2.6 A coaxial line with inner and outer conductor diameters of 0.5 cm and 1 cm,
respectively, is filled with an insulating material with εr = 4.5 and σ = 10−3 S/m.
The conductors are made of copper.
(a) Calculate the line parameters at 1 GHz.
(b) Compare your results with those based on CD Module 2.2. Include a printout
of the screen display.
Solution: (a) Given
a = (0.5/2) cm = 0.25 × 10−2 m,
b = (1.0/2) cm = 0.50 × 10−2 m,
combining Eqs. (2.5) and (2.6) gives
r
1
π f µc 1 1
′
R =
+
2π
σc
a b
s
π (109 Hz) (4π × 10−7 H/m)
1
1
1
=
+
2π
5.8 × 107 S/m
0.25 × 10−2 m 0.50 × 10−2 m
= 0.788 Ω/m.
From Eq. (2.7),
µ
b
4π × 10−7 H/m
ln
ln 2 = 139 nH/m.
L =
=
2π
a
2π
′
From Eq. (2.8),
G′ =
2π × 10−3 S/m
2πσ
=
= 9.1 mS/m.
ln (b/a)
ln 2
From Eq. (2.9),
2π × 4.5 × 8.854 × 10−12 F/m
2πε
2πεr ε0
C =
=
=
= 362 pF/m.
ln (b/a) ln (b/a)
ln 2
′
(b) Solution via Module 2.2:
Problem 2.8 Find α , β , up , and Z0 for the coaxial line of Problem 2.6. Verify your
results by applying CD Module 2.2. Include a printout of the screen display.
Solution: From Eq. (2.22),
p
γ = (R′ + jω L′ )(G′ + jω C ′ )
q
= (0.788 Ω/m) + j(2π × 109 s−1 )(139 × 10−9 H/m)
q
× (9.1 × 10−3 S/m) + j(2π × 109 s−1 )(362 × 10−12 F/m)
= (109 × 10−3 + j44.5) m−1 .
Thus, from Eqs. (2.25a) and (2.25b), α = 0.109 Np/m and β = 44.5 rad/m.
From Eq. (2.29),
s
s
R′ + j ω L ′
(0.788 Ω/m) + j(2π × 109 s−1 )(139 × 10−9 H/m)
=
Z0 =
G′ + jω C ′
(9.1 × 10−3 S/m) + j(2π × 109 s−1 )(362 × 10−12 F/m)
= (19.6 + j0.030) Ω.
From Eq. (2.33),
up =
ω
2π × 109
= 1.41 × 108 m/s.
=
β
44.5
Problem 2.15 Find α and Z0 of a distortionless line whose R′ = 2 Ω/m and
G′ = 2 × 10−4 S/m.
Solution: From the equations given in Problem 2.13,
√
α = R′ G′ = [2 × 2 × 10−4 ]1/2 = 2 × 10−2 (Np/m),
r
r
µ
¶1/2
L′
R′
2
Z0 =
=
=
= 100 Ω.
C′
G′
2 × 10−4