1

Prof. David R. JacksonDept. of ECE

Fall 2013

Notes 12

ECE 6340 Intermediate EM Waves

2

Wave Impedance

Assume TMz wave traveling in the +z direction

1ˆt tTM

H z EZ

ˆ ˆ ˆTMt tz E Z z z H

ˆTMt tE Z z H

ˆ TMt tz E Z H

TM z z

c

k kZ

k

From previous notes:

so

or

3

-z wave:

z zk k zz z j k zjk z jk ze e e so we can simply substitute

1ˆt tTM

H z EZ

1ˆt tTM

H z EZ

Summary:+ sign: +z wave

- sign: - z wave

Wave Impedance (cont.)

In this formula the wavenumber and the wave impedance are taken to be the same for negative and positive

traveling waves.

4

TEz wave:

TE

z z

kZ

k k

1ˆt tTE

H z EZ

Wave Impedance (cont.)

Note: 2TE TMZ Z

5

Transverse Equivalent Network (TEN)

Denote

(assume TMz wave) Also, write

0( , , ) ( , ) ( )t tE x y z E x y V z

( ) z zjk z jk zV z V e V e

0 ˆ( , ) ( , ) ( , )t TM tE x y e x y x y

ˆ( , , ) ( , ) ( , ) ( )t TM tE x y z e x y x y V z

where Note: V(z) behaves as a voltage function.

Note: We can assume that both the unit vector and the transverse amplitude function are both real (because of the corollary to the “real” theorem).

Hence we have

There is still flexibility here, since there can be any real scaling constant in the definition of t.

6

Transverse Equivalent Network (cont.)

1ˆˆ ( , )

1ˆˆ ( , )

z

z

jk zt TM tTM

jk zTM tTM

H z e x y V eZ

z e x y V eZ

1ˆt tTM

H z EZ

( ) zjk zV z V e

ˆ( , , ) ( , ) ( , ) ( )t TM tE x y z e x y x y V z

For the magnetic field we have

The result for the transverse magnetic field is then

1ˆ ˆt t tTM

H z E z EZ

Now use

7

Transverse Equivalent Network (cont.)

Define

ˆ ˆˆ( , ) ( , )TMTMh x y z e x y

ˆ( , , ) ( , ) ( , ) ( )t TM tH x y z h x y x y I z

and

Note: I(z) behaves as a transmission-line current function.

( )z zjk z jk z

TM TM

V e V eI z

Z Z

We then have

8

Transverse Equivalent Network (cont.)

Summary

ˆ ˆˆ( , ) ( , )TMTMh x y z e x y

ˆ( , , ) ( , ) ( , ) ( )t TM tH x y z h x y x y I z

ˆ( , , ) ( , ) ( , ) ( )t TM tE x y z e x y x y V z

( )z zjk z jk z

TM TM

V e V eI z

Z Z

( ) z zjk z jk zV z V e V e

9

TEN

( )

( )t

t

V z E

I z H

+z-z

zWaveguide

Z0 = ZTM or ZTE + V (z)-

I (z)

zkz = kz of waveguide mode

Transverse Equivalent Network (cont.)

10

Transverse Equivalent Network (cont.)Note on Power

*1

2TENP VI

* *

*

2*

1 1ˆ ˆ

2 21 ˆˆ ˆ( , ) ( , ) ( ) ( , ) ( , ) ( )21 ˆˆ ˆ( , )2

WGz t t

TM t TM t

TMt TM

S E H z E H z

e x y x y V z h x y x y I z z

VI x y e h z

ˆˆ ˆ ˆˆ ˆ ˆ ˆ ˆ 1TM TM TMTMe h z e z e z z z

2*1( , )

2WGz tS VI x yHence

TEN:

WG:

A B C B A C C A B Note :

11

Transverse Equivalent Network (cont.)

The total complex power flowing down the waveguide is then

2*1( , )

2WG WG

z t

S S

P S dS VI x y dS

2( , )WG TEN

t

S

P P x y dS or

If we choose2

( , ) 1t

S

x y dS

WG TENP Pthen

12

At z = 0:

Hence

0 0 0 0t t t tE E H H

(0 ) (0 ) (0 ) (0 )V V I I

Junction:

These conditions are also true at a TL junction.

TEN: Junction

(EM boundary conditions)

+

-

+

-00 0,TM

zZ k 01 1,TMzZ k(0 )V (0 )V

(0 )I (0 )I

z0 r

same cross sectional shape

z = 0

13

Example: Rectangular Waveguide

TE10 mode incident

01 00

01 00

TE TE

TE TE

Z Z

Z Z

1T

TEN:

T1

00 0,TEzZ k 01 1,TE

zZ k

b a

0

r

z

a

z = 0

x

y

#0

#1

14

0 000 2

0 20

0 0 0

2 2

0 0

/

1 1

TE

z

Zk

ka

k

k a k a

101 2

1

1

TEZ

k a

Similarly,

0 376.730313461771 [ ]

Example: Rectangular Waveguide (cont.)

15

Example (Numerical Results)X-band waveguide

f = 10 GHz

2.2225 [cm]

1.0319 [cm]

a

b

Choose 2.2 ( )r teflon

6.749 [GHz] ( )cf air - filled guide

00 01510.25 [ ] 285.18 [ ]TE TEZ Z

0.28295

0.71705T

The results are:

0 154.74 [rad/m]zk 1 276.87 [rad/m]zk

16

Example (cont.)

2 2%

%

100 100 0.28295 8.00

100 8.00 92.0

r

t

P

P

Note:2% 100tP T

This is because the impedances of the two guides are different.

%

%

8.00

92.0

r

t

P

P

(conservation of energy and orthogonality)

0 0z zP P

0 inc refz z zP P P

2ref incz zP P

0z tP P

conservation of energy

othogonality

17

2 22

% 01 012 2

2

0000

( , )2 2

100 1001

( , ) 22

t

tTE TES

t

TEtTES

V Tx y dS

Z ZP

Vx y dS ZZ

Alternative calculation:

% 92.0tP 2% 00

01

100TE

t TE

ZP T

Z

Example (cont.)

18

Example (cont.)

ˆ( , , ) ( , ) ( , ) ( )t TE tH x y z h x y x y I z

ˆ( , , ) ( , ) ( , ) ( )t TE tE x y z e x y x y V z

Fields

1ˆ( , , ) sin zjk zt

xE x y z y Te

a

1

01

1ˆ( , , ) sin zjk z

t TE

xH x y z x Te

a Z

0 0ˆ( , , ) sin z zjk z jk zt

xE x y z y e e

a

0 0

00

1ˆ( , , ) sin z zjk z jk z

t TE

xH x y z x e e

a Z

Air region:

Dielectric region:

ˆ ˆ,

ˆ ˆ,

, sin

TE

TE

t

e x y y

h x y x

xx y

a

0 0

0 0

1

1

00

01

1

1

z z

z z

z

z

jk z jk zair

jk z jk zairTE

jk zdiel

jk zdielTE

V z e e

I z e eZ

V z Te

I z TeZ

b a

0

r

z

a

z = 0

x

y

#0

#1

TL part

19

Matching Elements

A couple of commonly used matching elements:

Capacitive diaphragm

x

y

Inductive post (narrow)x

yTEN

Z0TE Z0

TE Lp

TEN

Z0TE Z0

TE Cd

00 10TE

z

Zk

20

Discontinuity

Rectangular waveguide with a post:

Top view:

Higher-order mode region

z

x

TE10TE10

Post

TE10

ap = radius of post

Assumption: only the TE10 mode can propagate.

21

Discontinuity (cont.)

TEN : Z0TE Z0

TE Lp

Note: The discontinuity is a approximately a shunt load because the tangential electric field of the dominant mode is approximately continuous, while the tangential magnetic field of the dominant mode is not (shown later).

Top view:

The TL discontinuity is chosen to give the same reflected and transmitted

TE10 waves as in the actual waveguide.

00 10

0

2

0

1

TE

z

Zk

k a

Higher-order mode region

z

x

TE10TE10

22

Discontinuity (cont.)

Top view:

Narrow strip model for post (w = 4ap)

Assume center of post is at x = x0

Flat strip model (strip of width w)

In this model (flat strip model) the equivalent circuit is exactly a shunt inductor (proof given later).

z

x

TE10TE10

w

z = 0- z = 0+

x0

23

Discontinuity (cont.)

Top view:

inc scay y yE E E

10

sin zjk zincy

xE e

a

scayE field produced by strip current

(that is, the field scattered by the strip)

z

x

TE10TE10

w

z = 0- z = 0+

x0

24

Discontinuity (cont.)

We assume a field representation in terms of TEm0 waveguide modes (therefore, there is only a y component of the electric field).

Note that TMm0 modes do not exist.

0

0

1

1

, , sin

, , sin

mz

mz

jk zscay m

m

jk zscay m

m

m xE x y z A e

a

m xE x y z A e

a

Transverse fields radiated by the post current (no y variation):

1/220 2

0mz

mk k

a

0

0

1 0

1 0

, , sin

, , sin

mz

mz

jk zsca mx TE

m m

jk zsca mx TE

m m

A m xH x y z e

Z a

A m xH x y z e

Z a

0

1ˆt tTE

m

H z EZ

By symmetry, the scattered field should have no y variation.

25

Discontinuity (cont.)

Top view:

0 0y yE z E z

From boundary conditions:

0 0sca scay yE z E z Hence

z

x

TE10TE10

w

z = 0- z = 0+

x0

10

sin zjk zincy

xE e

a

26

Discontinuity (cont.)

Equate terms of the Fourier series:

1 1

sin sinm mm m

m x m xA A

a a

m m mA A A 1 1 1A A A

10 ˆ sint

xE y V z

a

0 0V V Modeling equation: so

0 0sca scay yE z E z

10 100 0sca scay yE E

10 100 0y yE E

27

Discontinuity (cont.)

0 0V V

Z0TE Z0

TE jXp

This establishes that the circuit model for the flat strip must be a parallel element.

p pX L

28

Discontinuity (cont.)

0 0x x syH z H z J x

00

1sinm m syTE

mm

m xA A J x

Z a

From the Fourier series for the magnetic field, we then have:

Magnetic field:

0

sinsy mm

m xJ x j

a

0

2sin

a

m sy

m xj J x dx

a a

00

12 sinm syTE

mm

m xA J x

Z a

Represent the strip current as

or

29

Discontinuity (cont.)

1 10

2sin sinm mTE

m mm

m x m xA j

Z a a

Therefore

Hence0

2

TEm

m m

ZA j

In order to solve for jm, we need to enforce the condition that Ey = 0 on the strip.

30

Discontinuity (cont.)

0 0 022

0

1/

2 2

2

sy

w wJ x I x x x

wx x

Assume that the strip is narrow, so that a single “Maxwell” function describes accurately the shape of the current on the strip:

0

2sin

a

m sy

m xj J x dx

a a

0

0

/2

0 2/2 2

0

2 1/sin

2

x w

m

x w

m xj I dx

a awx x

Hence

I0 is the unknown

(I0 is the total current on the strip.)

31

Discontinuity (cont.)

Hence

where

0

2m mj I c

a

0

0

/2

2/2 2

0

1/sin

2

x w

m

x w

m xc dx

awx x

Note: cm can be evaluated in closed form (in terms of the Bessel function J0).

00 sin

2m

m xm wc J

a a

(Please see the Appendix.)

32

Discontinuity (cont.)

To solve for I0, enforce the electric field integral equation (EFIE):

0inc scay yE E on strip

10

0

1

, , sin

, , sin

z

mz

jk zincy

jk zscay m

m

xE x y z e

a

m xE x y z A e

a

0 01

sin sin2 2m

m

x m x w wA x x x

a a

Hence

(unit-amplitude incident mode)

0 00,2 2

w wz x x x

33

Discontinuity (cont.)

0 01

sin sin2 2m

m

x m x w wA x x x

a a

00 0

1

sin sin2 2 2

TEm

mm

Zx m x w wj x x x

a a

00 0 0

1

2sin sin

2 2 2

TEm

mm

Zx m x w wc I x x x

a a a

or

or

34

Discontinuity (cont.)

00 0 0

1

2sin sin

2 2 2

TEm

mm

Zx m x w wI c x x x

a a a

This is in the form

0 0 02 2

w wf x I g x x x x

To solve for the unknown I0, we can use the idea of a “testing function.” We multiply both sides by a testing function and then integrate over the strip.

0 0

0 0

/2 /2

0

/2 /2

x w x w

x w x w

T x f x dx I T x g x dx

35

Discontinuity (cont.)

0 0

0 0

/2 /2

00

1/2 /2

2sin sin

2

x w x wTEm

mmx w x w

Zx m xT x dx I c T x dx

a a a

Galerkin’s method: The testing function is the same as the basis function:

22

0

1/

2

T xw

x x

36

Discontinuity (cont.)

0

0

0

0

/2

2/2 2

0

/2

00 2

1 /2 2

0

1/sin

2

2 1/sin

2

2

x w

x w

x wTEm

mm x w

xdx

awx x

Z m xI c dx

a awx x

Hence

37

Discontinuity (cont.)

01 0

1

2

2

TEm

m mm

Zc I c c

a

Hence

10

20

1

22

TEm

mm

cI

Zc

a

10

20

1

TEm m

m

a cI

Z c

so

or

38

Discontinuity (cont.)

Summary

10

20

1

TEm m

m

a cI

Z c

0

2m mj I c

a

0

2

TEm

m m

ZA j

0

0

1

1

, , sin

, , sin

mz

mz

jk zscay m

m

jk zscay m

m

m xE x y z A e

a

m xE x y z A e

a

00 sin

2m

m xm wc J

a a

39

Discontinuity (cont.)

Reflection Coefficient:

1 11 1

1 1inc

A AA A

A

10

10

TEin

TEin

Z Z

Z Z

10

1010

TEpTE

in p TEp

jX ZZ Z jX

jX Z

From these equations, Xp may be found.

1A

Post Reactance:

where

jXp0 10TEZ Z 0 10

TEZ Z

40

Discontinuity (cont.)

Result:

10

1

2TE

pjX Z

101 1 2

TEZA j

1 0 1

2j I c

a

10

20

1

TEm m

m

a cI

Z c

0

0 sin2m

m xm wc J

a a

010 2

0

1

TEZ

k a

41

Appendix

In this appendix we evaluate the cm coefficients .

0

0

/2

2/2 2

0

1/sin

2

x w

m

x w

m xc dx

awx x

Use 0x x x

/20

2/2 2

1 1sin

2

w

m

w

m x xc dx

awx

42

Appendix (cont.)

/20

2/2 2

1 1sin

2

w

m

w

m x xc dx

awx

0/2

2/2 02

sin cos1 1

cos sin2

w

m

w

m xm x

a ac dx

m xm xwx a a

This term integrates to zero (odd function).

or

43

/20

2/2 2

1 1sin cos

2

w

m

w

m x m xc dx

a awx

/20

20 2

cos2

sin

2

w

m

m xm x a

c dxa w

x

Appendix (cont.)

or

44

Next, use the transformation sin2

cos2

wx

wdx d

/20

0

cos sin2 2

sin cos2cos

2

m

m wm x wa

cwa

Appendix (cont.)

45

/20

0

2sin cos sin

2m

m x m wc d

a a

Next, use the following integral identify for the Bessel function:

0

1cos sinnJ z z n d

so that

0

0

1cos sinJ z z d

0

0

1sin cos sin

2m

m x m wc d

a a

Appendix (cont.)

or

46

0

0

1sin cos sin

2m

m x m wc d

a a

00sin

2m

m x m wc J

a a

Hence

Appendix (cont.)