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.
4.4
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 2.30 Show that at the position where the magnitude of the voltage on the
line is a maximum, the input impedance is purely real.
Solution: From Eq. (2.70), dmax = (θr + 2nπ )/2β , so from Eq. (2.61), using polar
representation for Γ,
Ã
!
1 + |Γ|e jθr e− j2β lmax
Zin (dmax ) = Z0
1 − |Γ|e jθr e− j2β lmax
!
Ã
¶
µ
1 + |Γ|
1 + |Γ|e jθr e− j(θr +2nπ )
,
= Z0
= Z0
1 − |Γ|
1 − |Γ|e jθr e− j(θr +2nπ )
which is real, provided Z0 is real.
Problem 2.33 Two half-wave dipole antennas, each with an impedance of 75 Ω,
are connected in parallel through a pair of transmission lines, and the combination is
connected to a feed transmission line, as shown in Fig. P2.33.
λ
0.2
0.3λ
75 Ω
(Antenna)
Zin1
Zin2
Zin
0.2
λ
75 Ω
(Antenna)
Figure P2.33: Circuit for Problem 2.33.
All lines are 50 Ω and lossless.
(a) Calculate Zin1 , the input impedance of the antenna-terminated line, at the
parallel juncture.
(b) Combine Zin1 and Zin2 in parallel to obtain ZL′ , the effective load impedance of
the feedline.
(c) Calculate Zin of the feedline.
Solution:
(a)
Zin1
·
¸
ZL1 + jZ0 tan β l1
= Z0
Z0 + jZL1 tan β l1
½
¾
75 + j50 tan[(2π /λ )(0.2λ )]
= 50
= (35.20 − j8.62) Ω.
50 + j75 tan[(2π /λ )(0.2λ )]
(b)
ZL′ =
(c)
(35.20 − j8.62)2
Zin1 Zin2
=
= (17.60 − j4.31) Ω.
Zin1 + Zin2
2(35.20 − j8.62)
l = 0.3 λ
ZL'
Zin
Figure P2.33: (b) Equivalent circuit.
½
(17.60 − j4.31) + j50 tan[(2π /λ )(0.3λ )]
Zin = 50
50 + j(17.60 − j4.31) tan[(2π /λ )(0.3λ )]
¾
= (107.57 − j56.7) Ω.
Problem 2.50 Use the Smith chart to determine the input impedance Zin of the
two-line configuration shown in Fig. P2.50.
l1 = 3λ/8
C
B
Zin
Z01 = 100 Ω
l2 = 5λ/8
A
Z02 = 50 Ω
ZL = (75 − j50) Ω
Figure P2.50: Circuit for Problem 2.50.
Solution:
0.625 λ
1.0
1.4
4
0.1
1.6
7
0.3
3
0.1
1.8
0.6
60
0.5
50
2
0.4
0.2
40
0.3
3.0
0.6
1
0.2
9
0.2
30
4.0
1.0
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
50
4
0.
0.3
0.8
SWR Circle
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
20
0.4
0.1
10
0.6
Br
A
8
-20
0.
1.0
5.0
0.47
1.0
4.0
0.8
9
0.6
3.0
2.0
1.8
0.2
1.6
-60
1.4
-70
0.15
0.35
1.2
6
4
0.14
-80
0.36
1.0
0.1
0.3
0.9
7
3
0.8
0.1
0.3
0.7
0.308 λ
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
0.0
CA
1
P
8
A
2
0
CIT
I
V
0.4
ER
EA
3
CT
0.0
AN
7
CE
-1
30
CO
M
PO
N
EN
T
(-j
06
0.4
0.
0.3
19
0.
44
0.2
0.2
0
-4
4
0.
0.28
0.2
1
-30
0.3
-90
0.12
0.13
0.38
0.37
0.11
-100
0.433 λ
0.22
0.2
0.
0.22
0.0 —> WAVELEN
0.49
GTHS
TOW
ARD
0.48
— 0.0
0.49
GEN
D LOAD <
ERA
OWAR
0.48
± 180
HS T
T
TO
G
170
N
R—
-170
ELE
0.47
V
>
WA
0.0
6
160
0
—
6
4
.4
<
0
-1
0.4
4
6
0.0
IND
)
o
UCT
0.0
/Y
5
15
0
(-jB
I
5
5
V
0
E
1
0.4
ER
C
0.4
EA
AN
5
CT
5
PT
0.0
AN
CE
0.1
CE
US
ES
C
V
OM
I
14
40
0
CT
PO
-1
DU
N
EN
IN
R
T
O
(+
),
jX
o
Z
/Z
0.2
X/
8
0.3
2.0
0.2
20
0.
06
0.
44
6
0.3
31
0.
R
,O
o)
0
12
0.1
70
19
0.
3
0.4
0
13
0.35
80
Yo)
jB/
E (+
NC
TA
EP
SC
U
S
VE
TI
CI
PA
CA
2
0.4
7
0.15
0.36
90
0.7
0.0
0.0
110
0.14
0.37
0.38
1.2
1
0.4
0.8
0.4
9
0.0
8
0.39
100
0.9
0.1
0.13
0.12
0.11
0.4
0.39
Smith Chart 1
Starting at point A, namely at the load, we normalize ZL with respect to Z02 :
zL =
75 − j50
ZL
=
= 1.5 − j1.
Z02
50
(point A on Smith chart 1)
From point A on the Smith chart, we move on the SWR circle a distance of 5λ /8 to
point Br , which is just to the right of point B (see figure). At Br , the normalized input
impedance of line 2 is:
zin2 = 0.48 − j0.36
(point Br on Smith chart)
Next, we unnormalize zin2 :
Zin2 = Z02 zin2 = 50 × (0.48 − j0.36) = (24 − j18) Ω.
1.2
1.0
1.4
0.7
0.1
1.6
7
0.3
3
0.1
1.8
0.6
60
8
0.3
0.5
2.0
0.2
2
50
0.4
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
0.2
6
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
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
20
0.4
0.1
10
Bl
0.6
8
-20
0.
1.0
5.0
0.47
1.0
4.0
0.8
C
3.0
0.6
2.0
0.5
1.8
0.2
1.6
-60
1.4
-70
1.2
6
4
0.15
0.14
-80
0.35
0.36
1.0
0.1
0.3
0.9
7
3
0.7
0.1
0.3
0.8
8
2
0.6
0
0.1
0.3
0.1
0.4
1
-110
0.0
9
0
.4
2
CAP
-12 0.08
AC
0
ITI
V
0.4
E
RE
3
AC
0.0
TA
7
NC
-1
E
30
C
OM
PO
N
EN
T
(-j
06
0.4
31
0.
-5
0.
0.3
19
0.
44
0.2
0
-4
4
0.
0.28
9
0.2
1
-30
0.2
0.3
0.
0.22
0.2
0.470 λ
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
17
0
N
0
R—
-17
E
0.47
VEL
>
A
W
0.0
6
160
4
<—
0.4
-160
0.4
4
6
0.0
IND
o)
Y
U
0.0
/
CTI
5
15
jB
0
5
VE
0
E (0.4
-15
C
RE
N
0.4
5
AC
TA
5
TA
0.0
EP
0.1
NC
SC
U
EC
ES
V
O
I
1
M
40
40
CT
PO
-1
DU
N
EN
IN
R
T
(+
,O
jX
o)
Z
/
0.2
Z
X/
0.2
40
SWR Circle
20
06
0.
4
31
44
6
0.3
0.
0.
0.1
70
19
R
,O
o)
0.375 λ
0.35
80
0.
3
0.4
0
13
0.15
0.36
90
Yo)
0
0
jB/
12
E (+
NC
TA
EP
C
S
SU
VE
TI
CI
PA
CA
.42
7
0.0
110
0.14
0.37
0.38
0.9
1
0.4
0.8
9
0.0
8
0.0
0.39
100
0.4
0.13
0.12
0.11
0.1
0.345 λ
-90
0.12
0.13
0.38
0.37
0.11
-100
0.4
0.39
Smith Chart 2
To move along line 1, we need to normalize with respect to Z01 . We shall call this zL1 :
zL1 =
Zin2 24 − j18
=
= 0.24 − j0.18
Z01
100
(point Bℓ on Smith chart 2)
After drawing the SWR circle through point Bℓ , we move 3λ /8 towards the generator,
ending up at point C on Smith chart 2. The normalized input impedance of line 1 is:
zin = 0.66 − j1.25
which upon unnormalizing becomes:
Zin = (66 − j125) Ω.
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 .
Problem 2.68 A 50-Ω lossless line is to be matched to an antenna with
ZL = (75 − j20) Ω using a shorted stub. Use the Smith chart to determine the stub
length and distance between the antenna and stub.
1.0
1.2
1.4
1.6
7
3
0.1
1.8
0.6
0.3
0.2
2
0.4
0.2
40
0.3
3.0
0.6
1
0.2
9
4
0.
0.3
30
0.8
0.2
4.0
1.0
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
Y-LOAD-IN-1
0.6
10
0.1
0.4
Y-LOAD
20
0.2
10
5.0
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
Y-SHT
50
0.3
50
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
20
Z-LOAD
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
1.4
-70
0.15
0.35
1.2
6
4
0.173 λ
0.14
-80
0.36
0.9
0.1
0.3
1.0
3
0.8
7
0.3
0.7
0.1
0.6
8
0.1
0
-5
2
Y-STUB-IN-1
0.3
0.5
31
0.
0.1
0.4
1
-110
0.0
9
0
.4
2
CAP
-12 0.08
AC
0
I
T
IVE
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
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/
8
0.3
50
50
0.
06
0.1
60
31
0.
0.
44
4
19
0.
R
,O
o)
6
0.3
20
0
13
0
12
0.1
70
2.0
0
0.35
80
0.5
.43
0.15
0.36
90
Yo)
jB/
E (+
NC
TA
EP
SC
SU
E
V
TI
CI
PA
CA
2
0.4
0
110
0.14
0.37
0.38
0.7
0.0
.07
0.8
1
0.4
8
0.39
100
0.4
0.13
0.12
0.11
0.1
9
0.0
0.9
0.104 λ
-90
0.12
0.13
0.38
0.37
0.11
-100
0.4
0.39
Figure P2.68: (a) First solution to Problem 2.68.
Solution: Refer to Fig. P2.68(a) and Fig. P2.68(b), which represent two different
solutions.
ZL (75 − j20) Ω
zL =
=
= 1.5 − j0.4
Z0
50 Ω
and is located at point Z-LOAD in both figures. Since it is advantageous to work in
admittance coordinates, yL is plotted as point Y -LOAD in both figures. Y -LOAD is at
0.041λ on the WTG scale.
For the first solution in Fig. P2.68(a), point Y -LOAD-IN-1 represents the point
at which g = 1 on the SWR circle of the load. Y -LOAD-IN-1 is at 0.145λ on the
WTG scale, so the stub should be located at 0.145λ − 0.041λ = 0.104λ from the
load (or some multiple of a half wavelength further). At Y -LOAD-IN-1, b = 0.52,
so a stub with an input admittance of ystub = 0 − j0.52 is required. This point is
Y -STUB-IN-1 and is at 0.423λ on the WTG scale. The short circuit admittance
is denoted by point Y -SHT, located at 0.250λ . Therefore, the short stub must be
0.423λ − 0.250λ = 0.173λ long (or some multiple of a half wavelength longer).
1.0
0.9
1.2
1.6
7
3
0.1
1.8
0.6
0.5
2
50
0.4
0.314 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
10
0.1
0.4
Y-LOAD
20
0.2
10
5.0
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
Y-SHT
50
0.3
50
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
20
Z-LOAD
0.4
0.1
10
Y-LOAD-IN-2
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
0.15
0.35
1.2
6
4
0.14
-80
0.36
0.9
0.1
0.3
1.0
7
3
06
0.7
0.1
0.3
0.8
8
2
0.6
0
0.1
0.3
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
0.5
31
0.
-5
0.
0.4
9
0.
1
-90
0.12
0.13
0.38
0.37
0.11
-100
0.39
0.4
0.327 λ
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/
8
0.3
2.0
Y-STUB-IN-2
0.3
20
06
0.
0.1
60
0.2
31
44
4
0.
0.
6
0.3
19
R
,O
o)
0.1
70
0.
3
0.4
0
13
0.35
80
)
/Yo
0
0
12
(+jB
CE
AN
PT
CE
S
SU
VE
TI
CI
PA
CA
.42
7
0.15
0.36
90
1.4
0.0
0.0
110
0.14
0.37
0.38
0.7
8
0.8
0.4
9
0.0
1
0.4
0.39
100
0.13
0.12
0.11
0.1
Figure P2.68: (b) Second solution to Problem 2.68.
For the second solution in Fig. P2.68(b), point Y -LOAD-IN-2 represents the point
at which g = 1 on the SWR circle of the load. Y -LOAD-IN-2 is at 0.355λ on the
WTG scale, so the stub should be located at 0.355λ − 0.041λ = 0.314λ from the
load (or some multiple of a half wavelength further). At Y -LOAD-IN-2, b = −0.52,
so a stub with an input admittance of ystub = 0 + j0.52 is required. This point is
Y -STUB-IN-2 and is at 0.077λ on the WTG scale. The short circuit admittance
is denoted by point Y -SHT, located at 0.250λ . Therefore, the short stub must be
0.077λ − 0.250λ + 0.500λ = 0.327λ long (or some multiple of a half wavelength
longer).
Problem 2.74 A 25-Ω antenna is connected to a 75-Ω lossless transmission
line. Reflections back toward the generator can be eliminated by placing a shunt
impedance Z at a distance l from the load (Fig. P2.74). Determine the values of Z
and l.
l=?
B
Z0 = 75 Ω
A
ZL = 25 Ω
Z=?
Figure P2.74: Circuit for Problem 2.74.
Solution:
0.250 λ
1.0
1.2
1.6
7
0.1
1.8
0.6
3
2.0
0.5
2
0.2
40
0.3
3.0
0.6
1
9
0.8
0.2
30
SWR Circle
0.2
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
50
20
10
5.0
4.0
3.0
2.0
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
B
50
0.3
50
A
1.8
4
0.
0.3
1.0
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
4
1.4
-70
0.15
0.35
1.2
6
0.3
0.14
-80
0.36
0.9
0.1
1.0
7
3
0.7
0.1
0.3
0.8
8
2
0.6
0
0.1
0.3
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
0.5
31
0.
-5
-90
0.12
0.13
0.38
0.37
0.11
-100
0.39
0.4
06
0.4
9
0.
1
0.
0.3
0
-4
44
0.2
4
0.
0.2
0.2
1
-30
0.3
0.28
0.22
0.2
0.
0.22
0.0 —> WAVELEN
0.49
GTHS
TOW
ARD
0.48
— 0.0
0.49
GEN
D LOAD <
ERA
OWAR
0.48
± 180
HS T
T
TO
G
170
0
R—
-17
EN
0.47
VEL
>
A
W
0.0
6
160
4
<—
0.4
-160
0.4
4
.0
6
0
IND
o)
Y
U
0.0
/
C
5
15
TIV
0
(-jB
5
0
0.4
ER
-15
CE
N
0.4
EA
A
5
CT
5
PT
0.0
AN
CE
0.1
CE
US
S
CO
VE
14
0
M
TI
4
0
C
PO
-1
DU
N
EN
IN
R
T
O
(+
),
jX
o
Z
/Z
0.2
X/
8
0.3
50
0.4
1.6
06
0.
44
0.1
0.3
60
0.2
31
0.
4
0.
0.500 λ
6
0.3
19
R
,O
o)
0.1
70
0.
3
0.4
0
13
0.35
80
)
/Yo
0
0
12
(+jB
CE
AN
PT
CE
S
SU
VE
TI
CI
PA
CA
.42
7
0.15
0.36
90
1.4
0.0
0.0
110
0.14
0.37
0.38
0.7
8
0.8
0.4
9
0.0
1
0.4
0.39
100
0.9
0.1
0.13
0.12
0.11
0.750 λ
The normalized load impedance is:
zL =
25
= 0.33
75
(point A on Smith chart)
The Smith chart shows A and the SWR circle. The goal is to have an equivalent
impedance of 75 Ω to the left of B. That equivalent impedance is the parallel
combination of Zin at B (to the right of the shunt impedance Z) and the shunt
element Z. Since we need for this to be purely real, it’s best to choose l such that
Zin is purely real, thereby choosing Z to be simply a resistor. Adding two resistors in
parallel generates a sum smaller in magnitude than either one of them. So we need
for Zin to be larger than Z0 , not smaller. On the Smith chart, that point is B, at a
distance l = λ /4 from the load. At that point:
zin = 3,
which corresponds to
yin = 0.33.
Hence, we need y, the normalized admittance corresponding to the shunt
impedance Z, to have a value that satisfies:
yin + y = 1
y = 1 − yin = 1 − 0.33 = 0.66
1
1
z= =
= 1.5
y 0.66
Z = 75 × 1.5 = 112.5 Ω.
In summary,
λ
,
4
Z = 112.5 Ω.
l=