Problem 2.11 A 50-Ωmicrostrip line uses a 0.6-mm …etemadi/ee324/Homework/Fall...
Transcript of Problem 2.11 A 50-Ωmicrostrip line uses a 0.6-mm …etemadi/ee324/Homework/Fall...
Problem 2.11 A 50-Ω microstrip line uses a 0.6-mm alumina substrate withεr = 9.Use CD Module 2.3 to determine the required strip widthw. Include a printout of thescreen display.
Solution: According to the solution provided by CD Module 2.3, the required stripwidth is
w = 0.613 mm.
Problem 2.37 A lossless transmission line is terminated in a short circuit. Howlong (in wavelengths) should the line be for it to appear as an open circuit at its inputterminals?
Solution: From Eq. (2.84),Zscin = jZ0 tanβ l . If β l = (π/2+nπ), thenZsc
in = j∞ (Ω).Hence,
l =λ2π
(π2
+nπ)
=λ4
+nλ2
.
This is evident from Figure 2.19(d).
Problem 2.39 A 75-Ω resistive load is preceded by aλ/4 section of a 50-Ω losslessline, which itself is preceded by anotherλ/4 section of a 100-Ω line. What is theinput impedance? Compare your result with that obtained through two successiveapplications of CD Module 2.5.
Solution: The input impedance of theλ/4 section of line closest to the load is foundfrom Eq. (2.97):
Zin =Z2
0
ZL=
502
75= 33.33 Ω.
The input impedance of the line section closest to the load can be consideredas theload impedance of the next section of the line. By reapplying Eq. (2.97), thenextsection ofλ/4 line is taken into account:
Zin =Z2
0
ZL=
1002
33.33= 300Ω.
Problem 2.40 A 100-MHz FM broadcast station uses a 300-Ω transmission linebetween the transmitter and a tower-mounted half-wave dipole antenna. The antennaimpedance is 73Ω. You are asked to design a quarter-wave transformer to match theantenna to the line.
(a) Determine the electrical length and characteristic impedance of the quarter-wave section.
(b) If the quarter-wave section is a two-wire line withD = 2.5 cm, and the wiresare embedded in polystyrene withεr = 2.6, determine the physical length ofthe quarter-wave section and the radius of the two wire conductors.
Solution:(a) For a match condition, the input impedance of a load must match that of the
transmission line attached to the generator. A line of electrical lengthλ/4 can beused. From Eq. (2.97), the impedance of such a line should be
Z0 =√
ZinZL =√
300×73= 148Ω.
(b)λ4
=up
4 f=
c4√
εr f=
3×108
4√
2.6×100×106= 0.465 m,
and, from Table 2-2,
Z0 =120√
εln
(
Dd
)+
√(Dd
)2
−1
Ω.
Hence,
ln
(
Dd
)+
√(Dd
)2
−1
=
148√
2.6120
= 1.99,
which leads to (Dd
)+
√(Dd
)2
−1 = 7.31,
and whose solution isD/d = 3.73. Hence,d = D/3.73= 2.5 cm/3.73= 0.67 cm.
Problem 2.41 A 50-Ω lossless line of lengthl = 0.375λ connects a 300-MHzgenerator withVg = 300 V andZg = 50 Ω to a loadZL . Determine the time-domaincurrent through the load for:
(a) ZL = (50− j50) Ω(b) ZL = 50 Ω(c) ZL = 0 (short circuit)
For (a), verify your results by deducing the information you need from the outputproducts generated by CD Module 2.4.
Solution:
Vg Zin Z0ZL
~+
-
+
-
Transmission line
Generator Loadz = -l z = 0
Vg
Ii
Zin
~
Vi~~
+
-
⇓
(50-j50) Ω
l = 0.375 λ
= 50 Ω
50 Ω
Zg
Figure P2.41: Circuit for Problem 2.41(a).
(a) ZL = (50− j50) Ω, β l = 2πλ ×0.375λ = 2.36 (rad)= 135.
Γ =ZL −Z0
ZL +Z0=
50− j50−5050− j50+50
=− j50
100− j50= 0.45e− j63.43 .
Application of Eq. (2.79) gives:
Zin = Z0
[ZL + jZ0 tanβ lZ0 + jZL tanβ l
]= 50
[(50− j50)+ j50tan135
50+ j(50− j50) tan135
]= (100+ j50) Ω.
Using Eq. (2.82) gives
V+0 =
(VgZin
Zg +Zin
)(1
ejβ l +Γe− jβ l
)
=300(100+ j50)
50+(100+ j50)
(1
ej135 +0.45e− j63.43e− j135
)
= 150e− j135 (V),
IL =V+
0
Z0(1−Γ) =
150e− j135
50(1−0.45e− j63.43) = 2.68e− j108.44 (A),
iL(t) = Re[ILejωt ]
= Re[2.68e− j108.44ej6π×108t ]
= 2.68cos(6π ×108t −108.44) (A).
(b)
ZL = 50 Ω,
Γ = 0,
Zin = Z0 = 50 Ω,
V+0 =
300×5050+50
(1
ej135 +0
)= 150e− j135 (V),
IL =V+
0
Z0=
15050
e− j135 = 3e− j135 (A),
iL(t) = Re[3e− j135ej6π×108t ] = 3cos(6π ×108t −135) (A).
(c)
ZL = 0,
Γ = −1,
Zin = Z0
(0+ jZ0 tan135
Z0 +0
)= jZ0 tan135 = − j50 (Ω),
V+0 =
300(− j50)50− j50
(1
ej135 −e− j135
)= 150e− j135 (V),
IL =V+
0
Z0[1−Γ] =
150e− j135
50[1+1] = 6e− j135 (A),
iL(t) = 6cos(6π ×108t −135) (A).
From output of Module 2.4, atd = 0 (load)
I(d) = 2.68∠−1.89 rad,
which corresponds toI(d) = 2.68∠−108.29 .
The equivalent time-domain current atf = 300 MHz is
iL(t) = 2.68cos(6π ×108t −108.29) (A).
Problem 2.42 A generator withVg = 300 V andZg = 50 Ω is connected to a loadZL = 75 Ω through a 50-Ω lossless line of lengthl = 0.15λ .
(a) ComputeZin, the input impedance of the line at the generator end.
(b) ComputeIi andVi .
(c) Compute the time-average power delivered to the line,Pin = 12Re[Vi I∗i ].
(d) Compute VL , IL , and the time-average power delivered to the load,PL = 1
2Re[VL I∗L ]. How doesPin compare toPL? Explain.
(e) Compute the time-average power delivered by the generator,Pg, and the time-average power dissipated inZg. Is conservation of power satisfied?
Solution:
Vg Zin Z0~
+
-
+
-
Transmission line
Generator Loadz = -l z = 0
Vg
IiZg
Zin
~
Vi~~
+
-
⇓ l = 0.15 λ
= 50 Ω
50 Ω
75 Ω
Figure P2.42: Circuit for Problem 2.42.
(a)
β l =2πλ
×0.15λ = 54,
Zin = Z0
[ZL + jZ0 tanβ lZ0 + jZL tanβ l
]= 50
[75+ j50tan54
50+ j75tan54
]= (41.25− j16.35) Ω.
(b)
Ii =Vg
Zg +Zin=
30050+(41.25− j16.35)
= 3.24ej10.16 (A),
Vi = IiZin = 3.24ej10.16(41.25− j16.35) = 143.6e− j11.46 (V).
(c)
Pin =12Re[Vi I
∗i ] =
12Re[143.6e− j11.46 ×3.24e− j10.16 ]
=143.6×3.24
2cos(21.62) = 216 (W).
(d)
Γ =ZL −Z0
ZL +Z0=
75−5075+50
= 0.2,
V+0 = Vi
(1
ejβ l +Γe− jβ l
)=
143.6e− j11.46
ej54 +0.2e− j54 = 150e− j54 (V),
VL = V+0 (1+Γ) = 150e− j54(1+0.2) = 180e− j54 (V),
IL =V+
0
Z0(1−Γ) =
150e− j54
50(1−0.2) = 2.4e− j54 (A),
PL =12Re[VL I∗L ] =
12Re[180e− j54 ×2.4ej54 ] = 216 (W).
PL = Pin, which is as expected because the line is lossless; power input to the lineends up in the load.
(e)Power delivered by generator:
Pg =12Re[VgIi ] =
12Re[300×3.24ej10.16 ] = 486cos(10.16) = 478.4 (W).
Power dissipated in Zg:
PZg =12Re[IiVZg] =
12Re[Ii I
∗i Zg] =
12|Ii |2Zg =
12
(3.24)2×50= 262.4 (W).
Note 1:Pg = PZg +Pin = 478.4 W.
Problem 2.44 For the circuit shown in Fig. P2.44, calculate the average incidentpower, the average reflected power, and the average power transmittedinto the infinite100-Ω line. Theλ/2 line is lossless and the infinitely long line is slightly lossy. (Hint:The input impedance of an infinitely long line is equal to its characteristic impedanceso long asα 6= 0.)
Z0 = 50 Ω Z1 = 100 Ω
λ/250 Ω
2 V
Pavi
Pavr
Pavt
8
−
+
Figure P2.44: Circuit for Problem 2.44.
Solution: Considering the semi-infinite transmission line as equivalent to a load(since all power sent down the line is lost to the rest of the circuit),ZL = Z1 = 100Ω.Since the feed line isλ/2 in length, Eq. (2.96) givesZin = ZL = 100 Ω andβ l = (2π/λ )(λ/2) = π, soe± jβ l = −1. Hence
Γ =ZL −Z0
ZL +Z0=
100−50100+50
=13.
Also, converting the generator to a phasor givesVg = 2ej0 (V). Plugging all theseresults into Eq. (2.82),
V+0 =
(VgZin
Zg +Zin
)(1
ejβ l +Γe− jβ l
)=
(2×10050+100
)[1
(−1)+ 13(−1)
]
= 1ej180 = −1 (V).
From Eqs. (2.104), (2.105), and (2.106),
Piav =
∣∣V+0
∣∣2
2Z0=
|1ej180 |22×50
= 10.0 mW,
Prav = −|Γ|2Pi
av = −∣∣∣∣13
∣∣∣∣2
×10 mW= −1.1 mW,
Pav = Ptav = Pi
av+Prav = 10.0 mW−1.1 mW= 8.9 mW.
Problem 2.47 Use the Smith chart to find the reflection coefficient correspondingto 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) PointA is zL = 3+ j0. Γ = 0.5e0
(b) PointB is zL = 2− j2. Γ = 0.62e−29.7
(c) PointC is zL = 0− j2. Γ = 1.0e−53.1
(d) PointD is zL = 0+ j0. Γ = 1.0e180.0
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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.
Problem 2.48 Repeat Problem 2.47 using CD Module 2.6.
Solution:
Figure P2.48(a)
Figure P2.48(b)
Figure P2.48(c)
Figure P2.48(d)
Problem 2.49 Use the Smith chart to find the normalized load impedancecorresponding 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
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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) PointA′ is Γ = 0.5 atzL = 3+ j0.(b) PointB′ is Γ = 0.5ej60 atzL = 1+ j1.15.(c) PointC ′ is Γ = −1 atzL = 0+ j0.(d) PointD′ is Γ = 0.3e− j30 atzL = 1.60− j0.53.(e) PointE′ is Γ = 0 atzL = 1+ j0.(f) PointF ′ is Γ = j atzL = 0+ j1.