Lecture- EM Transmission Lines

7/27/2019 Lecture- EM Transmission Lines

1/100

1

Instructor:

Dr. Gleb V. Tcheslavski

Contact:

[email protected]

Office Hours:

Room 2030

Class web site:

www.ee.lamar.edu

/g leb /em /

Index.htm
mailto:[email protected]://www.ee.lamr.edu/gleb/em/Index.htmhttp://www.ee.lamr.edu/gleb/em/Index.htmhttp://www.ee.lamr.edu/gleb/em/Index.htmhttp://www.ee.lamr.edu/gleb/em/Index.htmmailto:[email protected]


2/100

2

In this topic, we model three electrical transmission systems that can be used to

transmit power: a coaxial cable, a strip line, and two parallel wires (twin lead).

Each structure (including the twin lead) may have a dielectric between two

conductors used to keep the separation between the metallic elements

constant, so that the electrical properties would be constant.


3/100

3


4/100

4


5/100

5


6/100

6

Instead of examining the EM field distribution within these transmission

lines, we will simplify our discussion by using a simple model

consisting of distributed inductors and capacitors. This model is valid if

any dimension of the line transverse to the direction of propagation is

much less than the wavelength in a free space.

The transmission lines considered here support the propagation of

waves having both electric and magnetic field intensities transverse to

the direction of wave propagation. This setup is sometimes called a

transverse electromagnetic (TEM) mode of propagation. We assumeno loss in the lines.


7/100

7

Distributed

transmission line

Its equivalent

circuit

z is a short distance containing the distributed circuit parameter.

and are distributed inductance and distributed capacitance.

Therefore, each section has inductance and capacitance

L C

L L z C C z (9.7.1)


8/100

8

(9.8.1)

(9.8.2)

(9.8.3)

Note: the equations for a microstrip line are simplified and do not includeeffects of fringing.

We can model the transmission line with an equivalent circuit consisting of an

infinite number of distributed inductors and capacitors.


9/100

9

The following simplifications were used:

1) No energy loss (resistance) was incorporated;

2) We neglected parasitic capacitances between the wires that

constitute the distributed inductances. We will see later that these

parasitic capacitances will lead to changes in phase velocity of thewave (dispersion);

3) Parameters of the line are constant.

We can analyze EM transmission lines either as a large number of distributedtwo-port networks or as a coupled set of first-order PDEs that are called the

telegraphers equations.


10/100

10

While analyzing the equivalent circuit of the lossless transmission line, it is

simpler to use Kirchhoffs laws rather than Maxwells equations.

Therefore, we will

consider theequivalent circuit of

this form:

For simplicity, we define the inductance and capacitance per unit length:

;L C

L Cz z

(9.10.1)

which have units of Henries per unit length and Farads per unit length, respectively.


11/100

11

The current entering the node at the location zis I(z). The part of this current

will flow through the capacitor, and the rest flows into the section. Therefore:

( , )( , ) ( , )

V z tI z t C z I z z t

t

( , ) ( , ) ( , )

I z z t I z t V z tC

z t

(9.11.1)

(9.11.2)

, the LHS of(9.11.2) is a spatial derivative. Therefore:0If z

( , ) ( , )

I z t V z tC

z t

(9.11.3)


12/100

12

Similarly, the sum of the voltage drops in this section can be calculated via the

Kirchhoffs law also:

( , )( , ) ( , )

I z tV z z t L z V z t

t

( , ) ( , ) ( , )

V z t V z z t I z t L

z t

(9.12.1)

(9.12.2)

, the LHS of(9.12.2) is a spatial derivative. Therefore:0If z

( , ) ( , )

V z t I z t L

z t

(9.12.3)

3


13/100

13

The equations (9.11.3) and (9.12.3) are two linear coupled first-order PDEs

called the telegraphers (Heaviside) equations. They can be composed in a

second-order PDE:

2 2

2 2

( , ) ( , )

0

I z t I z t

LCz t

2 2

2 2

( , ) ( , ) 0

V z t V z t LC

z t

(9.13.1)

(9.13.2)

We may recognize that both (9.13.1) and (9.13.2) are wave equations with

the velocity of propagation:

1

v

LC

(9.13.3)

14


14/100

14

Example 9.1: Show that a transmission line consisting of distributed linear

resistors and capacitors in the given configuration can be used to model diffusion.

We assume that the

resistance and the

capacitance per unit lengthare defined as

;R C

R Cz z

(9.14.1)

Potential drop over the resistorRand the current through the capacitorCare:

( , ) ( , )

( , )( , )

V z t I z t R z

V z tI z t C z

t

(9.14.2)

(9.14.3)

15


15/100

15

( , )0 ( , )

( , ) ( , )

V z tIf z I z t R

z

I z t V z tC

z t

The corresponding second-order PDE for the potential is:

2

2

( , )( , ) ( , )

( ,

)

I z t V z tR R

V z t V z t RC

zC

z t t

(9.15.2)

(9.15.1)

(9.15.3)

Which is a form of a diffusion equation with a diffusion coefficient:

1

DRC

(9.15.4)

16


16/100

16

Example 9.2: Show that a particular solution for the diffusion equation is given by2

41

( , )2

z

DtV z t eD t

Differentiating the solution with respect to z:2

43 2

( , ) 1

22

z

DtV z t z

ez DtD

22 2

4

2 3 2 2 5 2

( , ) 1

2 42

z

DtV z t z z

ez Dt D tD

Differentiating the solution with respect to t:

22

43 2 5 2

1 ( , ) 1 1

2 42

z

DtV z t t z

eD t D t DtD

(9.16.1)

(9.16.2)

(9.16.3)

(9.16.4)

17


17/100

17

Since the RHSs of(9.16.3) and (9.16.4) are equal, the diffusion equation is

satisfied.

The voltages at different times

are shown. The total area under

each curve equals 1.

This solution would be valid if a

certain amount of charge is

placed at z= 0 at some moment

in the past.

Note: the diffusion is significantly

different from the wave

propagation.

18


18/100

18

We are looking for the solutions of wave equations (9.13.1) and

(9.13.2) for the time-harmonic (AC) case. We must emphasize that

unlike the solution for a static DC case or quasi-static low-frequency

case (ones considered in the circuit theory) these solutions will be in

form oftraveling waves of voltage and current, propagating in either

direction on the transmission line with the velocity specified by (9.13.3).

We assume here that the transmission line is connected to a distant generator

that produces a sinusoidal signal at fixed frequency = 2f. Moreover, the

generator has been turned on some time ago to ensure that transient response

decayed to zero; therefore, the line is in a steady-state mode.

The most important (and traditional) simplification for the time-harmonic case

is the use of phasors. We emphasize that while in the AC circuits analysis

phasors are just complex numbers, for the transmission lines, phasors are

complex functions of the position zon the line.

19


19/100

19

( , ) Re ( ) ; ( , ) Re ( )j t j tV z t V z e I z t I z e

2

22

22

2

( )( ) 0

( )( ) 0

d V zk V zdz

d I zk I z

dz

Therefore, the wave equations will become:

Here, as previously, kis the wave number:

2k

v

Velocity of propagation

Wavelength of the voltage or current wave

(9.19.1)

(9.19.2)

(9.19.3)

(9.19.4)

20


20/100

20

A solution for the wave equation (9.13.3) can be found, for instance, in one of

these forms:

1 1

2 2

( ) cos sin

( ) jkz jkz

V z A kz B kz

V z A e B e

(9.20.1)

(9.20.2)

We select the exponential form (9.20.2) since it is easier to interpret in terms of

propagating waves of voltage on the transmission line.

Example 9.3: The voltage of a wave propagating through a transmission line was

continuously measured by a set of detectors placed at different locations alongthe transmission line. The measured values are plotted. Write an expression for

the wave for the given data.

21


21/100

21

Slo

peofthetraject

ory

The

data

22


22/100

22

We assume that the peak-to peak amplitude of the wave is 2V0. We also

conclude that the wave propagates in the +zdirection.

The period of the wave is 2s, therefore, the frequency of oscillations is Hz.

The velocity of propagation can be found from the slope as:5 1

41 0

v m s

The wave number is:12 0.5

4 4

k m

v

The wavelength is:2 4 2

8 mk

Therefore, the wave is:4

0( , )

zj t

V z t V e

(9.22.2)

(9.22.3)

(9.22.4)

(9.22.5)

Not in vacuum!

2 f (9.22.1)

23


23/100

23

Assuming next that the source is located far from the observation point (say, at z

= -) and that the transmission line is infinitely long, there would be only a

forward traveling wave of voltage on the transmission line. In this case, the

voltage on the transmission line is:

0( )jkz

V z V e

The phasor form of(9.12.3) in this case is

((

))

) (jkV z j LI zdV z

dz

0( ) ( )

jkzk kI z V z V eL L

Which may be rearranged as:

(9.23.1)

(9.23.2)

(9.23.3)

24


24/100

24

The ratio of the voltage to the current is a very important transmission line

parameter called the characteristic impedance:

( )

( )c

V z LZ

I z k

1

k and vv LC

Since

Then:

c

LZ

C

(9.24.1)

(9.24.2)

(9.24.3)

We emphasize that (9.24.3) is valid for the case when only one wave (traveling

either forwards or backwards) exists. In a general case, more complicated

expression must be used. If the transmission line was lossy, the characteristic

impedance would be complex.


25/100

26


26/100

26

The velocity of propagation does not depend on the dimensions of

the transmission line and is only a function of the parameters of the

material that separates two conductors. However, the characteristic

impedance DOES depend upon the geometry and physical

dimensions of the transmission line.

27


27/100

27

Example 9.4: Evaluate the velocity of propagation and the characteristic impedanceof an air-filled coaxial cable with radii of the conductors of 3 mm and 6 mm.

The inductance and capacitance per unit length are:

The velocity of propagation is:

The characteristic impedance of the cable is:

Both the vand the Zcmay be decreased by insertion of a dielectric between leads.

70 4 10 6 ln ln 0.14

2 2 3

H mb

L

a

1202 2 8.854 10 80

ln ln 6 3pF mC

b a

8

6 12

1 1

3 10 0.14 10 80 10 m sv LC

6 12 0.14 10 80 10 42cZ L C

(9.27.1)

(9.27.2)

(9.27.3)

(9.27.4)

28


28/100

28

So far, we assumed that the

transmission line was infinite. In the

reality, however, transmission lines

have both the beginning and the end.

The line has a real characteristic impedance Zc. We assume that the source of the

wave is at z= - and the termination (the end of the line) is at z= 0. The terminationmay be either an impedance or another transmission line with different parameters.

We also assume no transients.

29


29/100

The phasor voltage at any point on the line is:

2 2( )jkz jkz V z A e B e

The phasor current is:

2 2( )

jkz jkz

c

A e B eI z Z

At the load location (z= 0), the ratio of voltage to current must be equal ZL:

2 2

2 2

( 0)

( 0)L c

A BV zZ Z

I z A B

Note: the ratio B2 toA2 represents the magnitude of the wave incident on the

load ZL.

(9.29.1)

(9.29.2)

(9.29.3)

30


30/100

(9.30.1)

We introduce the reflection coefficient for the transmission line with a load as:

2

2

L c

L c

Z ZB

A Z Z

Often, the normalized impedance is used:

LL

c

Zz

Z

The reflection coefficient then becomes:

2

2

11

L

L

B zA z

(9.30.2)

(9.30.3)

31


31/100

Therefore, the phasor representations for the voltage and the current are:

0

0

( )

( )

jkz jkz

jkz jkz

c

V z V e e

VI z e e

Z

(9.31.1)

(9.31.2)

The total impedance is:( )

( )( )

V zZ z

I z

generally a complicated function of the position and NOT equal to Zc. However,

a special case ofmatched load exists when:

L cZ Z

In this situation: ( ) cZ z Z

(9.31.3)

(9.31.4)

(9.31.5)

32


32/100

Example 9.5: Evaluate the reflection coefficient for a wave that is incident fromz= - in an infinitely long coaxial cable that has r= 2 forz< 0 and r= 3 forz> 0.

The characteristic impedance is:

ln 2

c

b aLZ

C

The load impedance of a line is the characteristic impedance of the line forz> 0.

33


33/100

The reflection coefficient can be expressed as:

0 0

2 0 1 02 1

2 1 0 0

2 0 1 0

2 1

2 1

ln ln

2 2

ln ln2 2

1 1 1 1

3 20.1

1 1 1 13 2

r r

r r

r r

r r

b a b a

Z Z

Z Zb a b a

34


34/100

The reflection coefficient is completely determined by the value of the impedance ofthe load and the characteristic impedance of the transmission line. The reflection

coefficient for a lossless transmission line can have any complex value with magnitude

less or equal to one.

If the load is a short circuit (ZL = 0), the reflection coefficient = -1. The voltage at theload is a sum of voltages of the incident and the reflected components and must be

equal to zero since the voltage across the short circuit is zero.

If the load is an open circuit (ZL = ), the reflection coefficient = +1. The voltage at the

load can be arbitrary but the total current must be zero.

If the load impedance is equal to the characteristic impedance (ZL = Zc), the reflectioncoefficient = 0 line is matched. In this case, all energy of generator will be absorbed

by the load.

35


35/100

For the shorten transmission line:

0

0

1

( , ) Re

2 sin cos2

jkz jkz j tV z t V e e e

V kz t

(9.35.1)

(9.35.2)

For the open transmission line:

0

0

1

( , ) Re

2 cos cos

jkz jkz j tV z t V e e e

V kz t

(9.35.3)

(9.35.4)

In both cases, a standing wave is created.

The signal does not appear to propagate.

36


36/100

Since the current can be found as 0 0 cI V Z

ZL =

ZL = 0

Note that the current wave differs from the voltage wave by 900

37


37/100

Another important quantity is the ratio of the maximum voltage to the minimum

voltage called the voltage standing wave ratio:

max

min

1

1

VVSWR

V

Which leads to1

1

VSWR

VSWR

VSWR, the reflection coefficient, the load impedance, and the characteristic

impedance are related.Even when the amplitude of the incident wave V0 does not exceed the maximally

allowed value for the transmission line, reflection may lead to the voltage V0(1+||)

exceeding the maximally allowed.

Therefore, the load and the line must be matched.

(9.37.1)

(9.37.2)

38


38/100

Example 9.6: Evaluate the VSWR for the coaxial cable described in the Example

9.5. The reflection coefficient was evaluated as = -0.1.

1 1 0.11.2

1 1 0.1VSWR

Note: if two cables were matched, the VSWRwould be 1.

39


39/100

40


40/100

The ratio of total phasor voltage to total phasor current on a transmission line

has units of impedance. However, since both voltage and current consist of the

incident and reflected waves, this impedance varies with location along the line.

0

0

( )( )

( )

jkz jkz jkz jkz

c jkz jkz

jkz jkz

c

V e eV z e eZ z Z

VI z e ee eZ

Incorporating (9.30.1), we obtain:

2 co tan(

s 2 sin

2 cos 2 si) tan n

jkz jkzL c

L c L cc cjkz

L c

jkzL c c L

L

L

c

c

c

Z Ze e

Z Z Z kz j Z kzZ ZZ Z

Z jZ kzZ Z kzz Z Z jZ kzj Z kz

e eZ Z

(9.40.1)

(9.40.2)

The last formula is most often used to find the impedance at the line terminals.

41


41/100

Considering the transmission line shown, we assume that a load with theimpedance ZL is connected to a transmission line of length L having the

characteristic impedance Zcand the wave numberk.

The input impedance can be found as an impedance at z = -L:

tan( ) ( )

tanL c

in c

c L

Z jZ kLZ L Z z L Z

Z jZ kL

(9.41.1)

Or as the normalized input impedance: tan( )

1 tan

Lin

L

z j kLz L

jz kL

(9.41.2)

42


42/100

Example 9.7: A signal generator whose frequency f= 100 MHz is connected to a coaxialcable of characteristic impedance 100 and length of 100 m. The velocity of

propagation is 2108 m/s. The line is terminated with a load whose impedance is 50 .

Calculate the impedance at a distance 50 m from the load.

The normalized load impedance is 50 100 0.5L L cz Z Z

The wave number is

The normalized input impedance is

81

8

2 2 2 1 10

2 10

fk m

v

0.5 tan 50tan 1( 50 )

1 tan 1 0.5 tan 50 2L

in

L

jz j kLz z m

jz kL j

Therefore, the input impedance is

0.5 100 50in in c

Z z Z

43


43/100

The wave number can be expressed in terms of wavelength:

2k

Therefore:2

kL L

and 2tan tan tan ; integer kL L kL n n

If the length of the transmission line is one quarter of a wavelength,

2

4 2kL

tangent (9.42.3) approaches infinity and

2

44

L c cin c in

c L L

Z jZ ZZ z Z Z

Z jZ Z

(9.43.1)

(9.43.2)

(9.43.3)

(9.43.4)

(9.43.5)

Implying that the normalized input impedance zin of a /4 line terminated with the load ZLis numerically equal to the normalized load admittance yL = 1/zL.

44


44/100

The last example represents a one quarter-wavelength transmission line that is

useful in joining two transmission lines with different characteristic impedances or

in matching a load. One of the simplest matching techniques is to use a quarter-

wave transformer a section of a transmission line that has a particular

characteristic impedance Zc(/4).

This characteristic impedance Zc(/4)must be chosen such that the

reflection coefficient at the input of

the matching transmission line

section is zero.

This happens when

( 4)c c LZ Z Z (9.44.1)

One considerable disadvantage of this method is its frequency dependence since

the wavelength depends on the frequency.

45


45/100

When the load impedance equals the characteristic impedance, the load and theline are matched and no reflection of the wave occurs.

For the short circuit and the open circuit:

0( ) tan( )

( ) cot( )tan( )

L

L

in cZ

cin cZ

Z z L jZ kL

ZZ z L jZ kLj kL

In practice, it is easier to make short circuit

terminators since fringing effects may exist in open

circuits. In both cases, the input impedance will be a

reactance, Zin =jXin as shown for a short-circuited(a) and an open-circuited (b) transmission lines. The

value of the impedance depends on the length of

the transmission line, which implies that we can

observe/have any possible value of reactance that

is either capacitive or inductive.

(9.45.1)

(9.45.2)

46


46/100

Types of input

impedance of short-

circuited and open-

circuited lossless

transmission lines:

We introduce the characteristic admittance of the transmission line:

1c cY Z (9.46.1)

and the input susceptance of the transmission line:

1in inB X (9.46.2)

47


47/100

Assuming that a transmission line is terminated with

a load impedance ZL or load admittance YL that is not

equal to the lines characteristic admittance Yc.

Let the input admittance of the line be Yc+ jB at the

distance d1

from the load.

If, at this distance d1, we connect a susceptancejB

in parallel to the line, the total admittance to the left

of this point (d1) will be Yc.

The transmission line is matched from the insertionpoint (d1) back to the generator.

In practice, matching can be done by insertion of a

short-circuited transmission line of particular length.

48


48/100

Such transmission line used to match another (main) transmission line is called a

stub. The length of a stub is chosen to make its admittance be equaljB. This

process of line matching is called single-stub matching.

The length of the stub can be made adjustable. Such adjustable-length

transmission lines are sometimes called a trombone line.

Note that single-stub matching requires two adjustable distances: location of the

stub d1 and the length of the stub d2. In some situations, only the stubs length

can be adjusted. In these cases, additional stub(s) may be used.

The distances mentioned here are normalized to the wavelength. Therefore, this

method allows line matching at particular discrete frequencies.

49


49/100

Example 9.8: A lossless transmission line is terminatedwith an impedance whose value is a half of the

characteristic impedance of the line. What impedance

should be inserted in parallel with the load at the distance

/4 from the load to minimize the reflection from the load?

To minimize the reflection, the parallel combination of the ZQ and the input impedance at that

location should equal to the characteristic impedance of the line.

22

( 4)

2 2( 4)

22

2

2

ccQQ

Q in Q ccL

c

Q in Q cc cQ Q

L c

ZZ ZZZ Z Z ZZZ

ZZ Z Z ZZ Z

Z ZZ Z

2

1 22

Q

Q cQ c

Z

Z ZZ Z

50


50/100

The input impedance of a transmission line depends on the impedance of theload, the characteristic impedance of the line, and the distance between the load

and the observation point. The value of the input impedance also periodically

varies in space.

The input impedance can be found graphically via so called Smith chart.

tan( )

1 tanL

in

L

z j kLz L

jz kL

The normalized impedance at any location is complex and can be found as:

An arbitrary normalized load impedance is:

Lz r jx

Where: ;L L

c c

R Xr x

Z Z

Since the line is assumed to be lossless, its characteristic impedance is real.

(9.50.1)

(9.50.2)

(9.50.3)

51


51/100

The reflection coefficient will be complex:

1

1L

r i

L

zj

z

2 2

2 22 2

1 1 21

1 1 1 1

r i r i iL

r i r i r i

jz r jx j

j

Therefore, by equating the real and the imaginary parts:

2 2

2 1

1 1r i

r

r r

2 2

2 1 11r i

x x

(9.51.1)

(9.51.2)

(9.51.3)

(9.51.4)

(9.51.3) and (9.51.4) represent family of circles in a plane whose axes are rand i.

The center and radius of each circle are determined by the normalized resistance r

and reactancex. All circles are inside the unit circle since maximal = 1.

52


52/100

For the constant rcircles:1. The centers of all the constant r

circles are on the horizontal axis real

part of the reflection coefficient.

2. The radius of circles decreases

when rincreases.

3. All constant rcircles pass throughthe point r=1, i = 0.

4. The normalized resistance r= is

at the point r=1, i = 0.

For the constantx(partial) circles:

1. The centers of all the constantx

circles are on the r=1 line. The circles

withx> 0 (inductive reactance) are

above the raxis; the circles withx< 0

(capacitive) are below the raxis.

2. The radius of circles decreases when absolute value ofxincreases.

3. The normalized reactancesx= are at the point r=1, i = 0

53


53/100

The constant rcircles are orthogonal to the constantxcircles at every intersection.

The actual (de-normalized) load impedance is:

( )L cZ Z r jx The intersection of an rcircle and anxcircle specifies the normalized impedance.

The evenly spaced marks on the circumference indicate the fraction of a half-wavelength, since the

impedance repeats itself every half-wavelength.

Since the transmission line is lossless, the magnitude of the reflection coefficient is constant at

every point between the load and the signal generator.

The horizontal rand the vertical i axes have been removed from the chart.

Smith chart can be expressed in terms of either impedance or admittance.

(9.53.1)

54


54/100

Example 9.9: On the simplified Smith chart, locate the following normalizedimpedances:

) 1 0;

) 0.5 0.5

) 0 0;

) 0 1;

) 1 2;

)

a z j

b z j

c z j

d z j

e z j

f z

55


55/100

The reflection coefficient can also be expressed in the following form:

1

1Lj L

L

ze

z

The magnitude of the reflection coefficient

0 1 is determined by the value of the normalized load impedance zL and is constant

for all locations along the lossless transmission line.

L is a phase angle associated with the reflection coefficient .

(9.55.1)

(9.55.2)

The input impedance at any arbitrary point (say, -z) is

' ' 2 '

' ' 2 '

( ') 1( ')

( ') 1

jkz jkz j kz

in c cjkz jkz j kz

V z e e eZ z Z Z

I z e e e

(9.55.3)

56


56/100

The normalized input impedance at the point -zis

2 '

2 '

1( ') 1( ')

1 1

jj kz

inin j kz j

c

eZ z ez z

Z e e

Where 2 'L

kz

(9.56.1)

(9.56.2)

Compared to (9.51.2), we conclude that the only difference is in a phase shift that is linearly proportional to

coordinate zand can be translated to the Smith chart by rotating the initial value of the load impedance along a

circle with the radius equal to the magnitude of the reflection coefficient. A clockwise rotation is equivalent to

moving toward the signal generator; a counterclockwise rotation toward the load. The amount of rotation

depends on the distance 2kz = 4z/.

If the distance z = /4, the rotation will be equal to radians. In this situation, the numerical value of an

impedance will be converted into the numerical value of an admittance (refer back to our quarter wavelength lines

discussion).

57


57/100

For example, the impedancez= 0.5 +j0.5

corresponds to the admittance

11 1

0.5 0.5y j

j

Surprisingly, this calculation can be done

graphically using the Smith chart!

First, we locate the normalized

impedance on the chart.

Second, we draw a circle (or a semicircle

to generator) centered at the center of

the Smith chart and passing through theimpedance.

Third, we plot a straight line through the center of the Smith chart and through this

impedance. The intersection of the line with the semicircle yields the value for the

admittance.

58


58/100

This method suggests that the

normalized load impedance of 0

will yield a normalized load

admittance of.

The last observation implies that

the input impedance at thelocation that is /4 from a short

circuit will be an input

impedance of an open circuit!

Therefore, we may avoid using

open circuits since they are notvery attractive due to fringing,

and use short circuits only

59


59/100

Example 9.10: A load impedance 50 50LZ j

terminates a transmission line that is 5 m long and has Zc= 25 . Using the Smith

chart, find the impedance at the signal generator if the frequency of oscillation f=

105 Hz. The phase velocity for this transmission line is v= 2 Mm/s.

The wavelength is 6

52 10 201 10

v mf

Therefore, the distance between the load and the generator is: 5/20 = /4.

The normalized load impedance is

50 50 2 225

L jz j

We locate the impedance on the Smith chart, place a compass leg at the center of

the chart, and draw an arc a distance /4 (half a circle) in the clockwise direction,

which is toward the generator.

60


60/100

Finally, we draw a linepassing through the

impedance and the

center of the Smith chart.

The intersection of that

line with the arc

determines the

normalized load

impedance of

0.25 5inz j

Therefore, the loadimpedance is

25(0.25 5)

6.25 6.25

in c inZ Z z

j

j

61


61/100

An interesting property of the Smith chart is that it can be used equally well interms of an impedance or an admittance. Any constant coefficient circle centered at

the center of the Smith chart will pass the real part equals 1 circle at two

locations. The truth is that at any of these two locations, an admittance can be

added in parallel to the line to match it with the load!

Example 9.11: A load admittance is 0.2 5Ly j Find the locations where a matching admittance should

be placed. Also, find the value for the matching

admittance.

The input admittance will have the value 1iny jb

The value of a reflection coefficient can be

determined from the load admittance

Or we can simply use the Smith chart.

http://my.ece.ucsb.edu/sanabria/tattoo.html

62


62/100

We draw a circle centered

at the center of the Smith

chart and passing through

the input admittance we

have calculated.

Two intersections of thatcircle with the real part 1

circle determine the two

values of the matching

admittances. The location

where these admittances

are to be inserted can bedetermined by the angles

that correspond to each

arc.

63


63/100

Let us consider a transmissionline connected to the battery

with an internal impedance Zbthrough a switch. The line has

a characteristic impedance Zcand is connected to the load

ZL. We assume first that all

impedances are pure

resistances.

The signal propagates with the velocity vand will reach the load L/vseconds after

it was launched. The amplitude of the wave V1 that is launched on the line can be

determined by the voltage dividerrule:

1c

b

b c

ZV V

Z Z

(9.63.1)

Note: we are discussing a DC potential propagating along the line.

64


64/100

Assuming that the switch was engaged at the time t= 0, at the time = L/vthe

front of this propagating voltage step arrives at the load impedance ZL. At this

time, a portion of this incident voltage will be reflected from the load impedance,

and another portion will be absorbed by the load (transmitted to the load). The

reflection coefficient at the load is

2

1

L cL

L c

Z Z VZ Z V

(9.64.1)

The front of this propagating reflected wave (whose amplitude V2 can be either

positive or negative) reaches the battery impedance at a time 2= 2L/v. This front

will also be reflected from the battery impedance with the reflection coefficient

3

2

b cb

b c

Z Z V

Z Z V

(9.64.2)

Therefore, the battery impedance acts like a load impedance for the wave V2 and

the new voltage step V3 will be reflected towards the load

65


65/100

This process of wave reflections from two impedances may continue indefinitely

long. The front of the propagating voltage step bounces back and forth between

the load impedance and the battery impedance.

This graphical technique used to evaluate the voltage

at any location of the line as a function of time is

called a bounce diagram.

The horizontal axis represents the normalized

position; the vertical axis the normalized time. The

prediction of the temporal response at a given

location is obtained by inserting a vertical line on the

diagram at that location. The intersection of thetrajectory with that line indicates that the voltage at

that location will change its value over time since

more and more components (reflected waves) will be

added.

66


66/100

Assuming that the voltage of the battery is constant, we may conclude that thevoltage of each individual component will also be constant over time and will

have the same value as that of the front. The voltage at any location of the

transmission line is a sum of individual components:

1 2 3

2 2

... 1 ...

1 ... 1 ...

cL L b L b L b

b c

cL b L b L L b L b b

b c

ZV V V V V

Z ZZ

VZ Z

It can be seen that every next component in the sum in (9.66.1) has lower

amplitude than the previous components. Consequently, voltage on the line will

converge to its steady-state value:

The absolute value of the reflection coefficient cannot exceed one. Therefore, the

quantity in the brackets can be expressed as the closed form summation:

2 11 ... 11

for

(9.66.1)

(9.66.2)

67


67/100

Therefore, the steady-state (asymptotic) voltage will be:1

1c L

b

b c L b

ZV V

Z Z

or, using the definitions of reflection coefficients:

11

L c L ccb

b c L c L c b c b c

Z Z Z ZZV VZ Z Z Z Z Z Z Z Z Z

Lb

b L

ZV V

Z Z

Which simplifies to

The current flowing through the load is

b

L b L

VVI

Z Z Z

(9.67.1)

(9.67.2)

(9.67.3)

(9.67.4)

68


68/100

Example 9.12: A 12 V battery is connected via a switch to a transmission line thatis 6 m long. The characteristic impedance of the line is 50 , the battery

impedance is 25 , and the load impedance is 25 . The velocity of propagation is

2 106 m/s. Find and sketch the voltage at the midpoint of this line during the time

interval 0 < t< 9 s.

The amplitude of the wave that is launched on the transmission line is

1

5012 8

25 50c

b

b c

ZV V V

Z Z

The reflection coefficient at the load is

25 50 125 50 3

L cL

L c

Z ZZ Z

The reflection coefficient at the battery is

25 50 1

25 50 3b c

b

b c

Z Z

Z Z

(9.68.1)

(9.68.2)

(9.68.3)

69


69/100

We will use the bounce diagram to evaluate the

voltage at the midpoint of the line. We will identify

the amplitudes of the waves. The normalized time

is t/, where

3L v s

The voltage at the midpoint is zero until the firstwave arrives. At this moment, the amplitude

becomes 8. later on, the wave, reflected from the

load arrives and the voltage will be changed

The steady-state voltage found from (9.67.3) is

2512 6

25 25L

b

b L

ZV V V

Z Z

(9.69.1)

(9.69.2)

70


70/100

71


71/100

Example 9.13: A battery with no internal impedance has an open circuit voltage of100 V. At a time t= 0, this battery is connected to an air-felled 50 coaxial cable

via a 150 resistor. The cable is 300 m long and is terminated in a load of 33.3 .

a) Sketch a bounce diagram for the first 4 s after the switch is closed;

b) Plot the voltage on the load as a function of time;

c) Find the asymptotic value of the load voltage.

The reflection coefficients and the

voltage step propagating on the

line are:

150 50 1

150 50 2b c

b

b c

Z Z

Z Z

33.3 50 1

33.3 50 5L c

L

L c

Z Z

Z Z

1

50100 25

150 50c

b

b c

ZV V V

Z Z

(9.71.1)

(9.71.2)

72


72/100

Since the coaxial line is filled with air, the velocity of propagation

is 3108 m/s. Therefore, the signal takes 1 s to travel from one

end to another.

The asymptotic voltage and current computed from (9.67.3) and

(9.67.4) are

18.2

0.55

t

t

V V

I A

73


73/100

Up to this point, we were discussing propagation of a step voltage along thetransmission line. We learned that the front propagates with a certain velocity and

it takes specific time (proportional to L/v) for the signal to travel through the line of

length L.

Highly important is the case when a voltage pulse is propagating through the line.

We will consider the line

connecting a pulse

generator to the load.

As previously, we can

estimate the amplitude of thepulse launched on the line:

1c

g

g c

ZV V

Z Z

(9.73.1)

74


74/100

We assume next that the temporal width of the pulse is much less than the timeneeded for the pulse to travel through the line:

t L v

For instance, if the length of the transmission line is 3 m and the velocity of

propagation is c:9

8

310 10

3 10

Ls

v

the pulse duration must be less than 10 ns.

Note that the velocity of propagation along the transmission line is usually less than

that for vacuum and, therefore, the critical pulse duration will be slightly increased.The pulse of the voltage V1 is launched from the generator towards the load. A

portion of the pulse is absorbed by the load, and a portion of the pulse is reflected

back toward the generator. The amount of the energy that is absorbed (and

reflected) is determined by the ratio of the load impedance to the lines

characteristic impedance.

(9.74.1)

(9.74.2)

75


75/100

We can conclude that the energy of the incident pulse will be divided into theenergy of the reflected wave and the energy absorbed by the load:

.

inc ref abs

inc energy

P t P t P t

Or, in terms of impedances and reflection coefficients:

2

2 2

11 L

c c L

VV V

Z Z Z

Here, VL is the voltage that appears on the load. Incorporating the reflection

coefficient, we arrive at:

2

2 211 L c L c L

c c L

Z Z Z Z VV V

Z Z Z

(9.75.1)

(9.75.2)

(9.75.3)

76


76/100

The last equation can be solved for the load voltage:

2

1 1

2L c L c LL LL

c c L c

Z Z Z Z ZZ ZV V V

Z Z Z Z

At this point, we re-introduce the transmission coefficient as:

1

21L L

L c

V Z

V Z Z

(9.76.1)

(9.76.2)

77


77/100

Example 9.14: Evaluate the transmission coefficient for a wave that propagates inthe +zdirection through a coaxial cable. The dielectric constant in the region z< 0

is 2 and in the region z> 0 is 3. the physical dimensions of the cable are constant.

The characteristic impedance of the coax cable can be found as:

ln

2c

b aZ

We will consider the region 2 (z> 0) as a load for the line in the region 1. Thus, the

transmission coefficient given by (9.76.2) is:

0

0 0

2 3 ln 2 2 3 0.93 ln 2 2 ln 2 1 3 1 2

b ab a b a

78


78/100

Let us consider the joining of twotransmission lines having

different dimensions. Apparently,

the characteristic impedances in

two regions will differ and a

portion of the signal will be

reflected back from the secondtransmission line.

If we send a pulse to propagate through such a system of

two joined lines, we may observe a pulse reflected back

from a boundary (the second line in our case). If the

velocity of propagation on a transmission line vis known,and the time tneeded for a pulse to travel to and back

from the reflective boundary is measured, the distance to

the boundary is:2d v t (9.78.1)

Can be used for fault locations: time domain reflectometry.

79


79/100

Example 9.15: Using the reflection and transmission coefficients, show thatenergy is conserved at the junction between two lossless transmission lines.

The conservation of energy

implies that (9.75.1) or

(9.75.2) must be satisfied:

2 22

1 1 2

inc incinc

c c c

V VV

Z Z Z

Therefore:

2 2

2 1 2 12 22 2

2 22 1 2 1 2 2 1 2 1 2 1

1 1 2 1 2 1

2 2

2 22 1 22 2 2 2

2 1 2 1 1 2 1 2 1

2 2 2

2 2 1 2 1 1 2 1 2 1

2

2 21

42 4

2 21

2 2

c c c c

c c c c c c c c c c c

c c c c c c

c cc c c

c c c c c c c c c

c c c c c c c c c c

Z Z Z Z

Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z

Z ZZ Z Z

Z Z Z Z Z Z Z Z Z

Z Z Z Z Z Z Z Z Z Z

2 1

1

cZ

80


80/100

Example 9.16: A 1 V pulse propagates on a transmission line terminated in an opencircuit at z= 0. Four oscilloscopes are triggered by the same pulse generator and

are located at za = -6, zb = -4, zc= -2, zd= 0 m. Find the velocity of propagation and

interpret the signals on the oscilloscopes. Sketch the signals if the line is

terminated with a short circuit.

From the traces on the oscilloscopes A and B:

6

6

22 10

1 10

zv m s

t

The oscilloscope D is at the location of the open circuit, and the incident and the

reflected pulses add together.

81


81/100

The signals detected aftert= 4 s are the reflected pulses propagating backwardsto the generator. If the line was terminated with a short circuit, the oscilloscopes

would detect something like:

Note that the voltage across the

short circuit must be zero, as it isdepicted by the oscilloscope D.

Also, the reflected pulses will be

opposite to the incident ones.

82


82/100

Up to this point, our discussion was limited to lossless transmission linesconsisting of equivalent inductors and capacitors only. As a result, the

characteristic impedance for such lines is real. Let us incorporate ohmic

losses within the conductors and leakage currents between conductors.

In this case, the characteristic impedance becomes complex and the new

model will be:

83


83/100

The new first-order PDEs (telegraphers equations) are

( , ) ( , ) ( , )

( , ) ( , ) ( , )

I z t V z tC GV z t

z t

V z t I z t L RI z t

z t

(9.83.1)

(9.83.2)

Where the circuit elements are defines as

; ; ;L C R G

L C R Gz z z z

A time-harmonic excitation of the transmission line leads to a phasor notation:

( ) ( )

( ) ( )

I zG j C V z

z

V zR j L I z

z

(9.83.3)

(9.83.4)

(9.83.5)

84


84/100

The quantities in square brackets are denoted by distributed admittance anddistributed impedance leading to

YZ

( ) ( )

( ) ( )

dI zYV z

dz

dV zZI z

dz

(9.84.1)

(9.84.2)

Second-order ODEs can be derived for current and voltage:

2

2

2

2

( ) ( )

( )

( )

d I zZYI z

dz

d V z

ZYV zdz

(9.84.3)

(9.84.4)

The phasor form solutions:

1 2

1 2

( )

( )

z z

z z

V z V e V e

I z I e I e

(9.84.5)

(9.84.6)

85


85/100

Here, is the complex propagation constant:

j ZY R j L G j C (9.85.1)

As previously, we denote by V1 and I1 the amplitudes of the forward (in the +z

direction) propagating voltage and current waves; and V2

and I2

are the amplitudes

of the backward (in the -zdirection) propagating voltage and current waves.

Therefore, the time varying waves will be in the form:

1 2

1 2

( , ) cos cos

( , ) cos cos

z z

z z

V z t V e t z V e t z

I z t I e t z I e t z

(9.85.2)

(9.85.3)

We recognize that (9.85.2) and (9.85.3) are exponentially decaying propagating

waves: the forward wave (V1, I1) propagates and decays in the +zdirection, while

the backward wave (V2, I2) propagates and decays in thezdirection.

86


86/100

An example of a forward

propagating wave: it is

possible to determine the

values ofand from the

graph as shown.

87


87/100

For small losses, employing a binomial approximation (1x)n 1 nxforx


88/100

Example 9.17: Find the complex propagation constant if the circuit elements satisfythe ratio . Interpret this situation. R L G C

The complex propagation constant is:

RC Cj R j L G j C R j L j C R j L

L L

The attenuation constant is independent on frequency, which implies no distortion

of a signal as it propagates on this transmission line and a constant attenuation.

The characteristic impedance of this transmission line

c

Z R j L R j L LZY G j C C RC L j C

does not depend on frequency. Therefore, this transmission line is distortionless.

89


89/100

Example 9.18: The attenuation on a 50 distortionless transmission line is 0.01dB/m. The line has a capacitance of 0.110-9 F/m. Find:

a) Transmission line parameters: distributed inductance, resistance, conductance;

b) The velocity of wave propagation.

Since the line is distortionless, the characteristic impedance is

2 2 9 7

50 50 0.1 10 2.5 10

c c

LZ L Z C H m

C

The attenuation constant is:

0.01

0.01 1 20lg( ) 8.686 0.0012 8.686

CR dB m Np m e dB m Np mL

Therefore:

0.0012 50 0.0575

c

LR Z m

C

90


90/100

The distortionless line criterium is:

5

2 2

0.0575 2.3 10

50c

R G RC RG S m

ZL LC

The phase velocity is:

8

7 9

1 12 10

2.5 10 0.1 10v m s

LC

91


91/100

The losses of types considered previously (finite conductivity of conductors andnonzero conductivity of real dielectrics) lead to attenuation of wave amplitude as

it propagates through the line.

When the wavelength is comparable with the physical dimensions of the line or

when the permittivity of the dielectric depends on the frequency, another

phenomenon called dispersion occurs.

We will model dispersion by the

insertion of a distributed parasitic

capacitance in parallel to the distributed

inductance.

While developing the telegraphersequation for this case, we note that the

current entering the node will split into

two portions:

( , ) ( , ) ( , )L cI z t I z t I z t (9.91.1)

92


92/100

The voltage drop across the capacitor is:

( , )( , ) L

I z tV z tL

z t

The voltage drop across the parasitic (shunt) capacitor is:

( , ) 1

c

s

V z t I dtz C

Note: the units of this additional shunt capacitance are Fm rather than F/m.

The current passing through the shunt capacitor:

( , ) ( , )

I z t V z tCz t

The wave equation will be:

2 2 4

2 2 2 2

( , ) ( , ) ( , ) 0s

V z t V z t V z t LC LC

z t z t

(9.92.1)

(9.92.2)

(9.92.3)

(9.92.4)

93


93/100

Assume that there is a time-harmonic signal generator connected to theinfinitely long transmission line. The complex time-varying wave propagating

through the line will be

( )

0( , )j t zV z t V e

Combining (9.92.4) and (9.93.1) leads to the dispersion relation (the terms in thesquare brackets) relating the propagation constant to the frequency of the wave:

2 2 2 2( )

0 0j t z sV e j LC j LC j j

Therefore, the propagation constant is a nonlinear function of frequency:

2

1 s

LC

LC

(9.93.1)

(9.93.2)

(9.93.3)

94


94/100

dispersion

no dispersion

The propagation constant

depends on frequency this

phenomenon is called

dispersion.

The phase velocity is a function

of frequency also.

95


95/100

We can consider dispersion as a low pass filter acting on a signal. Thepropagation constant will be a real number for frequencies less than the particular

cutoff frequency 0.

0

1

sLC

This cutoff frequency is equal to the resonant frequency of the tank circuit. Abovethis frequency, the propagation constant will be imaginary, and the wave will not

propagate.

In the non-dispersive frequency range (below the cutoff frequency), the velocity of

propagation is

0

1

V

LC

The wave number below the cutoff frequency is:

00

0v

(9.95.1)

(9.95.2)

(9.95.3)

96


96/100

Dispersion implies thatthe propagation

constant depends on

the frequency.

There are positive and

negative dispersions.

Note that in the case of

negative dispersion,

waves of frequencies

less than a certain cutoff

frequency will propagate.

While for the positive dispersion, only the waves whose frequency exceeds the

cutoff frequency will propagate.

97


97/100

If two signals of different frequencies propagate through the same lineardispersive medium, we must employ the concept ofgroup velocity

if a narrow pulse propagates in a dispersive region, according to Fourier

analysis, such a pulse consists of a number of high frequency components.

Each of them will propagate with different phase velocity.

Let us assume that two waves of the same amplitude but slightly different

frequencies propagate through the same dispersive medium. The frequencies are:

1 0 2 0;

The corresponding propagation constants are:

1 0 2 0; The total signal will be a sum of two waves:

0 1 1 2 2

0 0 0

( , ) cos cos

2 cos cos

V z t V t z t z

V t z t z

(9.97.1)

(9.97.2)

(9.97.3)

98


98/100

Summation of two time-harmonicwaves of slightly different

frequencies leads to constructive

and destructive interference.

voltage

By detecting signals at two locations, we

can track a point of constant phase

propagating with the phase velocity:

0 0pv

and the peak of the envelope propagating with the group velocity: gv

(9.98.1)

(9.98.2)

In dispersive media, phase and group velocities can be considerably different!

99


99/100

Example 9.19: Find the phase and group velocities for:a) Normal transmission line;

b) A line in which the elements are interchanged.

a) The propagationconstant computed

according to (9.87.1)

will be:

j YZ j L j C j LC The phase velocity is 1

pv

LC

The group velocity is 11

gv

LC

100


100/100

The two velocities are equal in this case and both are independent on frequency.

b) The propagation constant computed according to (9.87.1) will be:

1 1 1

j YZj Lj C

j LC

The phase velocity is2

pv LC

The group velocity is2

1gv LC

The phase and group velocities both depend on frequency and are in the

opposite directions.

Lecture- EM Transmission Lines

Documents

Transcript of Lecture- EM Transmission Lines