prof. david r. jackson dept. of ece notes 9 ece 5317-6351 microwave engineering fall 2011 waveguides...
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Prof. David R. JacksonDept. of ECE
Notes 9
ECE 5317-6351 Microwave Engineering
Fall 2011
Waveguides Part 6:Coaxial Cable
2
To find the TEM mode fieldsWe need to solve
200 ; ( )
( ) 0t a V
b
0
0
( ) l
( ) l
n
n
C
C
D
or
Zero volt potential reference location (0 = b).
0
@
ln
a
aV C
b
0
ln
VC
ab
0
z
y
x
b
a
, ,
Coaxial Line: TEM Mode
3
0( ) lnln
V bba
Thus,
, , , ˆ zzjk zt
jk zE x y z x y ee
0ˆ, ,ln
zjk zVE x y z e
ba
1ˆH z E
Coaxial Line: TEM Mode (cont.)
Hence
z
y
x
b
a
, ,
0ˆ
ln
zjk zVH e
ba
: z ck k k jk TEM
c j
c
4
0
2 2
0 0
20
0
0
0
ˆˆ ˆ ˆ( )
ln
ˆ( )
( )
;
ln
2( )
ln
z
z
z
z
B B b
A A a
bjk z
a
sz
j
jk z
jk z
k z
V z E dr E d d zdz E d
Ve d
ba
V z V
I z J d H d a b
Ve d
e
VI z e
ba
ba
0
( )
( )
V zZ
I z
Coaxial Line: TEM Mode (cont.)
z
y
x
b
a
, ,
0 ln2
bZ
a
Hence
Note: This does not account for conductor loss.
5
Coaxial Line: TEM Mode (cont.)
z
y
x
b
a
, ,
d c
: d k TEM
0
(0)
2l
c
P
P
*0
0
*
2
0
1ˆRe
2
1Re
2
1Re
2
S z
P E H z dS
VI
Z I
Attenuation:
Dielectric attenuation:
Conductor attenuation:
2
0 0
1
2losslessP Z I (We remove all loss from the
dielectric in Z0lossless.)
6
Coaxial Line: TEM Mode (cont.)
z
y
x
b
a
, ,
1 2
2
0
2 22 2
0 0
2 22 2
0 0
2 22 22 2
0 0
2 2
2
(0)2
2 2
2 2 2 2
1 1
2 2 2 2
1 1
2 2 2 2
1
4
sl s
C C z
sa sbsz sz
sa sb
sa sb
sa sb
sa sb
RP J d
R RJ ad J bd
R RI Iad bd
a b
R RI ad I bd
a b
R RI I
a b
R RI
a b
Conductor attenuation:
2sR
7
Coaxial Line: TEM Mode (cont.)
z
y
x
b
a
, ,
2 1(0)
4sa sb
l
R RP I
a b
Conductor attenuation:
2sR
2
0 0
1
2losslessP Z I
0
(0)
2l
c
P
P
2
2
0
141
22
sa sb
clossless
R RI
a b
Z I
Hence we have
0
1 1
4sa sb
c lossless
R R
Z a b
or
8
Coaxial Line: TEM Mode (cont.)
z
y
x
b
a
, ,
2sR
Let’s redo the calculation of conductor attenuation using the Wheeler incremental inductance formula.
0
02
losslesss
c lossless lossless
R dZ
Z d
Wheeler’s formula:
In this formula, dl (for a given conductor) is the distance by which the conducting boundary is receded away from the field region.
The formula is applied for each conductor and the conductor attenuation from each of the two conductors is then added. lossless
9
Coaxial Line: TEM Mode (cont.)
z
y
x
b
a
, ,
2sR
0
0
0
2
ln2
losslesss
c lossless lossless
losslesslossless
R dZ
Z d
bZ
a
0
0
0
0
2
2
losslessa sac lossless lossless
losslessb sbc lossless lossless
R dZ
Z da
R dZ
Z db
d da
d db
0
0
1
2 2
1
2 2
losslessa sac lossless lossless
losslessb sbc lossless lossless
R
Z a
R
Z b
0
1
2 2
losslesssa sb
c lossless lossless
R R
Z a b
0
1 1
4sa sb
c lossless
R R
Z a b
Hence
so
or
10
Coaxial Line: TEM Mode (cont.)
z
y
x
b
a
, ,
where
We can also calculate the fundamental per-unit-length parameters of the coaxial line.
tanG C
0losslessL Z
0/ losslessC Z
02 losslesscR Z
0 ln2
losslesslossless b
Za
From previous calculations:
(Derived as a homework problem)
(Formulas from Notes 1)
11
Coaxial Line: Higher-Order Modes
z
y
x
b
a
, ,
We look at the higher-order modes of a coaxial line.
The lowest mode is the TE11 mode.
Sketch of field lines for TE11 mode
x
y
12
Coaxial Line: Higher-Order Modes (cont.)
z
y
x
b
a
, , TEz:
We look at the higher-order modes of a coaxial line.
2 20 0, , 0z c zH k H
2 2 2z ck k k
The solution in cylindrical coordinates is:
0
( ) sin( ),
( ) cos( )n c
zn c
J k nH
Y k n
0, , , zjk zz zH z H e
Note: The value n must be an integer to have unique fields.
13
Plot of Bessel Functions
0 1 2 3 4 5 6 7 8 9 100.6
0.4
0.2
0
0.2
0.4
0.6
0.8
11
0.403
J0 x( )
J1 x( )
Jn 2 x( )
100 x
x
Jn (x)
n = 0
n = 1
n = 2
(0)nJ is finite
2( ) ~ cos
2 4n
nJ x x
x
14
Plot of Bessel Functions (cont.)
0 1 2 3 4 5 6 7 8 9 107
6
5
4
3
2
1
0
10.521
6.206
Y0 x( )
Y1 x( )
Yn 2 x( )
100 x
x
Yn (x)
n = 0
n = 1
n = 2
(0)nY is infinite
2( ) ~ sin
2 4n
nY x x
x
15
z
y
x
b
a
, ,
We choose (somewhat arbitrarily) the cosine function for the angle variation.
0 , cos( ) ( ) ( )z n c n cH n AJ k BY k
0, , , zjk zz zH z H e
Wave traveling in +z direction:
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.
Coaxial Line: Higher-Order Modes (cont.)
16
z
y
x
b
a
, ,
Boundary Conditions:
, 0E a
2z
c
j HE
k
,
0a b
zH
( ) ( ) 0
( ) ( ) 0
c n c n c
c n c n c
k AJ k a BY k a
k AJ k b BY k b
Hence
, 0E b
Note: The prime denotes derivative with respect to the argument.
Coaxial Line: Higher-Order Modes (cont.)
17
z
y
x
b
a
, ,
( ) ( ) 0
( ) ( ) 0n c n c
n c n c
AJ k a BY k a
AJ k b BY k b
Hence
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.
( ) ( )
Det 0( ) ( )
n c n cc
n c n c
J k a Y k ak
J k b Y k b
( ) ( ) ( ) ( ) 0n c n c n c n cJ k a Y k b J k b Y k a
Coaxial Line: Higher-Order Modes (cont.)
18
z
y
x
b
a
, ,
Denote
( ) ( ) ( ) ( ) 0n c n c n c n cJ k a Y k b J k b Y k a
cx k a
( ; , / ) ( ) / / ( ) 0n n n nF x n b a J x Y x b a J x b a Y x
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.
The we have
Coaxial Line: Higher-Order Modes (cont.)
19
xn1 xn2
xn3 x
A graph of the determinant reveals the zeros of the determinant.
npx xc npk a x
; , /F x n b a
thnpx p zero
Coaxial Line: Higher-Order Modes (cont.)
cx k a
Note: These values are not the same as those of the circular waveguide, although the same notation for the zeros is being used.
20
Fig. 3.16 from the Pozar book.
n = 1
Approximate solution:
2
1 /ck ab a
Coaxial Line: Higher-Order Modes (cont.)
Exact solution
21
Coaxial Line: Lossless CaseWavenumber:
2 2z ck k k
2
ccf f
c c
k k
f k
22c c
c
r
k c k af
a
1 1
1 /c
r
cf
b aa
82.99792458 10 [m/s]c
TE11 mode:
22
Coaxial Line: Lossless Case (cont.)1 1
1 /c
r
cf
b aa
1 /
d
r
c
f
a b a
At the cutoff frequency, the wavelength (in the dielectric) is
d a b
2/ 2da b
so
or
Compare with the cutoff frequency condition of the TE10 mode of RWG:
2da
a
b
2b
23
Example
4
4
0.035 inches 8.89 10 [m]
0.116 inches 29.46 10 [m]
2.2r
a
b
Page 129 of the Pozar book:
RG 142 coax:
/ 3.31b a 1 1
1 /c
r
cf
b aa
16.8 [GHz]cf
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