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
1
April 13, 2005
Topic 4
27-Apr
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13-Apr
6-Apr
30-Mar
23-Mar
16-Mar
9-Mar
2-Mar
23-Feb
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9-Feb
2-Feb
26-Jan
19-Jan
Attendance
25
20
15
10
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

April 13, 2005
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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 
April 13, 2005
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Base Performance Equations
SignalPower
Eb 
BitRate
NoisePower  N0  Bandwidth
Eb SignalPower Bandwidth


N0 NoisePower
BitRate
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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

April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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

April 13, 2005
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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
April 13, 2005
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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

April 13, 2005
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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

April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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

April 13, 2005
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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
April 13, 2005
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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)
April 13, 2005
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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

April 13, 2005
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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

Topic 4
28
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 
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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?
April 13, 2005
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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 

April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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
April 13, 2005
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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

April 13, 2005
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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

Topic 4
45
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
Topic 4
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


April 13, 2005
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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

April 13, 2005
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50
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
April 13, 2005
51
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.
April 13, 2005
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52
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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53
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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54
April 13, 2005
Topic 4
55
EE320
Telecommunications
Engineering
James K Beard, Ph.D.
[email protected]
E&A 349
April 13, 2005
Topic 4
56
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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57
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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58
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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59
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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60
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
Topic 4
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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 

Topic 4
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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
April 13, 2005
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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
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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
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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 

Topic 4
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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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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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