Chapter 6 Transmission Lines - University of...

6-1

Chapter 6 Transmission Lines

ECE 3317Dr. Stuart A. Long

6-3

0

0

1

2

Voltage ( )

Current ( c

V( )

los

I( ) e d)

t

C

tC

z d

z d

C z

C

α

α

= •

=

⊥

•

∫

∫

E

H

s

s

p.133

( )a

( )b

General Definitions

http://www.cartoonstock.com

6-4

* *

0

Power

Characteristic infinite line with Impedance

1 1 ˆRe V( ) I ( ) Re 2 2

V( ) Z with I )

(

A

z z dA

zz

= × •

=

∫ E H z

no reflection

( )c

p.133

General Definitions

6-5Parallel Plate Waveguide (TEM mode)

-0

-0

-1 1 00

-02 20

ˆ

ˆ

ˆ V( )

ˆ I( )

jkz

jkz

d jkz

w jkz

E e

E e

z dx E d e

E wz dy e

η

α α

α αη

=

=

= • =

= • =

∫

∫

E

H

E

H

x

y

x

y

x

Co

w

d

y

Ct

6.1a

6.1b

6.2a

6.2b

The Transverse Electromagnetic Fields in a Parallel Waveguide are approximately as follows:

Using the General

Definitions

6-6

The time-average power transmitted is given by

x

Co

w

d

y

Ct

Parallel Plate Waveguide (TEM mode)

6-7

0

0

0 [ ]

V( )

I( )

Z

jkz

jkz

z E de

E wz e

dw

η

η

−

−

Ω

=

=

=

6.3a

6.3b

6.4

x

Co

w

d

y

Ct

Parallel Plate Waveguide (TEM mode)

6-8Coaxial Line

The fields inside a coaxial line for the TEM mode are given by

6.5a

6.5b

Using the General

Definitions

6.6a

6.6b

Co

ab

Ct

6-9


Co

ab

Ct

Coaxial Line

6-10

Coaxial Line

[ ]

Example: What is the ratio b/a for

an air-filled 50coax and line?

37750= ln( /2

b .3a

)

2

b aπ

Ω

=⇒

6.7

Coaxial Line

Co

ab

Ct

6-11Rectangular Waveguide (Dominant TE10 mode)

xCo

a

b

y

CtThe Electromagnetic fields in a Rectangular

Waveguide for the are TE10 mode are

6.8a

6.8b

6.8c

6-12Rectangular Waveguide (Dominant TE10 mode)

0

1 20

-1 0

2

-02

V( ) E

ˆ I( )

2

z

z

b

x a

jk z

C

jk zz

z dy

E be

z dx

E ak e

α

α

α

απωµ

==

=

= •

=

∫

∫

y

H

x

xCo

a

b

y

CtUsing the General

Definitions

6.9a

6.9b

6-13


xCo

a

b

y

Ct

Rectangular Waveguide (Dominant TE10 mode)

6-14

-0

2-0

0 [ ]

V( )2

2I( )8

Z 1

z

z

jk zz

jk zz

z

akz E beb

E akbz eak

ωµ

π ωµπωµ

Ω

=

=

=

xCo

a

b

y

Ct

Rectangular Waveguide (Dominant TE10 mode)

http://www.fmmicrowave.com/_images/hohlleit.jpg

6-15

00

-

V LI

-

V( ) ( )

Ijkz jkzx y

x y y xE -jωμH H jω

j j

E wz E de d E z e

εEz z

j

w

ω

H

z

ωµ ωε

η− −

∇× = ∇× =

= = = =

∂ ∂= =

∂ ∂

∂= −

∂

E H H E

I CV

L H F m

C m

d ww

jωz

dµ ε

= =

= −∂

∂

Transmission Line Equations (Parallel plate waveguide)

6-16

22

2

+ -

+ -0

0

wave equationV LCV 0

V V V

1I V VZ

LC

LZC

-jkz jkz

-jkz jkz

z

e e

e e

k

ω

ω

+

+

∂+ =

∂

= +

= −

=

= 6.18

6.16

6.17

6.15

Transmission Line Equations (Parallel plate waveguide)

6-17Standing Waves on Terminated Transmission Lines

+ -

+ -

0 0

+ -

+ -0

V( ) V V

V VI( )Z Z

V( ) ZI( )

Z( ) V VZ ( )Z V V

Impedan

Z(0) Z

ce

(note )

-jkz jkz

-jkz jkz

-jkz jkz

n -jkz jkz

L

z e e

z e e

z(z)z

z e eze e

+

+

+

+

= +

= −

=

+= =

−

=

6.22

6.23

6.24

6-18

-

+

Reflection coefficient

ΓV

V L =

ΓZ ( )Γ

LnL

jkz jkz-jkz jkz-e ez

e e

+

+

+=−

With the reflection coefficient now defined, we can rewrite 6.24 as

Standing Waves on Terminated Transmission Lines

6-19

GZLZ

Transmission LineGenerator Load

GV

6.27

Evaluating at the Load (Z 0)

Z 11Z( or0)Z ( 0) Z 1 Z 10

zn

LL

L L

z nznL Ln

=

−+ Γ==

= Γ =− +

=Γ

Evaluating at a point along line ( )Z

Z Z tanZ( ) 0Z ( )Z Z Z tan0 0

z ln

j klz l Lz ln j klL

= −

+= −= − = =

+

6.25


6-20

V whenmax

V when min

2 0 2

2 3

,

φ kz - π

φ

, ,...

, ..kz π π .

+ =

+ = − −

with

2 V V 1( )

( 2 )V V

1( )

Γ

Γ Γ

Γ

L

L L

L

j kzez

jφe

j φ kzez

+ +

++

=

=

+=


6-21Standing Wave Pattern

z

1+ LΓ

1

1- LΓ

+

V( )V

z

( 2 )V V 1( )

V( ) ( 2 )1V

Γ

Γ

L

L

j φ kzez

z j φ kze

++ +

++

=

=+

Fig. 6.6

6-22Crank Diagram

( 2 )1 ΓLj φ kze ++

Im

Re1ΓL

Fig. 6.6

( 2 )V V 1( )

V( ) ( 2 )1V

Γ

Γ

L

L

j φ kzez

z j φ kze

+

+

= +

+=

+

+

6-23

1- LΓ

Im

Re1 ΓL

V( ) ( 2 )1V

ΓLz j φ kze +++

=

1- LΓ 1+ LΓ

1+ LΓ

1

z

Crank Diagram

6-24

[ ]

[ ]

[ ][ ]

V Imax maxVSWR = =V Imin min

1+ ΓVSWR =

1- Γ

VSWR -1Γ =

VSWR +1

L

L

L

6.29

Z

Vmin

N

To dete

ote :

rmine ( ) or

also need position of

P2

ower ~

Power t

2V Γ

21o load ~ Γ

-

L L

PLP

L

⇒

∠Γ

−=

+

6.30

Standing Wave Ratio (SWR or VSWR)

6-25

[ ] [ ]Z 17.4 30 Z 500

Z 1 Z 0Γ 0.24 .55Z 1 Z 0

1.

an

99Γ 0

d

.6

or

jL

ZnL L jL ZnL L

jeL

= − =

− −= = = − −

+ +

−=

Ω Ω

Example

ZLZ0

+V 5. V25 =Ex. 6.6

Γ 0.6

Γ 1.99

L

Lφ

=

= ∠ =V when

max

V when min

2 0 2

2 3

,

φ kz - π

φ

, ,...

, ..kz π π .

+ =

+ = − −

where

z

1

+

V(z)V

1.6

0.4

-0.092λ-0.342λ-0.592λ

6-26

V 2min

( 1.99)

where

.0922 2(2 )

m

m

k z

zk

φ π

π φ π λ λπ

+ = −

− − − += = = −

z

8.4

2.1

-0.092λ-0.342λ-0.592λ

V

maxV

minV

Ex. 6.6

Example

6-27

( )

Vmax 8.4 4.0Vmin 2.1

-1 4 1Γ 0.61 4 1

2Γ 2 2 .092 1.99

1 ΓZ Z 17.4 300 1 Γ

VSWR

VSWRL VSWR

k zL m

L jLL

πφ π π λλ

= = =

−= = =

+ +

∠ = = − − = − − − = −

+ = = −−

Ω

Reverse problem

Ex. 6.6

To determine ( ) or also need position ofZ VminL L∠Γ

Example

6-28Open Circuit

00 0

0

Z Z tan 1Z( ) Z Z Z Z tan tan

L

L

j kllj kl j kl

+− = =

+

X

klπ 2π 3π

inductive

capacitive

0X Z cotj j kl= −

pure imaginary

ZL=+∞O/C

l

If the impedance is purely imaginary, Z = R+jX becomes Z = jX, where X is called the reactance.

6-29

ZL=+∞O/C

l

X

klπ

inductive

capacitive

0X Z cotj j kl= −

pure imaginary

( )at

(short circuit)

2 4

Z 0 4 near is inductive

near is ca

Z

Note:

0 Z pa citi ve

kl l

kl

kl

π λ

λ

π

= ⇒ =

− =

≤

≥

Open Circuit

6-30Smith Chart

• Shows the entire universe of complex impedances in one convenient circle.

• Invented at Bell Labs by Philip Smith in 1937.

• By 1975 about 9 million copies of his chart sold to microwave engineers all over the world.

• Its usefulness continues to this day as a method of displaying measured and calculated data produced by computer software and modern measurement instruments.

Philip Smith

6-31

Im( )Γ

Re( )Γ

Smith Chart

http://my.ece.ucsb.edu/sanabria/images/chris_tatoo2.jpg

6-32Constant R Circles

on Chart

Rn=0Rn=0.5

Rn=1

Rn=2 Rn=50

Z Chart

Rn: normalized resistance

nnnL [ ]

mpedance =

resistance + reactance

X Z i

Rj

j Ω= +

6-33

Xn=0

Xn=0.3

Xn=-0.3

Xn=1

Xn=-1

Xn=2

Xn=-2

Xn>0

Xn<0

Constant X Arcs

on Chart

Z Chart

nnnL [ ]

mpedance =

resistance + reactance

X Z i

Rj

j Ω= +

6-34nnnL

nL

[ ]

X R Z

[ ] 0.2 0.3 Z

j

j

Ω= +

Ω= −

Rn=0.2 circle

Xn=-0.3 arc x

6-35nnnL

nL

[ ]

X R Z

[ ] 0.2 0.3 Z

j

j

Ω= +

Ω= −

x

Using your compass draw the 0.2-j0.3 circle with center at zn=1.

(NOTE: this circle is not on the chart until you draw it.

1 Xcircle

j+

6-36

Find the value of reflection coefficient

Reflection Coeff magnitude Reflection Coeff magnitude Reflection Coeff MAGNITUDE

Reflection Coeff PHASE

6-37

S/C O/CPerfect Match

1 Xcircle

j+

=.707 45Z 1 2

L

nL jΓ ∠ °

= +

45

L

R=0 circle( 1)Γ =

L( 0)Γ =

0.707XXX

fig 6.13

Smith Chart

6-38

2

0 0 21Z( ) Z Z1

jkz jkz jkzL L

jkz jkz jkzL L

e e eze e e

− + +

− + +

+ Γ + Γ= =

−Γ −Γ

movement in negative direction clockwise motion on circle of constant toward generator

( )

ˆ LΓ⇔z

GZLZ

Transmission LineGenerator Load

GV

z

z = 0z = -l

minV

ΓL

Im( )Γ

Re( )Γ

To generator

Smith Chart

6-39

ΓL

Im( )Γ

Re( )Γ

360 =2 rad =2

1

complet

Y ( )= -Z

e circle ( )

can just replace by ( )

N

(

ote

)

:

n L Ln

zz

λπ

Γ⇒

°

Γ

V( ) 21V

ΓLz j kze= ++

Smith Chart

6-40

min L

ma

L+

L

2

x (RH Real axi

(LH Real axis)

s)

V =1- Γ

V( )= 1+Γ

V =1+ Γ

eV

jkzz

maxVminV

V( )z

ΓL

R VSWR=

Smith Chart

6-41

Z 1 2 nL j= +

minV maxVΓL

VLoad

nLZ

Smith Chart

=.707 45Γ ∠ °

max L

min LV =1

V =1+ Γ =1

- Γ

+0.707=1

1 .07

.7

0

0

2

7

7 0. 93= − =

VLoad

+

V(z)V

1.707

0.293

( )minz V ( )maxz Vz

45°

6-42

Z 1 2 nL j= +

Smith Chart

=.707 45Γ ∠ °

VLoad

+

V(z)V

1.707

0.293

( )minz V ( )maxz Vz

minV maxVΓL

VLoad

nLZ45°

360 on Smith Chart=2

180 on Smith Chart=4

4545 on Smith Chart=

3601

2 16

λ

λ

λ

°

°

° → =

4λ

16λ

6-43

Z 1 2 nL j= +

minV maxVΓL

VLoad45

nLZ

+

V(z)V

1.707

0.293

VLoad

516λ

16λ z

Smith Chart

=.707 45Γ ∠ °

6-44

VSWR=5.8

X

0.1875λ

Y .2 .4 nL j= −

=.707 45Z 1 2

L

nL jΓ ∠ °

= +

45

Vmax

0.25 0.1875 0.0625λ λ λ− =

Smith Chart

6-45Impedance Matching

2Usually want power to be absorbed by load (minimize ).To do so one adds pure reactances (or susceptances) to tune or match the network.

Γ

L

[ ]

mpedance = resistance + reactance

[S]

admittance = conductance + suscepta

nce

Z R X

Y

i

G B

j

j

j

j

Ω= +

= +

It is physically easier to add a shunt susceptance than series reactaNote:

nce.

6-46

2 2

Z 2 2 =0.62 30

0.

Example: Given

power reflected

cha

62 38%

Z Y =G + B

1 1

nge from

Y 0.2Z 2 2

nL

nL nL nL nL

nLnL

j

j

j

= + Γ ∠ °

Γ = =

= = =+

⇒

⇒

2 2

5 0.25

+ .25

Y 0.25 =0.6 0

0.6 3

add at load

power reflected

rotate tow

6

ard generator

%

t 1+ Bo cir

nL

j

j

j

−

= Γ ∠ °

Γ = =

⇒

could

instead cle and add a tB .- herej

Impedance Matching

6-47Smith Chart

0S

.ol

36utio

20.21

n: Add at 1.57

1.Or - 57 at 9j

jλ

λ+

0.170. 0.2198041 λλ λ=+0.320. 0.3632041 λλ λ=+

Z 2 2 nL j= +

0.178λ

Y 0.25 0. 5 2nL j= −

1 1.57j+

1 1.57j−

0.322λ

0.041λ

6-48Smith Chart

0S

.ol

36utio

20.21

n: Add at 1.57

1.Or - 57 at 9j

jλ

λ+

0.170. 0.2198041 λλ λ=+0.320. 0.3632041 λλ λ=+

Z 2 2 nL j= +

0.178λ

Y 0.25 0. 5 2nL j= −

1 1.57j+

1 1.57j−

0.322λ

0.363λ

0.219λ

0.041λ

6-49

http://www.ocf.berkeley.edu/~joydip/smithchart/index.html

Smith Chart Applet

http://www.ocf.berkeley.edu/%7Ejoydip/smithchart/index.html


ZL

z0.219λ− 0

nLZ

1.62

0.78

0.219λ

0.042λ1.55

( )

( )

max

min

1.55

1.620.38

0.78

0

0.219

loadV V

VV

V λ

=

=

=

=

=

−

Voltages: Unmatched Line


ZL

z0.219λ− 0

nLZ

0.219λ

0.550.38

62

1.45

Currents: Unmatched Line


ZL

z

-jB

0.219λ− 0nLZ

1.62

0.78

0.219λ

1.55

( )

max

min

1.551.620.38

0.780.219

loadVVV

V λ

=

=

=

=−

Voltages: Matched Line

6-53

ZL

z0

ZL

z

-jB

0.219λ− 0

1+ LΓ

1- LΓ

z

ΓL

2.08 2.00

0.71

0.49

1.86

0.042λ−0.219λ−

|V|

|I|

1+ LΓ

1- LΓ

z

ΓL

1.621.55

0.550.38

1.45

0.78

0.042λ−0.219λ−

|V|

|I|

0.38

0.292λ−

fig 6.19a

fig 6.19b

1

MATCHED

UNMATCHED

Standing Wave Pattern

( )

( )

max

min

max

min

1.551.620

0.551.55

.38

0.78

0.38

1.45

0.219

0.219

l

load

oad

II

I

VVV

V

I

λ

λ=

=

=

=

=

=

=

=

−

−

6-54How to Add Shunt Susceptance

For a transmission line usually add a short circuit section of a line placed perpendicular to the main line.

⇒

Yn(l)

l

Yn(l)= -j cotkl

ZLl

d

for kl varyingfrom 0 to π

fig 6.20

6.36

0

2

l λ

≤ ≤

⇓

all possible values- Bj j j∞ ≤ ≤ + ∞

S/C

S/C

6-55How to Add Shunt Susceptance

For a rectangular waveguide usually insert a metal iris inside the waveguide.

⇒

6.3720

-B =

Inductive

cot2

Irisg d

a aλ π

a

d

d

b

6.380

Capacitive I4B = ln csc

s

2

ri

g

b db

πλ

6-56S/C Stub – Smith Chart Solution

Rotate clockwise from to desirS/C ed Bj

S/C

0-j0.5

0-j1

0+j0.5

0+j1

0+j0

Y 0n j= ± ∞

Im( )Γ

Re( )Γ

0+j2

0-j2


Example: To add

B - 5 7 1.j j=

S/C

0 1.57j−

0.09λ

O/C Y cot

1.57 cot1cot 1.57 tan

1.57

2

analytically

;

[0.567

r

0.090

adian ]

3

s

n j klj j kl

kl kl

kl l

l

πλ

λ

= −− = −

= =

= =

=


Example: Given find everything

Smith chart Analytica

=

Z =0.5- 2

0.8246 -50.9

Y 0.11

= 0.82 -51

Y 0.12 0.47

l

76+ 0.47

ly

nL

n n

j

j j

Γ

== +

Γ ∠ °∠ °

1

1

11

1

2

1

10.5

0.131

B 2.9

06

10.46

0.1315

B 2.915

0.053 0.05

0.2

26

27

B

SWR

d

l

SWR

d

l

d

λ

λ

λ

λ

λ

=

=

= −

=

=

=

=

= −

=

2

2

2.9

0.447l λ

= +

=

1+ LΓ

1- LΓ

z

ΓL

1.82

1.65

0.18

0.179λ−0.429λ−

0.4293λ−

1.8246

1.649

0.1754

0.1793λ−

6-59Smith Chart

fig 6.18

Y 0.12 0. 7 4nL j= +1 2.9j+

Z 0.5 2nL j= −

0.071λ

0.202λ

0.298λ

1 2.9j−

1 0.131d λ=2 0.227d λ=

0.82 51Γ = ∠− °

6-60Smith Chart

2.9j

0.197λ

0.303λ

2.9j− 1l

S/C

2l

1

1

(0.303 0.25)0.053

ll

λλ

= −=

2

2

(0.197 0.25)0.447

ll

λλ

= +=

1 2

N0

:.5

otell λ+ =

Chapter 6 Transmission Lines - University of...

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