Problem 2.66 A 200-Ω transmission line is to be matched to a computer terminalwith ZL = (50− j25) Ω by inserting an appropriate reactance in parallel with theline. If f = 800 MHz andεr = 4, determine the location nearest to the load at whichinserting:

(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 forZL andZ0 into Module 2.6, we havezL

represented by the red dot in Fig. P2.66(a), andyL represented by the blue dot. Bymoving the cursor a distanced = 0.093λ , the blue dot arrives at the intersection pointbetween the SWR circle and theS= 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 ispositive, such as a capacitor. That is:

ωC = (1.54206)×Y0

=1.54206

Z0=

1.54206200

= 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)] leadsto

y(d) = 1.000001+ j1.520691,

atd2 = 0.447806λ .To cancel the imaginary part, we add an inductor in parallel such that

1ωL

=1.520691

200,

from which we obtain

L =200

1.52×2π ×8×108 = 2.618×10−8 H.

Figure P2.66(b)

Problem 2.68 A 50-Ω lossless line is to be matched to an antenna withZL = (75− j20) Ω using a shorted stub. Use the Smith chart to determine the stublength and distance between the antenna and stub.

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)

Z-LOAD

Y-LOAD

Y-LOAD-IN-1

Y-SHT

Y-STUB-IN-1

0.104 λ

0.173 λ

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 differentsolutions.

zL =ZL

Z0=

(75− j20) Ω50 Ω

= 1.5− j0.4

and is located at pointZ-LOAD in both figures. Since it is advantageous to work inadmittance coordinates,yL is plotted as pointY-LOADin both figures.Y-LOADis at0.041λ on the WTG scale.

For the first solution in Fig. P2.68(a), pointY-LOAD-IN-1 represents the pointat whichg = 1 on the SWR circle of the load.Y-LOAD-IN-1 is at 0.145λ on theWTG scale, so the stub should be located at 0.145λ − 0.041λ = 0.104λ from theload (or some multiple of a half wavelength further). AtY-LOAD-IN-1, b = 0.52,so a stub with an input admittance ofystub = 0− j0.52 is required. This point isY-STUB-IN-1 and is at 0.423λ on the WTG scale. The short circuit admittanceis denoted by pointY-SHT, located at 0.250λ . Therefore, the short stub must be0.423λ −0.250λ = 0.173λ long (or some multiple of a half wavelength longer).

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)

Z-LOAD

Y-LOAD

Y-LOAD-IN-2

Y-SHT

Y-STUB-IN-2

0.314 λ

0.327 λ

Figure P2.68: (b) Second solution to Problem 2.68.

For the second solution in Fig. P2.68(b), pointY-LOAD-IN-2 represents the pointat whichg = 1 on the SWR circle of the load.Y-LOAD-IN-2 is at 0.355λ on theWTG 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). AtY-LOAD-IN-2, b = −0.52,so a stub with an input admittance ofystub = 0+ j0.52 is required. This point isY-STUB-IN-2 and is at 0.077λ on the WTG scale. The short circuit admittanceis denoted by pointY-SHT, located at 0.250λ . Therefore, the short stub must be0.077λ − 0.250λ + 0.500λ = 0.327λ long (or some multiple of a half wavelengthlonger).

Problem 3.12 Two lines in thex–y plane are described by the expressions:

Line 1 x+2y = −6,Line 2 3x+4y = 8.

Use vector algebra to find the smaller angle between the lines at their intersectionpoint.

10 20 302515 35-10-15-20-25-30-35

-10

-15

-20

-25

-30

10

15

20

25

30

(0, 2)

(0, -3)

(20, -13)

BA

θAB

Figure P3.12:Lines 1 and 2.

Solution: Intersection point is found by solving the two equations simultaneously:

−2x−4y = 12,

3x+4y = 8.

The sum givesx = 20, which, when used in the first equation, givesy = −13.Hence, intersection point is(20,−13).Another point on line 1 isx = 0, y = −3. VectorA from (0,−3) to (20,−13) is

A = x(20)+ y(−13+3) = x20− y10,

|A| =√

202 +102 =√

500.

A point on line 2 isx = 0, y = 2. VectorB from (0,2) to (20,−13) is

B = x(20)+ y(−13−2) = x20− y15,

|B| =√

202 +152 =√

625.

Angle betweenA andB is

θAB = cos−1(

A ·B|A||B|

)

= cos−1(

400+150√500·

√625

)

= 10.3.