ELEN 3371 Electromagnetics Fall 2008 1 Lecture 9: EM Transmission Lines and Smith Chart Instructor:...

ELEN 3371 Electromagnetics Fall 2008

1

Lecture 9: EM Transmission Lines and Smith Chart

Instructor: Dr. Gleb V. Tcheslavski

Contact: [email protected]

Office Hours:

Room 2030

Class web site: www.ee.lamar.edu/gleb/em/Index.htm


2

Equivalent electrical circuits

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

Coaxial cable


4

Microstrip line


5

Twin lead


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 assume no loss in the lines.


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

ˆL CˆˆL L z C C z (9.7.1)


8


(9.8.1)

(9.8.2)

(9.8.3)

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

We can model the transmission line with an equivalent circuit consisting of an infinite number of distributed inductors and capacitors.


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 the wave (dispersion);3)Parameters of the line are constant.

We can analyze EM transmission lines either as a large number of distributed two-port networks or as a coupled set of first-order PDEs that are called the telegraphers’ equations.


10

Transmission line equations

While analyzing the equivalent circuit of the lossless transmission line, it is simpler to use Kirchhoff’s laws rather than Maxwell’s equations.

Therefore, we will consider the equivalent circuit of this form:

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

ˆˆ ;L C

L Cz z

(9.10.1)

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


11


The current entering the node at the location z is I(z). The part of this current will flow through the capacitor, and the rest flows into the section. Therefore:

( , )ˆ( , ) ( , )V z t

I z t C z I z z tt

( , ) ( , ) ( , )ˆI 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

( , ) ( , )ˆI z t V z tC

z t

(9.11.3)


12


Similarly, the sum of the voltage drops in this section can be calculated via the Kirchhoff’s law also:

( , )ˆ( , ) ( , )I z t

V z z t L z V z tt

( , ) ( , ) ( , )ˆV z t V z z t I z tL

z t

(9.12.1)

(9.12.2)

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

( , ) ( , )ˆV z t I z tL

z t

(9.12.3)


13


The equations (9.11.3) and (9.12.3) are two linear coupled first-order PDEs called the telegrapher’s (Heaviside) equations. They can be composed in a second-order PDE:

2 2

2 2

( , ) ( , )ˆˆ 0I z t I z t

LCz t

2 2

2 2

( , ) ( , )ˆˆ 0V z t V z t

LCz 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

ˆˆv

LC (9.13.3)


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 length are defined as

ˆˆ ;R C

R Cz z

(9.14.1)

Potential drop over the resistor R and the current through the capacitor C are:

ˆ( , ) ( , )

( , )ˆ( , )

V z t I z t R z

V z tI z t C z

t

(9.14.2)

(9.14.3)


15


( , ) ˆ0 ( , )

( , ) ( , )ˆ

V z tIf z I z t R

zI z t V z t

Cz t

The corresponding second-order PDE for the potential is:

2

2

( , )( , ) ( , )ˆ( ,ˆ ˆ)ˆˆI z t V z tR R

V z t V z tRC

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:

1ˆˆ

DRC

(9.15.4)


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 ze

z DtD

22 24

2 3 2 2 5 2

( , ) 1

2 42

z

DtV z t z ze

z Dt D tD

Differentiating the solution with respect to t:

224

3 2 5 2

1 ( , ) 1 1

2 42

z

DtV z t t ze

D t D t DtD

(9.16.1)

(9.16.2)

(9.16.3)

(9.16.4)


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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

Sinusoidal waves

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 of traveling 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 z on the line.


19

Sinusoidal waves

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

22

2

22

2

( )( ) 0

( )( ) 0

d V zk V z

dz

d I zk I z

dz

Therefore, the wave equations will become:

Here, as previously, k is 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

Sinusoidal waves

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 along the transmission line. The measured values are plotted. Write an expression for the wave for the given data.


21

Sinusoidal waves

Slo

pe o

f th

e tr

ajec

tory

…

The

dat

a


22

Sinusoidal waves

We assume that the peak-to peak amplitude of the wave is 2V0. We also conclude that the wave propagates in the +z direction.

The period of the wave is 2s, therefore, the frequency of oscillations is ½ Hz.

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

5 14

1 0v m s

The wave number is:12 0.5

4 4k m

v

The wavelength is:2 4 2

8mk

Therefore, the wave is:4

0( , )z

j 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

Sinusoidal waves

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( ) jkzV z V eThe phasor form of (9.12.3) in this case is

ˆ((

))

) (jkV z j LI zdV z

dz

0( ) ( )ˆ ˆ

jkzk kI z V z V e

L L

Which may be rearranged as:

(9.23.1)

(9.23.2)

(9.23.3)


24

Sinusoidal waves

The ratio of the voltage to the current is a very important transmission line parameter called the characteristic impedance:

ˆ( )

( )c

V z LZ

I z k

1

ˆˆk and v

v LC

Since

Then: ˆ

ˆc

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

Sinusoidal waves

If we know the characteristics of the transmission lane and the forward voltage wave, we may find the forward current wave by dividing voltage by the characteristic impedance.

Another important parameter of a transmission line is its length L, which is often normalized by the wavelength of the propagation wave.

Assuming that the dielectric between conductors has and


26

Sinusoidal waves

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

Sinusoidal waves

Example 9.4: Evaluate the velocity of propagation and the characteristic impedance of 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 v and the Zc may be decreased by insertion of a dielectric between leads.

70 4 10 6ˆ ln ln 0.142 2 3

H mb

La

12

02 2 8.854 10ˆ 80ln ln 6 3

pF mCb a

8

6 12

1 13 10

ˆˆ 0.14 10 80 10m sv

LC

6 12ˆˆ 0.14 10 80 10 42cZ L C

(9.27.1)

(9.27.2)

(9.27.3)

(9.27.4)


28

Terminators

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 termination may be either an impedance or another transmission line with different parameters. We also assume no transients.


29

Terminators

The phasor voltage at any point on the line is:

2 2( ) jkz jkzV 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 to A2 represents the magnitude of the wave incident on the load ZL.

(9.29.1)

(9.29.2)

(9.29.3)


30

Terminators

(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

1

1L

L

B z

A z

(9.30.2)

(9.30.3)


31

Terminators

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 of matched load exists when:

L cZ Z

In this situation: ( ) cZ z Z

(9.31.3)

(9.31.4)

(9.31.5)


32

Terminators

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

The characteristic impedance is:

ˆ lnˆ 2c

b aLZ

C

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


33

Terminators

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 ln

2 2

1 1 1 13 2 0.1

1 1 1 13 2

r r

r r

r r

r r

b a b a

Z Z

Z Z b a b a


34

Terminators

The reflection coefficient is completely determined by the value of the impedance of the 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 the load 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 reflection coefficient = 0 – line is matched. In this case, all energy of generator will be absorbed by the load.


35

Terminators

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

Terminators

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

Terminators

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 to 1

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

Terminators

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 VSWR would be 1.


39

Terminators


40

Impedance and line matching

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 jkzjkz 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 c

jkz

L c

jkzL c c L

L

L

c

cc

Z Ze e

Z Z Z kz j Z kzZ Z

Z ZZ jZ kz

ZZ kz

z ZZ jZ kzj Z kze e

Z Z

(9.40.1)

(9.40.2)

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


41


Considering the transmission line shown, we assume that a load with the impedance ZL is connected to a transmission line of length L having the characteristic impedance Zc and the wave number k.

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

tan( ) ( )

tanL c

in cc L

Z jZ kLZ L Z z L Z

Z jZ kL

(9.41.1)

Or as the normalized input impedance: tan( )

1 tanL

inL

z j kLz L

jz kL

(9.41.2)


42

Impedance and line matching (Ex)

Example 9.7: A signal generator whose frequency f = 100 MHz is connected to a coaxial cable 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

inL

jz j kLz z m

jz kL j

Therefore, the input impedance is

0.5 100 50in in cZ z Z


43


The wave number can be expressed in terms of wavelength:

2k

Therefore:2

kL L

and 2tan tan tan ; integerkL 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

44L c c

in c inc 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 ZL is numerically equal to the normalized load admittance yL = 1/zL.


44


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


When the load impedance equals the characteristic impedance, the load and the line 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 kL

j 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


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


Assuming that a transmission line is terminated with a load impedance ZL or load admittance YL that is not equal to the line’s 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 susceptance –jB 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 insertion point (d1) back to the generator.

In practice, matching can be done by insertion of a short-circuited transmission line of particular length.


48


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 equal –jB. 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 stub’s 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

Impedance and line matching (Ex)

Example 9.8: A lossless transmission line is terminated with 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 ccLc

Q in Q cc cQ Q

L c

ZZZZ

Z Z Z ZZZZ

Z Z Z ZZ ZZ Z

Z Z

21 2

2Q

Q cQ c

ZZ Z

Z Z


50

Smith chart

The input impedance of a transmission line depends on the impedance of the load, 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

inL

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

Smith chart

The reflection coefficient will be complex:

1

1L

r iL

zj

z

2 2

2 22 2

1 1 21

1 1 1 1r i r i i

Lr 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 r and i. The center and radius of each circle are determined by the normalized resistance r and reactance x. All circles are inside the unit circle since maximal = 1.


52

Smith chart

For the constant r circles: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 r increases.3.All constant r circles pass through the point r =1, i = 0.4.The normalized resistance r = is at the point r =1, i = 0.For the constant x (partial) circles:1.The centers of all the constant x circles are on the r =1 line. The circles with x > 0 (inductive reactance) are above the r axis; the circles with x < 0 (capacitive) are below the r axis.2. The radius of circles decreases when absolute value of x increases.3. The normalized reactances x = are at the point r =1, i = 0


53

Smith chart

The constant r circles are orthogonal to the constant x circles at every intersection.

The actual (de-normalized) load impedance is:

( )L cZ Z r jx

The intersection of an r circle and an x circle 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 r and 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

Smith chart (Example)

Example 9.9: On the simplified Smith chart, locate the following normalized impedances:

) 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

Smith chart

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

Smith chart

The normalized input impedance at the point -z’ is

2 '

2 '

1( ') 1( ')

1 1

jj kzin

in j kz jc

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 z’ and 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

Smith chart

For example, the impedancez = 0.5 + j 0.5corresponds 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 the impedance.

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

Smith chart

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 the location 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 not very attractive due to fringing, and use short circuits only…


59


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

5

2 1020

1 10

vm

f

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

The normalized load impedance is50 50

2 225L

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


Finally, we draw a line passing 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 load impedance is

25(0.25 5)

6.25 6.25

in c inZ Z z

j

j


61


An interesting property of the Smith chart is that it can be used equally well in terms 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


We draw a circle centered at the center of the Smith chart and passing through the input admittance we have calculated.Two intersections of that circle with the “real part 1” circle determine the two values of the matching admittances. The location where these admittances are to be inserted can be determined by the angles that correspond to each arc.


63

Transient effects and Bounce diagram

Let us consider a transmission line connected to the battery with an internal impedance Zb through a switch. The line has a characteristic impedance Zc and is connected to the load ZL. We assume first that all impedances are pure resistances.

The signal propagates with the velocity v and will reach the load L/v seconds after it was launched. The amplitude of the wave V1 that is launched on the line can be determined by the voltage divider rule:

1c

bb c

ZV V

Z Z

(9.63.1)

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


64


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

Z 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


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 the trajectory 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


Assuming that the voltage of the battery is constant, we may conclude that the voltage 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 Z

ZV

Z 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 ... 1

1for

(9.66.1)

(9.66.2)


67


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

1

1c L

bb c L b

ZV V

Z Z

or, using the definitions of reflection coefficients:

1

1L c L cc

bb c L c L c b c b c

Z Z Z ZZV V

Z 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


Example 9.12: A 12 V battery is connected via a switch to a transmission line that is 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

bb c

ZV V V

Z Z

The reflection coefficient at the load is

25 50 1

25 50 3L c

LL c

Z Z

Z Z

The reflection coefficient at the battery is

25 50 1

25 50 3b c

bb c

Z Z

Z Z

(9.68.1)

(9.68.2)

(9.68.3)


69


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 first wave 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

bb L

ZV V V

Z Z

(9.69.1)

(9.69.2)


70



71


Example 9.13: A battery with no internal impedance has an open circuit voltage of 100 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

bb c

Z Z

Z Z

33.3 50 1

33.3 50 5L c

LL c

Z Z

Z Z

1

50100 25

150 50c

bb c

ZV V V

Z Z

(9.71.1)

(9.71.2)


72


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

0.55t

t

V V

I A


73

Pulse propagation

Up to this point, we were discussing propagation of a step voltage along the transmission 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 the pulse launched on the line:

1c

gg c

ZV V

Z Z

(9.73.1)


74

Pulse propagation

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

t L vFor instance, if the length of the transmission line is 3 m and the velocity of propagation is c:

98

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 line’s characteristic impedance.

(9.74.1)

(9.74.2)


75

Pulse propagation

We can conclude that the energy of the incident pulse will be divided into the energy 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:

22 211 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:

22 211 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

Pulse propagation

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

Pulse propagation (Example)

Example 9.14: Evaluate the transmission coefficient for a wave that propagates in the +z direction 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 30.9

3 ln 2 2 ln 2 1 3 1 2

b a

b a b a


78

Pulse propagation

Let us consider the joining of two transmission 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 second transmission 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 v is known, and the time t needed 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

Pulse propagation

Example 9.15: Using the reflection and transmission coefficients, show that energy 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 22 1 2 12 22 2 2 2

2 1 2 1 2 2 1 2 1 2 1

1 1 2 1 2 1

2 22 2

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

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

22 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 ZZ 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


Example 9.16: A 1 V pulse propagates on a transmission line terminated in an open circuit 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:

66

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


The signals detected after t = 4 s are the reflected pulses propagating backwards to 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 is depicted by the oscilloscope D.Also, the reflected pulses will be opposite to the incident ones.


82

Lossy transmission lines

Up to this point, our discussion was limited to lossless transmission lines consisting 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


The new first-order PDEs (telegrapher’s equations) are

( , ) ( , )ˆ ˆ ( , )

( , ) ( , )ˆ ˆ ( , )

I z t V z tC GV z t

z tV z t I z t

L RI z tz t

(9.83.1)

(9.83.2)

Where the circuit elements are defines as

ˆ ˆˆ ˆ; ; ;L C R G

L C R Gz z z z

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

( ) ˆ ˆ ( )

( ) ˆ ˆ ( )

I zG j C V z

zV z

R j L I zz

(9.83.3)

(9.83.4)

(9.83.5)


84


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

YZ

( ) ˆ ( )

( ) ˆ ( )

dI zYV z

dzdV z

ZI zdz

(9.84.1)

(9.84.2)

Second-order ODEs can be derived for current and voltage:2

2

2

2

( ) ˆ ˆ ( )

( ) ˆ ˆ ( )

d I zZYI z

dz

d V zZYV z

dz

(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


Here, is the complex propagation constant:

ˆ ˆˆ ˆ ˆ ˆj 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 -z direction) 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 +z direction, while the backward wave (V2, I2) propagates and decays in the –z direction.


86


An example of a forward propagating wave: it is possible to determine the values of and from the graph as shown.


87


For small losses, employing a binomial approximation (1 – x)n 1 – nx for x << 1, the complex propagating constant is approximately

ˆ ˆˆ ˆˆ ˆˆ ˆ1 1 1

ˆ ˆ ˆ ˆ2 2

R G R Gj j L j C j LC j

j L Lj C C

The attenuation constant can be approximated as

ˆ ˆˆ ˆ1ˆ ˆˆ ˆˆ ˆ ˆ ˆ22 2

R G R GLC LC

L LC C

The attenuation constant does not depend on frequency.

(9.87.1)

(9.87.2)


88

Lossy transmission lines (Ex)

Example 9.17: Find the complex propagation constant if the circuit elements satisfy the ratio . Interpret this situation.ˆ ˆˆ ˆR L G C

The complex propagation constant is:

ˆ ˆˆˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆRC C

j R j L G j C R j L j C R j LL 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

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆˆ ˆˆ ˆc

Z R j L R j L LZ

Y G j C CRC L j C

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


89


Example 9.18: The attenuation on a 50 distortionless transmission line is 0.01 dB/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 7ˆ

ˆˆ50 50 0.1 10 2.5 10ˆc c

LZ L Z C H m

C

The attenuation constant is:

ˆ 0.01ˆ 0.01 1 20lg( ) 8.686 0.0012ˆ 8.686

CR dB m Np m e dB m Np m

L

Therefore:ˆ

ˆ 0.0012 50 0.0575ˆ c

LR Z m

C


90

Lossy transmission lines (Ex)

The distortionless line criterium is:

52 2

ˆ ˆˆ ˆ ˆ 0.0575ˆ 2.3 10ˆ ˆ ˆ 50c

R G RC RG S m

ZL LC

The phase velocity is:

8

7 9

1 12 10

ˆˆ 2.5 10 0.1 10v m s

LC


91

Dispersion and group velocity

The losses of types considered previously (finite conductivity of conductors and nonzero 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 telegrapher’s equation 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


The voltage drop across the capacitor is:

( , )( , ) ˆ LI z tV z tL

z t

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

( , ) 1ˆ c

s

V z tI dt

z C

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

The current passing through the shunt capacitor:

( , ) ( , )ˆI z t V z tC

z t

The wave equation will be:

2 2 4

2 2 2 2

( , ) ( , ) ( , )ˆ ˆˆ ˆ 0s

V z t V z t V z tLC LC

z t z t

(9.92.1)

(9.92.2)

(9.92.3)

(9.92.4)


93


Assume that there is a time-harmonic signal generator connected to the infinitely 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 the square brackets) relating the propagation constant to the frequency of the wave:

2 2 2 2( )0

ˆ ˆˆ ˆ 0j t zsV e j LC j LC j j

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

2

ˆˆ

ˆˆ1 s

LC

LC

(9.93.1)

(9.93.2)

(9.93.3)


94


dispersion

no dispersion

The propagation constant depends on frequency – this phenomenon is called dispersion.

The phase velocity is a function of frequency also.


95


We can consider dispersion as a low pass filter acting on a signal. The propagation constant will be a real number for frequencies less than the particular cutoff frequency 0.

0

1

ˆˆsLC

This cutoff frequency is equal to the resonant frequency of the “tank” circuit. Above this 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

ˆˆV

LC

The wave number below the cutoff frequency is:

00

0v

(9.95.1)

(9.95.2)

(9.95.3)


96


Dispersion implies that the 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


If two signals of different frequencies propagate through the same linear dispersive medium, we must employ the concept of group 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


Summation of two time-harmonic waves of slightly different frequencies leads to constructive and destructive interference.

volta

ge

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

Dispersion and group velocity (Ex)

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 propagation constant computed according to (9.87.1) will be:

ˆ ˆˆ ˆ ˆ ˆj YZ j L j C j LC

The phase velocity is 1

ˆˆpv

LC

The group velocity is 11

ˆˆgv

LC


100

Dispersion and group velocity (Ex)

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 1ˆ ˆˆ ˆ ˆˆ

j YZj Lj C j LC

The phase velocity is2 ˆˆ

pv LC

The group velocity is2 ˆˆ1gv LC

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

?? QUESTIONS ??

ELEN 3371 Electromagnetics Fall 2008 1 Lecture 9: EM Transmission Lines and Smith Chart Instructor:...

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