Problem 2.20 A 300-Ω lossless air transmission line is connected to a complex
load composed of a resistor in series with an inductor, as shown in Fig. P2.20. At
5 MHz, determine: (a) Γ, (b) S, (c) location of voltage maximum nearest to the load,
and (d) location of current maximum nearest to the load.
R = 600 Ω
Z0 = 300 Ω
L = 0.02 mH
Figure P2.20: Circuit for Problem 2.20.
Solution:
(a)
ZL = R + j ω L
= 600 + j2π × 5 × 106 × 2 × 10−5 = (600 + j628) Ω.
ZL − Z 0
ZL + Z0
600 + j628 − 300
=
600 + j628 + 300
◦
300 + j628
= 0.63e j29.6 .
=
900 + j628
Γ=
(b)
S=
(c)
1 + |Γ| 1 + 0.63
=
= 1.67.
4.405
1 − |Γ| 1 − 0.63
θr λ
for θr > 0.
4
π
µ
¶
29.6◦ π 60
,
=
180◦
4π
= 2.46 m
lmax =
µ
¶
3 × 108
λ=
=
60
m
5 × 106
(d) The locations of current maxima correspond to voltage minima and vice versa.
Hence, the location of current maximum nearest the load is the same as location of
voltage minimum nearest the load. Thus
µ
¶
λ
λ
lmin = lmax + ,
lmax < = 15 m
4
4
= 2.46 + 15 = 17.46 m.
Problem 2.23 A load with impedance ZL = (25 − j50) Ω is to be connected to a
lossless transmission line with characteristic impedance Z0 , with Z0 chosen such that
the standing-wave ratio is the smallest possible. What should Z0 be?
Solution: Since S is monotonic with |Γ| (i.e., a plot of S vs. |Γ| is always increasing),
the value of Z0 which gives the minimum possible S also gives the minimum possible
|Γ|, and, for that matter, the minimum possible |Γ|2 . A necessary condition for a
minimum is that its derivative be equal to zero:
∂
∂ |RL + jXL − Z0 |2
2
|Γ| =
0=
∂ Z0
∂ Z0 |RL + jXL + Z0 |2
=
∂ (RL − Z0 )2 + XL2 4RL (Z02 − (R2L + XL2 ))
=
.
2
∂ Z0 (RL + Z0 )2 + XL2
((RL + Z0 )2 + XL2 )
Therefore, Z02 = R2L + XL2 or
q
Z0 = |ZL | = (252 + (−50)2 ) = 55.9 Ω.
A mathematically precise solution will also demonstrate that this point is a
minimum (by calculating the second derivative, for example). Since the endpoints
of the range may be local minima or maxima without the derivative being zero there,
the endpoints (namely Z0 = 0 Ω and Z0 = ∞ Ω) should be checked also.
Problem 2.26 A 50-Ω lossless transmission line is connected to a load composed
of a 75-Ω resistor in series with a capacitor of unknown capacitance (Fig. P2.26). If at
10 MHz the voltage standing wave ratio on the line was measured to be 3, determine
the capacitance C.
RL = 75 Ω
Z0 = 50 Ω
C=?
Figure P2.26: Circuit for Problem 2.26.
Solution:
|Γ| =
S−1 3−1 2
=
= = 0.5
S+1 3+1 4
ZL = RL − jXC ,
where XC =
ZL − Z 0
ZL + Z0
·µ
¶µ ∗
¶¸
ZL − Z 0
ZL − Z 0
2
|Γ| =
ZL + Z0
ZL∗ + Z0
1
.
ωC
Γ=
|Γ|2 =
ZL ZL∗ + Z02 − Z0 (ZL + ZL∗ )
ZL ZL∗ + Z02 + Z0 (ZL + ZL∗ )
Noting that:
ZL ZL∗ = (RL − jXC )(RL + jXC ) = R2L + XC2 ,
Z0 (ZL + ZL∗ ) = Z0 (RL − jXC + RL + jXC ) = 2Z0 RL ,
|Γ|2 =
R2L + XC2 + Z02 − 2Z0 RL
.
R2L + XC2 + Z02 + 2Z0 RL
Upon substituting |ΓL | = 0.5, RL = 75 Ω, and Z0 = 50 Ω, and then solving for XC ,
we have
XC = 66.1 Ω.
Hence
C=
1
1
=
= 2.41 × 10−10 = 241 pF.
ω XC 2π × 107 × 66.1
1.4
Problem 2.19 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 2.21 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 − j0.5
1+Γ
= 150
= (90 − j120) Ω.
ZL = Z0
1−Γ
1 + j0.5
λ
8.
Problem 2.24 A 50-Ω lossless line terminated in a purely resistive load has a
voltage standing-wave ratio of 3. Find all possible values of ZL .
Solution:
S−1 3−1
=
= 0.5.
S+1 3+1
For a purely resistive load, θr = 0 or π . For θr = 0,
·
¸
·
¸
1+Γ
1 + 0.5
ZL = Z0
= 50
= 150 Ω.
1−Γ
1 − 0.5
|Γ| =
For θr = π , Γ = −0.5 and
·
¸
1 − 0.5
ZL = 50
= 15 Ω.
1 + 0.5
Problem 2.25 Apply CD Module 2.4 to generate plots of the voltage standingwave pattern for a 50-Ω line terminated in a load impedance ZL = (100 − j50) Ω.
Set Vg = 1 V, Zg = 50 Ω, εr = 2.25, l = 40 cm, and f = 1 GHz. Also determine S,
dmax , and dmin .
Solution: