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0 0
0 0
( , , ) ( , ) ( , )
( , , ) ( , ) ( , )
z j z z
z j z z
E x y z E x y e E x y e e
H x y z H x y e H x y e e
At z = 0 :*
0 0
1ˆ(0) ( )
2f
S
P E H z dS
At z = Dz :* 2
0 0
1ˆ( ) ( )
2z
f
S
P z E H e z dS
Attenuation Formula
Waveguiding system (WG or TL): zS
Waveguiding system
3
Attenuation Formula (cont.)
Hence
If
2
2
( ) (0)
Re ( ) Re (0)
zf f
zf f
P z P e
P z P e
2( ) (0) zf fz e P P
1z
( ) (0) (1 2 )
(0) 2 (0)
f f
f f
z z
z
P P
P P
so
4
Attenuation Formula (cont.)
From conservation of energy:
(0) ( )
2 (0)
f f
f
z
z
P P
P
(0) ( ) ( / 2)lf f dz z z P P P
( )ld zP
where
= power dissipated per length at point z
( ) (0) 2 (0)f f fz z P P P
so0z
S
z z
5
Attenuation Formula (cont.)
Hence
( / 2)
2 (0)
ld
f
z z
z
P
P
(0)
2 (0)
ld
f
P
P
As z 0:
Note: The point z = 0 is arbitrary.
6
Attenuation Formula (cont.)
General formula:
0
0
( )
2 ( )
ld
f
z
z
P
P
z0
0( )f zP
This is a perturbational formula for the conductor attenuation.
The power flow and power dissipation are usually
calculated assuming the fields are those of the mode with
PEC conductors.
7
Attenuation on Transmission Line
A BC C C
2
ld
f
P
P
z
AC
szJ
LBC
z
Attenuation due to Conductor Loss
The current of the TEM mode flows in
the z direction.
c
8
Attenuation on Line (cont.)
2
2
2
1 1
2
1 1( )
2
1
2
ld s sz
S
s sz
C
s sz
C
R J dSz
z R J dlz
R J dl
P
C= CA+ CB
Power dissipation due to conductor loss:
2
0
1
2f Z IP
Power flowing on line:
zS
A
B
CA
CB
I
(Z0 is assumed to be approximately real.)
10
R on Transmission Line
Ignore G for the R calculation ( = c):
R Dz LDz
CDz GDz
Dz
2
ld
c
f
P
P
2
2
0
1
21
2
ld
f
R I
Z I
P
P
I
11
R on Transmission Line (cont.)
so
02c
R
Z
Hence
0(2 )cR Z
Substituting for ac ,
2
2
1( )s sz
C
R R J l dlI
12
Total Attenuation on Line
Method #1
c d
d TEM
TEMz dk j k k jk
so d k
c k Hence,
If we ignore conductor loss, we have a TEM mode.
13
Total Attenuation on Line (cont.)
Method #2
The two methods give approximately the same results.
1/ 2
Re
Re ( )( )R j L G j C
0(2 )cR Zwhere
c
c
G C
14
Example: Coax
)2
)2
sz
sz
IA J
aI
B Jb
2 2
20
1
2A B
sc sz sz
C C
RJ dl J dl
Z I
Coax
I
I z
a b
r
A
B
15
Example (cont.)
Hence
0
1
2 ln
rs
c
bR a
bba
2 22 2
20 0 0
0
1
2 2 2
1 1
2 2 2
sc
s
R I Ia d b d
Z a bI
R
Z a b
00 ln
2 r
bZ
a
Also,
Hence
(nepers/m)
17
Example (cont.)
1 1
2 2R
a b
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
a
DC equivalent model of coax
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Skin Inductance
R Dz
C Dz G Dz
DL Dz L0 Dz
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
19
Skin Inductance (cont.)
R Dz
C Dz G Dz
DL Dz L0 Dz
21
2A B
I s sz
C C
P X J dl
Imaginary (reactive) power per meter consumed by the extra inductance:
21
2IP L I
Skin-effect formula:
I
Circuit model:Equate
20
Skin Inductance (cont.)
Hence:
2
2
2
2
1 1 1
2 2
1 1
2
1
2
A B
A B
s sz
C C
s sz
C C
L X J dlI
R J dlI
R
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Summary of High-Frequency Formulas for Coax
1
2HFaR
a
1
2HFbR
b
1
2HF HFa aX L
a
1
2HF HFb bX L
b
Assumption: << a
HF HF HFa bR R R
HF HF HFa bL L L
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Low Frequency (DC) Coax Model
At low frequency (DC) we have:
DC DC DCa bR R R
2
1DCaR
a
1
2DCbR
bt
a
b
c
t = c - b
0
8DCaL
DC DC DCa bL L L
42 2
02 2 22 2
ln3
2 4DCb
cc
b cbL
c bc b
Derivation omitted
24
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() and L().
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.
(Please see the Appendix for a discussion of the Tesche model.)
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Twin Lead
a
x
y
h
a
x
y
h
DC equivalent model
Twin Lead
Assume uniform current density on each
conductor (h >> a).
1 1
2 2R
a a
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Wheeler Incremental Inductance Rule
0
0
1s
LR R
n
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 (expanding the surface into the active field region).
2
2
1( )s sz
C
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.
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The boundaries are incremented a small amount n into the field region
Wheeler Incremental Inductance Rule (cont.)
PEC conductors
x
y
A B
n̂Sext
n
0
0
1s
LR R
n
L0 = external inductance (assuming perfect conductors).
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Derivation of Wheeler Incremental Inductance rule
Wheeler Incremental Inductance Rule (cont.)
2
2
1( )s sz
C
R R J l dlI
200 2
extS
L H dSI
2
0
2
0
1
41
4ext
H
H
S
W L I
W H dS
2 20 0 02 2 ( )sz
C C
LH dl J l dl
n I I
Hence 20
0 2C
L n H dlI
We then have
PEC conductors
x
y
A B
n̂Sext
n
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Wheeler Incremental Inductance Rule (cont.)
2
2
1( )s sz
C
R R J l dlI
2 20 0 02 2
0
1 1( ) ( )sz sz
C C
L LJ l dl J l dl
n nI I
PEC conductors
x
y
A B
n̂Sext
n
From the last slide,
0
0
1s
LR R
n
Hence
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Wheeler Incremental Inductance Rule (cont.)
Example 1: Coax
a
b
c
00 ln
2
bL
a
1 1
0 0 0 0 02
0
1 11
2 2
1 1
2
L L L b bb
n a b a a a a
a b
1 1
2 2sR Ra b
0
0
1s
LR R
n
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Example 2: Twin Lead
100 cosh
2
hL
a
a
x
y
h
Wheeler Incremental Inductance Rule (cont.)
01
1
cosh2
Cha
100 cosh
2
hZ
a
From image theory:
00 ln ,
hZ a h
a
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Example 2: Twin Lead (cont.)
100 cosh
2
hL
a
10 0 0 0 022 2
1 1 2cosh
2 21 1
2 2
hL L h h an a a a a ah h
a a
2
1 2
12
s
ha
R Ra h
a
0
0
1s
LR R
n
a
x
y
h
Wheeler Incremental Inductance Rule (cont.)
Note: By incrementing a, we increment both conductors
simultaneously.
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Example 2: Twin Lead (cont.)
2
1 2
12
s
ha
R Ra h
a
a
x
y
h
Wheeler Incremental Inductance Rule (cont.)
100 cosh
2
hZ
a
100 cosh
2
hL
a
01
1
cosh2
Cha
tanG C
Summary
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Attenuation in Waveguide
2
ld
c
f
P
P
2
2
1 1
2
1
2
c
lsd s
S
ss
C
R J dSz
R J dl
P
z
C
cSS
z
A waveguide mode is traveling in the
positive z direction.
We consider here conductor loss for a waveguide mode.
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Attenuation in Waveguide (cont.)
or21
ˆ2
ld s
C
R n H dl P
0 0ˆ( ) orWG WG TE TMt tE Z z H Z Z Z
Hence
*0
1ˆ ˆRe ( )
2WG
f t t
S
Z z H H z dS
P
Power flow:*1
ˆRe ( )2f t t
S
E H z dS P
Next, use
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Assume Z0WG
= real ( f > fc and no dielectric loss)
Hence
* * *
2
ˆ ˆ ˆ ˆ ˆ( ) ( ) ( )t t t t t t
t
z H H z z H H H z H z
H
2
0
1Re
2WG
f t
S
Z H dS P
2
0
1
2WG
f t
S
Z H dS P
Attenuation in Waveguide (cont.)
Vector identity: B C A B A C C A B
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Total Attenuation:
Attenuation in Waveguide (cont.)
c d
1/ 22 2z ck j k k
1/ 22 2Imd ck k
ck
Calculate d (assume PEC wall):
where
so
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Attenuation in dB
( ) (0) z j zV z V e e
10 10
( )dB 20log 20log ( )
(0)zV z
eV
10
lnlog
ln10
xx Use
z = 0 z
zS
Waveguiding system(WG or TL)
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Appendix: Tesche Model
CZa Zb G
z
L0
02
ln
rcCba
tan c
c
G
C
0
0 ln2
bL
a
0a bZ Z Z j L
Y G j C
The series elements Za and Zb account for the finite conductivity, and give us R and L for each conductor at any frequency.
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Appendix: Tesche Model (cont.)
Inner conductor of coax
Outer conductor of coax
a a aZ R j L
b b bZ R j L
The impedance of this circuit is denoted as
The impedance of this circuit is denoted as
aZ
bZ
DCaR
HFaR HF
aL
DCaL
DCbR
HFbR HF
bL
DCbL
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Inner conductor of coax
At low frequency the HF resistance gets small and the HF inductance gets large.
DCaR
HFaR HF
aL
DCaL
DCaR
HFaR HF
aL
DCaL
Appendix: Tesche Model (cont.)
46
Inner conductor of coax
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.
HFaR HF
aL
DCaR
HFaR HF
aL
DCaL
Appendix: Tesche Model (cont.)