l 0 (b) zl =( j z l j l solution: refer to fig. p2.47. (a ...faculty.uml.edu/xwang/16.360/homework 5...
TRANSCRIPT
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 = 0.5exp 0
(b) Point B is zL = 2− j2. Γ = 0.62exp−29.7
(c) Point C is zL = 0− j2. Γ = 1.0exp−53.1
(d) Point D is zL = 0+ j0. Γ = 1.0exp 180.0
0.1
0.1
0.1
0.2
0.2
0.2
0.3
0.3
0.3
0.4
0.4
0.4
0.5
0.5
0.5
0.6
0.6
0.6
0.7
0.7
0.7
0.8
0.8
0.8
0.9
0.9
0.9
1.0
1.0
1.0
1.2
1.2
1.2
1.4
1.4
1.4
1.6
1.6
1.6
1.8
1.8
1.8
2.0
2.0
2.0
3.0
3.0
3.0
4.0
4.0
4.0
5.0
5.0
5.0
10
10
10
20
20
20
50
50
50
0.2
0.2
0.2
0.2
0.4
0.4
0.4
0.4
0.6
0.6
0.6
0.6
0.8
0.8
0.8
0.8
1.0
1.0
1.0
1.0
20-20
30-30
40-40
50
-50
60
-60
70
-70
80
-80
90
-90
100
-100
110
-110
120
-120
130
-130
140
-140
150
-150
160
-160
170
-170
180
±
0.04
0.04
0.05
0.05
0.06
0.06
0.07
0.07
0.08
0.08
0.09
0.09
0.1
0.1
0.11
0.11
0.12
0.12
0.13
0.13
0.14
0.14
0.15
0.15
0.16
0.16
0.17
0.17
0.18
0.18
0.190.19
0.2
0.2
0.21
0.210.22
0.220.23
0.230.24
0.24
0.25
0.25
0.26
0.26
0.27
0.27
0.28
0.28
0.29
0.29
0.3
0.3
0.31
0.31
0.32
0.32
0.33
0.33
0.34
0.34
0.35
0.35
0.36
0.36
0.37
0.37
0.38
0.38
0.39
0.39
0.4
0.4
0.41
0.41
0.42
0.42
0.43
0.43
0.44
0.44
0.45
0.45
0.46
0.46
0.47
0.47
0.48
0.48
0.49
0.49
0.0
0.0
AN
GL
E O
F R
EF
LE
CT
ION
CO
EF
FIC
IEN
T IN
DE
GR
EE
S
—>
WA
VE
LE
NG
TH
S T
OW
AR
D G
EN
ER
AT
OR
—>
<—
WA
VE
LE
NG
TH
S T
OW
AR
D L
OA
D <
—
IND
UC
TIV
E R
EA
CT
AN
CE
CO
MPO
NEN
T (+jX
/Zo), O
R CAPACIT
IVE SUSCEPTANCE (+jB/Yo)
CAPACITIVE REACTANCE COM
PONEN
T (-jX
/Zo), O
R IN
DU
CT
IVE
SU
SCE
PT
AN
CE
(-j
B/
Yo
)
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
A
B
C
D
Figure P2.47: Solution of Problem 2.47.
2.49 Use the Smith chart to find the normalized load impedance corresponding to a
reflection coefficient of
(a) Γ = 0.5
(b) Γ = 0.5∠60
(c) Γ = −1
(d) Γ = 0.3∠−30
(e) Γ = 0
(f) Γ = j
0.1
0.1
0.1
0.2
0.2
0.2
0.3
0.3
0.3
0.4
0.4
0.4
0.5
0.5
0.5
0.6
0.6
0.6
0.7
0.7
0.7
0.8
0.8
0.8
0.9
0.9
0.9
1.0
1.0
1.0
1.2
1.2
1.2
1.4
1.4
1.4
1.6
1.6
1.6
1.8
1.8
1.8
2.0
2.0
2.0
3.0
3.0
3.0
4.0
4.0
4.0
5.0
5.0
5.0
10
10
10
20
20
20
50
50
50
0.2
0.2
0.2
0.2
0.4
0.4
0.4
0.4
0.6
0.6
0.6
0.6
0.8
0.8
0.8
0.8
1.0
1.0
1.0
1.0
20-20
30-30
40-40
50
-50
60
-60
70
-70
80
-80
90
-90
100
-100
110
-110
120
-120
130
-130
140
-140
150
-150
160
-160
170
-170
180
±
0.04
0.04
0.05
0.05
0.06
0.06
0.07
0.07
0.08
0.08
0.09
0.09
0.1
0.1
0.11
0.11
0.12
0.12
0.13
0.13
0.14
0.14
0.15
0.15
0.16
0.16
0.17
0.17
0.18
0.18
0.190.19
0.2
0.2
0.21
0.210.22
0.220.23
0.230.24
0.24
0.25
0.25
0.26
0.26
0.27
0.27
0.28
0.28
0.29
0.29
0.3
0.3
0.31
0.31
0.32
0.32
0.33
0.33
0.34
0.34
0.35
0.35
0.36
0.36
0.37
0.37
0.38
0.38
0.39
0.39
0.4
0.4
0.41
0.41
0.42
0.42
0.43
0.43
0.44
0.44
0.45
0.45
0.46
0.46
0.47
0.47
0.48
0.48
0.49
0.49
0.0
0.0
AN
GL
E O
F R
EF
LE
CT
ION
CO
EF
FIC
IEN
T IN
DE
GR
EE
S
—>
WA
VE
LE
NG
TH
S T
OW
AR
D G
EN
ER
AT
OR
—>
<—
WA
VE
LE
NG
TH
S T
OW
AR
D L
OA
D <
—
IND
UC
TIV
E R
EA
CT
AN
CE
CO
MPO
NEN
T (+jX
/Zo), O
R CAPACIT
IVE SUSCEPTANCE (+jB/Yo)
CAPACITIVE REACTANCE COM
PONEN
T (-jX
/Zo), O
R IN
DU
CT
IVE
SU
SCE
PT
AN
CE
(-j
B/
Yo
)
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
A’
B’
C’
D’
E’
F’
Figure P2.49: Solution of Problem 2.49.
Solution: Refer to Fig. P2.49.
(a) Point A′ is Γ = 0.5 at zL = 3+ j0.
(b) Point B′ is Γ = 0.5exp j60 at zL = 1+ j1.15.
(c) Point C ′ is Γ = −1 at zL = 0+ j0.
(d) Point D′ is Γ = 0.3exp− j30 at zL = 1.60− j0.53.
(e) Point E ′ is Γ = 0 at zL = 1+ j0.
(f) Point F ′ is Γ = j at zL = 0+ j1.
2.50 Use the Smith chart to determine the input impedance Zin of the two-line
configuration shown in Fig. P2.50.
Z01 = 100 Ω
l1 = 3λ/8 l2 = 5λ/8
Z02 = 50 ΩZin
BC A
ZL = (75 − j50) Ω
Figure P2.50 Circuit for Problem 2.50.
Solution:
0.1
0.1
0.1
0.2
0.2
0.2
0.3
0.3
0.3
0.4
0.4
0.4
0.5
0.5
0.5
0.6
0.6
0.6
0.7
0.7
0.7
0.8
0.8
0.8
0.9
0.9
0.9
1.0
1.0
1.0
1.2
1.2
1.2
1.4
1.4
1.4
1.6
1.6
1.6
1.8
1.8
1.8
2.0
2.0
2.0
3.0
3.0
3.0
4.0
4.0
4.0
5.0
5.0
5.0
10
10
10
20
20
20
50
50
50
0.2
0.2
0.2
0.2
0.4
0.4
0.4
0.4
0.6
0.6
0.6
0.6
0.8
0.8
0.8
0.8
1.0
1.0
1.0
1.0
20-20
30-30
40-40
50
-50
60
-60
70
-70
80
-80
90
-90
100
-100
110
-110
120
-120
130
-130
140
-140
150
-150
160
-160
170
-170
180
±
0.04
0.04
0.05
0.05
0.06
0.06
0.07
0.07
0.08
0.08
0.09
0.09
0.1
0.1
0.11
0.11
0.12
0.12
0.13
0.13
0.14
0.14
0.15
0.15
0.16
0.16
0.17
0.17
0.18
0.18
0.190.19
0.2
0.2
0.21
0.210.22
0.220.23
0.230.24
0.24
0.25
0.25
0.26
0.26
0.27
0.27
0.28
0.28
0.29
0.29
0.3
0.3
0.31
0.31
0.32
0.32
0.33
0.33
0.34
0.34
0.35
0.35
0.36
0.36
0.37
0.37
0.38
0.38
0.39
0.39
0.4
0.4
0.41
0.41
0.42
0.42
0.43
0.43
0.44
0.44
0.45
0.45
0.46
0.46
0.47
0.47
0.48
0.48
0.49
0.49
0.0
0.0
AN
GL
E O
F R
EF
LE
CT
ION
CO
EF
FIC
IEN
T IN
DE
GR
EE
S
—>
WA
VE
LE
NG
TH
S T
OW
AR
D G
EN
ER
AT
OR
—>
<—
WA
VE
LE
NG
TH
S T
OW
AR
D L
OA
D <
—
IND
UC
TIV
E R
EA
CT
AN
CE
CO
MPO
NEN
T (+jX
/Zo), O
R CAPACIT
IVE SUSCEPTANCE (+jB/Yo)
CAPACITIVE REACTANCE COM
PONEN
T (-jX
/Zo), O
R IN
DU
CT
IVE
SU
SCE
PT
AN
CE
(-j
B/
Yo
)
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
ABr
0.625 λ
0.308 λ
0.433 λ
SWR Circle
Smith Chart 1
Starting at point A, namely at the load, we normalize ZL with respect to Z02:
zL =ZL
Z02
=75− j50
50= 1.5− j1. (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 = Z02zin2 = 50× (0.48− j0.36) = (24− j18) Ω.
0.1
0.1
0.1
0.2
0.2
0.2
0.3
0.3
0.3
0.4
0.4
0.4
0.5
0.5
0.5
0.6
0.6
0.6
0.7
0.7
0.7
0.8
0.8
0.8
0.9
0.9
0.9
1.0
1.0
1.0
1.2
1.2
1.2
1.4
1.4
1.4
1.6
1.6
1.6
1.8
1.8
1.8
2.0
2.0
2.0
3.0
3.0
3.0
4.0
4.0
4.0
5.0
5.0
5.0
10
10
10
20
20
20
50
50
50
0.2
0.2
0.2
0.2
0.4
0.4
0.4
0.4
0.6
0.6
0.6
0.6
0.8
0.8
0.8
0.8
1.0
1.0
1.0
1.0
20-20
30-30
40-40
50
-50
60
-60
70
-70
80
-80
90
-90
100
-100
110
-110
120
-120
130
-130
140
-140
150
-150
160
-160
170
-170
180
±
0.04
0.04
0.05
0.05
0.06
0.06
0.07
0.07
0.08
0.08
0.09
0.09
0.1
0.1
0.11
0.11
0.12
0.12
0.13
0.13
0.14
0.14
0.15
0.15
0.16
0.16
0.17
0.17
0.18
0.18
0.190.19
0.2
0.2
0.21
0.210.22
0.220.23
0.230.24
0.24
0.25
0.25
0.26
0.26
0.27
0.27
0.28
0.28
0.29
0.29
0.3
0.3
0.31
0.31
0.32
0.32
0.33
0.33
0.34
0.34
0.35
0.35
0.36
0.36
0.37
0.37
0.38
0.38
0.39
0.39
0.4
0.4
0.41
0.41
0.42
0.42
0.43
0.43
0.44
0.44
0.45
0.45
0.46
0.46
0.47
0.47
0.48
0.48
0.49
0.49
0.0
0.0
AN
GL
E O
F R
EF
LE
CT
ION
CO
EF
FIC
IEN
T IN
DE
GR
EE
S
—>
WA
VE
LE
NG
TH
S T
OW
AR
D G
EN
ER
AT
OR
—>
<—
WA
VE
LE
NG
TH
S T
OW
AR
D L
OA
D <
—
IND
UC
TIV
E R
EA
CT
AN
CE
CO
MPO
NEN
T (+jX
/Zo), O
R CAPACIT
IVE SUSCEPTANCE (+jB/Yo)
CAPACITIVE REACTANCE COM
PONEN
T (-jX
/Zo), O
R IN
DU
CT
IVE
SU
SCE
PT
AN
CE
(-j
B/
Yo
)
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
Bl
C
0.375 λ
0.470 λ
0.345 λ
SWR Circle
Smith Chart 2
To move along line 1, we need to normalize with respect to Z01. We shall call this zL1:
zL1 =Zin2
Z01
=24− j18
100= 0.24− j0.18 (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.
2.61 Using a slotted line on a 50-Ω air-spaced lossless line, the following
measurements were obtained: S = 1.6 and |V |max occurred only at 10 cm and 24 cm
from the load. Use the Smith chart to find ZL.
0.1
0.1
0.1
0.2
0.2
0.2
0.3
0.3
0.3
0.4
0.4
0.4
0.5
0.5
0.5
0.6
0.6
0.6
0.7
0.7
0.7
0.8
0.8
0.8
0.9
0.9
0.9
1.0
1.0
1.0
1.2
1.2
1.2
1.4
1.4
1.4
1.6
1.6
1.6
1.8
1.8
1.8
2.0
2.0
2.0
3.0
3.0
3.0
4.0
4.0
4.0
5.0
5.0
5.0
10
10
10
20
20
20
50
50
50
0.2
0.2
0.2
0.2
0.4
0.4
0.4
0.4
0.6
0.6
0.6
0.6
0.8
0.8
0.8
0.8
1.0
1.0
1.0
1.0
20-20
30-30
40-40
50
-50
60
-60
70
-70
80
-80
90
-90
100
-100
110
-110
120
-120
130
-130
140
-140
150
-150
160
-160
170
-170
180
±
0.04
0.04
0.05
0.05
0.06
0.06
0.07
0.07
0.08
0.08
0.09
0.09
0.1
0.1
0.11
0.11
0.12
0.12
0.13
0.13
0.14
0.14
0.15
0.15
0.16
0.16
0.17
0.17
0.18
0.18
0.190.19
0.2
0.2
0.21
0.210.22
0.220.23
0.230.24
0.24
0.25
0.25
0.26
0.26
0.27
0.27
0.28
0.28
0.29
0.29
0.3
0.3
0.31
0.31
0.32
0.32
0.33
0.33
0.34
0.34
0.35
0.35
0.36
0.36
0.37
0.37
0.38
0.38
0.39
0.39
0.4
0.4
0.41
0.41
0.42
0.42
0.43
0.43
0.44
0.44
0.45
0.45
0.46
0.46
0.47
0.47
0.48
0.48
0.49
0.49
0.0
0.0
AN
GL
E O
F R
EF
LE
CT
ION
CO
EF
FIC
IEN
T IN
DE
GR
EE
S
—>
WA
VE
LE
NG
TH
S T
OW
AR
D G
EN
ER
AT
OR
—>
<—
WA
VE
LE
NG
TH
S T
OW
AR
D L
OA
D <
—
IND
UC
TIV
E R
EA
CT
AN
CE
CO
MPO
NEN
T (+jX
/Zo), O
R CAPACIT
IVE SUSCEPTANCE (+jB/Yo)
CAPACITIVE REACTANCE COM
PONEN
T (-jX
/Zo), O
R IN
DU
CT
IVE
SU
SCE
PT
AN
CE
(-j
B/
Yo
)
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.357 λ
SWR
Z-LOAD
Figure P2.61: Solution of Problem 2.61.
Solution: Refer to Fig. P2.61. The point SWR denotes the fact that S = 1.6.
This point is also the location of a voltage maximum. From the knowledge of the
locations of adjacent maxima we can determine that λ = 2(24 cm−10 cm) = 28 cm.
Therefore, the load is 10 cm28 cm
λ = 0.357λ from the first voltage maximum, which is at
0.250λ on the WTL scale. Traveling this far on the SWR circle we find point Z-LOAD
at 0.250λ + 0.357λ −0.500λ = 0.107λ on the WTL scale, and here
zL = 0.82− j0.39.
Therefore ZL = zLZ0 = (0.82− j0.39)×50 Ω = (41.0− j19.5) Ω.
2.66 A 200-Ω transmission line is to be matched to a computer terminal with ZL =(50− j25) Ω by inserting an appropriate reactance in parallel with the line. If f = 800
MHz and εr = 4, determine the location nearest to the load at which inserting:
(a) A capacitor can achieve the required matching, and the value of the capacitor.
(b) An inductor can achieve the required matching, and the value of the inductor.
Solution:
(a) After entering the specified values for ZL and Z0 into Module 2.6, we have zL
represented by the red dot in Fig. P2.66(a), and yL represented by the blue dot. By
moving the cursor a distance d = 0.093λ , the blue dot arrives at the intersection point
between the SWR circle and the S = 1 circle. At that point
y(d) = 1.026126− j1.5402026.
To cancel the imaginary part, we need to add a reactive element whose admittance is
positive, such as a capacitor. That is:
ωC = (1.54206)×Y0
=1.54206
Z0
=1.54206
200= 7.71×10−3,
which leads to
C =7.71×10−3
2π ×8×108= 1.53×10−12 F.
Figure P2.66(a)
(b) Repeating the procedure for the second intersection point [Fig. P2.66(b)] leads
to
y(d) = 1.000001+ j1.520691,
at d2 = 0.447806λ .
To cancel the imaginary part, we add an inductor in parallel such that
1
ωL=
1.520691
200,
from which we obtain
L =200
1.52×2π ×8×108= 2.618×10−8 H.
Figure P2.66(b)