Analytic Solution for Shunt Stub

00

00

022

0

00

20

20

002

20

2

2

0

00

;2

;/])[(

/1satisfy tochosen is

])([

))((

)(

)1(

1Let

tan where;)(

)(

is load thefrom lengtha down impedance The

1 impedance Load

ZRZ

X

ZRZR

ZXRZRX

t

ZYGd

tZXRZ

tZXtXZtRB

tZXR

tRG

jBGZ

Y

dttjXRjZ

tjZjXRZZ

dZ

jXRY

Z

LL

LL

LLLL

LL

LLL

LL

L

LL

LL

LLL

L

0for ])(tan[2

1

0for )(tan2

1

stub circuited-shorta for while

0for ])(tan[2

1

0for )(tan2

1

stub circuited-open anfor Then, .satisfy tochosen is

0for )tan(2

1

0for tan2

1

therefore,2

tantan

0

1

0

1

0

1

0

1

1

1

BY

B

BY

Bl

BY

B

BY

Bl

BBl

tt

ttd

ddt

O

O

S

Problem 1Problem 1: Repeat example 5.5 using analytic solution.

Example5.6: Design two single-stub (open circuit) series tuning networks to match this load ZL = 100+j 80 to a 50 line, at a frequency of 2 GHz?

1. The normalized load impedance ZL= 2-j1.6. 2. SWR circle intersects the 1+jx circle at both points z1 = 1.0-j1.33 z2 = 1.0+j1.33. Reading WTG can obtain: d1= 0.328-0.208=0.12 d2= 0.672-0.208=0.463.3. The stub length for tuning z1 to 1 requires l1 = 0.397,and for tuning z1 to 1 needs l2 = 0.103.

Solution

1. ZL = 100+j 80 at 2 GHz can find R= 100,L=6.37nH.

Analytic Solution for Series Stub

00

00

022

0

00

20

20

002

20

2

2

0

00

;2

;/])[(

/1satisfy tochosen is

])([

))((

)(

)1(

1Let

tan where;)(

)(

is load thefrom lengtha down admittance The

1 admittance Load

YGY

B

YGYG

YBGYGB

t

YZRd

tYBGY

tYBtBYtGX

tYBG

tGR

jXRY

Z

dttjBGjY

tjYjBGYY

dY

jBGZ

Y

LL

LL

LLLL

LL

LLL

LL

L

LL

LL

LLL

L

0for ])(tan[2

1

0for )(tan2

1

stub circuited-opena for while

0for ])(tan[2

1

0for )(tan2

1

stub circuited-short anfor Then, .satisfy tochosen is

0for )tan(2

1

0for tan2

1

therefore,2

tantan

0

1

0

1

0

1

0

1

1

1

BY

B

BY

Bl

XZ

X

XZ

Xl

BBl

tt

ttd

ddt

O

S

S

Problem 2Problem 2: Repeat example 5.6 using analytic solution.

Double-Stub Matching

Variable length of length d between load and stub to have adjustable tuning between load and the first stub.

Shunt stubs are easier to implement in practice than series stubs.

In practice, stub spacing is chosen as /8 or 3/8 and far away 0 or /2 to reduce frequency sensitive.

Original circuit

Equivalent circuit

adjustable tuning

Disadvantage is the double-stub tuner cannot match all load impedances. The shaded region forms a forbidden range of load admittances.

Two possible solutions

b1,b2 and b1’,b2’ with the same distance d.

Example5.7: Design a double-stub (open circuit) shunt tuning networks to match this load ZL = 60-j 80 to a 50 line, at a frequency of 2 GHz?

1. The normalized load impedance YL= 0.3+j0.4 (ZL= 1.2-j1.6). 2. Rotating /8 toward the load (WTL) to construct 1+jb circle can find two values of first stub b1 = 1.314 b’1 = -0.114. 3. Rotating /8 toward the generator (WTG) can obtain y2= 1-j3.38 y’2= 1+j1.38.

Solution

4. The susceptance of the second stubs should be b2 = 3.38 b’2 = -1.38.5. The lengyh of the open-circuited stubs are found as l1 = 0.146, l2 = 0.204,or l1 = 0.482, l2 = 0.350.6.ZL = 60-j 80 at 2 GHz can find R= 60, C=0.995pF.

Analytic Solution for Double Stub

1 1

1

1 02 0

0 1

The first stub with admittance

( )

where is the load, is the susceptance of the first stub.

The second stub with admittance

( ); where tan ,

( )

L L

L L L

L L

L L

Y G j B B

Y G jB B

G j B B Y tY Y t d Y

Y j G jB jB t

00

2 0

2 2 20 1

0 2 2 2 20

0

1 (1)

Let Re{ } Re{ } can derive

1 4 ( )[1 1 ] (2)

2 (1 )

Since is real, it gives the range of for matching a given .

1 0

LL

L L

L

Z

Y Y

t t Y B t B tG Y

t Y t

G G d

tG Y

20

2 2sin

Y

t d

2 2 20 0

1

2 2 20 0 0

2

1

0

After has been fixed, both stubs can be found as

(1 ) from (2)

(1 ) from (1)

For an open-circuited stub

1tan ( )

2

while for a shor

L LL

L L L

L

O

d

Y t G Y G tB B

t

Y t G Y G t G YB

G t

l B

Y

1 0

1 2

t-circuited stub

1tan ( )

2where or

Sl Y

BB B B

Problem 3Problem 3: Repeat example 5.7 using analytic solution.

Quarter-Wave transformer

It can only match a real load impedance.

The length l= /4 at design frequency f0.

The important characteristics

]2

1[cos

42 BandwidthFraction

d tolertatemaximunfor )2

(2

)sec2

(11

0

0

2

1

0

2

0

02

01

ZZ

ZZ

f

f

ZZ

ZZ

ZZZ

L

L

m

m

mm

mL

L

m

L

Example5.8: Design a quarter-wave matching transformer to match a 10 load to a 50 line? Determine the percent bandwidth for SWR1.5?Solution

%2929.0

]5010

50102

)2.0(1

2.0[cos

42

]2

1[cos

42

2.015.1

15.1

1

1

36.221050

2

1

0

0

2

1

0

01

ZZ

ZZ

f

f

SWR

SWR

ZZZ

L

L

m

m

m

L

Binomial Multi-section MatchingThe passband response of a binomial matching transformer is

optimum to have as flat as possible near the design frequency, and is known as maximally flat.

The important characteristics

])(2

1[cos

42

42

)(2 bandwidth Fraction

2 ;cos2t coefficien reflection Maximun

)4

( 2

at , !)!(

! , 2 where

;)1()( response Binomial

11

0

0

0

00

0

0

22

Nmm

mmNN

m

Nn

L

LN

Nnn

N

n

jnNn

Nj

Af

ff

f

f

A

lfnnN

NC

ZZ

ZZA

CAeCAeA

Binomial Transformer Design

If ZL<Z0, the results should be reversed with Z1 starting at the end.

00

01 ln2)2(22lnZ

ZCC

ZZ

ZZ

Z

Z LNn

NNn

L

LNn

n

n

Example5.9: Design a three-section binomial transformer to match a 50 load to a 100 line? and calculate the bandwidth for m=0.05?Solution

5.5400.4ln2lnln:2

7.7026.4ln2lnln:1

7.91518.4ln2lnln:0

3!2!1

!3 3

!1!2

!3 1

!0!3

!3

%707.0])0433.0

05.0(

2

1[cos

42

])(2

1[cos

42

0433.0ln2

1 2

3For

10

3223

20

3112

10

3001

32

31

30

311

11

0

01N

0

0

ZZ

ZCZZn

ZZ

ZCZZn

ZZ

ZCZZn

CCC

Af

f

Z

Z

ZZ

ZZA

N

LN

LN

LN

Nm

L

L

LN

Using table design for N=3 and ZL/Z0=2(reverse) can find coefficient as 1.8337, 1.4142, and 1.0907.