February 14, 2005Topic 3 1 Telecommunications Engineering Topic 3: Modulation and FDMA James K Beard, Ph.D. (215) 204-7932 [email protected] jkbeard

February 14, 2005 Topic 3 1

Telecommunications EngineeringTopic 3: Modulation and FDMAJames K Beard, Ph.D.(215) [email protected]://astro.temple.edu/~jkbeard/

Topic 3 2February 14, 2005

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Topics

The Survey Homework

Problem 3.30, p. 177, Adjacent channel interference Problem 3.35 p. 178, look at part (a); part (b) was

done in class Problem 3.36 p. 178, an intermediate difficulty

problem in bit error rate using MSK Topics

Why follow sampling with coding? Shannon’s information theory


Survey Thumbnail

Nine completed surveys Six incomplete or just looks Five did not open the survey Selection biases results?

Only 45% of class gives all resultsOther 55% can be best, worst, or more of the

same


Multiple Choice Questions

All are replies on 1-5 basis No answers were 1 or 5 Questions

My background is appropriate I understand sampled time/frequency domains I am comfortable with readings and text


Multiple Choice Summary

0% 20% 40% 60% 80% 100%

Background

Time/Frequency

Readings


Study Difficulties Reported

Problems2.5 p. 292.6 p. 332.14 p. 672.20 p. 772.22 p. 81

Examples2.17 p. 75Theme example 1 p. 82


Suggestions

More examples and homework problems worked through – already in progress

Go over more difficult homework problems before they are assigned

Warn and correct wrong answers – AIP Discuss WHY as well as HOW


Survey Summary

Everybody is OKBased on 40% sampleBut, nobody answered 5’s

NeedMore coverage of how and whyFill in for prerequisites


Homework

Problem 3.30, p. 177, Adjacent channel interference

Problem 3.35 p. 178, look at part (a); part (b) was done in class

Problem 3.36 p. 178, an intermediate difficulty problem in bit error rate using MSK


Problem 3.30 Page 177

Adjacent channel interference

2

2

power spectral density of input

Frequency response of channel filter

channel separation

G f

H f

f

G f H f f df

ACI f

G f H f df


Solution

Power of signal in correct filter

Power of signal in adjacent channel

2

CCP G f H f df

2

ACP G f H f f df


Problem 3.35 p. 178 (a)

Formulas in table 3.4 page 159 Begin BPSK with

Integrate over Rayleigh distribution

2

0

, be

EP erfc

N

0 0

1exp , 0p


Evaluate the Integral

Average BER is

Evaluation of integral is left as ETR

0

0

0

0 0

1 11

2 1 4

eP P p d


Problem 3.36 p. 178

Use MSK with a BER of 10-4 or better AWGN

Use Table 3.4 or Figure 3.32, pp. 159-160SNR requirement is about 8.3 dB

Rayleigh fadingUse Table 3.4 p. 159Solve for SNR of about 34 dB


Coding Follows Sampling

SamplingSimply converts base signal to elementary

modulation formFormatting for performance is left to coding

CodingRemoval of redundancy == source codingChannel coding == error detection and

correction capability added


Shannon’s Information Theory

First published in BSTJ article in 1948 Builds on Nyquist sampling theory Adds BER concepts to find maximum flow of bits

through a channel limited by Bandwith SNR

Channel capacity maximum is

/ 2log 1 TBits Sec

Noise

PC BW

P


Other Important Results

Channel-coding theorem Given

A channel capacity CB/S

Channel bit rate less than channel capacity

Then There exists a coding scheme that achieves an arbitrarily

high BER

Rate distortion theory – sampling and data compression losses exempt from channel-coding theorem


Concept of Entropy

Definition – Average information content per symbol

ImportanceFundamental limit on average number of bits

per source symbolChannel-coding theorem is stated in terms of

entropy


Equation for Entropy

1

20

source alphabet set

average number of bits per symbol

number of symbols in alphabet

probability of symbol k in message

1log (entropy)

(coding efficiency)

k

K

kk k

S

L

K

p

H S pp

H S

L


Study Problems and Reading Assignments Reading assignments

Read Section 4.6, Cyclic Redundancy ChecksRead Section 4.7, Error-Control Coding

Study examplesExample 4.1 page 197Problem 4.1 page 197


Problem 2.4 p. 28

4 GHz microwave link Towers 100 m and 50 m tall, 3 km apart Midway between, tower 70 m tall Radius of Fresnel zone, eq. (2.38) p. 27

Distance d1 = d2 = 1.5 km

Raise both towers

2

1 21

1 2

0.075 15007.5 5

3000

d d m mr m m

d d m


Problem 2.5 p. 29

Similar to 2.4 but LOS is clearly obstructed Fresnel-Kirchoff diffraction parameter eq.

(2.39) is

Diffraction loss is 24 dB For 400 MHz, v = 1.096, loss = 16 dB

1 2

1 2

23.465

d dh

d d


Term Projects Areas for coverage

Propagation and noise Free space Urban

Modulation & FDMACodingDemodulation and detection

Will deploy over Blackboard this week


Term Project Timeline

First week Parse and report your understanding Give estimated parameters including SystemView

system clock rate Second week

Block out SystemView Signal generator Modulator

Through mid-April Flesh out as class topics are presented Due date TBD


EE320 Digital Telecommunications

Quiz 1 Report

February 21, 2005


The Curve

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The Answers

See previous lectures/slidesQuestions 1, 4

See Excel spreadsheetsQuestions 2, 3, 5

See Mathcad spreadsheetFresnel integrals for diffraction loss


Question 2Name Wavelength Ae, Ex. 2.2 p. 16

Gain, Eq. (2.9) p. 16 Gain, dB

ABREFA-KODOM 0.06 3.534291735 12337.0055 40.91209758APEHAYA 0.06122449 3.773838175 12651.5224 41.02142788BARAKAT 0.0625 4.021238597 12936.2879 41.11809672BIRCH 0.063829787 4.276493 13190.1835 41.20250836BOADO 0.065217391 4.539601384 13412.2221 41.27500738CARDONE 0.066666667 4.810563751 13601.5486 41.33588357DO 0.068181818 5.089380099 13757.439 41.38537595GEDZAH 0.069767442 5.376050428 13879.3012 41.423676MADJAR 0.071428571 5.67057474 13966.6746 41.45093014P H NGUYEN 0.073170732 5.972953033 14019.2302 41.46724168T T NGUYEN 0.075 6.283185307 14036.7707 41.47267206T H NGUYEN 0.076923077 6.601271563 14019.2302 41.46724168PANG 0.078947368 6.927211801 13966.6746 41.45093014PATEL 0.081081081 7.261006021 13879.3012 41.423676ROIDAD 0.083333333 7.602654222 13757.439 41.38537595SCHOLL 0.085714286 7.952156404 13601.5486 41.33588357STRAKER 0.088235294 8.309512569 13412.2221 41.27500738TANUI 0.090909091 8.674722715 13190.1835 41.20250836TRAN 0.09375 9.047786842 12936.2879 41.11809672ZAYZAY 0.096774194 9.428704952 12651.5224 41.02142788


Question 3

Name i Wavel 1 Wavel 2Part I Eq. (2.36) p. 27 h

Nu or v, eq. (2.39) p. 28

Loss, Fig. 2.10 p. 29

Nu or v, 2nd freq 2nd Loss

ABREFA-KODOM 0 0.06 0.6 7.745966692 3 0.547722558 -10.639 0.173205 -7.501APEHAYA 1 0.059761 0.59761 7.73052108 3 0.548816909 -10.64753 0.173551 -7.503895BARAKAT 2 0.059524 0.595238 7.715167498 3 0.549909083 -10.65605 0.173897 -7.506789BIRCH 3 0.059289 0.592885 7.699905035 3 0.550999093 -10.66458 0.174241 -7.509684BOADO 4 0.059055 0.590551 7.684732794 3 0.55208695 -10.67311 0.174585 -7.512579CARDONE 5 0.058824 0.588235 7.669649888 3 0.553172667 -10.68163 0.174929 -7.515474DO 6 0.058594 0.585938 7.654655446 3 0.554256258 -10.69016 0.175271 -7.518368GEDZAH 7 0.058366 0.583658 7.639748605 3 0.555337735 -10.69868 0.175613 -7.521263MADJAR 8 0.05814 0.581395 7.624928517 3 0.55641711 -10.70721 0.175955 -7.524158P H NGUYEN 9 0.057915 0.579151 7.610194341 3 0.557494395 -10.71574 0.176295 -7.527053T T NGUYEN 10 0.057692 0.576923 7.595545253 3 0.558569602 -10.72426 0.176635 -7.529947T H NGUYEN 11 0.057471 0.574713 7.580980436 3 0.559642743 -10.73279 0.176975 -7.532842PANG 12 0.057252 0.572519 7.566499085 3 0.560713831 -10.74132 0.177313 -7.535737PATEL 13 0.057034 0.570342 7.552100405 3 0.561782876 -10.74984 0.177651 -7.538632ROIDAD 14 0.056818 0.568182 7.537783614 3 0.562849891 -10.75837 0.177989 -7.541526SCHOLL 15 0.056604 0.566038 7.523547939 3 0.563914887 -10.76689 0.178326 -7.544421STRAKER 16 0.056391 0.56391 7.509392615 3 0.564977876 -10.77542 0.178662 -7.547316TANUI 17 0.05618 0.561798 7.49531689 3 0.566038868 -10.78395 0.178997 -7.550211TRAN 18 0.05597 0.559701 7.481320021 3 0.567097875 -10.79247 0.179332 -7.553105ZAYZAY 19 0.055762 0.557621 7.467401274 3 0.568154908 -10.801 0.179666 -7.556


Question 5Name BER Qinv(BER) Eb/N0, dB

Rayleigh SNR

Rayleigh SNR, dB

ABREFA-KODOM 0.000316 3.41734328 7.663472171 789.8195 28.97528APEHAYA 0.000251 3.47953449 7.820122959 994.518 29.97613BARAKAT 0.0002 3.54076432 7.971640442 1252.218 30.9768BIRCH 0.000158 3.60107467 8.118342568 1576.643 31.97733BOADO 0.000126 3.66050453 8.26051902 1985.071 32.97776CARDONE 0.0001 3.71909027 8.398434441 2499.25 33.9781DO 7.94E-05 3.77686584 8.532331217 3146.564 34.97837GEDZAH 6.31E-05 3.83386302 8.66243188 3961.483 35.97858MADJAR 5.01E-05 3.89011158 8.788941206 4987.406 36.97875P H NGUYEN 3.98E-05 3.94563947 8.912048035 6278.966 37.97888T T NGUYEN 3.16E-05 4.00047298 9.031926873 7904.944 38.97899T J NGUYEN 2.51E-05 4.05463684 9.148739293 9951.929 39.97907PANG 2E-05 4.10815439 9.262635176 12528.93 40.97914PATEL 1.58E-05 4.16104764 9.373753798 15773.18 41.97919ROIDAD 1.26E-05 4.2133374 9.482224805 19857.46 42.97924SCHOLL 0.00001 4.26504337 9.588169073 24999.25 43.97927STRAKER 7.94E-06 4.31618421 9.691699474 31472.39 44.9793TANUI 6.31E-06 4.36677762 9.792921563 39621.58 45.97932TRAN 5.01E-06 4.41684041 9.891934194 49880.81 46.97933ZAYZAY 3.98E-06 4.46638855 9.988830071 62796.41 47.97935


The Quiz in the Text (1 of 2)

Question 1 Text pp 3-5 Lectures and slides several times

Question 2 Antenna gain equations (2.2), (2.3) pp 14 Also equations (2.9), example 2.2, pp 16-17

Question 3 Section 2.3.2 and exa,[;e 2.3 pp 24-29 Two lectures, example worked in class, practice quiz


The Quiz in the Text (2 of 2)

Question 4Problem 3.30 p. 177Given in classAnswer was to give equation that was given in

the problem statement Question 5

Problem 3.36, given in classUse table 3.4, figure 3.33, pp. 159-161


EE320 Convolutional Codes

James K Beard, Ph.D.


Bonus Topic: Gray Codes

Sometimes called reflected codes Defining property: only one bit changes

between sequential codes Conversion

Binary codes to Gray Work from LSB up XOR of bits j and j+1 to get bit j of Gray code Bit past MSB of binary code is 0

Gray to binary Work from MSB down XOR bits j+1 of binary code and bit j of Gray code to get bit j

of binary code Bit past MSB of binary code is 0


Polynomials Modulo 2

Definition Coefficients are ones and zeros Values of independent variable are one or zero Result of computation is taken modulo 2 – a one or

zero

The theory Well developed to support many DSP applications Mathematical theory includes finite fields and other

areas


Base Concept – Signal Polynomial Pose data as a bit stream Characterize data as impulse response of a filter

with weights 1 and 0 Characterize as z transform Substitute D for 1/z in transfer function Example

Signal 11011001 Filter is 1 + (1/z) +(1/z)3+(1/z)4+ (1/z)6

Signal polynomial is 1 + D + D2+D4+D6

Signal and filter polynomials provide base method for understanding convolutional codes


Base Concept –Modulo 2 Convolutions Scenario

Bitstream into convolution filterFilter weights are ones and zerosOutput is taken modulo 2 – i.e. a 1 or 0

Result: A modulo 2 convolution converts one bit stream into another


Benefits of Concept

Convolution is product of polynomials Conventional multiplication of polynomials is

isomorphic to convolution of the sequence of their coefficients

Taking the resulting coefficients modulo 2 presents us with the output of a bitstream into a convolution filter with output modulo 2

These special polynomials have a highly developed mathematical basis

Implementation in hardware and software is very simple


Error-Control Coding

Two categories of channel codingForward error-correction (EDAC)Automatic-repeat request (handshake)

CRC CodesHash codes of the messageError detection, but not correction


Topics in Convolutional Codes

The node diagram is a block diagram Polynomial representations

Represent signals, convolutions and special polynomials and polynomial operations

Give us a simple way to understand and analyze convolutions Trellis diagrams

Give us a mechanism to represent convolution operations as a finite state machine

Provide a first step in visulaization of the finite state machine Node diagrams

Provide a simple visualization of the finite state machine Provide a basis for very simple implementation


Convolutional Code Steps

Reduce the message to a bit stream Operate using modulo-2 convolutions

Convolution filter with short binary maskTake result modulo 2

Implemented with one-bit shift registers with multiplexer (see Figure 4.6 p. 196)


Example 4.1 Page 197

Output

1/z 1/z

Path 1

Path 2

Input

Haykin & MoherFigure 4.6 p. 196


Example 4.1 (1 of 4)

Response of Path 1

Response of Path 2

Mutiplex the outputs bit by bitOne side output, then the otherProduce a longer bit stream

1 21g D D

2 21g D D D



SignalMessage bit stream (10011)Message as a polynomial

Multiply the message polynomial by the Path 1 and Path 2 filter polynomialsObtain two bit streams from resulting

polynomialsMultiplex (interleave) the results

3 41m D D D



Polynomial multiplication results

Messages Path 1 (1011111) Path 2 (1111001) Multiplexing them (11, 10, 11, 11, 01, 01, 11)

1 2 3 4

2 3 4 5 6

2 2 3 4

2 3 6

1 1

1

1 1

1

c D D D D

D D D D D

c D D D D D

D D D D



Length of coded message isTwice the order of the product polynomials +12.(length of message + length of shift registers

- 1) = 2.(5 + 3 - 1)=2.7=14 Shift registers have memory

Simplest way to clear is to feed zerosNumber of clocks is number of stagesZeros between message words are tail zeros


Problem 4.1

Signal and polynomial 1

Signal and polynomial 2

Result

1 1,0,0,1,1 1,0,1 1,0,1,1,1,1,1c

2 1,0,0,1,1 1,1,1 1,1,1,1,0,0,1c

11,10,11,11,01,01,11c


Modulo 2 Convolution Diagrams

Poly 1 0 1 Poly 1 1 1Signal 1 0 0 1 1 Signal 1 0 0 1 1Resullt: 1 1 Resullt: 1 1

Poly 1 0 1 Poly 1 1 1Signal 1 0 0 1 1 Signal 1 0 0 1 1Resullt: 0 Resullt: 1 1



Poly 1 0 1 Poly 1 1 1Signal 1 0 0 1 1 Signal 1 0 0 1 1Resullt: 1 1 Resullt: 0 1 1

Poly 1 0 1 Poly 1 1 1Signal 1 0 0 1 1 Signal 1 0 0 1 1Resullt: 1 1 Resullt: 0 1 1



Trellis and State Diagrams

Trellis diagram Figure 4.7 p. 198, and state table 4.2 p. 199Horizontal position of node represents timeTop line represents the inputEach row represents a state of the two-path

encoder – a finite state machine Trace paths produced by input 1’s and 0’s

Paths produced by 0’s are solidPaths produced by 1’s are dotted


Advantages of Trellis and State Diagrams Once drawn, output for any message is simple

to obtain Allowed and non-allowed state transitions are

explicit State diagram follows directly

Figure 4.8 p. 200 Shows state transitions and causes

Coding output of state diagram simpler than that of trellis diagram


States of the Filter

We need the output states Ordered pair of bits from Path 1 and Path 2 Objective is tracing through states to get outputs

Output states not the same as register states Only four states can be defined from two outputs Total number of states is defined by the order of

convolution Current example is three taps Number of states is 2<order> or 8 We get to eight states by considering each pair of

consecutive input bits


Drawing the Trellis Diagram

Begin with a state table For each paths “state”

For the last bit 0 Draw the solid path for a 0 Draw the dotted path for a 1

For the last bit 1 Draw the solid path for a 0 Draw the dotted path for a 1

This is 16 lines total

State Table

0 0 0 0 01 0 1 0 00 1 0 0 11 1 1 0 10 1 1 1 01 1 0 1 00 0 1 1 11 0 0 1 1

State Paths 1, 2


For Output State [0,0]

For the last bit 0 Adding a zero, solid path label is (0,0)

From {0,0,0} to {0,0,0} Next state is [0,0]

Adding a one, dotted path label is (0,1) From {0,0,0} to {1,0,0} Next state is [1,1]





























Drawn Trellis Diagram

0,0

1,0

0,1

1,1

0,0

0,1

1,1

1,0

0,0

0,1

1,0

1,1

0,0

0,1

1,0

1,1

0,0

0,1

1,0

1,1


State Diagram

[01][00]

[11][10]

0,0 1,0

0,1 1,1

1,1

1,1

1,1 0,0

0,0

0,0

1,0

1,0

1,0

0,1

0,1

0,1


EE320 Decoding Convolutional Codes

James K Beard, Ph.D.


Where We are Going

Exploit the Channel Coding TheoremFor any required channel bit rate CR less than

the channel capacity C

A coding exists that achieves an arbitrarily low BER

Method is error-correcting codes

2log 1RC C B SNR


Hamming Weight and Distance

Hamming weight A property of the code Equal to the number of 1’s

Hamming distance Based on two codes Equal to the number of 1’s in an XOR

Used in definition of error correction An ECC makes the Hamming distance between

characters > 2 Overhead is increase in required bit rate


Hamming Distance and Error Correction Code error correction capability

Upper bound is half the Hamming distance between code vectors, dfree/2

The length extension due to convolutional codes can allow larger Hamming distance between input code vectors

NOTE: Gray codes are contrived to have a Hamming distance of 1 between adjacent characters

The constraint length K Is equal to the number of convolution delays plus 1 Bounds the error correction capability of two-convolution codes Table 4.3 p. 201 Our example has K=3, dfree=5, can theoretically correct 2 bits


Haykin & Moher Table 4.3Page 201

Constraint Length K Systematic Codes Non-Systematic Codes

2 3 3

3 4 5

4 4 6

5 5 7

6 6 8

7 6 10

8 7 (Note) 10

9 Not Available 12

NOTE: (1) From example polynomials 400, 671 in a non recursive code

Maximum Free Distance Attainable for Rate 1/2


Fundamental of Maximum Likelihood: Multivariate PDF Consider N Gaussian random variables

with mean zero and variance one,

The covariance of z is the identity matrix I

2

/ 2

1 1 1exp , exp

2 22 2

Tii z N

zp z p z z z

,T

i j i jz z z z I


Identities and Definitions

Determinant of product and inverse

Differential

Gradient

1 1,A B A B A

A

1 2d x dx dx

i

j

y y

x x


Variable Change to Correlated Variables Consider the variable change

The pdf of x is found from the differential and the Jacobian determinant

The covariance R of x is

x A z

T T T TR x x A z z A A A

11z z x

zp z d z p A x d x A p x d x

x


PDF of Correlated Variables

The pdf of x is

1

1/ 2/ 2

1

1 1exp

22

x z

T

N

p x p zA

x R xR


With a Mean…

The pdf of a Gaussian vector x of N elements with covariance R and mean a is

1

1/ 2/ 2

1exp

2

2

T

x N

x a R x ap x

R


Maximum Likelihood Estimators

Principle Given the pdf of a data vector y as available, for a

given set of parameters x is p(y|x) Find the set of parameters x hat that maximizes this

pdf for the given set of measurements y Properties

If a minimum variance estimator exists, this method will produce it

If not, the variance will approach the theoretical minimum – the Cramer-Rao bound – as the amount of relevant data increases


Observations on Maximum Likelihood All known minimum variance estimators can be

derived using the method of maximum likelihood; examples include Mean as average of samples Proportion in general population as proportion in a

sample Statistics and error bounds on estimators are

found as part of the derivation The method is simple to use


Our Example

Given a message vector m and its code vector c and a received vector r

Make an estimate m hat of the message vector Process

With the noisy c through the receiver channel estimate c hat

Select the code m hat that produces a code vector c tilde has the shortest Hamming distance to c hat


The MLE for c

Data

Log likelihood function

Solution is Nearest Neighbor

y c n

11ln | ln 2 ln

2

Tp y c N R y c R y c

c y


Assignment

Read 4.7, 4.8, 4.10, 4.11, 4.16 Do problem 4.1 p. 197 Do problem 4.2 p. 198 Do encoding in your term project


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Assignment

Read 5.2, 5.3, 5.5 Look at Problem 4.5 p. 243 Do problems 4.6, 4.7 p. 252


Interleaving and TDMA

The Viterbi Method Interleaving

Scatters the code data streamMakes low BER communication more robust

through fading medium and interference Noise performance TDMA


Viterbi Algorithm References

Viterbi, A.J., Error bounds for convolutional codes and an asymptotically optimum decoding algorithm, IEEE Trans. Inform. Theory, Vol. IT-13, pp. 260-269, 1967.

Forney, G.D. Jr. The Viterbi algorithm, Proceedings of the IEEE, Vol. 61, pp. 268-278, 1973.


The Viterbi Method

Use the trellis for decoding Dynamic programming

A method originally developed for a control theory problem by Richard Bellman

Based on working the problem from the end back to the beginning

Uses the Hamming distance as an optimization criteria

Crops the growing decision tree by taking the steps backward in time a few at a time


Interleaving

Based on the coherence time of a fading channel; fading due to motion leads to

Interleaving operations Block data over times much larger than TCOHERENCE

Reformat in smaller blocks Multiplex the smaller blocks into an interleaved data

stream

Maximum range rate0.3,

2COHERENCE DD

T ff


Interleaver Parameters

Object of interleavingData blackouts of TCOHERENCE cause loss of less

than dfree bits

FEC EDAC coding can bridge these gaps Make each row

Consist of at least dfree bits

Last TCOHERENCE or longer


Interleaver Methods

MethodsSimple sequential interleaving as just

describedPseudorandom interleaving

Combine with codesUse Viterbi interleaverUse multiplexer of convolutional codes as

interleaver


Noise performance

Compare AWGN channels using Figure 4.13 p. 213

Rayleigh fading performance given in Figure 4.14 p. 214


Turbo Codes

Revolutionary methodology Emerged in 1993 through 1995 Performance approaches Shannon limit

Technique Encoding blocked out in Figure 4.5 p. 215 Two systematic codes in parallel, one interleaved Excess parity bits trimmed or culled Decoding shown in Figure 4.17 p. 217

Performance shown in Figure 4.18 p. 219


Next Time

TDMA Chapter 4 examples Quiz postponed one week to March 30

Will cover Chapter 4Some topics in CDMA


TDMA

Time-Division Multiple Access (TDMA) Multiplex several users into one channel Alternative to FDMA Third alternative is CDMA, presented next

Advantages over FDMA Simultaneous transmit and receive aren’t required Single-frequency operation for transmit and receive Can be combined with interleaving Can be overlaid on FDMA, CDMA


Types of TDMA

Wideband Used in links such as satellite communications Frequency channels several MHz wide

Medium band Global System for Mobile (GSM) telecommunications Several broadband links

Narrow band TIA/EIA/IS-54-C standard in use for US cell phones Single frequency channel TDMA


Advantages of TDMA overlaying FDMA Cooperative channel allocation between

base stations with overlapping coverage (channel-busy avoidance)

Dropouts in some channels from frequency-dependent fading can be avoided

Equalization can mitigate frequency-dependent fading in medium and broad band TDMA/FDMA


Global System for Mobile (GSM) Internationally used TDMA/FDMA From Haykin & Moher 4.17 pp. 236-239 Overview given here Full description available on WWW

http://ccnga.uwaterloo.ca/~jscouria/GSM/gsmreport.html Organization

Time blocks Major frames are 60/13 = 4.615 milliseconds Eight Time Slots of 577 microseconds in each major frame 156.25 bits per time slot; 271 kBPS data rate

Frequency channels 124 channels 200 kHz wide, 200 kHz apart Frequency hopping with maximum of 25 MHz


GSM Characteristics

Frequency allocation (Europe) Uplink (to base station) 890 MHz to 915 MHz Downlink (to handsets) 935 MHz to 960 MHz

Design features counter frequency-selective fade Channel separation matches fading notch width Frame length matches fading duration EDAC combined with multilayer interleaving

Intrinsic latency is 57.5 milliseconds


Subframe Organization

Guard period of 8.25 bits begins the frame Three tail bits end guard and time slots Three data blocks

57 bits of data 26 bits of “training data” 57 bits of data

Flag bit precedes training and second data block Defines speech vs. digital or training data

Overall efficiency about 75% data


GSM Coding

Complex speech coder/decoder (CODEC) Concatenated convolutional codes Multi-layer interleaving GMSK channel modulation

About 40 dB adjacent channel rejection ISI effects are small

An international standard that defines affordable enabling technologies


Coherence Time Examples

Mobile terminal moving at 30 km/hr (19 mph) Frequency allocation about 1.9 GHz

Problem 4.4 p. 210 answer is 9.6 ms (!!!) Coherence time about 2.84 ms

Frequency of about 900 MHz European GSM allocation Coherence time of about 6 ms Velocity of 39 km/hr (24 mph) gives coherence time of

4.62 ms frame time


Problems 4.6, 4.7

Message is 10111… (1’s continue) Codes are (see p. 253)

Problem 4.6: (11)(10)Problem 4.7: (1111)(1101)

Find output code stream by polynomial method


Solution for Problem 4.6

Solution is simple enough to do by inspection:(11,10,11,01,01,01,…)Feed-through path embeds signal in the codeThis makes the code systematic


Solution for Problem 4.7

Message polynomial is

Generator polynomials are

2 3 4( ) 1m x x x x

(1) 2 3

(2) 3

( ) 1

( ) 1

g x x x x

g x x x


Solution for Problem 4.7 (concluded) Code polynomials are

Code bits are

Code output is

(1) (1) 3 4

(2) (2) 2 3 5 6

1

1

c x g x m x x x x

c x g x m x x x x x x

(1)

(2)

1,1,0,1,1,0,0,0,

1,1,1,1,0,1,1,1,

c x

c x

11,11,01,11,10,01,01,


Simulation of Problem 4.7(1 of 2)

program main !Execute a convolution code implicit none integer,dimension(4)::g1=(/1,1,1,1/),g2=(/1,0,1,1/) !Reverse order

integer,dimension(23)::message=(/0,0,0,1,0,1,1,1,1,1,1,1,1, & 1,1,1,1,1,1,1,1,1,1/)

integer::i,j,k=4

do i=1,10 print 1000,i,convolve(k,message,g1,i),convolve(k,message,g2,i) end do

1000 format(i3,": (",i2,",",i2,")")

contains…


Simulation of Problem 4.7(2 of 2)

integer function convolve(k,m,g,i) !Convolve m(i:i+k-1) with g(1:k) integer,intent(IN)::k,m(*),g(*),i integer::sumc

!Perform an ordinary convolution and take the result modulo 2

sumc=0

do j=1,k sumc=sumc+m(i+j-1)*g(j) end do

convolve=modulo(sumc,2)

end function convolve

end program main


Simulation Output

1: ( 1, 1) 2: ( 1, 1) 3: ( 0, 1) 4: ( 1, 1) 5: ( 1, 0) 6: ( 0, 1) 7: ( 0, 1) 8: ( 0, 1) 9: ( 0, 1) 10: ( 0, 1)

A few minutes with Fortran 95Your choice of language will do.


Interleaving and Coherence Time Problem 4.14 page 254 Coherence time

0.3 0.15

2 vehicle speedCOHERENCED

Tf

0.15

vehicle speed

COHERENCEBitLossBlockLength T BitRate

BitRate


Discussion Questions

One-Channel TDMAWhat about both transmit and receive on the

same frequency channel? Is it a good idea? Why?

What are the advantages and disadvantages of systematic and non-systematic codes?


Assignment

Read 5.2, 5.3, 5.5 Look at problem 5.1 p. 262 Do problem 5.2 p. 263 Next time

Spreading the spectrumGold codesGalois fields for binary sequences

Documents

February 14, 2005Topic 3 1 Telecommunications Engineering Topic 3: Modulation and FDMA James K Beard, Ph.D. (215) 204-7932 [email protected] jkbeard