1

Prof. David R. Jackson

Notes 20

Rectangular Waveguides

ECE 3317

Spring 2013

2

Rectangular Waveguide

Rectangular Waveguide

a

bx

y

,

Cross section

We assume that the boundary is a perfect electric conductor (PEC).

We analyze the problem to solve for Ez or Hz (all other fields come from these).

TMz: Ez onlyTEz: Hz only

3

TMz Modes

2 2E E 0z zk (Helmholtz equation)

H 0, E 0z z

E 0z on boundary (PEC walls)

0E , , E , zjk zz zx y z x y eGuided-wave assumption:

2 2 22

2 2 2

E E EE 0z z z

zkx y z

2 22 2

2 2

E EE E 0z z

z z zk kx y

TMz

4

TMz Modes (cont.)

We solve the above equation by using the method of separation of variables.

2 2

2 22 2

E EE 0z z

z zk kx y

Define:2 2 2c zk k k

2 22

2 2

E EE 0z z

c zkx y

Note that kc is an

unknown at this point.We then have:

We assume: 0E ,z x y X x Y y

2 220 0

02 2

E EE 0z z

c zkx y

Dividing by the exp(-j kz z) term, we have:

5

TMz Modes (cont.)

Hence, 2cX Y XY k XY

2c

X Yk

X Y

Hence 2c

X Yk

X Y

Divide by XY :

0E ,z x y X x Y ywhere

F x G yThis has the formBoth sides of the equation must be a constant!

2 220 0

02 2

E EE 0z z

c zkx y

6

TMz Modes (cont.)

Denote

2c

X Yk

X Y

constant

2x

Xk

X

constant

(0) 0 (1)

( ) 0 (2)

X

X a

( ) sin( ) cos( )x xX x A k x B k x

Boundary conditions:

General solution:

0 ( ) sin( )xB X x A k x

sin( ) 0xk a

(1)

(2)

7

TMz Modes (cont.)

sin( ) 0xk a The second boundary condition results in

This gives us the following result:

, 1, 2xk a m m

( ) sinm x

X x Aa

Hence

Now we turn our attention to the Y (y) function.

x

mk

a

8

TMz Modes (cont.)

We have

2 2c x

X Yk k

X Y

2 2x c

Yk k

Y

Hence

2 2 2y c xk k k

Then we have2y

Yk

Y

Denote

( ) sin( ) cos( )y yY y C k y D k y General solution:

9

TMz Modes (cont.)

( ) sin( ) cos( )y yY y C k y D k y

(0) 0 (3)

( ) 0 (4)

Y

Y b

Boundary conditions:

0 ( ) sin( )yD Y y C k y

sin( ) 0yk b

Equation (4) gives us the following result:

, 1, 2yk b n n

(3)

(4)

y

nk

b

10

TMz Modes (cont.)

( ) sinn y

Y y Cb

Hence

0E , ( ) sin sinz

m x n yx y X x Y y AC

a b

Therefore, we have

0E , sin sinz mn

m x n yx y A

a b

,

E , , sin sinm n

zjk zz mn

m x n yx y z A e

a b

New notation:

The TMz field inside the waveguide thus has the following form:

11

TMz Modes (cont.)

Recall that 2 2 2y c xk k k

2 2 2c x yk k k Hence

2 22c

m nk

a b

Therefore the solution for kc is given by

Next, recall that 2 2 2c zk k k

2 2 2z ck k k Hence

12

TMz Modes (cont.)

Summary of TMz Solution for (m,n) Mode (Hz = 0)

,

E , , sin sinm n

zjk zz mn

m x n yx y z A e

a b

2 2

, 2m nz

m nk k

a b

1,2,

1,2,

m

n

Note: If either m or n is zero, the entire field is zero.

13

TMz Modes (cont.)

Cutoff frequency

2, ,2m n m n

z ck k k

2 2

,m nc

m nk

a b

where

Set , 0m nzk

,m nck k

,m nck

,2 m ncf k

We start with

Note: The number kc is the

value of k for which the wavenumber kz is zero.

14

TMz Modes (cont.)

,2 m ncf k Hence

2 2

2m n

fa b

which gives us

,

2 2

2m nTM d

c

c m nf

a b

The cutoff frequency fc of the TMm,n mode is then

,

2 21

2m nTM

c

m nf

a b

This may be written as

1dc

15

TEz Modes

2 2

2 22 2

H HH 0z z

z zk kx y

We now start with

Using the separation of variables method again, we have

0H ,z x y X x Y y

( ) sin( ) cos( )y yY y C k y D k y

( ) sin( ) cos( )x xX x A k x B k x

2 2 2c x yk k k 2 2 2

z ck k k

where

and

E 0, H 0z z TEz

16

TEz Modes (cont.)

Boundary conditions:

E ( ,0) 0

E ( , ) 0x

x

x

x b

E (0, ) 0

E ( , ) 0

y

y

y

a y

a

bx

y

,

cross section

0H , cos cosz mn

m x n yx y A

a b

The result is

2 2 2 2z z z

xz z

H jk EjE

k k y k k x

2 2 2 2z z z

yz z

H jk EjE

k k x k k y

This can be shown by using the following equations:

0, 0,zHy b

y

0, 0,zHx a

x

17

TEz Modes (cont.)

Summary of TEz Solution for (m,n) Mode (Ez = 0)

,

H , , cos cosm n

zjk zz mn

m x n yx y z A e

a b

2 2

, 2m nz

m nk k

a b

0,1,2

0,1,2

m

n

Note: The (0,0) TEz mode is not valid, since it violates the magnetic Gauss law:

, 0,0m n

00ˆH , , jkzx y z z A e H , , 0x y z

Same formula for cutoff frequency as the TEz case!

18

Summary

,

H , , cos cosm n

zjk zz mn

m x n yx y z A e

a b

2 2

, 2m nz

m nk k

a b

0,1,2,

0,1,2,

, 0,0

m

n

m n

Same formula for both cases

,

E , , sin sinm n

zjk zz mn

m x n yx y z A e

a b

TMz

TEz

2 2

,

2m n d

c

c m nf

a b

Same formula for both cases

1,2,

1,2,

m

n

TMz TEz

19

Wavenumber

2 2z ck k k

2 2

c

m nk

a b

General formula for the wavenumber

TMz or TEz mode: with

2

2

2

1 /

1 /

1 /

z c

c

c

k k k k

k

k f f

c c

k

k

Recall:

21 /z ck k f f Hence

2 2

2d

c

c m nf

a b

Note: The (m,n) notation is suppressed here.

Above cutoff:

20

Wavenumber (cont.)

2 2 2 2z c ck k k j k k

Hence

2 2

2

2

2

1 /

1 /

1 /

z c

c c

c c

c c

k j k k

jk k k

jk

jk f f

21 /z c ck jk f f Hence

2 2

2d

c

c m nf

a b

Below cutoff:

21

Wavenumber (cont.)

zk j

Hence we have

21 / ,c ck f f f f

21 / ,c c ck f f f f

2 2

2d

c

c m nf

a b

Recal that

22

Wavenumber Plot

f

ck

cf

k

General behavior of the wavenumber

21 / ,c ck f f f f

21 / ,c c ck f f f f

2 2

2d

c

c m nf

a b

“Light line”

23

Dominant Mode

The "dominant" mode is the one with the lowest cutoff frequency.

2 2

2d

c

c m nf

a b

Assume b < a

a

bx

y

,

Cross sectionLowest TMz mode: TM11

Lowest TEz mode: TE10

0,1,2,

0,1,2,

, 0,0

m

n

m n

1,2,

1,2,

m

n

TMz

TEz

The dominant mode is the TE10 mode.

24

Dominant Mode (cont.)

Formulas for the dominant TE10 mode

10H , , cos zjk zz

xx y z A e

a

22

zk ka

2d

c

cf

a

21 / ,c ck f f f f

21 / ,c c ck f f f f

At the cutoff frequency:

/ /d d cc f c f

cf f

2a

25

Dominant Mode (cont.)

What is the mode with the next highest cutoff frequency?

2 2

2d

c

c m nf

a b

Assume b < a / 2

The next highest is the TE20 mode.

fcTE10 TE20

2,0 1,02c cf f

useful operating region

a

bx

y

,

Cross section

TE01

2

0,1 1

2 2d d

c

c cf

b b

2

2,0 2 2

2 2d d

c

c cf

a a

26

Dominant Mode (cont.)

Fields of the dominant TE10 mode

10H , , cos zjk zz

xx y z A e

a

Find the other fields from these equations:

2 2 2 2

H EE z z z

yz z

jkj

k k x k k y

2 2 2 2

E HH z z z

xz z

jkj

k k y k k x

2 2 2 2

H EE z z z

xz z

jkj

k k y k k x

2 2 2 2

E HH z z z

yz z

jkj

k k x k k y

27

Dominant Mode (cont.)

From these, we find the other fields to be:

10ˆE , , sin zjk zxx y z y E e

a

10ˆH , , sin zjk zzk xx y z x E e

a

x

y

a

b

E

H

10 102 2z

jE A

k k a

where

28

Dominant Mode (cont.)

2

2

E

H

1 /

1 /

y

x z

c

c

k

k f f

f f

2

E 1

H 1 /

yTE

x zc

Zk f f

Define the wave impedance:

Wave impedance:

Note: This is the same formula as for a TEz plane wave!

0

r

29

Example

Standard X-band* waveguide (air-filled):

* X-band: from 8.0 to 12 GHz.

a = 0.900 inches (2.286 cm)b = 0.400 inches (1.016 cm)

1,0

2c

cf

a

1,0 6.56 [GHz]cf

2,0 13.11 [GHz]cf

Find the single-mode operating frequency region.

Hence, we have

Use

Note: b < a / 2.

30

Example (cont.)

Standard X-band* waveguide (air-filled):

a = 0.900 inches (2.286 cm)b = 0.400 inches (1.016 cm)

1,0 6.56 [GHz]cf

Find the phase constant of the TE10 mode at 9.00 GHz.

Find the attenuation in dB/m at 5.00 GHz

21 / ,c ck f f f f

21 / ,c c ck f f f f

0 0 02 / 2 /k f c

At 9.00 GHz: = 129.13 [rad/m] At 5.00 GHz: = 88.91 [nepers/m]

At 9.00 [GHz]: k =188.62 [rad/m]

Recall:

/ 137.43 [rad/m]ck a

31

Example (cont.)

At 5.00 [GHz]: = 88.91 [nepers/m]

10E , , sin zy

xx y z A e

a

110dB/m 20log e

dB/m 772A very rapid attenuation!

dB/m 8.68589 [nepers/m] 8.68589

Note: We could have also used

Therefore,

32

Guide Wavelength

The guide wavelength g is the distance z that it takes for the wave to repeat itself.

2g

From this we have

2 2

2 221 / 1 /

g

c ck f f f f

Hence we have the result

21 /

g

cf f

0

r

(This assumes that we are above the cutoff frequency.)

33

Phase and Group Velocity

Recall that the phase velocity is given by

pv

Hence

2 2 2

1

1 / 1 / 1 /p

c c c

vk f f f f f f

21 /

dp

c

cv

f f

For a hollow waveguide (cd = c): 2

1 /p

c

cv

f f

We then have

vp > c ! (This does not violate relativity.)Hence:

34

Phase and Group Velocity (cont.)

The group velocity is the velocity at which a pulse travels on a structure.

The group velocity is given by

1g

dv

ddd

(The derivation of this is omitted.)

A pulse consists of a "group" of frequencies (according to the Fourier transform).

t

iV t

Vi (t) +- waveguiding system

gv

35

Phase and Group Velocity (cont.)

If the phase velocity is a function of frequency, the pulse will be distorting as it travels down the system.

Vi (t) +- waveguiding system

gv

A pulse will get distorted in a rectangular waveguide!

2

1 /

dp

c

cv f

f f

36

Phase and Group Velocity (cont.)

We then have the following final result:

2 2 2 2c ck k k

21 /g d cv c f f

vg < cFor a hollow waveguide:

To calculate the group velocity for a waveguide, we use

Hence we have

1/22 2

2 2 2 2

12

2 1 / 1 /c

c c c

dk

d k k k k k

37

Phase and Group Velocity (cont.)

k

For a lossless transmission line or a lossless plane wave (TEMz waves):

We then have

1p dv c

1 1g dv c

dd

p g dv v c Hence we have

For a lossless transmission line or a lossless plane wave line there is no distortion, since the phase velocity is constant with frequency.

j R j L G j C

j LC

j

jk

(a constant)

38

Plane Wave Interpretation of Dominant Mode

Consider the electric field of the dominant TE10 mode:

10

10

ˆE , , sin

ˆ2

z

x x

z

jk z

jk x jk xjk z

xx y z y E e

a

e ey E e

j

Separating the terms, we have:

10 10ˆ ˆE , ,2 2

x xz zjk x jk xjk z jk zE Ex y z y e e y e e

j j

This form is the sum of two plane waves.

1 ˆˆ z xk z k x k

/xk awhere

2 ˆˆ z xk z k x k

#1 #2

39

Plane Wave Interpretation (cont.)

2

/tan

1

x

zc

k a

k fk

f

At the cutoff frequency, the angle is 90o. At high frequencies (well above cutoff) the angle approaches zero.

Picture (top view):x

z

a

k1

k2

TE10 mode

1 ˆˆ z xk z k x k

2 ˆˆ z xk z k x k

40

Plane Wave Interpretation (cont.)

The two plane waves add to give an electric field that is zero on the side walls of the waveguide (x = 0 and x = a).

Picture of two plane waves

a

k2

k1

z

x

Crests of waves

g

cd

vp

cd

Waves cancel out

41

Waveguide Modes in Transmission Lines

A transmission line normally operates in the TEMz mode, where the two conductors have equal and opposite currents.

At high frequencies, waveguide modes can also propagate on transmission lines.

This is undesirable, and limits the high-frequency range of operation for the transmission line.

42

Waveguide Modes in Transmission Lines (cont.)

Consider an ideal parallel-plate transmission line (neglect fringing)

x

y

w

h

EH

PMC

PEC

A “perfect magnetic conductor” (PMC) is assumed on the side walls.

Assume w > h

Rectangular “slice of plane wave” (the region inside the waveguide)

(This is an approximate model of a microstrip line.)

43

Waveguide Modes in Transmission Lines (cont.)

Some comments about PMC

A PMC is dual to a PEC:

PEC PMC

E 0t

ˆE Enn

H 0t

ˆH Hnn

PEC

E

PMC

H

44

Waveguide Modes in Transmission Lines (cont.)

From a separation of variables solution, we have the following results for the TMz and TEz waveguide modes:

x

y

w

h PMC

PEC

,

H , , sin cosm n

zjk zz mn

m x n yx y z A e

w h

2 2

, 2m nz

m nk k

w h

,

E , , cos sinm n

zjk zz mn

m x n yx y z A e

w h

TMz

TEz

Assume w > h

45

Waveguide Modes in Transmission Lines (cont.)

The dominant mode is TE10

1,0

10H , , sin zjk zz

xx y z A e

w

x

y

w

h PMC

PEC

2

1,0 2zk k

w

The cutoff frequency of the TE10 mode is

2

2 2d d

c

c cf

w w

46

Waveguide Modes in Transmission Lines (cont.)

Example

x

y

w

h

Microstrip line

fc 32 [GHz]

2.2 (Rogers 5880 Duroid)r

1.575 [mm] (62 [mils])h

2 3.15[mm]w h

0 65.4 [ ]Z

2d

c

cf

w

8

3

3 10 / 2.2

2 3.15 10cf

103.2 10 [Hz]cf

Assume:

47

Waveguide Modes in Transmission Lines (cont.)

Dominant waveguide mode in coax (derivation omitted)

1 1

1 /c

r

cf

b aa

4

4

0.035 inches 8.89 10 [m]

0.116 inches 29.46 10 [m]

2.2r

a

b

Example: RG 142 coax

0/ 3.31 48.4[ ]b a Z

16.8 [GHz]cf

TE11 mode:

a

b

r