a jb a jb is cumbersome, and impractical for non integer exponents. Inpolar form , instead, the result is immediate

[ ] exp( ) expnn n z A j A jn =

In the case of roots , one should remember to consider + 2k asargument of the exponential, with k = integer, otherwise possibleroots are skipped:

2exp 2 expnn n

k z A j j k A j j n n

= =

The results corresponding to angles up to 2 are solutions of theroot operation.

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In electromagnetic problems it is often convenient to keep in mind

the following simple identities

exp exp2 2

j j j j = =

It is also useful to remember the following expressions fortrigonometric functions

exp( ) exp( ) exp( ) exp( )cos ; sin

2 2 jz jz jz jz

z z j

= =

resulting from Eulers identity

exp( ) cos( ) sin( ) jz z j z =

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Complex representation is very useful for time-harmonic functions

of the form]

]

]

cos Re exp

Re exp exp

Re exp

A t A j t j

j j t

A j t

= =

=

The complex quantity

exp A j

contains all the information about amplitude and phase of thesignal and is called the phasor of

cos A t

If it is known that the signal is time-harmonic with frequency , thephasor completely characterizes its behavior.

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Often, a time-harmonic signal may be of the form:

sin A t and we have the following complex representation

] ]

]

sin Re cos sin

Re exp

Re exp / 2 exp exp

Re exp / 2 exp

Re exp

A t jA t j t

jA j t j

j j j t

A j j t

A j t

= = =

= =

with phasor exp / 2 A A j This result is not surprising, since

cos( / 2) sin( )t t =

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Time differentiation can be greatly simplified by the use of phasors.Consider for instance the signal

0 0( ) cos expV t V t V V j = with phasor

The time derivative can be expressed as

}

0

0

0

( )

sin

Re exp exp

( )exp

V t

V t t

j V j j t

V t j V j j V t

=

=

= is the phasor of

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With phasors, time-differential equations for time harmonic signals

can be transformed into algebraic equations. Consider the simplecircuit below, realized with lumped elements

This circuit is described by the integro-differential equation

1( ) ( )

t d i t v t L Ri i t dt

dt C

C

R L

v t i (t)

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Upon time-differentiation we can eliminate the integral as

2

2( ) 1

( )d i t d v t d i

L R i t dt dt C dt

=

If we assume a time-harmonic excitation, we know that voltage andcurrent should have the form

0 0

0 0

( ) cos( ) exp( )( ) cos( ) epha

phxp( )o

a ors rsV V

I I

v t V t V V j i t I t I I j

= = = =

If V 0 and V are given,

I 0 and I are the unknowns of the problem.

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The differential equation can be rewritten using phasors

( ){ } ( ){ }

( ){ } ( ){ }

2Re exp Re exp

1Re exp Re exp

+

+ =

L I j t R j I j t

I j t j V j t C

Finally, the transform phasor equation is obtained as

1V R j L j I Z I C

= =

where

1 Z R j L

C =

Impedance Resistance

Reactance

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The result for the phasor current is simply obtained as

0 exp1 I V V

I I j Z

R j L j

C

= = =

which readily yields the unknowns I 0 and I .

The time dependent current is then obtained from

}

0

0

( ) Re exp exp

cos

I

I

i t I j j t

I t

= =

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The phasor representation of the circuit example above has

introduced the concept of impedance . Note that the resistance isnot explicitly a function of frequency. The reactance componentsare instead linear functions of frequency:

Inductive component proportional to Capacitive component inversely proportional to

Because of this frequency dependence, for specified values of L

and C , one can always find a frequency at which the magnitudes ofthe inductive and capacitive terms are equal

1 1r r

r L

C LC = =

This is a resonance condition. The reactance cancels out and theimpedance becomes purely resistive .

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The peak value of the current phasor is maximum at resonance

00 2

2 1

V I

R L C

=

I M

r

| I 0|

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Consider now the circuit below where an inductor and a capacitor

are in parallel

The input impedance of the circuit is

1

21

1in

j L Z R j C R

j L LC

= =

C

V

I L

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When

01

in

in

in

Z R

Z LC

Z R

= = =

=

At the resonance condition

1r

LC =

the part of the circuit containing the reactance componentsbehaves like an open circuit , and no current can flow. The voltageat the terminals of the parallel circuit is the same as the inputvoltage V .

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Power in CircuitsConsider the input impedance of a transmission line circuit, with anapplied voltage v(t) inducing an input current i (t) .

For sinusoidal excitation, we can write

[ ]/2 , /20

0

( ) cos( )

( ) cos( )

v t V t

i t I t

=

where V 0 and I 0 are peak values and is the phase differencebetween voltage and current. Note that = 0 only when the inputimpedance is real (purely resistive).

v(t)

i(t)

Z in

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The time-dependent input power is given by

[ ]

0 0

0 0

( ) ( ) ( ) cos( ) cos( )

cos ( ) cos(2 )2

P t v t i t V I t t

V I t

= =

=

The power has two (Fourier) components:

(A) an average value

0 0 cos( )2

V I

(B) an oscillatory component with frequency 2f

0 0 cos(2 )2

V I t

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The power flow changes periodically in time with an oscillation like(B) about the average value (A). Note that only when = 0 we havecos( ) = 1 , implying that for a resistive impedance the power isalways positive (flowing from generator to load).

When voltage and current are out of phase, the average value of thepower has lower magnitude than the peak value of the oscillatorycomponent. Therefore, during portions of the period of oscillationthe power can be negative (flowing from load to generator). Thismeans that when the power flow is positive, the reactive componentof the input impedance stores energy, which is reflected back to thegenerator side when the power flow becomes negative.

For an oscillatory excitation, we are interested in finding the

behavior of the power during one full period, because from this wecan easily obtain the average behavior in time. From the point ofview of power consumption, we are also interested in knowing thepower dissipated by the resistive component of the impedance.

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Using cos cos cos sin sin B A B A B= one can write

in phase withcurr

in quadraturewith current

0 0 0

ent

( ) cos( ) cos cos sin sinv t V t V t V t =

This gives an alternative expression for power:

0 0 0 0

2 0 00 0

0 0 0 0 0 0

( ) cos( ) cos( ) cos( ) cos( ) sin( ) sin( )

cos( ) cos ( ) sin( ) sin(2 )2

cos( ) cos( ) cos(2 ) sin(2 2 2

P t V t I t V t I t

V I V I t t

V I V I V I t

=

=

=

ReactReal Power

Real P

ive Powe

o

r

wer

) sin(2 )t

Reactive Power

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Transmission Lines


The real power corresponds to the power dissipated by the resistive

component of the impedance, and it is always positive.

The reactive power corresponds to power stored and then reflectedby the reactive component of the impedance. It oscillates frompositive to negative during the period.

Until now we have discussed properties of instantaneous power.Since we are considering time-harmonic periodic signals, it is veryconvenient to consider the time-average power

01

( ) ( )T

P t P t dt T

=

where T = 1 / f is the period of the oscillation.

To determine the time-average power, we can use either the Fourieror the real/reactive power formulation.

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Fourier representation

0 0 0 00 0

1 1( ) cos( ) cos(2 )

2 2T T V I V I

P t dt t dt T T

=

0 0

0

cos( )2

V I =

As one should expect, the time-average power flow is simply givenby the Fourier component corresponding to the average of theoriginal signal.

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Real/Reactive power representation

0 0 0 00 0

1( ) ( cos( ) cos( ) cos(2 )

2T T

t V I dt V I t dt T

= 0

0 00

)

1sin( ) sin(2 )

2T

V I t dt T

=

0 0

0

cos( )2

V I

=

This result tells us that the time-average power flow is the averageof the real power . The reactive power has zero time-average, sincepower is stored and completely reflected by the reactive componentof the input impedance during the period of oscillation.

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The maximum of the reactive power is

0 0 0 0max{ } max{ sin sin 2 } sin2 2reac

V I V I P t =

Since the time-average of the reactive power is zero, we often usethe maximum value above as an indication of the reactive power.

The sign of the phase tells us about the imaginary part of theimpedance or reactance :

> 0 The reactance is inductiveCurrent is lagging with respect to voltageVoltage is leading with respect to current

< 0 The reactance is capacitiveVoltage is lagging with respect to currentCurrent is leading with respect to voltage

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Transmission Lines


If the total reactance is inductive

V Z I R I j L I =

I

V j L I

R I

m

Re

Current lags > 0

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Transmission Lines


If the total reactance is capacitive

1V Z I R I j I

C = =

- j I / C

I

V

R I

m

Re

Voltage lags < 0

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The peak factor for sinusoidal signals is

0 2 1.4142rmsV

V =

For a symmetric triangular signal the peak factor is

0 3 1.732rms

V V

=

For a symmetric square signal the peak factor is simply

0 1rmsV

V =

V 0

t

t

V 0

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For a non-sinusoidal periodic signal , we can use a decomposition

into orthogonal Fourier components to obtain the r.m.s. value:

[ ]2 2

0 0

20 0

orthogona

1 2 3

l

0

2

20

1 1

( ) [ ( )]1 1

[ ( )] [ 2 ( ) ( )]

1

( ) ( )

1( ) [

( ) ( )

2 ( ) ( )

T T

k k

T T k i jk i j

av

T T k i jk

k

rms

k

V t dt V t dt T T

V t dt V t V t dt T T

V t dt

V t V V t V t V t V t

V t V t dt T T

V

= = =

=

=

=

=

=

21

2

2 2 22( )

] (

(

)

)( )

rms

rms rms rmsrms k avk

j k k i

V V V V V

V

= =

=

The final result holds for any decomposition into orthogonalfunctions and it is known in mathemics as Parsevals identity .

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In terms of r.m.s. values , the time-average power for a sinusoidal

signal is then

0 0( ) cos( ) cos( )2 2 rms rms

V I P t V I = =

Finally, we can relate the time-average power to the phasors ofvoltage and current. Since

0 0

0 0( ) cos( ) Re exp( )( ) cos( ) Re exp( )exp( )

v t V t V j t i t I t I j j t

= = = =

we have phasors

0

0 exp( )

V V

I I j

==

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The time-average power in terms of phasors is given by

*0 0

0 0

1 1( ) Re{ } Re{ exp( ) }

2 2

cos( )2

P t V I V I j

V I

= =

=

Note that one must always use the complex conjugate of the phasorcurrent to obtain the time-average power. It is important toremember this when voltage and current are expressed asfunctions of each other. Only when the impedance is purelyresistive, I = I* = I 0 since = 0 .

Also, note that the time-average power is always a real positivequantity and that it is not the phasor of the time-dependent power.It is a common mistake to think so.

Transmission Lines

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Now we consider power flow including explicitly the generator, to

understand in which conditions maximum power transfer to a loadcan take place.

*

1

1Re{ }

2

Rin G

G R

in G G R

in in in

Z V V Z Z

I V Z Z

P V I

=

=

=

Z G

V G V in

I in

Generator

Z R

Load

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To find the load resistance that maximizes power transfer to the

load for a given generator we impose( )

0d P t

d R=

from which we obtain

2

2

4

0

( )( ) 2 ( )

0( )

( ) 2 0

R

R G R

G R R G R

G R

G R R R G

d R

dR R R R R R R R

R R

R R R R R

=

=

= =

We conclude that for maximum power transfer the load resistancemust be identical to the generator resistance.

Transmission Lines

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Lets consider now complex impedances

R R G G G Z R jX Z R jX =

For maximum power transfer, generator and load impedances mustbe complex conjugate of each other:

* R G R G

G

Z Z R R

X X

= =

=

This can be easily understood by considering that, to maximize theactive power supplied to the load, voltage and current of the

generator should remain in phase . If the reactances of generatorand load are opposite and cancel each other along the path of thecurrent, the generator will only see a resistance. Voltage andcurrent will be in phase with maximum power delivered to the load.

V G RG

RG

X G jX G

Transmission Lines

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The total time-average power supplied by the generator in

conditions of maximum power transfer is2 2*1 1 1 1 1Re{ }

2 2 2 4tot G in G G R R P V I V V

R R= = =

The time-average power supplied to the load is

*

* *

2 22

1 1 1Re{ } Re2 2

1 1 1Re2 84

Rin in in G GG R G R

R RG G R

Z P V I V V Z Z Z Z

R jX V V R R

= =

= =

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The power dissipated by the internal generator impedance is

*

2 2 2

1Re { ( ) }

2

1 1 1 1 1 14 8 8

G G in in

G G G R R R

P V V I

V V V R R R

=

= =

We conclude that, in conditions of maximum power transfer, onlyhalf of the total active power supplied by the generator is actuallyused by the load . The generator impedance dissipates theremaining half of the available active power.

This may seem a disappointing result, but it is the best one can dofor a real generator with a given internal impedance!

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Transmission Line Equations

A typical engineering problem involves the transmission of a signalfrom a generator to a load. A transmission line is the part of thecircuit that provides the direct link between generator and load.Transmission lines can be realized in a number of ways. Common

examples are the parallel-wire line and the coaxial cable . Forsimplicity, we use in most diagrams the parallel-wire line torepresent circuit connections, but the theory applies to all types oftransmission lines.

Z R

Z G

V G Transmission line

LoadGenerator

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Examples of transmission lines

Microstrip

t

w

h

D

d

Coaxial cable

d

d D

Two-wire line

Transmission Lines

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If you are only familiar with low frequency circuits, you are used totreat all lines connecting the various circuit elements as perfectwires, with no voltage drop and no impedance associated to them(lumped impedance circuits). This is a reasonable procedure aslong as the length of the wires is much smaller than the wavelengthof the signal. At any given time, the measured voltage and current

are the same for each location on the same wire.

R R G

G R

Z V V

Z Z =

Z R Z G

V G

LoadGenerator

L

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Lets look at some examples. The electricity supplied to householdsconsists of high power sinusoidal signals, with frequency of 60Hzor 50Hz , depending on the country. Assuming that the insulatorbetween wires is air ( 0), the wavelength for 60Hz is:

862.999 10

5.0 10 5, 00060c

m km f

= = =

which is the about the distance between S. Francisco and Boston !Lets compare to a frequency in the microwave range, for instance60 GHz . The wavelength is given by

83

92.999 10

5.0 10 5.0

60 10

cm mm

f

= = =

which is comparable to the size of a microprocessor chip .

Which conclusions do you draw?

Transmission Lines

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For sufficiently high frequencies the wavelength is comparable withthe length of conductors in a transmission line. The signalpropagates as a wave of voltage and current along the line, becauseit cannot change instantaneously at all locations. Therefore, wecannot neglect the impedance properties of the wires (distributedimpedance circuits).

z z(z) j j V V e V e =

Z G

V G Z R

LoadGenerator

L

V( 0 ) V( z ) V( L )

Transmission Lines

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Note that the equivalent circuit of a generator consists of an idealalternating voltage generator in series with its actual internalimpedance. When the generator is open ( R Z ) we have:

0 i in n G V V I = = and

If the generator is connected to a load R Z

G

G in

G

nG

R

R

i R

I

V Z

V

V

Z Z

Z Z

=

=

If the load is a short ( 0 R

Z = )

0G

inG

in Z V

V I = = and

Z G

Generator

V in

I in

Z R

Load

V G

Transmission Lines

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The simplest circuit problem that we can study consists of avoltage generator connected to a load through a uniformtransmission line . In general, the impedance seen by the generatoris not the same as the impedance of the load, because of thepresence of the transmission line, except for some very particularcases:

in R Z Z in Z

[ integer ]2

n L n ==

Our first goal is to determine the equivalent impedance seen by thegenerator, that is, the input impedance of a line terminated by theload. Once that is known, standard circuit theory can be used.

Z RTransmission line

L

only if

Transmission Lines

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Z R

Z G

V G Transmission line

LoadGenerator

Equivalent LoadGenerator

Z in

Z G

V G

Transmission Lines

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A uniform transmission line is a distributed circuit that we candescribe as a cascade of identical cells with infinitesimal length.The conductors used to realize the line possess a certain seriesinductance and resistance . In addition, there is a shunt capacitance between the conductors, and even a shunt conductance if themedium insulating the wires is not perfect. We use the concept of

shunt conductance, rather than resistance, because it is moreconvenient for adding the parallel elements of the shunt. We canrepresent the uniform transmission line with the distributed circuitbelow (general lossy line)

L dz L dz R dz R dz

C dz C dzG dz G dz

dz dz

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Loss-less Transmission LineIn many cases, it is possible to neglect resistive effects in the line.In this approximation there is no Joule effect loss because onlyreactive elements are present. The equivalent circuit for the

elementary cell of a loss-less transmission line is shown in thefigure below.

L dz

C dz

dz

V z V z +d V

I (z) I z +d I

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The current flowing through the shunt capacitance determines thevariation of the current from input to output of the cell.

The circuit equation for the sub-circuit above is

d dz( d ) dz d dz I j C V V j CV j C V =

The second term (including dV dz ) tends to zero very rapidly in thelimit of infinitesimal length dz leaving a first order differentialequation for the current

ddz I

j C V

C dzV z +d V

I (z) I (z)+d I d I

Transmission Lines

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We have obtained a system of two coupled first order differentialequations that describe the behavior of voltage and current on theuniform loss-less transmission line. The equations must be solvedsimultaneously.

ddzddz

V j L I

I j C V

=

=

These are often called telegraphers equations of the loss-lesstransmission line.

Transmission Lines

l b f l d b

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One can easily obtain a set of uncoupled equations bydifferentiating with respect to the space coordinate. The first orderdifferential terms are eliminated by using the correspondingtelegraphers equation

d

dz

I j C V

22

2

2 22

d ddzdz

d ddzdz

V I j L j L j CV LC V

I V j C j C j L I LC I

= = =

= = =

ddzV

j L I

These are often called telephonists equations.

Transmission Lines

W h l d d d diff i l i f

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We have now two uncoupled second order differential equations forvoltage and current, which give an equivalent description of theloss-less transmission line. Mathematically, these are waveequations and can be solved independently.

The general solution for the voltage equation is

z z(z) j j V V e V e =

where the wave propagation constant is

LC =

Note that the complex exponential terms including have unitary magnitude and purely imaginary argument , therefore they onlyaffect the phase of the wave in space.

Transmission Lines

W h th f ll i f l l ti

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We have the following useful relations:

0 0

2 2

p p

r r r r

f v v

c

= = =

= = =

Here, pv f = is the wavelength of the dielectric mediumsurrounding the conductors of the transmission line and

0 0

1 1 p

r r v = =

is the phase velocity of an electromagnetic wave in the dielectric.

As you can see, the propagation constant can be written in manydifferent, equivalent ways.

Transmission Lines

The current distribution on the transmission line can be readily

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The current distribution on the transmission line can be readilyobtained by differentiation of the result for the voltage

z zddz

j j V j V e j V e j L I = =

which gives

z z z z

0

1(z) j j j j

C I V e V e V e V e

L Z

= =

The real quantity

0 L Z C =

is the characteristic impedance of the loss-less transmission line .

Transmission Lines

Loss Tr nsmission Line

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Lossy Transmission Line

The solution for a uniform lossy transmission line can be obtainedwith a very similar procedure, using the equivalent circuit for theelementary cell shown in the figure below.

L dz R dz

C dz

dz

G dzV z V z +d V

I (z) I z +d I

Transmission Lines

The series impedance determines the variation of the voltage from

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The series impedance determines the variation of the voltage frominput to output of the cell, according to the sub-circuit

The corresponding circuit equation is

( d ) ( dz dz)V V V j L R I = from which we obtain a first order differential equation for thevoltage

d( )

dzV

j L R I

L dz R dz

dz

V z V (z)+d V I (z)

Transmission Lines

The current flowing through the shunt admittance determines the

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The current flowing through the shunt admittance determines theinput-output variation of the current , according to the sub-circuit

The corresponding circuit equation is

d ( dz dz)( d )( ) dz ( )d dz

I j C G V V j C G V j C G V

= =

The second term (including dV dz ) can be ignored, giving a firstorder differential equation for the current

d( )

dz I

j C G V

C dzG dz V z +d V

I (z) I z +d I d I

Transmission Lines

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We have again a system of coupled first order differential equations that describe the behavior of voltage and current on the lossy transmission line

d ( )dzd

( )dz

V j L R I

I j C G V

=

=

These are the telegraphers equations for the lossy transmissionline case.

Transmission Lines

One can easily obtain a set of uncoupled equations by

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One can easily obtain a set of uncoupled equations bydifferentiating with respect to the coordinate z as done earlier

d( )

dz I

j C G V

2

2

2

2

d d( ) ( )( )

dzdz

d d( ) ( )( )dzdz

V I j L R j L R j C G V

I V j C G j C G j L R I

= =

= =

d( )dz

V j L R I

These are the telephonists equations for the lossy line.

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Transmission Lines

The current distribution on a lossy transmission line can be readily

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y yobtained by differentiation of the result for the voltage

z zd ( )dzV

j L R I V e V e = =

which gives

z z

z z0

( )(z) ( )

( )

1 ( )

j C G I V e V e

j L R

V e V e Z

=

=

with the characteristic impedance of the lossy transmission line

0( )( ) j L R

Z j C G =

Note: the characteristicimpedance is now complex !

Transmission Lines

For both loss-less and lossy transmission lines

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the characteristic impedance does not depend on the line length

but only on the metal of the conductors, the dielectric material surrounding the conductors and the geometry of the line cross-

section, which determine L , R , C , and G .One must be careful not to interpret the characteristic impedanceas some lumped impedance that can replace the transmission linein an equivalent circuit.

This is a very common mistake!

Z 0 Z R Z R Z 0

Transmission Lines

We have obtained the following solutions for the steady-state

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voltage and current phasors in a transmission line:

Loss-less line Lossy line

( )

z z

z z

0

(z)

1(z)

j j

j j

V V e V e

I V e V e Z

+

+

= +

= ( )

z z

z z

0

(z)

1(z)

V V e V e

I V e V e Z

+

+

= +

=

Since V (z) and I (z) are the solutions of second order differential(wave) equations , we must determine two unknowns , V + and V ,

which represent the amplitudes of steady-state voltage waves,travelling in the positive and in the negative direction, respectively.

Therefore, we need two boundary conditions to determine theseunknowns, by considering the effect of the load and of thegenerator connected to the transmission line.

Transmission Lines

Before we consider the boundary conditions, it is very convenient

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to shift the reference of the space coordinate so that the zeroreference is at the location of the load instead of the generator.Since the analysis of the transmission line normally starts from theload itself, this will simplify considerably the problem later.

We will also change the positive direction of the space coordinate,so that it increases when moving from load to generator along thetransmission line.

zd

0

Z R New Space Coordinate

Transmission Lines

We adopt a new coordinate d = z, with zero reference at the load

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location. The new equations for voltage and current along the lossytransmission line are


( )

d d

d d

0

(d)1

(d)

j j

j jV V e V e

I V e V e Z

+

+

= +

=

( )

d d

d d

0

(d)1

(d)

V V e V e

I V e V e Z

+

+

= +

=

At the load (d = 0) we have, for both cases,

( )0

(0)

1(0)

V V V

I V V Z

+

+

= +

=

Transmission Lines

For a given load impedance Z R , the load boundary condition is

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(0) (0) RV Z I =

Therefore, we have

( )0

R Z V V V V Z

+ + + =

from which we obtain the voltage load reflection coefficient

0

0

R R R

Z Z V

Z Z V

+

= =+

Transmission Lines

We can introduce this result into the transmission line equations as

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( )

( )

d 2 d

d2 d

0

(d) 1

(d) 1

j j R

j j

R

V V e e

V e

I e Z

+

+

= +

=

( )

( )

d 2 d

d2 d

0

(d) 1

(d) 1

R

R

V V e e

V e

I e Z

+

+

= +

=

At each line location we define a Generalized Reflection Coefficient

2 d(d)

j R e

=

2 d(d) R e

=

and the line equations become

( )

( )

d

d

0

(d) 1 (d)

(d) 1 (d)

j

j

V V e

V e I

Z

+

+= +

=

( )

( )

d

d

0

(d) 1 (d)

(d) 1 (d)

V V e

V e I

Z

+

+= +

=

Transmission Lines

We define the line impedance as

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( ) ( )( )

( )( )

d 1 ddd 1 d

V Z I

+= =

A simple circuit diagram can illustrate the significance of lineimpedance and generalized reflection coefficient:

Z eq=Z (d)R eq = (d)

d 0

Z R

Transmission Lines

If you imagine to cut the line at location d , the input impedance ofh f l d b h l d h h l

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the portion of line terminated by the load is the same as the lineimpedance at that location before the cut . The behavior of the lineon the left of location d is the same if an equivalent impedance withvalue Z (d ) replaces the cut out portion. The reflection coefficient ofthe new load is equal to (d)

( ) 00

Req Req

Req

Z Z d

Z Z

= =

+

If the total length of the line is L , the input impedance is obtainedfrom the formula for the line impedance as

( )( )

( )( )

L 1 L

L 1 Lin

in in

V V Z

I I

+= = =

The input impedance is the equivalent impedance representing theentire line terminated by the load.

Transmission Lines

An important practical case is the low-loss transmission line , whereth ti l t till d i t b t R d G t b

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the reactive elements still dominate but R and G cannot beneglected as in a loss-less line. We have the following conditions:

L R C G >> >>

so that

2

( )( )

1 1

1

j L R j C G

R G

j L j C j L j C

R G RG j LC

j L j C LC

= + +

= + +

+ +

The last term under the square root can be neglected, because it isthe product of two very small quantities.

Transmission Lines

What remains of the square root can be expanded into a truncatedTaylor series

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Taylor series

11

2

12

R G j LC

j L j C

C L R G j LC

L C

+ +

= + +

so that

12

C L R G LC

L C

= + =

Transmission Lines

The characteristic impedance of the low-loss line is a real quantityfor all practical purposes and it is approximately the same as in a

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for all practical purposes and it is approximately the same as in acorresponding loss-less line

0 R j L L

Z G j C C

=

and the phase velocity associated to the wave propagation is

1 pv LC

=

BUT NOTE:

In the case of the low-loss line, the equations for voltage andcurrent retain the same form obtained for general lossy lines.

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Transmission Lines

For all cases, the line impedance was defined as

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0(d) 1 (d)(d)(d) 1 (d)

V Z Z I

= =

By including the appropriate generalized reflection coefficient, wecan derive alternative expressions of the line impedance:

A) Loss-less line2 d

00 02 d

0

tan( d)1(d) tan( d)1

j R R

j R

Z jZ e Z Z Z jZ Z e

= =

B) Lossy line (including low-loss)

2 d 00 02 d

0tanh( d)1(d)

tanh( d)1 R R

R Z Z e Z Z Z Z Z e

= =

Transmission Lines

Lets now consider power flow in a transmission line , limiting thediscussion to the time-average power , which accounts for the

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g p ,active power dissipated by the resistive elements in the circuit.

The time-average power at any transmission line location is

{ }*1(d , ) Re (d) (d)2 P t V I =

This quantity indicates the time-average power that flows throughthe line cross-section at location d . In other words, this is thepower that, given a certain input, is able to reach location d andthen flows into the remaining portion of the line beyond this point .

It is a common mistake to think that the quantity above is the powerdissipated at location d !

Transmission Lines

The generator , the input impedance , the input voltage and the input

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g p p p g pcurrent determine the power injected at the transmission line input.

{ }*

1

1Re

2

inin G

G in

in GG in

in in in

Z V V

Z Z

I V Z Z

P V I

=+

= +

=

V in

I in

Generator Line

V G

Z G

Z in

Transmission Lines

The time-average power reaching the load of the transmission lineis given by the general expression

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{ }

( ) ( )( )

*

*

*0

1(d=0 , ) Re (0) (0)2

1 1Re 1 12 R R

P t V I

V V Z + +

=

= +

This represents the power dissipated by the load .The time-average power absorbed by the line is simply the

difference between the input power and the power absorbed by theload

(d 0 , ) = = line in P P P t

In a loss-less transmission line no power is absorbed by the line, sothe input time-average power is the same as the time-averagepower absorbed by the load. Remember that the internal impedanceof the generator dissipates part of the total power generated.

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Transmission Lines

*

*1 1(d=0, ) Re 1 1R R P t V V

+ + = + =

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02

0 02

02

0 02

0

1 Re

2

Re 1 Re Im 1 Re Im2

Re2

R

R R

R R R R

Z

V R jX j j

Z

V R jX

Z

+

+

= + + +

= +

+

2 2

22

0 02

02

2

0 0 02

0

Im 2Im

Re 1 2Im2

2 Im2

R R

R R

R R

j

V R jX j

Z

V R R X

Z

+

+

+ +

= + + =

=

Transmission Lines

Equivalently, using the complex characteristic admittance:*

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*0

2

0 0

2

0 0 1 Re

1(d=0, ) Re 1 12

Re 1 Re Im 1 Re Im2

Re2 R

R R

R R R R

P t V Y V

V G jB j j

V G jB

+ +

+

+

= + =

= + +

=

+

2 2

22

0 0

2

20 0 0

Im 2Im

Re 1 2Im2

2 Im2

R R

R R

R R

j

V G jB j

V G G B

+

+

+ +

= + =

= +

Transmission Lines

The time-average power, injected into the input of the transmissionline, is maximized when the input impedance of the transmission

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line and the internal generator impedance are complex conjugate ofeach other.

Z Z G in*

Z in

Z R

Z G

Transmission line

LoadGenerator

for maximum power transfer

V G

Transmission Lines

The characteristic impedance of the loss-less line is real and wecan express the power flow, anywhere on the line, as

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( )( )

*

d 2 d

** d 2 d

02 2 2

0 0

1(d , ) Re{ (d) (d) }21

Re 12

1( ) 1

1 12 2

j j

R

j j R

R

P t V I

V e e

V e e Z

V V Z Z

Incident Reflected wwave ave

+

+

+ +

=

= +

=

This result is valid for any location, including the input and the load,since the transmission line does not absorb any power.

Transmission Lines

In the case of low-loss lines, the characteristic impedance is againreal, but the time-average power flow is position dependent

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because the line absorbs power.

{ }( ){

( )

*

d d 2 d

** d d 2 d

02 2 22 d 2 d

0 0

1(d , ) Re (d) (d)

21

Re 12

1( ) 1

1 12 2

j R

j R

R

P t V I

V e e e

V e e e Z

V e V e Z Z

ReflecInciden ted wavet wave

+

+

+ +

=

= +

=

Transmission Lines

Note that in a lossy line the reference for the amplitude of theincident voltage wave is at the load and that the amplitude grows

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exponentially moving towards the input. The amplitude of theincident wave behaves in the following way

input inside the line

L d

loadV e V e V

The reflected voltage wave has maximum amplitude at the load, andit decays exponentially moving back towards the generator. Theamplitude of the reflected wave behaves in the following way

input inside the line lo

L d

ad

R R RV e V e V

Transmission Lines

For a general lossy line the power flow is again position dependent.Since the characteristic impedance is complex , the result has an

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additional term involving the imaginary part of the characteristicadmittance, B 0, as

{ }( ){

( ) }

*

d d

** * d d0

2 2 22 d 2 d0 0

2 2 d0

1(d , ) Re (d) (d)

21

Re 1 (d)2

( ) 1 (d)

2 2Im( (d))

j

j

P t V I

V e e

Y V e e

G GV e V e

B V e

+

+

+ +

+

=

= +

=

+

Transmission Lines

For the general lossy line , keep in mind that

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*0 0 0 0 00 0 0* 2 2 2

0 0 0 0 00

0 00 02 2 2 20 0 0 0

1 Z R jX R jX Y G jB

Z Z Z R X Z

R X G B R X R X

= = = = = ++

= =+ +

Recall that for a low-loss transmission line the characteristicimpedance is approximately real, so that

0 0 B and 0 0 01 Z G R .

The previous result for the low-loss line can be readily recoveredfrom the time-average power for the general lossy line.

Transmission Lines

To completely specify the transmission line problem, we still haveto determine the value of V+ from the input boundary condition .

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The load boundary condition imposes the shape of theinterference pattern of voltage and current along the line.

The input boundary condition, linked to the generator, imposesthe scaling for the interference patterns.

We have

01 (L)

(L)1 (L)

inin G in

G in

Z V V V Z Z

Z Z

with

+= = =+

00

00

00

tan( L)tan( L)

tanh( L)tanh( L)

Rin

R R

in R

Z jZ Z Z

jZ Z

Z Z Z Z

Z Z

loss - less line

lossy

o

lin

r

e

+=+

+=+

Transmission Lines

For a loss-less transmission line:

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[ ]L L 2 L(L) 1 (L) (1 ) j j j RV V e V e e + + = + = +

L 2 L1

(1 )inG j j

G in R

Z V V Z Z e e

+ = + +

For a lossy transmission line:

[ ] ( )L L 2 d(L) 1 (L) 1 RV V e V e e + + = + = +

L 2 L1

(1 )inG

G in R

Z V V Z Z e e + = + +

Transmission Lines

In order to have good control on the behavior of a high frequencycircuit, it is very important to realize transmission lines as uniform

ibl l g th i l gth th t th i d b h i f

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as possible along their length, so that the impedance behavior ofthe line does not vary and can be easily characterized.

A change in transmission line properties, wanted or unwanted,entails a change in the characteristic impedance, which causes areflection. Example:

1

Z R01 Z 02

Z 01 Z in 1 in 01in 01 Z Z

=

Transmission Lines

Special Cases

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0 R Z (SHORT CIRCUIT)

The load boundary condition due to the short circuit is V (0) = 0

0 2 0(d 0) (1 )

(1 ) 0

1

j j R

R

R

V V e e

V

= =

= = =

Z R = 0 Z 0

Transmission Lines

Since

RV

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RV

V V

=

=

We can write the line voltage phasor as

d d

d d

d d

(d)

( )

2 sin( d)

j j

j j

j j

V V e V e

V e V e

V e e

jV

===

=

Transmission Lines

For the line current phasor we have

d d1(d) ( )

j j I V e V eZ

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0

d d

0

d d

0

0

(d) ( )

1( )

( )

2 cos( d)

j j

j j

I V e V e Z

V e V e Z

V e e

Z

V Z

=

=

=

=

The line impedance is given by

00

(d) 2 sin( d)(d) tan( d)

(d) 2 cos( d) /

V jV Z jZ

I V Z

= = =

Transmission Lines

The time-dependent values of voltage and current are obtained as

(d ) Re[ (d) ] Re[2 | | sin( d) ] j t j j t

V t V e j V e e

= =

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( )(d, ) Re[ (d) ] Re[2 | | sin( d) ]

2| |sin( d) Re[ ]

2| |sin( d) Re[ cos( ) sin( )]

2| |sin( d) sin( )

j t V t V e j V e e

V j e

V j t t

V t

= = = = =

0( )

0

0

0

(d, ) Re[ (d) ] Re[2 | | cos( d) ] /

2| |cos( d) Re[ ] /

2| |cos( d) Re[ (cos( ) sin( )] /

| |2 cos( d)cos( )

j t j j t

j t

I t I e V e e Z

V e Z

V t j t Z

V t

Z

= =

=

=

=

Transmission Lines

The time-dependent power is given by

(d ) (d ) (d )P t V t I t=

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2

0

2

0

(d, ) (d, ) (d, )| |

4 sin( d)cos( d) sin( )cos( )

| |sin(2 d)sin (2 2 )

P t V t I t V

t t Z

V t

Z

==

=

and the corresponding time-average power is

0

2

00

1(d, ) (d, )

| | 1sin(2 d) sin (2 2 ) 0

T

T

P t P t dt T

V t

Z T

< > =

= =

Transmission Lines

R Z (OPEN CIRCUIT)

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The load boundary condition due to the open circuit is I (0) = 0

0 2 0

0

0

(d 0) (1 )

(1 ) 0

1

j j R

R

R

V I e e Z

V Z

= =

= = =

Z R Z 0

Transmission Lines

Since

RV

=

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RV

V V

=

=

We can write the line current phasor as

d d

0d d

0

d d

0 0

1(d) ( )

1( )

2( ) sin( d)

j j

j j

j j

I V e V e Z

V e V e Z

V jV e e Z Z

=

=

= =

Transmission Lines

For the line voltage phasor we have

d dj j

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d d

d d

d d

(d) ( )

( )

( )

2 cos( d)

j j

j j

j j

V V e V e

V e V e

V e e

V

==

==

The line impedance is given by

0

0

(d) 2 cos( d)(d)

(d) tan( d)2 sin( d) /

V V Z Z j

I jV Z

= = =

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Transmission Lines

The time-dependent power is given by

(d, ) (d, ) (d, )P t V t I t= =

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2

0

2

0

(d, ) (d, ) (d, )| |

4 cos( d) sin( d)cos( ) sin( )

| |sin(2 d)sin (2 2 )

P t V t I t V

t t Z

V t

Z

=

=

and the corresponding time-average power is

0

2

00

1(d, ) (d, )

| | 1sin(2 d) sin (2 2 ) 0

T

T

P t P t dt T

V t

Z T

< > =

= =

Transmission Lines

0 R Z Z (MATCHED LOAD)

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The reflection coefficient for a matched load is

0 0 0

0 0 0 0 R

R R

Z Z Z Z Z Z Z Z

= = = no reflection!

The line voltage and line current phasors are

d 2 d d

d 2 d d

0 0

(d) (1 )

( ) (1 )

j j j R

j j j R

V V e e V e

V V I d e e e

Z Z

= =

= =

Z R = Z 00

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Transmission Lines

The time-dependent power is

0

| |(d, ) | | cos( d ) cos( d )

V P t V t t Z

=

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02

2

0

( , ) | | ( ) ( )

| |cos ( d )

Z

V t

Z

=

and the time average power absorbed by the load is

2 20 0

2

0

1 | |(d) cos ( d )

| |2

t V P t dt

T Z

V Z

< > =

=

Transmission Lines

R Z jX (PURE REACTANCE)

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The reflection coefficient for a purely reactive load is

0 0

0 02 2

0 0 0 02 2 2 20 0 0 0

( )( )2( )( )

R R

R

Z Z jX Z Z Z jX Z

jX Z jX Z X Z XZ j jX Z jX Z Z X Z X

= = =

= =

Z R = j X Z 0

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Transmission Lines

Reactive impedances can be realized with transmission linesterminated by a short or by an open circuit. The input impedance of

a loss-less transmission line of length L terminated by a shortcircuit is purely imaginary

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Amarcord, 2006 Digital Maestro Series 107

p y g y

0 0 02 2

tan L tan L tan Lin p

f Z j Z j Z j Z

v= = =

For a specified frequency f , any reactance value (positive ornegative!) can be obtained by changing the length of the line from 0

to /2 . An inductance is realized for L < /4 (positive tangent)while a capacitance is realized for /4 < L < /2 (negative tangent).

When L = 0 and L = /2 the tangent is zero, and the inputimpedance corresponds to a short circuit. However, when L = /4 the tangent is infinite and the input impedance corresponds to anopen circuit.

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Transmission Lines

Short circuited transmission line Fixed frequency

L

L

= =

< < >

0 0

0 0

Z

Z

in sho rt circui t

inductance

Im k p

L

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Amarcord, 2006 Digital Maestro Series 109

L

L

L

L

L

L

L

< < >

=

< <

=

< < Z 0 , the voltage standing wavepattern starts with a maximum at the load.

If the load is real and Z R < Z 0 , the voltage standing wavepattern starts with a minimum at the load.

If the load is complex and Im (Z R ) > 0 (inductive reactance),the voltage standing wave pattern initially increases whenmoving from load to generator and reaches a maximum first.

If the load is complex and Im (Z R ) < 0 (capacitive reactance),the voltage standing wave pattern initially decreases whenmoving from load to generator and reaches a minimum first.

Transmission Lines

Since in all possible cases

d 1 the voltage standing wave pattern

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Amanogawa, 2006 -Digital Maestro Series 131

g g p

(d) 1 (d)V V

= cannot exceed the value 2 | V+ | in a loss-less transmission line.If the load is a short circuit, an open circuit, or a pure reactance,there is total reflection with

d 1= since the load cannot consume any power. The voltage standingwave pattern in these cases is characterized by

max min2 0V V V = =and .

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Transmission Lines

We can use a geometric construction to visualize the behavior ofthe voltage standing wave pattern

(d) 1 (d)V V =

i l b l ki g t t l t f |(1 + (d)) | |V+| i j t

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simply by looking at a vector plot of |(1 + (d)) | . |V +| is just ascaling factor, fixed by the generator. For convenience, we placethe reference of the complex plane representing the reflectioncoefficient in correspondence of the tip of the vector (1, 0) .

1+R

Im( )

Re ( )1

R

Example: Loadwith inductive

reactance

Transmission Lines

1+(d) R

Im( )

Maximum ofvoltage standing

wave pattern

2 d max

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maxd 2 d 0 = =

mind 2 d = =

1 Re ( ) d

1+(d)

1

Im( )

Re ( )

R

Minimum of

voltage standingwave pattern

(d)

2 d min

Transmission Lines

The voltage standing wave ratio (VSWR) is an indicator of loadmatching which is widely used in engineering applications

max

min

11

R

R

V VSWR

V = =

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When the load is perfectly matched to the transmission line

0 1 R VSWR= = When the load is a short circuit, an open circuit or a pure reactance

1 R VSWR=

We have the following useful relation

11 R

VSWRVSWR

=

Transmission Lines

Maxima and minima of the voltage standing wave pattern.

Load with inductive reactance

0

0Im 0 Im Im 0 R R R

R

Z Z Z

Z Z> = >

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Im

Re

The load reflection coefficientis in this part of the domain

1

0 R Z Z

The first maximum of the voltage standing wave pattern is closestto the load, at location

max maxd 2 d 0 d 4 = = =

Transmission Lines

Im( )The load reflection coefficient

Load with capacitive reactance

0

0Im 0 Im Im 0 R

R R R

Z Z Z Z Z < = <

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Im( )

Re( )

is in this part of the domain 1

The first minimum of the voltage standing wave pattern is closest tothe load, at location

min mind 2 d d 4

= = =

Transmission Lines

A measurement of the voltage standing wave pattern provides thelocations of the first voltage maximum and of the first voltageminimum with respect to the load.

The ratio of the voltage magnitude at these points gives directly thevoltage standing wave ratio (VSWR).

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g g ( )

This information is sufficient to determine the load impedance Z R ,if the characteristic impedance of the transmission line Z 0 isknown.

STEP 1: The VSWR provides the magnitude of the loadreflection coefficient

11 R

VSWRVSWR

=

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Transmission Lines

STEP 3: The load impedance is obtained by inverting theexpression for the reflection coefficient

0RZ Z

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0

0

0

exp

1 exp

1 exp

R R R

R

R R

R

Z Z j

Z Z

j Z Z

j

= =

=

dfdb17ab

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