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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010

Power Dividers and Couplers

A. Nassiri -ANL

Lecture 8



Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 2

Basic properties of dividers and couplersthree-port network (T-junction), four-port network (directional coupler),

directivity measurement

The T-junction power divider Lossless divider, lossy divider

The Wilkinson power divider Even-odd mode analysis, unequal power division divider,N-way Wilkinson divider

The quadrature (90°) hybrid branch-line coupler

Coupled line directional couplersEven- and odd-mode Z 0, single-section and multisection coupled line

couplers

The Lange coupler The 180° hybrid

- rat-race hybrid, tapered coupled line hybridOther couplers

reflectometer

Power dividers and directional couplers




• N-port network

Basic properties of dividers and couplers




Three-port network (T-junction)

Discussion1. Three-port network cannot be lossless, reciprocal and matched at al l ports .2. Lossless and matched three-port network is nonreciprocal

circulator




3 . Matched and reciprocal three-port network is lossy

resistive divider

4. Lossless and perfect isolat ion three-port network cannot be matched atall ports.




Four-port network (directional coupler)

Input port 1

Isolated port 4

Through port 2

Coupled port 3

Coupling:

Directivity:

Isolation:


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1. Matched, reciprocal and lossless four-network symmetrical(90°) directional coupler or antisymmetrical (180°) directional

coupler.

2. C = 3dB 90° hybrid (quadrature hybrid, symmetrical coupler),180° hybrid (magic-T hybrid, rate-race hybrid)

Discussion



• Lossless divider

Lossless divider has mismatched ports

“ not practical”

The T-junction power divider



Resistive (lossy) divider

matched ports



Wilkinson power divider

• Basic concept

Input port 1 matched, port 2 and port 3 have equal potential

Input port 2, port 1 and port 3 have perfect isolationlossy, matched and good isolation (equal phase) three-port divider

420 λ⇒ ,Z





If we inject a TEM mode signal at port 1, equal in-phase signals reachpoints a and b. Thus, no current flows through the resistor, and equalsignals emerge from port 2 and port 3. The device is thus a 3dB power

divider. Port 1 will be matched if the λ /4 sections have a characteristicimpedance .02Z

If w e now inject a TEM mode signal at port 2, w ith matched loads placedon port 1 and on port 3, the resi sto r is effect ively grounded at poin t b. Equalsignals flow tow ard port 1, and dow n into the resistor, w ith port 2 seeing a

match. Half the incident power emerges from port 1 and half is dissipated inthe resistor film.

Similar performance occurs when port 1 and port 2 are terminated inmatched loads, and a TEM mode signal is injected at por t 3.

If w e choose the terminal planes at 1.0 w avelengths from the three Tee junctions, the scattering matrix is

[ ]−−

−−=

00

00

0

2

1

j

j

j j

S




Wilkinson power divider for unequal power splits

0Z

02Z

03Z

P 1

P2

P 3

R2 = KZ 0

R3 = Z 0/K




Design a Wilkinson power divider wi th a power division ration of 3 dB and a sourceimpedance of 50 Ω

Solution:

( )dB P

P 350

2

3 .=

70702

1

2

32.=⇒== K

P

P K

( )( ) Ω=+=+= 0103

7075

50150

13

2

003 ...

.

K

K Z Z

( )( ) Ω=Ω== 5511035032

02 ..Z K Z

Ω=

+=

+= 1106

7070

1707050

10 .

..

K K Z R



The output impedances are

( ) Ω=== Ω===7270707050

3535707050

03

02

.. / ..

K Z R K Z R

Ω50Ω=10303Z

Ω= 55102 .Z

Ω= 500Z

Ω= 500Z

•

•

Ω= 1106 .R




Bethe-Hole Directional Coupler

This the simplest form of a waveguide directional coupler. A small hole in thecommon broad wal l between two rectangular guides provides 2 wave componentsthat add in phase at the coupler por t, and are cancel led at the isolat ion por t.

0

b

2b

y

INPUT1

43

2

z

Coupling hole

x

COUPLEDISOLATED

THROUGH

S



Let the incident wave at Port 1 be the dominant TE 10 mode:Top Guide

Bottom Guide

Port 3(coupled)

Port 1(input)

Port 2(through)

Couplinghole

Z

−10 A +

10 A −

10 A +10 A

z j y e

a x A E β−π= sin

z j x e

a x

Z A

H β−π−= sin10

z j z e a x

aZ A j H β−πβ π= cos

10

A = amplitude of electric field (V m -1)

( )20

00

21 a Z

λ−η= = wave impedance,

dominate mode, Ω

( )200 21 a λ−κ=β = phase constant rad/m

00 2 λπ=κ




In the bottom guide the ampli tude of the forward scattered wave is given by

while the ampli tude of the reversed scattered w ave is g iven by

πβπ+παµ−παεω−=+ a

s

a a

s

Z a

s

P

A j A m e

2

22

22

210

02

010

10 cossinsin

πβπ

−παµ

+π

αεω

−=−

a s

a a s

Z a

s

P

A A

m e

2

22

22

210

02

010

10 cossinsinwhere

1010

Z ab

P =

For round coupling hole or radius r 0, we have

20

3

2r e =α

20

3

4r m =α

electric polarizabili ty

magnetic polarizability




Let s = offset distance to holea

s

We can then show that

( )220

0

2 a a

s

−λλ=π

sin

The coupl ing factor for a single-hole Bethe Coupler is

( )dB A

A C −=

10

20 log

and its directivity is

( )dB A

A D +

−=

10

1020 log




Design procedure

1. Use to find position of hole.( )220

0

2 a a

s

−λλ

=π

sin

2. Use to determine the hole radius r 0 to

give the required coupling factor.

( )dB A

A C

−=

10

20 log




h l l l




B 0

B 0

B 1

B 1

B 2

B 2

BN

BN FNF0

F1

F1

F2

F2

FNF0

n=0 n=1 n=2 n=N

Port 1

Port 4(ISO)

Port 4Coupl

Port 2TRU

d d d

Let a wave of value be injected at Port 1. If the holes are small, there is

only a small fraction of the power coupled through to the upper guide so thatwe can assume that the wave amplitude incident on all holes is essentiallyunity. The hole n causes a scattered wave F n to propagate in the forwarddirection, and another scattered wave B n to propagate in the backwarddirection. Thus the output signals are:

01∠

UPPER GUIDE





Port 1 (input)and Port 4 (isolated)

( ) ( ) d n j N

n n e B B B β−

=∑== 2

0

41

Port 2 (through) ( )

waves scattered forward

N

n n

d jN

wave incident main

d jN Total F e e F

∑=β−β− +=

0

2

Port 3 (coupled) ( ) ∑=

β−=N

n n

d N j F e F 0

23

All of these waves are phase referenced to the n = 0 hole.

( ) ( )dB F F C N

n n ∑

=−=−=

0

3 2020 loglog

( )

( ) ( )dB

F

e B

F

B D

N

n n

N

n

nd j n

∑∑

=

=β−

−=−=

0

0

2

2

4

2020 loglog





We can rewrite this as

∑∑ ==

β−+−=

N

n n

N

n

nd j n F e B D

00

2

2020 loglog

∑=

β−−−=N

n

nd j n e B C

0

220 log

The coupling coefficients are proportional to the polarizability α e and α m ofthe coupling holes. Let r n = radius of the n th hole. Then the forward scatteringcoefficient from the n th hole is

( )n A F n += 10

And the backward scattering from the hole is

( )n A B n −= 10





Now let us assume the coupling holes are located at the midpoint acrosscommon broad wall, i.e. s = a/2. Then for circular holed, we have

3

2100

0

10

010

213

2n n r

Z P A j A F ε

µ−ωε−== +

But

( ) 2

20

2

0

202

10

0

00

121 −

η=

λ−

η=η

=ωε

f f a

Z and k

c

−−

η−==∴

2

100

03 1213

2

f f

P

A k j K r K F c

f n f n where

Let A = 1 v/m. Then

−

η−= 123

2 2

100

0

f f

P

k j K c

f





Likewise the backward scattering coefficient is

32

100

0

323

2

n b n c

b r K β f f

P

k K =−

η= and

Note that K f and K b are frequency-dependent constants that are the same forall aperture. Thus,

( )dB r K C

N

n n f ∑=−−=

0

3

2020 loglog

( )dB e r K C D N

n

nd j n b ∑

=

β−−−−=0

232020 loglog





Consider the following design problem:

Given a desired coupling level C, how do we design the coupler so that the directivity D isabove a value D min over a specified frequency band?

Note that if the coupling C is specified, then is known.∑=

N

n n F

0

We now assume that either (1) the holes scatter symmetrically (e.g. they areon the common narrow wall between two identical rectangular guides) or (2)

holes scatter asymmetrically (e.g. they are on the centerline of the commonbroad wall, i.e. s=a/2 ). Thus:

n n

n n

F B

F B

−==

or

In either case, we have

∑∑

=

β−

==N

n

nd j n

N

n n

e F

F D

0

2

020 log





Thus, keeping the directivity D > D min is equivalent to keeping

below a related minimum value. Let∑=

β−N

n

nd j n e F

0

2

d j j e e w and d

β−φ==β−=ϕ

2

2

We also introduce the function

( ) ( ) ∑∑=

φ

=

β− =φ⇒=βN

n

jn n

N

n

nd j n e F g e F d g

00

2

( ) ( )∏∑∑===

−===∴N

n n N

n N

n N

n N

N

n

n n w w F w

F

F F w F w g

100

Thus we have ( )

( )w g

g D

120 log=

( ) 20

0

101 / C N

n n F g −

=== ∑

Couplingfactor (dB)





From the previous two equations we can deduce that

( ) 20101 / max

minD g g −×=

The multi-hole coupler design problem thus reduces to finding a set of roots w nthat will produce a satisfactory g(w), and thus a satisfactory D(f) in the desired

frequency band under the constraint that .( ) max g w g ≤Example: Design a 7-hole directional coupler in C-band waveguide with a binomial

directivity response to provide 15 dB coupling and with D min =30 dB. Assume an operatingcenter frequency of 6.45 GHz and a hole spacing d = λg/4 (or λg + λg/4). Also assume broad-wall coupling with s = a/2.

Solution:

From , we have( ) ( )∏=

−=N

n n N w w F w g

1

( ) ( ) 1266 −==−= β− d j

n n e w where w w F w g

( ) ( ) 1615201561 234566

66 ++++++=+=∴ w w w w w w F w F w g





Thus,

( ) ( ) 062015

66

6 002780177801064111 F F F F g ==∴===+= − ..

By the binomial expansion we have

( ) ( ) n

n n w C w ∑

==+

6

0

661

where, is the set of binomial coefficients( )

( ) ( ) !!

!

!!

!

n n n n N

N C n

−=

−=

6

66

Thus

05557020

04168015

0166706

63

624

615

.

.

.

==

======

F F

F F F

F F F





−−

η

−==2

100

03 121

3

2

f

f

P

A k j K wherer K F c

f n f n

−

η−= 123

2 2

100

0

f f

P

k j K c

f

We now can compute the radii of the coupling holed from

and

We have – with f c = 4. 30 GHz for C-Band guide, f =6.45 GHz, k 0 =2 π /λ 0 =135.1 m -1,η 0 =376.7 Ω , P 10 = ab/Z 10 ,

245981456

3042

1008173763

11352 2

6 =

−

××××= − .

.

..

.f K

Ω×=Ω=

−η= − 26

10

2

010 1008145051 m P f f

Z c . ,.





The hole radii are:

cm r m K

r f

4830004830002780

6

31

0 ..

.

←==

=

cm r m K

r f

8780008780016670

5

31

1 ...

←==

=

cm r m K r f

19210119210041680

4

31

2 ...

←==

=

cm m K

r f

31101310055570

31

3 ...

←=

=





cm a 4853 .=cm r 96602 0 .= cm r 75612 1 .= cm r 38422 2 .= cm r 6222 3 .=

4.68 cm

Top view of C-Band guide common broad wall with coupling holes

The guide wavelength is

The nominal hole spacing is . However, the center holehas a diameter of 2.62 cm, so i t would overlap w ith adjacent holes. We can

increase the hole spacing to wi th no effect on electricalperformance.

m g 6240

456341

2

0 .

..

=

−

λ=λ

cm d g 5614

.=λ

=

cm d g 6844

3.=

λ=

The total length of the common broad wall section with coupling holes is ~30 cm, which is fairly large WG section.





We now plot the coupling and directivity vs. frequency

( ) ( ) ( )

6

26

6

2226

6

6

6

6 2211

φ=

+=+=+=

φφ−φφφ

cos

j j j j j

e F e e e F e F w F w g

( )66

66

217780

22

φ=φ=∴ cos.cosF w g

We then have

( ) ( )( ) g

d g

w g dB D

λπ−=−=

2120

120 cosloglog

where d=4.68 cm

( ) ( )28

20

0

1

103

21 f f

f

a c

g −×=

λ−λ=λ





Note that the directivity is better than Dmin = -30 dB over a bandwidth of 900 MHzcentered about 6.45 GHz.

-50

-40

-30

-20

-10

0

D ( d

B )

5.75 6.0 6.25 6.5 6.75 7.0 7.25 7.5Frequency (GHz)

Dmin

900 MHz


Even-odd mode analysis




Even odd mode analysis

d




Even-mode

Symmetry of port 2 and 3

Odd d




Odd-mode

symmetry of port 2 and 3,

open, short at bisection

port 1 matched

2332 S S =⇒

011 =⇒ S

Di i




3dB Wilk inson power divider has equal ampl itude and phase outputs at port2 and port 3.

3dB Wilkinson power combiner

Discussion




Unequal power division Wilkinson power divider




N-way Wilkinson power divider




y p

Th d t (90°) h b id




• Branch-line coupler Port 2 and port 3 have equal amplitude, but 90° phase different

[ ] −=

010

001100

010

21

j

j j

j

S

Input

1

4

2

3

Isolated

Z0 Z0

Z0 Z0

λ /4λ /4

20Z

20Z

Output

Output

• •

• •

•

•

•

•

1

1

1

1

21

21

1 1

1

3

2

4

A1=1

B1

B41

1

1

B2

B3

00 ∠⇒ dB

903 −∠−⇒ dB

903 −∠−⇒ dB

The quadrature (90°) hybrid

Quadrature Hybrids




Quadrature HybridsWe can analyze this circuit by using superposition of even-modes and 0dd-modes . We add the even-mode exci tat ion to the odd-mode exci tat ion to producethe original excitat ion of A 1=1 vol t at por t 1 (and no exci tation at the other por ts .)

21

• •

• •

•

•

••

••

1 1

1 1

1

1

1

1

21

21

21+

21+

Open circuit

• •

• •

•

•

•

•

1

1

1

1

21

1

1

1

3

2

4

1/2

1

1Line of symmetry I=0, v=max

Even Mode Excitation

• •

• •

•

•

•

•

1

1

1

1

21

1

1

1

3

2

4

1/2

1

1Line of symmetry v=0, I=max

1/2

-1/2

• •

• •

•

•

••

••

1 1

1 1

21

21

21+

21−

short circuit stubs

T0Γ0

Odd Mode Excitation(2 separate 2-ports)

(2 separate 2-ports)

Quadrature Hybrids




Quadrature HybridsWe now have a set of two decoupled 2-port networks. Let e and T e be thereflection and transmission coefficients of the even-mode excitation. Similarly Γoand T e for the odd-mode excitation.

Superposition:

Γ−Γ=

Τ−Τ=Τ+Τ=

Γ+Γ=

o e

o e

o e

o e

B

B

B

B

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

4

3

2

1Input

Through

Coupled

Isolated

Reflectedwaves

Quadrature Hybrids




Quadrature HybridsConsider the even-mode 2-port circuit:

Port 1

Input

Port 2

Coupled

open open

4λ

8λ8λ 21

We can represent the two open circuit stubs by their admittance:8λ

j

z

j

j z y

L

L

z L

=π+

π+

=∞→

41

4

1

tan

tanlim

The transmission line, with characteristic impedance has ab ABCD matrix

4λ 21

02

20

j

j

Quadrature Hybrids




Thus, the ABCD matrix for the cascade is

stub line stub e

j j

j

j D C

B A

84

8

1

01

02

20

1

01

λλλ

=−

−=1

1

2

1

j

j

Using the conversion table (next slide) to convert [S] parameters (with Z 0=1 asthe reference characteristic impedance).

2

1

222

10

0

−++−=+++= j j D CZ

Z

B A denominator

( )( )

0211

211

00

00

11 =−++−+++−=

+++

−−+==Γ

j j

j j

D CZ Z

B A

D CZ Z B

A

S e

( ) ( ) j j j D CZ Z

B A

S e +−=−++−=+++==Τ 12

1

211

22

00

21

Similarly for odd mode we have: ( ) j S and S −=Τ==Γ= 12

10 012011



Quadrature Hybrids




Quadrature Hybrids

Therefore we have

=−=

−==

0

2

1

2

0

4

3

2

1

B

B

j B

B

Scattered wavevoltages

(port 1 is matched)

Through (half power, -90 0 phase shift port 1 to 2)

(half power, -1800

phase shift port 1 to 3)

(no power to port 4)

The bandwidth of a single branch-line hybrid is about 10% - 20%, due to the

requirement that the top and bottom lines are λ /4 in length. We can obtainincreased directivity bandwidth (with fairly constant coupling) by using three or more sections.

Quadrature Hybrids




Next we consider a more general single section branch-line coupler:

Input Through Output

Isolated Output (coupled)

Z0

Z0

Z0

Z0

Z01

Z01

Z02 Z02

1

4

2

3

4λ

4λ

We can show that if the condition

( )2001

001

0

02

1 Z Z

Z Z

Z

Z

−=

is satisfied, then port 1 is matched; port 4 is decoupled from port 1.

Quadrature Hybrids

Quadrature Hybrids




Single section branch-line coupler

=

−=

−=

=

0

0

4

002

0013

0

012

1

B

Z Z Z Z

B

Z

Z j B

B

scattered wavevoltages

isolated

matched

(matched)

0∠

90−∠180−∠

Thus, the directivity is theoretically infinite at the design frequency. We canalso show that the coupling is given by

( )

( )dB

Z Z

C

−

=2

0011

110 log

For stripline + microstrip, we control Z 01 /Z0 by varying the strip width, in coaxby adjusting the ratio b/a, and in the rectangular guide by changing the bdimension.

Quadrature Hybrids

Quadrature Hybrids




Example:Design a one-section branch-line directional coupler to provide a coupling of 6 dB.

Assume the device is to be implemented in microstrip, with an 0.158 cm substratethickness, a dielectric constant of 2.2, and that the operating frequency is 1.0 GHz.

Solution:( )

( )dB Z Z

C 61

110 2

001

=−

= log

Ω=⇒=∴ 274386530 01001 .. Z Z Z

( ) Ω=⇒=−= 3186726311

022

001

001

0

02 .. Z Z Z

Z Z Z Z

cm r

2262022

300 ..

==ε

λ≅λ cm l 05655

4.=λ=

( ) 0813610

39112

1121

2...lnln = ε−+−ε−ε+−−−π=

r r r B B B

d w

cm w 4870 .=⇒

Quadrature Hybrids

Quadrature Hybrids




, Ω= 274301 .Z For

167511

83186

22

02 . ,. =

−

=Ω= A

A

e

e

d

w Z For cm w 18502 .=

cm w 60101 .=

Input(0 dB)

Isolated

Z0

Z0

Z0

Z0

Z01

Z01

Z02 Z02

Through(-1.26dB)

Coupled(-6dB)

0.4868 cm 0.6008 cm

0.1845 cm

5.0565 cm

5.0565 cm

With 0 dB power input at the upper left arm, the power delivered to a matched loadat the through arm is

( )( )

( )

dB Z

Z

B B P

P dB P

out

in

2612743

501010

11010

22

01

0

222

12

..

loglog

loglog *

−=

−=

−=

−=−=

Quadrature Hybrids

Coupled Line Directional Couplers




These are either stripline or microstrip 3-wire lines with close proximity of parallellines providing the coupling .

Co-planar stripline Side-stacked coupled stripline

Broadside-stacked coupler stripline Co-planar microstrip

open ckt

short ckt

even mode

odd mode

0V

0V

0V

0V

Theory of Coupled Lines





• •

•C 11

C12

C22

Three-wire coupled line Equivalent network

C12 capacitance between two strip conductors in absence of the ground conductor.

C11 ,C 22 capacitance between one conductor and ground

In the even mode excitation, the currents in the strip conductor s are equal inamplitude and in the same direction.

In the odd mode excitation, the currents in the strip conductors are equal inamplitude but are in opposi te directions.





eo y o Coup ed es

In the even mode, fields have even symmetry about centerline, and nocurrent flows between strip conductors. Thus C 12 is effectively open-circuited.

The resulting capacitance of either line to ground isC e = C 11 = C 22

The characteristic impedance of the even modes is

E

e e e

C C

L Z

ν==

10

In the odd mode, fields have an odd symmetry about centerline, and avoltage null exists between the strip conductors.

E

• •

•

2C 12

C 22C 11

2C 12

•

The effective capacitance betweeneither strip conductor and ground is

C 0 = C 11 + 2 C 12 = C 22 + 2 C 12

o o o C C

L Z

ν==

10





Example: An edge-coupled str ipl ine with ε r =2.8 and a ground plane spacing of 0.5cm is required to have even- and odd-mode characterist ic impedance of Z 0e 100 Ω

and Z 0o =50 Ω . Find the necessary strip widths and spacing.

Solution: b = 0.50 cm, r = 2.8, Z 0e=100 , Z 0o=50

odd

o r

even

e r Z Z 66833167 00 .. =ε=ε∴

from the graph, s/b ≈ 0.095, W/b ≈ 0.32S=0.95 b = 0.095 × 0.5 = 0.0425 cm

W=0.32b=0.32 × 0.5=0.16 cm

W W s

b=.5cm

r =2.8

0.5cm r= 2.80.16cm 0.16cm

0.0425cm

y p



Even Mode ABCD Analysis




Consider the equivalent ci rcui t for the even mode:


1 2

e Γ8

3 λ

4

λ

2

8

λ2

2

e Τ21

o.c.

o.c.

At port 1, the admittance looking in tothe λ /8 o.c. stub i s

( )28

2

2

101

j j j y y S =

λ

⋅λπ=β= tantan l

The ABCD matrix of a shunt admittance y S1 is

= 12

01

1

j D C

B A Y3

Y1 Y2

A=1+Y 2/Y3

C=Y 1 + Y2+Y1Y2/Y3

B=1/Y 3

D=1+Y 1/Y3





At port 2, the admittance looking into the 3λ/8 o.c. stub is

( )28

32

2

102

j j j y y S −=

λ⋅

λ

π=β= tantan l

The ABCD matrix of this shunt admittance is

−= 12

01

3

j D C

B A 2=o Z 2=o Z ys1 ys2

4λ

quarter-wave section

The ABCD matrix is

ββββ

=l l

l l

cossin

sincos

o

o

jy

jZ

D C

B A

2

212

242

==π=π⋅=β

o o y Z

λπ where

,

l =∴0

2

20

2

j j D C B A





We can now compute the ABCD matrix of the even mode cascade

1

8

3 λ

4λ

2

8

λ2

2

o.c.

o.c.

−==

−=

12

21

02

21

12

01

12

01

0

2

20

12

01

j

j j

j j

j j j

j D C

B A

e

2

Odd Mode ABCD Analysis




1

o Γ8

3 λ

4

λ

2

8

λ2

2

o Τ21

s.c.

s.c.

2The input admittance to a short-circuited lossless stub is

( )l β−= cot j y y in 0

Thus the input admittance to thes.c. λ /8 tub is

28

2

2

11

j j y

S −=

λ

⋅

λ

π−= cot

ABCD matr ix:

−= 1

2

01

1

j D C

B A

28

32

2

12

j j y S +=

λ

⋅λπ−= cot (Input admittance to s.c. 3λ /8 stub)

Odd Mode ABCD Analysis




and

= 12

01

3

j D C

B A

So the ABCD matrix of the odd-mode cascade is

−=

−=

12

21

12

01

02

20

12

01

j

j

j j

j

j D C

B A

o

22

22

j j

j j

o o

e e

−=Τ=Γ−=Τ−=Γ

,

,



Excitation at Port 4




Even mode: −=

12

21

j

j

D C

B A

e

D CZ Z B A

D CZ Z

B A S

o o

o o e +++

+−+−==Γ 22

2222

12211221 j

j j j j j e =−=−++ −−+−=Γ

( )D CZ Z B A

BC AD S

o +++−==Τ

012

2e

( )222

2

1221

212 j

j j j e −==−++

+−=Τ

Odd mode: −=12

21

j j

D C B A

o

222

2

1221

122122

j

j j j

j j S o −==

+++−+−+==Γ

( )222

2

1221

21212

j

j j j S o −==

+++−

+−==Τ

Excitation at Port 4 is expressed as :

Even212 =+V

214 =+V Odd

212 −=+V

214 =+V



Excitation at Port 4 of Rat-Race Coupler (cont.)

2

j e =Γ

2

j e −=Τ

2

j o −=Γ

2

j o −=Τ

Output waves

o e

o e

o e

o e

B

B

B

B

Γ+Γ=

Τ+Τ=

Γ−Γ=

Τ−Τ=

2

1

2

12

1

2

1

2

1

2

12

1

2

1

4

3

2

1

The resulting output vector for unit excitation at Port 4 is :

[ ]−

=

0

2

2

0

4 j

j B i

Documents

Lecture 8 coupler Lecture 8 couplerLecture 8 couplerLecture 8 couplerLecture 8 couplerLecture 8 couplerLecture 8 couplerLecture 8 couplerLecture 8 couplerLecture 8 couplerLecture 8