Impedancematchingandtuning

WatcharapanSuwansan4suk

#8

EIE/ENE450AppliedCommunica4onsandTransmissionLines

KingMongkut’sUniversityofTechnologyThonburi

Outline

•  Impedancematchingandtuning

•  Matchingwithlumpedelements(Lnetworks)•  Single-stubtuning•  Double-stubtuning(selfreading,Sec5.3)•  Thequarter-wavetransformer(selfreading,Sec5.4)

2

AVerthislecture,youwillbeableto

•  State the func4on of an impedance matching and tuning

circuit

•  Design a matching circuit using the resistor, capacitor, and

inductor(L-matchingnetwork)

•  Designasinglestubmatchingnetwork

•  UseasimplesoVware(Matlab)tohelpdesigningasinglestub

matchingnetwork

•  Computethereflec4oncoefficientofamatchingnetwork

•  Plot over a range of frequencies the magnitude of the

reflec4on coefficient and the return loss of a matching

network

•  Computeafrac4onalbandwidthofamatchingnetwork

3

5.IMPEDANCEMATCHINGANDTUNING

4

Whentheloadismatchedtotheline,

maximumpowerisdeliveredtotheload

•  Reason:Recallthatthepowerdeliveredtotheloadis

whichismaximumwhen,i.e.,when

•  From the power standpoint, it’s desirable for the load to

matchtheline

Load

ZL

Z0characteris4cimpedance

P =1

2

|V +0 |Z0

(1� |�|2)

� = 0 Zin = Z0

Zin

= ZL

5

Impedancematchingisadesignofa

networksothat.

•  Factorstoconsiderinadesign:–  complexity:ifpossible,wewantasimpledesign

– bandwidth: if possible,wewantanetwork that canmatch the

loadoverseveralbandsoffrequencies

Load

ZL

Z0

Zin

Matching

network

designthisbox

Zin = Z0

6

Recallsomecommonprefixes

7

Prefix Valuegiga(G) 10

9

mega(M) 106

kilo(k) 103

mili(m) 10−3

micro(μ) 10−6

nano(n) 10−9

pico(p) 10−12

Whatistheinputimpedance?

8

+

−

50 ⌦ 1 nH 10 pF

Z0 = 50 ⌦T-line

Zin

Zin

v(t) = 2 sin(109t+ 45�) V

(A) 50Ω (B) 50–jΩ

(C) 50+jΩ (D) 50–j99Ω

Whatisthereflec4oncoefficient?

9

+

−

50 ⌦ 1 nH 10 pF

Z0 = 50 ⌦T-line

Zin

v(t) = 2 sin(109t+ 45�) V

(A) 0.5–j0.5 (B) –0.5+j0.5

(C) 1+j (D) –1–j

� IfweuseaVNAtomeasurethescalering

parameter,wewillgetof...

10

S11

(vectornetworkanalyzer)

+

−

50 ⌦ 1 nH 10 pF

Z0 = 50 ⌦T-line

v(t) = 2 sin(109t+ 45�) V

(A) 0.5–j0.5 (B) –0.5+j0.5

(C) 1+j (D) –1–j

1-portnetwork

Whatisthereturnloss?

11

+

−

50 ⌦ 1 nH 10 pF

Z0 = 50 ⌦T-line

Zin

v(t) = 2 sin(109t+ 45�) V

(A) 1dB (B) 2dB

(C) 3dB (D) 4dB

0 5 10 150

0.7

1

Iftheangularfrequencychanges,the

magnitudewillchangeasinwhichgraph?

12

|�|!

(C)

(B)

(A)

|�|

! (Giga rad/s)

Students’version:

Removethe2bullets

angularfrequencyof

thegenerator

Magnitudeofthe

reflec4oncoefficient

Fromthelastslide,atwhatfrequency

doestheloadmatchtotheT-line?

13

(A) 1Gigarad/s (B) 7Gigarad/s

(C) 10Gigarad/s (D) 13Gigarad/sStudents’version:

Removethe2bullets

Theseareexamplesof4types

ofmatchingnetworks

1.  Matching with lumped elements: use capacitors and an

inductorsinthedesign

Matchingnetwork

eachblockisa

capacitororinductor

Matchingnetwork

LoadLoad

14

(con4nued)

2.  Single-stub tuning: connect a single transmission line (a

stub)withthefeedlineatcertaindistance

Load

opencircuit:

commonin

microstrip,

stripline

Load Load

LoadLoad

shortedcircuit:

commonin

coax,

waveguide

Load

shuntstub seriesstub

15

(con4nued)

3.  Double-stubtuning

16

LoadLoad

LoadLoad

LoadLoad

LoadLoad

(con4nued)

4.  Quarter-wavelengthtransformer

17

Z0 Z1

�0/4

ZL

5.1MATCHINGWITHLUMPEDELEMENTS(LNETWORKS)

18

Recalltheseterminologiesincircuit

•  Impedance:

•  Admilance:

resistance

Z = R+ j X

reactance

conductance

susceptance

Y =1

Z= G+ j B

19

AstructureofthematchingL-networkdependsonRLandZ0

LoadimpedanceZ0matching

L-network ZL = RL + jXL

(assumetobe

arealnumber)

RL > Z0

jX

impedance

jB

admilance

ZL

load

impedance

•  Case1:

(2solu4onsfromthe+/-sign)

20

(con4nued)

•  Case2: RL < Z0

jX

impedance

jB admilance ZL

load

impedance

= RL + jXL

X = BZ0RL �XL

(2solu4onsfromthe+/-sign)

21

WhatdoestheL-matchingnetwork

looklike?

22

Loadimpedancematching

L-network ZL = 20� j10 ⌦

Characteris4c

impedanceZ0 = 50 ⌦

jX

impedance

jB

admilance

(A)

jX

impedance

jB admilance

(B)

Student’sversion:Remove

•  the2blueboxesat50,20

•  Reason:(3lines)

Example1:designofanL-network

•  Ques4on:ConsideraseriesofRCload.Atafrequencyof500MHz,whatistheloadimpedance?

•  Answer:ZL=200–j100Ω– ConvertRandC to thephasordomainandgeta seriesof two

impedances

200Ω

3.18pF

1

j2⇡fC

3.18⇥ 10�12500⇥ 106

= �j100 ⌦

200 ⌦

RCloadinthe4medomain RCloadinthephasordomain

23

(con4nued)problemstatement

•  Design the L-sec4onmatching network tomatch a series of

RC load (shown above) to a 100 Ω line at a frequency 500

MHz

200Ω

3.18pF

matching

L-network

Z0=100Ω

At500MHz,theload

impedanceis

ZL=200–j100Ω

24

(con4nued)solu4on

•  Solu4on:Wewillshowthattherearetwosolu4ons,asshown

above

load

load

solu4on#1:

solu4on#2:

2.60 pF

25

(con4nued)solu4on

•  Solu4on:Since,theL-networkisabove•  Subs4tu4ngintotheformulas

wegettwosolu4ons:

Z0=100Ω

ZL = 200� j100 ⌦

RL = 200 > Z0 = 100

jX

impedance

jB

admilance

RL = 200, XL = �100, Z0 = 100

B = 0.0029, X = 122.4745

B = �0.0069, X = �122.4745

(solu4on1,takethe+sign)

(solu4on2,takethe–sign)

26

(con4nued)solu4on#1

•  Let’sturnsolu4on#1intocircuitelements:

•  Considertheimpedance

•  Equa4ng,wegettheinductance

•  Considertheimpedance

•  Equa4ng,wegetthecapacitance

B = 0.0029, X = 122.4745

Theimaginarypart>0,so

thecomponentisaninductor

jX = j2⇡fL

L =X

2⇡f

122.4745

500⇥ 106

= 3.8985⇥ 10�8 H = 38.8 nH

1

jB=

1

j0.00291

jB=

1

j2⇡fC

C =B

2⇡f

jX = j 122.4745

= j�1

0.0029

Theimaginarypart<0,so

thecomponentisacapacitor

0.0029

500⇥ 106

= 9.2277⇥ 10�13 F = 0.92 pF

27

(con4nued)solu4on#2

•  Let’sturnsolu4on#2intocircuitelements:

•  Considertheimpedance

•  Equa4ng,wegetthecapacitance

•  Considertheimpedance

•  Equa4ng,wegettheinductance

Theimaginarypart<0,so

thecomponentisacapacitor

500⇥ 106Theimaginarypart>0,so

thecomponentisaninductor

B = �0.0069, X = �122.4745

jX = j (�122.4745)

jX =1

j2⇡fC

C = � 1

2⇡fX

�122.4745

= 2.5990⇥ 10�12 F = 2.60 pF

1

jB=

1

j(�0.0069)= j

1

0.00691

jB= j2⇡fL

L = � 1

2⇡fB

500⇥ 106�0.0069

= 4.6139⇥ 10�8 H = 46.1 nH

2.60 pF

28

Example2:Reflec4oncoeffvsfrequency

•  Inbothsolu4ons,onlyatthefrequencyof500MHz

– because we use f=500 MHz to convert the impedances and

admilancesintothecapacitancesandinductances

•  Ques4on:Considersolu4on#1– SupposethatloadisaseriesofRandCasshown– Whatisthegraphofversusfrequency?

� = 0

solu4on#1:

200Ω

3.18pF

0.92pF

38.8nH

Z0=100Ω

|�|

29

Example2(solu4on)

•  Solu4on:Takingsteps1—4(nextslides),wegetthegraph

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

frequency (GHz)

|K|

30

Frac0onalbandwidthisara4oofthebandwidthtothedesignfrequency

•  Frac4onalbandwidth

31

|�|

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

frequency (GHz)

|K|

0

1

�m = 0.2

freq(GHz)0.50

f00.39 0.62

fmaxfmin

3.  Computethebandwidth:

BW = fmax

� fmin

BW=0.23GHz

2.  Findthefrequenciesandat

theboundaries

fmin fmax

4.  Letdenotethedesignfrequency

f0

=fmax

� fmin

f0

=0.23 GHz

0.50 GHz= 0.46 = 46%

1.  Let=themaximum

valueofacceptable

reflec4oncoefficient

�m

Forthisexample

Example2(solu4on)

1.  Convertthecircuitelementsintocompleximpedances

1

j(5.78⇥ 10�12)f⌦

j(2.43⇥ 10�7)f ⌦

1

j(2.00⇥ 10�11)f⌦

200 ⌦

j2⇡fLfrom

1

j2⇡fCfrom

32

Example2(solu4on)

2.  Findtheinputimpedance,whichdependsonf

1

j(5.78⇥ 10�12)f⌦

j(2.43⇥ 10�7)f ⌦

1

j(2.00⇥ 10�11)f⌦

200 ⌦

(1)series,sotheequivalent

impedanceisZa = 200 + 1j(2.00⇥10�11)f ⌦

Zb =

(3)series,sotheequivalent

impedanceis

Zin = j(2.43⇥ 10�7)f + Zc

Zin

(2)parallel,sotheequivalent

impedanceisZc =ZaZb

Za + Zb

33

Example2(solu4on)

3.  Obtainthereflec4oncoefficient:

4.  Plotforvariousfrequencies’susingMatlab

� =Zin � Z0

Zin + Z0

|�| f

100Ω(given)expressioninlastpage

34

Matlabcode(Solu4on1)

clear all;!% a series of RC load!

R_load = 200; % (Ohm) resistor at the load!C_load = 3.18 * 10^(-12); % (F) capacitor in the load!

!% circuit elements in the matching network!

L1 = 38.8 * 10^(-9); % (H) inductor at the matching network (solution 1)!C1 = 0.92 * 10^(-12); % (F) capactor at the matching network (solution 1)!

!

Z0 = 100; % (Ohm) characteristic impedance!f = linspace(0, 10^9); % (Hz) range of frequencies to plot!

ZL = R_load + 1./(j*2*pi*f*C_load); % load impedance! !

% Solution 1!ZC1 = 1./(j*2*pi*f*C1); % impedance of C in the matching network!

Zin1 = (j*2*pi*f*L1) + ZC1 .* ZL ./ ( ZC1 + ZL ); % input impedance!Gamma1 = ( Zin1 - Z0 ) ./ ( Zin1 + Z0 ); % reflection coefficient at the matching network!

!

plot( f/10^9, abs(Gamma1), 'Linewidth', 2 );!xlabel('frequency (GHz)');!

ylabel('|\Gamma|');!

35

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Hereisaplotforbothsolu4ons

•  There is not much difference in bandwidth of the two

solu4ons

solu4on2

solu4on1

f (GHz)

|�|

36

DERIVATIONOFEXPRESSIONSOFXANDB

37

Thisisaderiva4onforcase1:.

• Wewantrealnumbersandsuchthat,i.e.,

•  Separa4ngtheequa4onintorealandimaginaryparts,weget

2equa4ons:

jX

impedance

jB

admilance

ZL

load

impedance

Z0

Zin

X B Z0 = Zin

RL > Z0

= RL + jXL

Z0 = jX +1

jB + 1/(RL + jXL)

(realnumber)

38

(con4nued)devia4onofcase1:.

•  Solvingthesetwoequa4ons,wegettheunknowns

•  Thederiva4onofcase2issimilar(seethetextbook)

RL > Z0

Bisarealnumberbecause

RL>Z0,sotheargument

ofthesquarerootis≥0

39

5.2SINGLE-STUBTUNING

40

Ifwejoin2T-linesegmentstogetherasinthe

picture,theinputimpedancewillequal...

41

Zin

load

load

Zin

load

Zseg, in = 50 + j20 ⌦

load

Zseg, in = 50 + j20 ⌦

join

(inputimpedanceofthesegment)

(A) 50Ω (B) 25+j10Ω

(C) 50+j20Ω (D) 100+j40Ω

(con4nued)Reasons

42

load

load

Zin

sameinput

impedanceas

sameinput

impedanceas

Twocomplex-valuedresistorsinparallel:

Ifwenowjointhesegmentslikethis,

theinputimpedancewillequal...

43

Zin

load

load

Zin

(A) 50Ω (B) 25+j10Ω

(C) 50+j20Ω (D) 100+j40Ω

(con4nued)Reasons

44

load

load

Zinsameinput

impedanceas

sameinput

impedanceas

Twocomplex-valuedresistorsinseries:

Thereare2configura4ons

•  ThegoalistochooselanddsothattheinputimpedanceisZ0

1.Shuntstub 2.Seriesstub

characteris4c

admilance,i.e.,

Z0 = 1/Y0

ZL = 1/YL

loadadmilance,

i.e.,

distancetoplacethestub

(tobedesigned)

Z0 = 1/Y0

lengthofthestub

(tobedesigned)

characteris4c

admilance,i.e.,

(assumedtoberealnumber)

Zin Zin

Zin = Z0 = 1/Y0

45

SHUNTSTUB

46

Hereisasolu4onfortheshuntstub

(realnumber)

Opencircuit:

Shortcircuit:

`o

�=

�1

2⇡tan�1

✓B

Y0

◆

`s�

=1

2⇡tan�1

✓Y0

B

◆

t =

(XL±

pRL[(Z0�RL)2+X2

L]/Z0

RL�Z0, RL 6= Z0

�XL2Z0

, RL = Z0

ZL = 1/YL = RL + jXL

realandimaginarypartsofZL:

addtothelengthandiftheyarenega4ve�2

`o

`s

Solu4on

Solu4on

±yields2tuples

(d,l)’sofsolu4ons

47

keyresults

Example3:Singlestubshunttunning

•  Given– aloadimpedanceof

–  characteris4cimpedanceofthelineof50Ω

•  Designtwosingle-stub(shortcircuit)shunttuningnetwork

Load

ZL = 60� j80 ⌦

= 60� j80 ⌦ZLZ0 = 50 ⌦

48

Example3:Solu4on

•  Matlab(nexttwoslides)givesusthesolu4ons:

Solution #1 Distance of the stub: d/lambda = 0.110423 Short circuit: ls/lambda = 0.0949746 Open circuit: lo/lambda = 0.344975 Solution #2 Distance of the stub: d/lambda = 0.259445 Short circuit: ls/lambda = 0.405025 Open circuit: lo/lambda = 0.155025

ZL ZL

Solu4on#2Solu4on#1

49

Example3(Solu4on:Matlab)

% adjustable parameters!clear all;!ZL = 60 - j*80; % input impedance (Ohm)!Z0 = 50; % characteristic impedance (Ohm), must be real number! !% program starts here!RL = real(ZL); % load resistance!XL = imag(ZL); % load reactance!Y0 = 1/Z0; % characteristic admittance of the line! !% obtain t!if ( RL ~= Z0 )! % two solution. Put them in a vector 't'! t = ( XL + (-1).^[0 1] * sqrt( RL*( (Z0-RL)^2 + XL^2 )/Z0 ) ) / ( RL - Z0 );!else! % one solution! t = -XL/(2*Z0);!end! !% obtain B!B = ( RL^2*t - (Z0-XL*t).*(XL + Z0*t) ) ...! ./ ( Z0*(RL^2 + (XL + Z0*t).^2 ) );! !

50

Example3(Solu4on:Matlab,con4nued)

% obtain normalized length, norm_ls = ls/lambda, of the short-circuit stub!norm_ls = atan( Y0./B ) / (2*pi);!norm_ls( norm_ls < 0 ) = norm_ls( norm_ls < 0 ) + 1/2;! !% obtain normalized length, norm_lo = lo/lambda, of the open-circuit stub!norm_lo = -atan( B/Y0 ) / (2*pi);!norm_lo( norm_lo < 0 ) = norm_lo( norm_lo < 0 ) + 1/2;! !% obtain the normalized distance, norm_d = d/lambda, of the stub!% Note that t can be a vector, so we can have multiple solutions of norm_d!norm_d = atan( t ) / (2*pi);!norm_d( t<0 ) = norm_d( t< 0 ) + 1/2;! !% Print out the solution!nsol = length( norm_d ); % number of solution!fprintf(1, '[Single-stub shunt tuner] %d solution(s):', nsol );!for k=1:nsol! fprintf(1, '\nSolution #%d\n', k );! fprintf(1, ' Distance of the stub: d/lambda = %g\n', norm_d(k) );! fprintf(1, ' Short circuit: ls/lambda = %g\n', norm_ls(k) );! fprintf(1, ' Open circuit: lo/lambda = %g\n', norm_lo(k) );!end!

51

Example4:Reflec4oncoeffvsfrequency

•  Inexample3,assumethe load ismatchedat2GHzand isa

seriesofresistor(R=60Ω)andcapacitor(C=0.995pF)•  Thereflec4oncoefficientmagnitudeisshownabove

1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

solu4on2

solu4on1

f (GHz)

|�|

solu4on1hasabeler

bandwidththansolu4on2

(Matlabcodeon

thecoursewebsite)

52

DERIVATIONOFEXPRESSIONSOFAND(SHUNT)

d`

53

Zstub impedancedownalengthlofthestub

impedancedownalengthdfromtheload

Z

Todesigndandl,wesimplifythecircuit

•  Choosedandlsuchthat,i.e.,

load

impedance

d

`

Y0Y0

Y0shortedor

openstub

ZLZstub Z

Z0 = 1/Y0

Z0 = Zstub||Z

1

Z0=

1

Zstub+

1

Z

depends

onddepends

onl

theequivalenceimpedanceof

twoimpedancesand

inparallel

Zstub Z

(assumetobe

arealnumber)

54

TheexpressionsofZandZstub

are...

= RL + jXL

d

impedancedown

alengthdfromtheload

ZLZ0 = 1/Y0

`

shortedor

openstub

impedancedownalengthlofthestub

Zstub = �jZ0 cot�`

Zstub = jZ0 tan�`(open)

(short)Z = Z0ZL + jZ0 tan�d

Z0 + jZL tan�d

55

Wesolvetwoequa4onsfordandl

•  Bysubs4tu4ngexpressionsofimpedancesinto

•  andsepara4ngtheequa4onintorealandimaginaryparts,

•  andsolvefordandl(seethetextbookfordetails)

Zstub ZZ0 = 1/Y0

(assumetobe

arealnumber)

Zstub = �jZ0 cot�`

Zstub = jZ0 tan�`(open)

(short)

Z = Z0ZL + jZ0 tan�d

Z0 + jZL tan�d

1

Z0=

1

Zstub+

1

Z

56

SERIESSTUB

57

t =

(BL±

pGL[(Y0�GL)2+B2

L]/Y0

GL�Y0, GL 6= Y0

� BL2Y0

, GL = Y0

Hereisasolu4onfortheseriesstub

(realnumber)

Opencircuit:

Shortcircuit:

realandimaginarypartsofYL:

addtothelengthandiftheyarenega4ve�2

`o

`s

Solu4on

Solu4on

±yields2tuples

(d,l)’sofsolu4ons

YL = 1/ZL = GL + jBL`o

�=

1

2⇡tan�1

✓Z0

X

◆

`s�

=�1

2⇡tan�1

✓X

Z0

◆

58

keyresults

Example5:Singlestubseriestunning

•  Given– aloadimpedanceof

–  characteris4cimpedanceofthelineof50Ω

•  Designtwosingle-stub(opencircuit)seriestuningnetwork

ZLZ0 = 50 ⌦ = 100 + j80 ⌦

ZL = 100 + j80 ⌦

59

(corrected)

Example5:Solu4on

•  Matlab(nexttwoslides)givesusthesolu4ons:

Solution #1 Distance of the stub: d/lambda = 0.119744 Short circuit: ls/lambda = 0.147631 Open circuit: lo/lambda = 0.397631 Solution #2 Distance of the stub: d/lambda = 0.463373 Short circuit: ls/lambda = 0.352369 Open circuit: lo/lambda = 0.102369

ZL

Solu4on#2Solu4on#1

ZL

.398�

60

Example5(Solu4on:Matlab)

% adjustable parameters!clear all;!ZL = 100 + j*80; % input impedance (Ohm)!Z0 = 50; % characteristic impedance (Ohm), must be real number! !% program starts here %!YL = 1/ZL; % load ad!GL = real(YL); % conductance of the load!BL = imag(YL); % susceptance of the load!Y0 = 1/Z0; % characteristic admittance of the line! !% obtain t!if ( GL ~= Y0 )! % two solution. Put them in a vector 't'! t = ( BL + (-1).^[1 0] * sqrt( GL*( (Y0-GL)^2 + BL^2 )/Y0 ) ) / ( GL - Y0 );!else! % one solution! t = -BL/(2*Y0);!end! !% obtain X!X = ( GL^2*t - (Y0-BL*t).*(BL + Y0*t) ) ...! ./ ( Y0*(GL^2 + (BL + Y0*t).^2 ) );!

61

Example5(Solu4on:Matlab,con4nued)

% obtain normalized length, norm_ls = ls/lambda, of the short-circuit stub!norm_ls = -atan( X/Z0 ) / (2*pi);!norm_ls( norm_ls < 0 ) = norm_ls( norm_ls < 0 ) + 1/2;! !% obtain normalized length, norm_lo = lo/lambda, of the open-circuit stub!norm_lo = atan( Z0./X ) / (2*pi);!norm_lo( norm_lo < 0 ) = norm_lo( norm_lo < 0 ) + 1/2;! !% obtain the normalized distance, norm_d = d/lambda, of the stub!% Note that t can be a vector, so we can have multiple solutions of norm_d!norm_d = atan( t ) / (2*pi);!norm_d( t<0 ) = norm_d( t< 0 ) + 1/2;! !% Print out the solutionsf!nsol = length( norm_d ); % number of solution!fprintf(1, '[Single-stub series tuner] %d solution(s):', nsol );!for k=1:nsol! fprintf(1, '\nSolution #%d\n', k );! fprintf(1, ' Distance of the stub: d/lambda = %g\n', norm_d(k) );! fprintf(1, ' Short circuit: ls/lambda = %g\n', norm_ls(k) );! fprintf(1, ' Open circuit: lo/lambda = %g\n', norm_lo(k) );!end!

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1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Example6:Reflec4oncoeffvsfrequency

•  Inexample5,assumethe load ismatchedat2GHzand isa

seriesofresistor(R=100Ω)andinductor(L=6.37nH)•  Thereflec4oncoefficientmagnitudeisshownabove

solu4on2

solu4on1

f (GHz)

|�|

solu4on1hasabeler

bandwidththansolu4on2

(Matlabcodeon

thecoursewebsite)

63

DERIVATIONOFEXPRESSIONSOFAND(SERIES)

d`

64

Zstub

ZZ0

impedancedownalengthdfromtheload

Z

Todesigndandl,wesimplifythecircuit

•  Choosedandlsuchthat

load

impedance

d

`

shortedor

openstub

ZL

depends

onl

theequivalenceimpedanceof

twoimpedancesand

inseries

Zstub Z

(assumetobe

arealnumber)

Zstub

impedancedownalengthlofthestub

depends

ond

Z0 = Z + Zstub

Z0 Z0

Z0

65

TheexpressionsofZandZstub

are...

= RL + jXL

d

impedancedown

alengthdfromtheload

ZL

`

shortedor

openstub

impedancedownalengthlofthestub

Zstub = �jZ0 cot�`

Zstub = jZ0 tan�`(open)

(short)Z = Z0ZL + jZ0 tan�d

Z0 + jZL tan�d

Z0

66

Zstub

ZZ0

Wesolvetwoequa4onsfordandl

•  Bysubs4tu4ngexpressionsofimpedancesinto

•  andsepara4ngtheequa4onintorealandimaginaryparts,

•  andsolvefordandl(seethetextbookfordetails)

(assumetobe

arealnumber)

Zstub = �jZ0 cot�`

Zstub = jZ0 tan�`(open)

(short)

Z = Z0ZL + jZ0 tan�d

Z0 + jZL tan�d

Z0 = Z + Zstub

67

Summary

•  Impedancematchingandtuning

•  Matchingwithlumpedelements(Lnetworks)– Method

– Example,reflec4oncoefficientvsfrequency

– Deriva4on•  Single-stubtuning– Shuntvsseriestunning– Method

– Example,reflec4oncoefficientvsfrequency

– Deriva4on•  Frac4onalbandwidth

68