Microwave Magnetics 9 - Sharifee.sharif.edu/~mwmagnetics/Microwave Magnetics_9.pdfY-junction circulator 3 Y-junction microwave circulators Most widely used microwave component based

Microwave Magnetics

Graduate CourseElectrical Engineering (Communications)2nd Semester, 1389-1390Sharif University of Technology

Y-junction circulator 2

General information

� Contents of lecture 10:

• Y-junction microwave circulators

� Field analysis

� Resonances of an isolated system

� The Y-junction circulator

� The Green’s function

� The impedance matrix

� The circulation condition

� Basic circulator design

� Field profile

� Remarks


Y-junction microwave circulators

� Most widely used microwave component based on magnetic (ferrite) materials are Y-junction circulators

� Circulators based on Faraday effect in circular waveguides were discussed before

� The device passes the signal to the next port in a cyclic fashion, without the reverse being true

1

3 2� Planar circulators: 3-port junctions

are the most popular



� In terms of scattering parameters, a cyclic symmetric, 3-port network (or device) is described by

1

3 2� � �

� � �

� � �

� ��

S

� If the device is perfectly matched:

0

0 0

0

� �

� � �

� �

� ��

S



� If the device is lossless then

0 0

0 0 0

0 0

�

� �

�

� ��

S 1� �

� We will study the realization of this device using vertically magnetized ferrite disks as in the paper:

H. Bosma, "On stripline Y-circulation at UHF," IEEE Transactions on Microwave Theory and Technique, vol. MTT-12, pp. 61-72, Jan. 1964.



� Planar realizations based on a magnetic (ferrite) disk magnetized perpendicular to the disk

� Both stripline and microstrip versions have been in use, but we limit ourselves to the microstrip version

ferrite

Ground plane

ferrite

Ground plane

Ground plane

ferrite

0 0,M H


(i) Field analysis

� Maxwell equations in cylindrical coordinates:

Ferrite disk 0 0,M H�

rz

� �

1

1 1

jr z

jz r

r jr r r

�

�

�

��

�

��

��

� ��

� � ��

��

� �

zr

r z

rz

eeb

e eb

ee b � �

0

0

0

1

1 1

jr z

jz r

r jr r r

�

�

�

��

��

��

��

� ��

� ��

��

� �

zr

r z

rz

ehe

h he

hh e


(i) Field analysis

� We again assume uniformity of fields inside the disk in the vertical (z-) direction which is a good assumption if the disk is thin compared to its lateral dimensions:

� �

1

1 1

jr

jr

r jr r r

�

�

��

�

��

��

��

��

��

� �

zr

z

rz

eb

eb

ee b � �

0

0

0

1

1 1

jr

jr

r jr r r

�

�

��

��

��

��

��

� ��

��

� �

zr

z

rz

he

he

hh e


(i) Field analysis

� Inside the magnetic material we have

0

0

0

0 0 1

a

a

j

j� �

� �

� � �

� � � � � ��

r r

z z

b h

b h

b h

0 0,M H

�r

rh�h


(i) Field analysis

� This leads to two decoupled sets

� � 0

1

1 1

jr

jr

r jr r r

�

�

��

�

��

��

��

��

��

� �

zr

z

rz

eb

eb

hh e � �

0

0

1

1 1

jr

jr

r jr r r

�

�

��

��

��

��

��

� ��

��

� �

zr

z

rz

he

he

ee b


(i) Field analysis

� The 2nd set has no solution which can satisfy boundary conditions on the metal disk and ground plane

� 1st set leads to

2202 2

1 10r k

r r r r��

� �

� �� z z

z

e ee

� Solutions have to be periodic functions of the angle �There, try solutions of the type

� � � �, ( ) expnr f r jn� �� ze


(i) Field analysis

� Result: � �2

2 2 2 202

( ) ( )( ) 0n n

n

f r f rr r k r n f r

r r��

� ��

� �

( ) ( ) ( )n n nf r AJ k r BY k r� �� 0k k ��

� Finite field at the disk center requires 0B �

� �, ( ) exp( )n nr A J k r jn� �� ze

ze


(i) Field analysis

� Magnetic field:

� � � �

0

0

1

exp( )

a

nn an

j

r r

nJ k rA kJ k r jn

k r

��

��

��

�

��

� �

� ��

� ��

� �

z zr

e eh

� � � �

0

0

1 1

exp( )

a

nn an

jj r r

J k rA kn J k r jn

j k r

�

��

��

��

�

��

� �

� ��

� ��

� �

z ze eh


(i) Field analysis

� Note that in each mode n all the fields contain the exponential factor

exp( )jn��

� If we return to time domain, we observe that each mode represent a field rotating in the counter clockwise (n>0) or clockwise (n<0) direction

� Example: � �, ( ) exp( )nr J k r jn� �� ze

� � � �, , Re exp( )

( ) cos( )z

n

e r t j t

J k r j t jn

� �

� ��

�

� �ze


(ii) Resonances of an isolated system

� Let us first consider an isolated system. We again use the magnetic-wall boundary conditions by assuming that

• The ferrite disk is thin compared to its diameter

• There is negligible electric current on the top metal disk flowing perpendicular to the outer edge of the disk

�

Jr

J�

( , ) 0R� � �h

R

0� �h



� The solutions to this problem (without any sources or connections to the external ‘world’) define the resonances of the system

� Let us now impose the magnetic wall boundary condition on the solutions found

( , ) 0R� � �h

� � � ��

na

n

k R J k Rn

J k R

��

� �

�

��



� 1st case:

� �2 2

0 0 0 2 2M Hk k

� � ��

� ��

� ��

�

or 0 : realM H k� � � � � ��

�� M H� ��

k�

�



� �0 0J k R��

� Solution for the rotationally symmetric non-rotating mode n=0:

0 0: zero's of ( )pk R J z��

01��

k�

01 / R�02 / R�

03 / R�

01��



� To each zero correspond two resonances

� Electric field distribution is same at both

00

prJ

R

��

� �0pze

00

0

park

JR

��

�

�

� ��

0prh 0

00

p rkJ

j R�

�

��

�

� ��

0ph

� Magnetic field polarization will be different for the two frequencies since and will be different �� /a� �



� But n=0 is not the most interesting case. For other values of n, however, we have to solve

� � � ��

na

n

k R J k Rn

J k R

��

� �

�

��

� For each n, we have a number of solutions (as before indexed by p)

� But + n and – n will not resonate at the same frequency because the above equation depends on the sign of n

� Note: right hand side does not depend on the sign of n .

( ) ( 1) ( )nn nJ z J z� � �



� If the ratio would have been zero (no off-diagonal permeability element), the resonance frequencies would have been the same

� For a non-zero, but small , the two modes will have different but close resonance frequencies.

� Then the roots of the two equations are

/a� �

/a� �

2 2

npanp

np

k R nn

��

� ��

�� 0n npJ ��

� When the roots are:0a�� 0

npk R ��



� To find the resonance frequencies let us expand the equation

/ 0a� � �

� �0 npresk

R

��

2 2

( )( )

( )npa

npnp

k R nn

��

� � ��

� We expand around the solution when . That solution is defined by

� � � �kc

��



� Expansion:

� � � �00

0 0 02 2

( )resres

npres res res

np

dk n d

d R n d ��

� ��

� � ��

� ��

� � ��

� � � �kc

�� ( )

( )( )

a� ��

� ��

� � 0

00 ( )

/ /res

np resres res

np

c

dk d c d d�

� ��

� � ��

�

� ��

2 2

npnp

np

nc

R n

�

��

�



� Manipulating the terms: 12

dk k d

d d

��

�

� ��

� �

� ��

� ��

2

22 2 2

2 2

2M H M

M H

M Ha

M H

d

d

� � � ��

� � ��

�

� �

��

� ��

��

� ��

� ��

2 2 2 2

2 2 22 2

M ad

d

� � � � � ��

� � � ��

� �

��

� ��

��



� Numerical example: magnetic disk with

2 GHz, 6 GHz, 4 H Mf f ��

1, 1n p� � �

1, 1n p� � �

(mm)R

(GHz)resf 1, 2n p� � �

1, 2n p� � �

(mm)R



� 2nd case: 0 : imaginaryM H k� � � � ��

� � � ��

� � � ��

n na

n n

k R J k R q R I q Rn

J k R I q R

��

� � � �

� �

� �� k jq� ��

� The right hand side is always positive. The left hand side is negative for positive n and positive for negative n.

� Therefore, only solutions with n < 0 are possible: these are modes rotating in the clockwise sense



� Besides, fields drop exponentially towards the disk centre: they are concentrated near the edge of the disk

r

� These are similar to the edge-guided modes in a microstrip built on a vertically magnetized ferrite substrate (why?)

� �, ( ) exp( )n nr A I q r jn� �� ze


(iii) The Y-junction circulator

� In order to see how such a device can be built, we should now add the junctions to the system

� These are three microstrip lines connected to the metal disk on top of the magnetic (ferrite) disk

ferriteJ

� Electric currents will now flow through these microstrips



� To analyze this problem we again impose magnetic wall boundary conditions on the vertical wall of the ferrite disk, but only outside the microstrip junctions

� Below the microstrips we assume a given distribution of on the ferrite surface

J

Magnetic wall

�h

�2h

�1h

�3h

0� �h

0� �h

0� �h



� Returning to the general solution:

� �, ( ) exp( )n nn

r A J k r jn� ��

��

� ��ze

� � � � � �0

, exp( )n an n

nJ k rkr A J k r jn

k r

��

��

��

��

� ��

� ��rh

� � � � � �0

, exp( )nan n

J k rkr A n J k r jn

j k r�

��

��

��

��

� ��

� ��h



� We next impose the following boundary condition:

� �

� �

( )

2 2( )

3 3,

2 2( )

3 3

0 elsewhere

R

�

�

�

�

�

� �

� ��

��

� �

�

� ��

� � � � � ��

�

1

2

3

b

h

h

h

h

h

�2h

�1h

�3h

�

2�



� Fourier analysis:

� �

� � � �0

exp( )

2nna

n

jn dj

AJ k Rk

J k R nk R

�

��

� � ��

�

� �

��

�

� � ��

� �

� bh

� ��

� � � �0

( ) exp ( )

,2

n

n nan

J k r jn dj

rJ k Rk

J k R nk R

�

��

� � � ��

��

�

��

��

�

� � ��

� �

��

b

z

h

e


(iv) The Green’s function

� The electric field at the edge (why do we need this?) can be written as

� � � � � �,R G d�

��

� � � � ��

� � �� bze h

� � � ��

� �

0exp ( )

2 n n a

n

jnj RG

k R J k Rn

J k R

� ��

� ��

��

��

�

� ��

��

�

Green’s function



� : response of the electric field at the disk

edge at any angle � to a �-function magnetic source at the angle �0

� �0G � ��

��-function

source

( , )R �ze�� 0� � � � �� bh

� � � � � � � �0,R G d G�

��

� � � � � � ��

� � �� bze h



� �

� ��

� �

0 ( );

2exp ( )

( ) ( )

( )n n a

n

j RG

jn

k R J k Rn

J k R

��

��

� � ��

�

�

��

�

��

� ��

��

� Consider the frequency dependence of the Green’s function which is brought about by , , ,ak� � ��

� Poles of Green’s function in frequency are exactly the resonances of an isolated disk!



� Let us inspect the Green’s function in more detail: G(�)is, in fact, the electric field at the disk edge at an angle

�, generated by a source at �0 = 0. It can be written as

� �

� � � � � ��

0

0

221

1

2

cos / sin 2

/n a

n n a

j RG

U

U n j n n

U n

��

�

� � � �

� �

�

�

�

��

��

� ��

� � � ��

nn

n

k R J k RU

J k R� �

�

��



� �0

10

cos10 ( ) 2

2an n

nj RG

U U

��

�

��

�

��

� ��

� In a disk without any gyrotropic properties, excited field

will be symmetric with respect to the source at �0 = 0

-function source

1n �1n �

�



� But in our case this is not true. The imaginary part of the electric field is symmetric but the real part is anti-symmetric

� � � ��

022

10

cos1Im ( ) 2

2 /n

n n a

U nRG

U U n

��

� � �

��

�

� ��

�

� � � � � ��

022

1

2 / sinRe ( )

2 /a

n n a

n nRG

U n

� � ��

� � �

��

�

��

�

1n �

�


(v) The impedance matrix

� Neglecting the fringe, the magnetic field between the microstrip and the ground plane is approximately constant and transverse to the microstrip

� Hence we assume that is constant in the junction regions (better approximation: use the true magnetic field distribution below the microstrip)

J h

�h



� We approximate this field by

1I

h

, 1, 2,3k kIk

w� � � �h

2I3I

w

:kI Total current through the k-th microstrip



� Approximation justified as propagation mode of microstrip is quasi TEM

� Electric and magnetic fields then studied separately using electrostatics and magnetostatics

� Magnetic field of microstrip ~ field of a (~uniform) current sheet (microstrip) and its (negative) image in the ground plane

ˆ� � ximage

Jh

2

x

J

-J

x

ˆ� � xMS

Jh

2

ˆ ˆI

w� � � �x xMS imageh +h J



� Electric field at any point on the edge given by

� �

� �

� �

� �

1

2 / 32

2 / 3

2 / 33

2 / 3

,R

IG d

w

IG d

w

IG d

w

�

�

�

�

�

� � �

� � �

� � �

�

��

��

��

� ��

� ��

�

� ��

� ��

� ��

�

�

�

ze

��

2

3

��

�

2

3

��

2

3

��

2

3

��

1I

2I3I



� Considering the complex power, voltage on each port is found to be the “averaged” voltage on the port

1V

ze

3V

b

2V

ze

2

1

( , )2

k

k

k

bV R d

�

�

� �� ze

2k�1k�



� For simplicity assume that the microstrips are narrow compared to the disk circumference:

2w R� ��

� � 2 ( )G d G� � � ��

��

� ��

��

2

3

��

�

2

3

��

2

3

��

2

3

��

w

R� �23

2

3

2 ( 2 / 3)G d G

�

�

� � � � ��

��

� ��

� �23

2

3

2 ( 2 / 3)G d G

�

�

� � � � ��

� ��

� ��



� As a result: (use w = 2� R)

� � � � 31 2 2 2,

3 3

II IR G G G

R R R

� ��

� � � �ze

1 ( , 0)V b R� � ze

� �2 , 2 / 3V b R �� ze

� �3 , 2 / 3V b R �� ze

� Applying the same approximation to voltages:



� The impedance matrix: note that

� �1 1 2 3

2 20

3 3

b b bV G I G I G I

R R R

� ��

� �2 1 2 3

2 20

3 3

b b bV G I G I G I

R R R

� ��

� �3 1 2 3

2 20

3 3

b b bV G I G I G I

R R R

� ��

� � � �2G m G� � ��



� � � � � ��

0 2 / 3 2 / 3

2 / 3 0 2 / 3

2 / 3 2 / 3 0

G G Gb

G G GR

G G G

� �

� �

� �

��

Z

� We have

� �ki ikZ Z�

� �

� This result is more general and not limited to the approximations made (why?)

� � � � � �0 : imaginary 2 / 3 2 / 3G G G� ��


(vi) The circulation condition

� Now take 1 to be the input port. Let 2 and 3 be connected to infinite transmission lines. Replace the lines by their characteristic impedances.

0Z 0Z



� � �

� � �

� � �

� ��

Z

1I

3I 2I

3V2V0Z 0Z

� For circulator operation we should have

2 30 , 0I I� �

� Use the notation

� �

� �

� �

0

2 / 3

2 / 3

bG

Rb

GRb

GR

�

� �

� �

�

� �

�



1I

3I 2I

3V2V0Z 0Z

� Circulation condition:

2

0Z�

��

� �

� Input impedance:

2 2 21

01

in

VZ Z

I

� � ��

� � ��

� Matching condition:2 2

0 02inZ Z Z� ��

� � � �



� Since we consider a lossless system:

: imaginaryj� �

� � �

�

� �

� � �

� � �

� � �

� ��

Z

� �2

0j Z�

��

�

� �� 2

2

02 Z� ��

�

��

� 2nd condition is just the real part of the first: it is automatically satisfied if the 1st condition holds



� � 0cos 3 Z� � �exp( )j� � ��

� �sin 3� � ��

� � � �0 2 / 3b b

j G GR R

� � �� Remember that

� ��

022

10

cos12

2 /n

n n a

U nRj

U U n

��

��

�

� ��

�

� � � ��

022

1

2 / sin( )

2 /a

n n a

n nRG

U n

� � ��

� � �

��

�

� ��

�

Circulation condition



� �0

2210

21

2 /n

n n a

b U

U U n

��

� � �

��

�

� ��

�

� � � ��

� ��

022

1

022

10

2 / sin 2 / 3

2 /

2 cos 2 / 31

2 /

a

n n a

n

n n a

n nb

U n

U nbj

U U n

� � ��

� � �

��

��

�

��

�

� � ��

� ��

�

�

� � � ��

nn

n

k R J k RU

J k R� �

�

��


(vii) Basic circulator design

� � 0cos 3 Z� � � � �sin 3� � ��

� The circulation (and matching) conditions satisfied if

� A possible solution to this problem provided by

0: real 0 or 0, Z� � � � ��

� As a result

0j� �

� � ��

� �

� � � �

0

0

0

� �

� �

� �

��

Z



� But how can we realize this situation? Consider

� � � ��

� ��

022

1

022

10

2 / sin 2 / 3

2 /

2 cos 2 / 31

2 /

a

n n a

n

n n a

n nb

U n

U nbj

U U n

� � ��

� � �

��

��

�

��

�

� � ��

� ��

�

�

� �0

2210

21

2 /n

n n a

b U

U U n

��

� � �

��

�

� ��

��

� �n

nn

k R J k RU

J k R� �

�

��



� Each quantity is written as a sum over contributions by different modes n. Each term itself is a function of frequency. For example (n >0):

� �

� � � ��

� � � ��

22

2 1 1

/ //

1 1

/ /

n

n a n an a

n na a

n n

U

U n U nU n

k R J k R k R J k Rn n

J k R J k R

� � � ��

� � � ��

� �

� ��

� ��

� �

� Poles of the right and left term: resonances of the +n and –n modes, for different values of p



� We consider the case where both resonances corresponding to the left- and right turning fields exist.

� This is only possible if

0 or M H� � � � � ��

� We restrict ourselves to this situation.

� Since the lower range is limited in frequency, usually the second region is used. In this range, note that

0aM H

��

��



� Assume again that is small and that the two resonances are relatively close.

� Then consider one particular value of p. (Usually the lowest value 1).

� Let us call the corresponding resonance frequencies

/a� �

� In the absence of gyrotropic term resonance occurs when

0 0 ( ) npres resk

R

��

res� ��

/ 0na

n

k R J k Rn

J k R� ��

�

��

� � 0n npJ ��



� Expansion in frequency:

� � � ��

n a

n

k R J k RF n

J k R

��

� ��

�

��

� � � �0

0 0( )res

res res

dFF F

d �

� � � ��

� ��

� � � �0 0res resF n� � ��

00 0

2 2

resres res

np

np

ndF dk dR n

d d d ��

� ��

��

a��

�



� Expansion in frequency:

� � � �00

2 20 0( )

resres

npres res

np

n dk dF n R n

d d ��

� ��

� � ��

�

� ��

� ��

� At resonance: � � 0resF ��

� �� 00

02 2

0resres

resnp

np res res

nn dk dR n

d d ��

� ��

��

� ��

��

� � � �0

0( )

res

resres res

nF

� ��

� ��

� ��



� Similarly:

� � � �0

0( )

res

resres res

nF

� ��

� ��

� ��

� Bear in mind that these approximations are only valid when the frequency is close to the two resonances which, are also close to each other

� � 0resF ��



� Collecting the results:

� �

� �

22

0 0

0

2 1 1

/

1

n

n a

res res res res

res resres

U

F FU n

n

� �

� � � ��

� �

� �

� �

� ��

� ��

� Note that in the range we have M H� � ��

0 0res res� �� 0 0res res� ��



� Let us plot this function as function of frequency

� �0 0

0

1 res res res res

res resresn

� � � ��

� �

� �

� ��

res��res��

� Between the resonances, it becomes zero when

0res� ��



� In a similar fashion:

� �

� �

22

0 0

0

2 / 1 1

/

1

a

n a

res res res res

res resres

n

F FU n

n

� �

� �

� � � ��

� �

� �

� �

� ��

� ��

res��res��

� This term remain nearly flat when

0res� �� 0

1 1 2

resF F n� ��

� � �



� Now consider the mode n = 1 as an example. This is the mode usually used for Y-junction circulator design.

� Assume that we are operating close to the (first) left-and right resonances of this mode.

� Also assume that the other resonances (other n’s) are far away in frequency and do not contribute significantly

� � � ��

� ��

10 02 22 2

1 1

2 / sin 2 / 3 2 cos 2 / 3

2 2/ /a

a a

Ub bj

U U

� � � ��

� ��

� �

� �0 1

221

2

2 /a

b U

U

��

� � ��

�



� Now, when

� �0 2sin 2 / 3

2

b��

� ��0� �

0res� ��

� Satisfied by choosing right ferrite thickness (b), or changing material parameters in order to tune or .

� Circulation condition satisfied when

� �0

000

( )sin 2 / 3

( )res

res

bZ

��

��

��



� Disk radius is mainly determined by operation frequency

� General remark: since we considered the crossing through zero of the parameters, it is not strictly correct to neglect the terms due to other values of n.

� They can be included in the calculation, but the overall picture does not change much.

0( )np

res

Rk

�

��

�ferrite

J

R

b


(viii) Field profile

� How do fields look like at the circulation frequency?

� From the circulation condition it follows that

1I

3I 2I

3V2V0Z 0Z

1 2 3, 0I I I I� � �

� We next calculate the fields using the general expressions in terms of the (Fourier series) solution



� �

� � � �

� � � �

0

0

exp( )

2

1 exp( 2 / 3)

2

nna

n

nan

jn dj

AJ k Rk

J k R nk R

j I j nJ k Rk R

J k R nk R

�

��

� � ��

��

�

� �

��

�

�

��

�

� � ��

� �

��

� �

� bh

� �, ( ) exp( )n nn

r A J k r jn� ��

��

� ��ze



� � 1 1 1 1, ( ) exp( ) ( ) exp( )r A J k r j A J k r j� � �� ze

� As before, assume we are working close to the

resonances of the n =�1 modes. Neglecting other terms

� � � �0

11

1

1 exp( 2 / 3)

2 a

j I jA

J k Rk RJ k R

k R

��

�

�

��

�

��

� �

� � � �0

11

1

1 exp( 2 / 3)

2 a

j I jA

J k Rk RJ k R

k R

��

�

��

� ��

�

� ��

� �



� �

� ��

� �

� ��

� �

0 1

1

1 1

1 1

( ),

2 ( )

1 exp( 2 / 3) exp( ) 1 exp( 2 / 3) exp( )

a a

j I J k rr

J k R

j j j j

k R J k R k R J k R

J k R J k R

��

�

� � � ��

� �

�

� � � �

� �

� � �

� ��

ze

� � � �0

0

res

resres res

F� �

� ��

��

�

� � � �0

0

res

resres res

F� �

� ��

��

�



� � � �

� �

� �

0 10

1

0

0

( ),

( )2

1 exp( 2 / 3) exp( )

1 exp( 2 / 3) exp( )

res

res res

res

res res

res

j I J k rr

J k R

j j

j j

��

��

� ��

� �

� ��

� �

� �

�

�

�

�

�

� � �

� ��

��

� � � ��

ze

0res� ��

� � � � � � � �0 10

1

( ), sin sin 2 / 3

( )res

I J k rr

J k R

��

��

�

� � � �� ze



� � � ��

0 10

1

( ),

( )

sin sin 2 / 3

res

I J k rr

J k R

��

��

� � �

� �

�

� �

� ��

ze

Due to current at port 1 Due to current at port 2

I3 0I �

�

I

� � � � � �0 10

1

( ), sin / 3

( )res

I J k rr

J k R

��

��

�

� � �ze



� � � �

0 1

11

1 1

( )

( ) cos 2 / 3 ( ) sin 2 / 3

a k Ij

r r J k R

J k rj J k r

k r

��

� � � � �

�

� �

��

�

� ��

� ��

� �

z zr

e eh

� � � �

0 1

11

1 1 1

( )

( ) cos 2 / 3 ( )sin 2 / 3

a jk Ij

j r r J k R

J k rj J k r

k r

�

��

� � � � �

�

� �

��

�

� ��

��

z ze eh


(ix) Remarks

� Note that to arrive at the final design we made a number of assumptions and approximations

• We assumed uniformity of fields in the vertical direction

• We adopted the magnetic wall boundary condition

• We approximated the magnetic field of at the microstripports by a constant field (can be improved by using the true distribution)

• We assumed narrow microstrip lines (not a big issue, can be improved by performing the actual integrals)


(ix) Remarks

• We derived the circulation condition, but just looked at a particular solution (this is restrictive, in fact some papers have shown that choosing a different solution leads to higher bandwidth)

• We choose frequencies at which �� > 0 so that there are two close resonances corresponding to right- and left-turning fields (This is, in fact, not necessary: designs can be

made also when �� < 0, then there is just one resonance, but different n’s can be combined. The resulting designs have a much wider bandwidth.)


(ix) Remarks

• We just considered the mode n=1. Better calculations should include all modes.

��

Documents

Microwave Magnetics 9 - Sharifee.sharif.edu/~mwmagnetics/Microwave Magnetics_9.pdfY-junction circulator 3 Y-junction microwave circulators Most widely used microwave component based