SmithChartThequarter-wavetransformer

WatcharapanSuwansan8suk

#3EIE/ENE450AppliedCommunica8onsandTransmissionLines

KingMongkut’sUniversityofTechnologyThonburi

Wewillcoverthesetopics•  The Smith Chart •  The Quarter-Wave Transformer

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AOerthislecture,youwillbeableto•  Find the reflec8on coefficient, standing wave ra8o, and theinputimpedanceusingSmithchart

•  Designaquarter-wavetransformer

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2.4THE SMITH CHART

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SmithChartsource:hVp://sss-mag.com/smith.html

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SmithchartisagraphicalaidforsolvingT-lineproblems

•  DevelopedbyP.Smith(1939,Thaicalendar2482)•  Usage:– Past(WWIIera)formicrowave-systemdesign– Present:apartofcomputer-aiddesign(CAD)soOware

•  Smithchartcombines2graphstogether:– normalizedimpedanceand–  reflec8oncoefficient(inpolarcoordinate)

• Wewill consider the simplest case: Smith chart for losslesstransmissionlines

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SMITHCHARTDERIVATION

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ManyT-lineequa8onsareoftheform,orequivalently,

•  forcomplexnumbersandsuchthatand

•  Ex1:Anequa8onforthereflec8oncoefficientisofthisform

•  Ex2:Anequa8onfortheSWRisofthisform

z = 1+�1��� = z�1

z+1

� z z 6= �1 � 6= 1

meansthatwedon’tdividebyzerointheequa8onform

normalizedloadimpedance:

SWR =1 + |�|1� |�|

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(con8nued)

•  Ex3:TheT-lineimpedanceequa8onisofthisform

•  Smithcharthelpsussolvegraphicallythisformofsolu8on

ZL

Zin

`

zin =1 + �e�2j�`

1� �e�2j�`

normalizedinputimpedance:zin = Zin/Z0

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Variablesandarerelatedbycircle-to-circleintersec8on

•  Recallthatinthexy-plan,theequa8onisthecirclecenteredatandofradius

z �

(a, b) r

0x

y

(x� 2)2 + (y + 1)2 = 2.52

(2,�1)

2.5

10

(x� a)2 + (y � b)2 = r

2

(con8nued)

• Wecanrewriteas2circlesintheplane:✓�r �

zr1 + zr

◆2

+ �2i =

✓1

1 + zr

◆2

(�r � 1)2 +

✓�i �

1

zi

◆2

=

✓1

zi

◆2

�r, �i

1

axis

axis0

Intersec8onoftwocircles,shownfor

Given,thisisthecorresponding

� = z�1z+1

and

realpartof

imaginarypartof

realpartof

imaginarypartof� zz�

1

zi

zr1 + zr

zr > 0, zi > 0

�i

�r

z�

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ASmithchartsuperimposesthesecircles

•  InMatalb:commandsmithchartdrawsaSmithchart

0.2

0.5

1.0

2.0

5.0

+j0.2

−j0.2

+j0.5

−j0.5

+j1.0

−j1.0

+j2.0

−j2.0

+j5.0

−j5.0

0.0 '

Anarcofthecirclefortheimaginarypart

Anarcofthecirclefortherealpartzr = 0.2

Theloca8onofontheSmithchart

z = 0.2 + j1.0 zi = 1.0

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USINGTHESMITHCHART

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Example1:findattheload,givenZL

•  Solu8on1(Arithme8c):Fromthedefini8on,

ZL = 40 + j70 ⌦

� =ZL � Z0

ZL + Z0

= 0.59 ej104�

characteris8cimpedance

loadimpedance

Z0 = 100 ⌦

=(40 + j70)� 100

(40 + j70) + 100

= �0.14 + j0.57

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(con8nued)

•  Solu8on2(Smithchart):ThenormalizedloadimpedanceisTheintersec8onof2circles(“0.4”and“j0.7”)gives� = 0.59 ej104

�

0.4

0.7

1.0

+j0.4

−j0.4

+j0.7

−j0.7

+j1.0

−j1.0

0.0 '

0.59

104�

0.4 + j0.7

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TheactualSmithchartcontainsmarksforanglesandascaleforlength

Readoffthedegreefromthering“angleofreflec8oncoefficientindegrees”

UsetheRELCOEFForORIGINscalestoconvertthemeasuredlengthtothemagnitude

0.59

0.59(lengthfrompreviousslide)

RELCOEFF

ORIGIN16

Example1(cont’d):FindtheSWR•  Solu8on1(Arithme8c):

SWR =1 + |�|1� |�|

= 3.87

� = 0.59 ej104�

Subs8tute

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(con8nued)

•  Solu8on2(SmithChart):Rotatethepointontotheaxis,andreadoffthevalueontheaxis– ThevalueistheSWR,soSWR=3.9

+�r

� = 0.59 ej104�

0.4

0.7

1.0

3.9

+j0.4

−j0.4

+j0.7

−j0.7

+j1.0

−j1.0

+j3.9

−j3.9

0.0 '

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1.0

3.9

+j1.0

−j1.0

+j3.9

−j3.9

0.0 '

Thisrota8onyieldstheSWRbecause...

•  Thispointmeansthatforand•  So

z =1 + �

1� �

3.9 =1 + 0.59

1� 0.59

z = 0.39 � = |�| = 0.59

=1 + |�|1� |�|

= SWR

Defini8on19

Example1(cont’d):FindZinforalineoflength.

•  Solu8on1(Arithme8c):

` = 0.3�

Zin =1 + �e�2j�`

1� �e�2j�`Z0

� = 0.59 ej104�

Subs8tute 2�` = 2

✓2⇡

�

◆0.3�

= 1.2⇡ radian = 216�

ZL

Zin

` = 0.3�

= 40 + j70 ⌦

Subs8tute

Z0 = 100 ⌦

=1 + �e�j216�

1� �e�j216�100 = 37� j61 ⌦

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0.4

0.6

1.0

+j0.4

−j0.4

+j0.6

−j0.6

+j1.0

−j1.0

0.0 '

(con8nued)

•  Solu8on2(SmithChart):Rotateclockwisethepointfor– Theendpointisthenormalizedinputimpedance:– Sotheinputimpedanceis

� = 0.59 ej104�

zin = 0.37� j0.61

Zin = zinZ0 = 37� j61 ⌦

2�` = 216�

216�

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0.4

0.6

1.0

+j0.4

−j0.4

+j0.6

−j0.6

+j1.0

−j1.0

0.0 '

Thisrota8onyieldsZinbecause...

•  Thispointmeans:forand•  So

z =1 + �

1� �

Defini8on

z = 0.37� j0.61 � = �e�j216�

0.37� j0.61 =1 + �e�j216�

1� �e�j216�=

1 + �e�2j�`

1� �e�2j�`= zin

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TheactualSmithchartcontainsWTGscaleforanglerota8on

• Wavelengths toward generator (WTG) = moving away fromtheload

•  Aswe rotate clockwise on theWTG scale,we get the inputimpedanceattheposi8onfurtherawayfromtheload

ZL

Zin

WTGdirec8on

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ThisawaytousetheWTGscale1.Theangle104°mapstotheWTGscaleof0.106λ

0.106

2.Movingdowntheline0.3λbringsusto0.106λ+0.3λ=0.406λontheWTGscale

0.406

� = 0.59 ej104�

3.Drawaradicalline

4.ReadtheSmithchart:thispointis,whichisthenormalizedinputimpedance

zin = 0.37� j0.61

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THESLOTTEDLINE

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AsloVedlineisaT-lineconfigura8on(usuallyawaveguideoracoaxialcable)

•  Usefulfordeterminingtheloadimpedance•  is superseded by the modern equipment, a vector networkanalyzer(VNA)

asloVedline aVNA

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Recalltheserela8onships

�

|�| ✓ = angle of � in radiusSWR

SWR =1 + |�|1� |�|

|�| = SWR� 1

SWR+ 1

`min

ZL

ZL = Zin(` = 0)

Zin(`) = Z0ZL + jZ0 tan�`

Z0 + jZL tan�`

Zin(`)

Zin(`) =1 + �e�2j�`

1� �e�2j�`Z0

✓ = ⇡ + 2�`min

`min =

✓✓

4⇡� n

2� 1

4

◆�

foranyintegernthatmakestheright-side≥0

distancefromtheloadtothevoltageminimumreflec8oncoefficientattheload

� =ZL � Z0

ZL + Z0 ZL = Z01 + �

1� �

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TypicalstepstodeterminetheloadimpedanceZLareasfollows1.  Findthemagnitudeandangleof,so

2.  Determinedfrom

|�| ✓ � = |�|ej✓�

determinedfromtheSWR

determinedfrom`min

�ZL

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Example2:findλusingasloVedline

•  Givena50ΩcoaxialsloVedline•  Ashortcircuitisplacedattheload•  On the arbitrarily posi8oned scale on the sloVed line, thevoltageminimaoccurat

z=0.2cm,2.2cm,4.2cm•  Ques8on:Whatisthewavelengthλontheline?

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(con8nued)

•  Answer:thevoltageminimarepeatseveryλ/2,soλ=4cm

2cm

Reasonforλ/2:Ashortcircuitimpliesthat.Hence,

Sinceperiodof=halftheperiodof,thevoltageminimarepeatseveryλ/2

� = �1

sin�z| sin�z| =1

2· 2⇡�

=�

2

|V (z)| = |V +0 (e�j�z + �ej�z)|

= |V +0 | · |e�j�z � ej�z|

= 2|V +0 || sin�z|

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Example2(con8nued):findZL

•  Theshortcircuitisremoved&replacedbyanunknownload•  GiventheSWR=1.5•  Thevoltageminimaarerecordedat

z=0.72cm,2.72cm,4.72cm•  Ques8on:WhatistheloadimpedanceZL?

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(con8nued)•  Answer:

•  ForaloadZLandthecorrespondingreflec8oncoefficient,thevoltageisrepeatedinmagnitudevery:

•  Sowecaneffec8velyconsidertheloadterminalstobeatanyofvoltageminimainthe1ststep(shortcircuit)

43.7 + j19.7 ⌦

�/2����V

✓z +

�

2

◆���� = |� V (z)| = |V (z)|

AOersomealgebra

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(con8nued)

•  Assumewithoutlossofgeneralitythattheloadisat4.2cm•  Then,cm,givingus

4.2cm2.72cm

`min = 4.2� 2.72 = 1.48

✓ = ⇡ + 2�`min

= ⇡ + 2 · 2⇡�

· `min

= 86.4�

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(con8nued)•  FromtheSWR=1.5,wegetthemagnitude

•  So

•  Andtheloadimpedanceis

|�| = 1.5� 1

1.5 + 1= 0.2

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2.5THEQUARTER-WAVETRANSFORMER

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Wewillconsiderthefollowingsetup

•  A feed line of characteris8c impedanceZ0 is connected to alinesec8onoflengthλ/4

•  TheimpedancesZ0andRLareknownrealnumbers•  Ques8on: what is the characteris8c impedance Z1 thatmakes?

•  Answer:Wewillshowthatusing2methods:–  (1)Impedanceviewpointand(2)mul8-reflec8onviewpoint

RLZ0 Z1

� = 0

Z1 =p

Z0RL

�/4

meansthelinesec8onismatchedtothefeedingline(sothereisnostandingwaveonthefeedingline)

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Herearesomeremarksabouttheanswer

•  meansisageometricmeanofand•  Thereisnostandingwaveonthefeedlinealthoughthereisastandingwaveonthematchingsec8on

•  Thevalueisvalidforsec8onlengthandforodd

•  Thematchingsec8onisdesignedforawavelength– At other wavelengths (or other frequencies), the impedancemismatchedwilloccur

Z1 =p

Z0RL Z1 Z0 RL

RLZ0

�/4

Z1 =p

Z0RL

n�/4Z1 =p

Z0RL �/4

n = 1, 3, 5, . . .

�

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IMPEDANCEVIEWPOINT

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Hereisareasonfor.

•  Theinputimpedanceis

•  Toget,wemusthave,whichyieldsthestatedvalueof

RLZ0 Z1

�/4

Zin

�` =2⇡

�

�

4=

⇡

2

Takethelimit(tan=or)

�` ! ⇡2

+1 �1

� = 0 Zin = Z0

Z1

Z1 =p

Z0RL

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Example3:aneffectofthefrequency

•  A quarter-wavelength transformer is designed for a specificfrequency–  sothelengthofthematchingsec8onis–  for=thecorrespondingwavelength

•  Let=thefrequencyontheline•  Ques8on:Plotasafunc8onof

RLZ1

�0/4

= 100 ⌦Z0 = 50 ⌦

f0

�0

�0/4

=

phase velocity vpf0

f

|�| f/f0

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Example3(con8nued):Answer

•  Solu8on:Wewillshowthattheplotisgivenabove.•  No8cethatforthefrequencies|�| = 0

f = f0, 3f0, 5f0, . . .

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Example3(con8nued):Deriva8on•  Thenecessarilycharacteris8cimpedanceis

•  Thereflec8oncoefficienthasamagnitudeof

•  Subs8tu8onwillgiveustheplotof

|�| =����Zin � Z0

Zin + Z0

����

50Ω

100Ω

|�|

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Zin = Z1RL + jZ1 tan�`

Z1 + jRL tan�`

70.71Ω

MULTIPLE-REFLECTIONVIEWPOINT

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Infinitesetofwavestravelforwardandbackwardonthematchingsec8on

RLZ1Z0

T1

�2T2

�1

�3 T3

par8alreflec8oncoefficient

par8altransmissioncoefficient

...

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RL

Thecoefficientsequaltheseexpressions

Z1Z0

T1�1

�1 =Z1 � Z0

Z1 + Z0

(ThinkoftheincidentwaveseeingonlyanimpedanceZ1atthejunc8on)

T1 = 1 + �1

�3

�2T2

=RL � Z1

RL + Z1

�2 =Z0 � Z1

Z0 + Z1T2 = 1 + �2

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Thereflec8oncoeff.isaninfinitesum

T1

�2T2

�1

�3

...

� = �1

�T1�3T2

�3

�2T2

+T1�3�2�3T2

�/4

� · · ·

Theround-trippathupanddowntheλ/4sec8onresultsina180°phaseshiO(signchange)

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Wegetthesameconclusion:ifthen.

•  Thereflec8oncoefficientis

RLZ1Z0

Geometricseries

=2(Z2

1 � Z0RL)

(Z1 + Z0)(RL + Z1)(1 + �2�3)

Subs8tutetheexpressions

Z1 =pZ0RL � = 0

EqualszeroifwechooseZ1 =

pZ0RL

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Summary•  TheSmithChart– Smithchartderiva8on– UsingSmithchart– SloVedline

•  TheQuarter-WaveTransformer–  Impedanceviewpoint– Mul8ple-reflec8onviewpoint

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