Notes 20Rectangular Waveguides

1

ECE 3317Applied Electromagnetic Waves

Prof. David R. JacksonFall 2021

Rectangular Waveguide

Rectangular Waveguide

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

2

Cross section

x

y

a

b ,ε µ

TMz Modes

0, 0z zH E= ≠

2 2 0z zE k E∇ + = (Helmholtz equation)

0zE = on boundary (PEC walls)

( ) ( )0, , , zjk zz zE x y z E x y e−=Guided-wave assumption:

2 2 22

2 2 2 0z z zz

E E E k Ex y z

∂ ∂ ∂+ + + = ∂ ∂ ∂

2 22 2

2 2 0z zz z z

E E k E k Ex y

∂ ∂+ − + = ∂ ∂

3

TMz Modes (cont.)

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

( )2 2

2 22 2 0z z

z zE E k k Ex y

∂ ∂+ + − =

∂ ∂

Define: 2 2 2c zk k k≡ −

2 22

2 2 0z zc z

E E k Ex y

∂ ∂+ + =

∂ ∂Note that kc is an

unknown at this point.We then have:

2 220 0

02 2 0z zc z

E E k Ex y

∂ ∂+ + =

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

4

Please see Appendix A for the solution.

TMz Modes (cont.)

5

Solution from separation of variables method:

( ) ( ),

, , sin sinm n

zjk zz mn

m x n yE x y z A ea bπ π − =

( )2 2

, 2 2 2m nz c

m nk k k ka bπ π = − = − −

1,2,1,2,

mn==

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

2 22c

m nka bπ π = +

0 r rk kω µε µ ε= =

0 0 00

2kcω πω µ ε

λ= = =

Cutoff Frequency for Lossless Waveguide

Set ( ), 0m nzk =

,

2 2TM2 m n

c cf f

m nk fa bπ ππ µε

=

= = +

We start with

6

TMz Modes (cont.)

( )2 2

, 2m nz

m nk ka bπ π = − −

,

2 2TM

2m n d

cc m nf

a bπ π

π = +

( )dr

ccε

= nonmagnetic material

This defines the cutoff frequency.

Note: Cutoff frequency only

has a clear meaning in the lossless case.

Summary of TMz Solution

( ) ( ),

, , sin sinm n

zjk zz mn

m x n yE x y z A ea bπ π − =

1,2,1,2,

mn==

7

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

TMz Modes (cont.)

,

2 2TM

2m n d

cc m nf

a bπ π

π = +

( )2 2

, 2 2 2m nz c

m nk k k ka bπ π = − = − −

TEz Modes

( )2 2

2 22 2 0z z

z zH H k k Hx y

∂ ∂+ + − =

∂ ∂

We now start with

0, 0z zE H= ≠

8

( ) ( )0, , , zjk zz zH x y z H x y e−=Guided-wave assumption:

2 220 0

02 2 0z zc z

H H k Hx y

∂ ∂+ + =

∂ ∂

2 2 2c zk k k≡ −

Please see Appendix B for the solution.

Summary of TEz Solution

( ) ( ),

, , cos cosm n

zjk zz mn

m x n yH x y z A ea bπ π − =

( )2 2

, 2m nz

m nk ka bπ π = − −

0,1,20,1,2

mn==

( ) ( ), 0,0m n ≠

Same formula for cutoff frequency as the TMz case!

9

TEz Modes (cont.)

,

2 2TE

2m n d

cc m nf

a bπ π

π = +

Summary for Both Modes

( ) ( ),

, , cos cosm n

zjk zz mn

m x n yH x y z A ea bπ π − =

( )2 2

, 2m nz

m nk ka bπ π = − −

( ) ( )

0,1,2,0,1,2,

, 0,0

mnm n

==

≠

same formula for both modes

( ) ( ),

, , sin sinm n

zjk zz mn

m x n yE x y z A ea bπ π − =

TMz

TEz

( )2 2

,

2m n d

cc m nf

a bπ π

π = +

same formula for both modes

1,2,1,2,

mn==

TMz TEz

10

dr

ccε

=

Wavenumber

2 2z ck k k= −

2 2

cm nka bπ π = +

TMz or TEz mode: with

zk β=

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

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Above cutoff:2 2

2 m nka bπ πβ = − −

zk jα= −Below cutoff:2 2

2m n ka bπ πα = + −

Wavenumber Plot

β

f

ck

cf

α

2k fβ π µε= =

2 22 , c

m nk f fa bπ πβ = − − >

2 22 , c

m n k f fa bπ πα = + − <

12

“Light line”

( )2 2

,

2m n d

c cc m nf f

a bπ π

π = = +

dr

ccε

=

Guide Wavelength

Recall: The guided wavelength λg is the distance z that it takes for the wave to repeat itself.

2g

πλβ

=

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

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( )21 /d

g

cf f

λλ =−

After some algebra (see next slide):

0d

r

λλε

=

( )g dλ λ>Note :

Guide Wavelength

2 2 2 22

2 2 2g

ck km nka b

π π πλβ π π

= = =− − −

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( ) ( ) ( )2 2 2 22

2 2 221 / 1 /1 /

dg

c c ccd

k k k k k k kk k

π π π λλ πλ

= = = =− − −−

2k fω µε π µε= =

2c c ck fω µε π µε= =

( )21 /d

g

cf f

λλ =−

Derivation of wavelength formula:

Dominant Mode

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

2 2

2d

cc m nf

a bπ π

π = +

Assume b < a

Lowest TMz mode: TM11

Lowest TEz mode: TE10

( ) ( )

0,1,2,0,1,2,

, 0,0

mnm n

==

≠

1, 2,1, 2,

mn==

TMz

TEz

The dominant mode is the TE10 mode.

15

x

y

a

b ,ε µd

r

ccε

=

Dominant Mode (cont.)

Summary (TE10 Mode)

( ) 10, , cos zjk zz

xH x y z A eaπ − =

22

zk kaπ = −

2d

ccfa

=

22 , ck f f

aπβ = − >

22 , ck f f

aπα = − <

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dr

ccε

=

Fields of the Dominant TE10 Mode

( ) 10, , cos zjk zz

xH x y z A eaπ − =

Find the other fields from these equations:

2 2 2 2z z z

yz z

j H jk EEk k x k k yωµ ∂ ∂

= − − ∂ − ∂

2 2 2 2z z z

xz z

j E jk HHk k y k k xωε ∂ ∂

= − − ∂ − ∂

2 2 2 2z z z

xz z

j H jk EEk k y k k x

ωµ − ∂ ∂= − − ∂ − ∂

2 2 2 2z z z

yz z

j E jk HHk k x k k y

ωε − ∂ ∂= − − ∂ − ∂

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Dominant Mode (cont.)

From these, we find the fields to be:

10 102 2z

jE Ak k aωµ π = − −

where

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Dominant Mode (cont.)

x

yE

H

a

b

( ) 10, , sin zjk zy

xE x y z E eaπ − =

( ) 10, , sin zjk zzx

k xH x y z E eaπ

ωµ− = −

( ) 10, , cos zjk zz

xH x y z A eaπ − =

Dominant Mode (cont.)What is the mode with the next highest cutoff frequency?

2 2

2d

cc m nf

a bπ π

π = +

Assume b < a / 2

Then the next highest is the TE20 mode.

( ) ( )2,0 1,02c cf f=

19

( )2

0,1 12 2

d dc

c cfb bπ

π = =

( )2

2,0 2 12 2 / 2

d dc

c cfa aπ

π = =

x

y

a

b ,ε µ

fcTE10 TE20

Useful operating regionTE01

dr

ccε

=

Dominant Mode (cont.)What is the mode with the next highest cutoff frequency?

If b > a / 2

Then the next highest is the TE01 mode.

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( )2

0,1 12 2

d dc

c cfb bπ

π = =

( )2

2,0 2 12 2 / 2

d dc

c cfa aπ

π = =

x

y

a

b ,ε µThe useable bandwidth is now lower.

fcTE10 TE20

Useful operating region TE01

dr

ccε

=

Dominant Mode (cont.)

Power flow in waveguide (f > fc):

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2104z

abP Eβωµ

=

( ) 10, , sin zjk zy

xE x y z E eaπ − =

Make b larger to get more power flow for a given value of a.

Keep b smaller than a/2 to get maximum bandwidth.

(watts flowing down the waveguide)(derivation omitted)

The optimum dimension for b is a/2(gives maximum power flow without sacrificing bandwidth).

Example

Standard X-band* waveguide:

* X-band: from 8.0 to 12 GHz.

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

( )1,0

2ccfa

=

( )1,0 6.56 [GHz]cf =

( )2,0 13.11 [GHz]cf =

Find the single-mode operating frequency region for air-filled X-band waveguide.

Hence, we have:

Use

22

Note: b < a / 2

Example (cont.)

( )1,0 6.56 [GHz]cf =

0 0 0 02 / 2 /k k f cω µ ε π π λ= = = =

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

Recall:

/ 137.43 [rad/m]ck aπ= =

23

: 188.62 [rad/m]: 104.79 [rad/m]

kk==

At 9.0 GHz At 5.0 GHz

22 , ck f f

aπβ = − >

22 , ck f f

aπα = − <

Find the phase constant of the TE10 mode at 9.00 GHz. Find the attenuation in dB/m at 5.00 GHz

Example (cont.)

772 dB/m=Attenuation

This is a very rapid attenuation!

dB/m 8.68589α=

24

At 5.0 GHz: [ ]88.91 nepers/mα =

Recall:

Waveguide Components

25

Straight sections Flexible waveguides Waveguide bends

Waveguide adapters Waveguide couplers

https://www.pasternack.com

Waveguide terminations

Waveguide Modes in Transmission Lines

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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.

Waveguide Modes in Coax

27

Dominant waveguide mode in coax (derivation omitted):

1 11 /c

r

cfb aa πε

≈ +

4

4

0.035 inches 8.89 10 [m]0.116 inches 29.46 10 [m]2.2r

abε

−

−

= = ×

= = ×=

Example: RG 142 coax

0/ 3.31 48.4[ ]b a Z= ⇒ = Ω

11TE 16.8 [GHz]cf ≈

TE11 mode:

rε a

bz

00 ln

2 r

L bZC a

ηπ ε

= =

Appendix A: TMz Modes

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

( )2 2

2 22 2 0z z

z zE E k k Ex y

∂ ∂+ + − =

∂ ∂

Define: 2 2 2c zk k k≡ −

2 22

2 2 0z zc z

E E k Ex y

∂ ∂+ + =

∂ ∂Note that kc is an

unknown at this point.We then have:

We assume: ( ) ( ) ( )0 ,zE x y X x Y y=

2 220 0

02 2 0z zc z

E E k Ex y

∂ ∂+ + =

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

28

We want to solve:

Hence 2cX Y X Y k XY′′ ′′+ = −

2c

X Y kX Y′′ ′′+ = −

Hence 2c

X YkX Y′′ ′′= − −

Divide by XY :

( ) ( ) ( )0 ,zE x y X x Y y=

( ) ( )F x G y=This has the form Both sides of the equation must be a constant!

2 220 0

02 2 0z zc z

E E k Ex y

∂ ∂+ + =

∂ ∂

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Appendix A (cont.)

Denote

2c

X YkX Y′′ ′′= − − = constant

2x

X kX′′= − = constant

(1)(0( )

) 0( 0 2)

XX 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)30

x

y

a

b ,ε µ

Appendix A (cont.)

sin( ) 0xk a =

This gives us the following result:

, 1, 2xk a m mπ= =

( ) sin m xX x Aaπ =

Hence

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

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xmkaπ

=

From the last slide:

Appendix A (cont.)

We have

2 2c x

X Yk kX Y′′ ′′= − − = −

2 2x c

Y k kY′′= −Hence

2 2 2y c xk k k= −

Then we have 2y

Y kY′′= −

Denote

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

32

Appendix A (cont.)

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

(3)(0( )

) 0( 0 4)

YY 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 nπ= =

(3)

(4)

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ynkbπ

=

x

y

a

b ,ε µ

Appendix A (cont.)

( ) sin n yY y Cbπ =

The Y(y) function is then

( ) ( )0 , ( ) sin sinzm x n yE x y X x Y y AC

a bπ π = =

Therefore, we have

( )0 , sin sinz mnm x n yE x y A

a bπ π =

( ) ( ),

, , sin sinm n

zjk zz mn

m x n yE x y z A ea bπ π − =

New notation:

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

34

Appendix A (cont.)

Recall that 2 2 2y c xk k k= −

2 2 2c x yk k k= +Hence

2 22c

m nka bπ π = +

Therefore the solution for kc is given by

Next, recall that 2 2 2c zk k k= −

2 2 2z ck k k= −Hence

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Appendix A (cont.)

( )2 2

, 2m nz z

m nk k ka bπ π = = − −

Appendix B: TEz Modes

( )2 2

2 22 2 0z z

z zH H k k Hx y

∂ ∂+ + − =

∂ ∂

We now start with

Using the separation of variables method again, we have

( ) ( ) ( )0 ,zH 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

0, 0z zE H= ≠

36

Appendix B (cont.)

Boundary conditions:

( ,0) 0( , ) 0

x

x

E xE x b

==

(0, ) 0( , ) 0

y

y

E yE a y

=

=

( )0 , cos cosz mnm x n yH x y A

a bπ π =

The result is

2 2 2 2z z z

xz z

H jk EjEk k y k k x

ωµ ∂ ∂−= − − ∂ − ∂

2 2 2 2z z z

yz z

H jk EjEk k x k k yωµ ∂ ∂

= − − ∂ − ∂

This can be shown by using the following equations:

0, 0,zH y by

∂= =

∂

0, 0,zH x ax

∂= =

∂

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x

y

a

b ,ε µ

( ) ( ),

, , cos cosm n

zjk zz mn

m x n yH x y z A ea bπ π − =

( )2 2

, 2m nz

m nk ka bπ π = − −

0,1,20,1,2

mn==

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

( ) ( ), 0,0m n ≠

( ) 00ˆ, , jkzH x y z z A e−= ( ), , 0H x y z∇⋅ ≠

Same formula for cutoff frequency as the TEz case!

38

Appendix B (cont.)

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