Prof. David R. Jackson Dept. of ECE Notes 8 ECE 5317-6351 Microwave Engineering Fall 2012 Waveguides...

1

Prof. David R. JacksonDept. of ECE

Notes 8

ECE 5317-6351 Microwave Engineering

Fall 2012

Waveguides Part 5:Transverse Equivalent Network

(TEN)

2

Waveguide Transmission Line ModelOur goal is to come up with a transmission line model for a waveguide mode.

I

+- V

z

The waveguide mode is not a TEM mode, but it can be modeled as a wave on a transmission line.

z

a

bx

y

3

For a waveguide mode, voltage and current are not uniquely defined.

102sin zjk z

yc

jE A x e

k a a

0

10 02sin sinz z

Bjk z jk z

AB ycA b

jV z V z E dr E dy A b x e V x e

k a a a

The voltage depends on x!

x

y

a

b

A

B

TE10 Mode

yE x

z

a

bx

y

Waveguide Transmission Line Model (cont.)

4

For a waveguide mode, voltage and current are not uniquely defined.

2 2 2

1 1 1

102

10 2 12

01 2

sin

cos cos

cos cos2

z

z

z

x x xjk ztop z

sz xcx x x

jk zz

c

jk z

jkI z J x dx H x dx A x e dx

k a a

jk aA x x e

k a a a

Ix x e

a a

The current depends on the length of the interval!

102sin zjk zz

xc

jkH A x e

k a a

TE10 Mode

x

y

a

b xH x

x1 x2

Current on top wall:

Note: If we integrate around the entire boundary, we get zero current.

z

a

bx

y


5

( , , ) ( , ) z zjk z jk zt tE x y z e x Ay e eA

1ˆ( )t t

w

h z eZ

( , , ) ( , ) z zjk z jk zt tH x y z h x Ay e eA

Wave impedance


w TE TMZ Z Z or

Examine the transverse (x, y) fields:

The minus sign arises from:

Modal amplitudes

6

Introduce a defined voltage into field equations:

0 01

1( , , ) ( , ) z zjk z jk z

t tE x y z e x y e eV VC

0 01

1( , , ) ( , ) z zjk z jk z

t tH x y z h x y e eV VC


0

1

VA

C

where

0 01

V VC

A A

or

0

1

VA

C

We may use whatever definition of voltage we wish here.

7

0 0

0 02

1( , , ) ( , ) z zjk z jk z

t tH x y z h x y eV

eV

ZC Z


12

0

CC

Z

Introduce a defined current and then from this define a characteristic impedance:

0

VZ

I

0 01

1( , , ) ( , ) z zjk z jk z

t tH x y z h x y e eV VC

where

We may use whatever definition of current we wish here.

8

Summary:

0 01

( )

1( , , ) ( , ) z zjk z j

V

t t

z

k zE x y z e x y e eVC

V

0 0

0

( )

02

1( , , ) ( , ) z zjk z jk z

t

z

t

I

H x yV

z h x y eV

Ze

C Z


0 01

V VC

A A

10

2

CZ

C

9

Note on Z0:

0 01

( )

1( , , ) ( , ) z zjk z j

V

t t

z


V

0 0

0

( )

02

1( , , ) ( , ) z zjk z jk z

t

z

t

I

H x yV

z h x y eV

Ze

C Z


We can define voltage and current, and this will determine the value of Z0.

Or, we can define voltage and Z0, and this will determine current.

10

The transmission-line model is called the transverse equivalent network (TEN) model of the waveguide


I z

+- 0 , zZ k

z

V z

11

Power flow down the waveguide:

*

* **

1 2

1ˆ

2

1 1ˆ( , ) ( , )

2

WGt t

S

t t

S

P z E H z dS

V z I z e x y h x y z dSC C


**

1 2

1ˆ( , ) ( , )WG TL

t t

S

P z P z e x y h x y z dSC C

12

Set


* *1 2 ˆ( , ) ( , )t t

S

C C e x y h x y z dS

WG TLP z P z

Then

13

Summary of Constants (for equal power)


* *1 2 ˆ( , ) ( , )t t

S


10

2

CZ

C

The most common choice: 0 wZ Z

Once we pick Z0, the constants are determined.

14

We have two constants (C1 and C2)


Here are possible constraints we can choose to determine the constants:

We can define the voltage We can define the current We can define the characteristic impedance We can impose the power equality condition

Any two of these are sufficient to determine the constants.

15

Method 1: Define voltage Define current (This determines Z0)

Example: TE10 Mode of Rectangular Waveguide

Method 2: Choose Z0 = ZTE

Assume power equality

z

a

bx

y

16

ˆ sin

1ˆ sin

z

z

jk zt

jk zt

TE

E y A x ea

H x A x eZ a

Define:

0

/ 2, , / 2, , z

botjk z

y

top b

V z E a y z dr E a y z dy A b e

Example: TE10 Mode (cont.)

0 0 0

sin 1 1z z

a a ajk z jk ztop top

sz xTE TE

A A aI z J dx H dx x e dx e

Z a Z

10

ˆ sin

1ˆ sin

t

tTE

TEz

e y xa

h x xZ a

Zk

Method 1

z

a

bx

y

17

zjk zV z A b e

2zjk z

TE

A aI z e

Z

0 2TE

bZ Z

a

Since we have defined both voltage and current, the characteristic impedance is not arbitrary, but is determined.

Hence

10

ˆ sin

1ˆ sin

t

tTE

TEz

e y xa

h x xZ a

Zk


za

bx

y

0

V zZ

I z

18

10

2 2TE

C bZ Z

C a

01

V A bC b

A A

12

2 1 2

TE TE

C a aC

Z b Z

1

2

1 2

TE

C b

aC

Z

10

ˆ sin

1ˆ sin

t

tTE

TEz

e y xa

h x xZ a

Zk


za

bx

y

19

1

2

1 2

TE

C b

aC

Z

10

ˆ sin

1ˆ sin

t

tTE

TEz

e y xa

h x xZ a

Zk


**

1 2

1ˆ( , ) ( , )

WG TLp

p t t

S

P z P z R

R e x y h x y z dSC C

*

*

1 1

21 2p

TE

TE

abR

Zab

Z

4pR

z

a

bx

y

20

* *1 2 ˆ( , ) ( , )t t

S


* 21 2 *

2*

0 0

*

1sin

1sin

1

2

TES

a b

TE

TE

xC C dS

Z a

xdydx

Z a

ab

Z

10

2TE

CZ Z

C 10

ˆ sin

1ˆ sin

t

tTE

TEz

e y xa

h x xZ a

Zk

Method 2


za

bx

y

21

*1 2 *

1

2TE

abC C

Z

1

2TE

CZ

C

Solution:

1

2

2

1

2TE

abC

abC

Z


10

ˆ sin

1ˆ sin

t

tTE

TEz

e y xa

h x xZ a

Zk

Take the conjugate of the second one and multiply the two together.

The solution is unique to within a common phase term.

za

bx

y

22

10

ˆ sin

1ˆ sin

t

tTE

TEz

e y xa

h x xZ a

Zk


**

1 2

1ˆ( , ) ( , )

WG TLp

p t t

S

P z P z R

R e x y h x y z dSC C

*

*

1 1

212 2

pTE

TE

abR

Zab abZ

1pR

1

2

2

1

2TE

abC

abC

Z

(as expected)

za

bx

y

23

220

220

158.0 rad / m

304.1 rad / m

za

zb r

k ka

k ka

Example: Waveguide Discontinuity

a = 2.2856 cmb = 1.016 cmr = 2.54f = 10 GHz

b a0

rz

a

z = 0

x

y

A

B

For a 1 [V/m] (field at the center of the guide) incident TE10 mode E-field in guide A, find the TE10 mode fields in both guides, and the reflected and transmitted powers.

0a b

24

TEz

Zk

Example (cont.)

Convention: Choose Z0 = ZTE

Assume power equality

1

2

2

1

2TE

abC

abC

Z

0

0

499.7

259.6

aa TE

za

bb TE

zb

Z Zk

Z Zk

0 1

V / m1 ( , )

2

tA e x y

abV C A

since already has 1 [ ]

TEN

0V 0TV

0V

0 ,TEa zaZ k 0 ,TE

b zbZ k

10

ˆ sin

1ˆ sin

t

tTE

TEz

e y xa

h x xZ a

Zk

25

0

0

0

0

0

0

za za

zb

za za

zb

jk z jk za

jk zb

jk z jk za

a

jk zb

b

V z V e e

V z V Te

VI z e e

Z

V TV z e

Z

Equivalent reflection problem: 0.316ob oa

ob oa

Z Z

Z Z

Example (cont.)

0TEbZ0

TEaZ 1 0.684T

TEN

0V 0TV

0V

0 ,TEa zaZ k 0 ,TE

b zbZ k

Note: The above TL results come from enforcing the continuity of voltage and current at the junction, and hence the tangential electric and magnetic fields are continuous in the WG problem.

26

Recall that for the TE10 mode:

Example (cont.)

0

0

0.3162

0.6842

10.316

2

0.684

2

za za

zb

za za

zb

jk z jk za

jk zb

jk z jk za

a

jk zb

b

abV z e e

abV z e

abI z e e

Z

abV z e

Z

0 01

( )

1( , , ) ( , ) z zjk z j

V

t t

z


V

0 0

0

( )

02

1( , , ) ( , ) z zjk z jk z

t

z

t

I

H x yV

z h x y eV

Ze

C Z

ˆ sin

1ˆ sin

t

tTE

e y xa

h x xZ a

TEN

0V 0TV

0V

0 ,TEa zaZ k 0 ,TE

b zbZ k

27

Hence, we have that

Example (cont.)

2 0

11( , , ) ( , 0.31

2) 6za zajk z jk z

ta

a t

abe e

ZH x y z h x y

C

1

0.6841

( , )2

, ( , ) zbjtb

kt

zE x y z e xb

eC

ay

02

10.684

2

1( , , ) ( , ) zbjk z

btb tH x y z h x y e

ZC

ab

1

0.3161

( , , ) ( , )2

za zata

k j zt

j z kabE x y z e x

Cey e

TEN

0V 0TV

0V

0 ,TEa zaZ k 0 ,TE

b zbZ k

28

Substituting in, we have

Example (cont.)

1ˆ 0( , , ) si 3n . 6

2

2

1za zata

jk z jk zE x y z y xa

abe

ae

b

1

0.3161

( , , ) ( , )2

za zata

k j zt

j z kabE x y z e x

Cey e

29


Example (cont.)

0

00

1 1ˆ( , , ) s

10.316

2in

12

za zajk z jk zTEa

ta TEa

TEa

H x y z xa

xZ aab

be e

Z

Z

2 0

11( , , ) ( , 0.31

2) 6za zajk z jk z

ta

a t

abe e

ZH x y z h x y

C

0 0a

a TEZ Z Z

aw TEZ Z

(our choice)

(wave impedance)

30


Example (cont.)

1

0.6841

( , )2

, ( , ) zbjtb

kt

zE x y z e xb

eC

ay

1ˆ( , , ) sin

2

0.6842

zbtb

jk zE x y z y xaa

ae

b

b

31


Example (cont.)

2 0

1( , , ) ( , )

10.684

2zb

tb tjk z

TEb

abe

ZH x y z h x y

C

0

00

1 1ˆ

10.684( , , ) sin

12

2zbjk

tb TEb

TE

Eb

b

zT

H x y z x xZ aab

ae

Z

b

Z

0 0b

b TEZ Z Z

bw TEZ Z

(our choice)

32

Summary of Fields

Example (cont.)

0

1ˆ( , , ) sin 0.684 zbjk z

tb TEb

H x y z x x eZ a

ˆ( , , ) sin 0.684 zbjk ztbE x y z y x e

a

0

1ˆ( , , ) sin 0.316za zajk z jk z

ta TEa

H x y z x x e eZ a

ˆ( , , ) sin 0.316za zajk z jk ztaE x y z y x e e

a

0

0

499.7

259.6

a

b

Z

Z

158.0 rad / m

304.1 rad / m

za

zb

k

k

33

Power Calculations:

22*

0 0 0 *

1 1 1 1 1Re Re

2 2 2 2inc

a TE TEoa oa

abP V I V

Z Z

Example (cont.)

2

2 2*0 0

1 1 1Re

2 2 2ref

a TEoa

abP V I

Z

2

2 2*0 0

1 1 1Re 1 1

2 2 2trans

b TEoa

abP V I

Z

2*

2* 00 0 0

1 1 1 1Re Re 1 1 1

2 2 2 2trans trans trans

b TE TEoa oa

V abP V I V

Z Z

Alternative:

Note: In this problem, Z0 and are real.

34

Final Results:

1.161 mW

0.116 mW

1.045 mW

inca

refla

transb

P

P

P

Example (cont.)

a = 2.2856 cmb = 1.016 cmr = 2.54f = 10 GHz

For a 1 [V/m] incident TE10 mode E-field in guide A (field at the center of the guide) , find the TE10 mode fields in both guide, and the reflected and transmitted powers.

b a0

rz

a

z = 0

x

y

A

B

35

Discontinuities in Waveguide

inductive iris

capacitive iris

resonant iris

Rectangular Waveguide(end view)

Note: Planar discontinuities are modeled as purely shunt elements.

The equivalent circuit gives us the correct reflection and transmission of the TE10 mode.

36

Discontinuities in Waveguide (cont.)Inductive iris in air-filled waveguide

Top view:

z

x

TE10TE10

Higher-order mode region

Z0TE Z0

TE Lp

TEN Model

00 0 10

0

2

0

1

TE

z

Z Zk

k a

1T

Because the element is a shunt discontinuity, we have

T

1

37

Discontinuities in Waveguide (cont.)

Much more information can be found in the following reference:

N. Marcuvitz, Waveguide Handbook, Peter Perigrinus, Ltd. (on behalf of the Institute of Electrical Engineers), 1986.

Equivalent circuits for many types of discontinuities Accurate CAD formula for many of the discontinuities Graphical results for many of the cases Sometimes, measured results

Prof. David R. Jackson Dept. of ECE Notes 8 ECE 5317-6351 Microwave Engineering Fall 2012 Waveguides...

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