EKT 241/4 & 242/3 : ELECTROMAGNETIC ELECTROMAGNETIC THEORYTHEORY

PREPARED BY: Razel

CHAPTER 5 – TRANSMISSION LINESCO4 - Ability to analyze and evaluate transmission lines as circuit components and Smith’s chart for transmission line calculations and impedance matching design.

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Chapter Outline

General ConsiderationsLumped-Element ModelTransmission-Line EquationsWave Propagation on a Transmission LineThe Lossless Transmission Line Input Impedance of the Lossless LineSpecial Cases of the Lossless LinePower Flow on a Lossless Transmission LineThe Smith Chart Impedance MatchingTransients on Transmission Lines

General Considerations

• Transmission line – a two-port network connecting a generator circuit to a load.

So…What is the use of transmission line??

• A transmission line is used to transmit electrical energy/signals from one point to another – i.e. from one source to a load.

• Types of transmission line include: wires, (telephone wire), coaxial cables, optical fibers n etc…

The role of wavelength

• At low frequency, the impact is negligible• At high frequency, the impact is very significant

The impact of a transmission line on the current and voltage in the

circuit depends on the: frequency, f of the signal provided by

generator.

length of line, l

Propagation modesElectric field lines

Magnetic field lines

Propagation modes

A few examples of transverse electromagnetic (TEM) and higher order transmission line

Lumped- element model

• A transmission line is represented by a parallel-wire configuration regardless of the specific shape of the line, (in term of lumped element circuit model)– i.e coaxial line, two-wire line or any TEM line.

• Lumped element circuit model consists of four basic elements called ‘the transmission line parameters’ : R’ , L’ , G’ , C’ .

Series element Shunt element

Lumped- element model Lumped-element transmission line parameters:

– R’ : combined resistance of both resistance per unit length, in Ω/m

– L’ : the combined inductance of both inductor per unit length, in H/m

– G’ : the conductance of the insulation medium per unit length, in S/m

– C’ : the capacitance of the two conductors per unit length, in F/m

• For example, a coil of wire has the property of inductance. When a certain amount of inductance is needed in a circuit, a coil of the proper

dimension is inserted

Lumped- element model

Lumped- element model for 3 type of lines

Note: µ, σ, ε pertain to the insulating material between conductors

Propagation modes

A few examples of transverse electromagnetic (TEM) and higher order transmission line

Exercise 1:

• Use table 5.1 to compute the line parameter of a two wire air line whose wires are separated by distance of 2 cm, and, each is 1 mm in radius. The wires may be treated as perfect conductors with σc= .

R’ = ?, L’=?, G’=?, C’=?

Solution exercise 1:

a

RsR

'

1)2/()2/(ln' 2adadL

)1)2/()2/(ln

'2adad

C

o

ofRs

σc= 0

Rs

fRs o

0'R

σc= 0'G

)1)2/()2/(ln

'2adad

G

mmma

mcmd

001.01

02.02

1)2/()2/(ln' 2adadL

1)

)001.0(2

02.0()

)001.0(2

02.0(ln' 2

L

mHL /20.1'

)1)2/()2/(ln

'2adad

C

mmma

mcmd

001.01

02.02

1))001.0(2

02.0()

)001.0(202.0

(ln

'2

C

mpFC /29.9'

Exercise 2:

• Calculate the transmission line parameters at 1 MHz for a rigid coaxial air line with an inner conductor diameter of 0.6 cm and outer conductor diameter of 1.2 cm. The conductors are made of copper.

(μc=0.9991 ; σc=5.8x107)

f = 1MHzr1 = 0.006m/2 = 0.003mr2 = 0.012m/2 = 0.006m

Solution exercise 2:

ba

RsR

11

2'

o

fRs

47

10608.2108.5

)1( xRsx

MhzRs

mR /0208.0'

006.0

1

003.0

1

2

10608.2'

4

x

R

mb

ma

006.0

003.0

)/ln(2

' abL

003.0

006.0ln

2'

L

mHL /138.0'

(pg238)

B appendix from

const

r

o

ro

)33(

BARE IN UR MIND

From calculator

abC

/ln

2'

mmma

mcmd

001.01

02.02

003.0006.0

ln

2'

C

mpFC /3.80'

237) (pg

B appendix from

const

r

o

ro

)32(

BARE IN UR MIND

From calculator

a

bG

ln

2'

0'GBecause, the material

separating the inner and outer is perfect dielectric (air) with σ=0, thus G’ = 0

Because, the material separating the inner and outer is perfect dielectric (air) with σ=0, thus G’ = 0

G’ : the conductance of the insulation medium per unit length, in S/m

Transmission line equations

• Complex propagation constant, γ

• α – the real part of γ - attenuation constant, unit: Np/m

• β – the imaginary part of γ - phase constant, unit: rad/m

j

CjG'LjR'

''

Is used to describes the voltage and the current across the transmission line in term of propagation constant and

impedance

Is used to describes the voltage and the current across the transmission line in term of propagation constant and

impedance

Transmission line equations

• The characteristic impedance of the line, Z0 :

• Phase velocity of propagating waves:

where f = frequency (Hz) λ = wavelength (m) β = phase constant

''

''0 CjG

LjRZ

fu p

f 2

Example 1

An air line is a transmission line for which air is the dielectric material present between the two conductors, which renders G’ = 0.

In addition, the conductors are made of a material with high conductivity so that R’ ≈0.

For an air line with characteristic impedance of 50Ω and phase constant of 20 rad/m at 700MHz, find the inductance per meter and the capacitance per meter of the line.

Solution to Example 1• The following quantities are given:

• With R’ = G’ = 0,

Hz 107 MHz 700 rad/m, 20 ,50 80 fZ

'

'

''

''

''

''''

''''

0

2

C

L

CjG

LjRZ•

and

CL

CLCjLj

CjGLjR constant, npropagatio•

CZ

CZ

o

o

2222

• The ratio is given by:

• We get L’ from Z0

)''('

' 222 CLZC

Lo

Solution to Example 1

pF/m 9.90501072

20'

80

Z

C

nH/m 227109.9050''' 1220 LCLZ

''

''

0CL

CL

Z

CLZ

C

L'

'

'0

2

Lossless transmission line

• Lossless transmission line - Very small values of R’ and G’.

• We set R’=0 and G’=0, hence:

line) (lossless ''

line) (lossless 0

CL

Transmission line can be designed to minimize ohmic losses by selecting high conductivities and dielectric material, thus we assume :

Transmission line equations

• Complex propagation constant, γ

• α – the real part of γ - attenuation constant, unit: Np/m

• β – the imaginary part of γ - phase constant, unit: rad/m

j

CjG'LjR'

''0 0

Lossless transmission line

• Lossless transmission line - Very small values of R’ and G’.

• We set R’=0 and G’=0, hence:

line) (lossless ''

line) (lossless 0

CL

line) (lossless '

'

0,G' and 0R' since

''

''

0

0

C

LZ

CjG

LjRZ

Transmission line can be designed to minimize ohmic losses by selecting high conductivities and dielectric material, thus we assume :

Lossless transmission line

• Using the relation properties between μ, σ, ε :

• Wavelength, λ

Where εr = relative permittivity of the insulating material between conductors

(m/s) u

(rad/m)

p

1

rr

p

f

c

f

u

01

rGHz

x

1

1

103207.0

8

Exercise 3:

• For a losses transmission line, λ = 20.7 cm at 1GHz. Find εr of the insulating material.

λ=20.7cm 0.207m ; f=1 GHz

rr

p

f

c

f

u

01

207.0

1

1

103 8

GHz

xr 449.1r 1.2r

2

Exercise 4

• A lossless transmission line of length 80 cm operates at a frequency of 600MHz . The line parameters are &

Find the characteristic impedance, the phase constant and the phase velocity.

pF/m C 100 μH/m L 25.0

The condition apply that the line is lossless, So: R= 0 & G=0

• characteristic impedance :

• phase constant: With R n G = 0

C

LZ 0 pF/m C 100

μH/m L 25.0

50

10100

1025.012

6

0x

xZ

CL

CjGLjR

''

''''Im

)10100)(1025.0()10600(2 1266 xxx

= 18.85 rad/m

• phase velocity: fu p

smx

xv p

/102

85.18

)10600(2

8

6

f 2

Voltage Reflection Coefficient

• Every transmission line has a resistance associated with it, and comes about because of its construction. This is called its characteristic impedance, Z0.

• The standard characteristic impedance value is 50Ω. However when the transmission line is terminated with an arbitrary load ZL, in which is not equivalent to its characteristic impedance (ZL ≠ Z0), a reflected wave will occur.

Voltage reflection coefficient

• Voltage reflection coefficient, Γ – the ratio of the amplitude of the reflected voltage wave, V0

- to the amplitude of the incident voltage wave, V0

+ at the load.

• Hence,

0

0

0

0ZZ

ZZ

V

V

L

L

impedance sticcharacteri Z

impedance load Z

tcoefficien reflection Where

less)(dimention ZZ

ZZ

L

L

L

0

0

01/

1/

Voltage reflection coefficient

• The load impedance, ZL

Where;

= total voltage at the load

V0- = amplitude of reflected voltage wave

V0+ = amplitude of the incident voltage wave

= total current at the load

Z0 = characteristic impedance of the line

00

~VVVL

LV~

0

0

0

0~

Z

V

Z

VI L

LI~

L

LL

I

VZ ~

~

Voltage reflection coefficient

• And in case of a RL and RC series, ZL :

ZL = R + jL ; ZL = R -1/ jC

• A load is matched to the line if ZL = Z0 because there will be no reflection by the load (Γ = 0 and V0

−= 0.

• When the load is an open circuit, (ZL=∞), Γ = 1 and V0

- = V0+.

• When the load is a short circuit (ZL=0), Γ = -1 and V0

- = V0+.

What is the difference between an open and closed circuit?

• closed allows electricity through, and open doesn't.

• open circuit - Any circuit which is not complete is considered an open circuit. The open status of the circuit doesn't depend on how it became unclosed, so circuits which are manually disconnected and circuits which have blown fuses, faulty wiring or missing components are all considered open circuits.

• close circuit: A circuit is considered to be closed when electricity flows from an energy source to the desired endpoint of the circuit. A complete circuit which is not performing any actual work can still be a closed circuit. For example, a circuit connected to a dead battery may not perform any work, but it is still a closed circuit.

Example 2

• A 100-Ω transmission line is connected to a load consisting of a 50-Ω resistor in series with a 10pF capacitor. Find the reflection coefficient at the load for a 100-MHz signal.

Solution to Example 2

• The following quantities are given

• The load impedance is

• Voltage reflection coefficient is

Hz10MHz100 ,100 F,10 ,50 80

11LL fZCR

1595010102

150

/

118

LLL

jj

CjRZ

7.6076.0159.15.0

159.15.0

1/

1/

0L

0L

j

j

ZZ

ZZ

7.6076.0159.15.0

159.15.0

1/

1/

0

0j

j

ZZ

ZZ

L

L

59.15.1

tan59.15.1

59.15.0

tan59.15.0

159.15.0

159.15.0

122

122

j

j

7.4619.2

6.7257.13.11976.0

3.11976.0 je

In order to convert from –ve magnitude for Г by replacing the

–ve sign with e-j180 7.60

1803.119

76.0

)(76.0

j

jj

e

ee

7.6076.0 r ;

Math’s TIP…1

2

Exercise 5

• A 150 Ω lossless line is terminated in a load impedance ZL= (30 –j200) Ω. Calculate the voltage reflection coefficient at the load.

Zo = 150 ΩZL= (30 –j200) Ω 0

0ZZ

ZZ

L

L

150)20030(501)20030(

jj oje 95.72867.0

Standing Waves

• Interference of the reflected wave and the incident wave along a transmission line creates a standing wave.

• Constructive interference gives maximum value for standing wave pattern, while destructive interference gives minimum value.

• The repetition period is λ for incident and reflected wave individually.

• But, the repetition period for standing wave pattern is λ/2.

Standing Waves

• For a matched line, ZL = Z0, Γ = 0 and

= |V0+| for all values of z. zV

~

Standing Waves

• For a short-circuited load, (ZL=0), Γ = -1.

Standing Waves

• For an open-circuited load, (ZL=∞), Γ = 1.

The wave is shifted by λ/4 from short-circuit case.

Standing Waves

• First voltage maximum occurs at:

• If θr ≥ 0 n=0;

• If θr ≤ 0 n=1

• First voltage minimum occurs at:

024max n where

nl r

Where θr = phase

angle of Γ

4/ if 4/

4/ if 4/

maxmax

maxmaxmin

ll

lll

VSWR

• Voltage Standing Wave Ratio (VSWR) is ratio between the maximum voltage an the minimum voltage along the transmission line.

• VSWR provides a measure of mismatch between the load and the transmission line.

• For a matched load with Γ = 0, VSWR = 1 and for a line with |Γ| - 1, VSWR = ∞. impedance sticcharacteri Z

impedance load Z

tcoefficien reflection Where

ZZ

ZZ Where,

VSWR

L

L

L

0

0

||1

||1

The VSWR is given by:

Example 3

A 50- transmission line is terminated is terminated in a load with ZL = (100 + j50)Ω . Find the voltage reflection coefficient and the voltage standing-wave ratio (VSWR).

Solution to Example 3

• We have,

• VSWR is given by:

6.26

0

0 45.05050100

5050100 j

L

L ej

j

ZZ

ZZ

6.245.01

45.01

1

1

VSWR

Exercise 6:

• A 140 Ω lossless line is terminated in a load impedance ZL= (280 +j182) Ω, if λ = 72cm, find

a) Reflection coefficient, Г

b) The VSWR,

c) The locations of voltage maxima and minima

• a) Reflection coefficient, Г

0

0ZZ

ZZ

L

L

)140()182280(

)140()182280(

j

j

182420

182140

j

j

420

182tan182420

140

182tan182140

122

122

o

o

o

97.285.0

43.23457

4.52230

• b) The VSWR;

||1

||1

VSWR

o97.285.0

|97.285.0|1

|97.285.0|1

VSWR

35.01

5.01

VSWR

• The locations of voltage maxima and minima

024max n where

nl r

29.2

24

)72)(5.0(max

ncm

n

l

4/4/

4/4/

maxmax

maxmaxmin

l if l

l if ll

2n

4

72 )

2n2.9

ll

9.20

(

4/maxmin

l

cmcm

cm

4/

184/72

72

max

Input impedance of a lossless line

• The input impedance, Zin is the ratio of the total voltage (incident and reflected voltages) to the total current at any point z on the line.

• or

ljZZ

ljZZZ

ljZlZ

ljZlZZlZ

tan

tan

sincos

sincos

L0

0L0

L0

0L0in

zj

zj

in

e

eZ

zI

zVzZ

2

2

0 1

1

)(~

)(~

)(

Special cases of the lossless line

• For a line terminated in a short-circuit, ZL = 0:

• For a line terminated in an open circuit, ZL = ∞:

ljZ

lI

lVZ

sc

scscin tan~

~

0

ljZ

lI

lVZ cot~ 0

oc

ococin

Application of short-circuit and open-circuit measurements

• The measurements of short-circuit input impedance, and open-circuit input impedance, can be used to measure the characteristic impedance of the line:

• and

scinZ

ocinZ

ocin

scin ZZZo

ocin

scintan

Z

Zl

Length of line

• If the transmission line has length , where n is an integer,

• Hence, the input impedance becomes:

2/nl

0tan

2//2tantan

n

nl

2/for ZZ Lin nl

Quarter wave transformer

• If the transmission line is a quarter wavelength,

with ,

where , we have

, then the input impedance becomes:

2/4/ nl integer positiveany or 0n

24

2

l

2/4/for Z

ZZ

L

20

in nl

Example 4A 50-Ω lossless transmission line is to be matched to a resistive load impedance with ZL=100Ω via a quarter-wave section as shown, thereby eliminating reflections along the feedline. Find the characteristic impedance of the quarter-wave transformer.

Quarter wave transformer

• If the transmission line is a quarter wavelength,

with ,

where , we have

, then the input impedance becomes:

2/4/ nl integer positiveany or 0n

24

2

l

2/4/for Z

ZZ

L

20

in nl

Solution to Example 4

• Zin = 50Ω; ZL=100Ω

• Since the lines are lossless, all the incident power will end up getting transferred into the load ZL.

7.7010050

)100)(50(

02

202

202

Z

ZZ

ZZ

Lin

Matched transmission line

• For a matched lossless transmission line, ZL=Z0:

1) The input impedance Zin=Z0 for all locations z on the line,

2) Γ =0, and

3) all the incident power is delivered to the load, regardless of the length of the line, l.

Input Input Impedance, ZImpedance, Zinin

Input Input Impedance, ZImpedance, Zinin

Ratio of the total voltage to total

current on the line

Ratio of the total voltage to total

current on the line

When ZL=0(short circuit)When ZL=0(short circuit)

When ZL=(open circuit)When ZL=(open circuit)

ljZZ scin tan0

ljZZ ocin cot0

ApplicationBe used to measure the

characteristic impedance of the line :

ocin

scino ZZZ

ocin

scin

Z

Zl

tan

But, If the transmission line is

4

l 2

l

L2

0in ZZZ

2

l

0lLin ZZ

Special case

Power flow on a lossless transmission line

• Two ways to determine the average power of an incident wave and the reflected wave;– Time-domain approach– Phasor domain approach

• Average power for incident wave;

• Average power for reflected wave:

• The net average power delivered to the load:

(W) 2 0

2

0iav Z

VP

iav

2

0

2

02rav 2

PZ

VP

(W) Z

VPPP r

aviavav

2

0

2

01

2

• The time average power reflected by a load connected to a lossless transmission line is equal to the incident power multiplied by |Г|2

Power flow on a lossless transmission line

Exercise 7• For a 50Ω lossless transmission line terminated in

a load impedance ZL = (100 + j50)Ω, determine the percentage of the average power reflected over average incident power by the load.

Z0=50Ω; ZL = (100 + j50)Ω

(W) PP iav

rav

2

(W) P

Piav

rav 2

• Reflection coefficient, Г

0

0ZZ

ZZ

L

L

)50()50100(

)50()50010(

j

j

50150

5050

j

j

150

50tan50150

50

50tan5050

122

122

2.0

6.2645.0

4.181.158

457.70

2

o

o

o

the percentage of the average incident power reflected by the load = 20%

Exercise 8

• For the line of exercise previously (exercise 7), what is the average reflected power if |V0

+|=1V

iav

2

0

2

02rav 2

PZ

VP

mWPrav 2

)50(2

145.0

22

Smith Chart • Smith chart is used to analyze & design

transmission line circuits.• Reflection coefficient, Γ :

Гr = real part, Гi = imaginary part

• Impedances on Smith chart are represented by normalized value, zL :

• the normalized load impedance, zL is dimensionless.

0Z

Zz L

L

ije rj r

Smith Chart• Reflection coefficient, ΓA :0.3 + j0.4

Reflection coefficient, ΓB :-0.5 - j0.2

5.04.03.02/122

533.0/4.0tan 1 r

54.02.05.02/122

2022.0/5.0tan 1 r 158202360 rIn order to eliminate –ve part, thus

The complex Γ plane.

ΓA :0.3 + j0.4 ΓB :-0.5 - j0.2

Smith Chart• Reflection coefficient, Γ :

• Since , Γ becomes:

• Re-arrange in terms of zL:

rL = Normalized load resistance

xL = Normalized load admittance

1/

1/

0

0

ZZ

ZZ

L

L

0Z

Zz L

L 1

1

L

L

z

z

LL jxrz

1

1L

The families of circle for rL and xL.

Plotting normalized impedance, zL = 2-j1

6.262/1tan 1 r

45.013

11

1)12(1)12(

22

22

jj

Input impedance

• The input impedance, Zin:

• Γ is the voltage reflection coefficient at the load.

• We shift the phase angle of Γ by 2βl, to get ΓL. This will match zL to zin. The |Γ| is the same, but the phase is changed by 2βl.

• On the Smith chart, this means rotating in a clockwise direction (WTG).

lj

lj

in e

eZZ

2

2

0 1

1

Input impedance

• Since β = 2π/λ, shifting by 2 βl is equal to phase change of 2π.

• Equating:

• Hence, for one complete rotation corresponds to l = λ/2.

• The objective of shifting Γ to ΓL is to find Zin at an any distance l on the transmission line.

2

222 ll

Example 5

• A 50-Ω transmission line is terminated with ZL=(100-j50)Ω. Find Zin at a distance l =0.1λ from the load.

Normalized the load impedanceSolution:

jz

j

Z

Zz

L

LL

2

50

50100

0

Solution to Example 5

zin = 0.6 –j0.66

jA 2

l =0.1λ

de normalize (multiplying by Zo)

Zin = 30 –j33

VSWR, Voltage Maxima and Voltage Minima

zL=2+j1

VSWR = 2.6 (at Pmax).

lmax=(0.25-0.213)λ=0.037λ.

lmin=(0.037+0.25)λ =0.287λ

VSWR, Voltage Maxima and Voltage Minima

• Point A is the normalized load impedance with zL=2+j1.

• VSWR = 2.6 (at Pmax).

• The distance between the load and the first voltage maximum is lmax=(0.25-0.213)λ=0.037λ.

• The distance between the load and the first voltage minimum is lmin=(0.037+0.25)λ =0.287λ.

Impedance to admittance transformations

yL=0.25 - j0.6

zL=0.6 + j1.4

Example 6

• Given that the voltage standing-wave ratio, VSWR = 3. On a 50-Ω line, the first voltage minimum occurs at 5 cm from the load, and that the distance between successive minima is 20 cm, find the load impedance.

Solution:The distance between successive minima is equal to λ/2. the distance between successive minima is 20 cm, Hence, λ = 40 cm

)20(2

2/20

Solution to Example 6

Point A =VSWR = 3

125.040

5min l

8.06.0L jz

de normalize (multiplying by Zo)

Zin = 30 –j40

Solution to Example 6

• First voltage minimum (in wavelength unit) is at

on the WTL scale from point B.

• Intersect the line with constant SWR circle = 3.• The normalized load impedance at point C is:

• De-normalize (multiplying by Z0) to get ZL:

125.040

5min l

8.06.0L jz

40308.06.050L jjZ

Exercise

Normalized the load impedanceSolution:

4.06.00

jZ

Zz LL

a) reflection coefficient from smith Chart

12134.0 jj ee

34.04.06.1

4.04.0

1)4.06.0(

1)4.06.0(

22

22

j

j

121r

4.06.0 jzL

082.0

121

r

0.383

0.082 0.301•

0.301

:length

j0.62 -0.72•

in Z

0.168

0.082-0.25•

lmaxlmin

3) Move a distance 0.301λ towards the generator (WTG) (refer to Smith chart) 

• → 0.301λ + 0.082λ=0.383λ

• At 0.383λ, read the value of which at the point intersects with constant circle, we have = zin = 0.72- j0.62.

• Denormalized it, hence = Zin = 72- j62

4) Distance from load to the first voltage maximum, (refer to Smith chart)

→ 0.25λ-0.082λ=0.168λ

Impedance Matching

• Transmission line is matched to the load when Z0 = ZL.

• This is usually not possible since ZL is used to serve other application.

• Alternatively, we can place an impedance-matching network between load and transmission line.

Single- stub matching

• Matching network consists of two sections of transmission lines.

• First section of length d, while the second section of length l in parallel with the first section, hence it is called stub.

• The second section is terminated with either short-circuit or open circuit.

Single- stub matching

YL=1/ZL

Yd = Y0+jB

feed

line

stub

d

l

Single- stub matching

• The length l of the stub is chosen so that its input admittance, YS at MM’ is equal to –jB.

• Hence, the parallel sum of the two admittances at MM’ yields Y0, which is the characteristic admittance of the line.

Yd = Y0+jB

Single- stub matching

• Thus, the main idea of shunt stub matching network is

to:

• (i) Find length d and l in order to get yd and yl .

• (ii) Ensure total admittance yin = yd + ys = 1 for complete

matching network.

Example 7

50-Ω transmission line is connected to an antenna with load impedance ZL = (25 − j50)Ω. Find the position and length of the short-circuited stub required to match the line.

Solution: The normalized load impedance is:

jj

Z

Zz

5.0

50

5025

0

LL

(located at A).

Solution to Example 7

jA 5.0

admittance loady

j0.80.4

yB

L

L

Solution to Example 7

• Value of yL at B is which locates at position 0.115λ on the WTG scale.

• Draw constant SWR circle that goes through points A and B.

• There are two possible matching points, C and D where the constant SWR circle intersects with circle rL=1 (now gL =1 circle).

8.04.0L jy

C = 1+j1.58

A

B

115.0B

D = 1+j1.58

Solution to Example 7

First matching points, C.• At C, is at 0.178λ on WTG scale.• Distance B and C is• Normalized input admittance

at the juncture is:

E is the admittance of short-circuit stub, yL=-j∞.

Normalized admittance of −j 1.58 at F and position 0.34λ on the WTG scale gives:

58.11d jy 063.0155.0178.0 d

58.1

58.1101

s

s

sin

jy

jyj

yyy d

09.025.034.01 l

F

l1

d1 = 0.063λ

= 0.090λ

C = 1+j1.58

063.0115.0178.0 d

09.025.034.0 lA

B

115.0B

E

F = -j1.58

58.1

58.1101

s

s

sin

jy

jyj

yyy d

Short circuited stub

Open circuited stub

First matching points, C

• Thus, the values are:

• d1 = 0.063 λ

• l1 = 0.09 λ

• yd1 = 1 + j1.58 Ω

• ys1 = -j1.58 Ω

• Where Yin = yd + ys = (1 + j1.58) + (-j1.58) = 1

Solution to Example 7

Second matching point, D.• At point D,• Distance B and D is• Normalized input admittance at G. • Rotating from point E to point G, we get

58.11 jyd 207.0115.0322.02 d

58.1s jy

41.016.025.02 l

l2

d2= 0.207λ

= 0.41λ

G

207.0

115.0322.0

d

41.0

16.025.0

l

B

A

G = +j1.58

E

D = 1-j1.58

d1=0.063 λ

l1=0.09λ,

d2=0.207 λ

l2=0.41 λ

First matching points, D

• Thus, the values are:

• d2 = 0.207 λ

• l2 = 0.41 λ

• yd2 = 1 - j1.58 Ω

• ys2 = +j1.58 Ω

• Where Yin = yd + ys = (1 - j1.58) + (+j1.58) = 1