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.
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 2.22 Using a slotted line, the following results were obtained: distance
of first minimum from the load = 4 cm; distance of second minimum from the load
= 14 cm; voltage standing-wave ratio = 1.5. If the line is lossless and Z0 = 50 Ω,
find the load impedance.
Solution: Following Example 2.6: Given a lossless line with Z0 = 50 Ω, S = 1.5,
dmin(0) = 4 cm, dmin(1) = 14 cm. Then
dmin(1) − dmin(0) =
λ
2
or
λ = 2 × (dmin(1) − dmin(0) ) = 20 cm
and
β=
2π
2π rad/cycle
=
= 10π rad/m.
λ
20 cm/cycle
From this we obtain
θr = 2β dmin(n) − (2n + 1)π rad = 2 × 10π rad/m × 0.04 m − π rad
= −0.2π rad = −36.0◦ .
Also,
|Γ| =
So
ZL = Z0
µ
1+Γ
1−Γ
¶
= 50
S − 1 1.5 − 1
=
= 0.2.
S + 1 1.5 + 1
Ã
1 + 0.2e− j36.0
1 − 0.2e− j36.0
◦!
◦
= (67.0 − j16.4) Ω.
Problem 2.47 Use the Smith chart to find the reflection coefficient corresponding
to a load impedance of
(a) ZL = 3Z0
(b) ZL = (2 − j2)Z0
(c) ZL = − j2Z0
(d) ZL = 0 (short circuit)
Solution: Refer to Fig. P2.47.
◦
(a) Point A is zL = 3 + j0. Γ = 0.5e0
◦
(b) Point B is zL = 2 − j2. Γ = 0.62e−29.7
◦
(c) Point C is zL = 0 − j2. Γ = 1.0e−53.1
◦
(d) Point D is zL = 0 + j0. Γ = 1.0e180.0
-90
0.39
0.4
0.38
0.9
1.2
1.0
0.11
-100
0.12
0.13
0.35
0.14
-80
0.15
4
1.4
-70
6
0.37
0.36
1.6
0.1
0.4
1
-110
0.0
9
0.4
2
CAP
-12 0.08
A
0
C
ITI
VE
0.4
RE
3
AC
0.0
TA
7
NC
-1
EC
30
O
M
PO
N
EN
T
(-j
0.8
1.8
2.0
9
Figure P2.47: Solution of Problem 2.47.
0.8
0.2
C
1
0.3
0.2
0.1
1.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
50
20
10
5.0
4.0
1.0
1.0
5.0
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
0.4
0.6
B
1.0
0.28
0.7
8
0.22
0.6
0.
-20
0.2
0.3
3.0
10
20
50
0.6
0.1
0.4
0.2
20
10
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
COEFFICIENT IN
0.27
REFLECTION
DEGR
LE OF
EES
ANG
31
0.
0.3
0.2
20
19
0.
0.5
0.6
0.28
0.6
06
0.
44
0.
06
0.5
2.0
1.8
0.3
0.22
A
50
9
0.
0.2
0.3
1
44
0.6
1.6
60
0.3
0.
1.2
1.0
0.9
0.8
1.4
0.7
4
0.2
-30
4
0
-4
0.0 —> WAVELEN
0.49
GTHS
TOW
ARD
0.48
<— 0.0
0.49
GEN
RD LOAD
ERA
TOWA
0.48
± 180
THS
TO
G
170
0
N
R—
-17
E
0.47
VEL
>
A
W
0.0
6
160
4
<—
0.4
-160
0.4
4
.0
6
0
IND
Yo)
UCT
0.0
/
5
15
B
j
0
(IVE
5
0
0.4
-15
CE
R
N
0.4
E
5
AC
TA
5
TA
0.0
EP
0.1
NC
SC
SU
EC
E
V
OM
14
0
TI
4
0
C
PO
-1
DU
N
EN
IN
R
T
(+
,O
jX
o)
Z
/Z
0.2
X/
0.3
0.2
-60
70
0.2
0.47
0.8
30
0.2
0.35
0.
0.2
0.1
0.36
7
80
0.1
0.13
3
0.37
0.3
D
4
0.4
31
0.
40
R
,O
o)
90
0.3
0
0.38
8
0.1
0
-5
13
0.12
2
.43
)
2
/Yo
0
0.4
12
(+jB
CE
AN
PT
CE
S
SU
VE
TI
CI
PA
CA
19
0.
0.
8
0
0.
7
0.0
0
110
0.4
0.0
0.39
100
0.3
0.11
3.0
8
.41
0.4
4.0
0.1
5.0
9
50
0.0
0.14
0.15
0.1
6
0.3
3
7
0.1
2
0.1
8
4.0
Problem 2.59 A 75-Ω lossless line is 0.6λ long. If S = 1.8 and θr = −60◦ , use the
Smith chart to find |Γ|, ZL , and Zin .
1.2
1.0
0.9
6
0.3
4
0.1
1.6
7
0.3
3
0.1
1.8
0.6
60
0.5
2
0.4
0.2
40
0.3
3.0
0.6
1
0.2
9
4.0
0.28
1.0
5.0
0.2
20
8
0.
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
COEFFICIENT IN
0.27
REFLECTION
DEGR
LE OF
EES
ANG
0.6
10
0.1
0.4
20
0.2
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
0.9
0.8
0.7
0.6
0.5
0.4
1.0
50
0.3
50
SWR
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
20
0.4
0.1
Z-IN
10
Z-LOAD
0.6
8
-20
0.
1.0
0.47
5.0
1.0
4.0
0.8
9
0.6
3.0
2.0
1.8
0.2
1.6
-60
1.4
-70
1.2
6
4
0.15
0.35
0.14
-80
0.36
0.9
0.1
0.3
1.0
7
3
0.8
0.1
0.3
0.7
2
0.6
8
0.1
0
-5
θr
0.3
0.5
31
0.
0.1
0.4
1
-110
0.0
9
0.4
2
0.0
CAP
-1
8
A
2
0
C
ITI
VE
0.4
RE
3
AC
0.0
TA
7
NC
-1
EC
30
O
M
PO
N
EN
T
(-j
06
0.4
19
0.
0.
0.3
0
-4
44
0.2
4
0.
1
0.2
0.2
-30
0.3
0.28
0.22
0.2
0.
0.22
1.0
50
4
0.
0.3
30
0.8
0.2
0.0 —> WAVELEN
0.49
GTHS
TOW
ARD
0.48
<— 0.0
0.49
GEN
RD LOAD
ERA
TOWA
0.48
± 180
THS
TO
G
170
0
N
R—
-17
E
0.47
VEL
>
A
W
0.0
6
160
4
<—
0.4
-160
0.4
4
.0
6
0
IND
Yo)
UCT
0.0
/
5
15
B
j
0
(IVE
5
0
0.4
-15
CE
R
N
0.4
E
5
AC
TA
5
TA
0.0
EP
0.1
NC
SC
SU
EC
E
V
OM
14
0
TI
4
0
C
PO
-1
DU
N
EN
IN
R
T
(+
,O
jX
o)
Z
/Z
0.2
X/
50
31
0.
R
,O
o)
8
0.3
2.0
0.2
20
0.
06
0.1
70
19
0.
0.
44
0.35
80
)
/Yo
0
0
12
(+jB
CE
AN
PT
CE
S
SU
VE
TI
CI
PA
CA
.42
3
0.4
0
13
0.15
0.36
90
1.4
0.0
7
0.0
110
0.14
0.37
0.38
0.7
8
0.8
1
0.4
0.39
100
0.4
9
0.0
0.13
0.12
0.11
0.1
-90
0.12
0.13
0.38
0.37
0.11
-100
0.4
0.39
0.100 λ
Figure P2.59: Solution of Problem 2.59.
Solution: Refer to Fig. P2.59. The SWR circle must pass through S = 1.8 at point
SWR. A circle of this radius has
|Γ| =
S−1
= 0.29.
S+1
The load must have a reflection coefficient with θr = −60◦ . The angle of the reflection
coefficient is read off that scale at the point θr . The intersection of the circle of
constant |Γ| and the line of constant θr is at the load, point Z-LOAD, which has a
value zL = 1.15 − j0.62. Thus,
ZL = zL Z0 = (1.15 − j0.62) × 75 Ω = (86.5 − j46.6) Ω.
A 0.6λ line is equivalent to a 0.1λ line. On the WTG scale, Z-LOAD is at 0.333λ ,
so Z-IN is at 0.333λ + 0.100λ = 0.433λ and has a value
zin = 0.63 − j0.29.
Therefore Zin = zin Z0 = (0.63 − j0.29) × 75 Ω = (47.0 − j21.8) Ω.