Problem 2.38 The input impedance of a 31-cm–long lossless transmission line of
unknown characteristic impedance was measured at 1 MHz. With the line terminated
in a short circuit, the measurement yielded an input impedance equivalent to an
inductor with inductance of 0.064 µ H, and when the line was open-circuited, the
measurement yielded an input impedance equivalent to a capacitor with capacitance
of 40 pF. Find Z0 of the line, the phase velocity, and the relative permittivity of the
insulating material.
Solution: Now ω = 2π f = 6.28 × 106 rad/s, so
sc
Zin
= jω L = j2π × 106 × 0.064 × 10−6 = j0.4 Ω
oc = 1/ j ω C = 1/( j2π × 106 × 40 × 10−12 ) = − j4000 Ω.
and Zin
p
p sc oc
From Eq. (2.94), Z0 = Zin
Zin = ( j0.4 Ω)(− j4000 Ω) = 40 Ω. Using
Eq. (2.48),
up =
ω
ωl
p sc oc
=
β
tan−1 −Zin /Zin
=
6.28 × 106 × 0.31
1.95 × 106
´=
³ p
m/s,
(±0.01 + nπ )
tan−1 ± − j0.4/(− j4000)
where n ≥ 0 for the plus sign and n ≥ 1 for the minus sign. For n = 0,
up = 1.94 × 108 m/s = 0.65c and εr = (c/up )2 = 1/0.652 = 2.4. For other values
of n, up is very slow and εr is unreasonably high.
Problem 2.40 A 100-MHz FM broadcast station uses a 300-Ω transmission line
between the transmitter and a tower-mounted half-wave dipole antenna. The antenna
impedance is 73 Ω. You are asked to design a quarter-wave transformer to match the
antenna to the line.
(a) Determine the electrical length and characteristic impedance of the quarterwave section.
(b) If the quarter-wave section is a two-wire line with D = 2.5 cm, and the wires
are embedded in polystyrene with εr = 2.6, determine the physical length of
the quarter-wave section and the radius of the two wire conductors.
Solution:
(a) For a match condition, the input impedance of a load must match that of the
transmission line attached to the generator. A line of electrical length λ /4 can be
used. From Eq. (2.97), the impedance of such a line should be
√
√
Z0 = Zin ZL = 300 × 73 = 148 Ω.
(b)
up
3 × 108
λ
c
= √
= 0.465 m,
=
= √
4
4f
4 εr f
4 2.6 × 100 × 106
and, from Table 2-2,
Hence,
which leads to


µ ¶ sµ ¶2
D
120
D
+
− 1  Ω.
Z0 = √ ln 
d
d
ε


√
µ ¶ sµ ¶2
148 2.6
D
D


= 1.99,
−1 =
+
ln
d
d
120
µ ¶ sµ ¶2
D
D
+
− 1 = 7.31,
d
d
and whose solution is D/d = 3.73. Hence, d = D/3.73 = 2.5 cm/3.73 = 0.67 cm.
Problem 2.41 A 50-Ω lossless line of length l = 0.375λ connects a 300-MHz
generator with Veg = 300 V and Zg = 50 Ω to a load ZL . Determine the time-domain
current through the load for:
(a) ZL = (50 − j50) Ω
(b) ZL = 50 Ω
(c) ZL = 0 (short circuit)
For (a), verify your results by deducing the information you need from the output
products generated by CD Module 2.4.
Solution:
50 Ω
~
Vg
Transmission line
+
ZL (50-j50) Ω
Z0 = 50 Ω
Zin
-
l = 0.375 λ
Generator
z = -l
Zg
~
Vg
+
~
Ii
Load
z=0
⇓
+
~
Vi
Zin
Figure P2.41: Circuit for Problem 2.41(a).
(a) ZL = (50 − j50) Ω, β l =
Γ=
2π
λ
× 0.375λ = 2.36 (rad) = 135◦ .
◦
− j50
ZL − Z0 50 − j50 − 50
=
=
= 0.45 e− j63.43 .
ZL + Z0 50 − j50 + 50 100 − j50
Application of Eq. (2.79) gives:
·
¸
·
¸
ZL + jZ0 tan β l
(50 − j50) + j50 tan 135◦
Zin = Z0
= 50
= (100 + j50) Ω.
Z0 + jZL tan β l
50 + j(50 − j50) tan 135◦
Using Eq. (2.82) gives
!µ
Ã
¶
eg Zin
V
1
+
V0 =
Zg + Zin
e jβ l + Γe− jβ l
µ
¶
1
300(100 + j50)
=
50 + (100 + j50) e j135◦ + 0.45 e− j63.43◦ e− j135◦
= 150 e− j135
◦
(V),
◦
V+
◦
◦
150 e− j135
(1 − 0.45 e− j63.43 ) = 2.68 e− j108.44
IeL = 0 (1 − Γ) =
Z0
50
jω t
e
iL (t) = Re[IL e ]
= Re[2.68 e− j108.44 e j6π ×10 t ]
◦
8
= 2.68 cos(6π × 108t − 108.44◦ ) (A).
(b)
ZL = 50 Ω,
Γ = 0,
Zin = Z0 = 50 Ω,
µ
¶
◦
300 × 50
1
V0+ =
= 150 e− j135
50 + 50 e j135◦ + 0
V + 150 − j135◦
◦
e
= 3 e− j135 (A),
IeL = 0 =
Z0
50
(V),
iL (t) = Re[3 e− j135 e j6π ×10 t ] = 3 cos(6π × 108t − 135◦ ) (A).
◦
8
(c)
ZL = 0,
Γ = −1,
¶
µ
0 + jZ0 tan 135◦
= jZ0 tan 135◦ = − j50 (Ω),
Zin = Z0
Z0 + 0
¶
µ
◦
300(− j50)
1
+
V0 =
= 150 e− j135 (V),
◦
◦
j135
−
j135
50 − j50
e
−e
(A),
◦
V+
◦
150 e− j135
[1 + 1] = 6e− j135
IeL = 0 [1 − Γ] =
Z0
50
8
iL (t) = 6 cos(6π × 10 t − 135◦ ) (A).
From output of Module 2.4, at d = 0 (load)
which corresponds to
e = 2.68∠−1.89 rad ,
I(d)
e = 2.68∠−108.29◦ .
I(d)
The equivalent time-domain current at f = 300 MHz is
iL (t) = 2.68 cos(6π × 108t − 108.29◦ ) (A).
(A),