Wireless Telecommunication Systems [1 ed.] 9781118649053, 9781118625422

Citation preview

Wireless Telecommunication Systems

Wireless Telecommunication Systems Michel Terré Mylène Pischella Emmanuelle Vivier Series Editor Pierre-Noël Favennec

First published 2013 in Great Britain and the United States by ISTE Ltd and John Wiley & Sons, Inc.

Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms and licenses issued by the CLA. Enquiries concerning reproduction outside these terms should be sent to the publishers at the undermentioned address: ISTE Ltd 27-37 St George’s Road London SW19 4EU UK

John Wiley & Sons, Inc. 111 River Street Hoboken, NJ 07030 USA

www.iste.co.uk

www.wiley.com

© ISTE Ltd 2013 The rights of Michel Terré, Mylène Pischella and Emmanuelle Vivier to be identified as the authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. Library of Congress Control Number: 2013941767 British Library Cataloguing-in-Publication Data A CIP record for this book is available from the British Library ISBN: 978-1-84821-543-6

Printed and bound in Great Britain by CPI Group (UK) Ltd., Croydon, Surrey CR0 4YY

Table of Contents

Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ix

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xi

Chapter 1. Radio Propagation . . . . . . . . . . . . . . . .

1

1.1. Free-space loss link budget and capacity . . . 1.2. Link budget and free-space loss . . . . . . . . . 1.3. Linear expression of the Okumura–Hata model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4. Frequency, distance and propagation model . 1.5. Link budget and diffraction . . . . . . . . . . . . 1.6. Link budget and refraction . . . . . . . . . . . . . 1.7. Link budget and diffusion . . . . . . . . . . . . . 1.8. Frequency and time selectivity . . . . . . . . . . 1.9. Doppler effect . . . . . . . . . . . . . . . . . . . . . .

... ... . . . . . . .

9 11 13 15 18 20 21

Chapter 2. F/TDMA and GSM . . . . . . . . . . . . . . . . .

23

2.1. Maximum transmitter–receiver distance . 2.2. Extended maximum transmitter–receiver distance . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3. Reuse distance, interference reduction factor K and regular pattern . . . . . . . . . . . . . 2.4. Radio resources dimensioning in GSM. . . 2.5. Link budget in an isolated GSM cell . . . .

. . . . . . .

. . . . . . .

2 7

.....

24

.....

26

..... ..... .....

26 32 33

vi

Wireless Telecommunication Systems

2.6. Deployment of a GSM network along a highway . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7. GSM network dimensioning and planning in a rural area . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8. GSM network dimensioning and planning in an urban area . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9. SMS transmission in a GSM network . . . . . . 2.10. Frequency reuse pattern determination. . . . 2.11. Traffic and Erlang for GSM cell dimensioning . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12. Signal to noise plus interference ratio . . . . .

..

36

..

41

.. .. ..

44 46 48

.. ..

50 52

Chapter 3. CDMA and UMTS . . . . . . . . . . . . . . . . .

59

3.1. Spreading and CDMA . . . . . . . . . . . . . . . 3.2. Hadamard spreading codes: a perfect orthogonality between the users? . . . . . . . . . 3.3. Relation between Eb/N0 and the reception threshold in UMTS networks . . . . . . . . . . . . . 3.4. Required number of codes in CDMA . . . . . 3.5. UMTS link budget . . . . . . . . . . . . . . . . . 3.6. Cell breathing in UMTS networks . . . . . . 3.7. Intersite distance calculation in UMTS networks for different frequency reuse patterns . . . . . . . . . . . . . . . . . . . . . . . 3.8. Case study in UMTS networks . . . . . . . . .

....

63

....

64

. . . .

. . . .

69 70 71 77

.... ....

80 83

Chapter 4. OFDM and LTE . . . . . . . . . . . . . . . . . .

95

4.1. Useful throughput of an OFDM waveform . 4.2. OFDM and PAPR . . . . . . . . . . . . . . . . . . 4.3. Frequency selectivity and OFDM dimensioning . . . . . . . . . . . . . . . . . . . . . . . . 4.4. OFDM dimensioning . . . . . . . . . . . . . . . . 4.5. OFDM dimensioning for 4G networks and data rate evaluations. . . . . . . . . . . . . . . . . . . 4.6. LTE data rates evaluation . . . . . . . . . . . . 4.7. LTE link budget . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

.... ....

96 100

.... ....

104 106

.... .... ....

107 110 113

Table of Contents

vii

4.8. LTE link budget taking into account the number of users . . . . . . . . . . . . . . . . . . . . . . . . . . 120 4.9. Modulation-coding scheme relation, spectral efficiency and SINR in LTE networks . . . . . . . . . . . 123 Chapter 5. MIMO and Beamforming . . . . . . . . . . . 129 5.1. Beamforming and signal-to-noise ratio . . . . . . . 133 5.2. Space diversity and chi-square distribution . . . . 140 5.3. MIMO and capacity . . . . . . . . . . . . . . . . . . . . . 149 Chapter 6. UWB . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 6.1. Impulse UWB . . . . . . . . . . . . . . . . . . . . . . . . . 157 6.2. UWB and OFDM . . . . . . . . . . . . . . . . . . . . . . . 161 6.3. Link budget for UWB transmission . . . . . . . . . . 163 Chapter 7. Synchronization . . . . . . . . . . . . . . . . . . 167 7.1. Cramer–Rao bound . . . . . . . . . . . . . . . 7.2. Modified Cramer–Rao bound . . . . . . . . 7.3. Constant parameter estimation . . . . . . 7.4. Radio burst synchronization . . . . . . . . . 7.5. Phase estimation for QPSK modulation .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

168 170 170 174 176

Chapter 8. Digital Communications Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 8.1. Review of signal processing for signal-to-noise ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2. Review of digital modulations . . . . . . . . . . . . . 8.3. Review of equalization . . . . . . . . . . . . . . . . . . 8.4. Signal-to-noise ratio estimation . . . . . . . . . . . 8.5. ASK 2 modulation error probability. . . . . . . . . 8.6. Spectral occupancy, symbol rate and binary throughput . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7. Comparison of two linear digital modulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8. Comparison of two-PSK modulation and power evaluations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . .

179 179 180 181 184

. 187 . 189 . 191

viii

Wireless Telecommunication Systems

8.9. Zero-forcing linear equalization . . . . 8.10. Minimum mean square error linear equalization . . . . . . . . . . . . . . . . . . . . . 8.11. Noise factor in equipments . . . . . . . 8.12. Data rate calculations . . . . . . . . . .

........

194

........ ........ ........

196 200 203

Chapter 9. Erlang B Tables . . . . . . . . . . . . . . . . . .

205

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

209

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

211

Foreword

In the world of telecommunications, radio systems possess an essential characteristic that has led to their success: they are universally accessible and they allow mobility. This is a huge advantage compared with wired systems, which are limited to fixed access but, however, provide higher data rates and quality of service. Consequently, in order to decrease the gap between wired and wireless systems, the main research effort in radiocommunications aims to increase data rates while improving the quality of service. The main drawback of universal access is complexity, since very different techniques must be implemented and combined in order to efficiently exploit the frequency spectrum, while offering an easy and user-friendly interface. The technical challenges are, indeed, phenomenal. First, the radio transmit channel varies and needs to be estimated in real time to adapt the data rates accordingly. Second, coexistence of a large number of users and several services must be guaranteed. Finally, the frequency spectrum that can be exploited under favorable economic conditions and a tolerable respect level for the environment is limited, and its usage must be optimized. To cope with all these challenges,

x

Wireless Telecommunication Systems

we require theories, algorithms and the most evolved technologies, and must integrate them easily. Multidisciplinarity and complexity are two challenges present in radiocommunications teaching. Indeed both specialized and non-specialized students must be trained. Specialized students must be able to lead research, break new ground and complete developments in their area of expertise. After having understood and evaluated the available techniques, non-specialized students must define systems, integrate their subsets, carry out their deployments and follow their exploitations. The collection of solved problems proposed by Michel Terré, Mylène Pischella and Emmanuelle Vivier is the fruit of their experience in both industry and teaching. It is particularly suitable for non-specialists’ training, as it covers important and specific themes of radiocommunication networks, with an approach based on knowledge tests and integration checking. More generally, it can allow professionals, especially those who work as operators, to refresh their knowledge and remind them of some scale orders and key values. It can even encourage radiocommunication non-specialists to go into, or deepen their knowledge of, a particular area, starting from a concrete problem. Thus, through training, this book will contribute to the progress of wireless systems. Maurice BELLANGER Emeritus Professor at CNAM Member of the French Academy of Technologies

Preface

There is a strong interest for wireless systems within the population. In 20 years, these systems have undergone at least three major technological breakthroughs and it is now hard to imagine how life in society could be organized without them. There are many outstanding works that help to explain the fundamental concepts of wireless systems. We can find, for example, books on electromagnetism and antennas [LAH 11, PIC 09], digital signal processing [BEL 06], digital communications [BAU 07] or electronic radio receivers [PAL 10]. It seems to us that these books could be fruitfully complemented by a book of solved problems. Our purpose is to address the capacity offered by different wireless systems and to investigate transversal problems leveraging different scientific fundamental disciplines. Therefore, we propose a macroscopic approach for wireless systems and try to answer the questions of power, speed, multiple access cellular solutions and global organization of access networks.

xii

Wireless Telecommunication Systems

These questions are complex topics and raise several challenges. Regarding the power aspect, we can expect that it will be the total radiated power but it cannot be analyzed independently of the transmission band in which it will be used; we will then have to analyze the power spectral density emitted. Similarly, when considering transmission systems with multiple antennas, we must consider the total transmission power. The throughput definition is also very complex; is it an overall throughput of a wireless network, a raw data rate of a communication link, a final bit rate for a user, a theoretical capacity? It is clear that this question is difficult and can be approached in various ways. We propose a series of problems solved with the objective of establishing the elements of the link budget for different radio systems. This book should be considered a collection of practical studies, illustrating the fundamental principles of wireless systems; it is built, per chapter, on the major technological concepts of radio systems. Each chapter begins with a very brief introduction of the main theoretical results, followed by a series of problems solved. We tried to propose practical problems coming from the professional world using, to the greatest extent possible, the terminology and units of this specific field. More than 15 years of experience in teaching these subjects to students in colleges and universities has convinced us that this book could be useful not only for students but also for engineering companies in the telecommunications sector involved in the implementation of wireless systems.

Chapter 1

Radio Propagation

This chapter aims at establishing the link budget of a radio transmission. The objective is to link the cell coverage and the useful throughput of a wireless system with the transmission power. These calculations are performed based on the transmission power, and using an accurate model of the power scattering in the space between the transmitter and the receiver. This radio space is then called the propagation channel. Its definition may be complex, as we may, in some cases, choose to integrate some elements of the transmission and reception chains in it (especially the antenna systems). Consequently, the most important elements, which will be introduced in an incremental way in the proposed exercises, are mainly: the transmission power, the antenna gain, the noise power spectral density, the useful throughput and the communication channel’s capacity. In an environment with obstacles, the radiocommunication channel undergoes lots of attenuations. It is modeled by propagation losses, also called path losses. Given that the same signal is transmitted, the average power of the received signal decreases as a function of the distance, according to a distribution that depends on the

2

Wireless Telecommunication Systems

environment, called the path loss model. The path loss models are empirical and are obtained using radio measurements. We can distinguish between the urban environment model, the rural environment model or indoor environment models that are used inside buildings. The radio channel also undergoes a shadowing effect. In the multipath case, it introduces fading phenomena into the signal spectrum. The multipath channel may also create inter-symbol interference. Inter-symbol interference occurs depending on the channel selectivity. A radio channel is frequency-selective if its channel bandwidth is large with respect to its coherence bandwidth or, equivalently, if the maximum transmission delay introduced by the channel is large with respect to the modulation symbol time. In this case, equalization techniques must be used at the receiver in order to recover the transmission symbols. The channel is time-selective if it varies faster than the modulation symbol time. This phenomenon is due to the mobile’s speed and is called the Doppler effect. It happens when the coherence time of the channel is small with respect to the symbol time. The channel is then a “fast fading” channel. If the channel varies during the training sequence, it is no longer possible to estimate the channel impulse response, and equalization can no longer be performed. 1.1. Free-space loss link budget and capacity In this first simple introductory problem, we consider free-space propagation for an earth-to-satellite radio link. The carrier frequency is equal to 6 GHz, the transmitter power is equal to 4 W, the transmission bandwidth is equal to 200 kHz, the earth station antenna is a parabolic antenna with a diameter equal to 80 cm, the satellite antenna is a

Radio Propagation

3

parabolic antenna with a diameter equal to 40 cm and both antennas have an efficiency η = 0.5. We consider only the free-space loss. The distance between the Earth and the satellite is equal to 36,000 km. Modulation requires a + 20 dB Eb/N0 ratio. The utilized constants will be: – k = 1.38 × 10–23 JK–1, Boltzmann constant; – T = 300 K, receiver noise temperature. 1) Calculate the equivalent isotropic radiated power (EIRP) of the earth station. 2) Calculate the free-space loss. 3) Calculate the satellite-received power; the result should be in dBW. 4) Calculate the signal to thermal noise power ratio in the receiving band. 5) Calculate the maximal throughput of this link, considering only the Eb/N0 constraint. 6) Now considering the communication channel capacity, what could you conclude for this transmission? 7) How could we calculate the electric field from the received power? Solution 1) The transmitted power is equal to Pe = 4 W, it is often easier to give this power in dBW. For this purpose, we just have to consider 10 × log10( ) of the linear value of the power. In this case, we obtain: Pe,dBW

10

log 10 (4) 6 dBW

4

Wireless Telecommunication Systems

Another unit, often used, is the dBm: then we have to consider the logarithm of the power in mW:

Pe,dBm

10

log 10 (4,000) 36 dBm

The antenna gain Ge is given, for a parabolic antenna with a diameter equal to De, by:

Ge



πDe 

2

As for the power, it can be useful to give this gain in dB, we obtain then:

Ge,dBi

10

log 10 (Ge )

31 dBi

The EIRP, defined by the PeGe product, is then equal to: 6 dBW + 31 dBi = 37 dBW 2) The free-space loss, at a distance distance from the transmitter, for a link with a carrier frequency with a wavelength λ is defined by (see problem 1.2):  1 = L 4πd

2

This corresponds, always in dB, to: 2

 λ  LdB = −10 × log10   = 199.1 dB  4π d 

3) The receiving antenna gain is equal to Gr



πDr 2 , 

corresponds to 25 dBi. The received power is then given by:

Pr

Pe Ge Gr L

it

Radio Propagation

5

Finally, doing the derivation in dBW, we obtain:

Pr,dBW

6 dBW

31 dBi

25 dBi – 199.1 dB

–137.1 dBW

4) The thermal noise power spectral density, given in W/Hz, is defined by: N0 = kT Given in dB, in this case, we obtain:

N0,dBJ

log 10 (N0 )

10

–203.8 dBJ

The transmission bandwidth is equal to B = 200 kHz, and the thermal power in this band is then equal to N0B = kTB. The signal-to-noise power ratio Γ = then equal to: dB

10

log

Pr given in dB, is N0 B

Pr N0 B

This yields to: dB

–137.1 dBW

203.8 dBJ – 53 dBHz

13.7 dB

5) The useful throughput Rb corresponding to the received power and to the thermal noise power can then be obtained as follows: Pr = RbEb In this last equation, Eb stands for the useful received bit energy just before the decision process.

6

Wireless Telecommunication Systems

We can then calculate the

Pr N0

R b Eb N0

Pr ratio, giving: N0

Pe Ge Gr kT L

We obtain then the maximum useful throughput of the link:

Rb

Pe Ge Gr 1 kT L Eb ⁄N0

Given in dBHz, this throughput is equal to:

R b,dBHz

10

R b,dBHz

–137.1 dBW

log 10 (R b )

then: 203.8 dBJ – 20 dB

46.7 dBHz

This corresponds to a throughput (the similarity between the throughput in dB and in linear is due to the values used in this exercise):

Rb

46.7 kbits/s

6) The channel capacity C is defined as follows:

 R E  C = B × log 2  1 + b b  B N0   leading to: C = 923 kbits/s We can note that the channel capacity is higher than the maximal throughput given by the link budget. The solution proposed in this exercise is then correct and could be implemented. When the channel capacity is lower than the

Radio Propagation

7

maximal throughput given by the link budget, some parameters of the system have to be changed. It could typically be the case with a narrow transmission bandwidth associated with a high throughput. 7) The received power can be obtained by calculating the flow of the Poynting vector through the equivalent area of the receiving antenna. By definition, we have: ∧H

Far from the transmitting antenna, linked by:

and

fields are

E = 120π H The equivalent area Aeq of an antenna is linked to its gain Gr by:

Aeq

2 G 4π r

We can then obtain the electric field from the received power with: 2

Pr

S

Aeq

E 120π

2 G 4π r

1.2. Link budget and free-space loss From the propagation formula giving the received power, Pr, with respect to the transmitted power, Pe, calculate the free-space loss in dB, provide the loss in LdB and show that it is equal to:

LdB

32.44 20

log 10 (d)

20

log 10 (f)

8

Wireless Telecommunication Systems

where f is the frequency of the transmission in MHz and d is the transmitter–receiver distance in km. Solution The power density transmitted by an isotropic antenna is uniformly spread over a sphere surrounding the antenna. At a distance d from this antenna, this power density (1 W/m²) is given by:

Pe

 (d )

4πd2

Knowing that the equivalent area Aeq of an antenna is linked to its gain Gr by the following formula [PIC 09]:

Gr

4π

2

Aeq

where λ is the wavelength of the transmitted signal (λf = c, c is the light speed), integrating the antenna gain Ge, we obtain:

Pr

Pe Ge Gr

Pe A 4πd2 eq

4πd 

2

Pe Ge Gr d2

4πf

2

c

The free-space loss term L is then equal to:  4π  L = d2 f 2    c 

2

Having c = 3 × 108 m/s = 300 m/μs = 0.3 km/μs, we can express L with the following equation:  4π  L = d2 f 2    0.3 

2

with f in MHz and d in km.

Radio Propagation

9

Converting this result in dB, we obtain: 4π 0.3

2

LdB

10

log 10 d2 f 2

LdB

20

log 10 (d )

20

log 10 (f )

20

LdB

20

log 10 (d )

20

log 10 (f )

32.44

log 10

4π 0.3

REMARK 1.1.– This free-space loss equation is correct if the line of sight between the transmitter and the receiver is not obstructed by any obstacles. But it is also necessary to have no obstacle in the first Fresnel ellipsoid between the transmitter and the receiver. This first ellipsoid has a small diameter equal to √ . For instance, if d = 1 km, we need to have a 18 m freespace area around the line of sight between the transmitter and the receiver for a transmission with a carrier frequency equal to 900 MHz and 13 m for 1,800 MHz. We notice these requirements are not often verified and we have frequently a ground diffraction to consider. 1.3. Linear expression of the Okumura–Hata model The Okumura–Hata propagation model is a very common model because it easily and rapidly gives reliable results. It was defined by Hata in 1980, from empirical measures done by Okumura in 1968 in Tokyo area. In an urban area, the loss L can be approximated by the following formula (without considering the mobile height):

LdB = 69.55 + ( 44.9 − 6.55 × log10 ( hb ) ) × log10 ( d ) +26.16 × log10 ( f ) − 13.82 × log10 ( hb )

10

Wireless Telecommunication Systems

The validity assumptions for this model are the following: the base station antenna height hb is expressed in m and is between 30 and 200 m; the distance d between the transmitter and the receiver is expressed in km and is between 1 and 20 km; the carrier frequency f is expressed in MHz and is between 150 MHz and 1.5 GHz. Considering a carrier frequency set at 900 MHz and a base station height at 40 m, give the value for L (in linear) as a function of d. Solution Starting from the formula giving LdB, we insert the known inputs, f and hb:

LdB = 69.55 + ( 44.9 − 6.55 × log10 ( 40 ) ) × log10 ( d ) +26.16 × log10 (900) − 13.82 × log10 (40)

LdB = 131.36 + 34.4 × log10 ( d ) As LdB

10

log 10 (L), we have L

LdB = 10 10

and then:

131.36+34.4×log10 ( d ) 10 L = 10

L = 1013.13 × 10

3.44×log10 ( d )

L = 1013.13 × d 3.44

L = 1.35 × 1013 × d 3.44 The power of d is often called “propagation coefficient” (denoted as γ). Here, we have γ = 3.44. In free space, γ = 2.

Radio Propagation

11

1.4. Frequency, distance and propagation model A mobile terminal transmits with an EIRP equal to 21 dBm toward its base station. The Okumura–Hata model is used as a propagation model for an urban area, for smallto medium-sized cities. We consider the following parameters: – transmission frequency: 900 MHz (global system for mobile communications (GSM)); – base station antenna height: 30 m; – mobile antenna height: 1.5 m. 1) Calculate the path loss in dB if the distance between the mobile and the base station is 500 m or 1 km. Calculate the average received power. 2) Do the same calculation with a transmission frequency equal to 2 GHz (universal mobile telecommunications system (UMTS)). Compare them and conclude. REMARK 1.2.– In Hata’s formula, the frequency is expressed in MHz. This formula should normally not be used for frequencies more than 1.5 GHz. Nevertheless, we allow this approximation in this exercise, for the purposes of comparison. Solution The Hata propagation model in urban area, for small- to medium-sized cities, is given by [GOL 05]: LdB (d)=A+B× log10 (d)-a1 where the parameters are defined as follows: A = 69.55 + 29.16 × log10 ( f c ) − 13.82 × log10 ( hb )

12

Wireless Telecommunication Systems

B = 44.9 − 6.55 × log10 ( hb )

a1 = 1.1×log10 fc – 0.7 ×hm – 1.56×log10 fc –0.8 where d is the distance between the transmitter and the receiver in km, fc is the central carrier frequency, expressed in MHz, and hb and hm are the respective heights of the base station and the mobile, expressed in m. 1) For fc = 900 MHz, we get A = 126.42; B = 35.22 and a1 = 0.016; consequently, the path loss equation in dB is: LdB = 126.43 + 35.22× log10(d) Thus, the path loss at 500 m is equal to LdB(0.5) = 115.80 dB and the path loss at 1 km is equal to LdB(1) = 126.43 dB. The received power is equal to the EIRP minus the path loss: Pr,dBm =EIRPdBm – LdB (0.5)=21 – 115.8= –94.8 dBm for a distance of 500 m and: Pr,dBm =EIRPdBm – LdB (1)=21 – 126.4= –105.43 dBm for a distance of 1 km. 2) For fc = 2,000 MHz, we obtain A = 135.49, B = 35.22 and a1 = 0.047. The path loss equation in dB is then: LdB = 135.44 + 35.22× log10(d) Thus, the path loss at 500 m is equal to LdB(0.5) = 124.84 dB and the path loss at 1 km is equal to LdB(1) = 135.44 dB. Consequently, the received power is: Pr,dBm =EIRPdBm – LdB (0.5)=21–124.84= –103.84 dBm

Radio Propagation

13

for a distance d = 0.5 and: Pr,dBm = EIRPdBm – LdB (1)=21–134.91= –113.91 dBm for a distance d = 1. To conclude, the path loss increases when the carrier frequency increases. As a consequence, radio propagation is less effective when the carrier frequency increases, thus generating lower received power levels. 1.5. Link budget and diffraction We consider a microwave link transmission. The carrier frequency is equal to 6 GHz, we consider 20 cm diameter parabolic antennas with efficiencies equal to 0.5. Only the free-space loss is taken into consideration. The distance between the transmitter and the receiver is equal to 40 km, and the antennas are located on 25 m high “mats”. At an equal distance between the transmitter and the receiver, we find a 20 m high narrow “obstacle”. The transmitted power is Pe = 100 mW.

1) Does the obstacle interfere with the transmission? 2) What is the impact of this obstacle on the link budget? Calculate the received power in dBm. 3) Same questions for a 30 m high obstacle. Solution 1) A radio link is not obstructed if the line of sight is free of obstacles and if the first Fresnel ellipsoid has no interference. The maximal “rayon” ρmax of this first Fresnel ellipsoid is observed at equal distance between the transmitter and the receiver. Its value depends on the

14

Wireless Telecommunication Systems

distance d of the radio link and on the wavelength λ of the transmission [LAV 97]: ρmax =

1 d 2

In this problem, we have f = 6 GHz, then λ =

c = 5 cm f

and we have d = 40 km, then: ρmax = 22.36 m The obstacle interferes with the first Fresnel ellipsoid. The link cannot be considered as corresponding to a freespace loss context. 2) To calculate the diffraction impact due to a narrow obstacle, we have to calculate the parameter v defined by: v=h

2 1 1   +  λ  d1 d 2 

In this expression, h stands for the distance between the line of sight and the top of the obstacle. This distance is negative if the top of the obstacle is under the line of sight and positive in the other case. Here, we find v = –0.31. When the parameter v is greater than –1, we can approximate the diffraction fading with the function J(v): J ( v ) = 6.4 + 20 × log10  v 2 + 1 + v   

In our case, we will find: J ( v ) = 3.7 dB

Radio Propagation

15

This value has to be subtracted from propagation losses in the link budget. The received power is then equal to: Pr,dBm = Pe,dBm + Ge,dBi + Gr,dBi – LdB – J(v) The transmission parabolic antenna gain is given by: πD Ge =  

2

We obtain then: Ge,dBi = Gr,dBi = 19 dBi The free-space propagation loss at 40 km is equal to LdB = 140 dB, then finally: Pr,dBm = –85.7 dBm 3) If we consider now a 30 m high obstacle, the term J(v) becomes: J ( v ) = 9.1 dB

The received power is then equal to Pr,dBm = –91.1 dBm. 1.6. Link budget and refraction We consider a microwave link transmission. The carrier frequency is equal to 6 GHz, and we have parabolic antennas located at h meters from the ground. The distance between the transmitter and the receiver is equal to 50 km. We consider an index “gradient” g = –40 µN/km. The earth “rayon” RT is considered to be equal to RT = 6,370 km.

16

Wireless Telecommunication Systems

What should be the antenna height in order to avoid the ground diffraction? Solution We first calculate the half rayon of the Fresnel ellipsoid at equal distance between transmitter and receiver: ρmax =

1 d 2

here f = 6 GHz then λ =

c = 5 cm and d = 50 km, then: f

ρmax = 25 m We consider a horizontal propagation of electromagnetic waves. Then, we have to correct the earth rayon. The “fictive” earth rayon Rf is given by Rf = then:

1

, we find

1 +g RT

Rf = 8,548 km To avoid a contact between the first Fresnel ellipsoid and the ground, the distance between the horizontal line of sight and the ground must be at least equal to ρmax. We simply write: 2 2 d ( R f + h) 2 =   + R f + x 2

(

)

Radio Propagation

17

Figure 1.1. First Fresnel Ellipsoid

Considering that Rf>> h and Rf>>x, second-order terms can be neglected, the equation becomes then: R2f 1+2

h d = Rf 2

2

+R2f 1+2

x Rf

2

2hRf =

d + 2xRf 4

In the limit case, we have x = ρmax, then: 2

h=

d +ρ 8Rf max

we obtain: h = 61.5 m Then, we have to locate antennas at high positions and use high pylons in order to reach this 61.5 m value.

18

Wireless Telecommunication Systems

1.7. Link budget and diffusion We study a microwave link transmission, with a transmitter–receiver distance d = 50 km, and we consider Table 1.1 [BOI 83, LAV 97, MAR 98], representing the rain intensity (R) for an outage probability. Probability (%)

Temperate area (R in mm/h)

0.001

78

0.002

62

0.005

41

0.01

28

0.02

18

0.05

11

0.1

7.2

0.2

4.8

0.5

2.7

1

1.8

Table 1.1. Probability of exceeding a rain intensity

This table should be analyzed as follows: a 0.1% probability for a temperate area means that the rain intensity will be higher than 7.2 mm/h for 0.1% of the time. This corresponds to 520 min per year. We consider a vertical polarization with the parameters given in Table 1.2. f (GHz) 6

0.00155

1.265

12

0.0168

1.2

Table 1.2. Parameters for calculating the lineic attenuation for vertical polarization

Radio Propagation

19

Using the Lin formula, calculate the margin that we should apply on the link budget in order to have a reliability of the link equal to 99.99%. Carry out the derivations for f = 6 GHz and f = 12 GHz. Lin’s formula: k ( L, R ) =

1 R − 6.2 1+ L 2636

where L is the effective distance where a rain intensity R is observed. Solution The 99.99% reliability requirement leads us to take very unlikely events into consideration. Then, we have to introduce important power margins into the link budget and overdesign the transmitter power amplifier. Then, we have to consider the limit case of an important rain with an outage probability of 0.01% for a temperate area. The table analysis gives us R = 28 mm/h. The Lin formula gives k = 0.7, the effective distance of the link is then equal to d e = γ × d = av × R bv × d = 35.4 km. For the 6 GHz frequency, we have γ = 0.105. For the 12 GHz frequency, we have γ = 0.916. Rain fadings are then equal to the following: – For the 6 GHz frequency, fading = 3.7 dB; – For the 12 GHz frequency, fading= 32.4 dB.

20

Wireless Telecommunication Systems

These two fadings represent margins that have to be taken into consideration in the link budget. If the 3.7 dB value can be envisaged, the 32.4 dB value will be more difficult to consider as a power back-off of the transmitter amplifier. In conclusion, the required reliability will be difficult for the 12 GHz carrier frequency case. 1.8. Frequency and time selectivity Let us consider a radio channel with the following parameters: – the Doppler spread bandwidth is Bd = 150 kHz; – the maximum multipath delay is equal to Tm = 3 μs. 1) Calculate the coherence time.

channel

coherence

bandwidth

and

2) Explain in which cases the channel is not frequencyselective. 3) Explain in which cases the channel undergoes slow fading. Solution 1) The coherence time is approximately equal to Tco = 1/Bd = 6.67 μs. The coherence bandwidth is equal to Bc = 1/Tm = 333 kHz. 2) The channel is not frequency-selective if B < Bc, so the channel bandwidth is lower than 333 kHz. 3) The channel undergoes slow fading if Ts < Tco, so the symbol time is lower than 6.67 μs.

Radio Propagation

21

1.9. Doppler effect A GSM system is operating at a central carrier frequency fc = 1.8 GHz. It serves a mobile terminal that is located in a rural environment, where the multipath delay is equal to 10 μs. 1) Calculate the channel coherence bandwidth. 2) The GSM channel bandwidth is B = 200 kHz. Is the channel frequency-selective? 3) Calculate the coherence time for a mobile having a speed of 50 km/h. 4) The data rate in GSM is equal to R = 271 kbits/s. Calculate the corresponding symbol time. 5) Is it possible to recover the transmitted signal using equalization? Solution 1) Bc= 1/Tm = 100 kHz. 2) B > Bc, so the channel is frequency-selective. There will be intercell interference. 3) The maximum Doppler frequency is fd = fcv/c = 83.33 Hz. The coherence time is approximately equal to Tco = 1/(2fd) = 6 ms. 4) The symbol time is Ts = 1/R = 3.69 μs. 5) As Ts < Tco, the channel does not vary with respect to the time during the transmission of a symbol. Consequently, equalization can be used at the receiver in order to recover the transmitted signals by suppressing intercell interference. This is even more accurate as the channel remains constant over several consecutive symbols.

22

Wireless Telecommunication Systems

Starting from Chapter 2, some hardware characteristics are expressed in dB, according to the notation used in many industrial technical sheets. The reader should determine whether this is a power level (in which case dB means dBW) or a loss or a gain.

Chapter 2

F/TDMA and GSM

The second-generation of cellular systems, Global System for Mobile Communication (GSM), General Packet Radio Service (GPRS) and Enhanced Data rates for GSM Evolution (EDGE), use a multiple access technique based on time division multiple access (TDMA) and frequency division multiple access (FDMA). The users served by the same cell are multiplexed on several sub-bands called transceiver receiver (TRX), with a bandwidth 200 of kHz and coupled in uplink and downlink. In the time space, the transmission within a frame is separated into eight time slots. In GSM, a user transmits and receives in one time slot of a TRX, which is dedicated to it during its whole communication. In GPRS and EDGE, the same user may be allocated up to eight time slots in the same TRX depending on their data rate needs and on those of the other mobile terminals. Adjacent cells do not use the same TRX subsets, and are consequently separated in frequency, which allows us to avoid high intercell interference levels. However, because the total bandwidth is limited, it is necessary to reuse the same frequency bands beyond a given distance. The frequency reuse factor and the corresponding frequency scheme have

24

Wireless Telecommunication Systems

an influence on the network performances. Network dimensioning should depend not only on the coverage, but also on the ratio between the useful signal power and the interference power. Transmission in GSM is performed in circuit-switched mode, whereas transmission in GPRS and EDGE is performed in packet-switched mode. In circuit-switched mode, dimensioning should be based on Erlang distributions. In packet-switched mode, the sum data rate to be transmitted should be considered. GSM

Circuit-switched numerical system. (Licensed) frequency bands: 880–960 MHz (↓↑) 1710–1880 MHz (↓↑) Transmission power of the terminals: ≤ 2W Maximum transmitter–receiver distance: 35 km FDD Multiple access: FDMA/TDMA National coverage (handover) International roaming Provided services: Voice/data (up to 14.4 kbits/s)/SMS

GPRS

Packet-switched numerical system, using GSM radio access Provided services: data (up to 171.2 kbits/s in theory)

EDGE

Packet-switched numerical system (and also circuit-switched numerical system, but rarely used), using GSM/GPRS network architecture Provided services: data (up to 473.6 kbits/s in theory)

Table 2.1. Second-generation cellular networks’ characteristics

2.1. Maximum transmitter–receiver distance On its uplink beacon channel, each GSM base station periodically gives the mobiles the opportunity to access the network (e.g. for outgoing calls, answering incoming calls or

F/TDMA and GSM

25

transmitting SMS). This can be done by sending a so-called “access burst” to the network in determined time slots of duration 577 µs that constitute the random access channel (RACH) [LAG 99] logical channel. The access request must reach the base station in the appropriate time slot; actually, it cannot overflow into the next time slot (it would increase the collision probability). Mobiles wait for the base station beacon channel’s clock “top” that indicates the beginning of the 577 µs duration and send their 88 bit long access burst at 270.8 kbits/s (this message is submitted to slotted-Aloha protocol collision performances). Calculate the maximum distance between the base station and the mobile station. For this purpose, it will be useful to first calculate the time necessary to transmit the message over the air. Solution The mobile situated at the distance d (expressed in km) from the base station waits for τ = d/c = d/0.3 µs before receiving the base station beacon channel’s clock “top”. Then, assuming an immediate reception/transmission switch, the mobile transmits its 88 bit long access burst at 270.8 kbits/s, during τ′ = 88/270.8 ms = 325 µs. This message reaches the base station after τ µs.

Figure 2.1. Access burst reception in the appropriate time slot

26

Wireless Telecommunication Systems

Consequently, it is necessary to ensure: τ + τ′ + τ < 577 µs

2× d + 325 < 577 0.3 577 − 325 d < 0.3 × 2 d < 37.8 km 2.2. Extended maximum transmitter–receiver distance We wish to increase the size of cells (e.g. in seaside areas). For this purpose, the mobiles that attempt to access the network are authorized to use two time slots instead of only one (as seen in the previous exercise) to transmit the same 88 bit long access burst at 270.8 kbits/s. What is the value of the extended maximum distance between the base station and the mobile? Solution Inspired by the previous exercise. We now have 252 + 577 = 829 µs for radio propagation, that is 248.7 km for round-trip propagation at light speed. Consequently, the extended maximum distance between the base station and the mobile becomes 124.3 km. 2.3. Reuse distance, interference reduction factor K and regular pattern We call the smallest group of K cells that contains once and only once all the frequency carriers allocated to the network operator a cellular pattern. This pattern is repeated over the entire area to be covered. The minimum distance

F/TDMA and GSM

27

between two base stations using the same frequency carrier is the so-called “reuse distance”, usually denoted by D. The larger the pattern, the greater the reuse distance expressed in the number of cells. The purpose of this exercise is to illustrate that it is advantageous to use patterns of size K, so-called “regular” because their size K is such that K = i² + i × j + j² with i and j positive or null natural integers. The first values for K that verify this relation are 1, 3, 4, 7, 9, 12, 13, 16, 19, 21, 25, 27, etc. In this exercise, we focus on the following patterns: the regular K = 7 pattern and the non-regular K = 8 pattern. We assume that the base stations are situated at the center of hexagonal cells. The maximum base station–mobile distance is denoted by dmax. Consequently, each cell is composed of six equilateral triangles of side dmax and altitude

3 / 2 dmax (see Figure 2.2).

Figure 2.2. Distances in a hexagonal cell

Base stations are equipped with omnidirectional antennas. The propagation coefficient is denoted by γ (see exercise 1.3). 1) In a K = 7 pattern as illustrated below (the central cell is shaded vertically, surrounded by the pattern’s six other cells; the interfering cells are shaded horizontally), calculate the reuse distance D7 as a function of dmax.

28

Wireless Telecommunication Systems

Deduce an approximate value of the signal-to-interference ratio (SIR7) at the cell edge.

Figure 2.3. Pattern of size K=7: reuse distance

2) In a K = 8 pattern as illustrated below (the central cell is shaded vertically, surrounded by the pattern’s seven other cells; the interfering cells are shaded horizontally), calculate the reuse distance D8, D8’ and D8’’ as a function of dmax. Deduce an approximate value of SIR8 at the cell edge. 3) Draw conclusions on the benefits of using patterns of regular size. Consequently, it will always be possible to use the following formula:

SIR =

1 6

(

3K

γ

)

F/TDMA and GSM

29

Figure 2.4. Pattern of size K=8: reuse distance

Solution 1) We can easily check that the six nearest interfering base stations from the central base station are situated on a circle of rayon D7. They are situated on the so-called “first ring of interferences”. Using Pythagorean theorem, we have: 2

  3 + 4 × dmax  2  

D7 =

(3 dmax )

D7 =

(9 + 12)dmax =

21 dmax

2

30

Wireless Telecommunication Systems

We note that 21 = 3 × 7, and generally, we can show that for all patterns of regular size K, we have DK = 3 × K dmax. To calculate the SIR at the cell edge, it is necessary to calculate the power of the useful signal received by a mobile situated at the cell edge, and the power of the interfering signals received by this mobile. Assuming that the carrier frequency and the base stations’ height (identical for all the base stations) are known, the propagation model can be written as follows: L (d ) = α d γ

Therefore, the power of the useful signal received by a mobile situated at the cell edge is: P G G C = Tx Tx Rx γ α ( d max )

The power, received by this mobile, coming from the six interfering base stations of the first ring of interferences is, assuming that the six base stations–mobile distances can be approximated by D7: P G G I = 6 × Tx Tx Rx γ α ( D7 )

Consequently: γ

SIR 7 =

C 1  D7  = I 6  d max 

SIR 7 =

1 6

( 21)γ

F/TDMA and GSM

31

In general, for all patterns of regular size K, we have: SIR =

1 6

(

3× K

)γ

2) For a K = 8 pattern, one can easily check that the seven nearest interfering base stations from the central base station are no longer situated on a circle. Again, using Pythagorean theorem, we have: 2

2   9 3 84 D8 =  dmax  +  dmax  = dmax = 21 dmax 4 2   2  2

2   3 3 84 D8' =  dmax  +  5 × dmax  = dmax = 21 dmax 2 4 2   

D8'' =



(3 dmax )2 +  6 × 

2  117 3 dmax  = dmax 2 2 

Therefore, we obtain: PTx GTx GRx

I = 4×

α

(

21 d max

  = × γ   α ( d max )  PTx GTx GRx

)γ 4

+ 2×

γ

( 21)

+

PTx GTx GRx

γ  117  d max  α  2 

  γ  117  

2 × 2γ

(

)

and: SIR 8 =

1 4

+

γ 117 ) ( = ( 21 ) × γ γ 4 × ( 117 ) + 2 × ( 84 ) γ

2 × 2γ

( 21)γ ( 117 )γ

32

Wireless Telecommunication Systems

3) It has been seen in the previous question that the four nearest interferers with the K = 8 pattern are situated at the same distance from the central base station as the six interferers with the K=7 pattern: 21 dmax . This means that it will be necessary to take the same precautions (margins, geographical distance, etc.) toward these four interferers in the K = 8 pattern as in the K = 7 pattern. In addition, as these interferers are the nearest, they contribute to the main part of the interferences, even if there are “only” four instead of six. Consequently, when we do not choose a pattern with a regular size: (1) the SIR does not increase so much, (2) it is necessary to guard against the worst interferers as if we were in a pattern of lower but regular size and (3) the interferers are not uniformly spread on the interfering rings. As a conclusion, there is no major interest in using patterns of non-regular size. 2.4. Radio resources dimensioning in GSM An operator wants to deploy a GSM network. He/she has acquired two bandwidths: a 5 MHz large bandwidth, located in the 890–915 MHz frequency area, and a 5 MHz large bandwidth, located in the 935–960 MHz frequency area. The operator wants to serve at least 40 users per cell simultaneously. The first time slot of the first chosen TRX is dedicated to the broadcast channel (in downlink) and to the RACH channel (in uplink). Moreover, we assume that one time slot per TRX approximately is dedicated to signaling channels, both in downlink and uplink. 1) Calculate the number of TRX required per cell. 2) Deduce the necessary bandwidth per cell.

F/TDMA and GSM

33

The operator can deploy its network either with a frequency reuse factor K = 3 or with a frequency reuse factor K = 7. 3) In which case is the interference level received by the users higher? Justify your answer. 4) Calculate, in both cases, the required bandwidth over the whole network to simultaneously serve at least 40 users per cell. Conclude with the feasibility of both deployments. Solution 1) There are seven traffic channels (TCH) per TRX, except in the first TRX, which contains only six TCH channels.  (40 + 1)  Consequently,   = 6 TRX are required, where . is  7  the upper integer value. 2) One TRX bandwidth is equal to BTRX = 200 kHz in uplink and in downlink. The required bandwidth per cell is Bcell = 6 BTRX = 1.2 MHz in uplink and Bcell = 1.2 MHz in downlink. 3) Interference is higher when the frequency reuse factor is set to K = 3 because interfering cells are closer. 4) The total required bandwidth for the whole network is B = K × Bcell in uplink and in downlink. It is equal to B = 3.6 MHz if K = 3 and B = 8.4 MHz if K = 7. Since the operator only possesses a 5 MHz bandwidth, the K = 7 case cannot be deployed in practice. 2.5. Link budget in an isolated GSM cell Let us consider an isolated GSM cell (that undergoes no interference). The following equipment parameters are provided:

34

Wireless Telecommunication Systems

– base station parameters: - antenna gain: GAdB = 15 dBi; - cable loss: LOdB = 2 dB; - maximum transmission power: Pe,dBm = 43 dBm; - receiver power threshold: SRECdBm = –112.5 dBm. – mobile terminal parameters: - antenna gain: GAdB = 2 dBi; - cable loss: LOdB = 2 dB; - maximum transmission power: Pe,dBm = 21 dBm; - receiver power threshold: SRECdBm = –102 dBm. 1) Calculate the maximum uplink path loss. 2) Calculate the maximum downlink path loss. 3) Choose the smallest path loss to perform dimensioning. Deduce that one of the transmission powers can be decreased. Calculate the corresponding new power value. 4) The path loss model is given by: LdB (d) = 126.43 + 35.22 × log10 (d), where d is expressed in km. Calculate the maximum cell size (coverage). Solution The link budget in an isolated cell is obtained by applying the two following steps, in uplink and in downlink: – calculate the equivalent isotropically radiated power (EIRP). It is equal to the actual transmitter radiated power, considering its hardware gains and losses. It is given by the following formula: EIRPdBm = Pe,dBm + GATx,dB  LOTx,dB

F/TDMA and GSM

35

– calculate the maximum path loss allowing to reach the receiver power threshold: LdB,max = EIRPdBm – RECdBm The cell size is finally obtained by taking the minimum path loss between uplink and downlink, since transmissions on both ways must be provided. 1) In uplink, we get: EIRPdBm = Pe,dBm + GATx,dB  LOTx,dB = 21 + 2 - 2 =21 dBm

Consequently, the maximum path loss is: LdB,max = EIRPdBm – RECdBm = 21 + 112.5 = 133.5 dB 2) In the downlink, the EIRP is equal to: EIRPdBm = Pe,dBm + GATx,dB  LOTx,dB = 43 + 15  2 = 56 dBm

The maximum path loss is then: LdB,max = EIRPdBm – RECdBm = 56 + 102 = 158 dB 3) The smallest path loss is obtained in uplink, which is the most limiting way. Consequently, the base station transmission power can be decreased. Its value is obtained by balancing the link budget. The downlink EIRP is given by the minimum path loss: EIRPdBm = LdB,max + RECdBm = 133.5 – 102 = 31.5 dB The base station transmission power is then equal to: Pe,dBm = EIRPdBm – GATx,dB + LOTx,dB = 31.5 – 15 + 2 = 18.5 dBm

We can see that the base station transmission power is highly decreased.

36

Wireless Telecommunication Systems

4) The coverage is obtained by inverting the path loss formula when the path loss is at its maximum value:

d = 10(133.5−126.43) / 35.22 = 1.59 km 2.6. Deployment of a GSM network along a highway A cellular network operator wishes to deploy a GSM network along a highway only (not outside). Input data are the following: – carrier frequencies are in the 900 MHz GSM band; – only 16 bidirectional carrier frequencies are available; – four TRX are used in each cell; – the Quality-of-Service (QoS) (in terms of possible blocking probability, that is the rate of rejected new calls) is set to 3%; – the path loss L at the distance d from the transmitter is obtained using the Okumura–Hata formula. In dB, it gives (with hb = 30 m):

LdB = 69.55 + (44.9 − 6.55 × log10 (hb)) × log10 (d ) +26.16 × log10 ( f ) − 13.82 × log10 (hb) – the operating point is set to 10 dB; – an 8 dB margin is taken to fight against shadowing and fading; – the following base stations equipments characteristics are considered: - low noise amplifier gain: 2 dB; - antenna gain: 20 dBi; - cables, etc.: 2 dB;

F/TDMA and GSM

37

- reception diversity: 2 dB; - duplexer: 3dB; - power coupler/splitter: to be deduced from the number of TRX; - maximum transmission power: PTx = 60 W; - sensitivity after radio frequency (RF) chain1: –145 dB; – the following mobile equipment characteristics are considered: - antenna gain: 1.5 dBi; - cables and connectors: 1 dB; - sensitivity after RF chain: –105 dBm. 1) Calculate the coupler/splitter.

loss

introduced

by

the

power

2) Establish and balance the link budget. 3) Calculate the maximum distance dmax between a base station and a mobile. 4) Propose a location for the base stations along the highway, as well as a frequency plan. 5) We aim at calculating the SIR for a mobile situated at the cell edge. Only the co-channel interferences are considered. It is assumed that the cells overlapping areas are null. How many interferers have to be considered? Calculate both the useful signal power and the interfering power received by the mobile. Deduce the SIR. 6) What is the capacity of each cell (in Erl)?

1 In this exercise, the sensitivity after RF chain is considered. It does not take into account gains, losses and noise factor.

38

Wireless Telecommunication Systems

7) Each subscriber generates 20 mE. What is the corresponding communication duration (in s) at peak time? How many subscribers can there be in each cell? Solution 1) Each base station has four TRX, hence the corresponding loss is 10 × log10 4 = 6 dB (the power is uniformly spread over the four carrier frequencies). 2) The following link budget can be established (see Table 2.2). Transmission

Reception Base station

Antenna (GA)

20

20

Cables, etc. (LO)

2

2

Duplexer (LO)

3

Coupler (LO)

6

Power

PTx

EIRPbs (dBm)

PTx+9

Antenna (GA)

1.5

3 Splitter

6

Low Noise Amplifier (GA)

2

Reception diversity (GA)

2

Sensitivity (dB)

–145

Reception threshold RECbs (dB)

–158

Mobile Cables, etc. (LO)

1

Power (dB)

3

EIRPms (dB)

3.5

1.5 1 Sensitivity (dBm) Reception threshold RECms (dBm)

–105 –105.5

Table 2.2. GSM link budget along a highway

Hence, the maximum authorized uplink propagation path loss is: EIRPms – RECbs = 161.5 dB

F/TDMA and GSM

39

Similarly, the maximum authorized downlink propagation path loss is: EIRPbs – RECms = PTx + 114.5 dB with PTx in dBm Consequently, the link budget is balanced for: PTx = 161.5 – 114.5 = 47 dBm = 50 W (we can check that this transmission power is lower than the maximum: 60 W). The authorized propagation path loss is 161.5 dB, that is 153.5 dB when considering the 8 dB margin against shadowing and fading. 3) The Okumura–Hata formula gives:

LdB = 69.55 + (44.9 − 6.55 × log10 (30)) × log10 (d ) +26.16 × log10 (900) − 13.82 × log10 (30) LdB = 126.4 + 35.2 × log10 ( d )

We want LdB < 153.5, hence d < 10

153.5−126.4 35.2

, i.e. dmax = 5.9 km.

4) The base stations are located along the highway (it is possible to locate them alternatively one at each side of the highway). Each base station is allocated four TRX and the operator has 16 bidirectional carrier frequencies (numbered from f1 to f16); hence, it leads to a pattern of size 4: {f1, f5, f9, f13}, {f2, f6, f10, f14}, {f3, f7, f11, f15} and {f4, f8, f12, f16}. These four quadruples are illustrated in Figure 2.5 by different ellipsoid patterns.

Figure 2.5. Frequency planning along a highway

40

Wireless Telecommunication Systems

Without overlapping areas, sites are separated from each other by a distance of 2 × dmax, that is 11.8 km. 5) Only one interferer can be considered: actually, in Figure 2.5, it can be seen that the mobile at a cell edge is situated at the opposite side of the boresight of one of the two nearest interfering antennas (see Figure 2.6).

Figure 2.6. Interfering cells along a highway

Because of the results from question 2, the useful signal power received by a mobile at the cell edge is (assuming the mobile in the boresight base station antenna): C = EIRPbs – L(dmax) C = 56 – 153.5 = –97.5 dBm Using the Okumura–Hata formula, we find the main interfering signal power received by a mobile at the cell edge (still assuming the mobile in the boresight main interfering base station antenna): I = EIRPbs – L(5 × dmax) I = 56 − (126.4 + 35.2 × log10 (29.5)) = −122.1 dBm

The SIR ratio is then: SIR = C – I = 24.6 dB It is higher than the operating point, even with fading and shadowing margins.

F/TDMA and GSM

41

We can check that the second interferer influence is negligible: in the considered position, the mobile is at the opposite side of the boresight interfering antenna; hence, its gain is lowered by at least 15 dB. Consequently: I′ < EIRPbs – 15 – L(3 × dmax) = 129.3 dBm hence: – 122.1dBm < I + I ' < 10 × log10 (10

−12.21

+ 10−12.93 ) < −121.1dBm

and: SIR = C – (I + I’) > 23.6 dB 6) Each cell has four TRX, that is 32 channels. Among them, there is a beacon channel and, usually, two channels dedicated for communication establishment (stand-alone dedicated control channel (SDCCH)) (on average, one SDCCH is booked for every two TRX). Consequently, there remain 29 TCH. From the Erlang table at 3% QoS (see Table 9.1), it corresponds to 22.14 Erl. 7) One Erl corresponds to the occupation of a channel during 1 h, hence 20 mE = 1/5 Erl corresponds to the occupation of a channel during 1/5 hour, that is 12 min. Each cell can contain: 22.4/(20 × 10−3) = 1,107 subscribers. 2.7. GSM network dimensioning and planning in a rural area A rural area of 400,000 km² is considered, with 100,000 inhabitants. A company receives a GSM license to become an operator and, for this purpose, is allocated 50 bidirectional carrier frequencies. The following parameters are provided: – the penetration rate is 30% over the area;

42

Wireless Telecommunication Systems

– it is necessary to dedicate 1/8 of the radio resources to signaling; – the operating point is obtained for SIR = 9 dB. An additional 8 dB margin is considered to fight against shadowing and fading; – each subscriber generates 40 mE; – Erlang tables are used for a 2% QoS; – the propagation coefficient γ is set to 3. 1) Calculate the pattern size. 2) Deduce the available per cell.

number

of

bidirectional

frequencies

3) Calculate the overall traffic in the area. 4) Calculate the number of TCH available per cell. 5) Calculate the number of necessary cells for the overall traffic in the area. Deduce the size of these cells. 6) Deduce that the limiting factor is the range and not the traffic. 7) Recalculate the number of necessary cells to cover the area, deduce the overall traffic per cell and the number of TRX per cell. 8) Propose a new pattern size in order to decrease the interferences. Solution 1) Using the formula to give the pattern size K as a function of the propagation coefficient γ and the expected SIR:

SIR =

1 6

(

3K

γ

)

F/TDMA and GSM

43

We want SIR > 9 + 8 = 17 dB, that is SIR > 1017 /10 = 50.11, hence: 1 ( 3K )γ > 50.11 6 (6 × 50.11) 2 / γ 3 K > 14.95 K>

We choose the nearest higher regular size for K (such that two integers exist, i and j, checking that i² + i × j + j² = K): K = 16 (i = 4, j = 0). 2) There are 50 bidirectional carrier frequencies to share amongst K = 16 cells. This gives three TRX per cell and two bidirectional “joker” carrier frequencies (50 = 16 × 3 + 2). 3) The penetration rate being 30%, there are 30,000 subscribers in the area. Each subscriber generates 40 mE, which results in overall traffic of 1,200 Erl. 4) There are three TRX, hence 24 channels per cell, among which 7/8 are dedicated to traffic, hence 21 channels. Using the 2% Erlang table (see Table 9.1), 14 Erl represents the overall traffic per cell. 5) It is necessary to have 1,200/14 = 86 cells. The surface area of each cell is consequently 400,000/86 ≈ 4,650 km². 6) Assuming that the base stations are situated at the center of hexagonal cells, each cell’s surface area is 3 3 ( d max )2 , and we obtain: 2 d max =

2 × 4650 = 42 km 3 3

44

Wireless Telecommunication Systems

We can easily check that 42 km exceeds the 37.8 km maximum transmitter–receiver distance extracted from exercise 2.1. Consequently, the cell range must be determined with the maximum radio range and not with the overall traffic. 7) Considering a 30 km maximum cell range, the surface 3 3 area of each cell is ( 30 )2 = 2 340 km². Consequently, 2 400,000/2,340 = 171 cells are necessary. Hence, the overall traffic in each cell is 1,200/171 = 7 Erl. From the 2% Erlang table (see Table 9.1), 13 TCH are necessary, hence a total of 15 channels (8/7 × 13). Consequently, only two TRX per cell instead of three are sufficient. 8) Having 50 bidirectional frequency carriers and two TRX per cell, it is valuable to use the highest regular pattern lower than 25. Hence, we will choose K = 21 (i = 4, j = 1) or 25 (i = 5, j = 0). If we choose K = 21, we will have:

1 SIR = ( 3 × 21)3 = 83.3 = 19.2 dB 6 If we choose K = 25, we will have:

1 SIR = ( 3 × 25)3 = 108.3 = 20.3 dB 6 Consequently, the gain is, respectively, 2 or 3 dB compared to initial constraints (SIR > 17 dB). 2.8. GSM network dimensioning and planning in an urban area An urban area of 100 km² is considered, with 5,000,000 inhabitants. A company receives a GSM license to become

F/TDMA and GSM

45

the operator and, for this purpose, is allocated 50 bidirectional carrier frequencies. The following parameters are provided: – the penetration rate is 40% over the area; – it is necessary to dedicate 1/8 of the radio resources to signaling; – the operating point is obtained for SIR = 9 dB. An additional 8 dB margin is considered to fight against shadowing and fading; – each subscriber generates 80 mE; – Erlang tables are used for a 2% QoS; – the propagation coefficient γ is set to 3.3. 1) Calculate the pattern size. 2) Deduce the available per cell.

number

of

bidirectional

frequencies

3) Calculate the overall traffic in the area. 4) Calculate the number of TCH available per cell. Deduce the overall traffic per cell. 5) Calculate the number of necessary cells for the overall traffic in the area. Deduce the size of these cells. 6) Deduce that the limiting factor is the traffic and not the range. Solution 1) The same calculations as those made in the previous exercise lead to, with γ = 3.3:

(6 × 50.11) 2 / γ 3 K > 10.6 K>

46

Wireless Telecommunication Systems

Finally, we choose K = 12 (i = 2, j = 2). 2) Similarly, there are 50 bidirectional carrier frequencies to share among K = 12 cells. It gives four TRX per cell and two bidirectional “joker” carrier frequencies (50 = 12 × 4 + 2). 3) The penetration rate being 40%, there are 200,000 subscribers in the area. Each subscriber generates 80 mE, which results in overall traffic of 16,000 Erl. 4) There are four TRX, hence 32 channels per cell, among which 7/8 are dedicated to traffic, hence 28 channels. Using the 2% Erlang table (see Table 9.1), 20.15 Erl represents the overall traffic per cell. 5) It is necessary to have 16,000/20.15 = 794 cells. Consequently, the surface area of each cell is 0.126 km². Assuming that the base stations are situated at the center of 3 3 hexagonal cells, each cell’s surface area is ( d max )2 , and 2 we obtain:

d max =

2 3 3

× 0.126 = 220 m

6) We can easily check that 220 m is much lower than the 37.8 km maximum transmitter–receiver distance extracted from exercise 2.1. Consequently, the cell range must be determined with the overall traffic (we can note that this cell range is not so far from the cell ranges in “real” dense urban areas). 2.9. SMS transmission in a GSM network An SMS is composed of no more than 140 bytes, that is 160 non-specific characters, each one being transmitted in ASCII code over 7 bits.

F/TDMA and GSM

47

If the SMS exceeds 160 characters and/or uses specific characters (accented characters, commas, question or exclamation marks, etc.) that are coded over 8 bits, several SMS will be concatenated in order to be transmitted over the network. At the beginning of each SMS, an overhead of about 20 bytes is inserted. Recall that the SMS is transmitted on the following logical channels: slow associated control channel (SACCH) (if the mobile is already in communication mode) or SDCCH (if the mobile is in idle mode). The SACCH multiplexing is one burst every 120 ms (in the 26 frame-long multiframe) and the SDCCH multiplexing is four bursts every 235.38 ms (in the 51 frame-long multiframe). Each burst carries 114 “useful” bits that are protected against transmission errors following the scheme in Figure 2.6.

Figure 2.7. Correspondence between bits of an SMS and their transmission in a radio burst

Assuming that a user transmits a maximum range SMS, calculate the radio transmission duration when the user is in communication or idle mode. Solution The SMS content is 140 bytes. We add the 20 byte-long overhead to this content. Hence, the SMS to be transmitted

48

Wireless Telecommunication Systems

is around 160 bytes long, that is 160 × 8 ÷ 184 ≈ 7 blocks of 184 bits each, that is 28 radio bursts. If the user is in communication mode, he/she receives/transmits his/her SMS at the rhythm of one burst every 120 ms; hence, his/her SMS will be transmitted in 0.120 × 28 = 3.36 s. If the user is in idle mode, he/she receives/transmits his/her SMS at the rhythm of four bursts every 235.38 ms; hence, his/her SMS will be transmitted in 0.23538 × 7 = 1.64 s. 2.10. Frequency reuse pattern determination We consider the downlink of a cellular wireless communication system. The terminal sensibility after the radio frequency stage of the receiver is equal to –104 dBm, but knowing that this power follows a log-normal probability density function, we consider a 20 dB margin with respect to this sensibility. We will then consider a target at −104 dBm + 20 dB = –84 dBm. The base station transmission power is equal to 6 W, and the base station and the terminal antennas are isotropic (gain equal to 0 dBi). We consider the following propagation loss formula 110 + 34 × log(d) and the SIR requirement is given by: S/I ≥ 14 dB We consider that the wireless provider has 49 carriers (49 up carriers and 49 down carriers). 1) Calculate the EIRP in dBm. 2) Calculate the downlink maximum allowable path loss.

F/TDMA and GSM

49

3) Calculate the maximal cell ray. 4) Calculate the frequency reuse pattern size K. 5) Calculate the maximal number of available carriers by cell. 6) Considering that a carrier is equivalent to eight communication channels, determine the communication channel density par km². 7) Propose a solution to increase this density. Solution 1) The transmitted power is equal to Pe = 6 W or, given in dBW, Pe = 7.8 dBW or, given in dBm, Pe = 37.8 dBm. The antenna gain is equal to: Ge = 0 dBi. The EIRP: PeGe is then equal to: 37.8 dBm + 0 dBi = 37.8 dBm 2) Starting from 37.8 dBm (to avoid considering a power division at the base station level due to the fact that the base station power has to be split for all carriers), the maximal propagation loss could lead, in the worst case, to the receiver sensibility. The downlink’s maximum allowable path loss is then equal to: 37.8 dBm – (–84 dBm) = 121.8 dB 3) From the propagation loss formula proposed in this exercise, we obtain: 110 + 34 × log(d) = 121.8 then: d = 10

121.8 – 110 34

d = 2.22 km

50

Wireless Telecommunication Systems

S 1 = I 6

4) From the SIR 2

 1 γ ln  6 K= e 3

(

3K

γ

)

, we obtain:

S  I

We obtain then K = 6.37. Then, we have to find the first regular pattern index greater than this value. We easily obtain: K=7 5) The number of available carriers per cell is then equal to 7: (

49 = 7 ). 7

6) The number of channels available by cell is then equal to 56 (8 × 7 = 56). This yields to a density equal to 3.61 channels per km²

56

πR²

= 3.61 .

7) The channel density per km² is relatively weak, we can increase it by reducing the cell ray. We can then reduce the transmitted power and finally we will converge toward a base station densification with small cells (micro cells or femto cells). 2.11. Traffic and Erlang for GSM cell dimensioning We consider a TDMA frame GSM with, for each carrier couple (couple: one uplink carrier and one downlink carrier), seven time slots devoted to voice channels. 1) Calculate the traffic in Erlang for a reject probability equal to 1%.

F/TDMA and GSM

51

Two communication systems are linked to each other with two sets of 12 circuits. Considering a reject probability of 1%, we would like to know: 2) What is the authorized traffic on each set of 12 circuits and what is the line efficiency? 3) What is the traffic authorized by the two sets of 12 circuits? 4) We merge the two sets into one set of 24 circuits. Considering the same reject probability, what is the new authorized traffic and the new line efficiency? Solution 1) To have one regular time slot on one uplink and one downlink carrier is equivalent to having a permanent circuit. A carrier with seven time slots is then equivalent to a set of N = 7 circuits. Reading the Erlang B table for N = 7 and for a reject probability E(A,7) = 1% leads to an authorized traffic A = 2.5 Erlang. 2) The authorized traffic with N = 12 and E(A,12) = 1% is given by the Erlang B table: A = 5.876 Erlang. The line efficiency is then equal to 48.97% (the maximum traffic of a set of 12 lines is equal to 12 Erlang). 3) The total authorized traffic by the two sets is equal to: 11.752 Erlang = 2 × 5.876 Erlang 4) The traffic authorized by a set of 24 circuits is equal to A = 15.295 Erlang, and the return is only 15.295/24 = 63.73%. This means that splitting telecommunication resources into small circuits results in decreasing the network’s capacity. It is then more efficient to keep an

52

Wireless Telecommunication Systems

important set of users and to concentrate them on a big switch. Frequency pattern reuse in GSM leads to a resource splitting that is inefficient considering the Erlang tables. A trade-off then has to be done between cellular constraints and traffic efficiency. 2.12. Signal to noise plus interference ratio We consider a GSM cell with one base station with total available power equal to 8 W. The base station has to split this power between different carriers. We consider a provider that has 48 frequency couples (uplink and downlink). The provider chooses a frequency reuse pattern K = 12. The propagation loss at a distance d (in km) of the transmitter is equal to 100 + 38 × log(d). The terminal sensibility is equal to –104 dBm, but in order to keep a margin we will take a sensibility equal to –80 dBm. Base station antennas have a gain equal to 6 dBi. The terminal antenna is isotropic with a gain equal to 0 dBi. We do not consider any additional loss term in the link budget. The required E b ratio is equal to N0

+12 dB. Base stations are located at the center of the GSM cells. –1 NOTE.– Boltzmann constant k = 1.38 10–23 JK ; noise temperature T = 300 K; D is the distance between cells using the same frequency: D = R 3K , where K stands for the frequency reuse pattern size and R stands for the cell radius.

1) From the frequency reuse pattern size K, determine the number of carriers per cell and deduce from this result the mean power per carrier. 2) From the mean power per carrier from the previous question, determine the maximal authorized path loss on the link. 3) What is the cell radius?

F/TDMA and GSM

53

4) Calculate the thermal noise power seen by the mobile in the receiving GSM band (200 kHz). Give the result in dBm. 5) Calculate the signal-to-noise ratio: S/N. 6) From the frequency reuse pattern size, calculate the signal-to-interference ratio: S/I. 7) Calculate the maximum throughput of the downlink, from the propagation parameters and considering only the thermal noise. 8) Considering interferers and thermal noise, calculate the signal-to-noise-plus-interference ratio: S/(N+I). 9) Is this throughput compliant with the channel capacity? Use the S/(N+I) ratio in the capacity formula (considering that interferers are similar to Gaussian noise). Which throughput value finally has to be considered for this link (justify briefly)? 10) On each carrier, we keep two time slots for signalization. Others time slots are used for TCH. What is, with a reject probability of 2%, the traffic that could be accepted by the base station? 11) Considering users with a traffic equal to 50 mE, how many users can be served by the base station? Solution 1) The number of frequency couples is given by the division of the number of frequencies by the frequency 48 = 4 frequency pattern reuse size K. We directly obtain 12 couples per cell. The power per carrier is then equal to 8 = 2W . 4

54

Wireless Telecommunication Systems

2) The received power is given by: Pr,dBm = Pe,dBm + Ge,dBi + Gr,dBi – LdB We must go below the receiver sensibility, we must then have, in the worst case (maximal distance from the base station): Pr,dBm = –80 dBm, this leads to: –80 dBm = 33 dBm + 6 dBi + 0 dBi – LdB With this equation, we obtain: LdB = 119 dB 3) Then, we write 119 = 100 + 38 × log(Rmax), and we deduce: Rmax = 3.16 km 4) The thermal noise power is given by (B = 200 kHz in GSM): Pth = kTB Pth,dBW = 150.8 dBW Pth,dBm = –120.8 dBm 5) The signal-to-noise ratio is given by: S N S N S N

dB

= Pr,dBm – Pth,dBm = –80 dBm – (–120.8 dBm)

dB

= 40.8 dB dB

F/TDMA and GSM

55

γ S 1 S 3K and ratio is given by = I I 6 knowing that we have K = 12 and γ = 3.8 (see propagation loss in 38 × log(d)), we obtain:

(

6) Knowing that the

S I

)

= 21.8 dB dB

7) The maximal throughput Rb is, given by the link budget, obtained by: Pr Rb Eb Pr = = N0 kT N0 Rb =

Pr 1 kT Eb N0

We have to be careful in order to derive calculations with the same units. The error risk comes from the kT product, given in dBJ, that gives –203.8 dBJ. To use this value, it must be converted into dBW: Rb,dBHz = –110 dBW – (–203.8 dBJ) – 12 dB Rb,dBHz = 81.8 dBHz Rb = 1.524 108 bits/s Rb = 152.4 Mbits/s 8) We can calculate the approach: N I N + I = + S S S

S ratio following an inverse N+I

56

Wireless Telecommunication Systems

Then, replacing numerical values: N+I = 10–4.08 + 10–2.18 S then: S N+I

= 21.7 dB dB

9) The exercise proposes to calculate the channel capacity through:

S   C = B × log2 1+   N+I  then: C = 1.45 Mbits/s We note that the channel capacity is lower than the maximal throughput given by the link budget. Actually, the link budget proposes a 152.4 Mbits/s throughput in a 200 kHz frequency band. If possible, this would involve a very high, and quite impossible, spectral efficiency. It means that the authorized frequency bandwidth in this exercise is too small and the solution given by the link budget in question 7 is impossible to reach. We then have to increase the frequency band, this will S increase the thermal noise power and decrease the ratio. N S Nevertheless, we have seen that the ratio is much lower I

F/TDMA and GSM

57

S S ratio and the degradation will consequently N N S not be very important with regards to the ratio. N+I than the

10) The solution with four carriers per cell gives 4 carriers × 6 TCH = 24 circuits per cell; the Erlang B table at 2 % gives traffic equal to 16.631 Erlang. 11) The user density is then given by: 16.631 50 10–3

= 332.6 users/cell

Chapter 3

CDMA and UMTS

Code division multiple access (CDMA) is a multiple access technique that relies on the use of orthogonal codes. All users transmit in the same bandwidth throughout the whole frame duration. Each user multiplies their signal by a dedicated code, called a spreading code. The chosen codes have the following properties: two different codes have an almost-zero intercorrelation, and each code’s autocorrelation is high. As a result, one user’s signal can be retrieved at the receiver being multiplied by that user’s code. After the adapted filter, the useful signal can be more easily obtained, since the remaining multiuser interference is very low. CDMA thus turns a narrowband signal into a broadband signal. In the Universal Mobile Telecommunications System (UMTS) and High Speed Packet Access (HSPA), the spread bandwidth is equal to W = 3.84 MHz. The spreading code’s size then depends on the data rate required by each user and may be different in uplink and downlink. The third generation (3G) systems use a multi-spreading technique: in the downlink, each mobile terminal possesses a code, made from the combination of a

60

Wireless Telecommunication Systems

specific spreading code generated from the orthogonal variable spreading factor (OVSF) tree and a scrambling code that depends on the base station, and allows us to decrease inter-cell interference. In the uplink, the code used is a combination of a spreading code and a scrambling code that both depend on the mobile terminal, so that desynchronization is taken care of, as well as the fact that OVSF code allocation is not centralized, and consequently not optimal. UMTS standard compels each terminal to have only one code in each link direction. HSPA standard allows the same terminal to use several codes in the same frame, depending on its data rate requirement, and that of the other terminals. Thus, a terminal may use all the available radio resources at a given moment in HSPA, which leads to a great increase of the data rates achieved. UMTS standard provides transmission in both circuitswitched and packet-switched modes. In HSPA, only packetswitched transmission is possible. Another specific characteristic of 3G is soft handover: a mobile terminal may be connected and transmit to and from several base stations at the same time. This leads to a great improvement of the data rates located at the border of cells. The noise rise, NR, parameter is linked to the UMTS cell’s load with the following relation:

NR =

1 1 − charge

Then Table 3.2 can be established.

CDMA and UMTS UMTS

61

Numerical circuit-switched and packet-switched systems (licensed) bandwidth: 1,885–2,025 MHz 2,110–2,200 MHz + GSM “UMTS 900” bandwidth Terminal transmission power: Bsp, the channel per subcarrier is non-frequency selective. 3) The cyclic prefix duration must be at least that of the maximum multipath transmission delay, Tm, to absorb all delayed symbols. Thus, it must be higher than 1 μs. 4) The symbol time per subcarrier is Ts,sp = 1/Bsp = 102.4 μs. The OFDM symbol duration is consequently equal to Ts,OFDM = Ts,sp + TCP = 107.4 μs. 5) The cyclic prefix overhead in an OFDM symbol is: Overhead = 5/107.4 = 4.66% 6) Only 600 of 1,024 subcarriers are used to transmit data. The sum data rate in symbols per second is equal to the ratio of the number of useful subcarriers over the OFDM symbol time. It is consequently equal to: Rsymbols = 600⁄(107.4 × 10–6 ) = 5.59 Msymbols/s 7) If all mobile terminals have a QPSK modulation, each symbol carries 2 bits. The sum data rate is then equal to: RQPSK = 2 × Rsymbols = 11.18 Mbits/s 8) This assumption is valid because of link adaptation. On average, we can admit that in a 4G cell in the downlink, the mobile terminals located on the border of the cell use a QPSK modulation, the mobile terminals located close to the base station use a 64-QAM modulation and the mobile

110

Wireless Telecommunication Systems

terminals at mid-distance use a 16-QAM modulation. This, of course, depends on the radio conditions and on the propagation model. The proportions given in this problem have been set according to a complete link budget. to:

9) The sum data rate in downlink of the cell is thus equal R = 0.35´ RQPSK + 0.35´ R16-QAM + 0.3´ R64-QAM = 5.59 ´ (2 ´ 0.35 + 4 ´ 0.35 + 6 ´ 0.3) + 21.8 Mbits/s

10) We cannot extrapolate these results to the uplink, because the proportion of each modulation in the cell is then unknown. It is likely that there will be more mobile terminals in QPSK and less in 64-QAM. This proportion will depend on the mobile terminals’ ability to use all modulations; for instance, in LTE, some mobile terminals are unable to use 64-QAM. 4.6. LTE data rates evaluation Let us consider an LTE cell with bandwidth B = 20 MHz. According to the standard, the sampling frequency is then Fe = 30.72 MHz. 1) We assume that 20% of the transmitted symbols are either used for cyclic prefixes or for pilots. Explain why these symbols are not useful. 2) Calculate the number of useful symbols per second in bandwidth B. 3) Only 1,200 subcarriers of 2,048 are used for transmission. Taking into account this limitation, claculate the number of useful symbols per second in bandwidth B. 4) According to the standard, what is the number of RBs per cell? Deduce from that the number of useful symbols per second in the bandwidth of an RB.

OFDM and LTE

111

5) Calculate the number of OFDM symbols transmitted in a time slot of 0.5 ms. Is this value coherent with the standard? 6) Deduce, for each LTE modulation, the data rate in kbits/s for the data transmitted in one RB. 7) We assume that 34% of the RBs per cell correspond to a QPSK modulation, 33% of the RBs correspond to a 16QAM modulation, and 33% to a 64-QAM modulation. Calculate the sum useful data rate per cell under this assumption. 8) Is the assumption made in question 7 realistic in uplink? What consequences can we expect for the data rate in uplink? Solution 1) The cyclic prefix only contains copied symbols: it is redundant, since these symbols do not correspond to useful user data. The pilot symbols are required to perform synchronization and frequency-equalization. They do not carry any useful user data. 2) The sum data rate is equal to 30.72 Msymbols/s. The useful sum data rate is then: Rsymbols = 30.72 × 106 × 0.8 = 24.576 Msymbols/s 3) Considering the number of subcarriers that are not used, the useful sum data rate becomes: Rsymbols,u = 24.576 × 106 × 1.200/2,048 = 4.4 Msymbols/s 4) According to LTE standard, the number of RBs in a bandwidth of 20 MHz is NRB = 100. Consequently, there are RRB = 144,000 useful symbols per RB and per second.

112

Wireless Telecommunication Systems

5) Since RBs are composed of 12 subcarriers, the symbol rate per subcarrier is Rsymbols,sc= 144,000/12 = 12,000 symbols per second. As the time slot duration is 0.5 ms, the number of symbols per subcarrier and per time slot is: Nsymbols,sc = 12,000 × 0.5 × 10–3 = 6 This value is coherent with the standard. Indeed in the LTE standard, there are six or seven OFDM symbols per subcarrier and per time slot, depending on the cyclic prefix duration. 6) Three different modulations can be used in LTE: QPSK (2 bits/symbol) 16-QAM (4 bits/symbol) and 64-QAM (6 bits/symbol). Consequently, the data rate per RB depending on the modulation is: – for QPSK: RQPSK = 2 × RRB = 288 kbits/s; – for 16-QAM: R16-QAM = 4 × RRB = 576 kbits/s; – for 64-QAM: R64-QAM = 6 × RRB = 864 kbits/s. 7) Among the 100 RBs, 34 use QPSK, 33 use 16-QAM and 33 use 64-QAM. Consequently, the sum useful data rate is: Rutile = 34 × RQPSK + 33 × R16-QAM + 33 × R64-QAM = 57,312 Mbits/s

8) This assumption is not realistic in uplink. In this case indeed, the proportion of mobile terminals using 16-QAM and 64-QAM should be lower, because the mobile terminals’ transmission power is far lower than that of the base station. We can thus expect to get lower data rates. Moreover, only category 5 mobile terminals can use a 64-QAM uplink. This limitation should lead to even lower data rates.

OFDM and LTE

113

4.7. LTE link budget This problem proposes performing a link budget in LTE, assuming that each user gets one RB. In downlink, we assume that the base station only serves one mobile terminal per time slot. Consequently, all base station transmission power is provided to that mobile terminal. The influence of the number of mobile terminals will be evaluated in the next problem. The signal-to-interference-plus-noise ratio (SINR) in LTE for a user located at a distance d from the base station can be written, in both uplink and downlink:

Γ =

Pe × GATx × GARx ( N + Iint er ) × L(d ) × LOTx × LORx × FRx

where: – L(d) is the path loss between the base station and the mobile terminal; – Nis the noise power in the considered bandwidth; – Pe is the transmission power; – Iinter is inter-cell interference; – GA and LO are the antenna and cable gains and losses (indexed by Tx for the transmitter, and Rx for the receiver); – FRx is the receiver noise factor. To simplify notations, in the following, the inter-cell interference plus noise is written N × Finter, where Finter is the inter-cell interference factor, defined as follows: Finter = 1 + Iinter/N. We assume that is has the same value in uplink and downlink.

114

Wireless Telecommunication Systems

Finally, the SINR can be written as:

Γ =

Pe × GATx × GARx N × Finter × L( d ) × LOTx × LORx × FRx

The link budget is performed with the following set of parameters: – required SINR in an RB, depending on the modulation: – for QPSK: G QPSK ,db = 2.5 dB; – for 16-QAM: G16-QAM ,dB = 9 dB; – for 64-QAM: G 64-QAM ,dB = 16 dB. – Thermal noise power spectral density: N0,dBm = –174 dBm/Hz. Inter-cell interference factor: Finter,dB = 3 dB. – Base station parameters: - antenna gain: GAdB = 18 dBi; - cable loss: LO dB = 2 dB; - noise factor: FRx,dB = 5 dB; - maximum transmission power: Pe = 20 W; - power limitation to problems: LPAPR,dB = 12 dB.

avoid

power

amplification

– Mobile terminal parameters: - antenna gain: GAdB = 2 dBi; - cable loss: LO1,dB = 2 dB; - body loss: LO2,dB = 3 dB; - noise factor: FRx,dB = 9 dB; - maximum transmission power: Pe = 0.125 W.

OFDM and LTE

115

1) Justify the difference in the SINR, compared with UMTS. 2) Why is there a power limitation at the base station only? 3) An RB has a bandwidth equal to 180 kHz. Calculate thermal noise power in an RB, in dBm. 4) Calculate the base station and mobile terminal’s transmission power in dBm. 5) Perform an uplink link budget for each modulation. The path loss model is COST 231 extension to Hata’s model in dense urban areas, when the carrier frequency is equal to 2.6 GHz, LdB(d) = 144.68 + 35.22 × log10(d). 6) Perform downlink link budget. 7) Deduce the cell size (coverage). 8) Is it possible to add a soft handover gain, as in UMTS? 9) The previous results were obtained when the frequency reuse factor was equal to K = 3. What can we expect if the frequency reuse factor becomes K = 7? 10) The previous results were obtained when the carrier frequency was fc = 2.6 GHz. We now consider an LTE system with carrier frequency fc = 800 MHz.Calculate the new path loss model, using Hata’s model for large cities. 11) Perform link budget again with the new carrier frequency. Draw conclusions on the performances of the system, depending on the carrier frequency. Solution 1) There is no processing gain, because LTE does not use CDMA but OFDM. There is no intra-cell interference, because all users are multiplexed on subcarriers using OFDMA: no user in the same cell is allocated to the same

116

Wireless Telecommunication Systems

subcarrier. Contrary to CDMA, multiplexing is then perfect, if there is no desynchronization in frequency (which is assumed here). 2) The power limitation only concerns the base station, because in the LTE standard, OFDM is used in downlink and SC-FDMA is used in uplink. But in SC-FDMA, the PAPR is far lower, and it is thus not necessary to put any power limitation to avoid power saturations. 3) The RB bandwidth is 180 kHz. The thermal noise power is then:

N dBm = -174 + 10´ log10 (180´103 ) = -174 + 52.5 = -121.5 dBm 4) The base station transmission power is equal to: Pe,dBm = 10 × log10 (20) + 30 – LPAPR,dB = 31 dBm The mobile terminal transmission power is equal to: Pe,dBm = 10 × log10 (0.125) + 30 = 21 dBm 5) As in UMTS, link budget is performed in three steps: – Calculate the required received power threshold at the antenna entry to reach the target SINR. In our case, it is equal to: RECdBm = ΓdB + NdBm + Finter,dB – GARx,dB + LORx,dB + FRx,dB – Calculate the EIRP. It is equal to the actual radiated power, taking into account all hardware gains and losses. It is given by the following formula: EIRPdBm = Pe,dBm + GATx,dB – LOTx,dB

OFDM and LTE

117

– Calculate the maximum path loss leading to the receiver power threshold: LdB,max = EIRPdBm –RECdBm In uplink, the transmitter is the mobile terminal, and the receiver is the base station. We begin with QPSK modulation. The receiver power threshold is: RECdBm = 2.5 – 121.5 + 3 – 18 + 2 + 5 = –127 dBm The EIRP of the mobile terminal is equal to: EIRPdBm = 21 + 2 – 2 – 3 = 18 dBm Finally, the maximum path loss is: LdB,max,QPSK = 18 + 127 = 145 dB In the previous formula, we have taken into account the total losses at the transmitter, which are equal to the sum of cable loss and body loss. The same calculations for 16-QAM and 64-QAM modulations are performed, by replacing Γ value in the receiver power threshold formula. The following values are obtained: LdB,max,16-QAM = 18 + 120.5 = 138.5 dB LdB,max,64-QAM = 18 + 113.5 = 131.5 dB In downlink, the same steps are taken as in uplink, starting with QPSK modulation. The transmitter is the base station, and the receiver is the mobile terminal. Its receiver power threshold is equal to: RECdBm = 2.5 – 121.5 + 3 – 2 + 2 + 3 + 9 = –104 dBm

118

Wireless Telecommunication Systems

The base station EIRP is: EIRPdBm = 31 + 18 – 2 = 47 dBm The maximum path loss is consequently equal to: LdB,max,QPSK = 47 + 104 = 151 dB The same calculations for 16-QAM and 64-QAM modulations are performed, by replacing Γ value in the receiver power threshold formula. The following values are obtained: LdB,max,16-QAM = 47 + 97.5 = 144.5 dB LdB,max,64-QAM = 47 + 90.5 = 137.5 dB 6) The coverage is given by the most limiting way. In this case, and for all considered modulations, it is the uplink. The coverage in QPSK is given by inverting the path loss formula for its maximum value, LdB,max,QPSK = 145 dB: dQPSK = 10

145-144.68 ⁄35.22

= 1.021 km.

Similarly, the coverage for the two other modulations is: d16-QAM = 0.668 km d64-QAM = 0.422 km The cell coverage is given by that of QPSK. 7) It is not possible to add a soft handover, since in LTE, a mobile terminal is never connected to several base stations at the same time. 8) With a frequency reuse factor K = 7, inter-cell interference will be lower than with a frequency reuse factor K = 3. Consequently, the value of Finter,dB will decrease, and

OFDM and LTE

119

will be between 0 and 3 dB. The cell coverage and the coverage of each modulation will then increase. 9) Hata’s dense urban propagation model for large cities is: LdB (d) = A + B × log10 (d)–a1 + C where the parameters are defined as follows: – A = 46.3 + 33.9 ´ log10 ( f c ) - 13.82 ´ log10 ( hb ); – B = 44.9 - 6.55´ log10 ( hb ); – a1 = 3.2 × log10 (11.75 × hm )

2

–1.75;

– C = 3 dB for large cities; with d being the distance between the transmitter and the receiver in km, fc the central carrier frequency, expressed in MHz, and hb and hm the respective base station and mobile terminal heights, expressed in m. These values can be found from the path loss model at fc = 2.6 GHz given in the problem’s formulation. For a carrier frequency of fc = 800 MHz, the following model is obtained: LdB (d) = 125.08 + 35.22 × log10 (d) The path loss calculations are unchanged in the link budget. Only the final coverage calculation is different. By reusing the uplink path loss values from question 5, we get the following coverage values: dQPSK = 3.678 km; d16-QAM = 2.404 km and d64-QAM = 1.521 km. Coverage is consequently highly increased. REMARK 4.2.– We can notice that carrier frequency fc = 800 MHz will mostly be used in rural areas, whereas carrier

120

Wireless Telecommunication Systems

frequency fc = 2.6 GHz will be used in urban areas. Thus, in rural areas, the number of base stations required for the deployment will be decreased. 4.8. LTE link budget taking into account the number of users This problem is an extension of the previous one. We consider an LTE cell with the same parameters as used previously. The cell bandwidth is B = 5 MHz. We first assume that 25 mobile terminals are simultaneously served in uplink and downlink per time slot, and that they all get one RB per direction. 1) Is this assumption realistic? 2) Calculate the base station transmission power per RB. Calculate the mobile terminal transmission power per RB. 3) Perform link budget when the central carrier frequency is equal to 2.6 GHz, with the following path loss model: LdB (d) = 144.68 + 35.22 × log10 (d). 4) Deduce the cell coverage. We now assume that five users are simultaneously served in uplink and downlink per time slot, and that each of them obtains five RBs per direction. 5) Calculate the transmission power per base station and per user in each RB. Calculate the transmission power of each mobile terminal per RB. 6) Perform a link budget with the same parameters as used previously. 7) Deduce the cell coverage.

OFDM and LTE

121

Solution 1) This assumption is realistic, since the LTE standard specifies that there are 25 RBs per time slot when the total bandwidth is B = 5 MHz. 2) The base station transmission power is equally shared by all 25 RBs. It is thus equal to: Pe,dBm = 10 × log10 (20⁄25) + 30–LPAPR,dB = 17.03 dBm The mobile terminal transmission power per RB is equal to its total transmission power. It is equal to: Pe,dBm = 10 × log10 (0.125) + 30 = 21 dBm 3) The uplink link budget is the same as in question 5 of the previous problem, since the mobile terminal EIRP is unchanged. The following path loss values are consequently obtained in uplink: LdB,max,QPSK = 18 + 127 = 145 dB LdB,max,16-QAM = 18 + 120.5 = 138.5 dB LdB,max,64-QAM = 18 + 113.5 = 131.5 dB In downlink, however, the link budget must be done again. Only the base station EIRP changes. It is now equal to: EIRPdBm = 17.03 + 18–2 = 33.03 dBm The receiver power threshold (here, at the mobile terminal) is unchanged compared to the previous problem. For QPSK modulation, it is equal to RECdBm = –104 dBm. Consequently, the maximum path loss is: LdB,max,QPSK = 33.03 + 104 = 137.03 dB

122

Wireless Telecommunication Systems

Using the receiver power thresholds of the two other modulations, we similarly obtain the path loss values: LdB,max,16-QAM = 33.03 + 97.5 = 130.53 dB LdB,max,64-QAM = 33.03 + 90.5 = 123.53 dB 4) The cell coverage is given in the most limiting way. In the studied case, it is equal to the downlink. The cell coverage is obtained by inverting the path loss formula. For QPSK modulation, we get: dQPSK = 10

(137.03-144.68)⁄35.22

= 0.606 km

The coverage with the two other modulations is: d16-QAM = 0.396 km d64-QAM = 0.251 km The maximum cell coverage is consequently 606 m. 5) In downlink, the 25 RBs are used. Assuming that the base station transmission power is equally shared among all RBs, it is still equal to Pe,dBm =17.03 dBm. In uplink, each user has five RBs. We assume that the mobile terminal transmission power is equally shared among all RBs. It is consequently equal to: Pe,dBm = 10 × log10 (0.125⁄5) + 30 = 13.98 dBm 6) The downlink link budget is the same as in question 3, since the base station EIRP is unchanged. However, the uplink link budget changes. The mobile terminal EIRP is now equal to: EIRPdBm = 13.98 + 2 – 2 – 3 = 10.98 dBm

OFDM and LTE

123

The uplink receiver power threshold (at the base station) is the same as in previous problem. For QPSK modulation, it is equal to: RECdBm = –127 dBm. We deduce that the maximum path loss is: LdB,max,QPSK = 10.98 + 127 = 137.98 dB Similarly, using the receiver power thresholds of the two other modulations, the following path loss values are obtained: LdB,max,16-QAM = 10.98 + 120.5 = 131.48 dB LdB,max,64-QAM = 10.98 + 113.5 = 124.48 dB 7) After having compared uplink and downlink path loss values, we can see that downlink is still the limiting way. The cell coverage and the coverage areas of the three modulations are consequently unchanged compared to question 4. REMARK 4.3.– This problem presents an example where downlink is the limiting direction. This case rarely takes place, but can happen in an FDMA system where the base station power must be shared between several frequency sub-bands. The LTE link budget must not only consider the distance, but also the number of RBs allocated per mobile terminal. 4.9. Modulation-coding scheme relation, efficiency and SINR in LTE networks

spectral

It is recalled that the elementary allocated radio resource in LTE is the RB. Each RB is composed of 84 OFDM symbols at 180 kHz, every 0.5 ms. Each RB is associated with a modulation (QPSK, 16-QAM or 64-QAM) and a coding rate (1/2, 2/3 or 3/4) to ensure the data protection before their radio transmission.

124

Wireless Telecommunication Systems

The modulation + coding rate association is denoted by the modulation and coding scheme (MCS). Recall that the Shannon capacity formula gives the maximum binary throughput D that is available in a given frequency bandwidth B, as a function of the SINR:

D = B × log 2 (1 + SINR ) Complete Table 4.3 by determining the LTE spectral efficiency (expressed in bits/s/Hz) as a function of the MCS and by deducing from the Shannon formula the relation between spectral efficiency and SINR. Solution Each RB transmitting 84 OFDM symbols at 180 kHz, every 0.5 ms, the LTE spectral efficiency E is: E=

84 0, 5.10

Modulation and coding scheme (MCS)

−3

× 180.10

3

Spectral efficiency (bits/s/Hz)

=

168 14 OFDM symbols/s/Hz = 180 15

SINRmin – SINRmax

SINRmin – SINRmax (dB)

QPSK 1/2 QPSK 2/3 QPSK 3/4 16-QAM 1/2 16-QAM 2/3 16-QAM 3/4 64-QAM 2/3 64-QAM 3/4 Table 4.3. Spectral efficiency as a function of the MCS

OFDM and LTE

125

Let us express the relations OFDM symbols ⇔ number of “useful” bits ⇔ spectral efficiency as a function of the MCS, knowing that each QPSK symbol (respectively, 16-QAM symbol and 64-QAM symbol) carries 2 bits (respectively, 4 and 6) and that a k/n coding rate means that k “useful” bits are protected by n coded bits (see Table 4.4).

Modulation and coding scheme

Spectral efficiency E (bits/s/Hz)

(MCS)

Number of bits per modulated symbol

Number of “useful” bits per OFDM symbol

QPSK 1/2

2

2 × 1/2 = 1

QPSK 2/3

2

2 × 2/3 = 4/3

4 14 ´ = 1.24 3 15

QPSK 3/4

2

2 × 3/4 = 1.5

1.5 ´

16-QAM 1/2

4

4 × 1/2 = 2

2´

16-QAM 2/3

4

4 × 2/3 = 8/3

16-QAM 3/4

4

4 × 3/4 = 3

3´

64-QAM 2/3

6

6 × 2/3 = 4

4´

64-QAM 3/4

6

6 × 3/4 = 4.5

4.5´

1´

14 = 0.93 15

14 = 1.4 15

14 = 1.86 15

8 14 ´ = 2.48 3 15 14 = 2.8 15

14 = 3.73 15 14 = 4.2 15

Table 4.4. Spectral efficiency as a function of the MCS (Solution)

126

Wireless Telecommunication Systems

Modulation and Spectral coding scheme efficiency E SINRmin – SINRmax (bits/s/Hz) (MCS)

SINRmin – SINRmax (dB)

QPSK 1/2

0.93

0.9 – 1.36

–0.4 – 1.3

QPSK 2/3

1.24

1.36 – 1.64

1.3 – 2.1

QPSK 3/4

1.4

1.64 – 2.63

2.1 – 4.2

16-QAM 1/2

1.86

2.63 – 4.58

4.2 – 6.6

16-QAM 2/3

2.48

4.58 – 5.96

6.6 – 7.7

16-QAM 3/4

2.8

5.96 – 12.27

7.7 – 10.9

64-QAM 2/3

3.73

12.27 – 17.38

10.9 – 12.4

64-QAM 3/4

4.2

≥17.38

≥12.4

Table 4.5. Minimum SINR as a function of the MCS

From the Shannon formula, the maximum spectral efficiency that can be reached is expressed as follows:

Emax =

D = log 2 (1 + SINR ) B

Therefore, the minimum SINR required on the radio link in order to achieve the spectral efficient E in LTE is obtained as follows:

E ≤ log 2 (1 + SINR ) SINR ≥ 2 E − 1

(

)

SINRdB ≥ 10 × log10 2 E − 1 This gives:

– 0.93 bits/s/Hz ⇔ SINR ≥ 0.9, SINRdB ≥ –0.4; – 1.24 bits/s/Hz ⇔ SINR ≥ 1.36, SINRdB ≥ 1.3; – 1.4 bits/s/Hz ⇔ SINR ≥ 1.64, SINRdB ≥ 2.1;

OFDM and LTE

– 1.86 bits/s/Hz ⇔ SINR ≥ 2.63, SINRdB ≥ 4.2; – 2.48 bits/s/Hz ⇔ SINR ≥ 4.58, SINRdB ≥ 6.6; – 2.8 bits/s/Hz ⇔ SINR ≥ 5.96, SINRdB ≥ 7.7; – 3.73 bits/s/Hz ⇔ SINR ≥ 12.27, SINRdB ≥ 10.9; – 4.2 bits/s/Hz ⇔ SINR ≥ 17.38, SINRdB ≥ 12.4.

127

Chapter 5

MIMO and Beamforming

Mobile communication systems must be able to cope with fading and multiuser interference. From the introduction of GSM up to today, these problems have been considered from different angles and several approaches have been proposed to mitigate both impairments. The digital processing of the signal coming from an array of antennas (called smart antenna techniques) [JAK 74, RAP 01, YAC 93] played a very important role in the progress achieved in this area so far. Among the novel techniques in this area, we can cite beamforming, diversity and multiple inputs multiple outputs (MIMO) techniques as the most successful techniques. The smart antenna techniques can be applied either to the base station or to the mobile, and on the downlink or uplink. For technological and economic reasons, it is often more advantageous to only have an array of antennas at the base station, and a single antenna at the mobile. The main goal of beamforming is to increase the signal-tonoise ratio (SNR) at the desired mobile and to reduce the interference generated toward other mobiles present in the system. This is done by directing the radiated signal towards

130

Wireless Telecommunication Systems

the receiver. The chosen direction does not necessarily match the geographical one, but can correspond to the main path of the electromagnetic waves traveling from the base station to the receiver.

Figure 5.1. Beamforming directs the radiated signal toward the desired mobile

By forming a beam in the direction of the mobile, the transmission power (PTX) can thus be reduced in order to maintain the same bit error rate (BER). The amount of transmission power saved in the process is called antenna gain. In fact, the effect of beamforming can be seen as a shift of the BER curve to the left. The diversity approach treats the same problem from a different perspective. When looking closer at the propagation channel between the base station and mobile receiver, we note that this channel is generally formed by the sum of several smaller paths (called multipaths). Each multipath is characterized by its attenuation, delay and relative phase. These parameters vary in time not only due to the relative motion between the transmitter and the receiver, but also

MIMO and Beamforming

131

due to the movement of all reflectors and obstacles present in the surroundings.

Figure 5.2. The antenna gain provided by beamforming allows us to obtain the same BER for a reduced transmission power (PTX). The performance curve is thus shifted to the left

Hence, the overall propagation channel seen by the receiver is the result of the sum of all multipaths, which translates into a time variation of the signal power at the receiver. This effect is the so-called fading. When the phases of the multipaths are such that they lead to a destructive combination (a strong attenuation of the transmission power), we talk about deep fading. In practice, the performance of the mobile systems is highly degraded by the presence of deep fadings. The mitigation of these deep fadings is thus the main goal of the diversity techniques. The main idea is to describe the fact that the channel shows a low correlation by sending copies of the same signal, which will suffer uncorrelated attenuations. Thus, the probability that all these copies encounter deep fading at the same time is very low. Therefore, by combining these copies at the receiver, we can drastically reduce the probability that the received signal is in deep fading and,

132

Wireless Telecommunication Systems

even in the rare cases when it occurs, the duration of the fading is also decreased. To best profit from diversity, the copies transmitted by uncorrelated multipaths should also be uncorrelated among them. In this manner, the receiver can combine these copies in such a way that results in the addition of the multipaths’ power, leading to the best use of the transmitted power at every time instant.

Figure 5.3. Space diversity: the transmitter uses two uncorrelated multipaths to communicate with the mobile

The different multipaths correspond to the diverse uncorrelated modes over which the signal can be transmitted and the number of modes is called the diversity order of the channel. These modes are characterized in different domains, respectively, in the spatial domain by the direction of arrival (DOA), in the temporal domain by the delay and in the frequency domain by the selectivity. When the multipaths are uncorrelated in the temporal domain, we say that the channel provides time diversity. In the same way,

MIMO and Beamforming

133

when the multipaths are uncorrelated in the frequency domain, the channel provides frequency diversity. On the other hand, the use of an antenna array introduces a new processing domain and, therefore, a new kind of diversity that describes the spatial decorrelation among multipaths, called space diversity. The use of space diversity is of fundamental importance in mobile communication since it allows for an overall diversity order greater than 1, even when the channel is flat in frequency (no frequency diversity) and time invariant (no temporal diversity). The use of space diversity makes better use of the transmission power (PTX) in probabilistic sense, leading to a reduction of PTX for the same BER. The slope of the BER curve becomes steeper with the diversity order of the channel. The relation between the actual slope and the slope obtained without exploiting the channel diversity is called diversity gain.

Figure 5.4. The use of space diversity leads to the same BER with a reduced transmission power (PTX). The slope of the performance curve becomes steeper

5.1. Beamforming and signal-to-noise ratio We consider a linear antenna array with N isotropic sensors. The source is narrow-band and so far from the antenna that we can consider a plane wave front. The source is located in a direction with respect to the normal direction of the linear array.

134

Wireless Telecommunication Systems

Figure 5.5. Plane wave front arriving on an N sensors linear array antenna. The antenna is a line array with equal interelement spacings of d= /2 where λ stands for the carrier wavelength. We consider the complex envelope of the received signals (i.e. after the carrier cancellation and the complex decomposition of the I and Q components of the signal). We consider the first sensor on the right as a reference. Signal of this sensor will be noticed as x1 (n) = s(n). We consider that s(n) is randomly distributed, centered and we introduce its power ps =E |s(n)|2 .

1) Considering the plane wave front, write the phase difference between sensors. We consider now that the signal on each sensor is received with an additive white complex Gaussian noise variable. We introduce ni(t) as the noise variable on the ith sensor. This complex variable has a variance equal to 2σ² (σ² for each way). 2) Give

the

expression

of

the

X (n) = ( x1 ( n )

T

x2 ( n )  xN ( n ) ) vector representing signal samples on antenna sensors, with respect to the source signal, to and to the noise the source steering vector

B(n) = ( b1 ( n ) b2 ( n )  bN ( n ) ) terms of the

T

vector. You will give generic

vector.

We apply a spatial filter W = (1 1 …)T to the received samples and we obtain:

y ( n ) = W H X ( n)

MIMO and Beamforming

135

3) Give the SNR of the y(n) signal. Numerical application: N = 10, θ = 0, ps = 10–9, σ² = 10–10 We now choose to apply the spatial filter W = Dθ to the received samples. Then, we obtain:

y ( n ) = W H X ( n) 4) Give the SNR obtained for the z(n) signal. Numerical application:

N = 10, θ =

π 4

, ps = 10−9 , σ 2 = 10−10

Solution 1) With the source being far from the antenna, we could consider a plane wave front. The path difference between the first and the second sensor is equal to: ∆x = dsinθ

Figure 5.6. Path difference between two isotropic sensors with a plane wave front

This path difference then involves a time delay τ = between the first and second sensor.

dsinθ c

136

Wireless Telecommunication Systems

If we introduce s (t ) e j 2π f 0t as the modulated useful signal on the first sensor, then the same signal on the second sensor will be given by: s (t − τ ) e

j 2π f 0 ( t −τ )

With the sensors being close to each other, the time duration τ is very small. Then, the narrow-band hypothesis for s(t) yields to: s(t  τ)≈s(t) Then, the received signal on the second sensor becomes: s (t ) e

j 2π f 0 ( t −τ )

After down conversion, i.e. after multiplication of the , the useful signal of the second received signal by sensor will be given by: s (t ) e − j 2π f 0 τ

We can then introduce the phase shift φ = 2πf0 τ = 

2πdsinθ



.

Replacing d by , we obtain the phase shift: 2

φ = πsinθ 2) After sampling and considering additive white noise terms, the signal received on the first sensor will be given by: x1 (n) = s(n) + b1 (n) The signal of the second sensor will be given by: x2 (n) = s(n)ejφ + b2 (n)

MIMO and Beamforming

137

By generalization, the signal on the Nth sensor is given by: xN (n)=s(n)ej(N1)φ +bN (n) Regrouping these sensors signal in a common ( ) vector, we obtain:  x1 ( n)    x ( n)  X (n) =  2       x N ( n) 

1    b1 (n)      − jϕ  e   b2 (n)  X ( n ) = s ( n)  +         e− j ( N −1)ϕ   bN (n)    Then, introducing the steering vector Dθ of the useful source, we obtain:

X ( n ) = s(n) Dθ + B(n) with:

1     − jϕ  e  Dθ =       e− j ( N −1)ϕ   

138

Wireless Telecommunication Systems

and:  b1 ( n)    b (n)  B (n) =  2       bN ( n) 

3) We have:

y ( n ) = W H X ( n) y ( n ) = W H ( s ( n) Dθ + B (n) )

y ( n) =

(1 + e

− jϕ

)

− j N −1)ϕ s(n) + b1 (n) + b2 (n) +  + bN (n) + e− j 2ϕ +  + +e (

y (n) = e

−j

N −1 ϕ 2

 Nϕ  sin   2  ×  s (n) + b '(n) ϕ  sin   2

where b′(n) stands for a white additive complex noise term with variance equal to 2Nσ² (Nσ² for In phase and Quadrature). The SNR Γ is equal to:

=

Nφ 2 φ sin 2

sin

2

ps

2Nσ2

Numerical application: N = 10,θ = 0, ps = 10−9 , σ 2 = 10−10

MIMO and Beamforming

=

139

Nps 2σ2

Γ = 50 Given in dB:

Γ dB = 17 dB In this case, the useful source is in the normal direction with respect to the antenna array. The direct summation of sensors gives a gain equal to N compared to the SNR ratio of a unique sensor. 4) We have:

y ( n ) = W H X ( n) y ( n ) = Dθ H ( s (n) Dθ + B (n) ) or Dθ H Dθ =N, then:

y ( n ) = Ns (n) + b1 (n) + e jϕ b2 ( n) + ... + e j ( N −1)ϕ bN ( n)

y ( n ) = Ns(n) + b '(n) where b′(n) stands for a white additive complex noise term with variance equal to 2Nσ² (Nσ² for In phase and Quadrature). The SNR Γ is then equal to: =

N2 ps

2Nσ2

140

Wireless Telecommunication Systems

Numerical application:

N = 10, θ = =

π 4

, ps = 10−9 , σ 2 = 10−10

Nps 2σ2

Γ = 50 or given in dB:Γ dB =17 dB We note that the SNR does not depend on the angle . This is due to the fact that the W spatial filter has corrected phase differences between all sensors. All filter taps have a modulus equal to one and they have not changed the noise power. Then, we find a result similar to that of question 3. 5.2. Space diversity and chi-square distribution The aim of the problem presented in this section is to lead a comparison between SNRs for single-input single-output (SISO) context and for a 2x1 multiple-input single-output (MISO) context. SISO case If we consider a cyclic prefix OFDM waveform, then at the FFT output the received signal on a frequency subcarrier can be written as follows:

r (n) = h1s (n) + b1 (n) where: – s(n) is a transmitted symbol with a mean power

Pe ;

– h1 is a random complex Gaussian value with a mean value μ and a variance equal to 2σ2h , representing the channel response for the considered subcarrier;

MIMO and Beamforming

141

– b1(n) stands for an additive white complex Gaussian noise term with variance equal to 2σ². MISO case We consider a simple school case with two transmitting antennas and one receiving antenna. We consider an Alamouti scenario [ALA 98], where after the reception algorithm, we have:

(

2

r ( n ) = h1 + h2 where:

2

) s '(n) + h b (n) + h b (n + 1) * 1 1

* 2 2

P

– s′(n) is a modulated symbol with a mean power e (the 2 division by 2 comes from the transmission power Pe split on the two antennas); – h1 (respectively, h2) is a random Gaussian complex variable with a mean equal to μ and a variance equal to 2σ2h , representing the channel response for the considered subcarrier; – b1(n) (respectively, b2(n)) stands for a complex additive white Gaussian variable with variance equal to 2σ2. NOTE.– – If X is a complex Gaussian zero-mean random variable with variance equal to σ2, then we have E[X4] = 3σ4. – If X1, X2, …, Xn are n complex Gaussian zero-mean independent variables with variance equal to σ2, then the random variable Y= ∑ni=1 X2i is chi-square distributed with n degrees of freedom. Moreover, we have E[Y2] = (n2 + 2n)σ4. 1) We consider that propagation channels are deterministic, we then have h1 = h2 = μ and σh = 0. Calculate

142

Wireless Telecommunication Systems

the mean value  of the SNR for the SISO case and the mean value  of the SNR for the MISO case. 2) We consider now random complex Gaussian zero-mean propagation channels (Rayleigh case). We then have μ = 0 and σh≠ 0. Calculate the mean value  of the SNR for the SISO case and the mean value  of the SNR for the MISO case. 3) We consider that the link is lost if the instantaneous received power is lower than –100 dBm. We consider the two cases: SISO and MISO and for the sake of simplicity we consider an unmodulated signal, such that: ∀n, s(n)= Pe and s (n)=

Pe 2

We consider μh = 0 and σh = 10–6, calculate the transmitted power Pe in the two cases, such that the outage probability stays lower than 10–2. We give cumulative distribution for 2 (Figure 5.7) and 4 degrees of freedom (Figure 5.8) chi-square distribution.

Figure 5.7. Cumulative density of the 2 degrees of freedom chi-square distribution

MIMO and Beamforming

143

Figure 5.8. Cumulative density of the 4 degrees of freedom chi-square distribution

Solution 1) The signal-to-noise derivation, in the deterministic case, leads to: 1 =

E |μs(n)|2 Pe = μ2 2 2 E |b1 (n)| 2σ

and: 2 =

E 2μ2 s(n) *

2

E μb1 (n)+μb2 (n+1)

2

The two noise terms are independent and centered, then:

E b1 ( n) b2* ( n + 1)  = 0  

144

Wireless Telecommunication Systems

This yields to: Pe Pe 2 = 2 22 = μ2 2 2σ μ 4σ 4μ4

 , this result is quite normal, knowing We note that  that we used a propagation channel model with an attenuation equal to μ and a transmitted power Pe in the first case, while, in the second case, we used identical channel models with equal attenuation equal to μ but with two P

transmitted powers equal to e , in order to lead the 2 comparison with the same global transmitted power. 2) The signal-to-noise derivation, in the case of zero-mean random propagation channels, gives:

1 =

E |h1 s(n)|2 E |b1 (n)|2

Propagation channels and transmitted symbols being independent, we can write:

1 =

E |h1 |2 E |s(n)|2 E |b1 (n)|2

We then obtain a random variable Y=|h1 |2 that is equal to Re{h1}2 + Im{h1}2. We know that Re{h1} and Im{h1} are two complex zero-mean random variables with the same variance σ2h . The summation of these two variables yields a new random variable chi-square distributed with 2 degrees of freedom [PRO 08]. 2 We have E  h1  = 2 σ h 2 , and then:  

2σ2h Pe σ2h 1 = = 2 Pe 2σ2 σ

MIMO and Beamforming

145

The derivation for 2 leads to:

2 =

|h1 |2 +|h2 |2 s(n)

E

2

*

*

E h1 b1 (n)+h2 b2 (n+1)

2

Independence of propagation channels, symbols and additive noise terms leads to:

2 =

2 =

to:

|h1 |2 +|h2 |2

E *

E h1 b1 (n)

E

2

2

E |s(n)|2 *

+E h2 b2 (n+1)

|h1 |2 +|h2 |2

transmitted

2

2

E |s(n)|2

E |h1 |2 E |b1 (n)|2 +E |h2 |2 E |b2 (n)|2

A new random variable Z=|h1 |2 +|h2 |2 appears. It is equal Re{h1}2 + Im{h1}2 + Re{h2}2 + Im{h2}2.

We know that Re{h1}, Im{h1}, Re{h2} and Im{h2} are four zero-mean Gaussian random variables with same variance σ2h . The Z variable is then chi-square distributed with 4 degrees of freedom. We have E |Z|2 = n2 +2n σ4 with n = 4, and then E |Z|2 = 24σ4h : Pe 2 = 2 2 22 2 2σh σ +2σh σ 24σ4h

2 = 3

σ2h P σ2 e

3) For the SISO case, the instantaneous received power is a random variable equal to: Pr = |h1 |2 Pe

146

Wireless Telecommunication Systems

This random variable is chi-square distributed with 2 degrees of freedom. Its mean is equal to 2σ2h Pe and its variance is equal to 4σ4h P2e . We can analyze the normalized variable: Pr =

Pr

σ2h Pe

=

|h1 |2

σ2h

This random variable is chi-square distributed with 2 degrees of freedom. Its mean is equal to 2 and its variance is equal to 4. The cumulative density function is shown in Figure 5.9.

Figure 5.9. Cumulative density function (cdf)

From Figure 5.9, it appears that an outage probability of 10–2 leads to a random variable equal to 2 × 10–2. This value for P′r should correspond to the limit value Pr = –100 dBm.

MIMO and Beamforming

147

The minimal transmitted power Pe is such that: Pr

σ2h Pe

= 2 102

or: Pe =

1

2

Pr 2 2 10 σh

Given in dBm, this equation becomes:

Pe, dBm = −3dB + 20dB + 120dB − 100dBm then:

Pe,dBm = 37 dBm In the SISO case, the instantaneous received power is a random variable equal to: Pr = |h1 |2 +|h2 |2

Pe 2

This random variable is chi-square distributed with 4 degrees of freedom. Its mean is equal to 8σ2h variance is equal to 8σ4h

P2e 4

Pe 2

and its

.

We can analyze the normalized variable: Pr =

2Pr

σ2h Pe

=

|h1 |2 |+h2 |2

σ2h

This is chi-square distributed with the 4 degrees of freedom law. Its mean is equal to 4 and its variance is equal

148

Wireless Telecommunication Systems

to 8. The cumulative density function of this variable is shown in Figure 5.10.

Figure 5.10. Cumulative density function (cdf)

From this figure, it appears that the 10–2 requirement of this exercise for the probability yields to a random variable lower than or equal to 3 × 10–1. This value for P′r should correspond to the limit value Pr = –100 dBm. to:

Then, the minimal transmitted power Pe should correspond 2Pr

σ2h Pe or: Pe =

=3 101

1 1

3 × 10

2Pr

σ2h

Given in dBm, this equation becomes:

Pe, dBm = −4.8dB + 10 dB + 120 dB + 3dB − 100 dBm

MIMO and Beamforming

149

then:

Pe,dBm = 28.2 dBm In conclusion, it appears that due to the 4 degrees of freedom distribution, it is possible to decrease the transmitted power with the same outage probability. 5.3. MIMO and capacity We consider a MIMO transmission with N transmitting antennas and N receiving antennas. We introduce the X(n) vector representing the transmitted communication symbols:  x1 ( n)    x2 ( n)   X (n) =       x N ( n) 

We consider that all the transmitted communication symbols have the same power ps. We introduce the R(n) vector representing the received symbols:  r1 ( n)    r ( n)  R ( n) =  2       rN (n) 

The two vectors X(n) and R(n) are linked through the propagation matrix H given by:  h1,1 ... h1, N    H =  h  N ,1 ... hN , N

    

150

Wireless Telecommunication Systems

Then, we have:

R ( n ) = HX ( n ) + B ( n) with:  b1 ( n )    b2 ( n )   B ( n) =       bN ( n ) 

This represents a complex vector of additive white Gaussian noise terms received on each sensor of the antenna. Each bk(n) term is zero mean and has a variance equal to 2σ2. We can then decompose the matrix H into singular values:

H = UDV H where U and V are the unitary matrix and D is the diagonal matrix of singular values of H. A precoder can then be used before the transmission. It gives a new vector X(n) defined as follows: X(n)=VX(n) The received signal is then given by: Y(n)=HX(n)+B(n) The decoder, which has to be implemented after reception, is defined by the following equation:

Y (n) = U H Y (n)

MIMO and Beamforming

151

Figure 5.11. Precoder and decoder MIMO

1) Write the received signal Y(n) with respect to H, X(n), and B(n). Prove that the “MIMO channel” can be viewed as N virtual parallel SISO channels. Calculate the SNR observed on each of these virtual channels.

UH

2) Considering that all virtual parallel SISO channels have the same transmission quality, i.e. have the same mean attenuation (∀k, |dk |= constant), calculate the capacity of the MIMO propagation channel and compare it to the capacity of the SISO propagation channel (with the same global transmitted power). Solution We have:

(

Y ( n) = U H H X ( n) + B (n)

)

Y ( n) = U H ( H V X ( n) + B ( n ) ) Y (n) = D X (n) + U H B (n)

Matrix D being diagonal, we can consider that we have N virtual propagation channels. The xk(n) symbol transmitted on the kth channel will be received as: N

yk (n) = dk xk (n)+

u*j,k bj j=1

where dk represents the diagonal terms of the matrix D and uk,j represents the terms of the matrix U. Knowing that

152

Wireless Telecommunication Systems 2

∑N j=1 uj,k =1, we can introduce a new bk (n) noise term defined as follows: N

u*j,k bj

bk (n)= j=1

Its power remains unchanged compared with bk(n), E |bk (n)|2 = 2σ2 . The SNR Γk on the kth virtual channel is equal to:

k =

|dk |2 ps 2σ2

REMARK.5.1.– In this question, we can write: ∀k, |dk |2 =|d|2 . We have to compare two transmission solutions. The first solution uses a transmitted power Pe with a SISO propagation channel, the SNR is then given by:

=

|d|2 Pe 2σ2

The channel capacity

C1×1 (in bits/s/Hz) is then equal to: B

C1×1 = log 2 (1 + Γ ) B In the second solution, we have N virtual propagation channels with identical SNRs. Moreover, the summation of the noise terms leads to a variance decrease (similar to a beamforming effect) and, finally, we have: , k =

|d|2 ps |d|2 Pe = 2σ2 2σ2

MIMO and Beamforming

153

Then:

Γk =Γ The global channel capacity is then equal to the summation of all capacities, and we obtain:

CN × N = N log 2 (1 + SNR) B Capacity values are given in Figure 5.12 for different SNR values and for different SISO and MIMO structures.

Figure 5.12. Capacity versus SNR for different SISO and MIMO structures

The gain due to the MIMO approach is shown clearly in Figure 5.12. Using different virtual channels leads to an important capacity increase. Nevertheless, this result is totally dependent on the number of non-null singular values of the matrix H.

Chapter 6

UWB

Ultra Wide band (UWB) communication has been analyzed in depth since the publication of the Federal Communications Commission (FCC) report on this subject in 2002. The main justification for such communications comes from the Shannon capacity theorem:

C = B × log 2(1 + SNR) where C represents the channel capacity and B represents the transmission bandwidth. It is obvious that the logarithm function will always temper a signal-to-noise ratio (SNR) increase while a bandwidth increase will have a directly proportional effect on the capacity. Recent interest for UWB communications was initially due to the development of impulse-based waveforms, generally demodulated by a receiver with an ultra-precise analog frontend, able to generate ultra-short impulses. These waveforms have shown various advantages (carrier-free waveforms); nevertheless, it recently emerged that their transmission capacities were restricted and the UWB community orientated some parts of its work towards

156

Wireless Telecommunication Systems

orthogonal frequency division multiplex (OFDM)-based waveforms (see, for instance, the UWB ECMA standard). The First Report and Order on UWB was published in 2002. FCC considers that UWB could be authorized in licensed bands with the condition of respecting a power transmission mask. It is considered that this mask is sufficient to protect other telecommunication or localization systems in these bands. The main UWB frequency bands range from 3.1–10.6 GHz. The maximal mean value for the power spectral density is equal to –41.3 dBm/MHz and 0 dBm/50 MHz for the peak power spectral density. The indoor and outdoor masks are presented in Figures 6.1 and 6.2.

Figure 6.1. UWB outdoor mean power spectral density mask (FCC 2002)

The main notch in these masks between 960 MHz and 1.61 GHz is justified by the global positioning system (GPS) protection constraints. UWB is mainly devoted to short-distance communication systems. Low bit rate and high bit rate UWB systems are currently available. Low bit rate UWB communication

UWB

157

systems are devoted to transmission with a bit rate lower than 1 Mbps for sensor networks with low-cost transmission systems. Localization requirements are generally included in these contexts and the main standard is IEEE802.15.4a.

Figure 6.2. UWB indoor mean power spectral density mask (FCC 2002)

High bit rate UWB communication systems are devoted to transmission with bit rates higher than 400 Mbps and for distances less than 10 m. The corresponding standard is ECMA-368 High rate UWB PHY and MAC Standard. 6.1. Impulse UWB We consider an ultra-short impulse-based waveform (shorter is the impulse, wider is the corresponding bandwidth). Figure 6.3 represents a typical impulse that could be defined as the derivation of a Gaussian function [BEG 10]. We consider that the impulse lasts approximately 250 ps, corresponding to a frequency bandwidth equal to 1 250ps = 4GHz .

158

Wireless Telecommunication Systems

Figure 6.3. Transmitted impulse lasting approximately 250 ps

Useful information could be transmitted using an impulse comb with micro delays between comb impulses (pulseposition modulation). Figure 6.4 represents such an impulse comb with four impulses and an inter-impulses time delay equals to 100 ns.

Figure 6.4. Four impulses, time between impulses equal to 100 ns

UWB

159

Figure 6.5. Four received impulses after a typical four paths propagation channel, time between impulses equal to 100 ns

We consider that the maximal delay spread of the channel is equal to 60 ns. Having ultra-short impulses we consider that different channel paths are resolved and can be distinguished from each other. Gain of these paths is often modelized as a deceasing exponential function [SAL 87]. A typical representation of the received signal after a four-path propagation channel is represented in Figure 6.5. 1) Considering that we use a 10 impulse comb to transmit 1 bit (without considering any forward error correcting code), that the impulse lasts 250 ps and that we transmit one impulse every 100 ns, give the spectral efficiency of this waveform. Could you propose a solution to increase this spectral efficiency? 2) In a multiple user context, where each user starts their transmission randomly without taking care of other users, how many users could share the transmission bandwidth with the proposed UWB waveform of this exercise? 3) Could you conclude and propose an analysis of this multiple access UWB system?

160

Wireless Telecommunication Systems

Solution 1) We need 10 impulses in order to transmit 1 bit, impulses are transmitted every 100 ns, the bit rate is then equal to: Rb =

1 10 × 100 ns

then:

R b = 1 Mbits/s The frequency bandwidth is approximately equal to 4 GHz and the spectral efficiency is then equal to: =

106 = 0.2510−3 bits s / Hz 4109

The spectral efficiency can be increased by a reduction of the inter-impulse time. Nevertheless, considering a 60 ns delay spread for the channel, it is not possible to shorten this time by more than this limit. An amplitude modulation applied to impulses can also be introduced; nevertheless, it does not seem to be reasonable to go further than a 2 ASK modulation for a UWB transmission context (two amplitude states). These two solutions could be merged together, and it should probably be possible to multiply the bit rate by 4. It is clearly apparent that the obtained throughput is much closer to 1 Mbps than to 400 Mbps, indicating that impulse-based waveforms will mainly be devoted to low bit rate UWB applications. 2) The proposed waveform spreads the signal in an ultra-wide frequency band and is then naturally well suited for a multiple access scheme. Considering first a simple hypothesis that each user needs a 250 ps time slot to transmit their pulse in a 100 ns frame, we can then divide this frame in 100 × 10−9 250 × 10−12 = 400 time slots. The

(

)(

)

UWB

161

multiple access scheme then has a theoretical maximal capacity equal to 400 simultaneous users (we can note that 400 users at 1 Mbps leads to a global 400 Mbps throughput). Of course, this maximal value does not take collisions into consideration, which should be very important in such a case due to random transmission and different multipaths of users’ propagation channels. 3) It was shown that a multiple access system could be perfectly built with such an impulse-based waveform. Each user will be spread over a 4 GHz wide bandwidth and will only transmit during (1 400) th of the time. Its power spectral density will then be very low and compliant with the FCC mask. Useful throughput per user will be weak and this solution will be devoted to low rate applications like sensor networks. Simple localization techniques based on triangularization algorithms could easily be envisaged with these impulse-based waveforms. 6.2. UWB and OFDM The other technical solution for UWB transmission is based on OFDM and is referred to as Multiband OFDM Alliance (MBOA). The UWB frequency band (3.1–10.6 GHz) is then divided into several 500 MHz sub-bands (i.e. the minimum authorized value for an UWB transmission system), each of them divided into 128 subcarriers. 1) Complete main characteristics of the OFDM UWB approach presented in the Table 6.1. 2) Explain the role of the TGI parameter. 3) What is the useful throughput of this waveform considering that only 100 subcarriers among 128 are used for useful data transmission? We consider the Quadrature Phase Shift Keying (QPSK) modulation and a forward error code with r = 3 4 .

162

Wireless Telecommunication Systems

Parameters

Value

FFT/IFFT size

128

Transmission bandwidth

?

Inter subcarrier spacing

?

Time for 128 samples (TFFT)

242.42 ns

Cyclic prefix duration (TCP)

60.61 ns (samples)

Guard interval duration (TGI)

9.47 ns

OFDM symbol duration (TSYM)

?

Table 6.1. MBOA solution main parameters

Solution 1) The OFDM symbol represents 128 + 32 = 160 samples, these samples will be transmitted in 242.42 + 60.61 = 303.03 ns, giving a sample duration:

Te =

303.03 = 1.89 ns 160

This sample duration leads to a frequency band equal to: B≈

1 = 528MHz Te

Subcarrier spacing is then equal to:

528MHz = 4.125 MHz 128 2) Compared with an OFDM approach using only one 528 MHz channel, we have to consider a 528 MHz sub-band frequency hopping scheme. A guard time TGI is then inserted between OFDM symbols in order to process to these frequency hops. Integrating this guard time, the complete

UWB

163

OFDM symbol duration is then equal to 312.5 ns. Finally, we obtain Table 6.2. Parameters

Value

FFT/IFFT size

128

Transmission bandwidth

528 MHz

Inter subcarrier spacing

4.125 MHz

Time for 128 samples (TFFT)

242,42 ns

Cyclic prefix duration (TCP)

60.61 ns (32 samples)

Guard interval duration (TGI)

9.47 ns

OFDM symbol duration (TSYM)

312.5 ns (TFFT+TPC+TGI)

Table 6.2. Main parameters of the MBOA solution

3) The reachable useful throughput with 100 subcarriers using a QPSK modulation (2 bits per symbol) and with an r = 3/4 forward error correcting code, with an OFDM symbol duration of 312.5 ns, is then equal to:

3 4 = 480 Mbits/s Rb = 312.5ns 100× 2×

Finally, we obtain the following spectral efficiency:

η=

480 = 0.9 bits/s/Hz 528

6.3. Link budget for UWB transmission We wish to establish a UWB transmission being compliant with the “FCC 2002 outdoor” requirements in the frequency band (3.1–10.6 GHz). We decide to transmit in a 2 GHz sub-band. Distance between the transmitter and the receiver is equal to 10 m. We consider a 64.2 dB propagation loss at 10 m. Transmitter and receiver antenna gains are

164

Wireless Telecommunication Systems

equal to 0 dBi. A noise figure of 6.2 dB has to be taken into consideration in this link budget and we add a 2.6 dB margin for implementation loss. 1) Calculate the received power. 2) Calculate the SNR. 3) Considering an additive white Gaussian noise channel, calculate the transmission capacity of this waveform. Solution 1) We are using a 2 GHz bandwidth, corresponding to 33 dBMHz. The “FCC 2002 outdoor” requirement limits the equivalent isotropically radiated power (EIRP) to –41 dBm/MHz, the transmitter EIRP is then equal to:

-41dBm/MHz + 33dBMHz = -8dBm The received power, before implementation loss, but considering the 64.2 dB of propagation loss, is equal to:

Pr = −72.2dBm 2) The thermal noise power spectral density is equal to N0 = kT, where k stands for Boltzman constant (1.38 × 10–23 JK–1) and T for the receiver noise temperature chosen, equal to (300 K) in this exercise. Given in dBm, this thermal noise power spectral density is equal to –174 dBm/Hz. The received power in the 2 GHz ↔93 dBHz sub-band is then equal to:

−174 dBm / Hz + 93 dBHz = −81dBm

UWB

165

Including implementation loss and the noise figure, we obtain the SNR:

ΓdB = −72.2 dBm − 6.2 dB − 2.6 dB + 81dBm = 0 dB Corresponding, in linear, to: Γ =1

3) Channel capacity is given by (integrating receiver loss):

C = B × log 2 (1 + Γ) then:

C = 2109 × log 2 (1 + 1) = 2 Bbits / s Finally, we have used 2 GHz bandwidth for 2 Gbit/s throughput, this result yields to a spectral efficiency of 1 bit/s/Hz.

Chapter 7

Synchronization

The performance of modems is very strongly related to time, phase and frequency synchronization algorithms. The problem of synchronization is often approached in a theoretical manner, by an optimal approach based on the maximum likelihood criterion. Algorithmic optimizations are often used to exhibit suboptimal structures that are feasible in practical terms [VIT 83]. Thus, the most effective approach is to jointly estimate the received symbols, the phase of the carrier and the timing error, but it is simpler, from a practical point of view, to separate these different problems. Synchronization algorithms can use a feedback loop or can work in open loop without any analysis of their results. In other words, they are often classified into data-aided (DA), non-data-aided (NDA) or decision-directed (DD) synchronization algorithms. In the NDA case, these algorithms are mentioned as blind algorithms. The main functions of these algorithms are in general: – Time synchronization, which consists of sampling the received signal with the optimal timing in order to benefit from the maximal opening of an eye diagram, for instance.

168

Wireless Telecommunication Systems

– Frequency synchronization, which consists of locking the receiver oscillator exactly onto the transmitter oscillator. A frequency shift can easily appear between these two oscillators and this leads to severe degradation of the transmission. The frequency Doppler effect due to the mobility of mobile users belongs to these kinds of well-known problems. – Amplitude and phase synchronization, which consists of aligning the received samples with the correct modulation alphabet. This problem is also referred to as channel estimation and frequency equalization in an orthogonal frequency division multiplex (OFDM) context. There are many different synchronization problems and this chapter, due to reasons of space, will not address all of them. It will only introduce some simple exercises, illustrating how these problems can easily be taken into consideration in the design of a radiocommunication modem. One of the best theoretical tools for qualifying these algorithms is the Cramer–Rao bound [SAP 90]. 7.1. Cramer–Rao bound Let us consider n-samples {R1, R2,…, Rn} of random variables Ri depending on an unknown scalar parameter θ that we have to estimate. The Fisher information In(θ), given by n-samples, and concerning the unknown parameter θ, is defined by the following equation: In (θ) = E

∂Ln pR (r|θ)

2

∂θ

In this expression, pR (r|θ) stands for the product of the conditional probability densities with respect to the unknown parameter θ of Ri random variables.

Synchronization

169

n

pR (r|θ)=

i=1

pR (ri |θ)

With some particular conditions, we can demonstrate that: In (θ) = –E

∂2 Ln pR (r|θ) ∂θ2

 Then, we show that for all unbiased estimators θ (r ) of the unknown parameter θ, we have:  E θ ( r )  = θ

Then the estimator variance is lower bounded by the value 1 : I n (θ ) σ2θ = E θ(r) – E θ(r)

2

≥

1 In (θ)

1 represents the Cramer–Rao bound for the estimation I n (θ ) of unknown parameter θ. where

An estimator that reaches this bound is called “consistent”. For

the

case T

of

a

vectorial

parameter

θ,

defined

as

θ = θ1 ,θ2 ,…,θK , we introduce the (K × K) Fisher information matrix defined by its ith line and jth column general element: Ji,j (θ)=E

∂Ln pR (r|θ) ∂θi

×

∂Ln pR (r|θ) ∂θj

Introducing Ji,j as the ith line and jth column general element of the inverse Fisher matrix J–1, we show that, for all unbiased

170

Wireless Telecommunication Systems

estimators of parameter θi, the estimator variance is lower bounded by the value Ji,i: 2

σθ2 = E θi (r) – E θi (r)

≥ Ji,i

i

7.2. Modified Cramer–Rao bound A simplified modified Cramer–Rao bound (MCRB) can easily be used for time, phase and frequency estimation [D’AN 94]. Considering that we observe L communication symbols with a symbol duration equal to Ts and a signal-to-noise ratio given by E the s ratio, MCRBs are given by: N0

MCRB (Δ f ) = MCRB (Δϕ ) =

3Ts

3

2π ² ( L Ts )

×

Es

1 1 × 2 L Es N0

1 N0

The unit of measurement for MCRB(Δf) is Hz2 and the unit for MCRB(Δφ) is rad2. Figures 7.1 and 7.2 show these two MCRBs with respect to signal-to-noise ratio for L = 64 and Ts = 10–4s. 7.3. Constant parameter estimation We consider the following scalar n-samples: ri = θ + ni where ni represent independent additive white Gaussian noise terms having the same variance σ2. We have to estimate the unknown constant θ due to the following estimator: 1 θ= n

n

ri i=1

Is this estimator consistent?

Synchronization

Figure 7.1. Modified Cramer–Rao bound for the frequency MCRB(Δf)

Figure 7.2. Modified Cramer–Rao bound for the phase MCRB(Δφ)

171

172

Wireless Telecommunication Systems

Solution Let us first check that this estimator is unbiased: 1 Eθ = n 1 Eθ = n

n

E ri i=1 n

(θ + E ni ) i=1

As white additive noise terms are centered, we have E[ni] = 0; then: 1 Eθ = n

n

θ =θ i=1

Having an unbiased estimator, we can now calculate its variance: 2

σ2θ = E θ

– θ2

σ2θ = E

σ2θ = E

θ+

σ2θ

2

n

1 n

(θ + n i )

– θ2

i=1

∑ni=1 ni n

2

– θ2

∑ n ni ∑ n ni + i=1 = E θ + 2θ i=1 n n 2

2

– θ2

Noise terms being centered, this equation becomes:

σ2θ = E

∑ni=1 ni n

2

Synchronization

173

or:

σ2θ

1 = 2E n

n

n

n2i

+

ni nj i,j,i≠j

i=1

The variables ni being independent and having same variance, we finally obtain:

σ2θ =

σ2 n

Let us now check that this value is equal to the Cramer–Rao bound. For that purpose, we calculate: n

pR (r|θ)=

i=1

pR (ri |θ)

n

pR (r|θ)=

i=1

1 2πσ2

(r -θ)2 - i 2 e 2σ

then:

Ln ( pR ( r | θ ) ) = Ln

(

n 2 2 2πσ

n

) − i =1

( ri − θ )2 2σ 2

Finally: ∂2 Ln pR (r|θ) 2

∂θ

=–

n σ2

The Cramer–Rao bound is equal to: 1 1 =– In (θ) ∂2 Ln pR (r|θ) ∂θ2

=

σ2 n

174

Wireless Telecommunication Systems

The proposed estimator reaches the Cramer–Rao bound, its variance is then minimal and the estimator is consistent. 7.4. Radio burst synchronization Radio transmission standards often define synchronization channels used for the first step of receiver synchronization. In this section, we will consider a logical channel devoted to the frequency synchronization (frequency channel – FCH). More precisely, we will consider a known 64-communication-symbols burst. Let us consider that these symbols are received with a signal-to-noise ratio given by:

Es = 20 dB N0 We consider a 10 k symbols-per-second data rate and focus on the frequency fine-tuning at the receiver level. 1) Describe the effect of a frequency shift at the receiver level. (Could we consider that all known transmitted symbols belong to a Quadrature Phase Shift Keying (QPSK) alphabet and are all equal to 1 + j?) 2) Propose a frequency synchronization algorithm in order to estimate the frequency shift Δf. 3) Considering that we are able to exhibit an unbiased estimator for Δf, calculate the Cramer–Rao bound for this frequency estimation problem. 4) Considering that this bound is too high, what should be the best solution: multiplying the number of known symbols by 2 or multiplying the transmitted power of these known symbols by 2? Solution 1) The frequency shift involves a rotation of symbols around the ideal modulation alphabet. A typical example is presented in Figure 7.3.

Synchronization

175

E

Figure 7.3. Evolution of 1,000 QPSK symbols, with Ns =20 dB, 0

for a frequency error Δf = 10 Hz and t for a data rate equal to 10 ksymb/s

2) The frequency shift can be estimated by calculating the argument of the maximum Fourier transform modulus of received symbols (knowing that these symbols are unmodulated and all equal to 1 + j). To improve the resolution of the Fourier transform, which will basically be equal to:

Fs 10 kHz = 64 N we can propose a zero padding [BEL 06], completing received symbols by a set of values equal to 0. 3) The MCRB is given by:

MCRB (Δ f ) =

3Ts

3

2π ² ( L Ts )

×

Es

1 N0

176

Wireless Telecommunication Systems

With values proposed for this exercise: L = 64, Ts = 10–4 E and s = 20 dB, we obtain: N MCRB(Δf) ≈ 58 Hz2 This value corresponds to a standard deviation equal to 7.6 Hz. This frequency shift is very weak but it will nevertheless generate a rotation of the modulation alphabet. A regular phase correction could then be proposed in order to periodically correct this rotation. 4) Analyzing the MCRB, we obtain:

MCRB (Δ f ) =

3Ts

3

2π ² ( L Ts )

×

Es

1 N0

It clearly appears that the symbol number L, used for the estimation process, plays a major role due to its third power in the equation. It will then be much more efficient to increase the number of symbols than to increase their power. 7.5. Phase estimation for QPSK modulation We receive QPSK communication symbols with a phase shift residual error Δφ with respect to the ideal constellation. We have to fulfill a requirement asking us to reach a phase shift residual error lower than

π

20

.

Es = 6dB N and considering that the phase estimation will be processed due to a preamble, how many symbols are required for this preamble? 1) Knowing that the signal-to-noise ratio is equal to

2) Could you propose a phase shift estimator algorithm?

Synchronization

177

Solution 1) We propose translating the requirement due to the fact that the standard deviation has to be lower than a variance lower than

π2 100

π

20

, corresponding to

.

Knowing that the best estimator cannot have a variance lower than the Cramer–Rao bound, the estimator variance σ2φ should belong to the following interval:

MCRB(Δϕ ) ≤ σ ϕ 2 ≤

π2 100

We can estimate the minimal number of preamble symbols using the following equation: 1 1 π2 = 2L Es 400 N0 We then obtain: L=

With

L=

200 1 π2 Es N0

Es = 100,6 ≈ 4, we obtain: N 50

π2

Then: L≈6

178

Wireless Telecommunication Systems

2) Introducing xi as received symbols, di as known symbols of the preamble, ni as independent additive white Gaussian noise samples and, finally, ∆ as the phase shift, the received signal can be written as follows: xi = di ej∆φ + ni A simple phase estimation algorithm could then be given by: L *

∆φ = angle

xi × di i=1

Chapter 8

Digital Communications Fundamentals

8.1. Review of signal processing for signal-to-noise ratio The most useful signal processing tool [BEL 06] to perform signal-to-noise ratio calculations is the mathematically expected value E[.]. To calculate the expected value, it is necessary to know the probability densities of the signals, which unfortunately rarely happens. The expected values are then evaluated by the time-average on the signal’s observed values. Since most telecommunication signals are centered, the second-order moment (the expected value of the square of the signal) is often assumed to be equal to the variance. This second-order moment is usually called the “signal power”. 8.2. Review of digital modulations Digital modulations allow us to transmit several bits at the same time in a given symbol time. For this purpose, bits are grouped into symbols. Two types of modulations can be distinguished: linear modulations and frequency modulations.

180

Wireless Telecommunication Systems

In linear modulations, the information relative to the symbols does not depend on the frequency and is thus similarly contained in the band-pass signal and in the equivalent low-pass signal. Three main types of linear modulations exist: amplitude modulations (amplitude shift keying – ASK), phase modulations (phase shift keying – PSK) and amplitude and phase modulations (quadrature amplitude modulation – QAM). The latter lead to lower binary bit error rates for the same required binary energy, and are therefore the most commonly used in practice. In frequency modulations, information is contained in the frequency of the band-pass signal. Continuous phase frequency modulations are preferred in systems because they are more efficient regarding the spectral band. Global system for mobile communications (GSM) and general packet radio service (GPRS) use a binary frequency modulation called Gaussian-filtered minimum shift keying (GMSK). In enhanced data rates for GSM evolution (EDGE), the modulation is either GMSK, or a modified 8-PSK, depending on the radio conditions. In universal mobile telecommunications system (UMTS) and high-speed uplink packet access (HSUPA), the transmission symbols belong to a QPSK. In HSDPA, either QPSK or 16-QAM modulations can be used. Finally, in the long-term evolution (LTE) standard, three modulations can be used, depending on the radio conditions: QPSK, 16-QAM and 64-QAM. 8.3. Review of equalization In a band-limited channel with additive white Gaussian noise (AWGN), inter-symbol interference can be totally removed if the equivalent filter verifies the Nyquist criterion. However, in wireless systems, this constraint is not always fulfilled, because of the radio channel distortions. If the maximum delay of the multipath channel is larger than the

Digital Communications Fundamentals

181

symbol time, successive transmission symbols will interfere at the receiver. In this case, the receiver must estimate the transmit symbols despite inter-symbol interference. For this purpose, it must add an equalization step before detection. Linear equalization is used after sampling. The received symbols are filtered by an equalization filter with a finite impulse response. The latter either aims to remove all intersymbol interference (zero-forcing equalizer), without considering the influence of this step on the noise, or to minimize the mean-square error between each transmission symbol and its estimate (minimum mean square error (MMSE) equalizer). Nonlinear equalization is more complex but more efficient. Decision-feedback equalization can be combined with zero-forcing or MMSE criteria. Maximum likelihood equalization determines the set of transmission symbols of highest probability using Viterbi algorithm. This method is the most accurate, but also the most complex. 8.4. Signal-to-noise ratio estimation We consider two real signals x1(n) and x2(n), defined as follows: x1(n) = s(n) + b1(n) x2(n) = s(n) + b2(n) s(n) is a useful zero-mean random signal, its mean power is given by: E[s(n)2] = ps b1(n) and b2(n) signals are two random independent white noises with same variance σ2.

182

Wireless Telecommunication Systems

1) Calculate the signal-to-noise ratio for x1(n) and x2(n). 2) We decide to add these two signals and we obtain y(n) = x1(n) + x2(n), give the signal-to-noise ratio for y(n). 3) We consider now that the two signals for x1(n) and x2(n) have been provided by two different antennas. We formalize this difference considering that the useful signal s(n) is, in comparison to the noise, less important on the second antenna than on the first antenna. We will then write x2(n) = αs(n) + b2(n), with 0 ≤ α ≤ 1. The x1(n) signal remains unchanged. We decide to add x1(n) and x2(n) but with an adaptive tap a acting on x2(n), we obtain then z(n) = x1(n) + ax2(n). Which value of the parameter a maximizes the signalto-noise ratio of ( )? Solution 1) We have: x1 =

E s(n)

2

E b1 (n)

2

=

ps σ2

and: x2 =

E s(n)

2

E b2 (n)

2

=

ps σ2

2) We have: y =

E (2s(n))2 E (b1 (n)+b2 (n))2

Noise terms being zero-mean and independent, we have:

E [b1 (n) × b2 (n) ] = E [b1 (n) ] × E [b2 (n) ] = 0

Digital Communications Fundamentals

183

then: 2 2 2 E ( b1 ( n ) + b2 ( n ) )  = E b1 ( n )  + E b2 ( n )  = 2 σ 2      

y =

4ps

2σ2

=2

ps

σ2

Summation of these signals gives finally a 3 dB gain on the signal-to-noise ratio. 3) We have now: z (a)=

E (1+aα)s(n)

2

E (b1 (n)+ab2 (n))2

We obtain then: z (a)=

(1+aα)2 ps (1+a2 )σ2

This signal-to-noise ratio is a positive function of the parameter a. We have then to identify values that cancel its derivation: ∂z (a) 2α(1+aα)ps 1+a2 σ2 – 2aσ2 (1+aα)2 ps = 2 ∂a (1+a2 )σ2 Or, after factorization: ∂z (a) 2(1+aα)ps σ2 (α–a) = 2 ∂a (1+a2 )σ2 Two solutions cancel this expression: – either a =

−1

α

, this solution cancels the useful signal

and leads to a signal-to-noise ratio equal to 0, it corresponds then to a minimum of z (a);

184

Wireless Telecommunication Systems

– or a = α, this solution corresponds to a maximum of Гz(a) In this case, we obtain: z (α)=

1+α2 ps

σ2

If we consider α = 1, we obtain the result of question 2. In the opposite direction, if we consider α = 0, we obtain the result of question 1. The function

Γ z (α ) is represented in Figure 8.1, we Γ x1

observe that the combining gain varies between 0 and +3 dB. We can comment that, even for weak values of the parameter a, the combining gain increases the signal-tonoise ratio.

Figure 8.1. Signal-to-noise ratio obtained after signal combining

8.5. ASK 2 modulation error probability An s(t) signal is transmitted by two symbols taking the following values A or B, with {pA,pB} as the probability

Digital Communications Fundamentals

185

associated with transmission of these symbols. This s(t) signal is affected by an AWGN channel with a σ2 variance. The resulting y(t) signal will then have the probability density function as illustrated in Figure 8.2 (assuming pA < pB).

Figure 8.2. Probability density function of the received signal in a non-equiprobable binary transmission

On the reception side, the y(t) signal is compared to a threshold S and it is then possible to decide which symbol was transmitted: if y(t) > S, then it is decided that the symbol A was transmitted, if y(t) < S, then it is decided that the symbol B was transmitted. 1) Calculate the symbol decision error probability Pe. For this purpose, the following Marcum function will be used: Q ( x) =

u2

+∞ − 1 e 2 du 2π x



2) Deduce the optimum position for S, which minimizes the error probability.

186

Wireless Telecommunication Systems

Solution 1) We have Pe = pA peA + pB peB where peA and peB are the error probabilities associated with A and B symbol transmission. As: peA =

S

1

-∞ σ√2π

with u =

e

-(y-A)2 2σ2

y−A

σ

dy=

S-A 2 σ -u e2 -∞

1 √2π

du=

1 √2π

A-S 2 σ -v e2 +∞

(–dv)

and v = –u.

Then: peA =

1 √2π

+∞ -v2 e2 A-S σ

dv=Q

A-S σ

Similarly, as: peB = with u =

+∞ S

y−B

σ

Pe =pA Q

1

σ√2π

e

-(y-B)2 2σ2

dy=

1 √2π

+∞ -u2 e2 S-B σ

du=Q

S-B σ

, then: A-S S-B +pB Q σ σ

2) The optimum position for S is obtained for means that we must have: ∂ A-S – 0 = pA × – ×e ∂S σ

A–S σ

2

2

∂Pe ∂S

∂ S–B – + pB × – ×e ∂S σ

=0. It S–B σ

2

2

Digital Communications Fundamentals

0 = pA × e

–

A–S σ

2

2

p S–B ln A = – pB σ 2σ2 ln

–

– pB × e 2

2+

S–B σ

2

A–S σ

187

2

2

2

pA 2 = 2 S (B-A)+A – B2 pB

pA p p 2 2 ln B –A +B2 2σ2 ln B +A –B2 pB pA pA A+B 2 = = +σ 2 2 (A–B) 2 (B–A) 2 (A–B)

2σ2 ln S=

We can see that if pA = pB, then S is situated at the exact middle between A and B. If B transmission is more probable than A transmission, the value for the threshold S increases: it shifts to A to allow the receiver more chances to have y(t) ≤ S and to decide that B symbol was transmitted. 8.6. Spectral occupancy, symbol rate and binary throughput We consider information to be transmitted at a binary throughput Db set to 1 Mbits/s. The transmission channel (assumed to be Gaussian) bandwidth is equal to 400 kHz and Eb/N0 = 9 dB must be ensured. We suppose that the channel global filtering function has a raised cosine filter shape, with roll-off α being equal to 0.35. 1) What is the bandwidth occupied by this raised cosine filter as a function of the symbol rate Ds? 2) What is the Ds maximum value that enables information transmission in the considered channel? 3) Which minimum value must be adopted for the constellation size (i.e. number of possible modulated symbols) Mm in order to modulate this information?

188

Wireless Telecommunication Systems

4) We wanted to use an 8-PSK modulation. Using the same characteristics as previously for the transmission channel (α, B), what would be the maximum binary throughput value? 5) With this 8-PSK modulation, keeping Db = 1 Mbits/s and α = 0.35, what would the transmission channel total bandwidth value be? 6) With this 8-PSK modulation, keeping Db = 1 Mbits/s and B = 400 kHz, what would the transmission channel maximum roll-off be? Solution 1) The occupied band B’ is: B’ = (1 + α)Ds then: B’ = 1.35 × Ds 2) We want B’ ≤ B, hence 1.35 × Ds ≤ 400 × 103, thus Ds ≤ 296.3 kbauds (1 baud = 1 symbol/s). 3) We have Ds = Db/log2(M) ≤ 296.3 × 103, hence106/(296.3 × 103) ≤ log2(M) and, finally, log2(M) ≥ 3.37. As a result, Mm= 16 (the number of levels in a classic modulation is a power of 2 in order to ensure a constant number of transmitted bits per symbol). Because of bite error rate performances, we have the advantage of choosing a 16-QAM modulation rather than a 16-PSK modulation (or even better a 16-ASK modulation). 4) With an 8-PSK, Ds ≤ 296.3 kbauds leads to Db ≤ 888.9 kbits/s. 5) With an 8-PSK, Db = 1 Mbits/s leads to Ds = 0.333 Mbauds, hence B’ = 1.35 × Ds = 450 kHz.

Digital Communications Fundamentals

189

6) With an 8-PSK, B’ = (1 + α)Ds ≤ B leads to α ≤ –1 + B/Ds, hence α ≤ –1 + 400 × 103/(106/3), and finally α ≤ 0.2. 8.7. Comparison of two linear digital modulations This problem aims to compare 16-QAM and 16-PSK modulations. The minimum distance in a 16-QAM constellation is equal to 2B. The 16-PSK constellation points are located on a circle of radius A. 1) Give the number of bits per symbol. 2) Determine the energy per symbol and the energy per bit for both constellations. 3) With A set to a fixed value, express B as a function of A so that the energy per symbol with a 16-QAM modulation is equal to the energy per symbol with a 16-PSK modulation. 4) Then compare the modulations’ minimum distances. Calculate their values. What conclusion can be drawn regarding the performances? 5) Give the expression of the symbol error probability and the bit error probability with both modulations, depending Eb/N0 ratio. 6) Calculate them when Eb/N0 = 10 dB and when Eb/N0 = 20 dB. The following approximation can be used:

Q( x) = 0.208 × exp(−0.971× x 2 ) + 0.147 × exp(−0.525 × x 2 ) 7) Draw conclusions on the performances of both modulations. Solution 1) Since M=16, there are four bits per symbol. 2

2) The energy per symbol is equal to Es,16-PSK = A for 16PSK modulation, because all symbols are located on the

190

Wireless Telecommunication Systems

same circle of radius A. For 16-QAM, after some calculation, it is equal to Es,16-QAM =10 B2 . The energy per bit is equal to ¼ of the energy per symbol. It is therefore 2 2 equal to Eb,16-PSK =A /4 for 16-PSK and to Eb,16-QAM =5/2 B for 16-QAM. 3) Both energies per symbol are equal if B = A

10 .

4) The minimum distance of a 16-PSK modulation is: dmin,16-PSK = 2A×sin(π⁄16) The minimum distance of 16-QAM modulation is: dmin,16-QAM = 2B When both energies per symbol are equal, we thus have: dmin,16-PSK = √10×sin(π⁄16)dmin,16-QAM = 0.62dmin,16-QAM The minimum distance is lower with 16-PSK than with 16-QAM. Thus, it is likely that the bit error rate will be lower with 16-QAM. This, however, has to be checked, because both constellations do not have the same distribution. It is then not possible to compare them only on the basis of their minimum distances. 5) For 16-PSK modulation, the symbol error probability is:

 2 Pe,16− PSK = 2 Q  8 Eb N 0 × ( sin (π 16 ) )    For 16-QAM modulation, at high Eb/N0, the symbol error probability is approximately: Pe,16-QAM = 3Q

(4⁄5 Eb ⁄N0 )

Digital Communications Fundamentals

191

In both cases, the bit error rate is one quarter of the symbol error probability, with Gray coding and at high Eb/N0. 6) Using the approximate formula, we get: – For Eb/N0 = 10 dB: Pe,16–PSK = 0.084 and Pe,16–QAM = 0.007; – For Eb/N0 = 20 dB: Pe,16–PSK = 3.36 × 10–8 and Pe,16–QAM = 2.53 × 10–19. 7) 16-QAM modulation provides lower error rates than 16PSK modulation, and the difference increases when the Eb/N0 ratio increases. 8.8. Comparison of two-PSK modulation and power evaluations We consider a mobile terminal transmitting to a base station. The path loss is equal to 130 dB. The signal undergoes a white Gaussian noise with power spectral density N0,dBm = –174 dBm/Hz. It wants to transmit a signal with data rate D = 10 Mbits/s and bit error rate BER = 10–2. The total channel filter is rectangular in frequency domain. The performances of two-phase modulations: 4-PSK and 16-PSK are evaluated: 1) Calculate the required modulation, B4 and B16.

bandwidth

with

each

2) Calculate the noise power in both bandwidth. 3) Assuming that Gray coding is used, calculate the symbol error probability for both modulations. 4) Using Figure 8.3, calculate the corresponding minimum Eb/N0 ratio, for both modulations.

192

Wireless Telecommunication Systems

5) Calculate the signal-to-noise ratio considering power values, P/N, in both cases. 6) Deduce the received power in both cases. 7) Deduce the required power. Make conclusions regarding the feasibility of each modulation, considering that the maximum transmission power of the mobile terminal is 2 W. REMARK 8.1.– In this problem, we do not take into account the transmitter and receiver hardware gains and losses.

Figure 8.3. PSK symbol error probability

Solution 1) The required bandwidth is B = D/log2(M), where M is the constellation size. Indeed, since the channel filter is rectangular, the bandwidth is equal to the symbol time inverse. It is thus equal to: B4 = 5 MHz and B16 = 2.5 MHz. 2) The noise power in bandwidth B4 is: N4,dBW = –174 – 30 + 10 × log10(B4) = –137 dBW. It is therefore equal to: N4 = 10–137/10 = 2 × 1014 W. Similarly, the noise power in bandwidth B16 is equal to N16 = 1 × 1014 W.

Digital Communications Fundamentals

193

3) The symbol error probability is equal to the binary error probability multiplied by the number of bits per symbol. It is thus equal to Pe,4 = 2 × 10–2 for 4-PSK and to Pe,16 = 4 × 10–2 for 16-PSK. 4) By reading the figure’s curves, we obtain the required Eb/N0 ratio values: Eb/N0 = 4.5 dB for 4-PSK and Eb/N0 = 11.5 dB for 16-PSK. 5) The signal-to-noise ratio considering power values depends on the Eb/N0 ratio in the following way: P/N = log2(M)Eb/N0: – for 4-PSK modulation, it is equal to: Pr,4 ⁄N4 = 2×104.5

⁄10

= 5.63

– for 16-PSK modulation, it is equal to: Pr,16 ⁄N16 = 4×1011.5

⁄10

= 56.5

6) The received power is therefore: – for 4-PSK modulation: Pr,4 = 5.63 × N4 = 1.12 × 10–13 W; – for 16-PSK modulation: Pr,16 = 56.5 × N16 = 5.65 × 10–13 W. 7) As the path loss is equal to 130 dB, the transmission power is equal to: – P4 = Pr,4 × 1013 = 1.12 W; – P16 = Pr,16 × 1013 = 5.65 W. The feasibility of each modulation depends on the maximum mobile transmission power. If it is equal to 2 W, only 4-PSK modulation can be used.

194

Wireless Telecommunication Systems

8.9. Zero-forcing linear equalization We consider a channel whose global filter is F(z) = 1 + 0.4z–1 . Binary symbols ak = –1;1 are sent to this channel with the same probability, each symbol being independent. The pulse-shaping filter is rectangular in frequency domain. The channel undergoes a white Gaussian noise of variance σ2 = 0.1. 1) Give the expression of the zero-forcing equalizer of infinite length. 2) Calculate the finite impulse response zero-forcing equalizer with three coefficients. 3) Give the expression of the transfer function of the total filter. 4) Calculate the inter-symbol interference power after equalization. 5) Calculate the noise power after equalization. 6) Deduce the signal to interference plus noise ratio. 7) Compare this value with that of the signal-tointerference-plus-noise ratio without equalization and make conclusions. Solution 1) The infinite impulse response zero-forcing equalizer is equal to the inverse of the channel filter: HZF,∞ (z) = 1⁄F(z) = 1⁄ 1 + 0.4z–1 This infinite impulse response filter removes all intersymbol interference. However, it is not feasible in practice.

Digital Communications Fundamentals

195

2) We aim to obtain an equalizer filter of length 3, which can be written HZF (z)=w0 +w1 z–1 +w2 z–2 . After the equalizer, the total filter is: T(z) = F(z) × HZF (z) =w0 +(w1 +0.4w0 )z–1 +(w2 +0.4w1 )z–2 +0.4w2 z–3 We want to remove the first interference terms. For that purpose, the following set of equations must be solved: w0 = 1 w1 + 0.4 × w0 = 0 w2 + 0.4 × w1 = 0 Its solution is: w0 = 1, w1 = –0.4 and w2 = 0.16. The transfer function of the zero-forcing equalizer is therefore: HZF (z)=1–0.4 z–1 +0.16 z–2 3) The total transfer function is previous question, it is equal to:

( ). According to the

T ( z ) = 1 + 0.4 × w2 z −3 = 1 + 0.064 z −3 4) Let dk be the received symbol at time kT, where T is the symbol time, after equalization. It is then equal to dk = ak +0.064 ak–3 +bk . In this equation, bk is the noise after equalization; ak is the useful symbol that will be estimated afterwards and 0.064 ak–3 is the inter-symbol interference. The useful signal power after the equalizer is therefore: Putile = E |ak |2 = 1

196

Wireless Telecommunication Systems

The inter-symbol interference power after the equalizer is: PISI = E |0.064 ak–3 |2 = 0.00411 5) The noise after equalization is: bk = bk –0.4 bk-1 +0.16 bk-2 , where bk is the initial white Gaussian noise. Its power is equal to: P

= E |bk –0.4 bk–1 +0.16 bk–2 |2

= 1+0.42 +0.162 E |bk |2 = 0.1186 where we used the fact that E  bk 2  = σ 2 = 0.1 and that we   consider white noise, whose autocorrelation is equal to 0 except in 0. 6) The signal to interference plus noise ratio is: SINRZF = P

⁄(PISI +P

) = 8.15

7) Without equalization, inter-symbol interference would be equal to 0.4ak-1 . Its power would then amount to 0.16. The noise power is E  bk 2  = σ 2 = 0.1. Finally, the useful signal   power would be 1. The signal-to-interference-plus-noise ratio would then be equal to:

SINR = P

⁄(PISI +Pnoise ) = 3.85

To conclude, adding an equalizer has led to a strong increase the signal-to-interference-plus-noise ratio. This will allow us to decrease the bit error rate at the end of the communication chain. 8.10. Minimum mean square error linear equalization We consider a channel whose global filter is F(z) = 1 + 0.4z–1 . Binary symbols ak = –1 ;1 are sent on this

Digital Communications Fundamentals

197

channel with the same probability, each symbol being independent. The pulse-shaping filter is rectangular in the frequency domain. The channel undergoes a white Gaussian noise of variance σ2 =0.1. 1) Calculate the first three autocorrelation coefficients of the symbols after the channel, before equalization. They are denoted by vk. 2) Calculate the finite impulse response MMSE equalizer with three coefficients. 3) Give the expression of the transfer function of the total filter. 4) Calculate the inter-symbol interference power after equalization. 5) Calculate the noise power after equalization. 6) Deduce the signal to interference plus noise ratio. 7) Compare this value with that of the signal-tointerference-plus-noise ratio obtained in the previous problem with zero forcing and make conclusions. Solution 1) After the channel, after sampling and before equalization, the symbols are: vk =ak +0.4ak-1 +bk , where bk is the white Gaussian noise. The symbols autocorrelation at 0 is equal to: rvv (0) = E |vk |2 = 1 + 0.42 E |ak |2 + E |bk |2 = 1.26 The preceding equation was obtained by using the fact that ak symbols are independent, and independent of the noise.

198

Wireless Telecommunication Systems

The symbols’ autocorrelation at 1 is equal to: rvv (1) = E vk v*k-1 = 0.4E |ak-1 |2 = 0.4 Finally, the symbols’ autocorrelation at 2 is equal to: rvv (2) = E vk v*k-2 = 0 2) The MMSE filter coefficients are obtained by applying the following formula:

( )

w = MT

−1

vaT

where w is the column vector containing coefficients w0, w1 and w2 of the equalizer filter, M is the autocorrelation matrix after the channel and the intercorrelation va = E[v*k ak E[v*k–1 a ] E[v*k-2 a ] ] is k

k

vector between the transmission symbol we aim at estimating, ak, and the symbols after the channel. We can see that E  vk *ak  = 1 according to the equation   linking vk to ak. The two other intercorrelation coefficients are equal to zero. Thus, vk = [1 0 0]. Matrix M has been calculated in the previous question and is equal to: M

1.26 0.4 0

0.4 1.26 0.4

0 0.4 1.26

( )

Finally, by applying formula w = M T

 0.8398    w =  −0.3156   0.1002   

−1

vaT , we obtain:

Digital Communications Fundamentals

199

Therefore, the transfer function of the MMSE equalizer is:

H MMSE ( z ) = 0.8938 − 0.3156 z −1 + 0.1002 z −2 3) The total transfer function is: T(z) = F(z)×HMMSE (z) After some calculation, we get: T(z) = 0.8938+0.0419 z–1 – 0.026 z–2 + 0.0401 z–3 4) Let dk be the symbol received at time kT after equalization, where T is the symbol time. It is equal to: dk = 0.8938 ak + 0.0419ak-1 – 0.026 ak-2 + 0.0401 ak-3 + bk . In this equation, bk is the noise after equalization, 0.8938 ak is the useful symbol that will be estimated afterwards and 0.0419 ak–1 –0.026 ak–2 +0.0401 ak–3 are three intersymbol interference terms. The useful signal power is: P

= E |0.8938 ak |2 = 0.7989

The inter-symbol interference power is: PISI = E |0.0419 ak-1 – 0.026 ak–2 + 0.0401 ak–3 |2 = 0.004 where we used the fact that symbols ak-1 , ak-2 , ak-3 are independent and have a unitary power. 5) The noise after equalization is:

bk = 0.8938 × bk − 0.3156 × bk −1 + 0.1002 × bk −2

200

Wireless Telecommunication Systems

Since bk is a white Gaussian noise with power σ2 = 0.1, we have: Pnoise = E |0.8938 bk –0.3156 bk–1 +0.1002 bk–2 |2 = 0.9085σ2 = 0.09085 6) The signal to interference plus noise ratio is equal to: SINRMMSE = Puseful ⁄(PISI + Pnoise ) =8.42 7) The signal to interference plus noise ratio is higher with the MMSE equalizer than with the zero-forcing equalizer. 8.11. Noise factor in equipments We consider a receiver composed of three amplifiers. The following parameters are used: – receiver power threshold after the radio frequency (RF) chain: SdBm = –90 dBm; - signal bandwidth: B = 200 kHz; - required signal-to-noise ratio: ΓdB = 9 dB; - first amplifier gain: G1,dB = 20 dB; - second amplifier gain: G2,dB = 6 dB; - temperature: T = 300 K. 1) Calculate the total noise factor, SNRin/SNRout. 2) Calculate the noise factor per block when each block is the unique source of noise in the receiver. It is called the maximum noise factor per block. 3) Calculate the noise factor per block when each block has the same contribution in the total noise factor.

Digital Communications Fundamentals

201

Solution We must first calculate the noise power in the considered bandwidth:

(

)

N dBm = −174 + 10 × log10 200 × 103 = −121 dBm Because the thermal noise power spectral density at 300 K is N0,dBm = –174 dBm/Hz. 1) The input signal-to-noise ratio is the ratio between the useful signal, which is the receiver power threshold, and the noise power. Therefore, it is equal to SNRin = S/N, or equivalently, in dB: SNRin,dB = SdBm – NdBm = 31 dB The output signal-to-noise ratio is ΓdB = 9 dB. We can deduce the total noise factor: FdB = SNRin,dB – SNRout,dB = 22 dB 2) We use Friis formula. The noise factor of only the first amplifier (maximum noise factor of the first block) is: FdB = F1,dB = 22 dB In linear scale, the maximum noise factor of the second block is: F = 1 + (F2 – 1)/G1 Therefore, F2 – 1= F– 1 ×G1 . In dB, this is approximately equal to: F2,dB = FdB + G1,dB = 42 dB

202

Wireless Telecommunication Systems

In linear scale, the maximum noise factor of the third block is: F = 1 + (F3 – 1)/(G1G2) We can deduce that F3 – 1 = F – 1 × G1 × G2 . In dB, this is approximately equal to: F3,dB =FdB +G1,dB

G2,dB =48 dB

3) Friis formula for the total receiver is:

F = 1 + ( F1 − 1) + ( F2 − 1) G1 + ( F3 − 1) ( G1G2 ) If all three amplifiers contribute in the same way, we get the following expression: F1 –1= (F2 –1)⁄G1 = (F3 – 1)⁄(G1 G2 ) This implies that the total noise factor can be expressed as a function of F1: F = 1 + 3(F1 – 1) Therefore, we get: FdB 10×log10 (3)+10×log10 (F1 ) and we can deduce that: F1,dB = 22 – 4.77 = 17.23 dB Similarly, the noise factor can be expressed as a function of F2: F = 1 + 3(F2 – 1)/G1. We can deduce that: FdB =10×log10 (3)+10×log10 (F2 )–10×log10 (G1 ) and therefore: F2,dB = FdB – 4.77 + G1,dB = 22 – 4.77 + 20 = 37.23 dB

Digital Communications Fundamentals

203

Finally, for the third amplifier, F = 1 + 3(F3 – 1)/G1G2. Consequently: FdB =10×log10 (3)+10×log10 (F3 )–10×log10 (G1 )–10× log10 (G2 ) We obtain: F3,dB = FdB – 4.77 + G1,dB + G2,dB = 22 – 4.77 + 20 + 6 = 43.23 dB

8.12. Data rate calculations Let us consider a directory containing 2,000 bitmap photographs in black and white, whose size is 256 × 256 blocks of 8 pixels. A pixel corresponds to 256 different gray levels. We wish to transmit these photos on a radio channel, for instance from a computer to a mobile phone. For that purpose, we use, for each photo: Joint Photographic Experts Group (JPEG) encoding whose compression factor is 8.53: – a channel code with rate 4/7 (linear Hamming code); – the data rate on the transmission channel is denoted as R. 1) Give the expression of the size of the directory to be transmitted on the channel. 2) Calculate the time required to transmit that directory with a high-speed packet access (HSPA) radio link, when R = 1 Mbits/s. 3) Calculate the time required to transmit that directory with an LTE radio link, when R = 20 Mbits/s.

204

Wireless Telecommunication Systems

Solution 1) One pixel corresponds to 256 = 28 gray levels, and is thus coded on 8 bits. Therefore, the size of one photograph before compression is Nphoto= 256 × 256 × 8 × 8 = 4,194,304 bits. The size of source then is: Nsource = Nphoto × 2,000 bits Let Nsc be the directory size after source coding. Since the considered JPEG coding has a compression rate 8.53, Nsource/Nsc= 8.53. Therefore, the size of the directory after source coding is: Nsc=Nsource/8.53 The channel code has a rate of 4/7, so the size of the directory after channel coding is: Ncc =7/4 × Nsc= (7 × Nphoto × 2,000)/(4 × 8.53) 2) The required time to transmit that directory with R is: t = Ncc/R For R = 1 Mbits/s, t = 1,721 s, which is equal to 28 min and 41 s. 3) For R = 20 Mbits/s, t = 86 s, which is equal to 1 min and 26 s.

Chapter 9

Erlang B Tables

The unit commonly used in traffic theory is Erlang, or Erl, which represents the occupancy of a channel over an hour. For example: – when a subscriber makes a call for half an hour between 8 pm and 9 pm, the generated traffic is 0.5 Erl; – when four subscribers make a call for 20 min each between 9 am and 10 am, the result is 4 × 20 min of communication, that is 80 min and 80/60 = 1.33 Erl; – a total of 10 subscribers make a call for 15 min each between 2 pm and 3 pm, the result is 150 min of communication, that is 150/60 = 2.5 Erl. The evaluation of the busy hour traffic load is used to dimension the system components (equipment, links and radio resources). Dimensioning of the network elements (such as equipments, radio links and radio resources) is performed after having determined the trafic load at the peak hour. These elements represent significant investment costs, and,

206

Wireless Telecommunication Systems

radio resources being scarce, their dimensioning is performed by considering that a small proportion of users can be unsatisfied in busy periods. Calls from users that cannot be served due to the lack of available resources are called “blocked calls”, and the likelihood of a user having a blocked call is called the “blocking probability”. The blocking probability is determined by the Erlang B law, which is based on queuing theory. The assumptions on which this model was developed are: – the call’s random arrival: it follows a Poisson process (the interarrival calls follow an exponential distribution); – the call’s duration follows an exponential distribution; – an infinite number of traffic sources. Denoting c the number of resources (servers or radio channels for users in wireless systems) and ρ the system’s load, the blocking rate (probability of blocked calls) PB can be expressed as follows (Erlang B formula):

PB =

ρ c 

c



ρ i 

−1

c ! i = 0 i !   

Erlang tables derived from the above formula give the traffic in the system (i.e. its load), in Erl, as a function of the number of available channels and the authorized blocking rate. Table 9.1 can be interpreted as follows: given, for example, a “requested” traffic of 18.6 Erl in a 21 resources system, the associated blocking rate will be 10% and the effective traffic will be 90% × 18.6 = 16.7 Erl.

Erlang B Tables Number of resources 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

1% 0.010 0.153 0.455 0.869 1.36 1.91 2.50 3.13 3.78 4.46 5.16 5.88 6.61 7.35 8.11 8.88 9.65 10.4 11.2 12.0 12.8 13.7 14.5 15.3 16.1 17.0 17.8 18.6 19.5 20.3 21.2 22.0 22.9 23.8 24.6 25.5 26.4 27.3 28.1 29.0 29.9 30.8 31.7 32.5 33.4 34.3 35.2 36.1 37.0 37.9

Admissible blocked calls rate 2% 3% 5% 0.020 0.031 0.053 0.223 0.282 0.381 0.602 0.715 0.899 1.09 1.26 1.52 1.66 1.88 2.22 2.28 2.54 2.96 2.94 3.25 3.74 3.63 3.99 4.54 4.34 4.75 5.37 5.08 5.53 6.22 5.84 6.33 7.08 6.61 7.14 7.95 7.40 7.97 8.83 8.20 8.80 9.73 9.01 9.65 10.6 9.83 10.5 11.5 10.7 11.4 12.5 11.5 12.2 13.4 12.3 13.1 14.3 13.2 14.0 15.2 14.0 14.9 16.2 14.9 15.8 17.1 15.8 16.7 18.1 16.6 17.6 19.0 17.5 18.5 20.0 18.4 19.4 20.9 19.3 20.3 21.9 20.2 21.2 22.9 21.0 22.1 23.8 21.9 23.1 24.8 22.8 24.0 25.8 23.7 24.9 26.7 24.6 25.8 27.7 25.5 26.8 28.7 26.4 27.7 29.7 27.3 28.6 30.7 28.3 29.6 31.6 29.2 30.5 32.6 30.1 31.5 33.6 31.0 32.4 34.6 31.9 33.4 35.6 32.8 34.3 36.6 33.8 35.3 37.6 34.7 36.2 38.6 35.6 37.2 39.6 36.5 38.1 40.5 37.5 39.1 41.5 38.4 40.0 42.5 39.3 41.0 43.5 40.3 41.9 44.5

Table 9.1. Erlang B table

10% 0.111 0.595 1.27 2.05 2.88 3.76 4.67 5.60 6.55 7.51 8.49 9.47 10.5 11.5 12.5 13.5 14.5 15.5 16.6 17.6 18.7 19.7 20.7 21.8 22.8 23.9 24.9 26.0 27.1 28.1 29.2 30.2 31.3 32.4 33.4 34.5 35.6 36.6 37.7 38.8 39.9 40.9 42.0 43.1 44.2 45.2 46.3 47.4 48.5 49.6

207

20% 0.250 1.00 1.93 2.95 4.01 5.11 6.23 7.37 8.52 9.68 10.9 12.0 13.2 14.4 15.6 16.8 18.0 19.2 20.4 21.6 22.8 24.1 25.3 26.5 27.7 28.9 30.2 31.4 32.6 33.8 35.1 36.3 37.5 38.8 40.0 41.2 42.4 43.7 44.9 46.1 47.4 48.6 49.9 51.1 52.3 53.6 54.8 56.0 57.3 58.5

Bibliography

[ALA 98] ALAMOUTI S.M., “A simple transmit diversity technique for wireless communications”, IEEE Journal on Selected Areas in Communications, vol. 16, no. 8, pp. 1451–1458, October 1998. [BAU 07] BAUDOIN G., Radiocommunications 2nd ed., Dunod, Paris, 2007.

Numériques,

[BEG 10] BEGAUD X., Ultra Wide Band Antennas, ISTE, London, John Wiley and Sons, New York, 2010. [BEL 06] BELLANGER M., Traitement Numérique du Signal, 8th ed., Dunod, Paris, 2006. [BOI 83] BOITHIAS L., Propagation des Ondes Radioélectriques, Dunod, Paris, 2003. [D’AN 94] D’ANDREA A.N., MENGALI U., REGGIANNINI R., “The modified Cramer-Rao bound and its application to synchronisation problems”, IEEE Transactions on Communications, vol. 42, no. 2/3/4, pp. 1391–1399, February/ March/April 1994. [GOL 05] GOLDSMITH A., Wireless Communications, Cambridge University Press, Cambridge, 2005. [HOL 10] HOLMA H., TOSKALA A., WCDMA for UMTS–HSPA Evolution and LTE, 5th ed., Wiley, New York, 2010.

210

Wireless Telecommunication Systems

[JAK 74] JAKES W.C., Microwave Mobile Communications, IEEE Press, New York, 1974. [LAG 99] LAGRANGE X., GODLEWSKI P., TABBANE S., Réseaux GSMDCS, 4th ed., Hermès, Paris, 1999. [LAH 11] LAHEURTE J.M., Compact Antennas for Wireless Communications and Terminals, ISTE, London, John Wiley and Sons, New York, 2011. [LAV 97] LAVERGNAT J., SYLVAIN Propagation, Masson, Paris, 1997.

M.,

Introduction

à

la

[MAR 98] MARAL G., BOUSQUET M., Satellite Communications Systems, Wiley, New York, 1998. [PAL 10] PALICOT J., Radio Engineering, ISTE, London, John Wiley and Sons, New York, 2011. [PIC 09] PICON O., Les Antennes, Dunod, Paris, 2009. [PRO 08] PROAKIS J.G., SALEHI M., Digital Communications, 5th ed., McGraw-Hill, New York, 2008. [RAP 01] RAPPAPORT T.S., Wireless Communications: Principles & Practice, 2nd ed., Prentice Hall, New York, 2002. [SAL 87] SALEH A., VALENZUELLA R., “A statistical model for indoor multipath propagation”, IEEE Journal on Selected Areas in Communications, vol. 5, no. 2, pp. 128–137, 1987. [SAP 90] SAPORTA G., Probabilités Analyse des Données et Statistique, Technip, Paris, 1990. [SCH 93] SCHOLTZ R.A., “Multiple access with time-hopping impulse radio”, Proceedings of the IEEE Milcom Conference, Boston, MA, October 1993. [VIT 83] VITERBI A.J., VITERBI A.M., “Nonlinear estimation of PSKmodulated carrier phase with applications to burst digital transmission”, IEEE Transactions on Information Theory, vol. 29, no. 4, pp. 543–551, July 1983. [YAC 93] YACOUB M.D., Foundations of Mobile Radio Engineering, CRC Press, Boca Raton, 1993.

Index

A, C antenna gain, 4, 15, 36, 49, 84, 114 Code division multiple access (CDMA), 59, 61, 63, 70, 78, 115 coherence bandwidth, 105 E, F Equivalent isotropic radiated power (EIRP), 3, 4, 12, 13, 34, 35, 48, 75, 79, 102, 116–118, 121, 122 Frequency division multiple access (FDMA), 24, 96, 97, 116 free-space loss, 7 G, H, L, M, N Global system for mobile communications (GSM), 11, 21, 23, 24, 32, 36, 38, 41, 44, 46, 50, 52–54, 61, 75, 180 High speed packet access (HSPA), 61, 203 Long-term evolution (LTE), 95–97, 110–113, 115, 118, 120, 121, 123, 124, 180

Multiple input multiple output (MIMO), 61, 96, 97 noise factor, 37, 84 O, P, R Orthogonal frequency division multiplex (OFDM), 95, 96, 98, 100, 103, 106, 107, 111, 115, 123–125, 168 Orthogonal frequency division multiple access (OFDMA), 96, 97 Peak-to-average power ratio (PAPR), 98–100, 116, 121 Resource Block, 114, 121, 122 S, T, U sensibility, 48, 49, 52, 54 sensitivity, 37, 100, 102, 104 Signal-to-interference-plus-noise ratio (SINR), 63, 69, 77–80, 123, 124, 126, 127, 196, 200

212

Wireless Telecommunication Systems

Time division multiple access (TDMA), 23, 24, 50 Transceiver receiver (TRX), 23, 32, 33, 36, 37, 38, 39, 41, 42, 43, 44, 46, 71

Universal mobile telecommunications system (UMTS), 11, 59, 60–62, 69–71, 73, 74, 77, 80, 83, 87, 97, 115, 180 Ultra wide band (UWB), 156