Mobile Communications

Mobile Communications TCS 455

Dr. Prapun Suksompong [email protected]

Lecture 14 (Review) Office Hours: BKD 3601-7 Tuesday 14:00-16:00 Thursday 9:30-11:30 1

Announcements  Read  Chapter 3: 3.1 – 3.2, 3.5.1, 3.6, 3.7.2  Posted on the web  Appendix A.1 (Erlang B)  Chapter 9: 9.1 – 9.5

 Due date for HW3: Dec 18

2

Course Organization  Course Web Site:

http://www.siit.tu.ac.th/prapun/ecs455/  Lectures:  Tuesday 10:40-12:00 BKD 2601  Thursday 13:00-14:20 BKD 3215

 Textbook:

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 Wireless Communications: Principles and Practice  By Theodore S. Rappaport  2nd Edition, Prentice Hall PTR, 2002.  ISBN-13: 978-0130422323.  Call No. TK5103.2 R37 2002  Companion Site: http://authors.phptr.com/rappaport/

Course Web Site  Please check the course  

 

Web site regularly. Announcement References Handouts/Slides Calendar  Exams  HW due dates

www.siit.tu.ac.th/prapun/ecs455/ 4

Grading System  Coursework will be weighted as follows:

Assignments Class Participation and Quizzes Midterm Examination

5% 15% 40%

•09:00 - 12:00 on Dec 22, 2009

Final Examination (comprehensive) •09:00 - 12:00 on Mar 9, 2010

    5

Mark your calendars now! Late HW submission will be rejected. All quizzes and exams will be closed book. For grad. student, this is 2/3 of your final score.

40%

Midterm Exam  Not to torture you!

 Most questions are straightforward  A few difficult ones  Worth 1 to 2 points each

 Study  HW questions / quiz  Only small parts of HWs are graded.  Please take a careful look at the solution.

 Lecture notes  Textbook chapters

6

Midterm Exam  9 pages

 9 problems  Start at 9:00 AM  You may start at 9:09 AM if you want to.

 99 Points + 1 hidden point

7

Topics  Chapter 1 > 10%  Fourier transform, modulation

 Chapter 2 > 50%  Cellular System

 Chapter 3 > 30%  Erlang B derivation: Poisson Process and Markov Chain

 Chapter 4 < 10%  Duplexing: FDD and TDD

8

Provided Formula

Am ErlangB  m, A   mm! k A  k 0 k !

2cos 2 x  1  cos  2 x  2sin 2 x  1  cos  2 x  

G f  



g  t  e  j 2 ft dt



1 1 j   f  f c  e    f  f c  e  j 2 2

cos  2 f ct    g  t  t0 

e  j 2  ft 0 G  f 

e j 2 f0t g  t  9

m  t  cos  2 f ct 

G  f  f0  1 1 M  f  fc   M  f  fc  2 2

Chapter 1 Review & Introduction

Office Hours: BKD 3601-7 Tuesday 14:00-16:00 Thursday 9:30-11:30 10

Handout #1  Fourier Transform

 Modulation  More on HW1

11

Frequency-Domain Analysis

Shifting Properties: g  t  t0 

e  j 2 ft0 G  f 

Modulation: m  t  cos  2 f ct  12

e j 2 f0t g  t 

G  f  f0 

1 1 M  f  fc   M  f  fc  2 2

Overview of Mobile Communications  Wireless/mobile communications is the fastest growing

segment of the communications industry.  Cellular systems have experienced exponential growth over the last decade.  Cellular phones have become a critical business tool and part of everyday life in most developed countries, and are rapidly replacing wireline systems in many developing countries.

13

Mobile?  The term “mobile” has historically been used to classify all

radio terminal that could be moved during operation.  More recently,  the term mobile is used to describe a radio terminal that is

attached to a high speed mobile platform  e.g., a cellular telephone in a fast moving vehicle

 the term portable is used to describes a radio terminal that can

be hand-held and used by someone at walking speed  e.g., a walkie-talkie or cordless telephone inside a home.  802.11?

14

History of Wireless Communications  The first wireless networks

were developed in the Preindustrial age.  These systems transmitted information over line-of-sight distances (later extended by telescopes) using smoke signals, torch signaling, flashing mirrors, signal flares, or semaphore flags. 15

Semaphore

16

History of Wireless Comm. (2)  Early communication networks were replaced first by the

 

  

17

telegraph network (invented by Samuel Morse in 1838) and later by the telephone. In 1895, Marconi demonstrated the first radio transmission. Early radio systems transmitted analog signals. Today most radio systems transmit digital signals composed of binary bits. A digital radio can transmit a continuous bit stream or it can group the bits into packets. The latter type of radio is called a packet radio and is characterized by bursty transmissions

History of Wireless Comm. (3)  The first network based on packet radio, ALOHANET, was

developed at the University of Hawaii in 1971.  ALOHANET incorporated the first set of protocols for channel access and routing in packet radio systems, and many of the underlying principles in these protocols are still in use today.  Lead to Ethernet and eventually wireless local area networks

18

History of Wireless Comm. (3)  The most successful application of wireless networking has been

the cellular telephone system.  The roots of this system began in 1915, when wireless voice transmission between New York and San Francisco was first established.  In 1946 public mobile telephone service was introduced in 25 cities across the United States.  These initial systems used a central transmitter to cover an entire metropolitan area.  Inefficient!  Thirty years

after the introduction of mobile telephone service, the New York system could only support 543 users.

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History of Wireless Comm. (4)  A solution to this capacity problem emerged during the 50’s

and 60’s when researchers at AT&T Bell Laboratories developed the cellular concept.  Cellular systems exploit the fact that the power of a transmitted signal falls off with distance.  Thus, two users can operate on the same frequency at spatially-separate locations with minimal interference between them.  Frequency reuse

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History of Wireless Comm. (5)  The second generation (2G) of cellular systems, first deployed in









21

the early 1990’s, were based on digital communications. The shift from analog to digital was driven by its higher capacity and the improved cost, speed, and power efficiency of digital hardware. While second generation cellular systems initially provided mainly voice services, these systems gradually evolved to support data services such as email, Internet access, and short messaging. Unfortunately, the great market potential for cellular phones led to a proliferation of (incompatible) second generation cellular standards. As a result of the standards proliferation, many cellular phones today are multi-mode.

Chapter 2 Cellular System


Handout #2

23

Radio-frequency spectrum  Commercially exploited bands

24

25

Tessellating Cell Shapes Hexagonal cells:  Having largest area for a given distance between the center of a polygon and its farthest perimeter points  Approximating a circular radiation pattern for an omnidirectional base station antenna and free space propagation

26

Frequency Reuse (N = 4, N = 7)  Cluster: a group of N cells use the complete set of available

frequencies A B A

A

C B

C

C

A D

A B

A D

B C

A D

B C

27

D

B C

C

B

D

D

D

Activity 1  You have seen N = 3, 4, 7

 Find the next five lowest values of N.  In HW2, find the next fifteen lowest values of N.

28

Hexagon R

3R R

R

3 R 2

3R R

R

3R 2 R 2 2R

29

R

3 R 2

1 3 1  3 3 2 Area  6  2    R R  R  2.598R 2 2  2 2 2

Frequency Reuse  Cluster: a group of N cells using the complete set of

available frequencies 4-cell reuse

7-cell reuse

Atotal S C  Acell N 30

12-cell reuse

19-cell reuse

Co-channel Interference (N=19)

Method of locating co-channel cells in a cellular system. In this example, N = 19 (i.e., I = 3, j = 2). (Adapted from [Oet83] © IEEE.)

Center-to-center distance (D) D

i

  j 2

3R

3R



2



 2 i 3R

 j



3R cos 120 

 R 3  i 2  j 2  ij   R 3N

i 3R

j 3R

This distance, D, is called reuse distance.

120

D

32

Co-channel reuse ratio D Q   3N . R

Q and N Co-channel reuse ratio D Q   3N . R

33

SIR  Frequency reuse  co-channel interference

 K = the number of co-channel interfering cells  The signal-to-interference ratio (S/I or SIR) for a

mobile receiver which monitors a forward channel can be expressed as S S SIR 

I



K

I i 1

i

 S = the desired signal power from the desired base station  Ii = the interference power caused by the ith interfering co-

channel cell base station.

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SIR  The SIR should be greater than a specified threshold for proper

signal operation.

 In the first-generation AMPS system, designed for voice calls, the

desired performance threshold is SIR equal to 18 dB.  For the second-generation digital AMPS system (D-AMPS or IS54/136), a threshold of 14 dB is deemed suitable.  For the GSM system, a range of 7–12 dB, depending on the study done, is suggested as the appropriate threshold.  Only a relatively small number of nearby interferers need be

considered, because of the rapidly decreasing received power as the distance increases.  In a fully equipped hexagonal-shaped cellular system, there are always

six cochannel interfering cells in the first tier.

 Approximation: 35





S kR 1 D 1      I K   kD   K  R  K







3N

SIR: N = 7 More accurate calculation…

36

SIR: N = 3

Even more accurate calculation… 2

D3 2R

D2

7R 13R D1

7R D4 4R

 3 2 D1  D5  R 1   4   R 13  2  3R 2

13R D5

D6

2

5  3 D2  D4  R      R 4 2  2  D3  2 R 2

D6  4 R

R 2 37

Pt R  SIR     Pt Di 2 i

  7

1



2

  13



 2  4 

Improving Coverage and Capacity  As the demand for wireless service increases, the number of

channels assigned to a cell eventually becomes insufficient to support the required number of users.  At this point, cellular design techniques are needed to provide more channels per unit coverage area.  Easy!? C

38

Atotal S  Acell N

Sectoring (N = 7)

39

Sectoring (N = 7)

40

Sectoring (N = 3, 120)

S 1  I K



K=2

41





3N

Sectoring (N = 3 , 60)

S 1  I K



3N

K=1 42





60 Degree Sectoring

43

Sectoring

S 1  I K







3N

C

Atotal S  Acell N

 Advantages  Assuming seven-cell reuse, for the case of 120 sectors, the number

of interferers in the first tier is reduced from six to two.  This reduction lead to the increase of SIR.

 The increase in SIT can be traded with reducing the cluster size which

increase the capacity.

 Disadvantages  Increase number of antennas at each base station.  Decrease trunking efficiency due to channel sectoring at the base

station.

 The available channels in the cell must be subdivided and dedicated to a

specific antenna.

44

Estimating the number of users  Trunking

 Allow a large number of users to share the relatively small

number of channels in a cell by providing access to each user, on demand, from a pool of available channels.  Exploit the statistical behavior of users  Each user is allocated a channel on a per call basis, and upon termination of the call, the previously occupied channel is immediately returned to the pool of available channels.

45

Common Terms  Traffic Intensity: Measure of channel time utilization, which is the average

channel occupancy measured in Erlangs.

 This is a dimensionless quantity and may be used to measure the time utilization

of single or multiple channels.  Denoted by A.

 Holding Time: Average duration of a typical call. Denoted by H = 1/.  Blocked Call: Call which cannot be completed at time of request, due to

congestion. Also referred to as a lost call.  Grade of Service (GOS): A measure of congestion which is specified as the probability of a call being blocked (for Erlang B).  The AMPS cellular system is designed for a GOS of 2% blocking. This implies

that the channel allocations for cell sites are designed so that 2 out of 100 calls will be blocked due to channel occupancy during the busiest hour.

 Request Rate: The average number of call requests per unit time. Denoted by

.

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M/M/m/m Assumption  Blocked calls cleared  Offers no queuing for call requests.  For every user who requests service, it is assumed there is no setup time and the

user is given immediate access to a channel if one is available.  If no channels are available, the requesting user is blocked without access and is free to try again later.

 Calls arrive as determined by a Poisson process.  There are memoryless arrivals of requests, implying that all users, including

blocked users, may request a channel at any time.  There are an infinite number of users (with finite overall request rate).

 The finite user results always predict a smaller likelihood of blocking. So,

assuming infinite number of users provides a conservative estimate.

 The duration of the time that a user occupies a channel is

exponentially distributed, so that longer calls are less likely to occur.  There are m channels available in the trunking pool.  For us, m = the number of channels for a cell (C) or for a sector 47

AC Pb  CC ! k . A  k 0 k !

Erlang B

A 48

Example  How many users can be supported for 0.5% blocking

probability for the following number of trunked channels in a blocked calls cleared system? (a) 5 (b) 10  Assume each user generates 0.1 Erlangs of traffic.

49

AC Pb  CC ! k . A  k 0 k !

Erlang B

A 50

Example  Consider a cellular system in which  an average call lasts two minutes  the probability of blocking is to be no more than 1%.

 If there are a total of 395 traffic channels for a seven-cell

reuse system, there will be about 57 traffic channels per cell.  From the Erlang B formula, the may handle 44.2 Erlangs or 1326 calls per hour.

51

AC Pb  CC ! k . A  k 0 k !

Erlang B

A 52

Example  Now employing 120° sectoring, there are only 19 channels

per antenna sector (57/3 antennas).  For the same probability of blocking and average call length, each sector can handle 11.2 Erlangs or 336 calls per hour.  Since each cell consists of three sectors, this provides a cell capacity of 3 × 336 = 1008 calls per hour, which amounts to a 24% decrease when compared to the unsectored case.  Thus, sectoring decreases the trunking efficiency while improving the S/I for each user in the system.

53

AC Pb  CC ! k . A  k 0 k !

Erlang B

A 54

Erlang B Trunking Efficiency

55

Big Picture S = total # available duplex radio channels for the system Frequency reuse with cluster size N A S “Capacity” C  total  Acell N

Trunking



S kR  1 D 1      K Tradeoff I K   kD   K  R  m = # channels allocated to each cell.







3N

Omni-directional: K = 6 120 Sectoring: K = 2 60 Sectoring: K = 1

 = Average # call attempts/requests per unit time Am  Call blocking m ! A  tra ffic i ntens i ty or l oad [Erla ngs] = P  m i.  probability b A 1   H  Average call length 56 i 0 i ! Erlang-B formula 

Chapter 3 Poisson process and Markov chain


M/M/m/m Assumption  Blocked calls cleared  Offers no queuing for call requests.  For every user who requests service, it is assumed there is no setup time and the

user is given immediate access to a channel if one is available.  If no channels are available, the requesting user is blocked without access and is free to try again later.

 Calls arrive as determined by a Poisson process.  There are memoryless arrivals of requests, implying that all users, including

blocked users, may request a channel at any time.  There are an infinite number of users (with finite overall request rate).

 The finite user results always predict a smaller likelihood of blocking. So,

assuming infinite number of users provides a conservative estimate.

 The duration of the time that a user occupies a channel is

exponentially distributed, so that longer calls are less likely to occur.  There are m channels available in the trunking pool.  For us, m = the number of channels for a cell (C) or for a sector 58

Assumption (con’t) The call request process is Poisson with rate  If m = 3, this call will be blocked t

K(t)

The duration of calls are i.i.d. exponential r.v. with rate .

m=3 2 1

t K(t) = “state” of the system = the number of used channel at time t We want to find out what proportion of time the system has K = m.

59

Poisson Process? One of these is a realization of a two-dimensional Poisson point process and the other contains correlations between the points. One therefore has a real pattern to it, and one is a realization of a completely unstructured random process.

60

Poisson Process All the structure that is visually apparent is imposed by our own sensory apparatus, which has evolved to be so good at discerning patterns that it finds them when they’re not even there!

61

Example  Examples that are well-modeled as Poisson processes include  radioactive decay of atoms,  telephone calls arriving at a switchboard,  page view requests to a website,  rainfall.

62

Handout #3: Poisson Process

63

Poisson Process The number of arrivals N1, N2 and N3 during non-overlapping time intervals are independent Poisson random variables with mean =   the length of the corresponding interval. 1

2

N1 = 1

W1

3 N3 = 1

N2 = 2 W2 W3

W4

The lengths of time between adjacent arrivals W1, W2, W3 … are i.i.d. exponential random variables with mean 1/. 64

Time

Small Slot Analysis (Poisson Process)  Aka discrete time approximation 1 2 N1 = 1 W1

N3 = 1

N2 = 2 W2 W 3

3

Time

W4

In the limit, there is at most one arrival in any slot. The numbers of arrivals on the slots are i.i.d. Bernoulli random variables with probability p1 of exactly one arrivals =  where  is the width of individual slot.

Time D1

The number of slots between adjacent arrivals is a geometric random variable.

The total number of arrivals on n slots is a binomial random variable with parameter (n,p1)

In the limit, as the slot length gets smaller, geometric binomial 65

exponential Poisson

Poisson Process (Recap)  We spent a few lectures now studying Poisson process.  This is used to model call arrivals in M/M/m/m queue (which

gives Erlang B formula).  Along the way, we review many facts from probability theory.     

pmf – Binomial, Poisson, Geometric pdf - Exponential Independence Expectation, characteristic function Sum of independent random variables and how to analyze it by characteristic functions

 You have seen that Poisson process connects many concepts that

you learned from introductory probability class.

66

Handout #4: Erlang B & Markov Chain

67

Small Slot Analysis (Erlang B) Suppose each slot duration is .

   



Consider the ith small slot. Let Ki = k be the value of K at the beginning of this time slot. k = 2 in the above figure. Then, Ki+1 is the value of K at the end of this slot which is the same as the value of K at the beginning of the next slot. P[0 new call request] ≈ 1 -  P[1 new call request] ≈  k How do these events affect Ki+1 ? 1   P[0 old-call end] ≈    1  k 



P[1 old-call end] ≈ k  1   

 

68

k 1

 k 

Small slot Analysis (2) Ki+1 = Ki + (# new call request) – (# old-call end)

1    k    k 

k-1

  1  k    

k

k+1

1   1  k       k    1    k  The labels on the arrows are probabilities. 69

P[0 new call request] ≈ 1 -  P[1 new call request] ≈  P[0 old-call end] ≈ 1 - k P[1 old-call end] ≈ k

Small slot Analysis: Markov Chain  Case: m = 2

 1  

0



1 2

 1    

70

2

1  2

Markov Chain  Markov chains model many phenomena of interest.  We will see one important property: Memoryless  It retains no memory of where it has been in the past.  Only the current state of the process can influence where it goes

next.

 Very similar to the state transition diagram in digital circuits.  In digital circuit, the labels on the arrows indicate the input/control

signal.  Here, the labels on the arrows indicate transition probabilities. (If the system is currently at a particular state, where would it go next on the next time slot? )  We will focus on discrete time Markov chain.

71

Example: The Land of Oz  Land of Oz is blessed by many things, but not by good

weather.  They never have two nice days in a row.  If they have a nice day, they are just as likely to have snow as rain

the next day.  If they have snow or rain, they have an even chance of having the same the next day.  If there is change from snow or rain, only half of the time is this a change to a nice day.  If you visit the land of Oz next year for one day, what is the

chance that it will be a nice day? 72

State Transition Diagram 1/4 1/4

R

1/2

N

1/2

S

1/2

1/4

1/2

1/4

73

R = Rain N = Nice S = Snow

Markov Chain (2)  Let Ki be the weather status for the ith day (from today).  Suppose we know that it is snowing in the land of Oz today. Then

K0 = S   



where S means snow. Goal: We want to know whether K365 = N where N means nice. Of course, the weather are controlled probabilistically; so we can only find P[K365 = N]. From the specification (or from the state transition diagram), we know that 1 1 1 P  K1  R   , P  K1  N   , P  K1  S  4 4 2 Define vector p  i    P  K i  R  P  K i  N  P  K i  S

 Then, 74

1 p  0    0 0 1 and p 1   4

1 4

1 2 

The Land of Oz: Transition Matrix 1/4

p  i  1  p  i   P

1/4

R N S 1 R 2  1 P N 2 1 S 4

1 4 0 1 4

p  n   p  0  Pn p  2    0.3750 0.1875 0.4375

1 4  1 2 1  2

R

1/2

N

1/2

S

1/2

1/4

1/2 1/4

P  K i 1  R K i  N 

p  3   0.3906 0.2031 0.4063 p  5    0.3994 0.2002 0.4004 75

p  7    0.4000 0.2000 0.4000  p  8   p  9   p 10  

 p  365 

Finding Pn for “large” n 1 2  1 P  2 1  4

1 4 0 1 4

1 4  1 2 1  2

 0.4375 P 2  0.3750  0.3750  0.4063 P 3   0.4063  0.3906

0.1875 0.3750  0.2500 0.3750   0.1875 0.4375  0.2031 0.3906  0.1875 0.4063   0.2031 0.4063 

0.4004 0.2002 0.3994  P 5  0.4004 0.1992 0.4004    0.3994 0.2002 0.4004 

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0.4000 0.2000 0.4000  8 9 10 P 7  0.4000 0.2000 0.4000   P  P  P    0.4000 0.2000 0.4000 

Land of Oz: Answer  Recall that  So,

p  n   p  0  Pn

p  7   p  0  P7  0.4 0.2 0.4

 Note that the above result is true regardless of the initial p  0  365  p  365  p  0  P  0.4 0.2 0.4

P[K365 = N]

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Global Balance Equations  Easier approach for finding the long-term probabilities

Let pk be the long-term probability that K = k.

2/5

 2 / 5 3 / 5 P  1 / 2 1 / 2   3/5

B

A 1/2

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3 1 p A   pB  5 2

1/2

M/M/m/m Queuing Model

Small slot Analysis: Markov Chain  Case: m = 2 1  





0

1

1  2

2

 1    

Let pk be the long-term probability that K = k.

Global Balance equations

 p0   p1 p0  p1  p2  1

p0  79

2

1 A2 1 A  2

 p1  2  p2

, p1  Ap0 , p2 

1 2 A p0 2

pb  pm

Truncated birth-and-death process  Continuous-time Markov chain

 More general than M/M/m/m

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Chapter 4 Multiple Access


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Duplexing  Allow the subscriber to send “simultaneously” information to the

base station while receiving information from the base station.  Talk and listen simultaneously.

 We define forward and reverse channels as followed:  Forward channel or downlink (DL) is used for communication

from the infrastructure to the users/stations  Reverse channel or uplink (UL) is used for communication from users/stations back to the infrastructure.  Two techniques 1. 2.

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Frequency division duplexing (FDD) Time division duplexing (TDD)

Frequency Division Duplexing (FDD)  Provide two distinct bands of frequencies (simplex channels)

for every user.  The forward band provides traffic from the base station to the mobile.  The reverse band provides traffic from the mobile to the base station.  Used in cellular

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Time Division Duplexing (TDD)  Use time instead of frequency to provide both a forward and   

  85

reverse link. Each duplex channel has both a forward time slot and a reverse time slot. The UL and DL data are transmitted on the same carrier frequency at different times. If the time separation between the forward and reverse lime slot is small, then the transmission and reception of data appears simultaneous to the users at both the subscriber unit and on the base station side. Used in Bluetooth and Mobile WiMAX Each transceiver operates as either a transmitter or receiver on the same frequency

Problems of FDD  Because each transceiver simultaneously transmits and

receives radio signals which can vary by more than100 dB, the frequency allocation used for the forward and reverse channels must be carefully coordinated within its own system and with out-of-band users that occupy spectrum between these two bands.  The frequency separation must be coordinated to permit the use of inexpensive RF and oscillator technology.

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Advantages of FDD  TDD frames need to incorporate guard periods equal to the

max round trip propagation delay to avoid interference between uplink and downlink under worst-case conditions.  There is a time latency created by TDD due to the fact that communications is not full duplex in the truest sense.  This latency creates inherent sensitivities to propagation delays

of individual users.

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Advantages of TDD  Enable adjustment of the downlink/uplink ratio to efficiently

support asymmetric DL/UL traffic.

 With FDD, DL and UL always have fixed and generally, equal

DL and UL bandwidths.

 Assure channel reciprocity for better support of link

adaptation, MIMO and other closed loop advanced antenna technologies.  Ability to implement in nonpaired spectrum  FDD requires a pair of channels  TDD only requires a single channel for both DL and UL

providing greater flexibility for adaptation to varied global spectrum allocations.

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