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© AIRCOM International limited 2001 Microwave Link Planning

P-TR-005-M101-ver4 1

Microwave Link Planning

Training Notes

AIRCOM InternationalGrosvenor House

65-71 London road

Redhill, Surrey.

RH1 1 LQ

United Kingdom

Tel: +44 (0) 1737 775700

Email: [email protected]: www.aircom.co.uk






P-TR-005-M101-ver4 3

Table of Contents

1 System Description and Definitions of Terms ............................................................. 5

1.1 Introduction ........................................................................................................... 5

1.2 Definitions of terms............................................................................................... 61.3 The Block Diagram ............................................................................................... 7

1.4 Output of design process ....................................................................................... 91.5 Module 1: Self-Assessment Exercise ................................................................. 14

2 Antennas and The Link Budget.................................................................................. 15

2.1 The Microwave Antenna ..................................................................................... 15

2.2 Significant Parameters......................................................................................... 172.2.1 Beamwidth............................................................................................................................ 18 2.2.2 Gain ...................................................................................................................................... 18

2.3 Calculating the Received Power.......................................................................... 202.4 Linking Gain to Beamwidth ................................................................................ 25

2.5 Linking Gain and Antenna Diameter................................................................... 262.6 Linking Diameter and Beamwidth ...................................................................... 272.7 EIRP..................................................................................................................... 28

2.8 Feeders, Combiners and Splitters ........................................................................ 292.8.1 Splitters and Combiners........................................................................................................ 30

2.9 The Link Budget.................................................................................................. 312.10 Module 2: Self-Assessment Exercises ............................................................ 34

3 Noise Considerations.................................................................................................. 37

3.1 Introduction ......................................................................................................... 373.2 Noise Figure and Noise Temperature.................................................................. 38

3.3 Assessing the Receiver Threshold Level............................................................. 423.4 Threshold levels and the Link Budget................................................................. 43

3.5 Cascaded Systems. .............................................................................................. 453.5.1 Down Converters.................................................................................................................. 49

3.6 Shannon and Nyquist........................................................................................... 503.7 Module 3: Self-Assessment Exercises................................................................. 56

4 Fading......................................................................................................................... 59

4.1 Introduction ......................................................................................................... 594.2 Multipath Fading ................................................................................................. 59

4.2.1 Predicting the likelihood of a fade. ....................................................................................... 61 4.3 Rain Fading.......................................................................................................... 65

4.4 Accommodating Rain and Multipath Fading ...................................................... 674.5 Selective Fading in Digital Systems.................................................................... 70

4.6 Atmospheric Absorption...................................................................................... 764.7 Estimating Link Performance.............................................................................. 784.8 Conclusion. .......................................................................................................... 81

4.9 Module 4: Self-Assessment Exercises................................................................ 82

5 Diversity Techniques.................................................................................................. 855.1 Introduction ......................................................................................................... 85

5.2 The Theory Behind Diversity Systems................................................................ 86

5.3 Types of Diversity ............................................................................................... 87




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5.4 Improvement Factor ............................................................................................ 89

5.5 Improvement for Other Types of Fading............................................................. 935.6 Combining Diversity Techniques........................................................................ 94

5.7 Module 5: Self-Assessment Exercises................................................................. 96

6 Interference Issues...................................................................................................... 97

6.1 Introduction ......................................................................................................... 976.2 Quantifying the effect of interference. ................................................................ 97

6.3 The Theory Behind Diversity Systems................................................................ 98

6.4 Co-channel and Adjacent Channel Interference.................................................. 986.5 Interference Scenarios ....................................................................................... 100

6.6 Reduction Techniques ....................................................................................... 103

6.7 Anomalous Propagation .................................................................................... 1046.8 Intermodulation Effects ..................................................................................... 106

6.9 Module 6: Self-Assessment Exercises............................................................... 109

7 Repeatered Systems.................................................................................................. 1137.1 Introduction ....................................................................................................... 113

7.2 Active and Passive Repeaters ............................................................................ 1147.2.1 Back-to-back antennas........................................................................................................ 116 7.2.2 Reflector repeaters.............................................................................................................. 118


8 Clearance Requirements........................................................................................... 129

8.1 Introduction ....................................................................................................... 1298.2 Earth Bulge ........................................................................................................ 130

8.3 The Fresnel Parameter ....................................................................................... 133

8.4 ITU-R Recommendations.................................................................................. 1358.5 Diffraction Loss................................................................................................. 137

8.6 Fading due to Ground Reflections..................................................................... 1408.6.1 An explanation of Reflection-induced Fading .................................................................... 140

8.6.2 The Rayleigh Criterion. ...................................................................................................... 144 8.6.3 Protection against reflection fades ...................................................................................... 145


9 Performance Objectives ........................................................................................... 1579.1 Introduction: ...................................................................................................... 157

9.2 Propagation-related Unavailability.................................................................... 158

9.3 Equipment-related Unavailability...................................................................... 1599.3.1 Hot Standby........................................................................................................................ 161

9.4 Unavailability Objectives .................................................................................. 164

9.5 Performance Standards ...................................................................................... 16510 Solutions to Self-Assessment Questions............................................................... 169

10.1 Module 1: Self-Assessment Exercise............................................................ 169

10.2 Module 2: Self-Assessment Exercise............................................................ 17010.3 Module 3: Self-Assessment Exercises ........................................................... 173

10.4 Module 4: Self-Assessment Exercises. ......................................................... 177

10.5 Module 5: Self-Assessment Exercises ........................................................... 18110.6 Module 6: Self-Assessment Exercises ........................................................... 183

10.7 Module 7: Self-Assessment Exercises ........................................................... 186

10.8 Module 8: Self-Assessment Exercises. .......................................................... 190




P-TR-005-M101-ver4 5

1 System Description and Definitions of Terms

1.1 Introduction

Aims of CourseAims of Course

• To enable you to plan the radio elements of a point to point microwave

link against a performance requirement and to be able to predict the

performance of the link that you have planned.

• This will involve gaining an understanding of

Introductory Session

• Antennas

• Link Budgets

• Noise

• Fading

• Diversity Techniques

• Interference

• Radio Propagation

• Modulation Methods

• Performance Prediction

Methods.

A microwave link will often present a convenient, economic way of providing high speed

data communications between two points. The objective of this course is to provide you

with a sufficient information and understanding to specify the radio equipment and

configuration of a microwave link for a given purpose.

The emphasis will be on performing quantitative analyses so that specific answers can be

given to the questions: “How high?”; “How big?”; “How long?”; “How far?”; “How

good?”.

Microwave equipment is readily available through a number of manufacturers. The link

designer’s job is to select and configure equipment in the most effective and economic

manner.




P-TR-005-M101-ver4 6

Why MicrowaveWhy Microwave

Microwave radio links provide high speed (2 Mbps+)communication between two points.

They are known to be:

• fast to implement

• convenient

• economic

when compared with wire-based alternatives.

1.2 Definitions of terms

Microwave frequencies are usually taken to mean frequencies between 3 GHz and 30

GHz (wavelengths of 100 mm to 10 mm). Higher frequencies (up to 40 GHz) than this,

known as “millimetric frequencies” are being used to provide point to point

communications and these will be included in the scope of this course.




P-TR-005-M101-ver4 7

What does “Microwave” meanWhat does “Microwave” mean

Microwave refers to a section of the RF spectrum lyingbetween 3 and 30 GHz. It is also referred to as “Super

High Frequency” (SHF).

LF MF HF VHF UHF SHF mm

300 GHz

30 GHz

3 GHz

300 MHz

30 MHz

3 MHz30 kHz

300 kHz

The Microwave Band

Note that frequencies up to 40 GHz are being used for

“microwave” links although the definition suggests that this

frequency is in the “millimetric” band.

A microwave link is taken to mean a fixed, permanent (or “semi-permanent”) connection

between two points. The design of “mobile microwave” or “microwave broadcast”

systems is beyond the scope of this course.

1.3 The Block Diagram

A microwave link can be thought of as consisting of a few elements that can be

purchased “off the shelf”. However, a whole variety of specifications will be available

and the link designer must select the most appropriate equipment for the job in hand. The

saying “An engineer is someone who can do something for a pound (or dollar, or euro

etc..) that any fool can do for ten pounds” is very true of microwave engineering. Buying

the “biggest, highest, most powerful and most expensive of everything and then

employing every performance-enhancing technique in existence will almost certainly

result in a system that will perform well but it will similarly result in a system that costs

far more than necessary.




P-TR-005-M101-ver4 8

System Block DiagramSystem Block Diagram

The equipment layout is essentially very simple. The job of the

link planner is to specify and configure the equipment.

Transceiver

Antenna

Feeder

The block diagram looks relatively straightforward. The two ends of the system are very

similar to each other with both consisting of: one or more antennas; a transmitter and

receiver (commonly known as a “transceiver”) and something to connect these two

together – a “feeder”.

Transceiver

Antenna

Feeder




P-TR-005-M101-ver4 9

The antenna will have to be mounted on a mast and the required height will be dependent

on the length of the link and the characteristics of the intervening terrain, amongst other

things. The focus in this course is on the radio engineering, rather than mechanical

engineering, aspects of the system design process and we will afford mast and tower

design only the most superficial of looks.

1.4 Output of design process

In its simplest form the final output of the design process will include details of:

• frequency of operation;

• antenna sizes

• antenna heights

• feeder type and length

• transmit power

• capacity

• path length

• predictions of unavailability




P-TR-005-M101-ver4 10

Answers, Please!Answers, Please!

How big must

the antenna

be?

How high

must the

antenna be?

What will the

loss of the

feeder be?

What should

the transmit

power be?

What power

level will we

receive?

What

frequency will

we use?

At what data

rate must we

send?

How good will the

performance be?

The above parameters will be the subject of individual attention but the prediction of

unavailability parameter will benefit from a mention at this point. Unlike an optical fibre

or coaxial cable transmission system, the performance of a microwave link will vary with

time. The received power level will suffer fades, mainly due to atmospheric refraction

effects and rain. Methods exist whereby the system can be predicted to be unavailable

for a (hopefully very small) percentage of the time. The percentage unavailability will

form one of the requirements specified by the customer.




P-TR-005-M101-ver4 11

Percentage OutagePercentage Outage

Unlike an optical fibre or coaxial cable system, the received

power level of a microwave system will vary significantlywith time.

This is due to atmospheric effects and “hydrometeors” suchas rain and snow.

This will inevitably lead to the system suffering an outage for a small percentage of the time.

The link planner must be able to predict the outage periodsas a percentage on a particular system.

Quantitative AnalysisQuantitative Analysis

The link planner must be able to determine numericalparameters to define the microwave system.

The course will involve methodologies, procedures andtechniques for arriving at the correct numerical solutions.

However, all solutions should fit in with the expectations of an intuitive engineer.




P-TR-005-M101-ver4 12

All the parameters in the list interact with each other and it is necessary to gain

knowledge regarding each of them before an intelligent approach can be taken to

microwave link design. Nevertheless, an intuitive engineering consideration of the

parameters can establish certain expectations formed from an engineering base. For

example we can confidently say that:

1. If the path length is increased then we should increase the transmit power

or antenna sizes.

2. If the length of feeder is reduced we can reduce the transmit power.

The two statements above are “qualitative” rather than “quantitative” in nature. It is

necessary to be able to analyse the situation quantitatively so that we can accurately

specify the equipment required. The following sections provide information, analyses

and techniques that will enable you to plan, design and predict the performance of a

microwave link.

Intuitive ExpectationsIntuitive Expectations

• If the antenna is bigger, the receive power will increase.

• If the link is longer the receive power will decrease.

• We will need a higher power to transmit a higher data rate.

• The higher the power received, the lower the percentage

outage.

• The longer the feeder, the lower the receive power.




P-TR-005-M101-ver4 13

Next StepsNext Steps

• All the parameters affect each other in an interactive way.

• The next sections will deal with particular parameters

whilst keeping one eye on the final goal

• In the next section we shall concentrate on the antenna

and methods of predicting the receive signal power.




P-TR-005-M101-ver4 14

1.5 Module 1: Self-Assessment Exercise

Designing by guessing.

As intuitive engineers we should have some idea regarding what a

microwave link should look be like and what its values should be.

Try and picture a microwave link in your mind and imagine what the

relevant parameters might be. It will be interesting to refer to these

“guesstimates” as we gain knowledge regarding the design of microwave

links.

Name of DesignerFrequency of Operation

Rate of transmission (bits per

second)

Mast Height

Antenna Diameter

Path Length

Transmit Power

Receive Power

Feeder length (metres)Feeder loss (dB)




P-TR-005-M101-ver4 15

2 Antennas and The Link Budget

2.1 The Microwave Antenna

The microwave antennas used for point to point links fall into the category of “aperture”

antennas, the parabolic dish antenna being the most common example. A propagating

electromagnetic wave has a power density P d (in watts per square metre) associated with

it. The aperture (known for these purposes as the “effective aperture” Ae) of the antenna

is measured in square metres and the antenna serves to convert the power density into an

actual power Pr (the suffix “r” standing for “received”) in accordance with the formula

ed r =

The Microwave AntennaThe Microwave Antenna

• Parabolic antennas are a form of

“aperture” antenna.

• The antenna faces an incoming

electromagnetic wave that has a

power density P d .

• The antenna converts this to a

received power P r .

Antennas and The Link Budget




P-TR-005-M101-ver4 16

The effective aperture of a microwave antenna is typically 60% of its measured aperture.

For example, a parabolic dish of 1 metre has an effective aperture of approximately

46.0 π × square metres.


• The “aperture” can be thought of as

a hole through which energy passes.

• This energy is delivered to the

antenna output..


ed r A P P =

E

H

Pd

Pr




P-TR-005-M101-ver4 17


• The “effective aperture” is linked to

the physical aperture.

• For an antenna presenting a circular

cross section of diameter D when

viewed from the front


46.0

2 D Ae

π ×≈

D

2.2 Significant Parameters

Although Ae is a vital parameter of a microwave antenna, it is not quoted very often.

Much more popular are the parameters of “gain” and “beamwidth”. It is important to

appreciate that an antenna performs both as a transmitter and as a receiver. Indeed, the

laws of physics dictate that the gain and beamwidth of an antenna as a receiver and as a

transmitter are exactly the same.




P-TR-005-M101-ver4 18

2.2.1 Beamwidth

A major purpose of a microwave antenna is to form the microwave energy into a narrow

beam rather than spread it widely. The narrower the beam, the higher the power densitythat will be achieved. The beamwidth is measured in degrees between the two points

either side of the principal axis (the principal axis is the name given to the line from the

antenna on which the power density is a maximum) at which the power density is half is

maximum value. This is often known as the “3 dB beamwidth”.

2.2.2 Gain

The Isotropic AntennaThe Isotropic Antenna

• A hypothetical antenna that

distributes its transmitted power

equally in all directions.

• As the surface area of a sphere

radius r is the power density

produced at a distance r is given by


24 r π

24 r

P P t

d π

=

r




P-TR-005-M101-ver4 19

Antenna GainAntenna Gain

• A practical microwave antenna will

produce a higher power density byconcentrating the energy into a

narrow beam.

• For an antenna of gain Gt , the power

density produced is, by definition


r

24 r

G P P t t

d π

=

The connection between beamwidth and power density means that beamwidth and gain

are inter-linked. The narrower the beamwidth, the higher the gain.

The gain of an antenna is measured as the increase in power density achieved as a

multiple of the density that would be produced by a theoretic “isotropic” antenna that

distributes the power equally in all directions. Given that the surface area of a sphere of

radius r is equal to 4πr 2, it is possible to say that the power density Pd is related to the

power transmitted Pt by the equation24 r

P P t

d π

= . The power density at the same

distance produced by an antenna with gain Gt is 24 r

G P t t

π . Notice that this gain,

Gt, refers to the principal direction of the antenna and will be very sensitive to errors in

the pointing direction.




P-TR-005-M101-ver4 20

2.3 Calculating the Received Power

If the power density is known, and the effective aperture of the receive antenna equals

Ae, then we can calculate the power received, Pr, using the equation

et t

r Ar

G P P

24π =

This is a simplified form of “link budget” that will allow us to predict the received power

level given the other parameters.

Example

The transmitting antenna on a point-to-point microwave link has a gain of 500. The

receiving antenna has an effective aperture of 2 m2. If the transmit power is 0.5 W and

the link is 20 km long it is possible to determine the power received, Pr, from the

equation

W

Ar

G P P e

t t r

8

2

2

1095.9)2000(4

25005.0

4

−×=××

=

=

π

π

Note that we have quoted a “gain” for the transmit antenna and an “effective aperture”

for the receive antenna. Identical antennas are normally used for transmit and receive

purposes and catalogues will normally quote only the gain. It is important to be able to

convert gain to effective aperture. For the purpose of achieving this we will rely on the

fact that the gain of the antenna as a transmitter is exactly the same as when used as a

receiver. The gain as a receiver, relative to an isotropic antenna is its effective aperture




P-TR-005-M101-ver4 21

as a multiple of the effective aperture of an isotropic antenna. If it has ten times the

effective aperture, it will capture ten times the power. It can be shown that the effective

aperture of an isotropic antenna at wavelength λ isπ

λ 4

2. Thus the gain, G, of the

antenna is given by

2

4

λ

π e AG = and hence

π

λ

4

2

G Ae =

Expressing the above equation in decibels gives

=210

4log10

λ

π e AG dBi (the “i” standing for “isotropic”)

Our formula for receive power now becomes

2

2

22

4

444Pr

=

==

r GG P

G

r

G P A

r

G P

r t t

r t t

e

t t

π

λ

π

λ

π π

Expressing this in terms of decibels gives

)(log20)(log20)4(log20)dBi()dBi()dBm()dBm( 101010 λ π +−−++= r GG P P r t t r

Now (Pt-Pr = Path loss). So

r t GGr −−−+= )(log20)(log20)4(log20LossPath 101010 λ π dB




P-TR-005-M101-ver4 22

Changing the units of distance from metres to kilometres and using the formula

f 3.0=λ where f is the frequency in Gigahertz gives.

[ ] r t r t GG f kmd P P −−+−+=− log20)3.0log(20)(1000log20)4log(20 π

r t

r t r t

GG f d

GG f d P P

−−++=

−−++++=−

1010

1010

log20log204.92

log20log2046.106098.21

The expression 92.4 + 20log(d) + 20log(f) equals the path loss when the gain of the

antennas is unity (0 dBi). This is known as the Free Space Loss (FSL).

r t GG f d −−++= 1010 log20log204.92FSL dB

where d is the path length in kilometres and f is the frequency in GHz.

Calculating the received power Calculating the received power

• This equation allows us to calculate

the received power given the other

parameters.


24 r

G P P t t

d π

= ed r A P P =

et t

r Ar

G P P

24π =




P-TR-005-M101-ver4 23

Calculating the received power Calculating the received power

• Example.


watts1095.9)20000(4

25005.0P

watts5.0

m20000

m2

500

8

2r

2

−×=××

=

=

=

=

=

π

t

e

t

P

r

A

G

Pt

Gt

Pr

Aer

Antenna CharacteristicsAntenna Characteristics

• Radiation pattern, gain, and antenna properties in general

have the same same characteristics whether the antennais being used as a transmitter or receiver.

• Considering the antenna as a receiver. The gain equals

its effective aperture as a multiple of the effective aperture

of an isotropic antenna.

• Aperture of isotropic antenna


dBi4

log10

4

4

210

2

2

=

=

=

λ

π

π

λ

λ

π

e

e

e

AG

G A

AGπ

λ

4

2

=




P-TR-005-M101-ver4 24

Calculating Received Power Calculating Received Power

• Substituting Effective Aperture in terms of Gain.


2

2

22

4

444Pr

=

==

r GG P

Gr G P A

r G P

r t t

r t t e

t t

π

λ

π λ

π π

)(log20)(log20)4(log20 101010 λ π +−−++= r GG P P r t t r

Calculating Received Power Calculating Received Power

• Changing units from metres to kilometres and from

wavelength in metres to frequency in Gigahertz:


)(log20)(log20)4(log20 101010 λ π +−−++= r GG P P r t t r

)(log20)(log204.92 1010 d f GG P P r t t r −−−++=

)/3.0(log20)1000(log20)4(log20 101010 f d GG P P r t t r +−−++= π

FSLGG P P r t t r −++=

[ ] [ ])km(log20)GHz(log204.92 1010 d f FSL ++=




P-TR-005-M101-ver4 25

2.4 Linking Gain to Beamwidth

Let us assume that an aperture antenna has a narrow, conical beam of beamwidth θ

radians.

At a distance r, the diameter of the circle illuminated is r θ and the area is4

22θ π r

Remembering that the area illuminated by an isotropic antenna is 4πr 2, the gain is given

by

( )

degrees 230

radians 4

16

4

422

2

G

G

r

r G

≈

≈

=≈

θ

θ θ π

π

Remember, G is a ratio (not in dB).

θ

r




P-TR-005-M101-ver4 26

Linking Gain andLinking Gain and BeamwidthBeamwidth

• A practical microwave antenna will

produce a gain by concentrating the

energy into a narrow beam.

• For an antenna of gain Gt , the area

illuminated will be reduced compared

with that illuminated by an isotropic

antenna by a factor equal to its gain.


r

θ r θdegrees230

4

16

4

4

2

222

t

t

t

t

G

G

G

G

r r

≈

≈

≈

≈

θ

θ

θ

π θ π

2.5 Linking Gain and Antenna Diameter

Remembering that

4 and 4

2

2 D A AG ee π

λ π ≈= where D is the antenna diameter in metres.

Hence

f D

Df DG

1010

22

log20log204.20Gain(dB)

0.3

)GHz(

++=

=

=

π

λ

π

The above formula ignores inefficiencies and imperfections. A more realistic formula for

the gain is

)(log20)(log205.17Gain(dB) 1010 GHz f m D ++=




P-TR-005-M101-ver4 27

Linking Gain and Antenna Diameter Linking Gain and Antenna Diameter

• The above equation ignores inefficiencies in the

antenna system. A more realistic equation is


22

2

2

3.0

4

4

=

≈

≈=

Df DG

D A AG ee

π

λ

π

π

λ

π

f DG 1010 log20log204.20 ++≈

f DG 1010 log20log205.17 ++≈

2.6 Linking Diameter and Beamwidth

( )

22(degrees)Beamwidth(GHz)frequency(m)Diameter

22

3.0230

230

0.3

2

≈××

≈

×≈

≈

=

θ

π θ

θ

π

Df

Df

G

Df G




P-TR-005-M101-ver4 28

Linking Antenna Diameter andLinking Antenna Diameter and BeamwidthBeamwidth

• From the previous slides:


22

3.0230

230

3.0

2

≈

×≈∴

≈

≈

θ

π θ

θ

π

Df

Df

G

Df G

• Diameter (metres) x frequency (GHz) x Beamwidth (degrees) 22

2.7 EIRP

The term “Equivalent Isotropic Radiated Power” (EIRP) is applied to a transmitting

antenna. It is the power that would have to be transmitted by an isotropic antenna to

produce the same power density. Mathematically it is very simple to express:

(ratio)(watts))EIRP(watts

(dBi)(dBm)EIRP(dBm)

t t

t t

G P

G P

×=

+=




P-TR-005-M101-ver4 29

EIRPEIRP

• A commonly used term - “Equivalent Isotropic Radiated Power”.


)ratio()watts()watts(

)dBi()dBm()dBm(

t t

t t

G P EIRP

G P EIRP

×=

+=

2.8 Feeders, Combiners and Splitters

A microwave station includes what is often referred to as “plumbing”; usually lengths of

waveguide and connectors. The “plumbing” exists to connect the transmitter and

receiver to the antenna. For most systems its main component will be a length of feeder

known as “waveguide”. Waveguide is the most common transmission line over the 3 –

30 GHz range. Coaxial cable becomes very lossy at frequencies above 3 GHz.

Waveguide can look similar to a large coaxial cable, the most significant difference being

that it has no inner conductor: effectively it traps the energy in the form of a radio wave

and causes this energy to propagate along the waveguide. The size of the is very much

frequency dependent with the width being approximately 0.7λ (where λ is the

wavelength). You can find rectangular, circular or elliptical guide for various purposes.

One crucial parameter that affects the loss of the guide is the conductivity of the inner

coating. Copper is most commonly used due to its high conductivity. However,

occasionally, at the highest frequencies (where losses are greatest), silver-plated




P-TR-005-M101-ver4 30

waveguide is used (silver has the highest conductivity of any metal). Manufacturers

tables should be referred to in order to select a feeder for the frequency range used and to

predict the loss incurred.

Feeders and CombinersFeeders and Combiners


• Co-axial cable is not suitable at frequencies above about 3 GHz.

• A hollow metal tube known as “waveguide” is used over the

frequency range 3 - 30 GHz.

• The size of the waveguide depends on the frequency being used

and typically has a width of 0.7λ.

• Usually made of copper or brass with a copper plating inner

coating. Occasionally silver plated.

• Losses typically 0.1 dB per metre. The higher the frequency, the

higher the loss.

2.8.1 Splitters and Combiners

The same antenna is used for transmitting and receiving. A sophisticated form of

combiner known as a “diplexing filter” is used to ensure that the high power transmitter

does not interfere with the very sensitive receiver. If a combined transmitter and receiver

is purchased “off the shelf” this filter will be an integral part of its construction.

Occasionally, the same transceiver will receive from two antennas whilst transmitting

from only one antenna. This arrangement requires a sophisticated “diversity

splitter/combiner” enable its implementation.

Such splitters and combiners inevitably have an “insertion loss” associated with them.

These losses must be considered when calculating the received signal strength.




P-TR-005-M101-ver4 31

Feeders and CombinersFeeders and Combiners


• A diplexer is a sophisticated device that makes it possible totransmit and receive from the same antenna.

• The received signal is sometimes the combination of two

antennas.

• Combiners and splitters have an insertion loss that must be

considered when predicting the received signal level.

• All miscellaneous losses must be considered.

2.9 The Link Budget

Producing a link budget is a disciplined way of calculating the received signal power in a

way that minimises the risk of omitting essential parameters. All parameters are quoted

in dB so that calculations entail only addition and subtraction. As an example consider

an 11 GHz system using two, 35 dBi, antennas. The path length is 20 km. Feeder losses

amount to 1.5 dB at the transmitter and receiver ends. Combiner losses total 2 dB. The

transmit power is 500 mW (27 dBm). The link budge shows that the received power is

expected to be -–47.2 dBm.

TRANSMIT

Transmit Power 27 dBm

Antenna Gain 35 dBi

Feeder Loss 1.5 dB

EIRP 60.5 dBm

PATH LOSS




P-TR-005-M101-ver4 32

Path Length 20 km

Frequency 11 GHz

Free Space Loss 139.2 dB

RECEIVE

Antenna Gain 35 dBi

Feeder Losses 1.5 dB

Net Gain 33.5 dB

MISC

Combiner Losses 2 dB

RECEIVE

POWER

-47.2 dBm

It is a straightforward matter to convert the link budget to a spreadsheet thus making it

easy to assess the impact of changing different parameters.

The Link BudgetThe Link Budget


• The Link Budget is usually of the form of a table that ensures no

sources of losses or gains are forgotten.

• Expressing all powers, losses and gains in dB, dBi, dBm etc.

Allows us to simply add or subtract the relevant amounts.

• The simplified link budget equation is given below. Each element

would be arrived at by considering its constituent parts.

Received Power = EIRP - FSL + Rx antenna gain - Misc Losses




P-TR-005-M101-ver4 33

The Link BudgetThe Link Budget


• Being able to determine the received power level is a

significant achievement.

• However, the question “Is this power level sufficient?”

must be answered.

• To be able to answer this question requires an

understanding of system noise.




P-TR-005-M101-ver4 34

2.10 Module 2: Self-Assessment Exercises

1. An antenna operates at a frequency of 15 GHz. If it has a diameter of 1.8

metres, estimate its gain.

2. Two such antennas are to be used over a link of length 12 km. Determinethe path loss.

3. Repeat the calculation of question 2 for antennas of the same size butoperating at a frequency of 30 GHz.

4. Estimate the beamwidth of a 1.8 metre antenna at 7 GHz, 15 GHz and 30GHz.




P-TR-005-M101-ver4 35

5. A transceiver outputs a power of 27 dBm via a feeder of 3 dB loss to an

antenna of diameter 0.9 metres. If the frequency of operation is 12 GHz,estimate the EIRP from the antenna.

6. For the situation described in question 5, estimate the power that would

be gathered by an identical antenna at a distance of 4 km.






P-TR-005-M101-ver4 37

3 Noise Considerations

3.1 Introduction

Noise ConsiderationsNoise Considerations

• Thermal Noise forms the fundamental limitation of any

telecommunications system.

• The level of thermal noise is directly proportional to

bandwidth and absolute temperature.

• k is Boltzmann’s constant and equals 1.38x10-23

joules/kelvin.

Noise Considerations

kTBPower Noise = watts

The fundamental limitation of any telecommunications system is thermal noise. The

random motion of electrons develops an alternating voltage such that a certain amount of

power will be delivered into a load resistor (which will in turn, generate its own thermal

noise). The level of thermal noise generated by a resistor, for example, is proportional to

the system bandwidth ( B), measured in hertz and also the absolute temperature (T ),

measured in kelvins (0°C = 273 K). The governing equation that describes the power

that will be delivered into a matched load is

wattsPower kTB=

k is known as Boltzmann’s constant and has a value 1.38 x 10-23

joules per kelvin.

An antenna will gather thermal noise along with a wanted signal. The amount of thermal

noise gathered depends on where the antenna is looking. A high quality satellite station




P-TR-005-M101-ver4 38

antenna “looking” at deep space can have a thermal noise temperature as low as 40

kelvins. A typical television satellite receiver will typically have a noise temperature of

160 K. However, we will be dealing with terrestrial systems whereby the antenna will be

looking at the earth’s atmosphere which is generally assumed to be at the “standard”

temperature of 290 K (usually referred to as T 0). In this circumstance the thermal noise

level will be watts/Hz1000.42901038.1 2123 −− ×=××× B .

Noise ConsiderationsNoise Considerations

• An antenna can be thought of as a noise gathering device.

• The figure for “absolute temperature” depends on where

the antenna is looking.

• For terrestrial systems, “normal” values such as 290 K are

suitable.

• For high quality satellite systems, values of T as low as 40

K are achievable.

• Cheaper systems (e.g. Sky TV) have values of 160 K.


3.2 Noise Figure and Noise Temperature.

The Signal to Noise ratio at the output of any electronic device, such as an amplifier or

even a length of feeder, will be worse that that at the input. This is because all devices

will contribute some noise to add to the existing thermal noise. In order to be able to

quantify the noise performance of a device it is assumed that the device includes a noise

generator at its input. The output power of this noise generator is assumed to be BkT e

where eT is known as the “effective noise temperature” of the device. The lower the

value of T e the better the performance of the system.




P-TR-005-M101-ver4 39

Receiver Noise Figure and NoiseReceiver Noise Figure and NoiseTemperatureTemperature

• No radio receiver is perfect, they all add noise to the

system.

• The SNR at the output of any amplifier is worse than at the

input.

• This is accounted for mathematically by imagining a noise

generator at the input of the amplifier.

• This noise generator has a power output of kT e B where T eis the noise temperature of the amplifier.


Knowing T e makes it possible to calculate the noise power at the output of the amplifier.

The total effective noise power at the input equals ( ) BT T k BkT kTB ee +=+ which

means that, if the device is an amplifier of gain G, the noise power at the output of the

amplifier will be ( ) BGT T k e+ . Values of T e will vary from a few tens of kelvins to

several thousand. As a result, the parameter is not particularly intuitive and the term

“noise figure” is often preferred.

Noise figure gives a direct impression of the amount by which a device makes the signal

to noise ratio worse. For example, if a device has a noise temperature of 290 kelvins and

the thermal noise power at the input was k (290) B then the noise power at the output

would be double what it would be if the device was perfect. The noise figure of this

device is 2 (or 3 dB) as the SNR at the output is 3 dB worse than that at the input. It is

possible to convert from noise temperature to noise figure by equating the noise power at

the output in terms of both effective noise temperature and noise figure.




P-TR-005-M101-ver4 40

( )

( )

0

0

00

00

1

1

T

T F

F T T

F T T T

BFGkT BGT T k

e

e

e

e

+=

−=

=+

=+

However, the term Noise Figure has a serious drawback. The equations described above

involving Noise Figure are only valid if the noise power at the input of the device is

thermal noise equating to a temperature of 290 kelvins. Use of Noise Figure under any

other circumstances is a mistake. Abuse of the term Noise Figure is widespread. Be

careful.

Effective Noise Temperature, on the other hand can be used regardless of the level ofnoise at the input of a device.



kTB

kTeB

G K(T+Te)BG

k(T o+T e )BG =kToBGF

• F is known as the noise figure of the amplifier.

• If the value of T at the input equals the “standard” temperature, To, of

290 K, then noise at the output equals




P-TR-005-M101-ver4 41



kToB

kTeB

G k(To+Te)BG= kToBGF

o

e

oeo

T

T F

F T T T

+=

=+

1

( )1−= F T T oe


• Using F is “convenient”. It can be expressed in dB, rather than as a ratio.

• In dB form is represents the “amount by which the SNR

gets worse”.

• However, the equation is only valid if the noise at the input

equals kT o B.

• Abuse of Noise Figure is widespread. Be careful.





P-TR-005-M101-ver4 42

3.3 Assessing the Receiver Threshold Level.

The SNR directly affects the error ratio (BER). Thus receiver noise performance directly

affects the BER achieved. For example, suppose an SNR of 14 dB is required to achieve

a BER of 1 x 10-6 with a system that has an 8 MHz bandwidth. We can determine the

minimum required signal power (the “threshold” level) in order to achieve this if the

receiver noise figure is known. If the noise figure is, say, 4 dB (a typical value;

equivalent to a ratio of 2.5) we can say that the noise temperature is 290(2.5-1) = 438 K.

Thus the equivalent noise power referred to the input is

( ) ( ) dBm)-101(watts1004.81084382901038.1 14623 =×=×+×=+ −− BT T k e

As the required SNR is 14 dB, it is simple to calculated the required signal power to be –

87 dBm. This establishes the minimum signal level required. Consulting manufacturers

literature will usually reveal information regarding the required signal level for a given

BER.

Assessing the minimum signal levelAssessing the minimum signal level• The error ratio experienced on a system depends on the SNR.

• We need to establish a required SNR in order to determine the minimum

required receive power (known as the receiver “threshold”).

• For example, a minimum SNR of 14 dB is required in order to deliver a

BER of better than 1x10-6. The system bandwidth is 8 MHz and the

receiver noise figure is 4 dB.

• 4 dB is a ratio of 2.5. Noise temperature is therefore 290(2.5-1) = 438 K.

• k(T+Te)B = 1.38x10-23(290+438)8x106 = 8.04x10-14 watts (=-101 dBm)

• SNR required of 14 dB is a ratio of 25. 25 x 8.04 x10-14 =2.0x10-12 watts

or -87 dBm. This establishes the minimum signal level.

• Note: receiver manufacturers will often quote their own threshold level.





P-TR-005-M101-ver4 43

3.4 Threshold levels and the Link Budget

So far link budgets have been introduced as a method of predicting the received signal

power. Now that we have a method of determining a minimum value for this signal

power allows us to establish minimum values for the various items of equipment that we

use. For example we could determine the minimum antenna sizes required given the

system parameters listed below.

Transmit Power -6 dBm

Receiver Threshold -87 dBm

Feeder/misc losses 5 dB

Transmit Frequency 6 GHz

Path Length 40 km

The required calculation is summarised below.

FSL 92.4 + 20log(6) +20 log (40)=140 dB

Misc Losses 5 dB

Tx Power -6 dBm

Rx Threshold -87 dBm

Allowable losses 81 dB

FSL + Misc Losses 145 dB

Required Antenna Gains 64 dBi

Each Antenna Gain must be 32 dBi

Diameter ( )metres88.010 20

6.155.1732

=−−

The calculation shows that 90 centimetre diameter antennas should be suitable.






P-TR-005-M101-ver4 45

3.5 Cascaded Systems.

Suppose we have a system that comprises of two amplifiers in tandem, with the output of

one forming the input to the other. The amplifiers have gains and effective noise

temperatures 2211 ,,, ee T GT G respectively. Suppose the noise power at the input to the

first amplifier equals kTB then the noise at the output of this first amplifier will be

( ) 11 BGT T k e+ . This is fed into the second amplifier with the result that the noise power

at the output will be ( )[ ]{ } 2211 BGT GT T k ee ++ . This can be equated to ( ) 21G BGT T k e+

where Te is the overall noise temperature of the system. Equating these quantities results

in the overall noise temperature being linked to the individual parameters by

1

21 G

T

T T eee += . This can be extended to more than two amplifiers resulting in the

general equation

.................21

3

1

21

GG

T

G

T T T ee

ee ++=

Examining the above equation shows that the first amplifier in a system (often referred to

as a “low noise amplifier”) is most crucial in determining the overall noise temperature as

the noise temperature of any subsequent devices is divided by the gain of the first

amplifier.




P-TR-005-M101-ver4 46

Noise Figure and Noise Temperature of Noise Figure and Noise Temperature of “cascaded systems”“cascaded systems”


G1 ,Te1 G2 ,Te2

k(T+Te1 )BG1 k{[(T+Te1 ) G1 ]+ Te2 } B G2

k(T)B

k{[(T+Te1 ) G1 ]+ Te2 } B G2 =k(T+Te)B G1 G2

1

21

G

TeTeT +=

.............21

3

1

21

GG

Te

G

TeTeT ++=

The same reasoning can be applied to determining the noise temperature of an attenuator.

Remember that feeders in a microwave communications system will act as attenuators.

The signal will be attenuated at the output but, in a matched system, the noise power at

the output will be the same as at the input. Using the same equations as before, the noise

power at the output is ( ) BGT T k e+ . Equating this to the noise power at the input gives

( )

( )

( )

−=

−=

=+

=+

11

1

GT

G

GT T

T GT T

kTB BGT T K

e

e

e

Remember that, for an attenuator, G will be less than one.




P-TR-005-M101-ver4 47

Noise Figure and Noise Temperature of Noise Figure and Noise Temperature of attenuatorsattenuators (and feeders)(and feeders)

• For a matched attenuator receiving thermal noise at its

input, the noise at the output equals the noise at the input.


Ik(T)B k(T+Te)BG

( )

( )

( )11

)1(

−=

−=

+=

+=

GT

G

GT T

GT T T

BGT T k kTB

e

e

e

• Note that, for an attenuator, G will

be less than 1.

As an example consider the situation where an antenna is connected to a receiver via a

feeder of loss 2.5 dB. If the temperature of the feeder is 290 K and the Noise Figure of

the receiver is 4 dB it is possible to determine the noise figure of the combination.

Receiver Noise Figure 4 dB (ratio of 2.5)

Receiver Noise Temperature 1.5 x 290 = 435 K

Attenuation 2.5 dB G = 1/1.778 = 0.562

Noise Temperature of Attenuator 290(1.778-1)=226 K

Overall Noise Temperature 226+(435÷0.562) = 1000 K

Overall Noise Figure 1+(1000÷290) = 4.45 (or 6.5 dB)

Thus the presence of the attenuating feeder has worsened the noise figure from 4 dB to

6.5 dB.




P-TR-005-M101-ver4 48

The situation described above can be improved by installing a low noise, “mast head

amplifier” in between the antenna and the feeder. If this has a gain of 15 dB and a Noise

Figure of 3 dB we can determine the new Noise Figure in the usual manner.

Noise Temperature of MHA 290 K

Gain of MHA 31.6

Overall Noise Temperature 290+(1000÷31.6) = 321.6 K

Overall Noise Figure 1+(321.6÷290) = 2.11 (or 3.24 dB)

Notice that this new Noise Figure is lower than that of the receiver alone. It is the Noise

Figure of the MHA, rather than that of the receiver, that forms the lower limit of the

resultant Noise Figure.

Cascaded System ExampleCascaded System Example

• An antenna is connected to a receiver via a feeder of loss

2.5 dB. If the temperature of the feeder is 290 K and the

Noise Figure of the receiver is 4 dB, determine the noisefigure of the overall combination.


• Noise Figure 4 dB. Ratio of 2.5.

• Noise temperature = 1.5x290=435 K.

• G for attenuator is 0.562. Noise temperature of

attenuator is 290(0.778)=226 K

• Overall noise temperature is 226+435/0.562=1000 K

• Overall noise figure = 1+1000/290=4.45 (6.48 dB)




P-TR-005-M101-ver4 49

Cascaded System Example - Low NoiseCascaded System Example - Low NoiseAmplifier Amplifier

• To improve the previously described situation, a Low Noise

amplifier is connected between the antenna and the feeder.This has a gain of 15 dB and a Noise Figure of 3 dB.

Determine the new noise figure.


• Noise Figure 3 dB. Ratio of 2. Noise temperature = 290 K

• G for amplifier is 31.6.

• Overall noise temperature is 290+1000/31.6=321.6 K

• Overall noise figure = 1+321.6/290=2.11 (3.24 dB)

3.5.1 Down Converters

It has been shown that the loss of a feeder severely affects the noise performance of a

system. As frequencies rise, so does the loss of waveguide feeder. At frequencies above

about 20 GHz a length of waveguide feeder longer than a few metres is not practical from

a loss viewpoint. To this end a mast head amplifier is used that not only provides low

noise gain but also converts the signal to a much lower frequency thus allowing it to be

fed to the receiver with a much lower loss using a much less expensive feeder.




P-TR-005-M101-ver4 50

Cascaded Systems - The Down-converter Cascaded Systems - The Down-converter

• Waveguide itself becomes very lossy (~1 dB/m) as

frequencies of 40 GHz are approached.• This would lead to very poor noise performance.

• This problem is overcome by a low noise amplifier that not

only amplifies with a low noise figure but also modulates

the incoming signal with a sub-carrier that reduces the

frequency to a lower value (~1 GHz).

• Low loss coaxial cable is then used to carry the signal to

the receiver.


3.6 Shannon and Nyquist

We have previously made an assumption that we needed a SNR of 14 dB in order to

achieve a satisfactory BER. SNR and bandwidth are linked to the maximum capacity by

Shannon’s Theorem which states that.

( )SNR 1logBandwidthCapacityMaximum 2 +×=

A further limitation is imposed by Nyquist’s Theorem that states that the maximum

symbol rate possible equals twice the bandwidth. As an example if a channel of

bandwidth 7 MHz is available with a SNR of 12 dB (a ratio of 15.8), the maximum

theoretical capacity is ( )8.151log107 26 +× = 28 Mbps. Nyquists theorem imposes a

symbol rate limit of 14 Megasymbols per second. It must be borne in mind, however,

that these figures are theoretical maxima. It is very rare to see 50% of these figures

achieved in practice.




P-TR-005-M101-ver4 51

SNR RequirementsSNR Requirements

• The required Signal to Noise ratio is chiefly influenced by the

modulation scheme and the maximum permitted error ratio.• Shannon’s and Nyquist’s Theorems provide fundamental

limits.


• Shannon’s Theorem States that:

Maximum Capacity = Bandwidth x log2(1 + SNR)

• Nyquist’s Theorem States that:

Maximum Symbol Rate = 2 x Bandwidth

SNR Requirements: ExampleSNR Requirements: Example

• Bandwidth 7 MHz, SNR 12 dB.• Maximum Capacity = 7x106 log2(1+15.8) = 28 Mbps

• Maximum Symbol Rate = 14 Megasymbols per second

• Remember: these are theoretical maxima. It is very rare to

exceed 50% of the calculated value in practice.





P-TR-005-M101-ver4 52

It should also be noticed that Nyquist’s Theorem imposes a limit on the symbol rate

rather than the bit rate that can be sent over a given bandwidth. The link between symbol

rate and bit rate depends on the modulation scheme which determines how many bits can

be sent in each symbol. If we have a binary modulation scheme such as BPSK or

GMSK, we can send only one bit per symbol (or modulation “state”). If we use QPSK,

the fact that there are four states means that each symbol can carry 2 bits of information.

Similarly, 8PSK will carry 3 bits per symbol and 16QAM will carry 4 bits per symbol.

There is a general trend that the higher the number of bits per symbol, the higher the

required SNR. On the other hand, the higher the number of bits per symbol, the narrower

the bandwidth required for a given capacity.

Symbol Rate and Bit RateSymbol Rate and Bit Rate

• Binary modulation systems such as BPSK and FSK send only

one bit per symbol.

• More sophisticated modulation schemes such as 8PSK and

16QAM have 3 and 4 bits per symbol respectively.


BPSK 8PSK 16QAM




P-TR-005-M101-ver4 53

The table below compares the different systems.

Modulation Scheme C/I for BER 1x10-6

Bandwidth for 8 Mbps

BPSK 10 dB 12 MHz

4PSK 14 dB 6 MHz

8PSK 19 dB 3 MHz

16PSK 24.5 dB 2 MHz

SNR requirements of different systemsSNR requirements of different systems


ModulationScheme

C/I for BER1 x 10-3

C/I for BER1 x 10-6

BPSK 7 dB 10 dB

4PSK 10 dB 14 dB

8PSK 15 dB 19 dB

16PSK 21.5 dB 24.5 dB

• The variety of C/I requirements for different modulation

schemes leads to the parameter “Energy per bit” (Eb) being

used as having global relevance.

It is clear that it is not possible to think of a single value for a “good” SNR; it depends onthe modulation scheme being used. As a result, the term “energy per bit”, Eb, is

commonly used as it has global relevance.

In the table above, it uses an 8 Mbit/s system as a reference. The basic building block of

microwave transmission systems in the “pleisiosynchronous digital hierarchy” (PDH) is




P-TR-005-M101-ver4 54

the 2 Mbit/s system that can carry thirty 64 kbit/s voice-equivalent channels. In link

specifications it is common to see terms such as “2 x 2” used meaning that the link is

carrying two, 2 Mbit/s systems. Further multiplexing yields 8 Mbit/s, 34 Mbit/s and 140

Mbit/s links. The synchronous digital hierarchy uses higher rates of 155 Mbit/s and also

622 Mbit/s.

Bandwidth requirements of differentBandwidth requirements of differentsystemssystems


ModulationScheme

Bandwidthrequirement for 8 Mbps system

BPSK 12 MHz

4PSK 6 MHz

8PSK 3 MHz

16PSK 2 MHz




P-TR-005-M101-ver4 55

Signal in Noise ExampleSignal in Noise Example

• An 8 Mbps QPSK receiver has a bandwidth of 6 MHz and

requires a SNR of requires a signal to noise ratio of 14 dB.Determine its threshold receive level if it has a noise factor

of 4 dB.


• Assuming input noise is at the level k (290) B, the effective

noise power at the input is (4 dB is a ratio of 2.5)

1.38x10-23 x 290 x 6 x 106 x 2.5 = 6 x 10-14 watts

= -102 dBm

• To deliver a signal to noise ratio of 14 dB we need a

minimum level (the “threshold”) of -88 dBm.

Data Rates CarriedData Rates Carried

• The basic “building block” of digital microwave systems is a

2 Mbit/s link that will carry, if required, 30 individual 64 kbit/s

channels. The 64 kbit/s channel is the traditional “digitisedspeech (PCM)” channel.

• Systems are often quoted as “2x2” (i.e. 4 Mbit/s) etc..

• Further multiplexing leads to the “Pleisiosynchronous Digital

Hierarchy” where four 2 Mbit/s link form an 8 Mbit/s link, four

8 Mbit/s form a 34 Mbit/s link and four 34 Mbit/s systems

form a 140 Mbit/s link.

• The synchronous digital hierarchy (SDH) specifies higher

rates of 155.52 Mbit/s and 622 Mbit/s.





P-TR-005-M101-ver4 56


Note Boltzmann’s constant k = 1.38 x 10-23 J/K

1. An antenna has a noise temperature of 280 kelvins. Determine the noise power gathered if the noise bandwidth of the receiver is 14 MHz.

2. An amplifier has a noise bandwidth of 2 MHz and a noise temperature of350 kelvins. If the noise power at the input equals k(480)B watts and thesignal power at the input is 0.172 picowatts, determine the signal to noise

ratio at the output of the amplifier.

3. A microwave system has a bandwidth of 4 MHz. The receiver noise

figure is 3 dB. Determine the noise temperature of the receiver and the

minimum required signal power in order to deliver a SNR of 13 dB. Stateany assumptions made.




P-TR-005-M101-ver4 57

4. A 10 GHz microwave link of length 30 km has 1.2 m diameter antennas.

The minimum required receive power has been determined to be –84dBm. Miscellaneous losses total 6 dB. Determine a suitable transmit

power.

5. A microwave link has a receiver with a noise bandwidth of 3 MHz. Thenoise temperature of the antenna is 290 kelvins. The receiver consists of

a mast head amplifier with a gain of 15 dB and a Noise Figure of 1.2 dB,

a feeder of 4.5 dB loss and a demodulator with a Noise Figure of 3.5 dB.

Determine the SNR with and without the mast head amplifier if the power gathered by the antenna is –97 dBm.

6. A Microwave link provides a 2 MHz channel with a SNR of 12 dB. Use

Shannon’s theorem to determine the maximum possible capacity of the

channel. Note:

( )

2log

loglog

1logBandwidthCapacity

10

102

2

x x

SNR

=

+×=






P-TR-005-M101-ver4 59

4 Fading

4.1 Introduction

Early experimenters in microwave radio were surprised to note that, even when it was

possible to view the far end of a link with binoculars, the signal received was by no

means constant. This fact launched vast and ongoing investigations into atmospheric

effects on the propagation of electromagnetic waves of the order of a few centimetres in

wavelength. In the “early days”, when frequencies used were generally less than 10

GHz, rain effects were minimal and the main area of concern was that of “multipath

fading”. The result of such fading is that it is not possible to guarantee the signal level in

the same way that it is on cable systems. A considerable margin must be built into the

designed receive power level in order to compensate for this fading.

FadingFading

• Unfortunately, the strength of the received signal

will vary with time, often quite dramatically.

• The two main contributors to “fading” are:

multipath propagation and;

hydrometeors (e.g. rain)

• It is important to be able to predict the likely

extent of fading and build in a “margin” to allow

for this in our link design.

Fading

4.2 Multipath Fading






P-TR-005-M101-ver4 61

4.2.1 Predicting the likelihood of a fade.

In order to compensate for multipath fading it is necessary to build in a margin so that,

even when the signal is fading, the signal level is sufficient to deliver an acceptable BER.It is therefore necessary to be able to estimate the likelihood of a fade of a particular

depth. The ITU publish recommended prediction methods, the relevant recommendation

being ITU-R P.530. These are updated every two years, the latest (530-10) having been

released in 2001. The complicated nature of the physical process involved in multipath

fading is reflected in the equations and formulas that are put forward. These are

generally based on experimental evidence gathered over many years on many different

links. The main formula governing multipath fading is that for predicting the percentage

of the time that a fade will exceed a depth of A dB. This is given as

( ) 10001.0033.02.10.3 101 Ah f

pw L Kd p −−−

×+= ε

In the above equation, K is known as the “radio climactic factor” and will be examined

further later. D is the path length in km, f is the frequency in GHz and ε p is the path

inclination in milliradians. h L is the altitude of the lower of the two antennas, above sea

level.

MultipathMultipath FadingFading

• Multipath fading exhibits Rayleigh characteristics.

• The deeper the fade the lower the probability

• Percentage time that a fade of depth A dB is exceeded is

proportional to 10-A/10.

• ITU-R report 530-9 gives the formula for percentage as

Fading

( ) 10001.0033.02.10.3 101 Ah f pw

L Kd p −−−

×+= ε

d is the path length in km

f is the frequency in GHz

K is the “radio climactic factor”

ε p is path inclination in milliradians




P-TR-005-M101-ver4 62


• Formulas come into the following categories:

Deterministic Heuristic

Empirical

• The multipath formula is empirical. It is based on

experimental evidence and the formula is created to fit the

results.

Fading

4.2.1.1 Radio Climatic Factor

ITU-R P.530-10 gives a formula for the climatic factor, K as shown below.

10029.02.410

dN K

−−= where dN 1 is the maximum gradient in refractive index that is

likely to be experienced. This depends on geographic location and typically varies

between –100 and –700. This leads to variations in K between about 1x10-4

and 7x10-4

.

In the United Kingdom, a value of –200 is appropriate leading to a value for K of

approximately 2.4x10-4

.

As an example, consider a 20 km path at a frequency of 7 GHz. The path is assumed to

be horizontal with antennas at an elevation of 100 metres above sea level.




P-TR-005-M101-ver4 63

( )

10

101.07033.00.34

10001.0033.02.10.3

106.2

1020104.2

101

A

A

Ah f pw

L Kd p

−

−−×−

−−−

×=

×××=

×+= ε

This equation allows us to draw up a table showing the likelihood of a fade of a particular

depth being exceeded.

Depth of Fade (dB) Percentage of time

exceeded

10 0.260

15 0.08220 0.026

25 0.008

Of particular interest is the fade depth that will be exceeded for 0.01% of the time. This

is found to be 24 dB.


• ITU-R P.530 gives a formula for K

• dN 1 can be found from ITU-R P.453-8

• values for dN 1 vary between -700 and -100.

• Values for K vary between about 1.23x10-4 and

6.76x10-3

Fading

10029.02.410

dN K

−−=




P-TR-005-M101-ver4 64


dN 1 ~ -200 in the United Kingdom.

Fading

( ) 42000029.02.4 104.210 −−−− ×== K


εp is the slope of the path in milliradians

d is in kilometres.

hr,e is the height of the two antennas (a.s.l.) in

metres.

For a flat path εp equals zero.

Fading

d hh er

p−=ε




P-TR-005-M101-ver4 65


For a flat path, with antennas at an elevation of 100

metres, 20 km in length with an operating

frequency of 7 GHz the probability formula

becomes:

Fading

( ) 10001.0033.02.10.3 101 Ah f

pw L Kd p

−−−×+= ε

10

101.07033.00.34

1060.2

1020104.2

A

Aw p

−

−−×−

×=

×××=


The formula can be used to produce a table of depth of fade

against the percentage that the fade is exceeded.

Fading

Depth of fade in dB Percentage of timeexceeded

10 0.260

15 0.082

20 0.0260

25 0.0082

It can be seen that, if 99.99% availability is required, a “fade

margin” of 24 dB would have to be designed in.

4.3 Rain Fading.






P-TR-005-M101-ver4 67

Rain FadingRain Fading

• Rain and other “hydrometeors” will absorb power from the

propagating electromagnetic wave and cause an additional,

variable, insertion loss. Again, a “margin” will have to be

designed in to ensure that the required availability is

maintained.

• Not surprisingly, this component is very climate dependent.

The “rainfall rate exceeded for 0.01% of the time”

(measured in mm/hr) is a key parameter. Such information

can be found in ITU-R P.837. The parameter is designated

R0.01.

Fading

4.4 Accommodating Rain and Multipath Fading

Using the ITU Recommendations, it is possible to establish a margin for multipath fading

and a margin for rain fading. The link planner should not simply add these two together

to give a total required margin as rain fading and multipath fading are highly likely to

occur simultaneously. Rather, the emphasis should be to use the margin available to

predict the “outage time” (as a percentage) for both rain and multipath and add these two

percentages together to give an estimate of the total outage. This outage prediction can

then be assessed as acceptable or not and adjustments made accordingly.




P-TR-005-M101-ver4 68


• R0.01 is approximately 25 mm/hr for the UK.

• Next, ITU-R P.838 must be used to convert this to aattenuation rate in dB/km, γR .

• Rain attenuation is polarisation and frequency dependent

• For a flat, vertically polarised path at 7 GHz, k=0.00265,

α=1.312. Hence γR = 0.18 dB/km.

Fading

α γ kR R =

Rain FadingRain Fading• The longer the path, and the higher the level of rainfall, the less likely

it is that it will be raining along the entire length of the path.

• This is accounted for by introducing a parameter known as the

“effective path length” that is equal to

• Thus a 20 km path would have an effective length, for rainfall

attenuation purposes of 10.9 km.

• 0.01% attenuation rate would be (0.18x10.9) = 2 dB.

• Insignificant compared with multipath margin (at these frequencies).

Fading

2435

1

01.0015.00

0

==

+

− Red

d d

d




P-TR-005-M101-ver4 69


• For different percentages, p, the value for 0.01% can be

modified according to the formula.

Fading

)log043.0546.0(

01.0

1012.0 p p

p A

A +−=

Accommodating both Rain andAccommodating both Rain andMultipathMultipath FadingFading

• Note that it would be regarded as highly unusual to add the

rain and multipath margins together.

• A more common approach would be to decide on the

maximum unavailability then build in the larger of the two

calculated margins.

• The “cause of outage” requiring the lower margin would

then increase the unavailability by a very small amount.

• Rain and multipath fading would not be expected to occur

simultaneously.

Fading




P-TR-005-M101-ver4 70

4.5 Selective Fading in Digital Systems.

Problems, in the form of high Bit Error Rates, can occur in digital systems even if the

wideband receive power is high. This is usually due to multipath propagation with delayslonger than just one or two nanoseconds resulting in the distortion of the signal. Such

multipath propagation may not induce a deep fade in the wideband power but, rather,

produce a notch in the received spectrum at a particular frequency. Such fading is

referred to as “selective fading” with the type of fading studied up to this point being

known as “non-selective”.

Selective Fading in Digital SystemsSelective Fading in Digital Systems

• The multipath fading that we have discussed so far caused

an outage by reducing the signal strength below the

threshold.

• High error rates (hence a further “outage”) can occur indigital systems with the signal distorted by multipath without

the wideband power necessarily reducing significantly.

• A method of predicting the unavailability due to this

phenomenon is required.

• Again ITU-R P.530 offers guidance.

Fading




P-TR-005-M101-ver4 71


• The resilience of receivers to such distortion is measured by

means of introducing a two-ray system whereby the delay

and relative strength of the second signal can be adjusted.

• Attenuation is adjusted for a number of values of τ so that

the pre-decided minimum value of BER is reached.

• The result is a set of “signature curves”.

Fading

Tx Rx

I

Note: relative amplitude

of the two paths is given

the parameter b.


• Measurements produce the above “signature curves”.

• For a fixed BER the relative strength of second path

depends on the delay and the notch position.

Fading

0.1

0.2

0.3

0.4 τ=32 ns

τ=16 ns

τ=8 ns

1-b

0.0 2 4-2-4 Notch offset (MHz)

Contours for BER of 10-6




P-TR-005-M101-ver4 72


• From a set of curves the parameters, signature width,

signature depth and reference delay can be obtained.

These can be quoted by the manufacturer.

Fading

0.1

0.2

0.3

0.4 τ=32 ns

τ=16 ns

τ=8 ns

1-b

0.0 2 4-2-4 Notch offset (MHz)

Contours for BER of 10-6

Minimum phase and Non-minimum phaseMinimum phase and Non-minimum phase

• A slightly different set of curves is produced if the stronger

signal is delayed. This is known as the “non-minimum

phase” configuration.

• The same parameters must be measured for the minimum

and non-minimum phase configurations.

Fading

Tx Rx

I

Note: relative amplitude

of the two paths is given

the parameter b.




P-TR-005-M101-ver4 73

Manufacturers publish data that makes it possible to predict the effect of selective fading

on a link. These take the form of parameters from “signature curves” that describe the

level of multipath required to induce a particular error rate for a given delay. Three vital

parameters are provided: the signature width, W in GHz (usually related to the system

bandwidth); the delay,τ r , required to cause the bit error rate when the average depth of

notch caused by the multipath was B dB. In fact, these parameters are established for two

slightly different configurations known as the “minimum phase” and “non-minimum

phase” configurations. This leads to the suffixes M and NM being introduced to

differentiate between the two configurations. Once these parameters are obtained, it is

necessary to relate them to the link in hand by means establishing relevant link

parameters. Two such parameters are identified: the mean time delay τ m and; the

“multipath activity factor” η . The mean delay is related to the path length by the

following equation (p.530-10)

ns 50

7.0

3.1

= d

mτ and the multipath activity factor is given by

( ) 75.002.0

1 P

e−−=η where

( )100

101 001.0033.02.10.3

0

Lh f p Kd

P

−−×+

= ε

Once the necessary parameters have been calculated, the outage probability P s, can be

determined from

×+×= −−

NM r

m B NM

M r

m B M s

NM M W W P .

220

.

220

101015.2τ

τ

τ

τ η




P-TR-005-M101-ver4 74

Considering a 20 km, 7 GHz link at 100 metres altitude as before (hence we can take K to

be equal to 2.4 x 10-4

) we find that

( )

( )013.01

026.0100

101

ns0.250

7.0

75.002.0

001.0033.02.10.3

0

3.1

=−=

=×+

=

= =

−

−−

P

h f p

m

e

Kd P

d

L

η

ε

τ

If examination of the manufacturer’s data reveals that

ns;4GHz;008.0dB;5 r ===== τ NM M NM M W W B B then the probability of

outage can be calculated to be

6

2205

2205

105.2

4

2.010008.0

4

2.010008.0013150.2

−

−−

×=

×+××= s P

Determining the outage probability due toDetermining the outage probability due toselective fadingselective fading

• Step 1: estimate the mean time delay on the path

Fading

• Step 2: estimate the “multipath activity factor”,η for the path.

ns50

7.03.1

=

d mτ

( )

( )100

101

1

001.0033.02.10.3

0

75.02.0 0

Lh f p

P

Kd P

e

−−

−

×+=

−=

ε

η




P-TR-005-M101-ver4 75

Determining the outage probability due toDetermining the outage probability due toselective fadingselective fading

• Step 3: Obtain values for signature width (W ), signature

depth ( B dB) and reference delay τ from the manufacturers

data.

Fading

• Step 4: Calculate the outage probability P s.

×+×= −−

NM r

m B NM

M r

m B M s

NM M W W P ,

220

,

220 101015.2

τ

τ

τ

τ η

Determining the outage probability due toDetermining the outage probability due toselective fading - exampleselective fading - example

• Considering a 20 km, 7 GHz link at 100 m altitude as before.(Hence we can take K to be 2.4 x 10-4) Steps 1 and 2:

Fading

• Step 3: From manufacturers details W M =W NM =0.008 GHz (it

seems we have an 8 MHz system here); B M = B NM =5 dB; τ r =4 ns.

( ) 013.01

0260.01001020104.2

75.0026.02.0

100001.07033.00.340

=−=

=÷×××=

−

×−×−

e

P

η




P-TR-005-M101-ver4 76

Determining the outage probability due toDetermining the outage probability due toselective fading - exampleselective fading - example

Fading

• Step 4: Calculate the probability of outage.

6

2205

2205

105.2

4

2.010008.0

4

2.010008.0013.015.2

−

−−

×=

××+×××=S P

• Note that this probability is dependent on path length,

frequency and bandwidth, but NOT on received signal level.

4.6 Atmospheric Absorption

Resonances with oxygen and water molecules present in the atmosphere lead to energy

being absorbed by the atmosphere in a frequency-dependent way. This absorption loss

adds to the free space loss and, as a result, is not a “fade” as the loss is constant.

However, it does require a margin to be built into the link design. Generally speaking,

atmospheric absorption is negligible below 10 GHz, rising to approximately 0.1 dB/km at

20 GHz. There is a resonant peak of about 0.2 dB/km at about 24 GHz apart from which

the level is approximately 0.1 dB/km up to 40 GHz.




P-TR-005-M101-ver4 77

Atmospheric AbsorptionAtmospheric Absorption

Fading

• Resonances with oxygen and water molecules lead to

energy being absorbed in a frequency dependent way by theatmosphere. This adds to the path loss.

• Atmospheric absorption is not, strictly speaking, an example

of fading as it is a constant loss. Nevertheless it is

necessary to design a margin into the link in order to

compensate for such absorption.

• Atmospheric absorption is negligible below 10 GHz, rising to

approximately 0.1 dB/km at 20 GHz. It is approximately 0.1

dB/km between 20 GHz and 40 GHz apart from a resonant

peak of 0.2 dB/km at approximately 24 GHz.

Atmospheric AbsorptionAtmospheric Absorption

Fading

• Graph showing losses due to

water vapour and oxygen

absorption. Total atmospheric

absorption is obtained by

summing the two losses. L O S S

d B / K m

20

10

1

0.1

0.01

FREQUENCY GHz.

1 10 100

Additional Loss Due To Atmospheric Content.

W a t e

r v a p

o u r

O x y g e n




P-TR-005-M101-ver4 78

Estimating Link PerformanceEstimating Link Performance

Fading

• Now we appreciate the fading mechanisms and their effects,

we can look again at our 7 GHz, 20 km system. Suppose weuse a 100 milliwatt (20 dBm) transmitter. The threshold is

assumed to be -87 dBm with miscellaneous losses amounting

to 5 dB. 60 cm antennas are used.

• Step 1: Estimate antenna gains to be

17.5+20log(0.6)+20log(7) = 30 dBi

• Step 2: Free space loss = 92.4+20 log(20)+20 log(7)=135 dB

• Step 3: calculate unfaded receive level to be

20-5-135+30+30=-60 dBm

4.7 Estimating Link Performance

Now that we have established methods of predicting the outages from multipath fading,

rain fading and selective fading, it is possible to estimate the link performance

considering all these factors. As an example, we shall consider our 20 km, 7 GHz

system. We shall assume that the transmit power is 20 dBm and the threshold level is –

87 dBm. Further, we shall assume that 60 cm antennas are used.

Firstly, we need to predict the unfaded signal level. The antenna gains can be estimated

by using the formula dBi30)log(20)log(205.17 =++= f DG the free space loss is

dB.135)7log(20)20log(204.92 =++ If miscellaneous losses amount to 5 dB, the total

loss is 80 dB, resulting in an unfaded receive level of –60 dBm, thus providing a fade

margin of 27 dB. We have previously derived a formula for multipath fading for a 7

GHz, 20 km link that links fade depth with a percentage of time.




P-TR-005-M101-ver4 79

101060.2 AW p −×=

For a fade of 27 dB, the percentage is found to be 5.19x10-3

%. The rain fadingcalculations are such that the percentage fading will be less that 0.001%, the smallest

figure for which the figures in the recommendation are valid. The probability of an

outage occurring due to selective fading has been calculated to be 2.5x10-6 or 0.00025%.

Adding the two figures together suggests a total outage of 0.0052 + 0.00025 = 0.0055%.

Estimating Link Performance -Estimating Link Performance - multipathmultipath

fadingfading

Fading

• As the frequency is below 10 GHz, atmospheric absorption

can be ignored.

• The unfaded receive level can be seen to be 27 dB above the

threshold. This gives us a “fade margin” of 27 dB.

• We have previously derived a formula

for a link of this length and frequency

• For A = 27 dB, pW is found to be 5.19x10-3%

101060.2 A

W p −

×=




P-TR-005-M101-ver4 80

Estimating Link Performance - rain fadingEstimating Link Performance - rain fading

Fading

• We have previously shown that the rain fading margin for a 20

km, 7 GHz path for 0.01% of the time is 2 dB.• Although the likelihood is that rain fading can be ignored, we

can determine the percentage outage given a fade of 27 dB

from the formula

• For a value of Ap of 27 dB,

• Examining this equation it is found that the outage will be far

less than 0.001%, which is the valid range of the equation. We

can therefore ignore outages due to rain fading.

( ) p p p

A

A10log043.0546.0

01.0

12.0 +−=

( )5.11210log043.0546.0 =+− p

p

Estimating Link Performance - selectiveEstimating Link Performance - selectivefadingfading

Fading

• We have previously shown that the selective fading outage

probability for a 20 km, 7 GHz, 8 MHz bandwidth path is

0.0025%. This is not affected by the received power level.

• Summing the outages, we would predict a total outage of

0.0052 +0.00025 = 0.0055%.




P-TR-005-M101-ver4 81

4.8 Conclusion.

Methods of predicting the percentage outage due to multipath fading (both non-selectiveand selective) and rain fading have been examined and examples of implementation

shown. The link examined has produced estimates of outage that would, in practice, be

satisfactory. It is however, easy to visualise a link (longer in length, higher in frequency)

for which the initial prediction would suggest an unsatisfactory performance. In such

cases, diversity techniques can be used to improve the performance. The next section

introduces such methods and methods for predicting their effectiveness.

What’s next?What’s next?

Fading

• We have obtained encouraging estimates of outage. The link,

if implemented, would provide a high quality service.

• However, we must be able to accommodate situations where

the initial prediction is for an unsatisfactory performance.

• Diversity techniques can be used to improve the performance.

• The next session reviews and analyses diversity improvement

methods.




P-TR-005-M101-ver4 82

4.9 Module 4: Self-Assessment Exercises.

1. A 24 km microwave link is located in Sweden and operates at a

frequency of 18 GHz. One antenna is 1100 m above sea level and the

other is at 800 m above sea level. Estimate the percentage time for whicha fade exceeding 25 dB would occur.

2. A 21 km microwave link, located in Italy, operates at a frequency of 28GHz. Horizontal polarisation is used. Determine the rain-produced

attentuation for a 0.01% time period.




P-TR-005-M101-ver4 83

3. A horizontal 18 km, 14 GHz link, at an elevation of 200 m above sea

level, operating in the United Kingdom uses equipment for which the

relevant details are:

ns32

dB10

MHz34

r =

==

=

τ

NM M B B

W

for a BER of 10-6. Determine the probability of the BER exceeding this

value.

4. Estimate the atmospheric absorption on an 11.5 GHz link of path length

20 km.






P-TR-005-M101-ver4 85

5 Diversity Techniques

5.1 Introduction

The 20 kilometre, 7 GHz path that we have examined in the previous section would have

an outage prediction of less than 0.01% and would generally be regarded as giving very

good performance. If, however, the path length, bandwidth or operating frequency was

increased, the outage prediction may well exceed 0.01% and remedial action must be

taken. Sometimes it is possible to remedy such a situation by increasing the transmit

power or by adopting larger antennas. However, it must be borne in mind that selective

fading is not affected by signal strength and that sometimes the options described will not

be economic. In such circumstances, the adoption of diversity techniques would form the

best solution.

Diversity TechniquesDiversity Techniques

• Our 20 km, 7 GHz, 8 MHz bandwidth link just

meets the 0.01% unavailability requirement.

• It is sensible to assume that, if we made the path

longer, or increased the bandwidth, or increased

the operating frequency, we would struggle to

meet the requirements.

• Sometimes it is possible to improve the situation

by increasing the transmit power, or antenna size.

• Occasionally, these steps alone are not sufficient.

Diversity Techniques




P-TR-005-M101-ver4 86

5.2 The Theory Behind Diversity Systems

When a diversity protection system is implemented, effectively a separate link is

established to carry the same traffic. The output is either the better of the two signals or,

ideally, a combination of the two signals to provide an optimum output. Fading is a rare

event. In a diversity system, a simultaneous fade on both links should be even rarer. If

for example, each link had a fading probability of 1% (or 0.01), the probability of a

simultaneous fade on the two links would be 0.01 x 0.01 = 0.0001 or 0.01%.

This calculation would be correct if the two links were not correlated. However, because

the two links operate over the same route, the fact that there is a fade on one link means

that the probability of a fade occurring on the second link at that time would be greaterthan normal as it has been established that the conditions required for fading on the link

do exist. Thus the “diversity improvement” as it is known would not be as great as

indicated in the calculation given above. Nevertheless, a diversity improvement is

achievable. The ITU-R recommendation P.530-10 gives guidance on how this

improvement may be calculated.


• Diversity basically relies on establishing more than

one link and selecting the best performing link at any

one time or, ideally, combining the outputs from the

two links to provide the optimum output.

• Suppose we had estimated the unavailability to be 1%

on a particular link.

• If we established a separate, but virtually identical, linkthat would also have a 1% unavailability.

• The probability of both links being simultaneously

unavailable could be calculated to be 1%x1%=0.01%.





P-TR-005-M101-ver4 87


• Performing the calculation described would be valid

only if the two links established were independent of

each other (zero correlation between fading

characteristics).

• However, as they are very similar links between the

same two points, one would intuitively expect there to

be correlation between the two links.


5.3 Types of Diversity

All diversity systems involve establishing an alternative route for the traffic to take. The

major forms of diversity system are:

• Space Diversity: this consists of two antennas receiving at each end. These antennas

are usually positioned one above the other for maximum improvement. In this way

the signal can pass from one end to the other via either of the receiving antenna.

• Frequency Diversity: two transceivers operating at different frequencies carry the

same information over the same antenna. The improvement is afforded by the fact

that fading is a frequency-dependent phenomenon and will not occur with great

severity on two frequencies at the same time.




P-TR-005-M101-ver4 88

• Polarisation Diversity: the same information is transmitted at both horizontal and

vertical polarisations simultaneously with the hope being that a particular fade is not

as severe on both polarisations.

• Angle Diversity: by placing two feedhorns near the focus of the antenna it is possible

to have the energy split between two slightly different radiation patters, one with its

principal direction slightly offset from the other. It is hoped that, if the signal to one

feedhorn suffers a severe fade, then the signal to the second feedhorn will not be as

deep.

Diversity Techniques - most commonDiversity Techniques - most commontypes of diversity systemstypes of diversity systems

• Space diversity:- two receive antennas (usually one

above the other) at each end.

• Frequency diversity:- effectively two transceivers at

separate frequencies passing the same information

over the same antenna.• Polarisation diversity:- transmitting the same

information via two orthogonal feeders.

• Angle diversity:- usually achieved by having two

separate feedhorns near the focus of the antenna,

each providing a different radiation pattern.





P-TR-005-M101-ver4 89

5.4 Improvement Factor

The improvement factor I is the radio of the probability of a fade without diversity

protection to that probability with protection. For Space Diversity systems, ITU-RP.530-10 provides the equation

( ( ) 1004.10

48.012.087.0 1004.0exp1 V A pd f S I −−−×−−=

where S is the separation in metres, V is the difference in gain between the transmitting

and receiving antennas (usually zero) and p0 is the multipath occurrence factor, expressed

as a percentage. As an example, consider a 20 km, 7 GHz link for which the power

transmitted has been reduced to give a fade margin of 15 dB. It has previously been

calculated that for a fade of 25 dB, the probability was 0.008%. Thus for a fade of 15 dB,

the probability would be 0.08%. This makes p0 equal to 0.08 ( )

53.210 5.1 =× . If space

diversity is used with a separation of 5 metres then the improvement factor is given by

95.5

105.2207504.0exp1 5.104.148.012.087.0

=

××××−−= −− I

Thus, the outage probability would be reduced by a factor of 5.95 from 0.08% to 0.013%.

The ITU recommendation also gives details of the range of validity of the equations

published. In this case, the frequency range is 2 – 11 GHz, path lengths of 43 – 240

kilometres and antenna separations of 3 to 23 metres. Caution should be exercised when

using the equations outside this range. However, the following general rules should

remain true:

• The bigger the separation, the bigger the improvement.

• The longer the path length the bigger the improvement




P-TR-005-M101-ver4 90

• The improvement factor is not very frequency-dependent, exhibiting a slight decrease

with increasing frequency.

Diversity Techniques - estimatingDiversity Techniques - estimatingimprovementimprovement

• The ITU provide a recommended method of estimating the

improvement provided by a diversity technique.

• Essentially, this involves estimating the degree of correlation

between the fading of the two links.

• The term “Improvement Factor” ( I ) is used where

• p(A) is the probability of a fade without diversity; pd (A) is the

probability with diversity.


)()(

A p A p

I d

=




P-TR-005-M101-ver4 91

Diversity Techniques - space diversityDiversity Techniques - space diversity

• ITU-R P.530 gives the following equation for the improvement

factor.

• S is the vertical separation in metres. V is the difference in gain

between the Tx and Rx antennas (usually zero).


( )[ ] ( )( )

(%)factoroccurencemultipath

1004.0exp1

0

1004.10

48.012.087.0

=

×−−= −−−

p

where

pd f S I V A

Diversity Techniques - space diversityDiversity Techniques - space diversity• In our original link, we predicted a multipath (non-selective)

outage of 0.008% for a margin of 25 dB. To make the situation

more realistic for diversity purposes, let’s assume that the

transmit power was reduced so as to make the fade margin 15

dB. That would give an outage probability of 0.08%.

• Therefore the relevant parameters are: A=15; f =7; d =20; p0=2.5.

If the antennas are separated by 5 metres the improvement factor

is

• Thus the outage probability with diversity would be expected to

be 0.013%.


( )[ ] ( )( )

95.5

105.2207504.0exp1 101504.148.012.087.0

=

×−−= −− I




P-TR-005-M101-ver4 92


• The equation was produced by examining data produced on links

covering the frequency range 2 - 11 GHz; path lengths 43 - 240

km and antenna separations of 3 to 23 metres.

• Care must be taken when operating outside these parameters.

However, the equation doesn’t immediately “collapse” and the

general rules hold:

The bigger the separation the bigger the improvement

The longer the path length the bigger the improvement

Improvement factor is not very frequency-dependent exhibiting a

slight decrease with increasing frequency.



• The equation was produced by examining data produced on links

covering the frequency range 2 - 11 GHz; path lengths 43 - 240

km and antenna separations of 3 to 23 metres.

• Care must be taken when operating outside these parameters.

However, the equation doesn’t immediately “collapse” and the

general rules hold:

The bigger the separation the bigger the improvement

The longer the path length the bigger the improvement

Improvement factor is not very frequency-dependent exhibiting a

slight decrease with increasing frequency.





P-TR-005-M101-ver4 93

5.5 Improvement for Other Types of Fading

The improvement factor described above is the improvement factor for non-selective

multipath fading. ITU-R P.530-10 describes equivalent procedures for estimating the

improvement factor for selective fading and also for estimating the improvement factor

when other diversity techniques are involved. It should be noted that diversity techniques

do not provide a significant improvement in rain fading performance as the rain fade will

affect all elements simultaneously. For this reason rain fading forms the final limitation

on path length at the higher microwave frequencies.


• The equations considered so far have dealt with the

“non-selective” fading aspects of the unprotected

system.

• A separate procedure must be followed to determine

the new outage probability for the selective fading.

• These two must then be summed in order to obtain

the new outage estimate.





P-TR-005-M101-ver4 94

Diversity Techniques - other methodsDiversity Techniques - other methods

• ITU-R P.530 describes equivalent procedures for

estimating the improvement factor for Frequency,

Angle and Polarisation diversity techniques.


5.6 Combining Diversity Techniques

The link planner is not restricted to choosing just one diversity method. It is perfectly

legitimate to use two or more methods to increase the improvement afforded. It is

common to combine, for example, frequency and space diversity. Again, ITU-R P.530-

10 gives details regarding the improvement that is likely to be obtained.




P-TR-005-M101-ver4 95

Diversity Techniques - combiningDiversity Techniques - combiningmethodsmethods

• Greater improvement can be obtained by

implementing more than one technique; e.g. frequency

and space diversity.


TxRx f 1

f 2

f 1




P-TR-005-M101-ver4 96


1. A 28 km microwave link, located in Sweden operates at a frequency of

21 GHz. Identical antennas are used at both ends and the path is horizontal

with antennas at an elevation of 1200 metres above sea level. Without anydiversity it is found to have a fade margin of 16 dB. Determine the

probability of outage. Estimate the improvement factor provided if space

diversity is employed with an antenna separation of 4 metres.

2. Using the procedures of ITU-R P.530-10, page 28, evaluate the improvement possible

by utilising frequency diversity with a separation of 200 MHz instead of space diversity.




P-TR-005-M101-ver4 97

6 Interference Issues

6.1 Introduction

In carrying out link budgets we have always had to be aware that there is a minimum

level that the received signal must not drop below. The ultimate limiting factor is thermal

noise. However, in practice, interference will add to the effect of thermal noise raising

the minimum required receive signal power and thus rendering the receiver less sensitive.

Interference IssuesInterference Issues

• Interference is a problem because it “de-sensitises”

the receiver.

• It does this by effectively raising the noise floor.

• Remembering our 8 MHz bandwidth system, we

calculated a threshold of -87 dBm by deducing that

the noise floor was -101 dBm and that the SNR

requirement was 14 dB.

• If interference adds to this noise floor, then thethreshold will be raised and fade margins reduced.

Interference Issues

6.2 Quantifying the effect of interference.

The effect of interference can be calculated by adding the interfering power to the noise

floor to arrive at the increased “effective noise floor”. The threshold of the receiver

increased by the same amount that the noise floor is raised by. Unfortunately, quoting

power levels in dBm does not make it easy to add such powers together. It is necessary




P-TR-005-M101-ver4 98

to convert from dBm to milliwatts before adding. An example of this process is

illustrated below.

Interference Issues - adding powersInterference Issues - adding powers

• In order to add powers it is necessary to convert

from dBm to milliwatts.

• X dBm = 10X/10 milliwatts

• X dBm + Y dBm = 10log10(10X/10 + 10Y/10) dBm

• E.g. if an interfering signal of -98 dBm is added to

the noise floor of -101 dBm, the resultant power

level is 10log10(10-9.8 + 10-10.1) = -96.2 dBm

• The noise floor has effectively increased by 4.8 dB,

making the new threshold -82.2 dBm.

Interference Issues

6.3 The Theory Behind Diversity Systems

When a diversity protection system is implemented, effectively a separate link is

established to carry the same traffic. The output is either the better of the two signals or,

ideally, a combination of the two signals to provide an optimum output. Fading is a rare

event. In a diversity system, a simultaneous fade on both links should be even rarer. If

for example, each link had a fading probability of 1% (or 0.01), the probability of a

simultaneous fade on the two links would be 0.01 x 0.01 = 0.0001 or 0.01%.

6.4 Co-channel and Adjacent Channel Interference

It is obvious that a nearby transmitter operating at the same frequency will pose an

interference threat. However, this is not the only circumstance in which interference will




P-TR-005-M101-ver4 99

occur. No filter is perfect and power from other frequencies can cause problems.

Interference from one system to another at the same frequency is known as “co-channel”

interference. Interference at other frequencies is known as “adjacent channel”

interference. Just how “adjacent” a channel can be and cause interference depends on the

quality of the filters used in both transmitter and receiver. Basically, the greater the

separation between frequencies used by the interferer and the “victim”, the better.

Interference Issues : co-channel andInterference Issues : co-channel andadjacent channel interference.adjacent channel interference.

• The spectrum is divided into “slots” often referred to as

“channels”. The width of each slot determines the bandwidthof the system.

Interference Issues

7.000 7.008 7.016 7.024 7.0326.992 7.0487.040

Used ChannelAdjacent Channels

• Interference within the bandwidth of the channel being used

is known as “co-channel”. The slots either side are known as

“adjacent channels.

• Possible channel allocations

for a 7GHz system.

MHz




P-TR-005-M101-ver4 100

Interference Issues : co-channel andInterference Issues : co-channel andadjacent channel interference.adjacent channel interference.

• Co-channel interference is the most serious.

• Adjacent channel interference is reduced by the

selectivity of the filter at the receiver. Typically, it

will be attenuated by 20 dB.

• Interference at frequencies outside this region will

be attenuated further and is less likely to pose a

threat to the system.

Interference Issues

6.5 Interference Scenarios

The best defence that microwave systems have against interference is the fact that the

antennas only “look” at a very narrow beam. Any interference entering from outside this

beam will be severely attenuated. Indeed, the fact that, when transmitting, the narrow

beam results in very high power densities means that microwave systems are more likely

to cause interference than be victims. Some frequency bands are shared between satellite

and terrestrial systems. Satellite receivers operate on very low receive power levels. As

such they are very susceptible to interference. Care must be taken to ensure that any

terrestrial systems that are less than a specified distance away from a satellite earth

station are coordinated in such a way so as to avoid interference. This will include

ensuring that the main beam of the terrestrial system is not directed at the earth station.

Satellite system operators are generally given the opportunity to object to proposals to

implement terrestrial systems.




P-TR-005-M101-ver4 101

Interference Issues : possible scenariosInterference Issues : possible scenarios

• Off beam gain of a parabolic antenna is typically 45

dB down on main beam (Effective gain of -10 dBi).

This makes high interference levels unlikely.

• Terrestrial microwave links are more likely to cause

interference to satellite systems than be victims

themselves. This has licencing implications.

Interference Issues

Interference on terrestrial systems is most likely to occur when repeaters are used on a

long distance system. If the same frequency is used for two consecutive hops then the

final receiver can receive signals from both transmitting antennas. There will be a

somewhat unpredictable time delay between the two signals arriving, leading to

demodulation problems. This problem can be alleviated by either:

• Using different frequencies on consecutive hops

• Using orthogonal polarisations on consecutive hops

• Ensuring that consecutive hops are not co-linear but, rather, follow a “zigzag” pattern.




P-TR-005-M101-ver4 102


• Multi-hop paths present a possible interference

problem because of “overshoot”.

Interference Issues

• The effect can be reduced by using orthogonal

polarisations on consecutive hops and/or by

changing the direction between consecutive hops

by more than the antenna beamwidth.

Interference Issues : reductionInterference Issues : reductiontechniquetechnique

• Offsetting the direction of the hops.

Interference Issues

Interfering antennas no

longer “look at” each other.




P-TR-005-M101-ver4 103

One situation in which an interference assessment should be made occurs when one site

acts as a “hub” for a number of links. The assessment should allow the likely degree of

receiver de-sensitisation to be quantified. This would include consideration of the power

being received on each of the links together with allowances for frequency differences,

polarisation differences and the radiation patterns of the antennas involved.


• Microwave transmission systems often have a

“hub”.

• This hub receives signals from many different links.

Interference Issues

6.6 Reduction Techniques

It is possible to purchase microwave antennas that are referred to as “high performance”.

These will have the same gain in the principal direction as a standard antenna of the same

diameter. However, they will have a superior “off beam” performance and, in particular,

will have a higher “front to back” ratio. This is achieved by ensuring that “spill over”

from the feeder is reduced and, also, by using thicker metal for the reflector.

Additionally, it is important to appreciate that a frequency band for point to point

microwave communication will, in reality, consist of two bands. This is so that duplex




P-TR-005-M101-ver4 104

operation can be implemented. The two bands may be referred to as “go” and “return” or

simply as the “higher” and “lower” frequencies. When you have a site at the hub of a

network receiving signals from many links. The interference situation can be alleviated

by ensuring that the hub does not receive on the same frequency on all of the links. This

is achieved by judicious allocation of the higher and lower duplex frequencies on the

different links.

Interference Issues : reductionInterference Issues : reductiontechniquestechniques

• High performance antennas can be purchased.

These are less susceptible to “off-beam”interference.

• Frequency planning of the duplex links can also

help alleviate problems

Interference Issues

6.7 Anomalous Propagation

Under “normal” conditions the curvature of the earth will provide a shield from

interference when there is no line of sight from the interferer to the victim. However,

from time to time, atmospheric conditions will exist such that the electromagnetic wave

becomes trapped within a layer a few hundred metres in height. In this way, the level

received on a transhorizon path can very nearly equal that expected if there was clear line

of sight. The phenomenon by which energy is trapped within an atmospheric layer is

referred to as “ducting”.




P-TR-005-M101-ver4 105

Interference Issues : anomalousInterference Issues : anomalouspropagationpropagation

• Terrestrial microwave systems are very much “line

of sight” systems. The signal tends not to

propagate over the horizon.

• However, on rare occasions, interference occurs

from distant systems under conditions known as

“ducting”.

• Ducting falls into a category of propagation

conditions referred to as “anomalous” (“highlyunusual”; “noticeably different”).

Interference Issues

Interference Issues : ductingInterference Issues : ducting

Interference Issues

Normal conditions: no interference threat

Anomalous conditions: interference threat






P-TR-005-M101-ver4 107

Interference Issues :Interference Issues : IntermodulationIntermodulationproductsproducts

Interference Issues

• No amplifier is perfectly linear. For an input vi, the

output is generally:

• The “even numbered” terms are out of band

(harmonics), the “odd numbered” terms are “in band”

and therefore more serious.

.....432

0 ++++= iiii dvcvbvavv


Interference Issues

• If a number of signals at different frequency are

combined within an amplifier, the third, fifth andseventh order terms produce an interesting effect.

Original Signals

Intermodulation Products




P-TR-005-M101-ver4 108


Interference Issues

• If a broadband receiver is receiving multiple carriers, twodominant signals can severely interfere with a third carrier.

• If two signals at the input to an amplifier are at f 1 and f 2, the

most damaging intermodulation products will be at 2 f 2 - f 1 and

2 f 1 - f 2.

• A weak signal at these frequencies will be interfered with.

• Lesser effects occur at 3 f 2 -2 f 1 and 3 f 1 -2 f 2.

IntermodulationIntermodulation products (example)products (example)

Interference Issues

• A broadband receiver receives two signals. One at 10.02 GHz

and another at 10.035 GHz. Determine the frequencies of thefour most dominant intermodulation products.

• 2 f 2 - f 1 = 10.050 GHz

• 2 f 1 - f 2 = 10.005 GHz

• 3 f 2 - 2 f 1 = 10.065 GHz

• 3 f 1 - 2 f 2 = 9.990 GHz




P-TR-005-M101-ver4 109


1. A particular receiver has a receive power threshold of –87 dBm in orderto deliver a SNR of 14 dB. The antenna also receives two interference

signals: one at a level of –98 dBm and another at a level of –104 dBm.

Determine the degraded threshold of the receiver in the presence of theseinterfering signals




P-TR-005-M101-ver4 110

2. A mast is at the “hub” of a 12 GHz microwave network. Three links

converge on this hub. One is of 2 km length, one 5 km length and one 16

km in length. All the antennas are 1.2 metres in diameter and the transmitfrom all transmitters is 500 mW. The 16 km link is susceptible to

interference from the two shorter links. In the direction of the 2 km link, the

gain of the receiving antenna is 0 dBi, and in the direction of the 5 km linkthe gain is –5 dBi. Estimate the interference power gathered by the antenna

and compare it with that received from its wanted signal.




P-TR-005-M101-ver4 111

3. A receiver at the hub of a microwave network operates on a frequency of

12.260 GHz. Interfering signals are received at frequencies of 12.060 GHz,

12.200 GHz, 12.400 GHz and 12.540 GHz. Which of these interferingsignals requires most attention from the viewpoint of intermodulation

products causing interferenc






P-TR-005-M101-ver4 113

7 Repeatered Systems

7.1 Introduction

Examining the effects of fading, in particular rain fading at higher microwave

frequencies, it is clear that; the longer the link, the harder it will be to meet a performance

objective. Additionally, as links become longer, it will become necessary to build higher

and higher masts in order to maintain visibility. For the above reasons, it is often

necessary to design your link so that it has more than one “hop”. It is necessary to place

a repeater at the point where each hop is terminated. Shorter links will also require

repeaters to be used if the path from one end to the other is obstructed.

RepeateredRepeatered SystemsSystems

• Severe difficulties occur attempting to establish

single hops greater than about 50 km due to both

fading and visibility problems.

• Longer paths require repeaters.

• Shorter paths with visibility problems will also

require repeaters.

Repeatered Systems




P-TR-005-M101-ver4 114

RepeateredRepeatered SystemsSystems

• Longer paths require repeaters.

• Shorter paths with visibility problems will also

require repeaters.

Repeatered Systems

7.2 Active and Passive Repeaters

Repeaters are said to be either “active” or “passive”. An active repeater has an amplifier

that attempts to restore the signal to its original quality before re-transmitting. On digital

systems, a far better performance is achieved by demodulating the signal at each repeater

station and re-transmitting a restored baseband signal. The advantage that digital systems

have on repeatered systems comes from the ability to reproduce a noise-free signal at

each repeater. In that way the total error rate is approximately equal to the sum of the

error rates on the individual hops. It is impossible to recreate a noise free signal on an

analogue system. All the calculations regarding the link between signal to noise ratio and

error rate have assumed that the original signal is noise free. In an analogue system, the

noise will accumulate as the signal progresses from hop to hop. As a consequence of

this, analogue radio systems are vastly inferior to digital systems when multi-hop links

are used.




P-TR-005-M101-ver4 115

ActiveActive RepeateredRepeatered SystemsSystems

• Active repeaters have a transceiver at each

repeater station, demodulating and re-transmitting

the message.

• On digital systems the BER on the entire system is

approximately the sum of the individual BER’s.

• On analogue systems, the noise will accumulate,

causing serious problems.

Repeatered Systems

ActiveActive RepeateredRepeatered Systems (Analogue)Systems (Analogue)

• The Signal to Noise ratio on a point to point link is

calculated assuming that the signal is “clean” when

it leaves the transmitter.

• On the second hop the signal will be noisy as it

leaves the transmitter. Noise accumulates from

hop to hop.

• Analogue systems are vastly inferior to digital

systems when multi-hops are considered.

Repeatered Systems

Clean Signal Noisy Signal Noisier Signal




P-TR-005-M101-ver4 116

ActiveActive RepeateredRepeatered Systems (Analysis)Systems (Analysis)

• Repeatered digital microwave systems can be

analysed by regarding each hop as an individual

single hop system.

• The total unavailability can be approximated to be

the sum of the individual unavailabilites (provided

that the individual unavailabilities are fractions of a

percent).

Repeatered Systems

Clean Signal Noisy Signal Noisier Signal

When analysing a repeatered system in which all the repeaters are active, each hop can be

regarded as if it was a separate link. Sometimes, there may be enough margin in the link

budget to allow the use of a passive repeater. A passive repeater has no power supply but

merely re-directs the signal towards the receiver. There are two alternative methods of

implementing a passive repeater: back to back antennas and; billboard reflectors.

7.2.1 Back-to-back antennas

The construction of a passive back-to-back antenna repeater involves placing the two

antennas at the appropriate height on the mast as would be the case if two individual links

were being implemented. However, there would not be an feeders running up the tower

to the antennas, nor would there be a cabinet containing transceivers and associated

equipement. Rather, the one antenna would be connected directly to the other via a short

section of waveguide.




P-TR-005-M101-ver4 117

PassivePassive RepeateredRepeatered SystemsSystems

• On short, obstructed links, it is possible to avoid the

expense of a passive repeater and, instead use a

passive repeater.

• The diagram shows a back to back antenna

configuration of a passive repeater.

• Total path loss is the sum of the individual hops.

Repeatered Systems

PassivePassive RepeateredRepeatered Systems (Example)Systems (Example)

• Example: A 14 GHz microwave system is carried

over a 6 km path. The path is obstructed at its mid

point and a passive repeater is installed. The

antennas used have a 1.2 m diameter. Estimate

the path loss and compare with that of a single hop

of the same length.

Repeatered Systems

3 km 3 km f = 14 GHz




P-TR-005-M101-ver4 118

PassivePassive RepeateredRepeatered Systems (Solution)Systems (Solution)

• Antenna Gain ~ 17.5 +20log(1.2)+20log(14)= 42 dBi

• FSL (3 km) = 92.4+20log(3)+20log(14)=124.9 dB

• Loss per hop = 124.9 - 84 = 40.9 dB

• Total loss = 81.8 dB

• For a single (6 km) hop, FSL = 130.9 dB. Path loss = 130.9 - 84 = 46.9

dB.

• Passive repeaters increase the path loss substantially.

Repeatered Systems

3 km 3 km f = 14 GHz

7.2.2 Reflector repeaters

As illustrated by the slides shown below, reflectors form an attractive alternative to back

to back antennas when the repeater is placed well to the side of the line joining the two

ends of the link. One advantage of reflectors is that it is usually easier to construct a

large reflector than it is to construct large antennas. Additionally mounting reflectors on

the side of existing buildings is more likely to be a possibility. The analyses given show

that you get less path loss if the reflector is placed near one end of the link. When

conducting the calculations, a check should always be made to ensure that the predicted

path loss is greater than the loss for an unrepeatered link. If that is not the case the loss

for an unrepeatered link should be used. Such reflectors are not suitable for placing close

to the line joining the two ends of the link. In such situation, a double reflector may be

used. The design considerations for such systems are outlined below.




P-TR-005-M101-ver4 119

PassivePassive RepeateredRepeatered Systems (Reflectors)Systems (Reflectors)

• As an alternative to back-to-back antenna systems, “billboard

reflectors” can be used as passive repeaters. These simply reflect

the signal from one antenna to the other.

• Gain of the repeater depends on its size, the frequency of

operation and the angle between the paths.

Repeatered Systems

Reflector Systems (Analysis)Reflector Systems (Analysis)

• For a reflector of surface area A, the gain is given by:

• G = 42.8 + 40 log f(GHz) + 20 log A (m2) + 20 log [cos (θ/2)] dB

• Overall free space path loss is then FSL1 + FSL2 - G where FSL1

and FSL2 are the losses of the individual parts of the path.

Repeatered Systems

θ

FSL1 FSL2




P-TR-005-M101-ver4 120

Reflector Systems (Example)Reflector Systems (Example)

• Considering a 6 km, 14 GHz path as before with 1.2 m antennas,

determine the size of billboard required to limit the path loss to 81.8

dB.

Repeatered Systems

120 degrees

3 km 3 km

f = 14 GHz


• FSL1 = FSL2 = 124.9 dB.

• Path loss = 124.9 + 124.9 - 42 - 42 - G = 81.8 dBi

• G = 84 dB = 42.8 + 40 log 14 + 20 log A (m2) + 20 log [cos (60)]

• 1.4 dB = 20 log A

• A = 1.2 square metres.

Repeatered Systems

120 degrees

3 km 3 km

f = 14 GHz




P-TR-005-M101-ver4 121


• Comparison with non-symmetric split.

• G = 84 dB

• Path loss = 129.3 + 115.3 - 42 - 42 - 84 = 76.6 dBi (compared with 81.8 dBi)

• Conclusion is that placing the reflector near one of the sites is

advantageous.

• Limitation occurs when it is so close to one end that path loss equals that of

a single hop (always check to ensure your prediction for path loss is greater

than that for a single hop).

Repeatered Systems

120 degrees5 km

1.2 m2

1 km

Reflector Systems (Double Reflectors)Reflector Systems (Double Reflectors)

• Where the angle between the paths is greater than about 130 degrees, the

gain of the antenna reduces noticeably (120 degrees is the “-6 dB angle”;

130 degrees is the “-7.5 dB angle).

• Double reflector systems can be used for greater angles.

Repeatered Systems




P-TR-005-M101-ver4 122

Double Reflectors (Analysis)Double Reflectors (Analysis)

• Provided adequate clearance is provided (the 15λ clearance shown is taken

as sufficient), the gain of the double reflector is approximately equal to the

gain of the smaller of the two.

• If the direction of propagation is changed at the reflector then each reflector

will change the direction of propagation by a different amount.

Repeatered Systems

15λ

Double Reflectors (Analysis)Double Reflectors (Analysis)

• Remember

• G = 42.8 + 40 log f(GHz) + 20 log A (m2) + 20 log [cos (θ/2)] dB

• Compute G for both reflectors and take the smaller of the two.

Repeatered Systems

θ1

θ2 θθ = θ2+180 - θ1




P-TR-005-M101-ver4 123

Double Reflectors (Optimisation)Double Reflectors (Optimisation)

• G = 42.8 + 40 log f(GHz) + 20 log A

(m2) + 20 log [cos (θ/2)] dB

• θ2 and θ1 should be as small as

possible.

• E.g. if θ has to be 160 degrees. θ2 =

20 degrees and θ1 = 40 degrees will

be a better solution than θ2 = 60

degrees and θ1 = 80 degrees.

• However, the smaller the angle the

harder it is to ensure that the onereflector does not obstruct the other.

Repeatered Systems

θ1

θ2

θ

θ = θ2+180 - θ1




P-TR-005-M101-ver4 124


1. A 7 GHz, 20 km link is obstructed at its mid-point and requires a repeater

formed from two “back-to-back” parabolic antennas. The transmitter gives

an output power of 20 dBm into the antenna. The minimum required receivesignal level is –50 dBm. Determine the minimum antenna sizes required if:

a) The repeater is an active repeater

b) The repeater is a passive repeater




P-TR-005-M101-ver4 125

2. As an alternative to the back-to-back parabolic repeater, it is suggested

that a single flat, billboard reflector moved to the side of the path so that the

angle between the two paths is 120 degrees can provide the required signal

level. Determine the required size of the reflector. (Hint: the path length

from each antenna to the reflector will be greater than 10 km).




P-TR-005-M101-ver4 126

3. The size requirements for the billboard reflector are found to be excessive.

It is known that such reflectors are more effective if they are placed nearer to

one end of the link. Accordingly a suitable site is found 600 metres to the

side of one end of the link. The reflection angle is now 90 degrees and the

two “hops” are 20 km and 0.6 km in length. Re-calculate the size

requirements for the billboard reflector.




P-TR-005-M101-ver4 127

4.

A town is surrounded by a ridge of hills. In order to provide a 15 km, 14

GHz hop into the town from a neighbouring village, it is necessary to install

a double billboard reflector on the ridge, some 400 m from its terminal. 1.2

metre antennas are used on the two terminals. The reflector is configured

such that reflection angles of 30 degrees and 50 degrees are obtained.

Determine suitable reflector sizes if the maximum path loss is 70 dB.








P-TR-005-M101-ver4 130

Clearance RequirementsClearance Requirements

• We need to be able to calculate the “earth bulge”.

• Then, the terrain data needs to be extracted from

mapping information.

Clearance Issues

Earth Bulge

Clearance

8.2 Earth Bulge

The amount by which the bulge of the earth tends to obstruct the path between the

transmitter and the receiver can be determined using standard geometrical techniques.

However, the fact that the atmosphere is not uniform causes electromagnetic waves to

follow a curved path. In a standard atmosphere, the wave tends to follow the curvature of

the earth. This is helpful to radio engineers as it reduces the effective earth bulge to

typically 75% of its calculated value. However, the atmosphere is not static and

occasionally it will cause the radio wave to follow a path such that the effect of the

curvature of the earth is exaggerated. In carrying out calculations in such cases it is

necessary to increase the actual curvature of the earth by as much as 50%.




P-TR-005-M101-ver4 131

Earth BulgeEarth Bulge

Using the law of intersecting cords

If h is required in metres and R, d are

in kilometres:

Clearance Issues

d 2

d 1

h

2R

h - earth bulge

R - earth radius

d 1,2 - distances from hop ends

R

d d h

Rhd d

2

2

21

21

=

=

R

d d h

2

1000 21=

Earth BulgeEarth Bulge

Clearance Issues

R

d d h

2

1000 21=

• Earth bulge is a maximum where d 1=d 2=d/ 2.

• Then the earth bulge = Taking the earth radius tobe 6373 km:

R

d 2125

Path Length (km) Max Earth Bulge (m)

10 2.0

20 7.9

30 17.7

40 31.4

50 49.0

60 70.6




P-TR-005-M101-ver4 132

Modified Earth RadiusModified Earth Radius

Clearance Issues

• Radio signals will continue slightly beyond the horizon. This isbecause the refractive index of the atmosphere tends to reduce withheight causing the radio wave to bend in the direction of curvature of the earth.

• Thus the effect of the earth bulge does not have as big an effect asfirst calculated.

• The effective earth bulge can be calculated by assuming the earth’sradius is larger than its physical value.

Radio horizonVisible horizon

Modified Earth RadiusModified Earth Radius

Clearance Issues

• The actual Earth’s radius is multiplied by a factor given the value k (often referred to as the k -factor).

• For a “standard atmosphere”, k = 1.33 reducing the effective earth

bulge to 0.75 of its calculated value.

• k varies with atmospheric conditions.

k = 1.0k = 1.33

k = 2.0k = 4.0

k = ∞

k = 0.66




P-TR-005-M101-ver4 133

Variability ofVariability of k k -Factor -Factor

Clearance Issues

• Just as we need to know the extent of multipath fading for, say, 0.1%of the time, we also need to know the minimum value of the k-factor of the same percentage in order to establish the necessary clearance.

• The value exceeded for 99.9% of the time depends on the climate andon the path length (as very anomalous atmospheric structures willtend not to occur over large distances simultaneously.

10 20 40 80

k

0.3

0.5

0.7

0.9

Value of k exceeded for

99.9% of the worst month.

Path length (km)

8.3 The Fresnel Parameter

TheThe FresnelFresnel Parameter Parameter

Clearance Issues

• The amount of clearance required depends on the path length, theposition of the obstruction along the path and the frequency of operation.

• The Fresnel Parameter links these together to give a universallyapplicable parameter.

d 1

cb

h

21 d d cb +>+

221 λ ++=+ d d cb

• There exists a value of h such that

d 2




P-TR-005-M101-ver4 134

TheThe FresnelFresnel Parameter Parameter

Clearance Issues

• The locus of points for which this is true form an ellipsoid in threedimensions known as the “First Fresnel Zone” and the values of h atpoints along the path are known as the F 1 values.

• If h<<(d 1+d 2) then F 1 in metres is given approximately by

d 1

cbh

221 λ ++=+ d d cb

• There exists a value of h such that

d 2

( )21

211 3.17

d d f

d d F

+=

• f is in GHz, d 1, d 2 are measured in kilometres.

TheThe FresnelFresnel EllipsoidEllipsoid

Clearance Issues

( )21

211 3.17

d d f

d d F

+=

d 1 d 2

F 1




P-TR-005-M101-ver4 135

8.4 ITU-R Recommendations

Once the Fresnel zone radius has been established, it is possible to use ITU-R

recommendations in order to determine the amount of clearance that should be afforded

in any particular location.

Clearance RequirementsClearance Requirements

Clearance Issues

• The antennas should be sufficiently high to meet the more

onerous of the following requirements.

For k = 1.33, clearance of 1.0 F 1 should be obtained.

For k = “minimum exceeded for 99.9% of the time”, clearance of

0.3 F 1 should be obtained if the obstacle is rounded or zero if

there is a sharp single isolated obstacle.




P-TR-005-M101-ver4 136

Clearance Requirements (example)Clearance Requirements (example)

Clearance Issues

• A 30 km, 14 GHz path has an isolated obstacle 12 metresin height at a distance of 13 km from one end.

12 m

30 km

13 km

Clearance Requirements (example)Clearance Requirements (example)

Clearance Issues

• k = 1.33. Earth Bulge at 13 km from one end = 17.7/1.33 = 13.3 m

• Add 12 m obstacle height to give 25.3 m in total.

• F1 at 13 km from one end = 12.5 metres

• Total required clearance 37.8 metres.

• Each antenna should be 37.8 metres in height.

12 m

30 km

13 km








P-TR-005-M101-ver4 139

Diffraction over “average terrain”Diffraction over “average terrain”

Clearance Issues

• The description “knife-edge” may not apply to a particular

obstacle.

• An approximate formula for average terrain exists based

on the “normalized clearance” expressed as a multiple of

F 1.

• Path loss ~ 10 - 20 h/ F 1 dB.

• Note: valid for values of h larger than F 1 (obstructed paths

only).

Diffraction over “average terrain”Diffraction over “average terrain”(example)(example)

Clearance Issues

• A 30 km, 14 GHz path propagates over “average terrain”

of height equal to the height of the base of the antenna

towers.

• The antennas are 15 m above ground level.

• Determine the diffraction loss when the k-factor is 0.7.






P-TR-005-M101-ver4 141

Fading due to Ground ReflectionsFading due to Ground Reflections

Clearance Issues

• Multipath caused by ground reflection can cause severe fades.

• Smooth ground causes more severe fading than rough ground.

Diffracted Rays

Reflected Ray


• As the path length difference between the reflected and direct

ray alters, “constructive” and “destructive” interference is

experienced. Destructive interference can cause a severe

reduction in signal strength.

Clearance Issues




P-TR-005-M101-ver4 142


Clearance Issues

Direct Ray

Reflected Ray

Resultant

Constructive

Interference


Clearance Issues

Direct Ray

Reflected Ray

Resultant

DestructiveInterference




P-TR-005-M101-ver4 143

Protecting against Reflection FadesProtecting against Reflection Fades

Clearance Issues

• The effect is that an “interference pattern” develops in which

the strength of the received signal varies with height.

“Null” or “Trough”

“Peak”

The effect of surface roughness on the probability of fading is considered in ITU-R

P.530-10. A more accurate equation for the geoclimatic factor, K is given by

42.0003.09.3 110 −−−= adN s K where a s is the “standard deviation of terrain

heights” for the area of interest. Note that K reduces with increasing values of S a. A

minimum value for S a. of 6 metres should be adopted. As an example of the effect of this

refinement of the equation consider the situation where K is calculated for a value for

1dN of -200.




P-TR-005-M101-ver4 144

Approximate Formula

(independent of 1dN ) K = 2.4 x 10-4

6=a s K = 2.4x10

-4

12=a s K = 1.8 x 10-4

24=a s K = 1.3 x 10-4

42=a s K = 1.0 x 10-4

It can be seen that the approximate formula assumes a “worst case” for fading, agreeing

with the more accurate figure when the value for standard deviation of terrain heights is

at the lowest possible value. However, the value itself may not be particularly relevant

for a particular path as data is provided by making height measurements over a wide area.

The area where reflections take place on a particular path are subject to local variations

that may render the standard calculation methods irrelevant. A preliminary study into the

mechanics of reflection and possible counter measures is given here

8.6.2 The Rayleigh Criterion.

The likelihood of a fade occurring is influenced by the “coherence” of the reflected wave.

To provide a deep fade it must have the characteristics of a single sinusoid. This is only

the case if the surface is extremely smooth. But any definition of “smoothness” is related

to the wavelength. The effect of roughness is also influenced by the grazing angle

between the transmitter, receiver and reflection point. The Rayleigh criterion involves

evaluating the expression for the phase difference between two elements of a reflected (or

“scattered” wave). The expression incorporates the grazing angle, θ , the frequency f

(GHz) and the standard deviation of heights, s (metres) at the reflection point and is given

by θ sin42 f . If this expression is less than 0.1 then the reflections can be regarded as

mirror-like (or “specular”). If the expression is greater than 10 then the reflections will

be diffuse and troublesome fading from ground reflections are unlikely. Between these




P-TR-005-M101-ver4 145

two values a transition occurs and it is difficult to be conclusive about the likely

occurrence of reflection fading.

TheThe RayleighRayleigh CriterionCriterion

• The phase difference ∆θ betweentwo rays reflecting from twodifferent surfaces separated by

distance s is given by

s

Clearance Issues

θ λ

θ π φ sin42

sin4 sf

s≈=∆

θ

ition trans10sin420.1

diffuse 10sin42

specular 1.0sin42

<<

>

<

θ

θ

θ

sf

sf

sf

8.6.3 Protection against reflection fades

Once the possibility of problematic reflection fades, judging by the Rayleigh criterion,

has been identified, it is necessary to be aware of the procedures by which the probability

of fading can be reduced. ITU-R P.530-10 itemises four possible methods by which this

reduction can be achieved:

• Use of vertical polarisation

• Shielding of the reflection point• Moving of reflection point to poorer reflecting surface

• Optimum choice of antenna heights




P-TR-005-M101-ver4 146

8.6.3.1 Use of vertical polarisation

Vertically polarized waves will reflect less strongly from horizontal surfaces than will

horizontally polarized waves. Vertical polarisation should be the polarisation of choice

for paths where the possibility of ground reflection exists.

8.6.3.2 Shielding of the reflection point

Judicious use of the terrain and/or buildings can result in the reflected wave being

attenuated due to diffraction over an obstacle. Checks should be made to ensure that the

reflected path is obstructed over the complete range of likely values of earth bulge k

factor.


Clearance Issues

• Shielding of Reflection point

• Checks must be made to ensure that shielding occursthroughout the range of k-factors that will be experienced.

8.6.3.3 Moving of the reflection point.

It may be that a smooth reflecting surface (for example, a body of water) occurs at the

mid-point of a path. If equal antenna heights are adopted, the reflection point will be at

the mid-point. By adjusting antenna heights, the reflection point can be moved.

Lowering an antenna will cause the reflection point to move towards that antenna.




P-TR-005-M101-ver4 147


Clearance Issues

• Moving of Reflection point to

poorer reflecting surface

8.6.3.4 Optimisation of antenna heights.

Ground reflection will cause constructive and destructive interference. It is the

destructive interference that causes us most concern. Also, because the path length

differences are small (a few tens of centimeters) resulting in low relative time delays (less

than a nanosecond), reflection fading will be “flat” rather than “selective”. As the height

of the receiving antenna is varied, it will move through “peaks” and “troughs” that are

caused by the reflection. It is possible to place the receiving antenna in a peak thus

taking advantage of the reflection to elevate the signal level.




P-TR-005-M101-ver4 148


Clearance Issues

• Optimum choice of antenna heights.

“Null” or “Trough”

“Peak”

However, there is a problem with this strategy: if the reflection point is a body of water

whose level varies or; the k -factor of the earth changes significantly; the pattern of peaks

and troughs will move in a vertical direction. In this circumstance, it will be necessary to

implement a diversity system such as space diversity where the likelihood of both

antennas experiencing nulls simultaneously is very low.




P-TR-005-M101-ver4 149


Clearance Issues

• A problem – the interference pattern is not stationary.Changes in earth curvature k-factor and variations in the tidecause the pattern to move with time.

In order to determine the whether the pattern of peaks and troughs will move

significantly, it is necessary to estimate the amount by which the path length difference

between the direct and reflected waves will vary. If this varies by more than a

wavelength then diversity should be used.




P-TR-005-M101-ver4 150


• The pattern moves because thepath length difference changes.

2/d

Clearance Issues

2/d

h

• If the reflection point is at themidpoint, then (by Pythagoras):

( ) ( )2/2/2 22d hd d −+=∆

8.6.3.4.1 Ground reflection example

Consider a situation where a 20 km microwave path uses antennas that are 25 m above

sea level. At the mid-point of the path exists a tidal inlet. The height of the water surface

in the inlet varies from 6 metres below sea level to 4 metres above sea level. Assuming a

k-factor of 1.33, determine the maximum frequency that can be employed to limit the

variation in path length difference to 1 wavelength.




P-TR-005-M101-ver4 151


• Example:

• Antenna heights: 25 m a.s.l.

• Ground heights vary from 6 m below s.l. to 4 m a.s.l.

2/d

Clearance Issues

2/d

h

33.1km;20 == k d

Solution:

Earth bulge at midpoint = kRd /125 2 = 5.9 metres.

At 6 metres below sea level the path length difference (by Pythagoras)

063.0100001.2510000222

=

−+×= metres

At 4 metres above sea level the path length difference

023.0100001.15100002 22 =

−+×=

The difference between these two values equals 0.040 metres. This equals a wavelength

at frequency of98 105.7040.0103 ×=÷× Hz. This corresponds to a frequency of

7.5 GHz which should be regarded as the maximum frequency at which a non-diversity

system could be implemented.






P-TR-005-M101-ver4 153

Using Field MeasurementsUsing Field Measurements

Clearance Issues

• If the reflection coefficient of the ground is low, or thereflecting surface is very rough, the difference between the

peak and the trough will be very small.• One advantage of reflection fades is that the reflecting surface

is always there (unlike atmospheric ducts).

• It is therefore possible to measure the variation of signalstrength with height and assess the seriousness of theproblem.

• The difference between the peak and the null indicates thelikely depth of reflection fading that will be experienced.




P-TR-005-M101-ver4 154


1. Determine the earth bulge for a K-factor of 0.8 on a link of length 25 km.

2. Determine the radius of the first Fresnel zone at the midpoint of a 50 km

path at frequencies of 3 GHz, 10 GHz and 30 GHz.




P-TR-005-M101-ver4 155

3. A 30 km path has a sharp, knife-edge obstacle, 10 m in height, 8 km from

one end. The minimum k factor experienced (exceeded 99.9% of the

time) is 0.7. Determine suitable antenna heights.




P-TR-005-M101-ver4 156

4. Determine the diffraction loss at a k factor of 0.5 if the antenna heights

determined by the result of question 3 are adopted.




P-TR-005-M101-ver4 157

9 Performance Objectives

9.1 Introduction:

It has been demonstrated that 100% reliability is an unrealistic expectation from a

microwave radio system. Nevertheless, microwave links must form a viable alternative

to fixed line networks if they are to be considered for general use. This includes being

able to satisfy expectations regarding performance. In assessing reliability, the ITU-R

makes a distinction between “unavailability” and “outages” as described in the slide

below.

DefinitionsDefinitions

• Unavailability: System “not working” for 10

consecutive seconds.

“Not working” defined as BER worse that 1 x 10-3.

• Outages: Exist for less than 10 seconds and the

system is still regarded as “available” (even though

the user cannot access it).

Outages are subject to “performance objectives”.

Unavailability and Performance Objectives






P-TR-005-M101-ver4 159

Propagation ProblemsPropagation Problems

• Multipath problems: unlikely as multipath outages

tend to be short-lived (much less than 10 seconds).

• Diffraction loss: obeying clearance rules should

avoid this.

• Ducting: generally restricted to well-known

geographical regions. Can be combatted with

space diversity.

• Rain: the most likely cause of “propagation related”

unavailability at high (10 GHz+) microwave

frequencies.


9.3 Equipment-related Unavailability

The prevalence of unavailability due to equipment problems depends on the reliability of

the equipment (quantified by the Mean Time Between Failures: MTBF) and the average

length of time that the system is down for each failure (quantified as the Mean Time To

Repair: MTTR).




P-TR-005-M101-ver4 160

Equipment ProblemsEquipment Problems

• Definitions

MTBF: Mean Time Between Failures (usually several

thousand hours)

MTTR: Mean Time To Restore.

• Availability

• Unavailabity


%100×+

= MTTR MTBF

MTBF A

AU −=100

Equipment ProblemsEquipment Problems

• Example:

For a single transceiver and associated equipment MTBF = 50,000 hours

MTTR = 6 hours

Link MTBF = (Terminal MTBF) x 0.5 = 25,000 hours

• If we have a 12 hop link the total unavailability =

12x0.024=0.29% (approximately 25 hours per year).


%024.0

%976.99%100625000

25000

=

=×+

=

U

A




P-TR-005-M101-ver4 161

9.3.1 Hot Standby

The effective mean time to repair can be reduced to nearly zero by implementing a “hot

standby” facility for the equipment. It is possible for two receivers to be permanently

connected to the system so that either one can be utilised. This provides a seamless

continuity of service should one receiver fail.

Equipment Problems: Hot StandbyEquipment Problems: Hot Standby• A “hot standby” is a duplicate system permanently

powered up and ready to replace the active system

should a fault occur.

• Should a transmitter fail, for example, a replacement

is switched into its place. This can occur in as short

a time as 20 ms.

• With MTBF as long as 50000 hours, unavailability

due to transmitter or receiver failure becomes

negligible.





P-TR-005-M101-ver4 162

Implementing Hot StandbyImplementing Hot Standby

• Transmitter Hot Standby.

Cannot have both connected simultaneously.

An RF switch is required to connect the Hot Standby to the

antenna system in the event of the Main Transmitter failing.


Main

Transmitter

Hot

Standby

Implementing Hot StandbyImplementing Hot Standby• Receiver Hot Standby.

It is possible to connect two receivers to the antenna

system simultaneously, via a coupler.


Main

Receiverr

HotStandby




P-TR-005-M101-ver4 163

In the case of the transmitter, it is not possible to connect the main and the standby

transmitter simultaneously (as they would interfere with each other). It is necessary to

have a switch that automatically disconnects the main transmitter and connects the

standby transmitter to the antenna should the main transmitter fail. This does result in a

very brief period when the service is down.

The fact that it is possible to connect the main and standby receivers simultaneously leads

to the question “what fraction of the total receive power should be channeled through to

each receiver?” being asked. If an equal power is sent to each receiver, then an insertion

loss of approximately 4 dB will be incurred. This is seen as a waste of power as the use

of the standby receiver should be an extremely rare occurrence. It is more common to

split the received power so that the level received by the standby receiver is typically 10

dB below that received by the main receiver. This means that the power would be higher

most of the time with a 10 dB reduction in margin occurring when the main receiver fails.

Receiver Hot Standby: Coupler AnalysisReceiver Hot Standby: Coupler Analysis

• A symmetrical coupler will have an insertion loss of

at least 3 dB (usually nearer 4 dB) that must be

accounted for in the link budget.

• Asymmetrical couplers can put more insertion loss

in the standby leg and less in the main leg.


From Antenna From AntennaTo Main

To Standby

To Main

To Standby

Symmetrical Coupler Asymmetrical Coupler




P-TR-005-M101-ver4 164

Receiver Hot Standby: Coupler AnalysisReceiver Hot Standby: Coupler Analysis

• Suppose a coupler produces a 1 dB resistive loss.

• That means that, if we have 10 nW at the input, we will have a

total of 8 nW at the output.

• If this is divided equally, each arm of the coupler will receive 4

nW, equivalent to a loss of 10log10(2.5)=4dB.

• Alternatively, one arm could receive 7.27 nW and the other arm

0.727 nW.

• The losses would then be 1.4 dB and 11.4 dB respectively.

• The choice of having only 1.4 dB loss “permanently” and an

extra 10 dB degradation of fade margin during standby periods

is argued to be superior to having 4 dB loss in both “main” and “

standby” modes.


9.4 Unavailability Objectives

Some unavailability is inevitable. However, there are internationally agreed objectives that the link planner

should aim to meet. In forming the objectives, the ITU-T gives due consideration to the importance of the

link being planned and describe three different categories: High Grade; Medium Grade and Local Grade.

Connections between cellular mobile radio sites are classed as Local Grade.




P-TR-005-M101-ver4 165

Unavailability ObjectivesUnavailability Objectives

• ITU-T G.821 divides a hypothetical long-distance channel into

“High Grade”, “Medium Grade” and “Local Grade” services.

• Objectives for High Grade circuits of length L, where L isbetween 280 km and 2500 km are:

• Local Grade (e.g. GSM interconnect) objectives proposed vary

between 0.01% and 0.2%. This affects repair philosophy.


( )%2500

3.0100 L A ×−=

9.5 Performance Standards

When the system is available, it will still suffer outages. Performance Standards specify

the maximum amount of outages that will be tolerated. Again different categories of link

are defined with performance standards specified accordingly. Traditionally, the number

of one-second periods containing one or more errors would be reported. The “errored-

second ratio” (ESR) became a benchmark by which services were compared. However,

with the advent of high capacity services with data rates up to 155 Mbits/s it is apparent

that a single second will contain 155 million bits and a single error would probably not be

a serious issue. It was therefore decided to adopt a block of data as the standard unit of

transmission and define performance standards on this basis. It is then necessary to be

able to translate from a fade margin that has been calculated to a ratio for Severely

Errored Seconds (second periods during which 30% of blocks received contain errors).

An example is shown below whereby this is done when rain fading is considered.




P-TR-005-M101-ver4 166

Performance StandardsPerformance Standards

• These standards define the required system

performance when it is available.

• Causes of degradation in performance:

Multipath Fading (as previously analysed)

Background Errors (Gaussian noise has no absolute

maximum value and, hence some errors will occur)

Wind (causes misalignment of antennas)


Performance Criteria: High CapacityPerformance Criteria: High Capacity

ServicesServices• ES: Errored-second; any 1 second period in which an error occurs.

• Not an appropriate measure when 1 second can contain several

million bits. Instead, a block of data is considered and new terms

are introduced.

• EBR: Errored Block Ratio; refers to blocks containing one or more

errors. Block size is specified for each system rate.

• ESR: Errored Second Ratio; A 1-second period that contains one

or more errored blocks.

• SESR: Severely Errored Second Ratio; A 1-second period that

contains greater than 30% or errored blocks.

• BBE: Background Block Error; An errored block not occurring aspart of an SES.

• In-service measurements of block errors is possible.





P-TR-005-M101-ver4 167

Performance CriteriaPerformance Criteria

• Local Grade (including links between cellular

sites)

SESR should not exceed 0.00015 during the worst

month.

ESR should not exceed 0.012 during the worst

month


Linking SNR, BER, ES, ESR and SESRLinking SNR, BER, ES, ESR and SESR

• We have seen that SNR affects the BER. This will in turn affect

the other parameters.

• As an example, consider the procedure to predict the SESR

caused by rain attenuation.

• Step 1: for the system under consideration use ITU-R P.530-9 to

estimate the BER that will result is SES ( BERSES )

• Step 2: calculate the receive level without rain attenuation and

hence calculate the rain attenuation margin.

• Step 3: calculate the annual time percentage that the rain

attenuation will exceed the margin.

• Step 4: translate this to a worst month percentage (see ITU-RP.841)







P-TR-005-M101-ver4 169

10 Solutions to Self-Assessment Questions


Designing by guessing.

As intuitive engineers we should have some idea regarding what a

microwave link should look like and what its values should be.

Try and picture a microwave link in your mind and imagine what the

relevant parameters might be. It will be interesting to refer to these

“guesstimates” as we gain knowledge regarding the design of microwave

links.

Name of Designer Chris Haslett

Frequency of Operation 10 GHz

Rate of transmission (bits per

second)

8 Mbit/s

Mast Height 30 metres

Path Length 20 km

Antenna Diameter 1.2 metres

Transmit Power 1 watt

Receive Power 1 nanowattFeeder length (metres) 30 metres

Feeder loss (dB) 2 dB




P-TR-005-M101-ver4 170


1. An antenna operates at a frequency of 15 GHz. If it has a diameter of

1.8 metres, estimate its gain.

41000dBi46log20log205.17 ==++≅ D f G

Alternatively

dBi8.46480006.0

2

==

≅λ

π DG

2. Two such antennas are to be used over a link of length 12 km. Determine

the path loss.

dB45.592-137.5Loss

dBi92GainsAntenna

5.13715log2012log204.92

log20log204.92

==

=

=++=

++= f d SL




P-TR-005-M101-ver4 171

3. Repeat the calculation of question 2 for antennas of the same size but

operating at a frequency of 30 GHz.

dB39.5104-143.5Loss

dBi104GainsAntenna

5.14330log2012log204.92

log20log204.92dBi52log20log205.17

==

=

=++=

++= =++≅ f d FSL D f G

4. Estimate the beamwidth of a 1.8 metre antenna at 7 GHz, 15 GHz and 30

GHz.

)30(4.0);15(8.0);7(7.18.1

22GHz GHz GHz

f Beamwidth °°°=≈




P-TR-005-M101-ver4 172

5. A transceiver outputs a power of 27 dBm via a feeder of 3 dB loss to an

antenna of diameter 0.9 metres. If the frequency of operation is 12 GHz,

estimate the EIRP from the antenna.

dBm62.23-2738.2EIRP

dBi2.389.0log2012log205.17

=+==++=G

6. For the situation described in question 5, estimate the power that would

be gathered by an identical antenna at a distance of 4 km.

dBm-25.6

20log(12))20log(4)(92.4-38.262.2FSL-GainAntennaRxEIRPPowerRx

=

+++=+=




P-TR-005-M101-ver4 173


Note Boltzmann’s constant k = 1.38 x 10-23

J/K

1. An antenna has a noise temperature of 280 kelvins. Determine the noise power gathered if the noise bandwidth of the receiver is 14 MHz.

dBm102.7-watts105.4

watts10142801038.1

14

623

=×=

××××=−

−kTB

2. An amplifier has a noise bandwidth of 2 MHz and a noise temperature of

350 kelvins. If the noise power at the input equals k(480)B watts and the

signal power at the input is 0.172 picowatts, determine the signal to noise

ratio at the output of the amplifier.

dB8.8

5.7

102350)(48010.381

101.72

outputatSNR

watts102350)(480outputatPower Noise

watts101.72outputatPowerSignal

623-

13-

6

13-

=

=×+×

×

=

×+=

×=

Gk

G




P-TR-005-M101-ver4 174

3. A microwave system has a bandwidth of 4 MHz. The receiver noise

figure is 3 dB. Determine the noise temperature of the receiver and the

minimum required signal power in order to deliver a SNR of 13 dB.

State any assumptions made.

dBm-92

watts104.620104)580(

SNR required powernoise powersignalRequired

kelvins290isantennaof retemperatu NoiseAssuming

kelvins290)1(290

0.210

136

3.0

=

×=××=

×=

=−=

==

−k

F T

F

e

4. A 10 GHz microwave link of length 30 km has 1.2 m diameter antennas.

The minimum required receive power has been determined to be –84

dBm. Miscellaneous losses total 6 dB. Determine a suitable transmit

power.

dBm-14(-84)69.7PowerTx

dB69.778.2-6141.9lossPath

dB141.920log(10))2020log(3092.4FSL

dBi39.120log(10)20log(1.2)17.5antennam1.2of Gain

=+=

=+==++=

=++=




P-TR-005-M101-ver4 175

5. A microwave link has a receiver with a noise bandwidth of 3 MHz. The

noise temperature of the antenna is 290 kelvins. The receiver consists of

a mast head amplifier with a gain of 15 dB and a Noise Figure of 1.2 dB,

a feeder of 4.5 dB loss and a demodulator with a Noise Figure of 3.5 dB.

Determine the SNR with and without the mast head amplifier if the

power gathered by the antenna is –91 dBm.

dB16.5(-107.5)-91-SNR

dBm107.5-watts1078.1)290(

1411010359105273.92

MHAwith

dB10(-101)-91-SNR

dBm101-watts1057.7)290(

153910359527

:MHAwithout

527attenuator dB4.5

K 359dB3.5

K 92.3dB2.1

14

45.05.15.1

14

45.0

==

=×=+

=×÷+÷+=

==

=×=+

=×+=

⇒

⇒

⇒

−

−

BT k

T

BT k

T

e

e

e

e




P-TR-005-M101-ver4 176

6. A Microwave link provides a 2 MHz channel with a SNR of 12 dB. Use

Shannon’s theorem to determine the maximum possible capacity of the

channel. Note:

( )

2log

loglog

1logBandwidthCapacity

10

102

2

x x

SNR

=

+×=

Mbit/s1.8101.82log

8.16log102Capacity

8.1510

6

10

106

2.1

=×=××=

==SNR




P-TR-005-M101-ver4 177


1. A 24 km microwave link is located in Sweden and operates at a

frequency of 18 GHz. One antenna is 1100 m above sea level and the

other is at 800 m above sea level. Estimate the percentage time for whicha fade exceeding 25 dB would occur.

( )

( )

( )

%1009.1

105.121241012.9

101

1012.9

10

400

5.1224

8001100

3

10/25800001.018033.02.10.34

10/001.0033.02.10.3

4

0029.02.4

1

1

−

−×−×−−

−−−

−

−−

×=

×+×=

×+=

×=

=

−=

=−=

Ah f pw

dN

p

L Kd p

K

dN

ε

ε




P-TR-005-M101-ver4 178

2. A 21 km microwave link, located in Italy, operates at a frequency of 28

GHz. Horizontal polarisation is used. Determine the rain-produced

attentuation for a 0.01% time period.

dB8710.78.1nAttenuatio

km7.10

2.19241

24 length patheff

2.1935

1 length patheff

dB/km1.8)40(187.0

021.1

187.0Italyformm/hr40

01.0015.00

0

021.1

01.0

=×=

=+

=

==

+=

===

==

== ≈

− R

R

H

H

ed

d d

d

kR

k k R

α γ

α α






P-TR-005-M101-ver4 180

4. Estimate the atmospheric absorption on an 11.5 GHz link of path length

20 km.

From graph, water vapour and oxygen absorption both equal approximately

0.02 dB/km. Adding these two values gives 0.04 dB/km. Therefore a 20 km

path will suffer atmospheric absorption of approximately 0.8 dB.




P-TR-005-M101-ver4 181


1. A 28 km microwave link, located in Sweden operates at a frequency of

21 GHz. Identical antennas are used at both ends and the path is horizontal

with antennas at an elevation of 1200 metres above sea level. Without anydiversity it is found to have a fade margin of 16 dB. Determine the

probability of outage. Estimate the improvement factor provided if space

diversity is employed with an antenna separation of 4 metres.

From previous solution to Module 4 Question 1, K = 9.12 x 10-4

( )

( ){ }( ){ }

63.2

1023.62821404.0exp1

1004.0exp1

23.6

10

%156.010281012.9

101

101604.148.012.087.0

1004.10

48.012.087.0

10160

10161200001.021033.00.34

10001.0033.02.10.3

=

×−−=

−−=

=

=

=××=

×+=

−−

−−

−×−×−

−−−

A

w

Ah f pw

pd f S I

p p

Kd p Lε

Probability of fade reduced from 0.156% to 0.059%.




P-TR-005-M101-ver4 182

2. Using the procedures of ITU-R P.530-10, page 28, evaluate the

improvement possible by utilising frequency diversity with a separation of

200 MHz instead of space diversity.

008.0

1021

2.0

2821

80

1080

108

10

=

×=

∆= F

f

f

fd I

This suggests that frequency diversity is not suitable for links such as this.

Note that improvement would be gained if the link was shorter and operated

at a lower frequency and had a higher fade margin.




P-TR-005-M101-ver4 183


1. A particular receiver has a receive power threshold of –87 dBm in order

to deliver a SNR of 14 dB. The antenna also receives two interference

signals: one at a level of –98 dBm and another at a level of –104 dBm.Determine the degraded threshold of the receiver in the presence of these

interfering signals

dBm-81.61495.56-thresholdDegraded

dBm-95.56

mW1027.77floornoise plussinterferer of Total

mW103.98dBm104-

mW1015.85dBm98-

:sInterferer

mW107.94dBm101-14-87-Floor Noise

11-

11-

11-

11-

=+=

=

×=

×=×=

×===




P-TR-005-M101-ver4 184

2. A mast is at the “hub” of a 12 GHz microwave network. Three links

converge on this hub. One is of 2 km length, one 5 km length and one 16

km in length. All the antennas are 1.2 metres in diameter and the

transmit from all transmitters is 500 mW. The 16 km link is susceptible

to interference from the two shorter links. In the direction of the 2 km

link, the gain of the receiving antenna is 0 dBi, and in the direction of the

5 km link the gain is –5 dBi. Estimate the interference power gathered

by the antenna and compare it with that received from its wanted signal.

Wanted signal:

dBm29.6-81.4138-27PowerReceive

dBi40.71.2log2012log2017.5GainAntenna

dB13816log2012log204.92

=+=

=++=

=++= FSL

Signal on 2 km link

dBm3.5240.7120-27 powerReceive

dBi40.7gainsantennaTotal

dB120

−=+=

=

= FSL

Signal on 5 km link

dBm65.3-35.7128-27 powerReceive

dBi35.7gainsantennaTotaldB128

=+=

== FSL

Adding interfering powers

-65.3 dBm + - 52.3 dBm = -52.1 dBm.

This is 22.5 dB below the wanted signal power.




P-TR-005-M101-ver4 185

3. A receiver at the hub of a microwave network operates on a frequency of

12.260 GHz. Interfering signals are received at frequencies of 12.060

GHz, 12.200 GHz, 12.400 GHz and 12.540 GHz. Which of these

interfering signals requires most attention from the viewpoint of

intermodulation products causing interference?

If 12.400 GHz is regarded as f 1 and 12.540 GHz is regarded as f 2 then:

2 f 1 – f 2 = 24.800-12.540=12.260 GHz, the wanted signal frequency.




P-TR-005-M101-ver4 186


1. A 7 GHz, 20 km link is obstructed at its mid-point and requires a repeater

formed from two “back-to-back” parabolic antennas. The transmitter gives

an output power of 20 dBm into the antenna. The minimum required receivesignal level is –50 dBm. Determine the minimum antenna sizes required if:

a) The repeater is an active repeater

b) The repeater is a passive repeater

The link consists of two, 10 km hops. Total loss allowed is 70 dB. For the

active system this effectively means that each hop can suffer a loss of 70 dB.

metres60.010

log2016.917.530

20log20log717.5Gain

each)dBi(30dBi59.3gainsantennaCombineddB3.1297log2010log204.92

204.4

==

++=

++=

==++=

− D

D

D

FSL

For the passive system, each hop can suffer only 35 dB loss.

metres27.410

log2016.917.547

20log20log717.5Gain

each)dBi(47dBi94.3gainsantennaCombined

206.12

==

++=

++=

=

D

D

D




P-TR-005-M101-ver4 187

2. As an alternative to the back-to-back parabolic repeater, it is suggested

that a single flat, billboard reflector moved to the side of the path so that the

angle between the two paths is 120 degrees can provide the required signal

level. Determine the required size of the reflector if 1.35 metre diameter

antennas are used at each end of the link. (Hint: the path length from each

antenna to the reflector will be greater than 10 km).

2

205.46

m212

10

log205.46

6log208.338.421.117

)2/120log(cos20log20log4042.8117.1

117.1

74--261.170

dB130.5520log(7)5)20log(11.592.421

74-21dB70

GainsAntenna-21LossPath

km55.11sin(60)

10 nowlengthPath

=

=

=

−++=

+++=

==

=++==

+=

+=

==

A

A

A

A f

G

G

FSL FSL

-G FSL FSL

-G FSL FSL




P-TR-005-M101-ver4 188

3. The size requirements for the billboard reflector are found to be excessive.

It is known that such reflectors are more effective if they are placed nearer to

one end of the link. Accordingly a suitable site is found 600 metres to the

side of one end of the link. The reflection angle is now 90 degrees and the

two “hops” are 20 km and 0.6 km in length. Re-calculate the size

requirements for the billboard reflector.

2

206.22

m5.13

10

log206.22

3log208.338.422.96

)2/90log(cos20log20log4042.896.2

96.2

74--240.270

dB9.04120log(7)20log(0.6)92.42

dB135.320log(7)20log(20)92.41

74-21dB70

GainsAntenna-21LossPath

=

=

=

−++=

+++=

=

=

=++=

=++=

+=

+=

A

A

A

A f

G

G

FSL

FSL

-G FSL FSL

-G FSL FSL




P-TR-005-M101-ver4 189

4.

A town is surrounded by a ridge of hills. In order to provide a 15 km, 14

GHz hop into the town from a neighbouring village, it is necessary to install

a double billboard reflector on the ridge, some 400 m from its terminal. 1.2

metre antennas are used on the two terminals. The reflector is configured

such that reflection angles of 30 degrees and 50 degrees are obtained.

Determine suitable reflector sizes if the maximum path loss is 70 dB.

2m24.2

0.7log20

)50log(cos20log20)14log(408.428.91

anglelargerwithreflectorgConsiderin

dB8.91-84-107.4138.470

dBi0.2420log(14)20log(1.2)17.5GainAntenna

dB4.1074.0log2014log204.922

dB4.13815log2014log204.921

=

=

+++=

= +=

=++=

=++=

=++=

A

A

A

GG

FSL

FSL






P-TR-005-M101-ver4 191

3. A 30 km, 7 GHz path has a sharp, knife-edge obstacle, 10 m in height, 8

km from one end. The minimum k factor experienced (exceeded 99.9%

of the time) is 0.7. Determine suitable antenna heights.

For k factors of 1.33 and 0.7 we need to obtain the required clearance,

bearing in mind that the most significant obstacle could be either the knife

edge or earth bulge. Also, we need to note that, at k-factor of 0.7, the

clearance objective is 0 for the knife-edge or 0.3 F1 for the smooth earth

(depending on which is the most significant obstacle).

Considering earth bulge:

91.174

3.17midpointatradiuszoneFresnel

0.7)(k m25.2and1.33)(k m29.13

51

Bulge2

==

====

f

d k

d

Clearance requirement at k = 1.33 = 17.91+13.29 = 31.2 metres

Clearance requirement at k = 0.7 = 25.2+0.3x17.91 = 30.6 metres

Considering obstacle

m84.15

307

2283.171

:endonefrom8kmradiuszoneFresnel

0.7)(k 29.731.33);(k 20.38 :heightobstaclegConsiderin

)7.0(73.19);33.1(38.106373

)22)(8(500

:endonefromkm8atBulge

=

×

×=

==

=====

F

k k k

Clearance requirement at k=1.33: 15.84+20.38 = 36.22 metres

Clearance requirement at k=0.7 = 29.73 metres

Highest value is 36.22 metres and therefore antenna heights of 36.22 metres

should be adopted.




4. Determine the diffraction loss at a k factor of 0.5 if the antenna heights

determined by the result of question 3 are adopted.

Bulge at k=0.5:

112

metres1.38:nObstructio

m37.6 :10mAdd

m6.275.06373

)22)(8(500

21

+=

=×

d d

hv

λ

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