CHAPTER 4

TRANSMISSION LINES

2

TRANSMISSION LINES

4.1 INTRODUCTION

4.2 SMITH CHART

4.3 IMPEDANCE MATCHING

3

One of electromagnetic theory application: Power lines, telephone lines and cable

TV. Develop equation for wave propagation on

a transmission lines. Introduce Smith chart for the study of

transmission lines and use it to develop impedance matching networks.

4.1 INTRODUCTION

4

INTRODUCTION (Cont’d)

A sinusoidal voltage is dropped across the resistor.

If the supply and resistor is connected by an ideal (negligible length) conductor, it will be in same phase.

5

When a quarter wavelength is added

between the supply and the resistor, the

voltage at the resistor is 900 out of phase

with the supply voltage.

INTRODUCTION (Cont’d)

6

Transmission lines examples:

INTRODUCTION (Cont’d)

Twin lead

Coaxial Cable

7

INTRODUCTION (Cont’d)

Microstrip

8

The distributed

parameters for a

differential segment

of transmission line.

R’ (resistance/meter) L’ (inductance/meter)

G’ (conductance/meter) C’ (capacitance/meter)

* Primes indicate per unit length or distributed values.

INTRODUCTION (Cont’d)

9

,11

2

1'

ln2

'

ln

2'

,ln

2'

c

d

f

baR

abL

abC

abG

INTRODUCTION (Cont’d)

For example, the distributed parameters for a

coaxial cable can be determined by using these

formulas:

Where,

Conductance of dielectric

Conductor conductivity

dc

10

If the transmission line is modeled using

instantaneous voltage and current,

INTRODUCTION (Cont’d)

11

Telegraphist’s Equation is used to determine

the basic characteristics for transmission line,

which are:

'

'

'

''

''''

0

0

C

LZ

CjG

LjRZ

j

CjGLjR

Propagation constant

Characteristic impedance

Characteristic impedance for

lossless line

INTRODUCTION (Cont’d)

12

Section of transmission line for attenuation calculation:

INTRODUCTION (Cont’d)

13

in

out

P

PdBG log10)(

The power ratio can be expressed as a gain

G(dB) on logarithmic scale, called decibel scale.

mW

PdBG m 1

log10)(

Or in G(dBm) scale, where to represent

absolute power levels with reference to 1mW

INTRODUCTION (Cont’d)

14

Most of practical problems involving

transmission lines relate to what happens when

the line is terminated.

INTRODUCTION (Cont’d)

15

The load is simply the ratio of the voltage and

the current at the load.

00

000

VV

VVZZL

It can be rearrange as:

00

00 V

ZZ

ZZV

L

L

INTRODUCTION (Cont’d)

16

If the load is unequal to the characteristic

impedance of the line, the wave must be

reflected back to the load. The degree of

impedance mismatch is represented by the

reflection coefficient at the load,

0

0

0

0

ZZ

ZZ

V

V

L

LL

INTRODUCTION (Cont’d)

Range from 0 to 1

17

Generally, the reflection coefficient at any point

along transmission line is ratio of the reflected

wave to incident wave. Superposition of this two

waves creates a standing wave pattern or

voltage standing wave ratio (VSWR) or ratio of

max. to min. voltage amplitude.

L

LVSWR

1

1

INTRODUCTION (Cont’d)

Range from 1 to ∞

18

INTRODUCTION (Cont’d)

19

At any point along the transmission line, we can find the ratio of the total voltage to the total current.

INTRODUCTION (Cont’d)

20

lZZ

lZZZZ

L

Lin

tanh

tanh

0

00

It is known as input impedance,

For a special lossless case, it becomes:

ljZZ

ljZZZZ

L

Lin

tan

tan

0

00

INTRODUCTION (Cont’d)

21

EXAMPLE 1

A source with 50 source impedance drives

a 50 transmission line that is 1/8 of

wavelength long, terminated in a load

ZL = 50 – j25 . Calculate:

(i) The reflection coefficient, ГL

(ii) VSWR

(iii)The input impedance seen by the source.

22

SOLUTION TO EXAMPLE 1

It can be shown as:

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(i) The reflection coefficient,

076

0

0

242.0502550

502550 j

L

LL

ej

j

ZZ

ZZ

(ii) VSWR

64.11

1

L

LVSWR

SOLUTION TO EXAMPLE 1 (Cont’d)

24

(iii) The input impedance seen by the source, Zin

8.38.30

255050

50255050

tan

tan

0

00

j

j

jj

jZZ

jZZZZ

L

Lin

48

2 1

4tan

Need to calculate

Therefore,

SOLUTION TO EXAMPLE 1 (Cont’d)

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4.2 SMITH CHART

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• Graphical tool for use with transmission line circuits and microwave circuit elements.

• Only lossless transmission line will be considered.

• Two graphs in one ;

Plots normalized impedance at any point.

Plots reflection coefficient at any point.

SMITH CHART (Cont’d)

27

The transmission

line calculator,

commonly

referred as the

Smith Chart

SMITH CHART (Cont’d)

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HOW TO USE SMITH CHART?

The Smith Chart is a plot of normalized

impedance. For example, if a Z0 = 50 Ω

transmission line is terminated in a load

ZL = 50 + j100 Ω as below:

29

To locate this point on Smith Chart, normalize the

load impedance, ZNL = ZL/ZN to obtain ZNL = 1 + j2

Ω

SMITH CHART (Cont’d)

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The normalized load

impedance is located

at the intersection of

the r =1 circle and the

x =+2 circle.

SMITH CHART (Cont’d)

31

The reflection coefficient has a magnitude

and an angle : j

L e

L

Where the magnitude can be measured using a

scale for magnitude of reflection coefficient

provided below the Smith Chart, and the angle

is indicated on the angle of reflection

coefficient scale shown outside the

circle on chart.

1L

SMITH CHART (Cont’d)

32

SMITH CHART (Cont’d)

Scale for magnitude of reflection coefficient

Scale for angle of reflection coefficient

33

For this example,

0457.0 j

jL

e

e

SMITH CHART (Cont’d)

34

After locating the normalized impedance

point, draw the constant circle. For

example, the line is 0.3λ length:

jL e

SMITH CHART (Cont’d)

35

• Move along the constant circle is

akin to moving along the transmission line.

Moving away from the load (towards

generator) corresponds to moving in the

clockwise direction on the Smith Chart.

Moving towards the load corresponds to

moving in the anti-clockwise direction on

the Smith Chart.

jL e

SMITH CHART (Cont’d)

36

• To find ZIN, move towards the generator by:

Drawing a line from the center of chart to

outside Wavelengths Toward Generator

(WTG) scale, to get starting point a at

0.188λ

Adding 0.3λ moves along the constant

circle to 0.488λ on the WTG scale.

Read the corresponding normalized input

impedance point c, ZNIN = 0.175 - j0.08Ω

jL e

SMITH CHART (Cont’d)

37

Denormalizing, to find an input impedance,

475.80

jZ

ZZZ

IN

NININ

SMITH CHART (Cont’d)

VSWR is at point b,

9.5VSWR

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For Z0 = 50Ω ,

a ZL = 0 (short cct)

b ZL = ∞ (open cct)

c ZL = 100 + j100

Ω

d ZL = 100 - j100 Ω

e ZL = 50 Ω

SMITH CHART (Cont’d)

39

Take out your Smith Chart, pencil and compass!

LETS TRY!!

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EXAMPLE 2

Repeat Example 1 using the Smith Chart.

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SOLUTION TO EXAMPLE 2

(i) Locate the normalized load, and label it as point a, where it corresponds to

(ii) Draw constant circle.

(iii)It can be seen that

5.01 jZNL

jL e

076245.0 j

L e and 66.1VSWR

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(iv) Move from point a (at 0.356λ) on the WTG

scale, clockwise toward generator a distance

λ/8 or 0.125λ to point b, which is at 0.481λ.

We could find that at this point, it corresponds

to

Denormalizing it,

07.062.0 jZNIN

5.331 jZ IN

SOLUTION TO EXAMPLE 2 (Cont’d)

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EXAMPLE 3

The input impedance for a 100 Ω

lossless transmission line of length

1.162 λ is measured as 12 + j42Ω.

Determine the load impedance.

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SOLUTION TO EXAMPLE 3

(i) Normalize the input impedance:

(ii) Locate the normalized input impedance and

label it as point a

42.012.0100

4212

0

jj

Z

Zz inin

46

(iii) Take note the value of wavelength for point a at

WTL scale.

At point a, WTL = 0.436λ

(iv) Move a distance 1.162λ towards the load to point b

WTL = 0.436λ + 1.162λ

= 1.598λ

But, to plot point b, 1.598λ – 1.500λ = 0.098λ Note: One complete rotation of WTL/WTG =

0.5λ

SOLUTION TO EXAMPLE 3 (Cont’d)

47

(v) Read the point b:

7.015.0 jZNL

Denormalized it:

70150

j

ZZZ NLL

SOLUTION TO EXAMPLE 3 (Cont’d)

49

EXAMPLE 4

On a 50 lossless transmission line,

the VSWR is measured as 3.4. A

voltage maximum is located 0.079λ

away from the load (towards

generator). Determine the load.

50

(i) Use the given VSWR to draw a constant

circle.

(ii) Then move from maximum voltage at

WTG = 0.250λ (towards the load) to point

a at WTG = 0.250λ - 0.079λ = 0.171λ.

(iii)At this point we have ZNL = 1 + j1.3 Ω,

or ZL = 50 + j65 Ω.

SOLUTION TO EXAMPLE 4

jL e

52

EXAMPLE 5 (TRY THIS!)

Use Smith Chart to determine the input

impedance Zin of the two line configuration shown

as below:

53

ZIN = 65.7 – j 124.7Ω

ANSWER FOR EXAMPLE 5

54

4.3 IMPEDANCE MATCHING

• The transmission line is said to be matched

when Z0 = ZL which no reflection occurs.

• The purpose of matching network is to

transform the load impedance ZL such that

the input impedance Zin looking into the

network is equal to Z0 of the transmission

line.

55

Adding an impedance matching networks ensures that all power make it or delivered to the load.

IMPEDANCE MATCHING (Cont’d)

56

Techniques of impedance matching :

Quarter-wave transformer

Single / double stub tuner

Lumped element tuner

Multi-section transformer

IMPEDANCE MATCHING (Cont’d)

57

QUARTER WAVE TRANSFORMER

The quarter wave transformer matching

network only can be constructed if the load

impedance is all real (no reactive component)

58

To find the impedance looking into the quarter

wave long section of lossless ZS impedance

line terminated in a resistive load RL:

ljRZ

ljZRZZ

LS

SLSin

tan

tan

, 24

2 lBut, for quarter wavelength,

l tan

QUARTER WAVE TRANSF. (Cont’d)

59

So,

0

2

ZR

ZZ

L

Sin

Rearrange to get impedance matched line,

LS RZZ 0

QUARTER WAVE TRANSF. (Cont’d)

60

• It much more convenient to add shunt elements rather than series elements Easier to work in terms of admittances.

• Admittance:Z

Y1

IMPEDANCE MATCHING (Cont’d)

61

Adding shunt elements using admittances:

With Smith chart, it is easy to find normalized

admittance – move to a point on the opposite

side of the constant circle. jL e

IMPEDANCE MATCHING (Cont’d)

63

SHUNT STUB MATCHING NETWORK

The matching network has to transform the real part

of load impedance, RL to Z0 and reactive part, XL to

zero Use two adjustable parameters – e.g. shunt-stub.

64

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 ytot = yd + yl = 1 for

complete matching network.

SHUNT STUB MATCHING NET. (Cont’d)

65

• Locate the normalized load impedance ZNL.

• Draw constant SWR circle and locate YNL.

• Move clockwise (WTG) along circle to intersect with 1 ± jB value of yd.

• The length moved from YNL towards yd is the through line length, d.

• Locate yl at the point jB .

• Depends on the shorted/open stub, move along the periphery of the chart towards yl (WTG).

• The distance traveled is the length of stub, l .

SHUNT STUB USING SMITH CHART

jL e

66

SHORTED SHUNT STUB MATCHING

Generic layout of the shorted shunt stub

matching network:

67

EXAMPLE 6 (TRY THIS!)

Construct the shorted shunt stub

matching network for a 50Ω line

terminated in a load ZL = 20 –

j55Ω

68

1. Locate the normalized load impedance,

ZNL = ZL/Z0 = 0.4 – j1.1Ω

2. Draw constant circle.

3. Locate YNL. (0.112λ at WTG)

4. Moving to the first intersection with the

1 ± jB circle, which is at 1 + j2.0 yd

5. Get the value of through line length, d from 0.112λ to 0.187λ on the WTG

scale, so d = 0.075λ

SOLUTION TO EXAMPLE 6

jL e

69

6. Locate the location of short on the Smith Chart

(note: when short circuit, ZL = 0, hence YL =

∞)

on the right side of the chart with

WTG=0.25λ

7. Move clockwise (WTG) until point jB, which

is at 0 - j2.0, located at WTG= 0.324λ yl

8. Determine the stub length, l

0.324λ – 0.25λ = 0.074 λ

SOLUTION TO EXAMPLE 6 (Cont’d)

70

Thus, the values are:

d = 0.075 λ

l = 0.074 λ

yd = 1 + j2.0 Ω

yl = -j2.0 Ω

Where YTOT = yd + yl = (1 + j2.0) + (-j2.0) = 1

SOLUTION TO EXAMPLE 6 (Cont’d)

72

OPEN END SHUNT STUB MATCHING

Generic layout of the open ended shunt

stub matching network:

73

EXAMPLE 7 (TRY THIS!)

Construct an open ended shunt stub

matching network for a 50Ω line

terminated in a load ZL = 150 + j100

Ω

74

SOLUTION TO EXAMPLE 7

1. Locate the normalized load impedance,

ZNL = ZL/Z0 = 3.0 + j2.0Ω

2. Draw constant circle.

3. Locate YNL. (0.474λ at WTG)

4. Moving to the first intersection with the

1 ± jB circle, which is at 1 + j1.6 yd

5. Get the value of through line length, d from 0.474λ to 0.178λ on the WTG

scale, so d = 0.204λ

jL e

75

6. Locate the location of open end on the Smith

Chart (note: when short circuit, ZL = ∞, hence

YL = 0) on the left side with WTG = 0.00λ

7. Move clockwise (WTG) until point jB, which

is at 0 – j1.6, located at WTG= 0.339λ yl

8. Determine the stub length, l

0.339λ – 0.00λ = 0.339 λ

SOLUTION TO EXAMPLE 7 (Cont’d)

76

Thus, the values are:

d = 0.204 λ

l = 0.339 λ

yd = 1 + j1.6 Ω

yl = -j1.6 Ω

Where YTOT = yd + yl = (1 + j1.6) + (-j1.6) = 1

SOLUTION TO EXAMPLE 7 (Cont’d)

78

In both previous example, we chose the first

intersection with the1 ± jB circle in designing

our matching network. We could also have

continued on to the second intersection.

Thus, try both intersection to determine which

solution produces max/min length of through

line, d or length of stub, l.

IMPORTANT!!

79

Determine the through line length and stub

length for both example above by using

second intersection.

For shorted shunt stub (example 6):

d = 0.2 λ and l = 0.426 λ

For open ended shunt stub (example 7):

d = 0.348 λ and l = 0.161 λ

EXERCISE (TRY THIS!)

CHAPTER 4

END