Problem 2.10 Using a slotted line, the voltage on a lossless transmission line was
found to have a maximum magnitude of 1.5 V and a minimum magnitude of 0.6 V.
Find the magnitude of the load’s reflection coefficient.
Solution: From the definition of the Standing Wave Ratio given by Eq. (2.73),
S=
|Ve |max 1.5
= 2.5.
=
0.6
|Ve |min
Solving for the magnitude of the reflection coefficient in terms of S, as in
Example 2-5,
|Γ| =
S − 1 2.5 − 1
=
= 0.43.
S + 1 2.5 + 1
Problem 2.12 A 50-Ω lossless transmission line is terminated in a load with
impedance ZL = (30 − j50) Ω. The wavelength is 8 cm. Find:
(a) the reflection coefficient at the load,
(b) the standing-wave ratio on the line,
(c) the position of the voltage maximum nearest the load,
(d) the position of the current maximum nearest the load.
(e) Verify quantities in parts (a)–(d) using CD Module 2.4. Include a printout of
the screen display.
Solution:
(a) From Eq. (2.59),
Γ=
◦
ZL − Z0 (30 − j50) − 50
=
= 0.57e− j79.8 .
ZL + Z0 (30 − j50) + 50
(b) From Eq. (2.73),
S=
1 + |Γ| 1 + 0.57
=
= 3.65.
1 − |Γ| 1 − 0.57
(c) From Eq. (2.70)
dmax =
θr λ nλ
−79.8◦ × 8 cm π rad n × 8 cm
+
=
+
4π
2
4π
180◦
2
= −0.89 cm + 4.0 cm = 3.11 cm.
(d) A current maximum occurs at a voltage minimum, and from Eq. (2.72),
dmin = dmax − λ /4 = 3.11 cm − 8 cm/4 = 1.11 cm.
(e) The problem statement does not specify the frequency, so in Module 2.4 we
need to select the combination of f and εr such that λ = 5 cm. With εr chosen as 1,
f=
3 × 108
c
=
= 3.75 GHz.
λ
8 × 10−2
The generator parameters are irrelevant to the problem.
The results listed in the output screens are very close to those given in parts (a)
through (d).
Figure P2.19(a)
Figure P2.19(b)
Problem 2.13 On a 150-Ω lossless transmission line, the following observations
were noted: distance of first voltage minimum from the load = 3 cm; distance of first
voltage maximum from the load = 9 cm; S = 3. Find ZL .
Solution: Distance between a minimum and an adjacent maximum = λ /4. Hence,
9 cm − 3 cm = 6 cm = λ /4,
or λ = 24 cm. Accordingly, the first voltage minimum is at dmin = 3 cm =
Application of Eq. (2.71) with n = 0 gives
θr − 2 ×
2π λ
× = −π ,
λ
8
which gives θr = −π /2.
|Γ| =
S−1 3−1 2
=
= = 0.5.
S+1 3+1 4
Hence, Γ = 0.5 e− jπ /2 = − j0.5.
Finally,
·
¸
·
¸
1+Γ
1 − j0.5
ZL = Z0
= 150
= (90 − j120) Ω.
1−Γ
1 + j0.5
λ
8.