Lect7

EE 41139Microwave Technique1

Lecture 7

Power DividerQuadrature (90o) HybridCoupled Line Directional Couplers The 180o Hybrid


Resistive Divider

a three-port power divider can be matched at all ports using lumped resistors

consider the circuit diagram below:

Zo

Zo

Zo

Port 1

Port 2

Port 3V1

+

-

V

V2

V3

Zo/3Zo/3

Zo/3

Zin


Resistive Divider

the input impedance ZI at V is equal to

where Z is the impedance looking into a Zo/3 resistor followed by a 50 transmission linethe factor of 2 is due to two parallel lines of equal impedance

Z Z Zin o / /3 2


Resistive Divider

the input impedance at V1 is therefore given by

Which is matched to the transmission line

Z Z Z Zin o o o / /3 2 3


Resistive Divider

due to symmetry, all 3 ports are matched, i.e.,

input power at Port 1 will be equally divided between Port 2 and Port 3

S S S11 22 33 0


Resistive Divider

if the voltage at Port 1 is equal to V1, the voltage V at the junction is equal to

Zo/3

Zo/3+Zo

2V1 V

+

V V ZZ Z

Voo o

1 12 33 2 3

2 3// /

/


Resistive Divider

from voltage division again, the voltages at Port 2 and Port 3 are

the transmission from Port 1 to 2 is therefore given by

V V V ZZ Z

V Voo o

2 31

334 2

/

S21 1 2 /


Resistive Divider

Due to symmetry,

the scattering matrix is given by

S S S S S S31 23 13 32 12 21 1 2 /

S

0 12

12

12

0 12

12

12

0


Resistive Divider

note that the matrix is reciprocal due to symmetry, it is not a unitary matrix due to the resistive loss

the input power at Port 1 is given by

while the output power at Port 2 and Port 3 are both

, half of the input power is dissipated by the three resistors

P VZin

o

12

12

P VZ

VZin

o o 1

22 1

81

212( / )


The Wilkinson Power Divider

note that the S23 and S32 in the resistive divider are nonzero, i.e., input power from Port 2 can be coupled to Port 3 and vice versa

It can be shown that the Wilkinson power

divider can be matched at all ports with port isolation, i.e., S23 and S32 are both zero



the Wilkinson power divider can be made to give arbitrary power division, however, we will concentrate on the equal power division

ZoZo

Zo

2Zo

Zo2

Zo2

Zo 2Zo

Zo2Zo

Zo

Zo2


The Wilkinson Power Divider it is convenient to normalize the characteristic

impedance to 1 so that the Wilkinson power divider circuit is given by

Z

Port 1

Port 2

Port 3

+V 2

+V 3

+V 1

2

2 1

1

r/2

r/2

Vg2

Vg3



note that the transmission line at Port 1 is replaced by two parallel resistors with a normalized value of 2 each

Z

Port 1

Port 2

Port 3

+V 2

+V 3

+V 1

2

2 1

1

r/2

r/2

Vg2

Vg3



it will be shown that Z is equal to and r=2

to analyze this circuit, it is convenient to employ the even and odd symmetry

the final answer is obtained by combining the results from even- and odd-mode analysis

2


Even Mode Analysis

when Vg2=Vg3, there is no current going through the resistor r/2 as V2 and V3 have the same potential; therefore, these resistors can be removed


Even Mode Analysis

we can simplify the circuit by only consider half of the circiut

Z

Port 1

Port 2

Port 3

+V 2

+V 3

+V 1

2

2 1

1

r/2

r/2

Vg2

Vg3

i2

i3

Z

Port 1

Port 2

+V 2

+V 1

2

1

Vg2


Even Mode Analysis

looking into Port 2, we have

Patch 2 is matched, when and therefore Z = ; here the transmission line acts as a quarter-wave transformer

Z ZZ

Z

Zine

2 24

2 24

2

2tan

tan

Zine 1

2


Even Mode Analysis all the input power at Port 2 will be delivered to

Port 1, i.e., S22 = 0

to find S12, let us consider the transmission line

2 Z V2

V1

x=0


Even Mode Analysis The voltage alone the line is given by

at x = 0, V(x) =V2 and at x=/4, V=V1 the reflection coefficient is given by and

V x V e eoj x j x( ) ( )

22

ZZ

4

90


Even Mode Analysis

substituting Z = , we have

V V V V j j jVo o o2 11 1 ( ), ( ) ( )

2

S V

Vj j je

e

e121

2

11

2 2 2 22 2 2 2 2


Even Mode Analysis

due to symmetry, we also have and

From voltage division,

Se33 0 S je

13 2 /

V V ge1 2 2 /


Odd Mode Analysis

when Vg2=-Vg3, the voltage will change from Vg2 at Port 2 to -Vg2 at Port 3

the voltage must be zero at the point on the plane of symmetry


Odd Mode Analysis

we can simplify the circuit by grounding the circuit at the plane of symmetry

ZPort 1

Port 2

+V 2

+V 1

2

1

r/2 Vg2i2


Odd Mode Analysis

looking into Port 2, we have a short-circuited /4 line in parallel with a r/2 resistor, the input impedance reads

Port 2 is matched when and therefore, r=2; here the transmission line converts a short circuit to an open circuit

Z

Z r

Z Zino

1

1 12

02

' /

, ' /

Z ino 1


Odd Mode Analysis

all the input power at Port 2 will be delivered to the r/2 resistor, and none to Port 1, i.e., = 0, due to symmetry, we also have

From voltage division, the scattering matrix can be obtained from the

even- and odd-mode results

So22

So33 0

V V ge1 2 2 /


Odd Mode Analysis

since Ports 2 and 3 are matched, they are also zero for both even and odd mode S S22 33 0

SV V

V V

j V gV g V g

e o

e o121 1

2 2

2 2 2 02 2 2 2

( / ) /

/ /


The Quadrature (90o) Hybrid

quadrature hybrids are 3 dB directional couplers with a 90o phase difference in the outputs

1 2

34

input

isolated

output

output

Zo

2 Zo

2 Zo



with all the ports matched, power entering Port 1 will be equally divided between Port 2 and Port 3 with a 90o phase difference between the two

no power is coupled to Port 4


The Quadrature (90o) Hybrid the scattering matrix is given by

the scattering matrix can be obtained easily by using even-odd mode analysis

S

jj

jj

12

0 1 00 0 1

1 0 00 1 0



the circuit of the 90o hybrid is given below

the actual response can be obtained by the sum of the even and odd excitations

A1=1

B1

B4

B2

B3

1 1

1

11

1/

1

1/

2

2



At the plan of symmetry,

for even symmetry, the stubs terminate at A and B with an open circuit

for odd symmetry, the stubs terminate at A and B with a short circuit

the length of the stubs are /8

1 1

1

1/

1

2

A B



define the even and odd reflection and transmission coefficients for a two-port network as e,o and Te,o respectively

the scattering parameters are given by B B T T B T T Be o e o e o e o1 2 3 4

12

12

12

12

, , ,

B B T T B T T Be o e o e o e o1 2 3 412

12

12

12

, , ,



the analysis is conveniently presented by cascading ABCD matrices

A B

shunt stub

series stub



the shunt stubs are , the admittance at A is

, tan = 1

for even symmetry, YL = 0, YA = j (normalized) for odd symmetry, YA = -j (normalized)

Y Y Y jYY jYo

L oo L

tantan



for even mode, the ABCD matrix of the open circuit shunt stub is

A BC D Y j

1 0

11 0

1


The Quadrature (90o) Hybrid the ABCD matrix of the series stub is

A BC D

j

j

jj

cos sin

sin cos

/ ,

12

2

0 22 0

24

90

A BC D

j

j

jj

cos sin

sin cos

/ ,

12

2

0 22 0

24

90A BC D

j

j

jj

cos sin

sin cos

/ ,

12

2

0 22 0

24

90

A BC D

j

j

jj

cos sin

sin cos

/ ,

12

2

0 22 0

24

90A BC D

j

j

jj

cos sin

sin cos

/ ,

12

2

0 22 0

24

90

A BC D

j

j

jj

cos sin

sin cos

/ ,

12

2

0 22 0

24

90A BC D

j

j

jj

cos sin

sin cos

/ ,

12

2

0 22 0

24

90A BC D

j

j

jj

cos sin

sin cos

/ ,

12

2

0 22 0

24

90



the ABCD matrix from A to B is given by

A BC D j

jj j

jj

1 0

10 2

2 01 0

112

11

/

A BC D j

jj j

jj

1 0

10 2

2 01 0

112

11

/



ABCD matrix can be converted into scattering parameters

e

e

A B C DA B C D

j jj j

TA B C D j

j

1 11 1

0

2 22 2 2

12

( )

( ) /( )



for the odd mode, the ABCD matrix from A to B is given by

A BC D j

jj j

jj

1 0

10 2

2 01 0

112

11

/

A BC D j

jj j

jj

1 0

10 2

2 01 0

112

11

/


The Quadrature (90o) Hybrid the scattering parameters are given by

Port 1 is matched, half power transmitted to Port 2 with -90o phase shift

Port 4 isolated, half power transmitted to Port 3 with -180o phase shift

B B T T je o e o1 2

12

0 12 2

,

B T T Be o e o3 412

12

12

0 ,



due to the quarter-wave transformer, the bandwidth of the 90o hybrid is limited to 10-20%

this design can be modified for unequal power division


Coupled Line Directional Couplers

when two unshielded transmission lines are close together, power can be coupled between the lines

Wh

S

d

C12

C11 C22



C11 and C22 are the self capacitance in the absence of the other line

C12 is the mutual capacitance between the two lines in the absence of the ground plane



for the even mode, the electric field has even symmetry and the field lines of one transmission line repel those of the other line, therefore, C12 is effectively open-circuited

C C Ce 11 22



the characteristic impedance for the even mode is

for the odd mode, the electric field have an odd symmetry about the symmetry plane and a voltage null exists between the two strip conductors

this is effectively putting a ground plane between the conductors

Z LC vCoe

e e

1


Coupled Line Directional Couplers the effective capacitance between either strip

conductor and ground is

the characteristic impedance for the odd mode is

the transmission lines are assumed TEM lines, this is true for stripline but only approximately true for microstrip line

C C C C Co 11 2 12 22 2 12

Z LC vCoo

o o 1



a single-section coupled line coupler is shown below

1 2

3 4

input through

isolatedcoupled



the input impedance at Port 1 of the coupler is given by

1 2

3 4

I1 I2I3 I4Zoe, Zoo

Zo

V1 V2

V3 V4

V

Z VI

V VI Iin

e oe o

11

1 11 1



the input impedance for the even and odd modes are given by

Z Z Z jZZ jZ

Z Z Z jZZ jZin

eoe

o oeoe o

ino

ooo oooo o

tantan

, tantan

Z Z Z jZZ jZ

Z Z Z jZZ jZin

eoe

o oeoe o

ino

ooo oooo o

tantan

, tantan



by voltage division, we have

V V Z

Z ZV V Z

Z ZI V

Z ZI V

Z Zo

ino

ino

oe

ine

ine

oo

ino

oe

ine

o1 1 1 1

, , ,

V V Z

Z ZV V Z

Z ZI V

Z ZI V

Z Zo

ino

ino

oe

ine

ine

oo

ino

oe

ine

o1 1 1 1

, , ,



the input impedance is given by

Z Z Z Z Z Z Z

Z Z ZZ Z Z Z

Z Z Zin

ino

ine

o ine

ino

o

ine

ine

oo

ino

ine

o

ine

ine

o

( ) ( ) ( )

2

2

2

2

Z Z Z Z Z Z Z

Z Z ZZ Z Z Z

Z Z Zin

ino

ine

o ine

ino

o

ine

ine

oo

ino

ine

o

ine

ine

o

( ) ( ) ( )

2

2

2

2



note that the input impedance should be matched to Zo , we have to choose

and so that this condition is satisfied

Zino

Zine



Let , the even and odd mode characteristic impedances become

Z Z Zo oe oo

Z ZZ j ZZ j Z

Z ZZ j ZZ j Zin

eoe

oo oeoe oo

ino

oooe oooo oe

tantan

,tantan

Z ZZ j ZZ j Z

Z ZZ j ZZ j Zin

eoe

oo oeoe oo

ino

oooe oooo oe

tantan

,tantan



It can be shown that which leads to

Port 1 is matched, due to symmetry, all other ports are matched

Z Z Z Z Zine

ino

oe oo o 2

Z Zin o



the voltage at Port 3 is given by

V V V V V V Z

Z Z

Z

Z Z

V Z jZZ j Z Z

Z jZZ j Z Z

e o e oine

ine

o

ino

ino

o

o oeo oe oo

o ooo oe oo

3 3 3 1 1

3 2 2

tan( ) tan

tan( ) tan

V j Z ZZ j Z Z

oe ooo oe oo

3 2

( ) tan

( ) tan



we can now define a coupling factor C so that

The voltage at 3 is given by

C Z ZZ Z

oe oooe oo

V V jC

C j3 21

tan

tan



At Port 4, we have

At Port 2, we have

V V V V Ve o e o4 4 4 2 2 0

V V V V C

C je o2 2 2

2

21

1

cos sin



when is small, virtually all the power will be delivered to Port 2. none coupled to Port 3, Port 4 is always isolated

for = the coupler is long and and

V V C3 /

V V j C221/



the results satisfy power conservation, there is a 90o phase shift between the two output port voltages which can be used as a quadrature hybrid



if the characteristic impedance and the coupling coefficient are specific, we use the design formulas to obtain the even and odd mode characteristic impedance

Z Z CC

Z Z CCoe o oo o

11

11

,



we have assumed that the even and odd modes have the same propagation velocities which is not valid at higher frequencies for microstrip lines

the coupling of a single-section coupled line coupler is limited in bandwidth due to the quarter-wave length requirement, we can improve the coupler’s performance vs frequency by using multisections



for weak coupling, i.e., C << 1, owe have VV

jC

C j

jCj

jC e j31 21 1

tan

tan

tantan

sin

VV

C

C je j2

1

2

21

1

cos sin



using these results, we can cascade the multi-sections so that

+V jC e V jC e V e jC e V ej j j

Nj j N

3 1 1 2 12

12 1 ( sin ) ( sin ) ( sin ) ( )

V jC e V jC e V e jC e V ej j jN

j j N3 1 1 2 1

21

2 1 ( sin ) ( sin ) ( sin ) ( )



assuming that the coupler is symmetric so that , etc., for an odd number of segments, we have

+ at center frequency, we can define a coupling

factor C VVo

31 2 /

V jV e C N C N NjN3 1 1 22 1 3 1

21

2 sin [ cos( ) cos( ) ( )]

V jV e C N C N NjN3 1 1 22 1 3 1

21

2 sin [ cos( ) cos( ) ( )]


The 180o Hybrid

a 180 hybrid in the form of a ring (or rat-race) hybrid is shown below:

1

2

34

3

Zo

Zo

Zo

Zo

2 Zo

(

(


The 180o Hybrid

a signal applied to Port 1 will be evenly split into two in-phase components at Ports 2 and 3, Port 4 will be isolated

if the input is applied to Port 4, it will split into two signals with 180o phase difference at Ports 2 and 3, Port 1 will be isolated


The 180o Hybrid

when input signals applied at Ports 2 and 3, the sum of the inputs will be formed at Port 1 while the difference will be formed at Port 4

Ports 1 and 4 are referred to as the sum and difference ports, respectively


The 180o Hybrid

the scattering matrix for the ideal 3dB 180o

hybrid has the following form:

S j

2

0 1 1 01 0 0 11 0 0 10 1 1 0


The 180o Hybrid

the hybrid can be analyzed using the even- and odd-mode analysis, the amplitude of the scattered waves from the ring hybrid will be

B B T T B B T Te o e o e o e o1 2 3 412

12

12

12

, , ,


The 180o Hybrid

for equivalent circuits for the even and odd mode with excitation at Port 1 are given below:

1 2

2

2

2

I/2

T 1 2

2

2

2

I/2

T

e

e

o

o

open circuit short circuit


The 180o Hybrid

the even and odd mode ABCD matrices are given by


The 180o Hybrid

the positive sign is for the even mode while the negative sign is for the odd mode

we can convert the ABCD matrices back to the scattering matrices

e

e

A B C DA B C D

j jj j

j

TA B C D j j

j

1 2 2 11 2 2 1 2

2 21 2 2 1 2

( )


The 180o Hybrid

when the input is at Port 1, the outputs at Port 2 and 3 are in phase, Port 4 is isolated, Port 1 is matched

o

o

A B C DA B C D

j jj j

j

TA B C D j j

j

1 2 2 11 2 2 1 2

2 21 2 2 1 2

B B j B j B1 2 3 402 2

0

, , ,


The 180o Hybrid

when the input is at Port 4, we have the following equivalent circuits:

1 2

2

2

2

I/2 1 2

2

2

2

I/2

eT e

o

T o

open circuit short circuit


The 180o Hybrid

the even and odd mode ABCD matrices are given by

B T T B B T T Be o e o e o e o1 2 3 412

12

12

12

, , ,

A BC D j

jj j

jje o

, / // / /1 0

2 10 2

2 01 0

2 11 22 13 8 8

A BC D j

jj j

jje o

, / // / /1 0

2 10 2

2 01 0

2 11 22 13 8 8

A BC D j

jj j

jje o

, / // / /1 0

2 10 2

2 01 0

2 11 22 13 8 8

j

A BC D j

jj j

jje o

, / // / /1 0

2 10 2

2 01 0

2 11 22 13 8 8

A BC D j

jj j

jje o

, / // / /1 0

2 10 2

2 01 0

2 11 22 13 8 8

A BC D j

jj j

jje o

, / // / /1 0

2 10 2

2 01 0

2 11 22 13 8 8


The 180o Hybrid

the positive sign is for the even mode while the negative sign is for the odd mode

we can convert the ABCD matrices back to the scattering matrices

e

e

A B C DA B C D

j jj j

j

TA B C D j j

j

1 2 2 11 2 2 1 2

2 21 2 2 1 2


The 180o Hybrid

o

o

A B C DA B C D

j jj j

j

TA B C D j j

j

1 2 2 11 2 2 1 2

2 21 2 2 1 2

( )

B B j B j B1 2 3 402 2

0

, , ,


The 180o Hybrid

when the input is at Port 4, the outputs at Port 2 and 3 are different in phase by 180o , Port 4 is matched and Port 1 is isolated

the remaining elements in the scattering matrix can be found by enforcing symmetry

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