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

0.1

0.1

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-50

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-60

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-80

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-90

100

-100

110

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180

±

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0.210.22

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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.

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

0.1

0.1

0.1

0.2

0.2

0.2

0.3

0.3

0.3

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0.4

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1.2

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1.4

1.4

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1.6

1.6

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1.8

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10

10

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50

50

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30-30

40-40

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-60

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-70

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-80

90

-90

100

-100

110

-110

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-120

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-130

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-150

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0.04

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0.190.19

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0.210.22

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0.230.24

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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) 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.