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

Lecture 8

Power Dividers and Couplers A. Nassiri -ANL Massachusetts Institute of Technology

RF Cavities and Components for Accelerators

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Power dividers and directional couplers  Basic properties of dividers and couplers three-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 couplers  Even- and odd-mode Z0, single-section and multisection coupled line  couplers  The Lange coupler  The 180° hybrid  - rat-race hybrid, tapered coupled line hybrid  Other couplers  reflectometer Massachusetts Institute of Technology


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Basic properties of dividers and couplers • N-port network

Massachusetts Institute of Technology


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Three-port network (T-junction)

Discussion 1. Three-port network cannot be lossless, reciprocal and matched at all ports. 2. Lossless and matched three-port network is nonreciprocal circulator



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3. Matched and reciprocal three-port network is lossy resistive divider 4. Lossless and perfect isolation three-port network cannot be matched at all ports.



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Four-port network (directional coupler)

Input port 1

Through port 2

Isolated port 4

Coupled port 3

Coupling:

Directivity:

Isolation:



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Discussion 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)



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The T-junction power divider • Lossless divider

“not practical”

Lossless divider has mismatched ports



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Resistive (lossy) divider

matched ports



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Wilkinson power divider • Basic concept

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

⇒ 2Z 0 , λ 4 Input port 2, port 1 and port 3 have perfect isolation lossy, matched and good isolation (equal phase) three-port divider



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Wilkinson power divider The 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. Port 2

λ/4 Port 1

2Z 0

Z0

Z0 Film resistor R = 2Z0

Z0

2Z 0

λ/4

• 2Z 0 • 2Z •b •

a

Port 1

Z0


• •

0

Port 2

Port 3

Z0 R = 2Z0

Z0


Port 3

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Wilkinson power divider  If 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 2Z 0 .  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

 0 1  [S ] = − j 2 − j



−j 0 0

− j 0   0 

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Wilkinson power divider for unequal power splits

P2

Z 02

P1

R2 = KZ0

Z0 Z 03

P3 R3 = Z0/K



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Wilkinson power divider Design a Wilkinson power divider with a power division ration of 3 dB and a source impedance of 50 Ω

Solution:

P3 = 0.5(3dB ) P2 P3 1 2 K = = ⇒ K = 0.707 P2 2

Z 03 = Z 0

1+ K 2

K3

1 + 0.5 = 50 = 103.0 Ω (.5)(.707 )

Z 02 = K Z 03 = (.5)(103Ω ) = 51.5Ω 2

1 1    R = Z 0  K +  = 50 0.707 +  = 106.1Ω K 0.707    Massachusetts Institute of Technology


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Wilkinson power divider The output impedances are

R 2 = Z 0K = 50(0.707 ) = 35.35Ω R 3 = Z 0 K = 50 / 0.707 = 70.72Ω •

50Ω

Z 03 = 103Ω

R = 106.1Ω

Z 02 = 51.5Ω

•


Z 0 = 50Ω

Z 0 = 50Ω


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Bethe-Hole Directional Coupler This 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.

3

y

Coupling hole

COUPLED

4

ISOLATED

x

THROUGH 2b 1

S 2

b INPUT

z

0



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Bethe-Hole Directional Coupler Let the incident wave at Port 1 be the dominant TE10 mode: Top Guide Port 3 (coupled)

E y = A sin Hx = −

πx

a

e − jβ z

πx A sin e − jβz Z 10 a

+ A10

− A10 − A10

Port 1 (input)

Coupling hole

+ A10

Port 2 (through)

Bottom Guide

Z

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

Z0 =

η0 1 − (λ 0 2a )

2

πx − jβz jπA cos e Hz = βaZ 10 a β = κ 0 1 − (λ 0 2a )2

= wave impedance, dominate mode,Ω

= phase constant rad/m

κ 0 = 2π λ 0 Massachusetts Institute of Technology


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Bethe-Hole Directional Coupler In the bottom guide the amplitude of the forward scattered wave is given by

jωA  2 πs µ 0 α m + − A10 = − ε 0αe sin 2 P10  a Z 10

2  2 πs  π π s 2  sin  + cos 2 2   a a β a  

while the amplitude of the reversed scattered wave is given by

µ 0α m ωA  2 πs − + A10 = − ε 0αe sin 2 P10  a Z 10 where

P10 =

ab Z 10

2  2 πs  π π s 2  sin  − cos 2 2   a a β a  

For round coupling hole or radius r0, we have

2 2 αe = r0 3 4 2 α m = r0 3

electric polarizability

magnetic polarizability



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Bethe-Hole Directional Coupler Let s = offset distance to hole

a s

We can then show that

sin

πs

a

=

(

λ0

2 λ20 − a 2

)

The coupling factor for a single-hole Bethe Coupler is

A C = 20log − (dB ) A10 and its directivity is − A10 D = 20log + (dB ) A10 Massachusetts Institute of Technology


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Bethe-Hole Directional Coupler Design procedure: 1. Use

2. Use

sin

πs

a

=

C = 20 log

(

λ0

2 λ20 − a 2 A − A10

(dB )

)

to find position of hole.

to determine the hole radius r0 to

give the required coupling factor.



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Bethe-Hole Directional Coupler Typical x-Band -20 dB coupler 0

C C ( dB) - 20 D (dB)

D

- 40 - 60 6

7

8

9

10

11

Frequency (GHz) 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. Massachusetts Institute of Technology


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Bethe-Hole Directional Coupler UPPER GUIDE





Port 4 (ISO)

B0

F0 B1

F1 B2

F2 BN

Port 1

B0

F0 B1

F1

F2

B2

FN

BN

FN



Port 4 Coupl Port 2 TRU

 n=0

n=1 d

n=2 d

n=N d

Let a wave of value 1∠0 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:



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Bethe-Hole Directional Coupler Port 1 (input) and Port 4 (isolated) Port 2 (through)

B (1) = B (4 ) = (2 ) = FTotal

N

− j 2 nβd B e ∑ n

n =0 − jNβ d

e

+

e − jNβd

N

∑ Fn

main incident wave = 0  n  N forward scattered waves ( ) j N d 3 − 2 β Port 3 (coupled) F =e Fn n =0 All of these waves are phase referenced to the n = 0 hole. N C = −20 log F (3) = −20 log F n dB n =0 N − j 2β nd B e n ( 4) B D = −20 log (2 ) = −20 log n = 0 N dB

∑

∑

( )

∑

F

∑ Fn

( )

n =0 Massachusetts Institute of Technology


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Bethe-Hole Directional Coupler We can rewrite this as N

N

n =0

n =0

D = −20 log ∑ B ne − j 2βnd + 20 log ∑ F n

= −C − 20log

N

− j 2βnd B e ∑ n

n =0

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

+ (n ) F n = A10 And the backward scattering from the hole is

− (n ) B n = A10



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Bethe-Hole Directional Coupler Now 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

Fn = But

ωε 0 =

+ A10

k0

2ωε 0 A  2µ 0  3 =−j 1 −  rn 2 3P10  ε 0 Z 10 

and

η0

∴ F n = K f rn

3

2 Z 10

=

η02  1 − (λ 2a )2  0  

=

η02

f 1−  c  

 f 

2

2   f − j 2 k0 A   where K f = 1 − 2 1 −  c f     3η0P10   

Let A = 1 v/m. Then

− j 2 k0 Kf = 3η0P10 Massachusetts Institute of Technology

  f 2  2 c  − 1   f  


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Bethe-Hole Directional Coupler Likewise the backward scattering coefficient is

2 k0 Kb = 3η0P10

  f 2  2 c  − 3   f  

and

βn = K b rn3

Note that Kf and Kb are frequency-dependent constants that are the same for all aperture. Thus, N

C = −20 log K f − 20 log

3 r ∑n

(dB )

n =0

N

D = −C − 20 log K b − 20 log ∑ rn3e − j 2βnd

(dB )

n =0



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Bethe-Hole Directional Coupler Consider 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? N

∑ Fn

Note that if the coupling C is specified, then

is known.

n =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: or

Bn = F n B n = −F n

N

∑ Fn

In either case, we have

D = 20log

n =0 N

∑ F ne − j 2βnd



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Bethe-Hole Directional Coupler N Thus, keeping the directivity D > Dmin is equivalent to keeping

− j 2βnd F e ∑ n

n =0

below a related minimum value. Let

ϕ = −2βd and w = e jφ = e − j 2βd We also introduce the function

g (βd ) =

N

− j 2β nd ⇒ g (φ) = F e ∑ n

n =0

∴ g (w ) =

N

n F w ∑ n

n =0

Thus we have

D = 20log

g (1) =

N

N

jnφ F e ∑ n

n =0

N Fn n = FN ∑ w = F N ∏ (w − w n ) F n =0 N n =1 N

g (1) g (w )

−C / 20 F = 10 ∑ n

Coupling factor (dB)



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Bethe-Hole Directional Coupler From the previous two equations we can deduce that

g

max

= g (1) × 10 − D min / 20

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

g (w ) ≤ g

max . 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 broadwall coupling with s = a/2.

Solution: From g (w ) = F N

N

∏ (w − w n )

, we have

n =1 6

g (w ) = F 6 (w − w n ) where w n = e − j 2βd = −1

(

)

∴ g (w ) = F 6 (w + 1)6 = F 6 w 6 + 6w 5 + 15w 4 + 20w 3 + 15w 2 + 6w + 1 Massachusetts Institute of Technology


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Bethe-Hole Directional Coupler Thus,

g (1) = F 6 (1 + 1)6 = 64 F 6 = 10 −15 20 = 0.1778

∴ F 6 = 0.00278 = F 0

By the binomial expansion we have

(w where,

+ 1) = 6

C n(6 ) =

6

(6 )w n C ∑ n

n =0

6! N! = (N − n )! n! (6 − n )! n!

is the set of binomial coefficients

Thus

F 5 = F1 = 6 F 6 = 0.01667 F 4 = F 2 = 15 F 6 = 0.04168 F 3 = 20 F 6 = 0.05557



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Bethe-Hole Directional Coupler We now can compute the radii of the coupling holed from

F n = K f rn

3

where

2   f − j 2 k0 A   Kf = 1 − 2 1 −  c f     3η0P10   

and 2  − j 2 k0   f c  Kf = 2  − 1 3η0P10   f  

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,

Z 10 = η0

Kf =

f  1−  c  f 

2

= 505.4Ω , P10 = 1.08 × 10 − 6 m 2 Ω

2 × 135.1 3 × 376.7 × 1.08 × 10 − 6


  4.30  2   2 − 1 = 24598   6.45    


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Bethe-Hole Directional Coupler The hole radii are: 13

 0.00278   r0 =   K f  

= 0.00483m = r6 ← 0.483cm

13

 0.01667   = 0.00878m = r5 ← 0.878cm r1 =    Kf  13  0.04168   = 0.011921m = r4 ← 1.192cm r2 =    Kf  13

 0.05557   r3 =    Kf 

= 0.0131m ← 1.31cm



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Bethe-Hole Directional Coupler 4.68 cm

2r0 = 0.966cm 2r1 = 1.756cm

a = 3.485cm

2 r3 = 2.62cm

2 r2 = 2.384cm

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

λ0

λg =

2

= 0.624 m

 4.3  1−    6.45  λg = 1.56cm . However, the center hole The nominal hole spacing is d = 4 has a diameter of 2.62 cm, so it would overlap with adjacent holes. We can 3λ

g increase the hole spacing to d = = 4.68cm 4 performance.

with no effect on electrical

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



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Bethe-Hole Directional Coupler We now plot the coupling and directivity vs. frequency 6

φ  φ  j φ j φ   − j j 6 φ   2 2 6 jφ  g (w ) = F 6 (w + 1) = F 6 e + 1 = F 6 e e + e 2  = F 6 2e 2 cos    2     

(

)

6

φ φ 6 ∴ g (w ) = 2 F 6 cos = 0.1778 cos 2 2

6

6

We then have

g (w ) 2πd D (dB ) = −20 log = −120 log cos g (1) λg where d=4.68 cm

λg =

λ0 1 − (λ 0 2a )2


=

(3 ×108 f ) 1 − (f c f

)2


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Bethe-Hole Directional Coupler 0 -10

D (dB)

-20

900 MHz -30 -40

Dmin -50

5.75

6.0

6.25 6.5 6.75 Frequency (GHz)

7.0

7.25

7.5

Note that the directivity is better than Dmin= -30 dB over a bandwidth of 900 MHz centered about 6.45 GHz. Massachusetts Institute of Technology


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Even-odd mode analysis



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Even-mode

Symmetry of port 2 and 3



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Odd-mode

symmetry of port 2 and 3,

open, short at bisection

⇒ S 32 = S 23

port 1 matched

⇒ S11 = 0



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Discussion 3dB Wilkinson power divider has equal amplitude and phase outputs at port 2 and port 3. 3dB Wilkinson power combiner



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Unequal power division Wilkinson power divider



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N-way Wilkinson power divider



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The quadrature (90°) hybrid • Branch-line coupler Port 2 and port 3 have equal amplitude, but 90° phase different Input

1

Z0

Output

Z0

2

Z0

2

λ/4

λ/4 4

Z0

Isolated

A1=1

1 1

•

B1 ⇒ 0dB∠0



1

•

2

1 4 1

•

1

•

•

B4


2

•

2

B2

0  −1  j [S ] = 2 1 3 ⇒ −3dB∠ − 90 •  0 1

1 B3

3

⇒ −3dB∠ − 90

2

•

Z0

Output

1

1

1

Z0

1


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j 0 0 1

1 0 0

j

0 1  j  0 43

Quadrature Hybrids We can analyze this circuit by using superposition of even-modes and 0ddmodes. 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.)

•

2

1 1/2

4

•

1

•

•

1

•

1

• 1

+1 2 1

2

1

1

• 3

2

 +1 2 1

1

• 1 •1 •1 • 1

Even Mode Excitation

1 1

•

1 2

•

•

1 -1/2

4

•

1

1

•

1

• 1

+1 2 1

•

1

• 3

2



1

Line of symmetry v=0, I=max

• 1• 1• •

1

•

1

2

•

Open circuit (2 separate 2-ports)

Line of symmetry I=0, v=max

1/2

2

Γ0 −1 2

• • • •

1

2 

•

1/2

1



1 1

• • • •

1

1

•

T0

•

1 2 1 short circuit stubs (2 separate 2-ports)

Odd Mode Excitation Massachusetts Institute of Technology


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Quadrature Hybrids We 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 Γo

and Te for the odd-mode excitation. Superposition:

1 1  B1 = Γe + Γo  2 2  1 1 Through B 2 = Τe + Τo  2 2  Reflected  1 1 waves Coupled B3 = Τe − Τo  2 2  1 1  Isolated B 4 = Γe − Γo  2 2  Input



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Quadrature Hybrids Consider the even-mode 2-port circuit: Port 1

λ 4

Input

λ8 1 2 λ8 open

Port 2 Coupled

open

λ 8 open circuit stubs by their admittance: π 1 + j tan z 4 y = lim L =j j π z L →∞ 1+ tan zL 4

We can represent the two

The λ 4 transmission line, with characteristic impedance 1 ABCD matrix

 0  j 2


j

2

has ab

2  0 


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Quadrature Hybrids Thus, the ABCD matrix for the cascade is

j   1 0 1 − 1 j  A B   1 0  0   2 = = C D   j 1  j 1 2  j − 1  e     j 2   0     λ 8 stub

λ 4 line

λ 8 stub

Using the conversion table (next slide) to convert [S] parameters (with Z0=1 as the reference characteristic impedance).

B j j 1 1 + CZ 0 + D = − + + − Z0 2 2 2 2 B − CZ 0 − D A+ Z0 ( − 1 + j + j + 1) 2 Γe = S11 = = =0 B ( ) −1+ j + j −1 2 + CZ 0 + D A+ Z0 2 2 1 (1 + j ) Τe = S 21 = = =− B 2 A+ + CZ 0 + D (− 1 + j + j − 1) 2 Z0 1 (1 − j ) Similarly for odd mode we have: S 11 = Γ0 = 0 and S 12 = Τ0 = 2 denominator


=A+


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Conversion between two-port network parameters S

Z

Y

ABCD

D. Pozar Massachusetts Institute of Technology


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Quadrature Hybrids Therefore we have

Scattered wave voltages

B1 = 0  B 2 = −   B3 = −  B = 0  4

(port 1 is matched)

j 2 1 2

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.



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

Z0

Z01

Z0

Through Output

1

2

Z02

λ 4

Z02

4

3 Isolated

Z0

Z01

Z0

Output (coupled)

λ 4 We can show that if the condition

Z 02 Z 01 Z 0 = Z0 1 − (Z 01 Z 0 )2 is satisfied, then port 1 is matched; port 4 is decoupled from port 1.



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Quadrature Hybrids Single section branch-line coupler

B1 = 0  Z B 2 = − j 01 scattered wave  Z0 voltages  B = − Z 01 Z 0  3 Z 02 Z 0  isolated B 4 = 0

matched

∠0

∠ − 90 ∠ − 180 (matched)

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

  1 (dB ) C = 10log  2 1 − (Z 01 Z 0 )  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.



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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 substrate thickness, a dielectric constant of 2.2, and that the operating frequency is 1.0 GHz.

Solution:

  1 C = 10 log  =6 2 1 − (Z 01 Z 0 ) 

(dB )

∴ Z 01 Z 0 = 0.8653 ⇒ Z 01 = 43.27Ω Z 02 Z 01 Z 0 = = 1.7263 ⇒ Z 02 = 86.31Ω 2 Z0 1 − (Z Z ) 01

λ≅

0

λ0 30 = = 20.226 cm εr 2.2

l=

λ = 5.0565 cm 4

εr −1   w 2  0.61     = 3.081 = B − 1 − ln(2B − 1) + ln1B − 1 + .39 − ε r   d π 2ε r  

⇒ w  = 0.487 cm Massachusetts Institute of Technology


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52

Quadrature Hybrids

For Z 01 = 43.27Ω, w 1 = 0.601cm w2 8e A For Z 02 = 86.31Ω , = 2A = 1.1675 d e −1

w 2 = 0.185cm

0.4868 cm

Input (0 dB)

Through (-1.26dB)

0.6008 cm

Z01

Z0 5.0565 cm

Z0

Z02

Z02

0.1845 cm

Z01

Isolated

Z0

Z0

Coupled (-6dB)

5.0565 cm

With 0 dB power input at the upper left arm, the power delivered to a matched load (in ) at the through arm is

P2 (dB ) = −10 log

P1 1 10 log = − B 2B 2* P2(out ) 2

2  Z0  50    = −10 log  = −10 log   = −1.26dB Z 43 27 .    01  Massachusetts Institute of Technology


USPAS 2010

53

Coupled Line Directional Couplers These 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

0V

even mode 0V short ckt

0V

odd mode 0V



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54

Theory of Coupled Lines C12

•

C22

C11

Three-wire coupled line

C12 C11 ,C22

• •

Equivalent network

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



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55

Theory of Coupled Lines



In the even mode, E 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

Z 0e =



L 1 = C e νC e

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

•

2C12

•

C11

2C12

The effective capacitance between either strip conductor and ground is

•

C0 = C11 + 2 C12 = C22 + 2 C12 C22

•


Z 0o RF Cavities and Components for Accelerators

L 1 = = C o νC o USPAS 2010

56

Theory of Coupled Lines Example: 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 Ω and Z0o=50 Ω. Find the necessary strip widths and spacing.

εr =2.8

Solution: b = 0.50 cm, εr = 2.8, Z0e=100 Ω, Z0o=50 Ω

∴ ε r Z 0e = 167.3  even

ε r Z 0o = 83.66 

b=.5cm

odd

from the graph, s/b ≈ 0.095, W/b ≈ 0.32

W

W s

S=0.95 b = 0.095 ×0.5 = 0.0425 cm W=0.32b=0.32×0.5=0.16 cm

0.16cm

0.5cm



0.16cm

εr=2.8

0.0425cm

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57

Waveguide magic-T

∆

4

2 3 1

Σ

4

2

4

3

3

2

Input at port 1

Input at port 4

Port 4:0

Port 1:0

Ports 2 and 3: equal amplitude and phase

Ports 2 and 3: 1800 phase difference



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58

Even Mode ABCD Analysis Consider the equivalent circuit for the even mode:

1 2

Γe

λ 4 2

1

2

λ

2

3λ 8

2

8

Τe

At port 1, the admittance looking into the λ/8 o.c. stub is

y S 1 = y 0 ( j tan βl ) =

j 1  2π λ  j = tan ⋅   λ 8 2 2

o.c. o.c. The ABCD matrix of a shunt admittance yS1 is A=1+Y2/Y3

Y3

B=1/Y3

Y1

Y2

C=Y1 + Y2+Y1Y2/Y3

A B   1 C D  =  j  1  2

0 1 

D=1+Y1/Y3 Massachusetts Institute of Technology


USPAS 2010

59

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

y S 2 = y 0 ( j tan βl ) =

j 1  2π 3λ  j = − tan ⋅   λ 8  2 2

The ABCD matrix of this shunt admittance is

0 1 

Zo = 2

The ABCD matrix is

 cos βl A B  C D  =  jy sin βl 2  o 

where

jZ o sin βl 

ys2

Zo = 2

quarter-wave section

cos βl 

βl = 2π λ ⋅ π 4 = π 2

Zo = 2,y o = 1 2


ys1

{

 1 A B  C D  = − j   3  2

λ 4

 0 A B  ∴ = j  C D  2  2 


USPAS 2010

j 2  0  

60

Even Mode ABCD Analysis We can now compute the ABCD matrix of the even mode cascade

λ 4

1

λ

2

2

2 2

8

3λ 8

o.c. o.c.

 1 A B  C D  =  j  e  2

0  0  j  1   2

j 2 1

 1 = j  2

0  1  j  1   2

j 2


 j 0  −  2

 = 1 0  j 2 


0 1 

j 2  −1  USPAS 2010

61

Odd Mode ABCD Analysis

The input admittance to a shortcircuited lossless stub is

y in = y 0 (− j cot βl ) Thus the input admittance to the s.c. λ/8 tub is

1 2

Γo

j 1  2π λ  y S1 =  − j cot ⋅  = − λ 8 2 2

λ 4

1

λ

2

8

2

2 2

yS2

s.c.

0 1 

j 1  2π 3λ  =  − j cot ⋅  = + λ 8 2 2


3λ 8

s.c.

ABCD matrix:

A B   1 C D  =  − j  1  2

Τo

(Input admittance to s.c. 3λ/8 stub)


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62

Odd Mode ABCD Analysis and

 1 A B  C D  =  j   3  2

0 1 

So the ABCD matrix of the odd-mode cascade is

 1 A B  C D  =  − j  o  2

0  0  j  1   2

 −1 = j 2

j 2

Γe = − j Γo = j

0   2

0 1 

 1 

2 , Τe = − j 2 , Τo = − j


j 2 1  j

2 2


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63

Excitation at Port 4

We have derived the ABCD matrices for the Even (e) and Odd (o) modes:

 1 A B  C D  =  e  j 2 

j 2  −1 

and

 −1 A B  C D  =   o  j 2

j 2  1 

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

1

2

1

Te

2

Γe Even Mode

O.C.

Γe Odd Mode

S.C. S.C.

O.C.



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Excitation at Port 4 Even mode:

Γe = S 22 =

 1 A B  =  C D   e  j 2 −A +B − CZ o + D

Zo

A + B Z + CZ o + D

j 2  −1 

Γe =

j −1+ j 2 − j 2 −1 −2 = = j2 2 1+ j 2 + j 2 −1 2

Τe =

j 2(− 1 + 2 ) 2 = =− 1+ j 2 + j 2 −1 j 2 2 2

o

Τe = S12

2(AD − BC ) = A + B Z o + CZ 0 + D

 −1 A B  =  C D   o  j 2

Odd mode:

j 2  1 

j 1+ j 2 − j 2 +1 2 = =− −1+ j 2 + j 2 +1 j 2 2 2 j 2(− 1 + 2 ) 2 = = =− −1+ j 2 + j 2 +1 j 2 2 2

Γo = S 22 =

Τo = S12

Excitation at Port 4 is expressed as :

Even

V 2+ = 1 2

V 4+

=1 2


Odd

V 2+ = −1 2

V 4+ = 1 2


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65

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

Γe =

Γo = −

j

j

Τe = −

2

j 2

Τo = −

1 1 2 2 1 1 B 2 = Γe − Γo 2 2 1 1 B3 = Τe + Τo 2 2 1 1 B 4 = Γe + Γo 2 2

B1 = Τe − Τo

2

Output waves

j 2

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

[B i ]4 Massachusetts Institute of Technology

0    j 2   = 2 − j   0  


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66

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