ECE 5317-6351
Microwave Engineering
Fall 2011
Prof. David R. Jackson
Dept. of ECE
Notes 9
Waveguides Part 6:
Coaxial Cable
1
Coaxial Line: TEM Mode
y
To find the TEM mode fields
We need to solve
t2 0 ;
b
(a)  V0
(b)  0
   


0


   
a
z
  (  )  C ln   D
 
or  (  )  C ln 


 0
 ,  ,

0

@  a
a
V0  C ln  
b
C 
Zero volt potential reference location (0 = b).
x
V0
a
ln  
b
2
Coaxial Line: TEM Mode (cont.)
y
Hence
(  ) 
V0
b
ln  

b
ln    
a
b
a
Thus,
 ,  ,
z
E  x, y, z    t   x, y   e
E  x, y, z   ˆ
H
V0
e
b
 
 ln  
a
 zˆ  E 


1
jk z z
  
   ˆ
e




x
jkz z
TEM : kz  k    c  k   jk 
jk z z
c    j




H   ˆ
V0
e
b
 ln  
a

c
jk z z
3
Coaxial Line: TEM Mode (cont.)
y
B
B
A
A


b
ˆ   E d 
V ( z )   E  dr    ˆ E   ˆ d   ˆ  d  zdz
a
b
V0
e
b


a  ln
 
a
V ( z )  V0 e jk z z

2
I ( z) 
J
0
sz
 d 
jk z z
d
a
z
2
  H  ˆ   d
0
2
V0
e
b


0
 ln  
a
2 V0
I ( z)  
e jk z z
b
 ln  
a


b
jk z z
 d
;
a  b
 ,  ,
x
V  ( z)
Z0  
I ( z)
Hence
 b
Z0 
ln  
2  a 
Note: This does not
account for
conductor loss.
4
Coaxial Line: TEM Mode (cont.)
y
Attenuation:
   d  c
b
a
Dielectric attenuation:
z
TEM : d  k 
 ,  ,
x
  1


P0  Re     E  H *   zˆ dS 


z 0 
 S  2
Conductor attenuation:
Pl (0)
c 
2 P0
1

 Re  VI * 
2

1
2
 Re Z 0 I
2


1
2
P0  Z 0lossless I
2
(We remove all loss from the
dielectric in Z0lossless.)
5
Coaxial Line: TEM Mode (cont.)
Conductor attenuation:
Pl (0) 
Rs
2
R
 sa
2

Rsa
2

2
C1  C2

J sz
0
2

0
b
d
z 0
2
R
 I sa
2
2
Js
y
2
R
ad  sb
2
2
2

2
z
0
R
I
ad  sb
2 a
2
2
a
J sz bd
2

0
I
bd
2 b
 ,  ,
x
2
2
2 Rsb
 1 
ad


I
0  2 a 
2
2
2
 1 
0  2 b  bd
Rs 

2
Rsa  1 
2 Rsb  1 

I




2  2 a 
2  2 b 
R 
2  1  R
 I   sa  sb 
b 
 4  a
 I
2
6
Coaxial Line: TEM Mode (cont.)
Conductor attenuation:
c 
y
Pl (0)
2 P0
b
1
2
P0  Z 0lossless I
2
a
z
R 
2  1  R
Pl (0)  I   sa  sb 
b 
 4  a
Hence we have
R 
2  1  R
I   sa  sb 
4
a
b 
 c   
2
1
2  Z 0lossless I 
2

 ,  ,
Rs 
x

2
or
 1   1   Rsa Rsb 
 c   lossless   


Z
4

a
b



 0

7
Coaxial Line: TEM Mode (cont.)
Let’s redo the calculation of conductor
attenuation using the Wheeler
incremental inductance formula.
y
b
Wheeler’s formula:

 dZ 0lossless
Rs
 c   lossless lossless 
 2Z 0 
 d
a
z
 ,  ,
The formula is applied for each conductor and the conductor
attenuation from each of the two conductors is then added.
x
Rs 

2
 lossless 


In this formula, dl (for a given conductor) is the distance by which the
conducting boundary is receded away from the field region.
8
Coaxial Line: TEM Mode (cont.)
y

 dZ
R
 c   losslesss lossless 
 2Z 0 
 d
 lossless  b 
lossless
Z0

ln  
2
a
lossless
0

 dZ

Rsa
   lossless lossless  

2
Z

da

 0

lossless



R
dZ
b
sb
0
 c   lossless lossless  

2
Z

db

 0

lossless
0
a
c
b
a
 d  da 
d

   lossless  1  
Rsa
   lossless lossless   
  
 2Z 0 
  2  a  

   lossless  1  
Rsb
b
 c   lossless lossless  
 
2
Z

2

 b 

0


 ,  ,
x
Rs 
 db 
Hence
a
c
z

2
so

   lossless   Rsa Rsb 
1
 c   lossless lossless  



b 
 2Z 0 
  2   a
or
 1   1   Rsa Rsb 
 c   lossless   


Z
4

a
b

 0
  
9
Coaxial Line: TEM Mode (cont.)
y
We can also calculate the fundamental per-unit-length
parameters of the coaxial line.
b
a
From previous calculations:
(Formulas
from Notes 1)
 ,  ,
z
LZ
lossless
0
x
 
C    / Z0lossless
G  C  tan 
R   c  2 Z 0lossless 
where
Z
lossless
0
 lossless  b 

ln  
2
a
(Derived as a homework problem)
10
Coaxial Line: Higher-Order Modes
We look at the higher-order modes of a coaxial line.
y
The lowest mode is the TE11 mode.
y
b
a
z
 ,  ,
x
x
Sketch of field lines for TE11 mode
11
Coaxial Line: Higher-Order Modes (cont.)
y
We look at the higher-order modes
of a coaxial line.
b
TEz:
a
 H z 0   ,   k H z 0   ,   0
2
2
c
kz2  k 2  kc2
z
 ,  ,
H z   , , z   H z 0   ,   e
x
jkz z
The solution in cylindrical coordinates is:
 J n ( kc  ) 
H z0   ,   

Yn (kc  ) 
 sin(n ) 


cos(n ) 
Note: The value n must be an integer to have unique fields.
12
Plot of Bessel Functions
1
1
J n (0) is finite
n=0
0.8
n=1
0.6
n=2
0.4
J 0( x)
Jn (x)
J 1( x)
0.2
J n( 2  x)
0
0.2
0.4
 0.403 0.6
0
1
2
3
0
2
n  

J n ( x) ~
cos  x 
 
x
2
4

4
5
x
6
7
8
9
10
10
x
13
Plot of Bessel Functions (cont.)
0.521
1
n=0
n=1
0
1
n=2
2
Yn (x)
Y0( x)
Y1( x)
Yn (0) is infinite
3
Yn( 2  x)
4
5
6
 6.206
7
0
1
2
3
0
2
n  

Yn ( x) ~
sin  x 
 
x 
2 4
4
5
x
6
7
8
9
10
10
x
14
Coaxial Line: Higher-Order Modes (cont.)
y
We choose (somewhat arbitrarily) the cosine
function for the angle variation.
b
Wave traveling in +z direction:
Hz   ,, z   Hz 0   ,  e jkz z
a
z
 ,  ,
x
H z 0   ,   cos(n)  AJ n (kc  )  BYn (kc  )
The cosine choice corresponds to having the transverse electric field E being an even
function of, which is the field that would be excited by a probe located at  = 0.
15
Coaxial Line: Higher-Order Modes (cont.)
y
Boundary Conditions:
E  a,   0
E b,   0
b
j H z
E  2
kc 

a
z
 ,  ,
x
H z
0

  a ,b
Hence
Note: The prime denotes derivative
with respect to the argument.
kc  AJ n ( kc a )  BYn( kc a )   0
kc  AJ n ( kcb)  BYn( kcb)   0
16
Coaxial Line: Higher-Order Modes (cont.)
y
AJ n (kca )  BYn(kca )  0
AJ n (kcb)  BYn(kcb)  0
b
a
In order for this homogenous system of
equations for the unknowns A and B to have
a non-trivial solution, we require the
determinant to be zero.
z
 ,  ,
x
J n (kc a ) Yn(kca )
Det  kc  
0
J n (kcb) Yn(kcb)
Hence
J n (kca)Yn(kcb)  J n (kcb)Yn(kca)  0
17
Coaxial Line: Higher-Order Modes (cont.)
y
J n (kca)Yn(kcb)  J n (kcb)Yn(kca)  0
b
Denote
x  kca
a
z
 ,  ,
x
The we have
F ( x; n, b / a )  J n ( x )Yn  x  b / a    J n  x  b / a   Yn( x )  0
For a given choice of n and a given value of b/a, we can
solve the above equation for x to find the zeros.
18
Coaxial Line: Higher-Order Modes (cont.)
A graph of the determinant reveals the zeros of the determinant.
 p
xnp
F  x; n, b / a 
th
zero
Note: These values are not the same
as those of the circular waveguide,
although the same notation for the
zeros is being used.
xn3
xn1

x  xnp
xn2
x
x  kca

kca  xnp
19
Coaxial Line: Higher-Order Modes (cont.)
Approximate solution:
2
kc a 
1 b / a
n=1
Exact solution
Fig. 3.16 from the Pozar book.
20
Coaxial Line: Lossless Case
Wavenumber:
k z  k 2  kc2
k
f  fc
 kc
 2 f c   kc
 kc a 
fc 



2  a  r  2 
kc
c
c  2.99792458 108 [m/s]
TE11 mode:
 1  1 
fc 
 

a  r    1  b / a 
c
21
Coaxial Line: Lossless Case (cont.)
fc 
 1  1 
 

a  r    1  b / a 
c
At the cutoff frequency, the wavelength
(in the dielectric) is
d 
Compare with the
cutoff frequency
condition of the
TE10 mode of RWG:
c
f r
  a 1  b / a 
so
a
d    a  b
d
2
b
a
or
2 a  b 
d
 /2
2b
22
Example
Page 129 of the Pozar book:
RG 142 coax:
a  0.035 inches  8.89  10 4 [m]
b  0.116 inches  29.46  10 4 [m]
 r  2.2
 b / a  3.31
 1  1 
fc 
 


1

b
/
a
a  r  

c
f c  16.8 [GHz]
23