Problem 2.53 A lossless 50-Ω transmission line is terminated in a load with
ZL = (50 + j25) Ω. Use the Smith chart to find the following:
(a) The reflection coefficient Γ.
(b) The standing-wave ratio.
(c) The input impedance at 0.35λ from the load.
(d) The input admittance at 0.35λ from the load.
(e) The shortest line length for which the input impedance is purely resistive.
(f) The position of the first voltage maximum from the load.
1.0
0.9
0.35
0.1
6
70
0.3
4
1.4
0.1
7
1.6
)
2
/Yo
0
0.4
12
(+jB
CE
AN
PT
CE
S
SU
VE
TI
CI
PA
CA
60
0.1
0.4
0.106 λ
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
Z-LOAD
10
0.1
0.4
20
0.2
10
5.0
SWR
4.0
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
0.3
50
Z-IN
50
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
20
0.4
0.1
10
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
6
4
1.4
-70
0.15
0.35
1.2
0.1
0.3
0.14
-80
0.36
0.9
7
1.0
0.1
3
0.8
0.350 λ
0.3
0.7
2
0.6
8
0.1
0
-5
0.3
0.5
31
0.
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
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/
2
31
0.
R
,O
o)
50
19
0.
13
8
0.3
0.5
0
0.2
20
0
3
2.0
.43
0.3
1.8
0.6
7
0.0
0.
06
0.15
θr
0.7
0.0
0.
44
80
1.2
0
8
110
0.36
90
0.8
.41
0.14
0.37
0.38
0.39
100
0.4
0.13
0.12
0.11
0.1
9
0.0
-90
0.12
0.13
0.38
0.37
0.11
-100
0.4
0.39
Figure P2.53: Solution of Problem 2.53.
Solution: Refer to Fig. P2.53. The normalized impedance
zL =
(50 + j25) Ω
= 1 + j0.5
50 Ω
is at point Z-LOAD.
◦
(a) Γ = 0.24e j76.0 The angle of the reflection coefficient is read of that scale at
the point θr .
(b) At the point SW R: S = 1.64.
(c) Zin is 0.350λ from the load, which is at 0.144λ on the wavelengths to generator
scale. So point Z-IN is at 0.144λ + 0.350λ = 0.494λ on the WTG scale. At point
Z-IN:
Zin = zin Z0 = (0.61 − j0.022) × 50 Ω = (30.5 − j1.09) Ω.
(d) At the point on the SWR circle opposite Z-IN,
Yin =
yin (1.64 + j0.06)
=
= (32.7 + j1.17) mS.
Z0
50 Ω
(e) Traveling from the point Z-LOAD in the direction of the generator (clockwise),
the SWR circle crosses the xL = 0 line first at the point SWR. To travel from
Z-LOAD to SWR one must travel 0.250λ − 0.144λ = 0.106λ . (Readings are on the
wavelengths to generator scale.) So the shortest line length would be 0.106λ .
(f) The voltage max occurs at point SWR. From the previous part, this occurs at
z = −0.106λ .