Telecommunications Engineering Topic 2: Modulation and FDMA
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Transcript Telecommunications Engineering Topic 2: Modulation and FDMA
Telecommunications
Engineering
Topic 4: Spread
Spectrum and CDMA
James K Beard, Ph.D.
[email protected]
http://astro.temple.edu/~jkbeard/
April 13, 2005
Topic 4
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April 13, 2005
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Attendance
25
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5
0
2
Essentials
Text: Simon Haykin and Michael Moher, Modern
Wireless Communications
SystemView
Web Site
Use the full version in E&A 603A for your term project
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
April 13, 2005
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Topics
Today we explore the third tool
FDMA,
uses separate channels for each user
TDMA, uses time multiplexing to time
multiplex the channel between users
Now, CDMA with spread spectrum enables
multiple simultaneous users of the channel
Direct-sequence modulation
Spreading codes
Code synchronization
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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
s1 t s2 t
S1 S2 d
2
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Base Performance Equations
SignalPower
Eb
BitRate
NoisePower N0 Bandwidth
Eb SignalPower Bandwidth
N0 NoisePower
BitRate
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Performance in Noise
Base equation (Hayken & Moher
equations (5.12) page 263, (E.11) page
518)
Eb
2 Eb
Pe 0.5 erfc
Q
N
N
0
0
Adding spreading function -- Eb and N0 are
invariant through matched filter
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Performance in Interference
Consider a tone as interference
In base coded signal
Matched
filter spreads tone over channel
Tone energy becomes part of noise floor
In spread spectrum signal
Matched
filter spreads tone over channel
Effective additional noise reduced by
spreading factor
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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
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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
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The Symbol-Shaping Function
Q
g k t ck t g c t q Tc
q 1
Q number of pulses in spreading code
T
Tc
Q
ck t spreading code k, one of orthogonal set
g c t puse of width Tc
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Walsh-Hadamard Sequences
A simple way to formulate orthogonal code
sequences
Based on recursive augmentation of
Walsh-Hadamard matrices
1 1
H1
1 1
H i H i
Hi 1
H
H
i
i
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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 properties
Excellent
for zero lag
Poor for other lags
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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
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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
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Galois Field Vector Extensions
of Order 2m
Polynomials modulo 2 of order m-1
Arithmetic is done modulo a generating
polynomial of the form
gg x x m 1 other powers of x
Proper selection of generating polynomial
Sequence
of powers produces all 2m
elements
Set is closed on multiplication
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An Important Isomorphism
Shift registers with feedback
Bits
in shift register are isomorphic with
polynomial coefficients
Shift is isomorphic with multiplication by x
Modulo 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
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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 msequence generators
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Code Synchronization
Two phases
Recover
timing
Recover phase
Timing must be recovered first
To recover timing
Use
code bits known to be 1’s
Matched filter for symbol-shaping function
Step timing in increments of Tc until match is
found
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Assignment
Read 5.2, 5.3, 5.5, 5.7, 5.11, 5.15
Do problem 5.7 p. 273
Next time
Power
control
Frequency hopping
An example
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Chinese Remainder Theorem
Over numbers from 0 to 2.3.5=30
p
N
x mod x, pi
pi
i 1
The method works when N has no
repeated prime factors
Arithmetic advantages?
April 13, 2005
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A Finite Field
Integers mod a prime
A reciprocal of a positive integer always
exists
Addition, subtraction, multiplication,
division, all defined and commutative
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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 closelyspaced handsets or near base stations
Solution is power control
Reduce
handset power to make received power
constant
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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)
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EE320 March 28
April 13, 2005
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Topics
Term Project
Problem 5.1 p. 262
Problem 5.17 p. 299
Theme Example: WCDMA
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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
April 13, 2005
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Problem 5.1 p. 262
What is the equation for the spectrum of
the spreading sequence given by Eq. (5.5)
p. 261? Q
g t c q gc t q Tc
q 1
The chips c(q) are +1 or -1 and the chip
shape gc(t) is
April 13, 2005
1
, 0 t Tc
gc t T
0, otherwise
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Use the Convolution Theorem
The spreading sequence is
Q
g t c q t q Tc gc t
q 1
The Fourier transform of each term in the
sum is
1
1 T
2
Gq f Gc f sinc f Tc
Q
Q Q
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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
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WCDMA (1 of 3)
From Theme Example 4 pp.323-328
Cell phone technology generations
First:
analog cell phones
Second: TDMA, IS-95, GSM
Third: Universal Mobile Terrestrial
Telecommunications systems (UMTS)
WCDMA is a UMTS
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WCDMA (2 of 3)
Functional differences
Simultaneous
voice and data transmission
Other data such as real-time TV
Performance improvements
Three
times the bandwidth
Four times the maximum spreading factor
Optional turbo codes
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WCDMA (3 of 3)
Other differences
Multiple
simultaneous CDMA downlink
Downlink power control
Asynchronous base stations
Bottom line
Broadband
or ISDN in a cell phone
Near-far problems mitigated
Higher density of base stations and users
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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?
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Problem 5.19 page 309 (2 of 3)
The spectral efficiency for
K
users in the cell
Each transmitting 2/Q bits/second/Hz
2K
Q
From Eq. (5.85) page 304
N0
Q 1 f SINR K 1
I0
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Problem 5.19 page 309 (3 of 3)
Rolling up these two equations gives nu as
2
N0
1 f SINR 1
I0
The spreading factor Q influences
The
interference factor f
The interference to noise ratio I0/N0
April 13, 2005
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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
April 13, 2005
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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
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Channel Protocol of IS-95
Making an IS-95 call – the Mobile Terminal
Searches for Pilot
Locks to the Sync
channel and synchronizes with it
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
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Channel Protocol of IS-95
Receiving an IS-95 call – the Base Station
Transmits
a short message on the paging
channel
Accepts Mobile Terminal request for call
Differences
Request
for call has the phone number to
initiate a call
Paging channel has Mobile Terminal phone
number in the paging message
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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
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The Four Downlink Channels
Separated by use of Walsh-Hadamard
codes of length 64
Pilot
used Walsh #0
Sync uses Walsh #32
Paging using Walsh #1
Traffic uses one of the other codes
See Figure 5.29 page 314
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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
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Problem 5.2 Page 263
Filtering with an integrate-anddump 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-anddump is 1/T.
April 13, 2005
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Parseval’s Theorem
For Fourier transform pair see Table A.2 p. 482
H f
h t exp j 2 f t dt
t
h t rect
T
, H f T sinc f T
For Parseval’s theorem see Eq. (A.36) p. 491
h t
April 13, 2005
2
dt
H f
2
df
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Noise Bandwidth
Definition: ratio of
The
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
BN
April 13, 2005
H f
2
H 0
df
2
T
1
2
T
T
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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
K
xc t k sk t w t
k 1
k propagation loss on path k
s t bk Eb g k t signal k
w t receiver noise
bk g k t data sequence k X spreading code k
April 13, 2005
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SINR of First User
More detail in 5.4.1 pages 279-283
T
K
0
k 1
y x t g1* t dt 1 b1 Eb 1 Eb k k R1k
y 1 b1 Eb
K
y y N0 Eb R1k
k 2
2
2
y
y
SINR
April 13, 2005
2
y
2
2
k
12 Eb
K
1
N0 Eb k2
Q
k 2
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2
K
1
N0 Eb k2
Q
k 2
Eb
Dg
N0
2
1
48
Degradation in Multi-User
Performance
1
Dg
2
2
K
k
K 1 1 Eb 1
1
K 1 k 2 1
Q
N0
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FEC Coding and CDMA
Haykin & Moher Section 5.8 pp. 297-299
Direct Sequence Spread Spectrum (DSSS) spreads spectrum without added
redundancy
Use of FEC spreads spectrum and adds
redundancy
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Spreading Rate and
Degradation
The maximum spreading rate is
1
Q QDS , r FEC code rate
r
Degradation in multi-user
performance is 1
1
K 1 Es
K 1 Eb / r
Dg 1
1
Q / r N0
QDS N0
1
K 1 Eb
1
Q N0
Topic 4
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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 rate1/2 convolutional code. Compare the
degradation Db with and without FEC
coding at a BER of 10-5 when there are
seven interfering users.
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Base Parameters Without FEC
Encoding
Q
1
10
5
2 Eb
4.265043367=
N0
Eb
4.265043367
10 log
20 log
2
N0
9.59846903
8 1 4.265043367 2
Dg 1
32
2
0.33449332
April 13, 2005
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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
April 13, 2005
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April 13, 2005
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EE320
Telecommunications
Engineering
James K Beard, Ph.D.
[email protected]
E&A 349
April 13, 2005
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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
April 13, 2005
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Crunch Time
We have about four weeks left
Last
day of class is Monday May 2
Final exam is Monday May 11, 11:00 AM -1:00
PM
Some of you are in trouble
Some
quiz grades are low
Not everyone will pass
Department has been notified
April 13, 2005
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Watch for a Warning
If you are heading toward a grade lower
than C
You
will receive a warning with your Quiz 2
grade
The cover page of your quiz
Your minimum Final Exam grade will be given
Final Exam
By
the book
Two hours, no talking
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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
S g f A sinc
April 13, 2005
2
f f T
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Problem 5.3 page 265
Jammer spectral density at baseband is
S f A , f
bandwidth
The noise bandwidth of an integrate-and-dump
is 1/T (see problem 5.2)
The noise variance is
2
S f S f df
p
p
1
A
T
April 13, 2005
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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
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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 Q 1
R jj k c q c q k mod Q
Q q 0
1, k 0
1
,k 0
Q
April 13, 2005
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Proof of Autocorrelation
Property of m-Sequences
By the Shift property, the circular
autocorrelation, a modulo-2 sum of an msequence and a circularly-shifted version
of itself, is another circularly-shifted
version of iteslf
From the Balance property an msequence has one more 1 than zeros.
QED
April 13, 2005
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Course Material Overview
The problem
Make
a cell phone system work
Deal with mobile terminals
Deal with urban fading
The solution
Construct
the network layer infrastructure
Dispense the data link layer mobile terminals
Exploit the physical layer successfully
April 13, 2005
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Designing the Data Link Layer
The problem
Addressing
multiple users
Data push – making calls
Data pull – accepting calls
SINR
Dealing
with urban fading
The solution
and other 2nd generation standards
Third generation standards
IS-95
April 13, 2005
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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)
April 13, 2005
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Meeting the Challenge of
Fading
Forward Error Correction Codes (FECs)
Allows
Spread spectrum
Allows
robust operation with high bit error rates (BER)
higher BER in frequency-selective fading
Interleaving
Helps
bridge dropouts from fading and interference
over intervals of a few milliseconds
April 13, 2005
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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.
April 13, 2005
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Problem 5.8 Page 274
To show that scramblers based on msequences 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
April 13, 2005
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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 msequence 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
April 13, 2005
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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
April 13, 2005
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RAKE Signal Flow
Each channel
Multiplied
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
April 13, 2005
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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 preceding
April 13, 2005
K 1 Eb
Dg 1
Q N
0
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1
76
Example 5.5 (Continued)
The BER of 10-5 indicates single-user
Eb Es
9.12 10 dB
N0 N0
The degradation factor is
1
8 1
Dg 1
9.12 0.33 -4.8 dB
32
April 13, 2005
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Example 5.5 (Continued)
With rate ½ constraint length 7FEC encoding
Spreading
factor Qs=16, total Q=32
Single-user
Eb Es
2.82 4.5 dB
N0 N0
Degradation factor is
1
8 1
Dg 1
2.82 0.62 -2.1 dB
32
April 13, 2005
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Example 5.5 (Concluded)
Degradation improves 2.7 dB with FEC
Degradation vs. loading for rate ½ codes
shown in Figure 5.23 page 298
No
improvement for single user
Gains of about 2 dB for K near Q
Improvement doesn’t vary much with BER
Conclusion: BER is important powerbandwidth tradeoff with multiple users
April 13, 2005
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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
Use
a known pilot signal, as with IS-95
training sequences (standard messages) for
channel tracking, as with WCDMA
April 13, 2005
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EE320
Telecommunication
Engineering
Wireless Architectures
April 13, 2005
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81
Open System Interconnection
(OSI) Model
Seven-layer model
Physical
layer (modem)
Data link layer
Network layer
Transport layer (packetizing, ACK/NAK)
Session layer (Service selection and access)
Presentation layer (encryption, compression)
Application layer (HMI)
Layers designed together as a system
April 13, 2005
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Power Control Architectures
Open Loop
Closed Loop
Mobile terminals measure strength of pilot channel
Transmit power decreased for strong pilot channels
Fast and simple, but must be approximate
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
April 13, 2005
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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
April 13, 2005
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Next Time
Assignment: Read parts of Chapter 7
7.3,
OSI
7.6, Power Control
7.7, Handover
7.8, Network Layer
April 13, 2005
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