
82 Microwave Engineering
Solution
Given Z
R
= 200 Ω Z
0
= 50 Ω L = 0.4 μH/m
R = 0.1 Ω/m C = 40 pF/m G = 10
−5
℧/m
Γ = (Z
R
– Z
0
)/(Z
R
+ Z
0
) = (200 – 50)/(200 + 50) = 150/250 = 0.6
S = 1 +|Γ|/1 − |Γ| = (1 + 0.6)/(1 – 0.6) = 1.6/0.4 = 4
ω
= 2
π
f = 2
π
× 50 × 10
6
= 3.1415927 × 10
8
rad/s
Q =
ω
L/R = 3.1415927 × 10
8
× 0.4 × 10
−6
/0.1 = 3.1415927 × 400 = 1256.637
υ
= 1/√LC = 1/√(0.4 × 10
−6
× 40 × 10
−12
)
= 1/√16 × 10
−18
= 1/4 × 10
−9
= 0.25 × 10
9
m/s
Q can also be obtained by using the relation Q =
β
/2
α
β
=
ω
/
υ
= 3.1415927 × 10
8
/0.25 × 10
9
= 0.31415927/0.25 = 1.256637 rad/m
α
≈ 0.0005 nepers/m
from Equation 2.75b and the relation
α
≈ R/2Z
0
for smal