Telecommunications Engineering Topic 4: Spread Spectrum and CDMA

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April 13, 2005 Topic 4 1 Telecommunications Engineering Topic 4: Spread Spectrum and CDMA James K Beard, Ph.D. [email protected] http:// astro.temple.edu/ ~jkbeard/

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Telecommunications Engineering Topic 4: Spread Spectrum and CDMA. James K Beard, Ph.D. [email protected] http://astro.temple.edu/~jkbeard/. Attendance. Essentials. Text: Simon Haykin and Michael Moher, Modern Wireless Communications SystemView - PowerPoint PPT Presentation

Transcript of Telecommunications Engineering Topic 4: Spread Spectrum and CDMA

April 13, 2005 Topic 4 1

Telecommunications EngineeringTopic 4: Spread Spectrum and CDMAJames K Beard, Ph.D.

[email protected]

http://astro.temple.edu/~jkbeard/

Topic 4 2April 13, 2005

Attendance

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Topic 4 3April 13, 2005

Essentials Text: Simon Haykin and Michael Moher, Modern

Wireless Communications SystemView

Use the full version in E&A 603A for your term project Web Site

URL http://astro.temple.edu/~jkbeard/ Content includes slides for EE320 and EE521 SystemView page A few links

Office Hours E&A 349 Hours Tuesday afternoons 3:00 PM to 4:30 PM MWF 10:30 AM to 11:30 AM Others by appointment; ask by email

Topic 4 4April 13, 2005

Topics

Today we explore the third toolFDMA, uses separate channels for each userTDMA, uses time multiplexing to time

multiplex the channel between usersNow, CDMA with spread spectrum enables

multiple simultaneous users of the channel Direct-sequence modulation Spreading codes Code synchronization

Topic 4 5April 13, 2005

Direct Sequence Modulation

We begin with BPSK or QPSK We replace the simple pulse shape

Each “pulse” is a more complex wide band pulse The bandwidth of the resulting signal is that of the

new wide band pulse Spectrum of new signal is given by the

convolution theorem

1 2 1 2

1

2s t s t S S d

Topic 4 6April 13, 2005

Base Performance Equations

0

0

b

b

SignalPowerE

BitRate

NoisePower N Bandwidth

E SignalPower Bandwidth

N NoisePower BitRate

Topic 4 7April 13, 2005

Performance in Noise

Base equation (Hayken & Moher equations (5.12) page 263, (E.11) page 518)

Adding spreading function -- Eb and N0 are invariant through matched filter

0 0

20.5 erfc b b

e

E EP Q

N N

Topic 4 8April 13, 2005

Performance in Interference

Consider a tone as interference In base coded signal

Matched filter spreads tone over channelTone energy becomes part of noise floor

In spread spectrum signalMatched filter spreads tone over channelEffective additional noise reduced by

spreading factor

Topic 4 9April 13, 2005

Spreading Codes and CDMA

Common method is to use a code for each pulse in a signal

This is the spreading code CDMA is achieved when the spreading

code is one of an orthogonal set for each user of the channel

Topic 4 10April 13, 2005

Spreading Codes and CDMA

Use a coded pulse for each bit in the message The coded pulse is the symbol-shaping function Make the code one of an orthogonal set for each

user of the same broadened channel Result

BER performance is unchanged for each user Users of other spreading codes look like the noise

floor

Topic 4 11April 13, 2005

The Symbol-Shaping Function

1

number of pulses in spreading code

spreading code , one of orthogonal set

puse of width

Q

k k c cq

c

k

c c

g t c t g t q T

Q

TT

Q

c t k

g t T

Topic 4 12April 13, 2005

Walsh-Hadamard Sequences

A simple way to formulate orthogonal code sequences

Based on recursive augmentation of Walsh-Hadamard matrices

1

1

1 1

1 1

i ii

i i

H

H HH

H H

Topic 4 13April 13, 2005

Properties of Walsh-Hadamard Sequences Matrices are symmetrical Matrices are self-orthogonal Each matrix has rows or columns are a

sequence of orthogonal sequences of length 2k

Cross-correlation propertiesExcellent for zero lagPoor for other lags

Topic 4 14April 13, 2005

Maximal-Length Sequences

Bit sequence is essentially random Pseudo-random noise (PRN) code Codes Construction

Shift registers with feedback Recursive modulo-2 polynomial arithmetic

PRN codes are then selected for good cross-correlation properties

Topic 4 15April 13, 2005

Desirable PRN Code Properties

Maximal length – 2m codes before repeating

Balance – equal number of (+1) and (-1) pulses

Closed on circular shifts Contain shorter subsequences Good autocorrelation properties

Topic 4 16April 13, 2005

Galois Field Vector Extensions of Order 2m

Polynomials modulo 2 of order m-1 Arithmetic is done modulo a generating

polynomial of the form

Proper selection of generating polynomialSequence of powers produces all 2m elementsSet is closed on multiplication

1 other powers of xmgg x x

Topic 4 17April 13, 2005

An Important Isomorphism

Shift registers with feedbackBits in shift register are isomorphic with

polynomial coefficientsShift is isomorphic with multiplication by xModulo the generating polynomial is

isomorphic to multiple-tap feedback Shift registers with feedback can produce

a Galois field in sequence of powers of x These codes are also called m-sequences

Topic 4 18April 13, 2005

Gold Codes

R. Gold, optimal binary sequences for spread spectrum multiplexing, IEEE Trans. Inform. Theory, Vol. IT-14, pp. 154-156, 1968.

Based on summing the output of two m-sequence generators

Topic 4 19April 13, 2005

Code Synchronization

Two phasesRecover timingRecover phaseTiming must be recovered first

To recover timingUse code bits known to be 1’sMatched filter for symbol-shaping functionStep timing in increments of Tc until match is

found

Topic 4 20April 13, 2005

Assignment

Read 5.2, 5.3, 5.5, 5.7, 5.11, 5.15 Do problem 5.7 p. 273 Next time

Power controlFrequency hoppingAn example

Topic 4 21April 13, 2005

Chinese Remainder Theorem

Over numbers from 0 to 2.3.5=30

The method works when N has no repeated prime factors

Arithmetic advantages?

1

mod ,p

ii i

Nx x p

p

Topic 4 22April 13, 2005

A Finite Field

Integers mod a prime A reciprocal of a positive integer always

exists Addition, subtraction, multiplication,

division, all defined and commutative

Topic 4 23April 13, 2005

Power Control and CDMA

The near-far problem The spreading loss will vary up to 70 dB over the

coverage area Code rejection factors are usually less than this Result is that interference can occur between closely-

spaced handsets or near base stations

Solution is power control Reduce handset power to make received power

constant

Topic 4 24April 13, 2005

Frequency Hopping

Definition: Changing from channel to channel at regular intervals

Mitigates these problems The near-far problem between handsets Narrow band interference

But, non-coherent detection is necessary Advantages also include

Full and best use of available spectrum for QoS Can be combined with spread spectrum (FH-SS)

April 13, 2005 Topic 4 25

EE320 March 28

Topic 4 26April 13, 2005

Topics

Term Project Problem 5.1 p. 262 Problem 5.17 p. 299 Theme Example: WCDMA

Topic 4 27April 13, 2005

The Term Project

Continue with the start that you turned in with the first quiz backup Input

Frequency sweep 1000 Hz to 3500 Hz Noise to obtain 20 dB SNR

Sampling to obtain good performance Do NOT pitch your beginning and pick up the

ADC to bitstream modules as a template Sample and encode/decode as instructed Measure BER vs. Eb/N0 as instructed Compare hard decoding with soft decoding

Topic 4 28April 13, 2005

Problem 5.1 p. 262

What is the equation for the spectrum of the spreading sequence given by Eq. (5.5) p. 261?

The chips c(q) are +1 or -1 and the chip shape gc(t) is

1

Q

c cq

g t c q g t q T

1

, 0

0, otherwise

cc

t Tg t T

Topic 4 29April 13, 2005

Use the Convolution Theorem

The spreading sequence is

The Fourier transform of each term in the sum is

1

Q

c cq

g t c q t q T g t

21 1sincq c c

TG f G f f T

Q Q Q

Topic 4 30April 13, 2005

Problem 5.17 p. 299

Do you expect FEC codes to have a greater or lesser benefit in Rayleigh-fading channels? Discuss your answer

Rayleigh fading channels have higher BER than otherwise similar Gaussian channels – more opportunity for improvement

Interleavers are necessary to make sure that dfree or fewer bits are exposed in a coherency interval

Topic 4 31April 13, 2005

WCDMA (1 of 3)

From Theme Example 4 pp.323-328 Cell phone technology generations

First: analog cell phonesSecond: TDMA, IS-95, GSMThird: Universal Mobile Terrestrial

Telecommunications systems (UMTS) WCDMA is a UMTS

Topic 4 32April 13, 2005

WCDMA (2 of 3)

Functional differencesSimultaneous voice and data transmissionOther data such as real-time TV

Performance improvementsThree times the bandwidthFour times the maximum spreading factorOptional turbo codes

Topic 4 33April 13, 2005

WCDMA (3 of 3)

Other differencesMultiple simultaneous CDMA downlinkDownlink power controlAsynchronous base stations

Bottom lineBroadband or ISDN in a cell phoneNear-far problems mitigatedHigher density of base stations and users

Topic 4 34April 13, 2005

Problem 5.19 page 305 (1 of 3)

Define the cellular spectral efficiency nu, in bits/second/Hz/cell; this is the total number of bits/second/Hz transmitted by all users in a cell. For a QPSK base modulation, assume that the spectral efficiency of a single CDMA user is 1/Q bits/second/Hz, where Q is the length of the spreading code. Suppose the receiver requires a specified SINR. Using Eq. (5.85) page 304, develop an expression for nu that depends on the received I0/N0, SINR, and f. Whay does the result not depend explicitly on Q? How does it depend implicitly on Q?

Topic 4 35April 13, 2005

Problem 5.19 page 309 (2 of 3)

The spectral efficiency forK users in the cellEach transmitting 2/Q bits/second/Hz

From Eq. (5.85) page 304

2K

Q

0

0

1 1N

Q f SINR KI

Topic 4 36April 13, 2005

Problem 5.19 page 309 (3 of 3)

Rolling up these two equations gives nu as

The spreading factor Q influencesThe interference factor fThe interference to noise ratio I0/N0

0

0

2

1 1N

f SINRI

Topic 4 37April 13, 2005

Theme Example 1: IS-95

Section 5.12 Page 311 Wireless cellular generations

Analog systems Initial digital systems – GSM, IS-54, IS-95 Integrated voice and data systems

Cell bands Uplink 869-894 MHz, downlink 24 MHz lower Uplink 1930-1990 MHz, downlink 80 MHz lower

Topic 4 38April 13, 2005

IS-95 Specifications and Usage

Most CDMA cell phones use the IS-95 standard Data rate is 9.6 kbps

Mainly voice Some data, trend is increasing amounts

Direct sequence spread to 1.2288 megachips per second

Channel bandwidth is 1.25 MHz Emerging standard based on IS-95 is

CDMA2000

Topic 4 39April 13, 2005

Channel Protocol of IS-95

Making an IS-95 call – the Mobile Terminal Searches for Pilot channel and synchronizes with it Locks to the Sync channel that is synchronized with

the Pilot channel, and gets system information (spreading code) of the access and paging channels

Sends a request to set up a call to the Access channel

Listens to Paging channel for traffic channel assignment

Transmits up assigned uplink channel, receives on assigned downlink channel

Topic 4 40April 13, 2005

Channel Protocol of IS-95

Receiving an IS-95 call – the Base StationTransmits a short message on the paging

channelAccepts Mobile Terminal request for call

DifferencesRequest for call has the phone number to

initiate a callPaging channel has Mobile Terminal phone

number in the paging message

Topic 4 41April 13, 2005

What The Pilot Channel Is

Shared by all users of the base station Transmitted at higher power than the data channels –

about 20% of total power Unmodulated signal – no CDMA here Provides fast synch and reliable channel tracking to

support coherent demodulation and robust CDMA Mobile terminal

Tracks the pilot channel of the current cell Searches for other pilot channels Switches cells when another pilot signal is stronger Transparent to the user

Topic 4 42April 13, 2005

The Four Downlink Channels

Separated by use of Walsh-Hadamard codes of length 64Pilot used Walsh #0Sync uses Walsh #32Paging using Walsh #1Traffic uses one of the other codes

See Figure 5.29 page 314

Topic 4 43April 13, 2005

The Traffic Channel

Multiplexed with control bits for power control

Rate ½ FEC encoded and interleaved Scrambling with long code sequence

follows interleaving (42 bits) Block diagram in Figure 5.30 page 315

Topic 4 44April 13, 2005

Problem 5.2 Page 263

Filtering with an integrate-and-dump filter is equivalent to convolving with a rectangular pulse of length T. Show, by using Parseval’s theorem, that the noise bandwidth of an integrate-and-dump is 1/T.

Topic 4 45April 13, 2005

Parseval’s Theorem

For Fourier transform pair see Table A.2 p. 482

For Parseval’s theorem see Eq. (A.36) p. 491

2 2h t dt H f df

exp 2

rect , sinc

H f h t j f t dt

th t H f T f T

T

Topic 4 46April 13, 2005

Noise Bandwidth

Definition: ratio ofThe variance of the output of a transfer

function to a white noise with two-sided power spectral density N0/2

The power spectral density N0

Equation

2

2 2

1

0N

H f dfT

BT TH

Topic 4 47April 13, 2005

Power Control: The Near-Far Problem Haykin & Moher Section 5.7 pp. 294-297 Received signal from K CDMA

transmitters is, from Eq. (5.38) p. 279

1

propagation loss on path k

signal k

receiver noise

data sequence k X spreading code k

K

c k kk

k

k b k

k k

x t s t w t

s t b E g t

w t

b g t

Topic 4 48April 13, 2005

SINR of First User

More detail in 5.4.1 pages 279-283

*1 1 1 1 1

10

1 1

2 22 2 20 1 0

2 2

22

2112

2 00

2

1

1

T K

b b k k kk

b

K K

y b k k b kk k

b bgK

yb k

k

y x t g t dt b E E R

y b E

y y N E R N EQ

y E ESINR D

NN E

Q

Topic 4 49April 13, 2005

Degradation in Multi-User Performance

221

20 1

1

1 11

1

gK

b k

k

DEK

Q N K

Topic 4 50April 13, 2005

FEC Coding and CDMA

Haykin & Moher Section 5.8 pp. 297-299 Direct Sequence Spread Spectrum (DS-

SS) spreads spectrum without added redundancy

Use of FEC spreads spectrum and adds redundancy

Topic 4 51April 13, 2005

Spreading Rate and Degradation The maximum spreading rate is

Degradation in multi-user performance is

1, FEC code rateDSQ Q r

r

1 1

0 0

1

0

/1 11 1

/

11

s bg

DS

b

E E rK KD

Q N Q r N

EK

Q N

Topic 4 52April 13, 2005

Example 5.5 Pages 298-299

Suppose a system has an information rate Rb=4800 bps and Q=Rc/Rb=32. The system is error protected by a rate-1/2 convolutional code. Compare the degradation Db with and without FEC coding at a BER of 10-5 when there are seven interfering users.

Topic 4 53April 13, 2005

Base Parameters Without FEC Encoding

1 5

0

0

2

g

210 4.265043367=

4.26504336710 log 20 log

2

9.59846903

4.2650433678 1D 1

32 2

0.33449332

b

b

EQ

N

E

N

Topic 4 54April 13, 2005

Base Parameters With FEC Encoding With rate-1/2 constratin-length 7

convolutional FEC encoding BER improved to 10-5 with Eb/N0

decreased to 4.5 dB Result is Dg increased to 0.62 Improvement is about 2.7 dB

Topic 4 55April 13, 2005

April 13, 2005 Topic 4 56

EE320 Telecommunications Engineering

James K Beard, [email protected]&A 349

Topic 4 57April 13, 2005

Quiz 2

Not a difficult quiz Some problems were a slight variation of

the text material such as substitution of one code for another

I allowed 2 ½ hours for a 50-minute quiz The curve from this quiz should be

definitive

Topic 4 58April 13, 2005

Crunch Time

We have about four weeks leftLast day of class is Monday May 2Final exam is Monday May 11, 11:00 AM -

1:00 PM Some of you are in trouble

Some quiz grades are lowNot everyone will passDepartment has been notified

Topic 4 59April 13, 2005

Watch for a Warning

If you are heading toward a grade lower than CYou will receive a warning with your Quiz 2

gradeThe cover page of your quizYour minimum Final Exam grade will be given

Final ExamBy the bookTwo hours, no talking

Topic 4 60April 13, 2005

The Problem

EE320 Telecommunications Engineering is… A tough course A required course Material packed with new concepts and technology

But the Perception of some is… An easy course A required course that everyone will pass Watch the slides, read the text before each quiz, and

everything will be OK

Topic 4 61April 13, 2005

The Solution

Take notes Regular class notebook On the slides The act of taking notes helps retention

Study a little Even the best student needs to do two or three

homework problems per chapter The study guide can help you pick them

Do well on the Final Examination A good grade there can bring up your final grade Don’t wait until Study Day to catch up on four courses

Topic 4 62April 13, 2005

Problem 5.3 page 265

Fill in the missing details of Eq. (5.19) A non-spread link The jammer is on for T seconds Spectrum

After multiplying by de-spreading sequence Development as in Eq. (5.17) Eq. (5.17) with Tc->T, Q->1 because of no spreading

2sincgS f A f f T

Topic 4 63April 13, 2005

Problem 5.3 page 265

Jammer spectral density at baseband is

The noise bandwidth of an integrate-and-dump is 1/T (see problem 5.2)

The noise variance is

2

1

p pS f S f df

AT

, bandwidthS f A f

Topic 4 64April 13, 2005

Properties of m-Sequences

Length property: Each m-sequence is of length 2m-1 Balance property: Each m-sequence has 2m-1 ones and

2m-1-1 zeros Shift property: The modulo-2 sum of an m-sequence and

any circularly-shifted version of itself produces another circularly-shifted version of itself

Subsequence property: Each m-sequence contains a subsequence of 1, 2, 3,…,m-1 zeros and ones

Autocorrelation property: See Equations (3.30) and (3.31) pages 272, 272

Topic 4 65April 13, 2005

Problem 5.7 page 272

Prove the autocorrelation property of Eq. (5.31) for m-sequences (Hint: use preceding Properties 1 through 5 as needed.)

Eq. (5.30) and (5.31) pages 271 and 272

1

0

1mod

1, 0

1, 0

Q

jjq

R k c q c q k QQ

k

kQ

Topic 4 66April 13, 2005

Proof of Autocorrelation Property of m-Sequences By the Shift property, the circular

autocorrelation, a modulo-2 sum of an m-sequence and a circularly-shifted version of itself, is another circularly-shifted version of iteslf

From the Balance property an m-sequence has one more 1 than zeros.

QED

Topic 4 67April 13, 2005

Course Material Overview

The problemMake a cell phone system workDeal with mobile terminalsDeal with urban fading

The solutionConstruct the network layer infrastructureDispense the data link layer mobile terminalsExploit the physical layer successfully

Topic 4 68April 13, 2005

Designing the Data Link Layer

The problemAddressing multiple users

Data push – making calls Data pull – accepting calls SINR

Dealing with urban fading The solution

IS-95 and other 2nd generation standardsThird generation standards

Topic 4 69April 13, 2005

How IS-95 Meets the Challenges 64 Walsh-Hadamard codes Uplinks 45 MHz below downlinks Synchronization

The pilot channel (Walsh code 0) allows coherent detection

The synchronization channel (Walsh code 32) provides spreading codes of access and paging channels

Paging channel (Walsh code 1) assigns access channel

Access channels (other Walsh codes)

Topic 4 70April 13, 2005

Meeting the Challenge of Fading Forward Error Correction Codes (FECs)

Allows robust operation with high bit error rates (BER)

Spread spectrum Allows higher BER in frequency-selective fading

Interleaving Helps bridge dropouts from fading and interference

over intervals of a few milliseconds

Topic 4 71April 13, 2005

Meeting the Challenge of Higher Traffic and New Uses Use of CDMA to allow channel sharing Use of power control to limit SINR at the base

station In next-generation standards such as Universal

Mobile Terminal Terrestrial Telecommunication Systems (UMTSs) Base station power control to limit SINR at the mobile

station Higher bandwidths and data rates More sophisticated coding to approach Shannon

channel limit More versatile data formats for text, video, etc.

Topic 4 72April 13, 2005

Problem 5.8 Page 274

To show that scramblers based on m-sequences are not very good encryption devices, determine the minimum number of consecutive bits that would need to be known to reconstruct the initial state.

The generating polynomial is known. Use Figure 5.10, f(x)=x7+x3+1 as an

example

Topic 4 73April 13, 2005

Solution to Problem 5.8

If the generating polynomial is known then the entire m-sequence is known

The problem is reduced to determining how many successive bits in the m-sequence are necessary to uniquely determine the position in the sequence

From Figure 5.10 m, the number of lags – seven for this example – bits in a row determines the state of the shift register

Topic 4 74April 13, 2005

RAKE Receiver

RAKE Not an acronym Based on signal flow diagram that looks like a garden

rake

Receiver architecture used for CDMA systems Concept addresses multipath environments Consists of

An array of up to Q parallel receiver Timing between these receivers varies in steps of Tc

Topic 4 75April 13, 2005

RAKE Signal Flow

Each channelMultiplied by spreading code g(t) Integrate and dump filter of length T=Q.Tc

Weight by expected corresponding multipath channel amplitude

All are then summed into a single-channel processor

Result is “matched filter” to multipath channel

Topic 4 76April 13, 2005

Example 5.5 Pages 298-299

Given BPSK Information rate Rb=4800 bps

Spreading factor Q=Rc/Rb=32

Rate ½ convolutional code BER is 10-5

Number of interfering users is 7 (K=8) Compare degradation Dg with and without FEC

Use Eqs. Page 272 and (5.72) just preceding1

0

11 b

g

EKD

Q N

Topic 4 77April 13, 2005

Example 5.5 (Continued)

The BER of 10-5 indicates single-user

The degradation factor is

0 0

9.12 10 dBb sE E

N N

1

8 11 9.12 0.33 -4.8 dB

32gD

Topic 4 78April 13, 2005

Example 5.5 (Continued)

With rate ½ constraint length 7FEC encoding Spreading factor Qs=16, total Q=32

Single-user

Degradation factor is

0 0

2.82 4.5 dBb sE E

N N

1

8 11 2.82 0.62 -2.1 dB

32gD

Topic 4 79April 13, 2005

Example 5.5 (Concluded)

Degradation improves 2.7 dB with FEC Degradation vs. loading for rate ½ codes

shown in Figure 5.23 page 298No improvement for single userGains of about 2 dB for K near Q Improvement doesn’t vary much with BER

Conclusion: BER is important power-bandwidth tradeoff with multiple users

Topic 4 80April 13, 2005

Problem 5.49 Page 336

Describe how the use of a rate ¼ FEC doe would affect the implementation and performance of a RAKE Receiver

Effects of change in FEC code Delay-line parallelism is not affected Channel tracking (see 5.6 pages 292-294) is affected

because algorithm operates before FEC and Eb/N0 is lower with better codes

Measures for use with better codes include Use a known pilot signal, as with IS-95 Use training sequences (standard messages) for

channel tracking, as with WCDMA

April 13, 2005 Topic 4 81

EE320 Telecommunication Engineering

Wireless Architectures

Topic 4 82April 13, 2005

Open System Interconnection (OSI) Model Seven-layer model

Physical layer (modem)Data link layerNetwork layerTransport layer (packetizing, ACK/NAK)Session layer (Service selection and access)Presentation layer (encryption, compression)Application layer (HMI)

Layers designed together as a system

Topic 4 83April 13, 2005

Power Control Architectures

Open Loop Mobile terminals measure strength of pilot channel Transmit power decreased for strong pilot channels Fast and simple, but must be approximate

Closed Loop Base station measures mobile terminal signal strength Mobile station receives signal strength by downlink Accurate but delay and averaging must be smaller than channel

coherence time Outer Loop Control

Base station uses expected signal strength in control algorithm Complexity can result in a slow loop

Topic 4 84April 13, 2005

Power Control: Summary

Power control minimizes SINR in busy cells Handset power control minimizes SINR in the

base station but not at the mobile terminal Methods still evolving Next generation standards will implement

Newer techniques such as outer-loop control Base station power control for SINR control at the

mobile station

Topic 4 85April 13, 2005

Next Time

Assignment: Read parts of Chapter 77.3, OSI7.6, Power Control7.7, Handover7.8, Network Layer