1.1 Transmission line basic concepts: - UPCommons

1

Design and Analysis of RF and Microwave SystemsEuropean Master of Researchon Information TechnologyEuropean Master of Researchon Information Technology

1.1 Transmission line basic concepts:Introduction to narrow-band matching networks

March 2010

Francesc Torres, Lluís Pradell, Jorge Miranda


For any lossless transmission line:

)()()( zVzVzV

1

Voltage and current in the transmission line

V(z I(z) ZL≠ Z0 Z0

VL IL

V+(z)V-(z)

zjeVzV 0)(

zjeVzV 0)(

)()(1

)(0

zVzVZ

zI

where

LVVVV 00)0(

LIVVZ

I

0

1)0(

At z=0

nnz

nznzzz

2

2,2)(Periodicity of V(z) and I(z): wavelength

I(z)

z=0 -z

V (z)

Reflection coefficient: load

0

0

ZZ

ZZ

V

V

L

LL

Impedance: load

L

L

L

L

L

LL Z

VV

VVZ

VV

VVZ

I

VZ

1

1000

2


At any point z of the transmission line the impedance is computed as:

Reflection coefficient in the transmission line

)(1

)(1

)(

)()( 0

zZ

zVzZ

0

0 )(

)()(

ZzZZz

Modulus

)(1)()( 0 zzI 0

0 )()(

ZzZ

zjL

zjzj eeeV

V

zV

zVz 222 )0(

)(

)()(

At any point z of the transmission line the reflection coefficient is:

Lzj

Lez 2)( Constant in z

LinearIncreasing with +z (towards load)Periodicity: half a wavelength

Phase radz2 degº180

2

z

22

22,2)(22

nnn

z

nznzzz


The transmission line as impedance transformer

zjz L 2)( jL 2)(

ZL≠ Z0 Z0, β

z=0

|Γ(z) |=|ΓL|

z

jL )(

)( i

Example: Compute the input impedance Zi of a circuit formed by a transmission line of length is λ/8 and loaded with ZL=0 (sc).

10

0

ZZ

ZZ

L

LL

jeeejj

jLi

28

22

2)(

00 1

1jZZZ

i

ii

3


Standing waves in the transmission line (i)

zjjL

zjjL

zj

zjjL

zjjL

zj

eeZ

Veee

Z

VzI

eeVeeeVzV

LL

LL

22

22

11)(

11)(

LL ZZ 00

)(

At any point z where the term (1+ΓL) is real, voltage and current are real an their magnitude is either maximum or minimum:

radzL 22 max

L

L

Z

VzII

VzVV

1)(

1)(

0minmin

maxmax

zz Z0

L

L

Z

VzII

VzVV

1)(

1)(

0maxmax

minmin radzL min2

4minmax zz


Standing waves in the transmission line (ii)At any point z where the voltage is maximum the impedance is real an maximum, if the voltage is minimum the impedance is real and minimum:

11)( eVzV zj

max0

0

max

maxmax 1

1

1

1

)(

)()( ZZ

eZV

eV

zI

zVzZ

L

L

Lzj

L

min0

0

min

minmin 1

1

1

1

)(

)()( ZZ

eZV

eV

zI

zVzZ

L

L

Lzj

Lzj

)(

)(

minmaxmin

maxminmax

zIV

zIV

The voltage standing wave ratio (VSWR or S) is defined as

1

1,

1

1

min

max

S

S

V

VS L

L

L

SL

1

10

Matched load: Z=Z0, Γ=0, SWR=1

4


Standing waves in the transmission line (iii)Maximum and minimum impedances at in the transmission line:

)(zVλ/4 λ/4

Vmax

Vmin

z=0 zmin zmax zmin

VSWR is easy to measure and it is widely used to specify mismatch

SZZI

VZ

L

L00

min

maxmax 1

1

S

Z

ZI

VZ

L

L 0

0max

minmin 1

11


At any point z of the transmission line the net power is computed as:Power in the transmission line (i)

**

0

* )(11

)(1)()·()( zeVZ

zeVezIzVezP zjzj

222

0

)(1)(1 zPzZ

V

o

Since the modulus of the reflection coefficient is constant in z, the net transmitted power is constant at any z and equals the power delivered to the load:

LL PPPPzPzP 221)(1)(

jxraaaae ,1112*

LL PPPPzPzP 1)(1)(

PPZ

VP

Z

VP L

2

0

2

0

2

,,

Where the power associated to the “positive” (incident) and “negative” (reflected) waves is:

5


Example: P+ =1 Watt in transmission line of impedance Z0=50Ω

Power in the transmission line (ii)

WPPj

jjZ LLL 5.0

2

1,

2

1,

1,10050

2

)()()(),(·log20, 10

2dBRLdBWPdBWPRL

P

PRL LL

j 221

Return loss definition:

In this case

dddd dBWdBdBWPdBRL 330,3

8.51

1

L

LS

The standing wave ratio (SWR) in the transmission line


If we have a mismatched load

Matching networks

PPP LL2

1

ZL≠ Z0Z0

P+

• A fraction of the incident power P+

is not delivered to the load• A fraction of the incident power

returns to the generator

A matching network must be•Simple (passive)

L≠ 0 Z0

ΓL≠ 0

ZL≠ Z0

Z0 Matching Network

P-≠0

P+

P-=0

2

LPP

Simple (passive)•Lossless (L, C, Transformer, transmission line, waveguide,…)

PPP iLIf lossless: All power is delivered to the load

Zi= Z0 Γi= 0

P 0

6


An impedance can be normalized to a reference impedance Z0

Normalized impedances and admittances

LLL

LL YY

YZ

Z

ZZ

1; 0

0

ZP+

ZL≠ 1

In this case, the reflection coefficient:

LL YYZ0

L

L

L

L

L

LL Y

Y

Z

Z

ZZ

ZZ

1

1

1

1

0

0

Z0

ΓL≠ 0

ZL≠ 1

Z0=1 Matching Network

P-≠0

P+

P-=0

ZL≠ 1

Working with normalized impedances is equivalent to work with reference transmission lines of Z0=1

LLL 0

Zi= 1 Γi= 0


Example: lossless narrowband matching network (i)

Z 4 j2Z 1Z 1P+

Zx =jX

YB =jB

In this case, the reflection coefficient:

jXZ 11

ZL=4-j2 Z0=1

Z1

Z0=1

Zi= 1

P-=0 YB =jB

YL=0.2+j0.1

)1.0(2.0

1111 BjjBY

jXZL

By equalling the real parts By equalling the imaginary parts

5.0

3.0

)1.0(2.0

2.01

2

122 B

B

B

3

2

)1.0(2.0

)1.0(

2

122 X

X

B

BX

7


Example: lossless narrowband matching network (ii)Solution 1

Zx =-j2 Zx =j2

Solution 2

Sometimes a shunt-series solution does not exist and a series-shunt network must be used:

ZL=4-j2 Z0=1

Zi= 1

YB =-j0.5 ZL=4-j2 Z0=1

Zi= 1

YB =j0.3

Z0=1

Zi= 1

ZL Z0=1

Zi= 1


The Smith ChartThe Smith ChartZ(z)→ Impedance at any point z of the transmission line

Re 0,

Im ,

Z z

Z z

P.H. Smith, in 1939, developed a chart to represent any impedance Z(z) as a function of its related reflection coefficient ρ(z). This graphic tool is based in the fact that |ρ|≤1 which allows to represent all impedances in a finite area. The Smith chart is currently used as a universal tool to represent impedances.

P.H. SMITH

8


Relation Relation Z(z) -- ρ(z)

An impedance can be normalized in relation to a reference impedance Zo as

1ZZ r j x

1 jZe j

Mathematically, this correspond to a bilinear transformation which translates a circle in the impedance domain into a circle in the reflection coefficient domain.

0 1Z r j x

Z

1 r ie jZ

ixr≥0

r

≤1

rZ


Relation Relation Z(z) -- ρ(z)Now, if ρ=ρr+jρi is substituted in the expression of the normalized impedance, the equations that relate the loci r and x constants as a function of the components ρr and ρi are obtained. This is a

set of circumferences in the complex domain :

2 2 2

222

1 1 1; 1

1 1r i r i

r

r r x x

11,

1Constant reactance circle: CENTRE RADIUS

,01

r

r

1

1r Constant resistance circle: CENTRE RADIUS

r

x r=const

r

i

r

x

x=const.

r

ix

x

9


The reflection coefficient in the complex domainThe reflection coefficient in the complex domain

i90ºz=-ℓ z=0

ZLℓ= 360º ℓ= 180º ℓ= 90º

r

|=1

0º180º

LeZe

L

L|

ℓℓ

L

4( )Lj z

Towardsload

Rationale

ZL L Zee

L e

270º4

( ( ))2( )Ljj

e L Lz e e

0 1L|

e

L|( )2( )Lj zj z

L Lz e e

( 0) LjL Lz e

Towardsgenerator


x = ∞

x = 0.5

x = 2

x = 1

Chart of impedances

r = 0 r = 0.5 r = 1 r = 2 r = ∞

x = 0

x = - 2

x = - 1

Circles of constant reactance

Circles of constant resistance

x = - 0.5

10


At a point placed in thetransmission line of Z0 = 50Ω we measure animpedance 100+j·150 ΩWhat is the value of ρ atthis point? | ρ | = 0.75

x = +3

p

100 1502 3

50

jZ j

0.75 26º

| ρ | 0.75

r = 2

φ = 26º


If , what is itsrelated normalizedimpedance ? How doesthis impedance changeif the point is moved alongthe transmission line?

1 3 90º

ZZ φ = 90º

r = 0.8

x = +0.620.33j

e

0.8 0.6Z j

As we move along atransmission line, themodulus of the reflection

| ρ | = 0.33

modulus of the reflectioncoefficient is constant:

The normalized impedance varies as given by this circle.

0.33

Z

11


transformation: transformation: ρρ ↔ Z↔ Z

SWR = S (standing wave ratio)

RET’N LOSS dB =x| ρ |

Z

26º

¯

2 3Z j

20 log RET N LOSS, dB =

REFL. COEFF. P =

REFL.COEFF, E OR I =

φ

0.75 26º

20 log 2

0.75

7S

2.6 dBretL

72.6


Input impedanceInput impedance

0.45·λ

VZL= 60 – j·90

S 3 6

0.8 1.2LZ j

2 1.6Z j ZL

Ze

mín

máx

mín 0.28

V

I

Z

máx

mín

máx 3.6

V

I

Z x

x

ZLZe

l = 0.45·λ

Z0 = 75 Ω

S=3.61/S=0.28

2 1.6eZ j

150 120eZ j

270máxmáxZ R

21mínmínZ R

S=3.6

12


AdmittanceAdmittance

Z R j X

Y G j B

1

ZLx

1

1

cL

cCX

BL

X

C

L

B C

1 1 1

1 1 1

Z Y Y

Z Y Y

50Z x

Rotation: 180º

0 500.2 0.5

10 25 LL

ZZ j

Z j

00

0.7 1.7

0.014 0.034

LL L L

L

L

YY Y Y Y

Y j

Y j

Z

YL


Input admitanceInput admitance

ZL= 10 + j·15ZL

x

Zex

0.1·λ

Ye Z0 = 50 Ω

l = 0.1·λ

ZL

0 2 0 3Z j YL

x

x

0

10 15

50LZ j

Z

0.2 0.3LZ j

0.3 0.7eY j

0.006 0.0145eY j

0

10.02

50Y

Ye

x

0.1·λ

13


ShortShort--circuited linecircuited line

Ze, Ye s.c.

0.1·λ

Open circuit lineOpen circuit line

0.73

1.4e

e

Z j

Y j

l = 0.1·λ

Short circuit Open circuit xx

Ze, Ye o.c

l = 0.15·λ

0.73

1.4e

e

Z j

Y j

0.15·λ


Change of reference Change of reference impedanceimpedance

11’2’ 2

Z1x

0.1·λ

0.15·λ

Ze 50 Ω

0.15·λ

ZL100 Ω 150 Ω

0.1·λ 0.15·λ Z1’

x

Change of referenceimpedance

1

1 1

750.5

1501 0.7

150 150 105

LZ

Z j

Z Z j

1'

150 1051.5

1001 8 0 9

jZ j

Z j

ZL

x

Ze

Z2

xx

Z2’

x

2

2 2

1.8 0.9

100 180 90

Z j

Z Z j

2'

3

3

180 903.6 1.8

500.28 0.52

50 14 26e

j

j

Z j

Z

j

Z

Z

0.15·λ

14


Z

MATCHING. Transmission MATCHING. Transmission line plus reactance in line plus reactance in

seriesseries

l

ZL

j·XSZL

xx

0.114·λ

MATCHING. Transmission line plus reactance inseries

Z’

¿ l ?

1 2.1Z j

0.2 0.5LZ j

Ze=1 Z=1+j·XXS = -X

2.1Sj X j 1

Ze=1

x

0

12.1 Z

C

' 1 2.1Z j ' 2.1Sj X j

02.1L Z

0.234·λ


MATCHING. MATCHING. Transmission line plus Transmission line plus reactance in parallelreactance in parallel

YL

¿ l ?

ZLj·BPY

0.289·λ

ZLx

x

MATCHING. Transmission line plus reactance inparallel

1 2.4Y j

0.4 1.4LZ j

2.4Pj B j

¿ l ?

Y=1+j·BBP = -B

Ye=1

0.2 0.65LY j

Y’

Ye=1

x

0

12.4 Y

L

' 1 2.4Y j ' 2.4Pj B j

02.4C Y

0.402·λYL

x

15


Example A)

In a Smith chart referencedto an impedance Z0 = 50Ω, represent thefollowing sets ofimpedance loci:

a) The impedances thatcause a standing waveratio SWR=2

)

x

ROE=2

R>100

b) Impedances with realpart larger than 100

1002

50r


c) Set of impedances thatdisipate 20 mW when avoltage of 2 V rms isapplied

+

xx

LY

ZL

+

-

VL=2 V

PL=20 mW

2* * *

3

2

Re Re

20 100.005

4

L L L L L L L L

LL

L

P V I V V Y V G

PG

V

x

gL=0.25

x

rL=4

xx

ZL

d) Impedances that have thesame real and imaginaryparts

00

0.25LL L

Gg G Z

Y

r=x

16


Example B) A generator at f=300 MHz feeds an unknown load ZL by menas of atransmission line of impedance Z0=70 . In the line it is measured |Vmax|=5.2 V and |Vmin|=1.1V. Furthermore when the load is substituted by a short circuit the positions of all minimumvoltages move 15 cm towards the load. ¿What is the impedance ZL?

maxmin

14.72, 0.212, 300 1pvV

ROE Z f MHz mV ROE f

SS

ZL

0

ZLZLZL

2

3

2

Z iZ i

VminVmin

z=0z=-ℓ z(- 0.5- 1-1.5

|V|min

|V|max

0

|V(z)|0.15

0.15

minV ROE fS

22 ZminZmin

c.c.

z=0z=-ℓ

c.cc.cc.c

Vmin Vmin

0

- 0.5- 1-1.5 0

|V(z)|

z(

|V|max


Example B)

min 0.212

0.15

Z

z

0.52 1.2LZ j

minZ

x

Z 1 2j

0.15z

0.52

0.15

LZ 1.2j

0· 36.4 84L LZ Z Z j

17


Example B) The same result is obtained if the position Vmin moves 35 cmtowards the generatormax

min

14.72, 0.212, 300 1pvV

ROE Z f MHz mV ROE f

SS

ZLZLZLZL

2

3

2

Z iZ i

VminVmin

z=0z=-ℓ 0.15 0.35

0.35 0.15

- 0.5- 1-1.50

|V|min

|V|max

|V(z)|

z(

minV ROE fS

22 ZminZmin

c.c.

z=0z=-ℓ

c.cc.cc.c

Vmin Vmin

0

- 0.5- 1-1.5 0

|V(z)|

z(

|V|max


Example B

min 0.212

0.35

Z

z

0.52 1.2LZ j

minZ

x

Z 1 2j

0.35z

0.52

0.15

LZ 1.2j

0· 36.4 84L LZ Z Z j

0.35

18


Example C

An antenna has and input impedance of 75 at 400 MHz. It is fed by means of a parallel wirestransmission line with an impedance Z0=150 .

a) Design a matching network compounded of transmission line plus a shunt capacitor. The dielectric) g g p p pconstant of the transmission line is εr=2.2.

YL

ℓ

ZLj·Bc

Y=1+j·BB = BY =1 B = -Bc

Bc>0Ye=1

¿ ℓ, C ?


YL

ℓ

ZLj·Bc

Y=1+j·BB BY 1

x

1 0.7Y j

75 10.5 2

150L LL

Z YZ

B = -Bc

Bc>0Ye=1

x

xLY

1 0.7Y j

Solution with capacitor1 0.7Y j

0.25 1eY

ℓ=0.088

8

6

/ 3 10 / 2.20.506

400 10rc

mf

0.348

ℓ=0.088

0.7CB ℓ=4.45 cm

00.7 /1.85

2

ZC pF

f

19


Compute the length of atransmission line, ended inshort-circuit, that cansubstitute the capactior.

x

0.7cY j

YL

ZLYc=j·0.7

x

CC

ℓ

Y=1-j·0.7Ye=1 ℓ=0.347

ℓ=0.347 =cm


b) Matching network formed by a quarter wavelength transformer .

ZLZ0’

ℓ=

ZL

Ze= Z0

0

20' 0'e

e L e

Z Z ZZ Y Z

Z Z Z

In the case of a quarter wavelength transformer

0'e e

L LZ Z Z

0'

0 150e L

e

Z Z Z

Z Z

0' 106.07

12.654

Z

cm

20


Matching networks utilities and examples

Lecturer: Francesc Torres ([email protected])


There are a number of on-line tools for RF design and/or educational purpose

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EXERCISES: www.amanogawa.com:

1) Microstrip impedance computation: GlassSubstrate thickness: 1.59 mm Dielectric permittivity: 4.15Strip thickness: 0.1 mm Copper conductivity: 5.8 107 S/mSubstrate conductivity: 2.3 10-4 S/m

a) Calculate the substrate width W (mm) in order to have Zo=50 Ω at f=2.5 GHz

b) In the previous case, compute the return loss (RL) referred to Zo=50 Ω at f=1 GHz and f=5 GHz: RL degradation due to frequency dependence Zo(f)

c) Compute the return loss (RL) referred to Zo=50 Ω at f=2.5 Ghz if the strip thickness is neglected (t=0)

2) Narrowband matching networks

Select the narrow band matching structure that provides the best bandwidth Select the narrow band matching structure that provides the best bandwidth (VSWR<1.5) to adapt an impedance ZL=100+j120 Ω, referred to Zo=50 Ω at f=10 GHz, εr=2.4.

• Quarter wavelength adapter• Double stub adapter• Single stub (short/open) adapter

Gives line length in mm.

28


Layout and picture of a microstrip two-stage amplifier

GNDOuput matching network

Interstage matching network: coupled lines (DC block)

INTRT1

TRT2

IN OUT

o.c.

s.c.

λ/4

λ/4

λ/4

GNDBias resistor

Bias network

Input matching network

Gate bias voltage Drain bias voltage

Low frequency s.c.


Other utilities: http://www.hp.woodshot.com/•Simple tool for transmission line calculations, bias circuits, smith chart,..

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Exercises1) Derive the expression of the input impedance of a transmission line of

impedance Zo, length λ/4 and loaded with an impedance ZL.2) Demonstrate that |Γ|≤1 for any load Z=R+jX if R≥03) What is the return loss of a load ZL=75Ω when connected to a L

transmission line of Z0=50Ω ? What fraction (%) of the incident power is delivered to the load?

4) What is the tolerance (±x Ω) of a resistor of nominal impedance R=50Ω when connected to a transmission line of impedance Z0=50Ω if VSWR ≤1.1?

1.1 Transmission line basic concepts: - UPCommons

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