Transmission Lines - UPpaginas.fe.up.pt/~mines/EE/EE_lines_2.pdfP1 P2. EE 0708 Lines 7 Admittance Faculdade de Engenharia ... Z Z 2 =Z 0 d 2 =l p 4 0 Z 1 d Z R 1 0 z L d 1 r 1

Faculdade de Engenharia

Transmission Lines

ELECTROMAGNETIC ENGINEERINGMAP – TELE 2007/2008

EE 0708Lines 2

Faculdade de EngenhariaLast week

general transmission line equations

propagation constant and characteristic impedance

finite transmission lines

reflection coefficient

stationary wave à voltage maxima and minima, SWR

transmission lines as circuit elements

the Smith chart

impedance and reflection coefficient determination

admittance

location of voltage and current maxima and minima

( )( )CjGLjR ωωγ ++=

CjGLjR

Zωω

++

=0

0

0

ZZZZ

L

LL +

−=Γ

EE 0708Lines 3

Faculdade de EngenhariaToday

the Smith chart

review

impedance matching

λ/4 transformer

reactive elements

single-stub

double-stub

transients

EE 0708Lines 4

Faculdade de EngenhariaSmith chart

reΓ

imΓ

1

xL constant

rL constant

EE 0708Lines 5


Γθ

reΓ

imΓ

Smith chart

1

LΓ

LZ

•from:

point in chart ( intersection of curves corresponding to rL and xL )

ΓθandLΓ

rL and xL

Lx

Lr

LΓ

•from:

EE 0708Lines 6

Faculdade de EngenhariaInput impedance

1. draw the point corresponding to the normalized load impedance zL à point P1

2. draw the circle centered at the origin with radius OP1

3. draw the straight line from O to P1

4. draw the straight line from O that corresponds to a rotation of l toward the generator

5. intersection of this line with previous circle à point P2

6. obtain , where zin is read from P2

reΓ

imΓ

1

0ZzZ inin ⋅=

P1

P2

EE 0708Lines 7

Faculdade de EngenhariaAdmittance

reΓ

imΓ

1

( ) ( )( )'tan

'tan'

0

00 zjZZ

zjZZZzZ

L

L

ββ

++

= ( )( ) LL

L

ZZ

jZZjZZ

ZzZ20

0

00 2tan

2tan4

' =++

=

=

ππλ

( )LZ

ZZ

Z 0

0

4=

λ( ) Lyz =4λ

º3602 ⇔λ

º1804 ⇔λ

1. draw zL

2. rotate 180º

Ly

Lz

EE 0708Lines 8

Faculdade de EngenhariaMaxima and minima location

reΓ

imΓ

1

voltage maxima where ( ) πnz 2' =Γ∠

voltage minima where ( ) ( )π12' +=Γ∠ nz

voltage maxima

voltage minima

Note:

1. maxima and minima where

impedance is real

2. maxima (minima) points are

separated by nλ/2

EE 0708Lines 9

Faculdade de EngenhariaProblem

EE 0708Lines 10

Faculdade de EngenhariaImpedance matching

lossless line: ( ) ( )

ΓΓ−Γ−

+=

ΓΓ−Γ+

+= −

−−

−−

−

ljLg

zjLzj

g

ljg

ljLg

zjLzj

g

ljg

ee

eZZ

eVzI

ee

eZZ

eVZzV β

ββ

β

β

ββ

β

2

2

02

2

0

0

11

and1

1

( ) ( ) ( ){ }zIzVzPav*Re

21

=

( ) ( )LgLgLg

L

g

gav

lZZ

VZP

θθβ −−ΓΓ−ΓΓ+

Γ−

+=

2cos21

1

2 22

2

20

2

0

±gV

gZ

LZ0Z

0=Γg

given

often Zg=Z0

real, 0ZZ g

EE 0708Lines 11


( ) ( )LgLgLg

L

g

gav

lZZ

VZP

θθβ −−ΓΓ−ΓΓ+

Γ−

+=

2cos21

1

2 22

2

20

2

0

±gV

gZ

LZ0Z

0=Γg

given

often Zg=Z00=Γg

( )2

0

2

181

Lg

av Z

VP Γ−=

power delivered by the source

is maximum when 0=ΓL 0ZZ L =line should be

matched to the load

EE 0708Lines 12


±gV

gZ

LZ0Z

given

0

2

81

Z

VP

gav =

matching section

0ZZin =

matched line

(no reflections)

maximum power delivered by the source

mismatched section

(reflections)

( )2

0

2

181

Lg

av Z

VP Γ−=

maximum power delivered to the load

EE 0708Lines 13

Faculdade de Engenhariaλ /4 transformer

( ) ( )( )'tan

'tan'

0

00 zjZZ

zjZZZzZ

L

L

ββ

++

=( )( ) LL

L

ZZ

jZZjZZ

ZzZ20

0

00 2tan

2tan4

' =++

=

=

ππλ

4λ

LZ

transmission line with length λ/4 can be used to match impedances

EE 0708Lines 14

Faculdade de Engenhariaλ /4 adaptor

( )LZ

ZZ

20/

=

LZ0Z

4λ

/

0ZLZ0Z

mismatched line

00/ ZZZ L=

0Z=

matched line

0ZZ L ≠

Note: ZL must be real if both lines are lossless

EE 0708Lines 15

Faculdade de Engenhariaλ /4 adaptor – complex ZL

( )1

20/

RZ

Z =

4λ

/

0Z LZ0Z

010/ ZRZ =

1d

11 RZ =0Z=matched line

0Z

EE 0708Lines 16

Faculdade de Engenhariaλ /4 adaptor – SWR

0ZZ =

4λ

010/ ZRZ = LZ0Z

1d

11 RZ =

0ZL

LSWRΓ−Γ+

=11

0

0

ZZZZ

L

LL +

−=Γ

0≠ΓL001

01 ≠+

−=Γ

ZR

ZRL

mismatched sections

000

00 =+−

=ΓZZZZ

L

1≠SWR1≠SWR1=SWR

matched section

EE 0708Lines 17

Faculdade de Engenhariaλ /4 adaptor – Smith chart

LZ

0ZZ =4λ

010/ ZRZ =0Z

1d11 RZ =

0Z

reΓ

imΓ

1

0ZZ

z LL =Lz

1d

1r 011 ZrR =

010/ ZRZ =

EE 0708Lines 18


reΓ

imΓ

1reΓ

imΓ

1

λ /4 adaptor – different frequency

LZ

02 ZZ =42 pd λ=

0Z

1d11 RZ =

0Z

Lz1d

1r

design: pff =

mismatched line

pp f

v=λ

dd f

v=λ

010/ ZRZ =

4/2 dd λ≠

1zLz

1d

11 rz ≠

010/ ZRZ ≠

02 ZZ ≠

pd fff ≠=different frequency:

EE 0708Lines 19


matched line

aa jB

jX1

=

Impedance matching with reactive elements

LZ0Z

Note:elements in parallel à use admittances

ajBYY += 12

1d

10

11

jBZ

Y +=

02 ZZ =0

21Z

Y =

1BBa −=

design adaptor à find and aX 1d

such that { }0

11

ReZ

Y =such that { } 0Im 2 =Y

( )( )10

10

01 tan

tan1djZZdjZZ

ZY

L

L

ββ

++

=

0>aXω

= aXLinductor

0<aXaX

Cω

= 1capacitor

EE 0708Lines 20


aa jB

jX1

=

Matching with reactive elements – Smith chart

LZ0Z

ajBYY += 12

1d

10

11

jBZ

Y +=

normalized admittances:

0ZZ

z =0

01YZ

ZZ

zy ===

11 1 jby +=

aa jby =

ajbyy += 12

matching condition: 02 ZZ = 12 =y

12 =y

11 1 jby +=ajbyy += 12

1bba −=

1d such that ( ) 1Re 1 =y

EE 0708Lines 21

Faculdade de EngenhariaMatching with reactive elements – Smith chart

reΓ

imΓ

1

Ly

1bba −=

1d such that ( ) 1Re 1 =y

LZ0Z

11 1 jby +=

aa jby =

ajbyy += 12

1d

1=r

/1P

//1P

/1d

//1d

curve r =1 intersected at two points

two possible solutions

/1P //

1Pand 11 1 jby +=

1jbba −=a

a bZ

X 0−=

EE 0708Lines 22

Faculdade de EngenhariaTransmission lines as reactive elements

l

( ) ( )( )ljZZ

ljZZZlZ

L

L

β+β+

=tantan

0

00

0=LZ

( )ljZ β= tan0

short-circuited line

( ) ( )( )ljZZ

ljZZZlZ

L

L

β+β+

=tantan

0

00 ( )lj

Zβ

=tan

0

0Z

input impedances arepurely imaginary

short-circuited or open-circuited transmission linescan be used as reactive elements in impedance matching

l

∞=LZ

open-circuited line

0Z

STUBs

EE 0708Lines 23

Faculdade de EngenhariaLines as reactive elements – input admittance

l

0=LZ

∞=Ly

short-circuited line

0Z

l

∞=LZ

open-circuited line

0Z

iy

reΓ

imΓ

1

ii jby =

∞=Ly

l

0=Ly

reΓ

imΓ

10=Ly

ii jby =l

iy

EE 0708Lines 24


design:

Impedance matching with a single stub

LZ0Z

d

l

2y

ss jby =

( ) ( )12 ReRe yy =imaginarysy

matched line 12 =y

( ) 1Re 1 =y

syyy += 12stub in parallel 1y

( ) ( )1ImIm yy s −=

1. find d such that ( ) 1Re 1 =y

2. find l such that ( ) ( )1ImIm yy s −=

11 1 jby +=

1bbs −=

or

op

en-c

ircu

ited

EE 0708Lines 25


reΓ

imΓ

1

Impedance matching with single stub – Smith chart

Notes

LZ0Z

d

l

12 =y

1jby s −=

11 1 jby += /d

/1jb

∞=y

1=r

Ly

1. find d such that 11 1 jby +=

2. find l such that 1jby s −=

curve r = 1 intersected at two points à two solutions(only one represented in figure)

/l/1jb−

for open-circuited stub, l is measured from y = 0

design

/1y

EE 0708Lines 26

Faculdade de EngenhariaImpedance matching with single stub – Smith chart

reΓ

imΓ

1

/d

/1jb

Ly

/scl

/1jb−

/1y /

ocl

reΓ

imΓ

1

//d

//1jb−

//1y

Ly

//scl //

1jb

//ocl

solution 1 solution 2

EE 0708Lines 27


1

/d

/1jb

Ly

/l/1jb−

Impedance matching with single stub – summary

LZ0Z

d

l

12 =y

1jbys −=

11 1 jby +=

1. draw yL

11 1 jby +=

3. find l such that 1jby s −=

design

2. find d such that

problem: it is not always possible to place stub in the desired location

EE 0708Lines 28

Faculdade de EngenhariaImpedance matching with double stub

LZ0Z

d

Bl

3y

sAsA jby =

2y

or

op

eb-c

ircu

ited

sBsB jby =

Al

1y

or

op

en-c

ircu

ited

when the location of the stub is predefined the transmission line should be matchedusing a double stub

design of adaptor à obtain lA and lB

for a given d

EE 0708Lines 29

Faculdade de EngenhariaImpedance matching with double stub

LZ0Z

d

Bl

3y

sAsA jby =

2y

or

op

en-c

ircu

ited

or

op

en-c

ircu

ited

sBsB jby =

Al

1ysAL yyy +=1stubs in parallel

sByyy += 23

sy ( ) ( )Lyy ReRe 1 =imaginary

( ) ( )23 ReRe yy =

matched line 13 =y

d“rotating” 2y 1y

design stub A from y1

22 1 jby +=

2bbsB −= design stub B from y2

{ } { }LsAsAL ybbbyb ImIm 11 −=⇔+={ } 11 Re jbyy L +=

EE 0708Lines 30

Faculdade de EngenhariaImpedance matching with double stub – design using Smith chart

How to find y1 and y2 ?

sAL yyy +=1

( ) ( )Lyy ReRe 1 =

d“rotating” 2y 1y

2bbsB −=22 1 jby +=

y1 in circle r = 1, rotated by “d” toward load

y1 in circle r =Re{ yL }

y2 in circle r = 1

LZ0Z

d

Bl

3y

sAsA jby =

2y

sBsB jby =

Al

1yIntersection of these two circles à y1

EE 0708Lines 31

Faculdade de EngenhariaImpedance matching with double stub – Smith chart

r = 1, rotated by “d” toward loadIntersection of r =Re{ yL } and

LZ0Z

d

Bl

3y

sAsA jby =

2y

sBsB jby =

Al

1y

d

Ly

r = 1, rotated by “d”toward load

r =Re{ yL }

/1y

//1y

find y1

Important:

y1 might not exist

EE 0708Lines 32


rotate y1 by “d” toward generator

LZ0Z

d

Bl

3y

sAsA jby =

2y

sBsB jby =

Al

1y/1y

//1y

find y2

//2y

/2y

d

EE 0708Lines 33


length of stubs

LZ0Z

d

Bl

3y

sAsA jby =

2y

sBsB jby =

Al

1y

from y1 and y2

22 1 jby +=

2bbsB −=

{ }LsA ybb Im1 −=

{ } 11 Re jbyy L +=

find lA and lB

EE 0708Lines 34

Faculdade de EngenhariaImpedance matching with double stub – design

2. find y2

LZ0Z

d

Bl

3y

sAsA jby =

2y

sBsB jby =

Al

1y

design

3. find length of stubs

/1y

//1y

//2y

/2y

d

1. find y1

EE 0708Lines 35


/1y

//1y

//2y

/2y

d

Impedance matching with double stub – summary

2. draw circle r=1 rotated by d toward load

LZ0Z

d

Bl

3y

sAsA jby =

2y

sBsB jby =

Al

1y

design

1. draw yL

3. find y1 as intersection of this circle with { }Lyr Re=

4. find y2 rotating y1 by d toward generator

5. find lB from 2bbsB −=

6. find lA from { }LsA ybb Im1 −=

Notes:

• y1, y2, ySA e ySB might have two solutions

•for each stub it might be necessary to consider solutions

in short-curcuit and in open-circuit

EE 0708Lines 36


Ly

d

Impedance matching with double stub – load vs distance d

r = 1 rotated by “d” toward loadintersection of r =Re{ yL } with find y1

this intersection might not be possiblefor given loads and distances d

{ }Lyr Re=

rotated,1=r

Is it possible to match the line under theseconditions using a double stub?

first stub not located next to load

EE 0708Lines 37


LZ0Z

d

Bl

3y

sAsA jby =

2y

sBsB jby =

Al

1yLd4y

Ly

drotated,1=r

Impedance matcthing with double stub – load vs distance d

locating the stub at distance dL from the load

Ld

{ }4Re yr =4y

/1y

//1y

EE 0708Lines 38

Faculdade de EngenhariaImpedance matching – final conclusions

•impedance matching is only exact at the design frequency

•only the main line is matched

•a given method will not always work:

λ/4 linesà sometimes it is not possible to find lines with desired

single stub à the desired location for the stub might not be accessible

doule stub à it might not exist a solution for y1 and it might not be possible to

/0Z

locate the stub at a distance dL from the load

EE 0708Lines 39


EE 0708Lines 40


assume short-circuited stubs

EE 0708Lines 41


EE 0708Lines 42

Faculdade de EngenhariaTransients in lossless lines with resistive loads

lossless lines à

LCv 1=

CL

RZ == 00

±gV

gR

0

LR+

−LV

LI

zl−

0=t

0R

−1V

reflected at the load

+1V

incident

+2V

reflected at source

at a given location of the line, the voltage (current) at a given

instant is equal to the sum of all the voltage (current) waves that have reached that location

need to know the amplitude and location of all waves as function of time

EE 0708Lines 43

Faculdade de EngenhariaAmplitude of voltage waves

±gV

gR

0

LR+

−LV

LI

zl−

0=t

0R

−1V

reflected at load

+1V

incident

+2V

reflected at generator

reflection coefficient at the load

0

0

RRRR

L

LL +

−=Γ

gg

VRR

RV

0

01 +

=+

2nd wave +− Γ= 11 VV L

3rd wave −+ Γ= 12 VV g+ΓΓ= 1VLg

4th wave +− Γ= 22 VV L ( ) +2ΓΓ= 1VLg

5th wave −+ Γ= 23 VV g ( ) +2ΓΓ= 1VLg

reflection coefficient at the generator 0

0

RR

RR

g

gg +

−=Γ

M

1st wave

amplitudes:

EE 0708Lines 44

Faculdade de EngenhariaLocation of voltage waves

±gV

gR

0

LR+

−LV

LI

zl−

0=t

0R

−1V

reflected at load

+1V

incident

+2V

reflected at source

propagation time (end to end)

vl

T =

M

propagationvelocity v

load, at Tt =1st wave

times of arrival:

2nd wave source, at Tt 2=

3rd wave load, at Tt 3=

4th wave source, at Tt 4=

5th wave load, at Tt 5=

EE 0708Lines 45

Faculdade de EngenhariaVoltage reflection diagram

±gV

gR

0

LR+

−LV

LI

zl−

0=t

0R

voltage reflection diagram à graphical representation of the voltage waves

z0l−

t

T2

T

T3

T5

T4

gg

VRR

RV

0

01 +

=+

vl

T =

+1V

+− Γ= 11 VV L

−1V

+ΓΓ= 1VLg−+ Γ= 12 VV g

+2V

+− Γ= 22 VV L ( ) +2ΓΓ= 1VLg

−2V

−+ Γ= 23 VV g ( ) +2ΓΓ= 1VLg

+3V

M

−3V

EE 0708Lines 46

Faculdade de EngenhariaVoltage in the line at a given instant

±gV

gR

0

LR+

−LV

LI

zl−

0=t

0R

z0l−

t

T2

T

T3

T5

T4

+1V

−1V

+2V

−2V

+3V

−3V

0tt =

0zz =

−+ + 11 VV

+−+ ++ 211 VVV

0z z0l−

( )zV

EE 0708Lines 47

Faculdade de EngenhariaVoltage at a point in the line

±gV

gR

0

LR+

−LV

LI

zl−

0=t

0R

z0l−

t

T2

T

T3

T5

T4

+1V

−1V

+2V

−2V

+3V

−3V

1t

0zz =

4t

2t

5t

3t

−+ + 11 VV

+−+ ++ 211 VVV

1t t0

( )0zV

+1V

2t 3t 5t4t

−+−+ +++ 2211 VVVV

+−+−+ ++++ 32211 VVVVV

EE 0708Lines 48

Faculdade de EngenhariaVoltage at the load

±gV

gR

0

LR+

−LV

LI

zl−

0=t

0R

z0l−

t

T2

T

T3

T5

T4

+1V

−1V

+2V

−2V

+3V

−3V

−+ + 11 VV

T t0

LV

T3 T5

−+−+ +++ 2211 VVVV

−+−+−+ +++++ 332211 VVVVVV

EE 0708Lines 49

Faculdade de EngenhariaFinal voltage in the line

( ) L++++++=∞→ −+−+−+332211 VVVVVVtV

z0l−

t

T2

T

T3

T5

T4

+1V

−1V

+2V

−2V

+3V

−3V

±gV

gR

0

LR+

−LV

LI

zl−

0=t

0R

gg

VRR

RV

0

01 +

=+

+− Γ= 11 VV L

−+ Γ= 12 VV g+ΓΓ= 1VLg

+− Γ= 22 VV L ( ) +2ΓΓ= 1VLg

−+ Γ= 23 VV g ( ) +2ΓΓ= 1VLg

M

( ) ( ) ( ) ( )[ ]L+ΓΓ+ΓΓ+ΓΓ+Γ+=∞→ + 321 11 LgLgLgLVtV

EE 0708Lines 50

Faculdade de EngenhariaFinal voltage in the line

z0l−

t

T2

T

T3

T5

T4

+1V

−1V

+2V

−2V

+3V

−3V

±gV

gR

0

LR+

−LV

LI

zl−

0=t

0R( ) ( ) ( ) ( )[ ]L+ΓΓ+ΓΓ+ΓΓ+Γ+=∞→ + 32

1 11 LgLgLgLVtV

1<ΓΓ Lgif

( )Lg

LVtVΓΓ−

Γ+=∞→ +

11

1

( ) ggL

L VRR

RtV

+=∞→

0

0

0

0

0

01

RRRR

RR

RR

VRR

RV

L

LL

g

gg

gg

+−

=Γ

+

−=Γ

+=+

final value does not depend on R0

EE 0708Lines 51


EE 0708Lines 52


Documents

Transmission Lines - UPpaginas.fe.up.pt/~mines/EE/EE_lines_2.pdfP1 P2. EE 0708 Lines 7 Admittance Faculdade de Engenharia ... Z Z 2 =Z 0 d 2 =l p 4 0 Z 1 d Z R 1 0 z L d 1 r 1