ECE 6340 Intermediate EM Waves - University of Houstoncourses.egr.uh.edu/ECE/ECE6340/Class...

Prof. David R. Jackson Dept. of ECE

Fall 2016

Notes 15

ECE 6340 Intermediate EM Waves

1

0 0

0 0

( , , ) ( , ) ( , )( , , ) ( , ) ( , )

z j z z

z j z z

E x y z E x y e E x y e eH x y z H x y e H x y e e

γ β α

γ β α

− − −

− − −

= =

= =

At z = 0 : *0 0

1 ˆ(0) ( )2f

S

P E H z dS= × ⋅∫

At z = ∆z : * 20 0

1 ˆ( ) ( )2

zf

S

P z E H e z dSα− ∆∆ = × ⋅∫

Attenuation Formula

Waveguiding system (WG or TL): zS

Waveguiding system

2

Attenuation Formula (cont.) Hence

If

2

2

( ) (0)

Re ( ) Re (0)

zf f

zf f

P z P e

P z P e

α

α

− ∆

− ∆

∆ =

∆ =

2( ) (0) zf fz e α− ∆∆ =P P

1zα ∆

( ) (0) (1 2 )

(0) 2 (0)

f f

f f

z z

z

α

α

∆ ≈ − ∆

= − ∆

P P

P P

so

3

Attenuation Formula (cont.)

From conservation of energy:

(0) ( )

2 (0)f f

f

z

zα

− ∆≈

∆

P P

P

(0) ( ) ( / 2)lf f dz z z− ∆ ≈ ∆ ∆P P P

( )ld z z= power dissipated per length at pointP

where

( ) (0) 2 (0)f f fz zα∆ ≈ − ∆P P P

so

4

0z =

S

z z= ∆

Attenuation Formula (cont.)

Hence

( / 2)

2 (0)

ld

f

z z

zα

∆ ∆≈

∆

P

P

(0)

2 (0)

ld

f

α =P

P

As ∆z 0:

Note: Where the point z = 0 is located is arbitrary.

5

Attenuation Formula (cont.) General formula:

0

0

( )

2 ( )

ld

f

z

zα =

P

P

z0

0( )f zP

This is a perturbational formula for the conductor attenuation.

6

The power flow and power dissipation are usually

calculated assuming the fields are those of the mode with

PEC conductors.

Attenuation on Transmission Line

A BC C C= +

2

ld

f

α =P

P

z

AC

szJ

LBC

z∆

Attenuation due to Conductor Loss

The current of the TEM mode flows in

the z direction.

7

cα α=

Attenuation on Line (cont.)

( )

( )

2

2

2

1 12

1 1( )2

12

ld s sz

S

s szC

s szC

R J dSz

z R J dlz

R J dl

=∆

= ∆∆

=

∫

∫

∫

P

C= CA+ CB

Power dissipation due to conductor loss:

20

12f Z I=P

Power flowing on line:

∆z S

A

B

CA

CB

I

(Z0 is assumed to be approximately real.)

8

Hence

( ) 2

20

12

A B

sc sz

C C

R J dlZ I

α+

=

∫

Attenuation on Line (cont.)

9

R on Transmission Line

Ignore G for the R calculation (α = αc):

R ∆z L∆z

C∆z G∆z

∆z

2

ld

cf

α =P

P

2

20

1212

ld

f

R I

Z I

=

=

P

P

I

10

R on Transmission Line (cont.) We then have

02cRZ

α =

Hence

0(2 )cR Zα=

Substituting for αc ,

22

1 ( )s szC

R R J l dlI

=

∫

11

Total Attenuation on Line

Method #1

c dα α α= +

d TEMα α=

TEMz dk j k k jkβ α ′ ′′= − = = −

so d kα ′′=

c kα α ′′= +Hence,

When we ignore conductor loss to calculate αd, we

have a TEM mode.

12

Total Attenuation on Line (cont.)

Method #2

The two methods give approximately the same results.

( )Re

Re ( )( )R j L G j C

α γ

ω ω

=

= + +

( )

0(2 )c

c

c

R Z

G C

α

εωε

=

′′= ′

where

13

Example: Coax

)2

)2

sz

sz

IA Ja

IB Jb

π

π

=

−=

( ) ( )2 2

20

12

A B

sc sz sz

C C

R J dl J dlZ I

α = +

∫ ∫

Coaxial Cable

I

I z

a b

rε

A

B

14

Example (cont.)

Hence

0

1

2 ln

rs

c

bR a

bba

εα

η

+ =

2 22 2

20 0 0

0

12 2 2

1 12 2 2

sc

s

R I Ia d b dZ a bI

RZ a b

π π

α φ φπ π

π π

− = +

= +

∫ ∫

00 ln

2 r

bZa

ηπ ε

=

Also,

Hence

(nepers/m)

15

Example (cont.)

0

00

(2 )

1 1 (2 )2 2 2

1 121 1 1

2

c

s

s

R Z

R ZZ a b

Ra b

a b

α

π π

π

πσδ

=

= +

= +

= +

Calculate R:

16

Example (cont.)

1 12 2

Ra bπ σδ π σδ

= +

This agrees with the formula obtained from the “DC equivalent model.”

(The DC equivalent model assumes that the current is uniform around the boundary, so it is a less general method.)

b

δ

a

δ

DC equivalent model of coax

17

Internal Inductance

R ∆z

C ∆z G ∆z

∆L ∆z L0 ∆z

This extra (internal) inductance consumes imaginary (reactive) power.

The “external inductance” L0 accounts for magnetic energy only in the external region (between the conductors). This is what we

get by assuming PEC conductors.

An extra inductance per unit length ∆L is added to the TL model in order to account for the internal inductance of the conductors.

0L L L= + ∆

18

Internal inductance

Skin Inductance (cont.)

R ∆z

C ∆z G ∆z

∆L ∆z L0 ∆z

( ) 212

A B

I s szC C

P X J dl+

= ∫

Imaginary (reactive) power per meter consumed by the extra inductance:

( ) 212IP L Iω= ∆

Skin-effect formula:

I

Circuit model: Equate

19


Hence:

( )

( )

2

2

2

2

1 1 12 2

1 12

12

A B

A B

s szC C

s szC C

L X J dlI

R J dlI

R

ω+

+

∆ =

=

=

∫

∫

20


Hence

RLω

∆ =

1 12 2

L Rω∆ =

X R∆ =

or

21

Summary of High-Frequency Formulas for Coax

12

HFaR

aπ σδ=

12

HFbR

bπ σδ=

( ) 12

HF HFa aX L

aω

π σδ∆ = ∆ =

( ) 12

HF HFb bX L

bω

π σδ∆ = ∆ =

Assumption: δ << a

HF HF HFa bR R R= +

HF HF HFa bL L L∆ = ∆ + ∆

22

Low Frequency (DC) Coax Model

At low frequency (DC) we have:

DC DC DCa bR R R= +

( )2

1DCaR

aσ π=

( )1

2DCbR

btσ π=

a

b

c

t = c - b 0

8DCaL µ

π∆ =

DC DC DCa bL L L∆ = ∆ + ∆

( ) ( )

42 2

02 2 22 2

ln3

2 4DCb

ccb cbL

c bc b

µπ

− ∆ = +

− −

Derivation omitted

23

Tesche Model

This empirical model combines the low-frequency (DC) and the high-frequency (HF) skin-effect results together into one result by using an approximate circuit model to get R(ω) and ∆L(ω).

F. M. Tesche, “A Simple model for the line parameters of a lossy coaxial cable filled with a nondispersive dielectric,” IEEE Trans. EMC, vol. 49, no. 1, pp. 12-17, Feb. 2007.

Note: The method was applied in the above reference for a coaxial cable, but it should work for any type of transmission line.

24

(Please see the Appendix for a discussion of the Tesche model.)

Twin Lead

a

δ x

y

h

a

x

y

h

DC equivalent model

Twin Lead

Assume uniform current density on each

conductor (h >> a).

1 12 2

Ra aπ σδ π σδ

≈ +25

Twin Lead

a

x

y

h

Twin Lead

1 1 12 2

Ra a aπ σδ π σδ π σδ

≈ + =

sRRaπ

≈

or

26

(A more accurate formula will come later.)

Wheeler Incremental Inductance Rule

0

0

1s

LR Rnµ

∂= − ∂

x

y

A B

n̂

L0 is the external inductance (calculated assuming PEC conductors) and ∂n is an increase in the dimension of the conductors (expanded into the active field region).

22

1 ( )s szC

R R J l dlI

=

∫

Wheeler showed that R could be expressed in a way that is easy to calculate (provided we have a formula for L0):

H. Wheeler, "Formulas for the skin-effect," Proc. IRE, vol. 30, pp. 412-424, 1942.

27

The boundaries are expanded a small amount ∆n into the field region.

Wheeler Incremental Inductance Rule (cont.)

PEC conductors

x

y

A B

n̂ Field region

∆n

28

0

0

1s

LR Rnµ

∂= − ∂

L0 = external inductance (assuming perfect conductors).

Derivation of Wheeler Incremental Inductance rule


22

1 ( )s szC

R R J l dlI

=

∫

200 2

extS

L H dSIµ

=

∫

20

2

0

1414

ext

H

HS

W L I

W H dSµ

=

= ∫

2 20 0 02 2 ( )sz

C C

L H dl J l dln I I

µ µ∆= − = −

∆ ∫ ∫

Hence ( ) 20

0 2C

L n H dlIµ

∆ = − ∆ ∫We then have

PEC conductors

x

y

A B

n̂ Field region (Sext)

∆n

29


22

1 ( )s szC

R R J l dlI

=

∫

2 20 0 02 2

0

1 1( ) ( )sz szC C

L LJ l dl J l dln nI I

µµ

∂ ∂= − ⇒ = −

∂ ∂∫ ∫

PEC conductors

x

y

A B

n̂ Field region (Sext)

∆n

30

From the last slide,

0

0

1s

LR Rnµ

∂= − ∂

Hence


Example 1: Coax

a

b 0

0 ln2

bLa

µπ

=

( ) ( )1 1

0 0 0 0 02

0

1 112 2

1 12

L L L b bbn a b a a a a

a b

µ µπ π

µπ

− −∂ ∂ ∂ − = + − = − ∂ ∂ ∂ = − +

1 12 2sR R

a bπ π = +

0

0

1s

LR Rnµ

∂= − ∂

31

Example 2: Twin Lead

100 cosh

2hLa

µπ

− =


32

01

1

cosh2

Cha

πε−

=

100 cosh

2hZa

ηπ

− =

From image theory (or conformal mapping):

00 ln ,hZ a h

aηπ

≈ <<

0L C µε=

a

x

y

h

0 0,ε µ

Example 2: Twin Lead (cont.)

100 cosh

2hLa

µπ

− =

10 0 0 0 022 2

1 1 2cosh2 2

1 12 2

hL L h h an a a a a ah h

a a

µ µ µπ π π

−

∂ ∂ ∂ − = = = = − ∂ ∂ ∂ − −

2

1 2

12

s

haR R

a ha

π

= −

0

0

1s

LR Rnµ

∂= − ∂


Note: By incrementing a, we increment both conductors

simultaneously.

33

a

x

y

h

0 0,ε µ

Example 2: Twin Lead (cont.)

2

1 2

12

s

haR R

a ha

π

= −

a

x

y

h


34

100 cosh

2hZa

ηπ

− =

100 cosh

2hLa

µπ

− =

01

1

cosh2

Cha

πε−

=

( ) tanG Cω δ=

Summary

Attenuation in Waveguide

2

ld

cf

α =P

P( )

2

2

1 12

12

c

lsd s

S

ssC

R J dSz

R J dl

=∆

=

∫

∫

P

z

C

cSS

z∆

A waveguide mode is traveling in the

positive z direction.

35

We consider here conductor loss for a waveguide mode.

Attenuation in Waveguide (cont.) or

21 ˆ2

ld s

C

R n H dl= ×∫P

( )0 0ˆ( ) orWG WG TE TMt tE Z z H Z Z Z= − × =

Hence

*0

1 ˆ ˆRe ( )2

WGf t t

S

Z z H H z dS− = × × ⋅ ∫P

Power flow: *1 ˆRe ( )

2f t tS

E H z dS= × ⋅∫P

Next, use

36

Assume Z0WG

= real ( f > fc and no dielectric loss)

Hence

* * * *

2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) ( )t t t t t t t t

t

z H H z H z H z z H H H z H z

H

× × ⋅ = − × × ⋅ = − ⋅ + ⋅ ⋅

= −

20

1Re2

WGf t

S

Z H dS = ∫P

20

12

WGf t

S

Z H dS= ∫P

Attenuation in Waveguide (cont.)

Vector identity:

37

( ) ( ) ( )A B C B A C C A B× × = ⋅ − ⋅

Then we have

2

20

ˆ

2s C

c WGt

S

n H dlRZ H dS

α

× =

∫

∫


C

S n̂

x

y

38

Total Attenuation:


c dα α α= +

2 2z ck j k kβ α= − = −

2 2Imd ck kα = − −

( )0 1 tanc r rk k jω µε µ ε δ′= = −

Calculate αd (assume PEC wall):

where

39

so

TE10 Mode


40

( ) ( )

2

21 2

Re 1 /s c

c

c

R fba fb f f

αη

= +

−

22Imd k

aπα = − −

01

rc

η ηε

=rcε

a

bx

y

0 rck k ε=

1rµ =

Attenuation in dB

( ) (0) z j zV z V e eα β− −=

10 10( )dB 20log 20log ( )(0)

zV z eV

α−= =

10lnlogln10

xx =Use

z = 0 z

zS

Waveguiding system (WG or TL)

41

so ln( )dB 20

ln10( )20ln10

ze

z

α

α

−

=

−=

20Attenuation [dB/m]ln10

α =

Hence

Attenuation in dB (cont.)

42

( )Attenuation 8.6859 [dB/m]α=

or

Attenuation in dB (cont.)

43

Appendix: Tesche Model

C Za Zb G

∆z

L0

02

ln

rcCba

πε ε ′=

tan c

c

GC

εδω ε

′′= =

′0

0 ln2

bLa

µπ

=

0a bZ Z Z j Lω= + +

Y G j Cω= +

The series elements Za and Zb (defined on the next slide) account for the finite conductivity, and give us an accurate R and ∆L for each conductor at any frequency.

44

Appendix: Tesche Model (cont.)

Inner conductor of coax

Outer conductor of coax

( )a a aZ R j Lω= + ∆

( )b b bZ R j Lω= + ∆

The impedance of this circuit is denoted as

The impedance of this circuit is denoted as

aZ

bZ

45

DCaR

HFaR HF

aL∆

DCaL∆

DCbR

HFbR HF

bL∆

DCbL∆


At low frequency the HF resistance gets small and the HF inductance gets large.

46

DCaR

HFaR HF

aL∆

DCaL∆

DCaR

HFaR HF

aL∆

DCaL∆



At high frequency the DC inductance gets very large compared to the HF inductance, and the DC resistance is small compared with the HF resistance.

47

DCaR

HFaR HF

aL∆

DCaL∆


HFaR HF

aL∆

The formulas are summarized as follows:

( )2

1DCaR

aσ π=

( )1

2DCbR

btσ π=

0

8DCaL µ

π∆ =

( ) ( )

42 2

02 2 22 2

ln3

2 4DCb

ccb cbL

c bc b

µπ

− ∆ = +

− −

( ) 12

HF HFa aR L

aω

π σδ= ∆ = ( ) 1

2HF HFb bR L

bω

π σδ= ∆ =

48


ECE 6340 Intermediate EM Waves - University of Houstoncourses.egr.uh.edu/ECE/ECE6340/Class...

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