Telecommunications
Engineering
Topic 3: Modulation and
FDMA
James K Beard, Ph.D.
(215) 204-7932
jkbeard@temple.edu
http://astro.temple.edu/~jkbeard/
February 14, 2005
Topic 3
1
February 14, 2005
Topic 3
27-Apr
20-Apr
13-Apr
6-Apr
30-Mar
23-Mar
16-Mar
9-Mar
2-Mar
23-Feb
16-Feb
9-Feb
2-Feb
26-Jan
19-Jan
Attendance
25
20
15
10
5
0
2
Topics


The Survey
Homework
 Problem
 Problem
3.30, p. 177, Adjacent channel interference
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
February 14, 2005
Topic 3
3
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 results
 Other 55% can be best, worst, or more of the
same
February 14, 2005
Topic 3
4
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
February 14, 2005
Topic 3
5
Multiple Choice Summary
Readings
Time/Frequency
Background
0%
February 14, 2005
20%
40%
Topic 3
60%
80%
100%
6
Study Difficulties Reported

Problems
 2.5
p. 29
 2.6 p. 33
 2.14 p. 67
 2.20 p. 77
 2.22 p. 81

Examples
 2.17
p. 75
 Theme example 1 p. 82
February 14, 2005
Topic 3
7
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

February 14, 2005
Topic 3
8
Survey Summary

Everybody is OK
 Based
on 40% sample
 But, nobody answered 5’s

Need
 More
coverage of how and why
 Fill in for prerequisites
February 14, 2005
Topic 3
9
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

February 14, 2005
Topic 3
10
Problem 3.30 Page 177

Adjacent channel interference
G  f   power spectral density of input
H  f   Frequency response of channel filter
 f  channel separation

ACI  f  
 G f  H  f  f 


 G f  H  f 
2
2
 df
 df

February 14, 2005
Topic 3
11
Solution

Power of signal in correct filter

PCC 
 G f  H  f 
2
 df


Power of signal in adjacent channel

PAC 
 G f  H  f  f 
2
 df

February 14, 2005
Topic 3
12
Problem 3.35 p. 178 (a)
Formulas in table 3.4 page 159
 Begin BPSK with

Pe     erfc

 
Eb
 ,  
N0
2
Integrate over Rayleigh distribution
  
p      exp    , 0    
0
 0 
1
February 14, 2005
Topic 3
13
Evaluate the Integral

Average BER is

P   0    Pe     p     d 
0
0
1 
  1 
2 
1  0


1
 
 4 0
Evaluation of integral is left as ETR
February 14, 2005
Topic 3
14
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. 159-160
 SNR requirement is about 8.3 dB

Rayleigh fading
 Use
Table 3.4 p. 159
 Solve for SNR of about 34 dB
February 14, 2005
Topic 3
15
Coding Follows Sampling

Sampling
 Simply
converts base signal to elementary
modulation form
 Formatting for performance is left to coding

Coding
 Removal
of redundancy == source coding
 Channel coding == error detection and
correction capability added
February 14, 2005
Topic 3
16
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
CBits / Sec
February 14, 2005

PT 
 BW  log 2 1 

 PNoise 
Topic 3
17
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 channelcoding theorem
February 14, 2005
Topic 3
18
Concept of Entropy
Definition – Average information content
per symbol
 Importance

 Fundamental
limit on average number of bits
per source symbol
 Channel-coding theorem is stated in terms of
entropy
February 14, 2005
Topic 3
19
Equation for Entropy
S  source alphabet set
L  average number of bits per symbol
K  number of symbols in alphabet
pk  probability of symbol k in message
 1 
H  S    pk  log 2   (entropy)
k 0
 pk 
H S 

(coding efficiency)
L
K 1
February 14, 2005
Topic 3
20
Study Problems and Reading
Assignments

Reading assignments
 Read
Section 4.6, Cyclic Redundancy Checks
 Read Section 4.7, Error-Control Coding

Study examples
 Example
4.1 page 197
 Problem 4.1 page 197
February 14, 2005
Topic 3
21
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
r1 

   d1  d 2 
d1  d 2
d1 = d2 = 1.5 km

 0.075m   1500m 
3000m
Raise both towers
February 14, 2005
Topic 3
2
 7.5m  5m
22
Problem 2.5 p. 29
Similar to 2.4 but LOS is clearly obstructed
 Fresnel-Kirchoff diffraction parameter eq.
(2.39) is

2   d1  d 2 
  h
 3.465
  d1  d 2
Diffraction loss is 24 dB
 For 400 MHz, v = 1.096, loss = 16 dB

February 14, 2005
Topic 3
23
Term Projects

Areas for coverage
 Propagation
and noise
Free space
 Urban

 Modulation
& FDMA
 Coding
 Demodulation

and detection
Will deploy over Blackboard this week
February 14, 2005
Topic 3
24
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
 Due date TBD
February 14, 2005
topics are presented
Topic 3
25
EE320 Digital
Telecommunications
Quiz 1 Report
February 21, 2005
February 14, 2005
Topic 3
26
The Curve
100
90
80
70
60
50
40
30
20
10
0
1
2
3
February 14, 2005
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
Topic 3
27
The Answers

See previous lectures/slides
 Questions

See Excel spreadsheets
 Questions

1, 4
2, 3, 5
See Mathcad spreadsheet
 Fresnel
February 14, 2005
integrals for diffraction loss
Topic 3
28
Question 2
Name
Wavelength Ae, Ex. 2.2 p. 16
ABREFA-KODOM
APEHAYA
BARAKAT
BIRCH
BOADO
CARDONE
DO
GEDZAH
MADJAR
P H NGUYEN
T T NGUYEN
T H NGUYEN
PANG
PATEL
ROIDAD
SCHOLL
STRAKER
TANUI
TRAN
ZAYZAY
0.06
0.06122449
0.0625
0.063829787
0.065217391
0.066666667
0.068181818
0.069767442
0.071428571
0.073170732
0.075
0.076923077
0.078947368
0.081081081
0.083333333
0.085714286
0.088235294
0.090909091
0.09375
0.096774194
February 14, 2005
3.534291735
3.773838175
4.021238597
4.276493
4.539601384
4.810563751
5.089380099
5.376050428
5.67057474
5.972953033
6.283185307
6.601271563
6.927211801
7.261006021
7.602654222
7.952156404
8.309512569
8.674722715
9.047786842
9.428704952
Topic 3
Gain, Eq.
(2.9) p. 16
Gain, dB
12337.0055
12651.5224
12936.2879
13190.1835
13412.2221
13601.5486
13757.439
13879.3012
13966.6746
14019.2302
14036.7707
14019.2302
13966.6746
13879.3012
13757.439
13601.5486
13412.2221
13190.1835
12936.2879
12651.5224
40.91209758
41.02142788
41.11809672
41.20250836
41.27500738
41.33588357
41.38537595
41.423676
41.45093014
41.46724168
41.47267206
41.46724168
41.45093014
41.423676
41.38537595
41.33588357
41.27500738
41.20250836
41.11809672
41.02142788
29
Question 3
Name
ABREFA-KODOM
APEHAYA
BARAKAT
BIRCH
BOADO
CARDONE
DO
GEDZAH
MADJAR
P H NGUYEN
T T NGUYEN
T H NGUYEN
PANG
PATEL
ROIDAD
SCHOLL
STRAKER
TANUI
TRAN
ZAYZAY
February 14, 2005
i
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Wavel 1
Wavel 2
0.06
0.059761
0.059524
0.059289
0.059055
0.058824
0.058594
0.058366
0.05814
0.057915
0.057692
0.057471
0.057252
0.057034
0.056818
0.056604
0.056391
0.05618
0.05597
0.055762
0.6
0.59761
0.595238
0.592885
0.590551
0.588235
0.585938
0.583658
0.581395
0.579151
0.576923
0.574713
0.572519
0.570342
0.568182
0.566038
0.56391
0.561798
0.559701
0.557621
Part I Eq.
(2.36) p. 27
7.745966692
7.73052108
7.715167498
7.699905035
7.684732794
7.669649888
7.654655446
7.639748605
7.624928517
7.610194341
7.595545253
7.580980436
7.566499085
7.552100405
7.537783614
7.523547939
7.509392615
7.49531689
7.481320021
7.467401274
h
Nu or v, eq.
(2.39) p. 28
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
0.547722558
0.548816909
0.549909083
0.550999093
0.55208695
0.553172667
0.554256258
0.555337735
0.55641711
0.557494395
0.558569602
0.559642743
0.560713831
0.561782876
0.562849891
0.563914887
0.564977876
0.566038868
0.567097875
0.568154908
Topic 3
Loss, Fig. Nu or v,
2.10 p. 29 2nd freq
-10.639
-10.64753
-10.65605
-10.66458
-10.67311
-10.68163
-10.69016
-10.69868
-10.70721
-10.71574
-10.72426
-10.73279
-10.74132
-10.74984
-10.75837
-10.76689
-10.77542
-10.78395
-10.79247
-10.801
0.173205
0.173551
0.173897
0.174241
0.174585
0.174929
0.175271
0.175613
0.175955
0.176295
0.176635
0.176975
0.177313
0.177651
0.177989
0.178326
0.178662
0.178997
0.179332
0.179666
2nd Loss
-7.501
-7.503895
-7.506789
-7.509684
-7.512579
-7.515474
-7.518368
-7.521263
-7.524158
-7.527053
-7.529947
-7.532842
-7.535737
-7.538632
-7.541526
-7.544421
-7.547316
-7.550211
-7.553105
-7.556
30
Question 5
Name
ABREFA-KODOM
APEHAYA
BARAKAT
BIRCH
BOADO
CARDONE
DO
GEDZAH
MADJAR
P H NGUYEN
T T NGUYEN
T J NGUYEN
PANG
PATEL
ROIDAD
SCHOLL
STRAKER
TANUI
TRAN
ZAYZAY
February 14, 2005
BER
0.000316
0.000251
0.0002
0.000158
0.000126
0.0001
7.94E-05
6.31E-05
5.01E-05
3.98E-05
3.16E-05
2.51E-05
2E-05
1.58E-05
1.26E-05
0.00001
7.94E-06
6.31E-06
5.01E-06
3.98E-06
Qinv(BER) Eb/N0, dB
3.41734328
3.47953449
3.54076432
3.60107467
3.66050453
3.71909027
3.77686584
3.83386302
3.89011158
3.94563947
4.00047298
4.05463684
4.10815439
4.16104764
4.2133374
4.26504337
4.31618421
4.36677762
4.41684041
4.46638855
7.663472171
7.820122959
7.971640442
8.118342568
8.26051902
8.398434441
8.532331217
8.66243188
8.788941206
8.912048035
9.031926873
9.148739293
9.262635176
9.373753798
9.482224805
9.588169073
9.691699474
9.792921563
9.891934194
9.988830071
Topic 3
Rayleigh Rayleigh
SNR
SNR, dB
789.8195
994.518
1252.218
1576.643
1985.071
2499.25
3146.564
3961.483
4987.406
6278.966
7904.944
9951.929
12528.93
15773.18
19857.46
24999.25
31472.39
39621.58
49880.81
62796.41
28.97528
29.97613
30.9768
31.97733
32.97776
33.9781
34.97837
35.97858
36.97875
37.97888
38.97899
39.97907
40.97914
41.97919
42.97924
43.97927
44.9793
45.97932
46.97933
47.97935
31
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
February 14, 2005
Topic 3
32
The Quiz in the Text (2 of 2)

Question 4
 Problem
3.30 p. 177
 Given in class
 Answer was to give equation that was given in
the problem statement

Question 5
 Problem
3.36, given in class
 Use table 3.4, figure 3.33, pp. 159-161
February 14, 2005
Topic 3
33
EE320 Convolutional
Codes
James K Beard, Ph.D.
February 14, 2005
Topic 3
34
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
February 14, 2005
Topic 3
35
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
February 14, 2005
Topic 3
36
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
February 14, 2005
Topic 3
37
Base Concept –Modulo 2
Convolutions

Scenario
 Bitstream
into convolution filter
 Filter weights are ones and zeros
 Output is taken modulo 2 – i.e. a 1 or 0

Result: A modulo 2 convolution converts
one bit stream into another
February 14, 2005
Topic 3
38
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
February 14, 2005
Topic 3
39
Error-Control Coding

Two categories of channel coding
 Forward
error-correction (EDAC)
 Automatic-repeat request (handshake)

CRC Codes
 Hash
codes of the message
 Error detection, but not correction
February 14, 2005
Topic 3
40
Topics in Convolutional Codes


The node diagram is a block diagram
Polynomial representations



Trellis diagrams



Represent signals, convolutions and special polynomials and
polynomial operations
Give us a simple way to understand and analyze convolutions
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
February 14, 2005
Topic 3
41
Convolutional Code Steps
Reduce the message to a bit stream
 Operate using modulo-2 convolutions

 Convolution
filter with short binary mask
 Take result modulo 2

Implemented with one-bit shift registers
with multiplexer (see Figure 4.6 p. 196)
February 14, 2005
Topic 3
42
Example 4.1 Page 197
Path 2

Output
Input
1/z
1/z

Path 1
Haykin & Moher
Figure 4.6 p. 196
February 14, 2005
Topic 3
43
Example 4.1 (1 of 4)

Response of Path 1
g 1  D   1  D 2

Response of Path 2
g

 2
 D   1  D  D2
Mutiplex the outputs bit by bit
 One
side output, then the other
 Produce a longer bit stream
February 14, 2005
Topic 3
44
Example 4.1 (2 of 4)

Signal
 Message
bit stream (10011)
 Message as a polynomial
m  D  1 D  D
3

4
Multiply the message polynomial by the
Path 1 and Path 2 filter polynomials
 Obtain
two bit streams from resulting
polynomials
 Multiplex (interleave) the results
February 14, 2005
Topic 3
45
Example 4.1 (3 of 4)

Polynomial multiplication results
c 1  D   1  D 2 1  D 3  D 4 
 1  D 2  D3  D 4  D5  D 6
c
2
 D   1  D  D 2 1  D 3  D 4 
 1  D  D 2  D3  D 6

Messages
 Path
1 (1011111)
 Path 2 (1111001)
 Multiplexing them (11, 10, 11, 11, 01, 01, 11)
February 14, 2005
Topic 3
46
Example 4.1 (4 of 4)

Length of coded message is
 Twice
the order of the product polynomials +1
 2.(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 zeros
 Number of clocks is number of stages
 Zeros between message words are tail zeros
February 14, 2005
Topic 3
47
Problem 4.1

Signal and polynomial 1
c1  1,0,0,1,1  1,0,1  1,0,1,1,1,1,1

Signal and polynomial 2
c

 2
 1, 0, 0,1,1  1,1,1  1,1,1,1, 0, 0,1
Result
c  11,10,11,11,01,01,11
February 14, 2005
Topic 3
48
Modulo 2 Convolution
Diagrams
February 14, 2005
Poly
Signal
Resullt: 1
1 0 1
1 0 0 1 1
1
Poly
Signal
Resullt: 1
1 1 1
1 0 0 1 1
1
Poly
Signal
Resullt: 0
1 0 1
1 0 0 1 1
Poly
Signal
Resullt: 1
1 1 1
1 0 0 1 1
1
Poly
Signal
Resullt: 1
1 0 1
1 0 0 1 1
1
Poly
Signal
Resullt: 1
1 1 1
1 0 0 1 1
1
Poly
Signal
Resullt: 1
1 0 1
1 0 0 1 1
1
Poly
Signal
Resullt: 1
1 1 1
1 0 0 1 1
1
Poly
Signal
Resullt: 1
1 0 1
1 0 0 1 1
1
Poly
Signal
Resullt: 0
1 1 1
1 0 0 1 1
1 1
Poly
Signal
Resullt: 1
1 0 1
1 0 0 1 1
1
Poly
Signal
Resullt: 0
1 1 1
1 0 0 1 1
1 1
Poly
Signal
Resullt: 1
1 0 1
1 0 0 1 1
1
Poly
Signal
Resullt: 1
1 1 1
1 0 0 1 1
1
Topic 3
49
Trellis and State Diagrams

Trellis diagram Figure 4.7 p. 198, and
state table 4.2 p. 199
 Horizontal
position of node represents time
 Top line represents the input
 Each 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 solid
 Paths produced by 1’s are dotted
February 14, 2005
Topic 3
50
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
February 14, 2005
Topic 3
51
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
February 14, 2005
Topic 3
52
Drawing the Trellis Diagram
State Table


Begin with a state table
For each paths “state”
 For


Draw the solid path for a 0
Draw the dotted path for a 1
 For



the last bit 0
the last bit 1
Draw the solid path for a 0
Draw the dotted path for a 1
State
0 0 0
1 0 1
0 1 0
1 1 1
0 1 1
1 1 0
0 0 1
1 0 0
Paths 1, 2
0 0
0 0
0 1
0 1
1 0
1 0
1 1
1 1
This is 16 lines total
February 14, 2005
Topic 3
53
For Output State [0,0]

For the last bit 0

Adding a zero, solid path label is (0,0)



Adding a one, dotted path label is (0,1)



From {0,0,0} to {0,0,0}
Next state is [0,0]
From {0,0,0} to {1,0,0}
Next state is [1,1]
For the last bit 1

Adding a zero, solid path label is (1,0)



From {1,0,1} to {0,1,0}
Next state is [1,0]
Adding a one, dotted path label is (1,1)

From {1,0,1} to {1,1,0}
Next state is [0,1]

February 14, 2005
Topic 3
54
For Output State [0,1]

For the last bit 0

Adding a zero, solid path label is (0,0)



Adding a one, dotted path label is (0,1)



From {0,1,1} to {0,0,1}
Next state is [1,1]
From {0,1,1} to {1,0,1}
Next state is [0,0]
For the last bit 1

Adding a zero, solid path label is (1,0)



From {1,1,0} to {0,1,1}
Next state is [0,1]
Adding a one, dotted path label is (1,1)

From {1,1,0} to {1,1,1}
Next state is [1,0]

February 14, 2005
Topic 3
55
For Output State [1,0]

For the last bit 0

Adding a zero, solid path label is (0,0)



Adding a one, dotted path label is (0,1)



From {0,1,0} to {0,0,1}
Next state is [1,1]
From {0,1,0} to {1,0,1}
Next state is [0,0]
For the last bit 1

Adding a zero, solid path label is (1,0)



From {1,1,1} to {0,1,1}
Next state is [0,1]
Adding a one, dotted path label is (1,1)

From {1,1,1} to {1,1,1}
Next state is [1,0]

February 14, 2005
Topic 3
56
For Output State [1,1]

For the last bit 0

Adding a zero, solid path label is (0,0)



Adding a one, dotted path label is (0,1)



From {0,0,1} to {0,0,0}
Next state is [0,0]
From {0,0,1} to {1,0,0}
Next state is [1,1]
For the last bit 1

Adding a zero, solid path label is (1,0)



From {1,0,0} to {0,1,0}
Next state is [1,0]
Adding a one, dotted path label is (1,1)

From {1,0,0} to {1,1,0}
Next state is [0,1]

February 14, 2005
Topic 3
57
Drawn Trellis Diagram
0,0
 0, 0 
1,1
0,1
1,0
1,0
 0,1
1,0
1,1
 0,1
 0, 0 
1,1
February 14, 2005
 0,1
1,0
 0, 0 
1,1
1,1
 0, 0 
1,0
 0,1
Topic 3
58
State Diagram
1,1
 0, 0 
[00]
 0, 0 
1,1
1,0
[10]
February 14, 2005
 0, 0 
1,1
 0,1
1,1
[01]
 0,1
1,0
 0,1
1,0
 0, 0 
1,0
Topic 3
[11]
 0,1
59
EE320 Decoding
Convolutional Codes
James K Beard, Ph.D.
February 14, 2005
Topic 3
60
Where We are Going

Exploit the Channel Coding Theorem
 For
any required channel bit rate CR less than
the channel capacity C
CR  C  B  log2 1  SNR 
 A coding
exists that achieves an arbitrarily low
BER

Method is error-correcting codes
February 14, 2005
Topic 3
61
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
February 14, 2005
Topic 3
62
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
February 14, 2005
Topic 3
63
Haykin & Moher Table 4.3
Page 201
Maximum Free Distance Attainable for Rate 1/2
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
February 14, 2005
Topic 3
64
Fundamental of Maximum
Likelihood: Multivariate PDF

Consider N Gaussian random variables
with mean zero and variance one,
 zi2 
1
1
 1 T 
p  zi  
 exp    , pz  z  
 exp    z  z 
N /2
2
 2

 2 
 2

The covariance of z is the identity matrix I
zz
February 14, 2005
T
  zi  z j    i , j   I
Topic 3
65
Identities and Definitions
Determinant of product and inverse
1
1
A B  A  B , A 
A
 Differential

d x  dx1  dx2 

Gradient
 yi 


 x  x j 
y
February 14, 2005
Topic 3
66
Variable Change to Correlated
Variables
Consider the variable change
x  A z
 The pdf of x is found from the differential
and the Jacobian determinant

z
1
pz  z   d z  pz  A  x  
d x  A  px  x   d x
x
1

The covariance R of x is
R  x x
T
February 14, 2005
 A  z  z  AT  A  AT
T
Topic 3
67
PDF of Correlated Variables

The pdf of x is
1
px  x    pz  z 
A
 1 T 1 

 exp    x  R  x 
1/ 2
N /2
 2

 2   R
1
February 14, 2005
Topic 3
68
With a Mean…

The pdf of a Gaussian vector x of N
elements with covariance R and mean a is
T
 1

1
exp     x  a   R   x  a  
2


px  x  
1/ 2
N /2
 2   R
February 14, 2005
Topic 3
69
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
February 14, 2005
Topic 3
70
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
February 14, 2005
Topic 3
71
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
February 14, 2005
Topic 3
72
The MLE for c


Data
y cn
Log likelihood function
  
ln p y | c



1
  N  ln  2   ln  R   y  c
2

T

 R 1  y  c

Solution is Nearest Neighbor
ĉ  y
February 14, 2005
Topic 3
73
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

February 14, 2005
Topic 3
74
Curve for Backup Quiz
100
90
80
70
60
50
40
30
20
10
0
1
2
February 14, 2005
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
Topic 3
75
Assignment
Read 5.2, 5.3, 5.5
 Look at Problem 4.5 p. 243
 Do problems 4.6, 4.7 p. 252

February 14, 2005
Topic 3
76
Interleaving and TDMA
The Viterbi Method
 Interleaving

 Scatters
the code data stream
 Makes low BER communication more robust
through fading medium and interference
Noise performance
 TDMA

February 14, 2005
Topic 3
77
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. 268278, 1973.

February 14, 2005
Topic 3
78
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
February 14, 2005
Topic 3
79
Interleaving

Based on the coherence time of a fading
channel; fading due to motion leads to
TCOHERENCE

Maximum range rate
0.3

, fD 
2  fD

Interleaving operations
 Block
data over times much larger than TCOHERENCE
 Reformat in smaller blocks
 Multiplex the smaller blocks into an interleaved data
stream
February 14, 2005
Topic 3
80
Interleaver Parameters

Object of interleaving
 Data
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
February 14, 2005
Topic 3
81
Interleaver Methods

Methods
 Simple
sequential interleaving as just
described
 Pseudorandom interleaving

Combine with codes
 Use
Viterbi interleaver
 Use multiplexer of convolutional codes as
interleaver
February 14, 2005
Topic 3
82
Noise performance
Compare AWGN channels using Figure
4.13 p. 213
 Rayleigh fading performance given in
Figure 4.14 p. 214

February 14, 2005
Topic 3
83
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
February 14, 2005
Topic 3
84
Next Time
TDMA
 Chapter 4 examples
 Quiz postponed one week to March 30

 Will
cover Chapter 4
 Some topics in CDMA
February 14, 2005
Topic 3
85
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
February 14, 2005
Topic 3
86
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
February 14, 2005
Topic 3
87
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 frequencydependent fading in medium and broad
band TDMA/FDMA

February 14, 2005
Topic 3
88
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
February 14, 2005
Topic 3
89
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
February 14, 2005
Topic 3
90
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
February 14, 2005
Topic 3
91
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
February 14, 2005
Topic 3
92
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
February 14, 2005
Topic 3
93
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
February 14, 2005
Topic 3
94
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 code
 This makes the code systematic
February 14, 2005
Topic 3
95
Solution for Problem 4.7

Message polynomial is
m( x )  1  x 2  x 3  x 4 

Generator polynomials are
g ( x )  1 x  x  x
(1)
2
3
g (2) ( x )  1  x  x 3
February 14, 2005
Topic 3
96
Solution for Problem 4.7
(concluded)

Code polynomials are
c (1)  x   g (1)  x   m  x   1  x  x 3  x 4
c (2)  x   g (2)  x   m  x   1  x  x 2  x 3  x 5  x 6 

Code bits are
(1)
c
  x   1,1,0,1,1,0,0,0,
(2)
c
  x   1,1,1,1,0,1,1,1,

Code output is
11,11,01,11,10,01,01,
February 14, 2005
Topic 3
97
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
…
February 14, 2005
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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
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Simulation Output
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
February 14, 2005
(
(
(
(
(
(
(
(
(
(
1,
1,
0,
1,
1,
0,
0,
0,
0,
0,
1)
1)
1)
1)
0)
1)
1)
1)
1)
1)
A few minutes with Fortran 95
Your choice of language will do.
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Interleaving and Coherence
Time
Problem 4.14 page 254
 Coherence time

TCOHERENCE
0.3
0.15  


2  fD
vehicle speed
BitLossBlockLength  TCOHERENCE  BitRate
0.15    BitRate

vehicle speed
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Discussion Questions

One-Channel TDMA
 What
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 nonsystematic codes?
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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 spectrum
 Gold codes
 Galois fields for binary sequences
February 14, 2005
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