Page 475 - Engineering Electromagnetics, 8th Edition

CHAPTER 13 Guided Waves 457

Now, using the relation RC = /σ (see Problem 6.6), the conductance is
2πσ
G = (10)
ln(b/a)
where σ is the conductivity of the dielectric between the conductors at the operating
frequency.
The inductance per unit length was computed for the coaxial cable as Eq. (50) in
Section 8.10,
µ
L ext = ln(b/a) (11)
2π
Again, this is an external inductance, for the small skin depth precludes any appre-
ciable magnetic flux within the conductors.
Fora circular conductor of radius a and conductivity σ c ,we let Eq. (90) of
Section 11.4 apply to a unit length, obtaining
1
R inner =
2πaδσ c
There is also a resistance for the outer conductor, which has an inner radius b.We
assume the same conductivity σ c and the same value of skin depth δ, leading to
1
R outer =
2πbδσ c
Because the line current flows through these two resistances in series, the total resis-
tance is the sum:
1 1 1
R = + (12)
2πδσ c a b
Finally, the characteristic impedance, assuming low losses, is

L ext 1 µ b
Z 0 = = ln (13)
C 2π a
13.1.2 Coaxial (Low Frequencies)
We now obtain the coaxial line parameter values at very low frequencies where there
is no appreciable skin effect and the current is assumed to be distributed uniformly
throughout the conductor cross sections.
We first note that the current distribution in the conductor does not affect either
the capacitance or conductance per unit length. Therefore,

2π
C = (14)
ln(b/a)
and
2πσ
G = (15)
ln(b/a)
The resistance per unit length may be calculated by dc methods, R = l/(σ c S), where
l = 1m and σ c is the conductivity of the outer and inner conductors. The area of the