Post on 20-May-2020
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
Skin Inductance (cont.)
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
Skin Inductance (cont.)
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
Wheeler Incremental Inductance Rule (cont.)
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
Wheeler Incremental Inductance Rule (cont.)
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
Wheeler Incremental Inductance Rule (cont.)
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
− =
Wheeler Incremental Inductance Rule (cont.)
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
∂= − ∂
Wheeler Incremental Inductance Rule (cont.)
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
Wheeler Incremental Inductance Rule (cont.)
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
× =
∫
∫
Attenuation in Waveguide (cont.)
C
S n̂
x
y
38
Total Attenuation:
Attenuation in Waveguide (cont.)
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
Attenuation in Waveguide (cont.)
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∆
Inner conductor of coax
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∆
Appendix: Tesche Model (cont.)
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.
47
DCaR
HFaR HF
aL∆
DCaL∆
Appendix: Tesche Model (cont.)
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
Appendix: Tesche Model (cont.)