ECE 6340
Intermediate EM Waves
Fall 2016
Prof. David R. Jackson
Dept. of ECE
Notes 15
1
Attenuation Formula
Waveguiding system (WG or TL):
z
S
Waveguiding system
E ( x, y, z )  E 0 ( x, y) e z  E 0 ( x, y) e j z e z
H ( x, y, z )  H 0 ( x, y) e z  H 0 ( x, y) e j z e z
At z = 0 :
1
*
Pf (0)   ( E 0  H 0 )  zˆ dS
2
S
1
*
2 Dz
 zˆ dS
At z = Dz : Pf (Dz )   ( E 0  H 0 ) e
2
S
2
Attenuation Formula (cont.)
Hence
Pf (Dz )  Pf (0) e2 Dz
Re Pf (Dz )  Re Pf (0) e2 Dz
so
Pf (Dz )  Pf (0) e2 Dz
If
 Dz
1
Pf (Dz )  Pf (0) (1  2 Dz )
 Pf (0)  2 Dz Pf (0)
3
Attenuation Formula (cont.)
Pf (Dz )  Pf (0)  2 Dz Pf (0)
so

z 0
Pf (0)  Pf (Dz )
2 Dz Pf (0)
z  Dz
S
From conservation of energy:
Pf (0)  Pf (Dz )  Dz Pd (Dz / 2)
l
where
Pd l ( z )  power dissipated per length at point z
4
Attenuation Formula (cont.)
Hence

Dz Pd l (Dz / 2)
2 Dz Pf (0)
As Dz  0:

Pd l (0)
2 Pf (0)
Note: Where the point z = 0 is located is arbitrary.
5
Attenuation Formula (cont.)
General formula:
l

Pd ( z0 )
2 Pf ( z0 )
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.
z0
Pf ( z0 )
6
Attenuation on Transmission Line
Attenuation due to Conductor Loss
  c
The current of the
TEM mode flows in
the z direction.

J sz
Pd l
2 Pf
z
CA
C  CA  CB
CB
L
Dz
7
Attenuation on Line (cont.)
Power dissipation due to conductor loss:
Pd
1
Z0 I
2
2
(Z0 is assumed to be
approximately real.)
1 1
2

Rs J sz dS

Dz S 2
1
1

(Dz )  Rs J sz 
Dz
2
C
Power flowing on line:
Pf 
l
1
  Rs J sz 
2
C

2

2
dl
dl
S
Dz
CA
A
I
C= CA+ CB
B
CB
8
Attenuation on Line (cont.)
Hence
 Rs   1
c  
  2  J sz 
 2Z 0   I CA CB

2

dl 

9
R on Transmission Line
R Dz
LDz
I
CDz
GDz
Dz
Ignore G for the R calculation ( = c):
c 
Pd l
2 Pf
1
2
Pd  R I
2
1
2
Pf  Z 0 I
2
l
10
R on Transmission Line (cont.)
We then have
R
c 
2 Z0
Hence
R   c (2 Z 0 )
Substituting for c ,
 1
R  Rs  2
 I

C

J sz (l ) dl 

2
11
Total Attenuation on Line
Method #1
  c  d
 d  TEM
When we ignore conductor
loss to calculate d, we
have a TEM mode.
k zTEM    j d  k  k   jk 
so
Hence,
 d  k 
   c  k 
12
Total Attenuation on Line (cont.)
Method #2
  Re 
 Re

( R  j L)(G  j C )

where
R  c (2 Z 0 )
  c 
G  C   
  c 
The two methods give approximately the same results.
13
Example: Coax
Coaxial Cable
z
I
a
A
I
b
r
B
 Rs  1 

c  
 2   J sz 
 2Z 0  I 
C A
A)
B)

J sz 
2
dl 

CB
J sz 

2


dl 


I
2 a
I
J sz 
2 b
14
Example (cont.)
Hence
2
2
2

 Rs  1  2 I
I
c  
a d  
b d 
 2 
2 b
 2 Z 0  I  0 2 a
0

 Rs   1
1 




 2 Z 0   2 a 2 b 
Also,
0
b
Z0 
ln  
2  r  a 
Hence

 b 
 r 1   

Rs
 a 
c  
b 
b 
2 ln
 0  a  
(nepers/m)
15
Example (cont.)
Calculate R:
R   c (2 Z 0 )
  Rs   1
1 
 

  (2 Z 0 )

  2 Z 0   2 a 2 b  
 Rs   1 1 

  
 2   a b 
1 1

  
2  a b 
1
16
Example (cont.)
1
1
R

2 a 2 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


17
Internal Inductance
An extra inductance per unit length DL is added to the TL model in
order to account for the internal inductance of the conductors.
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.
L  L0  DL
Internal inductance
L0 Dz
DL Dz
R Dz
C Dz
G Dz
18
Skin Inductance (cont.)
Imaginary (reactive) power per meter consumed by the extra inductance:
1
PI   DL  I
2
Circuit model:
Skin-effect formula:
L0 Dz
2
1
PI  X s  J sz 
2 C A  CB
DL Dz
R Dz
C Dz
Equate

2
dl
I
G Dz
19
Skin Inductance (cont.)
Hence:
1
1
1
 DL  X s 2
2
2
I
1
1
 Rs 2
2
I

J sz 


J sz 

2
dl
C A  CB
2
dl
C A  CB
1
 R
2
20
Skin Inductance (cont.)
1
1
 DL  R
2
2
Hence
DX  R
or
DL 
R

21
Summary of High-Frequency Formulas
for Coax
Assumption:  << a
HF
a
1

2 a
HF
b
1

2 b
R
R
DX
DX
HF
a
HF
b
1
   DL  
2 a
R HF  RaHF  RbHF
HF
a
1
   DL  
2 b
HF
DLHF  DLHF

D
L
a
b
HF
b
22
Low Frequency (DC) Coax Model
At low frequency (DC) we have:
R DC  RaDC  RbDC
DC
DLDC  DLDC

D
L
a
b
DC
a
R

1
  a
Derivation omitted
t=c-b
a
b
DLDC
b
c
2

DC
b
R
DL
DC
a
1

  2 bt 
0

8
 4 c

c ln  
2
2 

0
b

3
c
b 



2
2
2
2   c 2  b 2  4  c  b  


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() and DL().
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.)
24
Twin Lead
y
Twin Lead
a
x
h
Assume uniform current
density on each
conductor (h >> a).
DC equivalent model
y
a
x

h
1
1
R

2 a 2 a
25
Twin Lead
y
Twin Lead
a
x
h
1
1
1
R


2 a 2 a  a
or
Rs
R
a
(A more accurate formula will come later.)
26
Wheeler Incremental Inductance Rule
y
n̂
x
A
 1
R  Rs  2
 I

C

J sz (l ) dl 

2
B
Wheeler showed that R could be expressed in a way that is easy to
calculate (provided we have a formula for L0):
 1 L0 
R  Rs  

 0 n 
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).
H. Wheeler, "Formulas for the skin-effect," Proc. IRE, vol. 30, pp. 412-424, 1942.
27
Wheeler Incremental Inductance Rule (cont.)
The boundaries are expanded a small amount Dn into the field region.
n̂
y
Field region
x
Dn
A
B
PEC conductors
L0 = external inductance (assuming perfect conductors).
 1 L0 
R  Rs  

 0 n 
28
Wheeler Incremental Inductance Rule (cont.)
Derivation of Wheeler Incremental Inductance rule
 1
R  Rs  2
 I

C
y
n̂

J sz (l ) dl 

Field region (Sext)
2
x
Dn
B
A
1
2
WH  L0 I
4
2
1
WH  0  H dS
4 Sext

L0   02
 I
Hence

Sext

H dS 

2
DL0    Dn 
We then have
DL0

  02
Dn
I

C
H dl  
2
0
I
2
PEC conductors

C
0
I
2

2
H dl
C
2
J sz (l ) dl
29
Wheeler Incremental Inductance Rule (cont.)
 1
R  Rs  2
 I

C
y
n̂

J sz (l ) dl 

Field region (Sext)
2
x
Dn
B
A
PEC conductors
From the last slide,
L0

  02
n
I

2
J sz (l ) dl
C

1
I
2

J sz (l ) dl  
C
2
1 L0
0 n
Hence
 1 L0 
R  Rs  



n
 0

30
Wheeler Incremental Inductance Rule (cont.)
Example 1: Coax
0  b 
L0 
ln  
2  a 
L0 L0
L
 b

  1 0  0  
n
a
b 2  a 
 1 1
 0   
2  a b 
 1 L0 
R  Rs  



n
 0

a
b
1
1
 1  0  b   1 
b
  2      
 a  2  a   a 
1 
 1
R  Rs 


2

a
2

b


31
Wheeler Incremental Inductance Rule (cont.)
y
Example 2: Twin Lead
 0 , 0
a
x
h
From image theory (or conformal mapping):
1
C   0
1  h 
cosh  
 2a 
L0C  
0
1  h 
L0 
cosh  

 2a 
0
1  h 
Z 0  cosh  

 2a 
Z0 
0  h 
ln   , a  h
 a
32
Wheeler Incremental Inductance Rule (cont.)
y
Example 2: Twin Lead (cont.)
0
1  h 
L0 
cosh  

 2a 
Note: By incrementing a, we
increment both conductors
simultaneously.
 0 , 0
a
x
h


 h 


 
L0 L0 0 
0 
0
1
  h   1 
 2a 
1  h 


cosh   



  2   a2 
n
a
 a
 a  h 2
 2a     h  2






1


  1
  2a 

 2a 


 1 L0 
R  Rs  

 0 n 


h






 1
 2a  
R  Rs 
 a  h 2 

1 




 2a 
33
Wheeler Incremental Inductance Rule (cont.)
y
Example 2: Twin Lead (cont.)
a
x
Summary
0
1  h 
Z 0  cosh  

 2a 


h






 1
 2a  
R  Rs 
 a  h 2 



1
 


 2a 
h
C   0
L0 
1
1  h 
cosh  
 2a 
0
 h 
cosh 1  

 2a 
G  C  tan 
34
Attenuation in Waveguide
We consider here conductor loss for a waveguide mode.
S
c 
A waveguide mode
is traveling in the
positive z direction.
Dz
C
Pd
Sc
z
l
2 Pf
Pd
l
1 1
2

Rs J s dS

Dz Sc 2
1
  Rs J s 
2
C

2
dl
35
Attenuation in Waveguide (cont.)
or
Pd 
l

C
1
2
ˆ
Rs n  H dl
2
Power flow:
1
Pf  Re  ( Et  Ht* )  zˆ dS
2
S
Next, use
Et   Z 0WG ( zˆ  H t )
Hence
Z
WG
0
 Z TE or Z TM 
 1 WG
*
Pf   Re  Z 0 ( zˆ  H t )  H t   zˆ dS
2

S
36
Attenuation in Waveguide (cont.)
Vector identity:
A  B  C   B  A  C   C  A  B 
( zˆ  H t )  H t*   zˆ    H t*  ( zˆ  H t )   zˆ   zˆ ( H t  H t* )  H t ( zˆ  H t* )   zˆ
  Ht
Hence
Pf
2
2
 1 WG
  Re  Z 0 H t  dS
2

S
Assume Z0WG = real ( f > fc and no dielectric loss)
1 WG
2
Pf  Z0  H t dS
2
S
37
Attenuation in Waveguide (cont.)
Then we have
 Rs
 c   WG
 2Z 0
 nˆ  H 2 dl 

  C


2
   H t dS 
 S

y
n̂
S
x
C
38
Attenuation in Waveguide (cont.)
Total Attenuation:
  c  d
Calculate d (assume PEC wall):
kz    j  k  k
2
2
c
so
d   Im k 2  kc2
where
k    c  k0 r r 1  j tan  
39
Attenuation in Waveguide (cont.)
TE10 Mode
y
r  1
k  k0  rc
b
 rc
x
a
c 
Rs
b  Re  1   f c / f 
 
 d   Im k 2   
a

2
  0
1
 rc
2

b  fc  
1  2   
a f  


2

40
Attenuation in dB
z
S
Waveguiding system
(WG or TL)
z=0
z
V ( z )  V (0) e z e j z
dB  20log10
Use
V ( z)
 20log10 (e z )
V (0)
ln x
log10 x 
ln10
41
Attenuation in dB (cont.)
so
ln(e  z )
dB  20
ln10
( z )
 20
ln10
Hence
 20 
Attenuation  

 ln10 
[dB/m]
42
Attenuation in dB (cont.)
or
Attenuation  8.6859
[dB/m]
43
Appendix: Tesche Model
The series elements Za and Zb (defined on the next slide) account for
the finite conductivity, and give us an accurate R and DL for each
conductor at any frequency.
Za
Zb
L0
C
G
Z  Z a  Zb  j L0
Y  G  jC
Dz
C
2 0 rc
b
ln  
a
 
G
 tan   c
C
 c
0  b 
L0 
ln  
2  a 
44
Appendix: Tesche Model (cont.)
Za
Inner conductor of coax
DLDC
a
The impedance of this
circuit is denoted as
RaDC
DLHF
a
RaHF
Zb
Outer conductor of coax
The impedance of this
circuit is denoted as
DL
DC
b
RbDC
RbHF
Za  Ra  j  DLa 
DLHF
b
Zb  Rb  j  DLb 
45
Appendix: Tesche Model (cont.)
 At low frequency the HF resistance gets small and the HF inductance
gets large.
DLDC
a
RaDC
DLHF
a
RaHF
Inner conductor of coax
DLDC
a
RaDC
RaHF
DLHF
a
46
Appendix: Tesche Model (cont.)
 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.
DLDC
a
RaDC
RaHF
DLHF
a
Inner conductor of coax
RaHF
DLHF
a
47
Appendix: Tesche Model (cont.)
The formulas are summarized as follows:
DC
a
R

DL
  a
DC
a
HF
a
R
1
2
0

8

DC
b
R
DLDC
b
1

  2 bt 
 4 c

c ln  
2
2 

0
b

3
c
b 



2
2
2
2   c 2  b 2  4  c  b  


1
   DL  
2 a
HF
a
HF
b
R
1
   DL  
2 b
HF
b
48