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Transcript of Lecture 10
Microwave Physics and Techniques UCSB –June 20031
Lecture 10
Power Dividers and Couplers
June 19,2003A. Nassiri - ANL
Microwave Physics and Techniques UCSB –June 20032
Power dividers and directional couplers
Basic properties of dividers and couplersthree-port network (T-junction), four-port network (directional coupler), directivity measurement
The T-junction power dividerLossless divider, lossy divider
The Wilkinson power dividerEven-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 Z0, single-section and multisection coupled linecouplers
The Lange couplerThe 180° hybrid
- rat-race hybrid, tapered coupled line hybridOther couplersreflectometer
Microwave Physics and Techniques UCSB –June 20033
Basic properties of dividers and couplers• N-port network
Discussion
1. Matched ports Sij=02. Reciprocal network symmetric property Sij = Sji3. Lossless network unitary property
Microwave Physics and Techniques UCSB –June 20034
Three-port network (T-junction)
Discussion1. Three-port network cannot be lossless, reciprocal and matched at all ports.2. Lossless and matched three-port network is nonreciprocal
circulator
Microwave Physics and Techniques UCSB –June 20035
3. Matched and reciprocal three-port network is lossyresistive divider
4. Lossless and perfect isolation three-port network cannot be matched at all ports.
Microwave Physics and Techniques UCSB –June 20036
Four-port network (directional coupler)
Input port 1
Isolated port 4
Through port 2
Coupled port 3
Coupling:
Directivity:
Isolation:
Microwave Physics and Techniques UCSB –June 20037
Discussion
1. Matched, reciprocal and lossless four-network symmetrical(90°) directional coupler or antisymmetrical (180°) directionalcoupler.
2. C = 3dB 90° hybrid (quadrature hybrid, symmetrical coupler), 180° hybrid (magic-T hybrid, rate-race hybrid)
Microwave Physics and Techniques UCSB –June 20038
The T-junction power divider• Lossless divider
Lossless divider has mismatched ports
“not practical”
Microwave Physics and Techniques UCSB –June 20039
Resistive (lossy) divider
matched ports
Microwave Physics and Techniques UCSB –June 200310
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
42 0 λ⇒ ,Z
Microwave Physics and Techniques UCSB –June 200311
Wilkinson power dividerThe Wilkinson power divider has these advantages:1. It is lossless when output ports are matched.2. Output ports are isolated.3. It can be designed to produce arbitrary power division.
λ/4
λ/4
Port 1
Port 2
Port 3
02Z
02Z
0Z
0Z
0ZFilm resistor
R = 2Z0
••
••••
Port 1
Port 2
Port 3
λ/4
λ/4
0Z02Z
02Z
0Z
0ZR = 2Z0
a
b
Microwave Physics and Techniques UCSB –June 200312
Wilkinson power dividerIf we inject a TEM mode signal at port 1, equal in-phase signals reach
points a and b. Thus, no current flows through the resistor, and equal signals 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 characteristic impedance .02Z
If we now inject a TEM mode signal at port 2, with matched loads placed on port 1 and on port 3, the resistor is effectively grounded at point b. Equal signals flow toward port 1, and down into the resistor, with port 2 seeing a match. Half the incident power emerges from port 1 and half is dissipated in the resistor film.
Similar performance occurs when port 1 and port 2 are terminated in matched loads, and a TEM mode signal is injected at port 3.
If we choose the terminal planes at 1.0 wavelengths from the three Tee junctions, the scattering matrix is
[ ]
−−
−−=
0000
0
21
jj
jjS
Microwave Physics and Techniques UCSB –June 200313
Wilkinson power divider for unequal power splits
λ/4
λ/4
0Z
02Z
03Z
P1
P2
P3
R2 = KZ0
R3 = Z0/K
Microwave Physics and Techniques UCSB –June 200314
Wilkinson power dividerDesign a Wilkinson power divider with a power division ration of 3 dB and a source impedance of 50 Ω
Solution:
( )dBPP 350
2
3 .=
707021
2
32 .=⇒== KPP
K
( )( ) Ω=+
=+
= 01037075
5015013
2
003 ....
K
KZZ
( )( ) Ω=Ω== 5511035032
02 ..ZKZ
Ω=
+=
+= 1106
707017070501
0 ..
.K
KZR
Microwave Physics and Techniques UCSB –June 200315
Wilkinson power divider
The output impedances are
( )Ω===
Ω===7270707050
3535707050
03
02
../..
KZR
KZR
λ/4
λ/4
Ω50Ω=10303Z
Ω= 55102 .Z
Ω= 500Z
Ω= 500Z
•
•
Ω= 1106.R
Microwave Physics and Techniques UCSB –June 200316
Bethe-Hole Directional CouplerThis the simplest form of a waveguide directional coupler. A small hole in the common broad wall between two rectangular guides provides 2 wave components that add in phase at the coupler port, and are cancelled at the isolation port.
0
b
2b
y
INPUT1
43
2
z
Coupling hole
xCOUPLED ISOLATED
THROUGH
S
Microwave Physics and Techniques UCSB –June 200317
Bethe-Hole Directional CouplerLet the incident wave at Port 1 be the dominant TE10 mode:
Top Guide
Bottom Guide
Port 3(coupled)
Port 1(input)
Port 2(through)
Coupling hole
Z
−10A
+10A
−10A +
10A
zjy e
ax
AE β−π= sin
zjx e
ax
ZA
H β−π−= sin
10
zjz e
ax
aZAj
H β−πβπ
= cos10
A = amplitude of electric field (V m-1)
( )20
00
21 aZ
λ−
η= = wave impedance,
dominate mode,Ω
( )200 21 aλ−κ=β = phase constant rad/m
00 2 λπ=κ
Microwave Physics and Techniques UCSB –June 200318
Bethe-Hole Directional Coupler
In the bottom guide the amplitude of the forward scattered wave is given by
while the amplitude of the reversed scattered wave is given by
πβπ
+παµ
−π
αεω
−=+
as
aas
Zas
PAj
A me
222
22
210
020
1010 cossinsin
πβπ
−παµ
+π
αεω
−=−
as
aas
Zas
PA
A me
222
22
210
020
1010 cossinsin
where
1010 Z
abP =
For round coupling hole or radius r0, we have
203
2re =α
203
4rm =α
electric polarizability
magnetic polarizability
Microwave Physics and Techniques UCSB –June 200319
Bethe-Hole Directional Coupler
Let s = offset distance to holea
s
We can then show that
( )220
0
2 aas
−λ
λ=
πsin
The coupling factor for a single-hole Bethe Coupler is
( )dBAA
C −=10
20 log
and its directivity is
( )dBA
AD +
−=
10
1020 log
Microwave Physics and Techniques UCSB –June 200320
Bethe-Hole Directional Coupler
Design procedure:
1. Use to find position of hole.( )220
0
2 aas
−λ
λ=
πsin
2. Use to determine the hole radius r0 to
give the required coupling factor.
( )dBA
AC −=
1020 log
Microwave Physics and Techniques UCSB –June 200321
Bethe-Hole Directional CouplerTypical x-Band -20 dB coupler
- 60
- 40
- 20
0
6 7 8 9 10 11Frequency (GHz)
C ( dB)
D (dB)
C
D
Note: Coupling very broad band, directivity is very narrow band (for single-hole coupler)
We can achieve improved directivity bandwidth by using an array of equispaced holes.
Microwave Physics and Techniques UCSB –June 200322
Bethe-Hole Directional Coupler
B0
B0
B1
B1
B2
B2
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 that we can assume that the wave amplitude incident on all holes is essentially unity. The hole n causes a scattered wave Fn to propagate in the forward direction, and another scattered wave Bn to propagate in the backward direction. Thus the output signals are:
01∠
UPPER GUIDE
Microwave Physics and Techniques UCSB –June 200323
Bethe-Hole Directional CouplerPort 1 (input)and Port 4 (isolated)
( ) ( ) dnjN
nneBBB β−
=∑== 2
0
41
Port 2 (through) ( )
4434421321
wavesscatteredforward
N
nn
djN
waveincidentmain
djNTotal FeeF
∑=
β−β− +=0
2
Port 3 (coupled) ( ) ∑=
β−=N
nn
dNj FeF0
23
All of these waves are phase referenced to the n = 0 hole.
( ) ( )dBFFCN
nn ∑
=
−=−=0
3 2020 loglog
( )
( ) ( )dB
F
eB
F
BD N
nn
N
n
ndjn
∑
∑
=
=
β−
−=−=
0
0
2
2
42020 loglog
Microwave Physics and Techniques UCSB –June 200324
Bethe-Hole Directional CouplerWe can rewrite this as
∑∑==
β− +−=N
nn
N
n
ndjn FeBD
00
2 2020 loglog
∑=
β−−−=N
n
ndjneBC
0
220 log
The coupling coefficients are proportional to the polarizability αe and αm of the coupling holes. Let rn = radius of the nth hole. Then the forward scattering coefficient from the nth hole is
( )nAF n+= 10
And the backward scattering from the hole is
( )nAB n−= 10
Microwave Physics and Techniques UCSB –June 200325
Bethe-Hole Directional CouplerNow let us assume the coupling holes are located at the midpoint across common broad wall, i.e. s = a/2. Then for circular holed, we have
32
100
0
10
010
213
2nn r
ZPA
jAF
εµ
−ωε
−== +
But
( ) 2
20
20
202
100
00
121
−
η=
λ−
η=
η=ωε
ffa
Zandk
c
−−
η−
==∴2
100
03 1213
2f
fP
AkjKrKF c
fnfn where
Let A = 1 v/m. Then
−
η−
= 123
2 2
100
0ff
Pkj
K cf
Microwave Physics and Techniques UCSB –June 200326
Bethe-Hole Directional CouplerLikewise the backward scattering coefficient is
32
100
0 323
2nbn
cb rKβ
ff
Pk
K =
−
η= and
Note that Kf and Kb are frequency-dependent constants that are the same for all aperture. Thus,
( )dBrKCN
nnf ∑
=
−−=0
32020 loglog
( )dBerKCDN
n
ndjnb ∑
=
β−−−−=0
232020 loglog
Microwave Physics and Techniques UCSB –June 200327
Bethe-Hole Directional CouplerConsider the following design problem:Given a desired coupling level C, how do we design the coupler so that the directivity D is above a value Dmin over a specified frequency band?
Note that if the coupling C is specified, then is known.∑=
N
nnF
0
We now assume that either (1) the holes scatter symmetrically (e.g. they are on the common narrow wall between two identical rectangular guides) or (2) holes scatter asymmetrically (e.g. they are on the centerline of the common broad wall, i.e. s=a/2 ). Thus:
nn
nn
FB
FB
−==
or
In either case, we have
∑
∑
=
β−
==N
n
ndjn
N
nn
eF
F
D
0
2
020 log
Microwave Physics and Techniques UCSB –June 200328
Bethe-Hole Directional CouplerThus, keeping the directivity D > Dmin is equivalent to keeping
below a related minimum value. Let ∑=
β−N
n
ndjneF
0
2
djj eewandd β−φ ==β−=ϕ 22 We also introduce the function
( ) ( ) ∑∑=
φ
=
β− =φ⇒=βN
n
jnn
N
n
ndjn eFgeFdg
00
2
( ) ( )∏∑∑===
−===∴N
nnN
nN
n N
nN
N
n
nn wwFw
FF
FwFwg100
Thus we have ( )( )wg
gD
120 log=
( ) 20
0101 /C
N
nnFg −
=
== ∑Coupling factor (dB)
Microwave Physics and Techniques UCSB –June 200329
Bethe-Hole Directional CouplerFrom the previous two equations we can deduce that
( ) 20101 /max
minDgg −×= The multi-hole coupler design problem thus reduces to finding a set of roots wnthat will produce a satisfactory g(w), and thus a satisfactory D(f) in the desired
frequency band under the constraint that . ( ) maxgwg ≤Example: Design a 7-hole directional coupler in C-band waveguide with a binomial directivity response to provide 15 dB coupling and with Dmin =30 dB. Assume an operating center 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
nnN wwFwg
1
( ) ( ) 1266 −==−= β− dj
nn ewwherewwFwg
( ) ( ) ( )1615201561 234566
66 ++++++=+=∴ wwwwwwFwFwg
Microwave Physics and Techniques UCSB –June 200330
Bethe-Hole Directional CouplerThus,
( ) ( ) 062015
66
6 002780177801064111 FFFFg ==∴===+= − ..
By the binomial expansion we have
( ) ( ) n
nn wCw ∑
=
=+6
0
661
where, is the set of binomial coefficients( )( ) ( ) !!
!!!
!nnnnN
NC n −
=−
=6
66
Thus
05557020
04168015
0166706
63
624
615
...
==
===
===
FF
FFF
FFF
Microwave Physics and Techniques UCSB –June 200331
Bethe-Hole Directional Coupler
−−
η−
==2
100
03 1213
2f
fP
AkjKwhere rKF c
fnfn
−
η−
= 123
2 2
100
0ff
Pkj
K cf
We now can compute the radii of the coupling holed from
and
We have – with fc = 4. 30 GHz for C-Band guide, f =6.45 GHz, k0 =2π/λ0 =135.1 m-1, η0 =376.7 Ω, P10 = ab/Z10,
2459814563042
100817376311352 2
6 =
−
××××
= − ..
...
fK
Ω×=Ω=
−η= − 26
10
2
010 1008145051 mPff
Z c .,.
Microwave Physics and Techniques UCSB –June 200332
Bethe-Hole Directional CouplerThe hole radii are:
cmrmK
rf
48300048300027806
31
0 ... ←==
=
cmrmK
rf
87800087800166705
31
1 ... ←==
=
cmrmK
rf
192101192100416804
31
2 ... ←==
=
cmmK
rf
3110131005557031
3 ... ←=
=
Microwave Physics and Techniques UCSB –June 200333
Bethe-Hole Directional Coupler
cma 4853.=cmr 96602 0 .= cmr 75612 1 .= cmr 38422 2 .= cmr 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 hole has a diameter of 2.62 cm, so it would overlap with adjacent holes. We can
increase the hole spacing to with no effect on electrical performance.
mg 6240
456341
20 .
..
=
−
λ=λ
cmd g 5614
.=λ
=
cmd 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.
Microwave Physics and Techniques UCSB –June 200334
Bethe-Hole Directional CouplerWe now plot the coupling and directivity vs. frequency
( ) ( ) ( )6
26
6
2226
66
66 2
211
φ=
+=+=+=
φφ−
φφφ cos
jjjjj eFeeeFeFwFwg
( )66
66
217780
22 φ
=φ
=∴ cos.cosFwg
We then have
( ) ( )( ) g
dg
wgdBD
λπ
−=−=2120
120 cosloglog
where d=4.68 cm
( )( )
( )28
20
0
1
103
21 ff
f
a cg
−
×=
λ−
λ=λ
Microwave Physics and Techniques UCSB –June 200335
Bethe-Hole Directional Coupler
Note that the directivity is better than Dmin= -30 dB over a bandwidth of 900 MHz centered about 6.45 GHz.
-50
-40
-30
-20
-10
0D
(dB)
5.75 6.0 6.25 6.5 6.75 7.0 7.25 7.5Frequency (GHz)
Dmin
900 MHz
Microwave Physics and Techniques UCSB –June 200336
Even-odd mode analysis
Microwave Physics and Techniques UCSB –June 200337
Even-mode
Symmetry of port 2 and 3
Microwave Physics and Techniques UCSB –June 200338
Odd-mode
symmetry of port 2 and 3,
open, short at bisection
port 1 matched2332 SS =⇒
011 =⇒ S
Microwave Physics and Techniques UCSB –June 200339
Discussion
3dB Wilkinson power divider has equal amplitude and phase outputs at port 2 and port 3.
3dB Wilkinson power combiner
Microwave Physics and Techniques UCSB –June 200340
Unequal power division Wilkinson power divider
Microwave Physics and Techniques UCSB –June 200341
Microwave Physics and Techniques UCSB –June 200342
N-way Wilkinson power divider
Microwave Physics and Techniques UCSB –June 200343
The quadrature (90°) hybrid
• Branch-line couplerPort 2 and port 3 have equal amplitude, but 90° phase different
[ ]
−
=
010001
100010
21
jj
jj
S
Input
1
4
2
3Isolated
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
o00 ∠⇒ dB
o903 −∠−⇒ dB
o903 −∠−⇒ dB
Microwave Physics and Techniques UCSB –June 200344
Quadrature HybridsWe can analyze this circuit by using superposition of even-modes and 0dd-modes. We add the even-mode excitation to the odd-mode excitation to produce the original excitation of A1=1 volt at port 1 (and no excitation at the other ports.)
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
1/21
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)
Microwave Physics and Techniques UCSB –June 200345
Quadrature HybridsWe now have a set of two decoupled 2-port networks. Let Γe and Te be the reflection and transmission coefficients of the even-mode excitation. Similarly Γoand Te for the odd-mode excitation.
Superposition:
Γ−Γ=
Τ−Τ=
Τ+Τ=
Γ+Γ=
oe
oe
oe
oe
B
B
B
B
21
21
21
21
21
21
21
21
4
3
2
1Input
Through
Coupled
Isolated
Reflected waves
Microwave Physics and Techniques UCSB –June 200346
Quadrature HybridsConsider the even-mode 2-port circuit:
Port 1
Input
Port 2
Coupled
open open
4λ
8λ 8λ 21
We can represent the two open circuit stubs by their admittance:8λ
j
zj
jz
y
L
L
z L
=π
+
π+
=∞→
41
41
tan
tanlim
The transmission line, with characteristic impedance has abABCD matrix
4λ 21
0220
jj
Microwave Physics and Techniques UCSB –June 200347
Quadrature HybridsThus, the ABCD matrix for the cascade is
32144 344 21321stublinestub
e jj
j
jDCBA
848
101
022
0101
λλλ
=
−
−=
11
21
jj
Using the conversion table (next slide) to convert [S] parameters (with Z0=1 as the reference characteristic impedance).
21
2221
00
−++−=+++=jj
DCZZB
Adenominator
( )( )
0211211
00
00
11 =−++−+++−
=+++
−−+==Γ
jjjj
DCZZB
A
DCZZB
ASe
( )( )j
jjDCZZB
ASe +−=
−++−=
+++==Τ 1
21
21122
00
21
Similarly for odd mode we have: ( )jSandS −=Τ==Γ= 12
10 012011
Microwave Physics and Techniques UCSB –June 200348
S Z Y ABCD
D. Pozar
Conversion between two-port network parameters
Microwave Physics and Techniques UCSB –June 200349
Quadrature Hybrids
Therefore we have
=
−=
−=
=
02
12
0
4
3
2
1
B
B
jB
B
Scattered wave voltages
(port 1 is matched)
Through (half power, -900 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 obtain increased directivity bandwidth (with fairly constant coupling) by using three or more sections.
Microwave Physics and Techniques UCSB –June 200350
Quadrature HybridsNext 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 ZZ
ZZZZ
−=
is satisfied, then port 1 is matched; port 4 is decoupled from port 1.
Microwave Physics and Techniques UCSB –June 200351
Quadrature HybridsSingle section branch-line coupler
=
−=
−=
=
0
0
4
002
0013
0
012
1
B
ZZZZ
B
ZZ
jB
B
scattered wave voltages
isolated
matched
(matched)
o0∠
o90−∠
o180−∠
Thus, the directivity is theoretically infinite at the design frequency. We can also show that the coupling is given by
( )( )dB
ZZC
−= 2
0011110 log
For stripline + microstrip, we control Z01/Z0 by varying the strip width, in coax by adjusting the ratio b/a, and in the rectangular guide by changing the b dimension.
Microwave Physics and Techniques UCSB –June 200352
Quadrature HybridsExample: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 substrate thickness, a dielectric constant of 2.2, and that the operating frequency is 1.0 GHz.
Solution:( )
( )dBZZ
C 61
110 2001
=
−= log
Ω=⇒=∴ 274386530 01001 .. ZZZ
( )Ω=⇒=
−= 318672631
1022
001
001
0
02 .. ZZZ
ZZZZ
cmr
2262022
300 ..==
ελ
≅λ cm l 056554
.=λ
=
( ) 081361039112
11212 ...lnln =
ε
−+−ε−ε
+−−−π
=rr
r BBBdw o
cmw 4870.=⇒ o
Microwave Physics and Techniques UCSB –June 200353
Quadrature Hybrids, Ω= 274301 .ZFor
167511
83186 22
02 .,. =−
=Ω= A
A
e
ed
wZFor cmw 18502 .=
cmw 60101 .=
Input(0 dB)
IsolatedZ0
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 load at the through arm is
( )( )
( )
dBZZ
BBP
PdBP out
in
2612743
501010
11010
22
01
0
222
12
..
loglog
loglog *
−=
−=
−=
−=−=
Microwave Physics and Techniques UCSB –June 200354
Coupled Line Directional CouplersThese are either stripline or microstrip 3-wire lines with close proximity of parallel lines 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
Microwave Physics and Techniques UCSB –June 200355
Theory of Coupled Lines• •
•C11
C12
C22
Three-wire coupled line Equivalent network
C12 capacitance between two strip conductors in absence of the ground conductor.
C11 ,C22 capacitance between one conductor and ground
In the even mode excitation, the currents in the strip conductors are equal in amplitude and in the same direction.
In the odd mode excitation, the currents in the strip conductors are equal in amplitude but are in opposite directions.
Microwave Physics and Techniques UCSB –June 200356
Theory of Coupled LinesIn the even mode, fields have even symmetry about centerline, and no current flows between strip conductors. Thus C12 is effectively open-circuited. The resulting capacitance of either line to ground is
Ce = C11 = C22
The characteristic impedance of the even modes is
Er
eee CC
LZ
ν==
10
In the odd mode, fields have an odd symmetry about centerline, and a voltage null exists between the strip conductors.
Er
• •
•
2C12
C22C11
2C12
•
The effective capacitance between either strip conductor and ground is
C0 = C11 + 2 C12 = C22 + 2 C12
ooo CC
LZ
ν==
10
Microwave Physics and Techniques UCSB –June 200357
Theory of Coupled LinesExample: An edge-coupled stripline with εr=2.8 and a ground plane spacing of 0.5 cm is required to have even- and odd-mode characteristic impedance of Z0e 100 Ωand Z0o=50 Ω. Find the necessary strip widths and spacing.Solution: b = 0.50 cm, εr = 2.8, Z0e=100 Ω, Z0o=50 Ω
44 344 2144 344 21
odd
or
even
er ZZ 66833167 00 .. =ε=ε∴
from the graph, s/b ≈ 0.095, W/b ≈ 0.32 S=0.95 b = 0.095 ×0.5 = 0.0425 cm
W=0.32b=0.32×0.5=0.16 cm
W Ws
b=.5cm
εr =2.8
0.5cm εr=2.80.16cm 0.16cm
0.0425cm
Microwave Physics and Techniques UCSB –June 200358
Waveguide magic-T
Input at port 1
Port 4:0
Ports 2 and 3: equal amplitude and phase
Input at port 4
Port 1:0
Ports 2 and 3: 1800 phase difference
2 3
4
2 3
4
23
1
4 ∆
Σ
Microwave Physics and Techniques UCSB –June 200359
Consider the equivalent circuit for the even mode:Even Mode ABCD Analysis
1 2
eΓ8
3λ
4λ
28λ
2
2
eΤ21
o.c.
o.c.
At port 1, the admittance looking into the λ/8 o.c. stub is
( )28
22
101
jjjyy S =
λ
⋅λπ
=β= tantan l
The ABCD matrix of a shunt admittance yS1 is
=
1
2
01
1
jDCBA
Y3
Y1 Y2
A=1+Y2/Y3
C=Y1 + Y2+Y1Y2/Y3
B=1/Y3
D=1+Y1/Y3
Microwave Physics and Techniques UCSB –June 200360
Even Mode ABCD Analysis
At port 2, the admittance looking into the 3λ/8 o.c. stub is
( )28
322
102
jjjyy S −=
λ
⋅λπ
=β= tantan l
The ABCD matrix of this shunt admittance is
−=
1
2
01
3
jDCBA 2=oZ 2=oZys1 ys2
4λ
quarter-wave sectionThe ABCD matrix is
ββββ
=
ll
llcossinsincos
o
o
jyjZ
DCBA
2
212
242
==
π=π⋅=β
oo yZ
λπwhere
,
l
=
∴ 0
2
20
2j
j
DCBA
Microwave Physics and Techniques UCSB –June 200361
Even Mode ABCD Analysis
We can now compute the ABCD matrix of the even mode cascade
1
83λ
4λ
28λ
2
2
o.c.
o.c.
−=
=
−
=
1221
02
211
2
01
12
01
02
201
2
01
jj
jj
j
jjj
jDCBA
e
2
Microwave Physics and Techniques UCSB –June 200362
Odd Mode ABCD Analysis
1
oΓ8
3λ
4λ
28λ
2
2
oΤ21
s.c.
s.c.
2The input admittance to a short-circuited lossless stub is
( )lβ−= cotjyy in 0
Thus the input admittance to the s.c. λ/8 tub is
282
21
1j
jy S −=
λ
⋅λπ
−= cot
ABCD matrix:
−=
1
2
01
1
jDCBA
2832
21
2j
jy S +=
λ
⋅λπ
−= cot (Input admittance to s.c. 3λ/8 stub)
Microwave Physics and Techniques UCSB –June 200363
Odd Mode ABCD Analysis
and
=
1
2
01
3
jDCBA
So the ABCD matrix of the odd-mode cascade is
−=
−=
1221
12
01
02
201
2
01
jj
jjj
jDCBA
o
22
22
jj
jj
oo
ee
−=Τ=Γ
−=Τ−=Γ
,
,
Microwave Physics and Techniques UCSB –June 200364
Excitation at Port 4
We have derived the ABCD matrices for the Even (e) and Odd (o) modes:
−=
1221
jj
DCBA
eand
−=
12
21j
jDCBA
o
For excitation at Port 4 instead of Port 1 the ABCD matrices remain the same. What changes are the definitions of Γ and T for each mode and their relations to B1 – B4 .
Te
O.C.
O.C.
Γe
1 2 1 2Te
Γe
S.C.
S.C.
Even Mode
Odd Mode
Microwave Physics and Techniques UCSB –June 200365
Excitation at Port 4
Even mode:
−=
1221
jj
DCBA
e
DCZZBA
DCZZBA
So
o
oo
e+++
+−+−==Γ 22 222
212211221 j
jjjjj
e =−
=−++−−+−
=Γ
( )DCZZBA
BCADS
o +++−
==Τ0
122
e( )
2222
1221212 j
jjje −==−++
+−=Τ
Odd mode:
−=
12
21j
jDCBA
o
2222
12211221
22j
jjjjj
So −==+++−+−+
==Γ
( )222
21221
21212
jjjj
So −==+++−
+−==Τ
Excitation at Port 4 is expressed as :
Even 212 =+V
214 =+VOdd
212 −=+V
214 =+V
Microwave Physics and Techniques UCSB –June 200366
Excitation at Port 4 of Rat-Race Coupler (cont.)
2j
e =Γ2j
e −=Τ
2j
o −=Γ2j
o −=Τ
Output waves
oe
oe
oe
oe
B
B
B
B
Γ+Γ=
Τ+Τ=
Γ−Γ=
Τ−Τ=
21
21
21
21
21
21
21
21
4
3
2
1
The resulting output vector for unit excitation at Port 4 is :
[ ]
−=
02
20
4 jj
B i