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ENE 428Microwave Engineering

Lecture 8 Rectangular waveguides and cavity resonator

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TM waves in rectangular waveguides• Finding E and H components in terms of z, WG geometry,

and modes.

From

Expanding for z-propagating field for the lossless WG gets

where

2 2 2( ) 0, xy z u zE E

2 22 2

2 2 ( ) 0z zu z

E EE

x y

( , ) j zz zE E x y e

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Method of separation of variables (1)

Assume

where X = f(x) and Y = f(y).

Substituting XY gives

and we can show that

2 22 2

2 2 ( ) 0u

d X d YY X XYdx dy

( , )zE x y XY

2 22 2

2 2

1 1. u

d Y d XY Xdy dx

for lossless WG.

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Method of separation of variables (2)

Let

and

then we can write

We obtain two separate ordinary differential equations:

22

2

1 x

d XX dx

22

2

1 y

d YY dy

2 2 2 . u x y

22

2 0 x

d XX

dx

22

2 0 y

d YY

dy

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General solutions

1 2

3 4

( ) cos sin

( ) cos sin

x x

y y

X x c x c x

Y y c y c y

Appropriate forms must be chosen to satisfy boundary conditions.

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Properties of wave in rectangular WGs (1)

( 0,1,2,3,...) xa m m

1. in the x-direction

Et at the wall = 0 Ez(0,y) and Ez(a,y) = 0and X(x) must equal zero at x = 0, and x = a.

Apply x = 0, we found that C1 = 0 and X(x) = c2sin(xx).Therefore, at x = a, c2sin(xa) = 0.

. xma

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Properties of wave in rectangular WGs (2)

( 0,1,2,3,...) yb n n

2. in the y-direction

Et at the wall = 0 Ez(x,0) and Ez(x,b) = 0and Y(y) must equal zero at y = 0, and y = b.

Apply y = 0, we found that C3 = 0 and Y(y) = c4sin(yy).Therefore, at y = a, c4sin(yb) = 0.

. ynb

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Properties of wave in rectangular WGs (3)

2 22 /

u

m nrad m

a b

0 sin sin / j zz

m x n yE E e V m

a b

and

therefore we can write

2 2m n

ha b

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Every combination of integers m and n defines possible mode for TMmn mode.

m = number of half-cycle variations of the fields in the x- direction

n = number of half-cycle variations of the fields in the y- direction

For TM mode, neither m and n can be zero otherwise Ez

and all other components will vanish therefore TM11 is the lowest cutoff mode.

TM mode of propagation

TM11 field lines

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Side view

End view

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Cutoff frequency and wavelength of TM mode

2 2

,

1

2 2

c mn

h m nf Hz

a b

, 2 2

2

c mn mm na b

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Ex2 A rectangular wg having the interior dimension a = 2.3cm and b = 1cm filled with a medium characterized by r = 2.25, r = 1a) Find h, fc, and c for TM11 mode

b) If the operating frequency is 15% higher than the cutoff frequency, find (Z)TM11, ()TM11, and (g)TM11. Assume the wg to be lossless for propagating modes.

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TE waves in rectangular waveguides (1)

• Ez = 0

From

Expanding for z-propagating field for a lossless WG gets

where

2 2 2( ) 0xy z u zH H

2 22 2

2 2 ( ) ( , ) 0z zu z

H HH x y

x y

( , ) j zz zH H x y e

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TE waves in rectangular waveguides (2)

• In the x-direction

Since Ey = 0, then from

we have

2 2 2 2z z

yu u

H Ej jE

x y

0zHx

at x = 0 and x = a

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TE waves in rectangular waveguides (3)• In the y-direction

Since Ex = 0, then from

we have

0zHy

at y = 0 and y = b

2 2 2 2z z

xu u

H Ej jE

y x

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Method of separation of variables (1)

Assume

then we have

( , )zH x y XY

1 2

3 4

( ) cos sin

( ) cos sin

x x

y y

X x c x c x

Y y c y c y

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Properties of TE wave in x-direction of rectangular WGs (1)

1. in the x-direction

at x = 0,

at x = a,

0zHx

1 2

( )sin cos 0x x x x

dX xc x c x

dx

2 0.c

0zHx

1

( )sin 0x x

dX xc x

dx

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Properties of TE wave in x-direction of rectangular WGs (2)

( 0,1,2,3,...) xa m m

. xma

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Properties of TE wave in y-direction of rectangular WGs (1)

2. in the y -direction

at y = 0,

at y = b,

0zHy

4 0c

0zHy

3 4

( )sin cos 0y y y y

dY yc y c y

dy

3

( )sin 0y y

dY yc y

dy

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Properties of TE wave in y-direction of rectangular WGs (2)

( 0,1,2,3,...)yb n n

.y

nb

For lossless TE rectangular waveguides,

0 cos cos /j zz

m x n yH H e A m

a b

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Cutoff frequency and wavelength of TE mode

2 2

,

1

2 2

c mn

h m nf Hz

a b

, 2 2

2

c mn mm na b

TE10 field lines

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Side view

End view

Top view

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A dominant mode for TE waves

• For TE mode, either m or n can be zero, if a > b, is a smallest eigne value and fc is lowest when m = 1 and n = 0 (dominant mode for a > b)

ha

10

1( )

22p

c TE

uf Hz

aa

10( ) 2c TE a m

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A dominant mode for TM waves

• For TM mode, neither m nor n can be zero, if a > b, fc is lowest when m = 1 and n = 1

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2 21 1 1

( )2

c TMf Hza b

11 2 2

2( )

1 1c TM m

a b

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Ex1 a) What is the dominant mode of an axb rectangular WG if a < b and what is its cutoff frequency?

b) What are the cutoff frequencies in a square WG (a = b) for TM11, TE20, and TE01 modes?

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Ex2 Which TM and TE modes can propagate in the polyethylene-filled rectangular WG (r = 2.25, r = 1) if the operating frequency is 19 GHz given a = 1.5 cm and b = 0.6 cm?

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Rectangular cavity resonators (1)

At microwave frequencies, circuits with the dimension comparable to the operating wavelength become efficient radiators An enclose cavity is preferred to confine EM field, providelarge areas for current flow. These enclosures are called ‘cavity resonators’.

There are both TE and TM modesbut not unique.

a

b

d

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Rectangular cavity resonators (2)

z-axis is chosen as the reference.

“mnp” subscript is needed to designate a TM or TE standingwave pattern in a cavity resonator.

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Electric field representation in TMmnp modes (1)

The presence of the reflection at z = d results in a standingwave with sinz or cozz terms.

Consider transverse components Ey(x,y,z), from B.C. Ey = 0 at z = 0 and z = d

\ 1) its z dependence must be the sinz type

2)

similar to Ex(x,y,z).

( 0,1, 2,...)p

pd

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Electric field representation in TMmnp modes (2)

From

Hz vanishes for TM mode, therefore

2 2 2 2

2 2 2 2

z zx

u u

z zy

u u

H Ej jE

y x

H Ej jE

x y

2

2

zx

zy

EjE

xhEj

Eyh

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Electric field representation in TMmnp modes (3)

If Ex and Ey depend on sinz then Ez must vary according to cosz, therefore

0 sin sin cos /m x n y p z

E V ma b d

( , , ) ( , ) cos /z z

p zE x y z E x y V m

d

2 2 2

2p

mnp

u m n pf resonant frequency Hz

a b d

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Magnetic field representation in TEmnp modes (1)

Apply similar approaches, namely

1) transverse components of E vanish at z = 0 and z = d

- require a factor in Ex and Ey as well as Hz.

2) factor indicates a negative partial derivative with z.

- require a factor for Hx and Hy

fmnp is similar to TMmnp.

sinp zd

cosp zd

( , , ) ( , )sin /z z

p zH x y z H x y A m

d

0 cos cos sin /m x n y p z

H A ma b d

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Dominant mode

• The mode with a lowest resonant frequency is called ‘dominant mode’.

• Different modes having the same fmnp are called degenerate modes.

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Resonator excitation (1)

For a particular mode, we need to

1) place an inner conductor of the coaxial cable where the electric field is maximum.

2) introduce a small loop at a location where the flux of the desired mode linking the loop is maximum.

source frequency = resonant frequency

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Resonator excitation (2)

For example, TE101 mode, only 3 non-zero components are Ey, Hx, and Hz.

insert a probe in the center region of the top or bottom face where Ey is maximum or place a loop to couple Hx maximum inside a front or back face.

Best location is affected by impedance matching requirements of the microwave circuit of which the resonator is a part.

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Coupling energy method

place a hole or iris at the appropriate location

field in the waveguide at the hole must have a component that is favorable in exciting the desired mode in the resonator.

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Ex3 Determine the dominant modes and their frequencies in an air-filled rectangular cavity resonator for

a) a > b > d

b) a > d > b

c) a = b = d