chapter 8: transmission baseband
TRANSCRIPT
Chapter 8
: Base
band D
igita
l Transm
ission
6/8/20
101
Intro
ductio
n
Tran
smissio
n of d
igital sig
nal
Over a b
aseban
d ch
annel (C
hap
ter 8)
local co
mmunicatio
ns
Over a b
and‐pass ch
annel u
sing m
odulatio
n (C
hap
ter 9)
netw
ork
Chan
nel‐in
duced
transm
ission im
pairm
ents
Chan
nel n
oise, o
r receiver noise
Interferen
ce: sometim
es treated as n
oise
Intersym
bol in
terference (IS
I)
Digital d
ata has a b
road ban
dwidth w
ith a sig
nifican
t low‐freq
uen
cy conten
t
Man
y chan
nels are b
andwidth lim
ited: d
ispersive, u
nlik
e low‐pass filter
Each
received pulse is affected
by n
eighborin
g pulses
ISI
Majo
r source o
f bit erro
rs in m
any cases
Solutio
ns to
be stu
died in th
is chap
ter
Noise: m
atched filter
maxim
ize the sig
nal n
oise level at th
e receiver
ISI:
Pulse sh
aping
minim
ize the IS
I at the sam
plin
g points
Equalizatio
n
compen
sate the resid
ual d
istortio
n fo
r ISI
6/8/20
102
6/8/20
103
Transm
ission Im
pairm
ent: N
oise
Therm
al noise: g
enerated
by th
e equilib
rium
fluctu
ations o
f the electric cu
rrent in
side th
e receiver circu
it Due to
the ran
dom th
ermal m
otio
n of th
e electrons
Modeled
as an Additive w
hite G
aussian
noise (A
WGN)
Noise sp
ectral den
sity: No = KT(w
atts per h
ertz), where K
is the B
oltzm
ann’s co
nstan
t K = 1.38
0×10−23, an
d T
is the
receiver system noise tem
peratu
re in kelvin
s ([K] = [°C
] +
273.15)
If b
andwidth is B
Hz, th
en th
e noise p
ower
is N = BKT
Alw
ays exists
Other so
urces o
f noise: in
terference
6/8/20
104
Additive N
oise
6/8/20
105
http
://files.amourau
x.web
node.co
m/20
0000047‐3fd
9440d34
/resint_eeg
2.jpg
received sig
nal
Transm
ission Im
pairm
ent: ISI
Line co
des
Map
ping 1’s an
d 0’s to
symbols
Ran
dom process, sin
ce 1’s and 0’s are ran
dom
Power sp
ectrum (S
ection 5.8
)Rep
resentatio
n in th
e frequen
cy domain
The n
ominal b
andwidth of th
e signal is th
e same o
rder o
f mag
nitu
de as 1/T
ban
d is cen
tered aro
und th
e orig
in
Mism
atch betw
een sig
nal b
andwidth B
san
d ch
annel
ban
dwidth B
c
If B
c ≥Bs , n
o problem
If B
c < B
s , the ch
annel is d
ispersive, th
e pulse sh
ape w
ill be
chan
ged an
d th
ere will b
e ISI
6/8/20
106
baseb
and
Power Sp
ectra of Several Lin
e Codes
6/8/20
107
•Freq
uen
cy axis norm
alized w
ith T
b
•Averag
e power is n
orm
alized to unity
Transm
ission Im
pairm
ents d
ue to
Lim
ited Channel B
andwidth
Each
received sym
bol m
ay be w
ider
than th
e transm
itted one, d
ue to
loss o
f high freq
uen
cy componen
ts
Overlap
betw
een ad
jacent sym
bols: IS
I
Lim
it on data rate: u
se guard
time b
etween
adjacen
t symbols
Or n
eed to sh
ape th
e pulses to
cancel IS
I at samplin
g points
6/8/20
108
From D
ata Communica
tions a
nd Netw
orking, B
ehrouz A
. Forouzan
Match
ed Filter –
The P
roblem
Meth
odology:
Cope w
ith th
e two typ
es of im
pairm
ents sep
arately
First assu
me an
ideal ch
annel an
d only co
nsid
er noise
e.g
., low data rate o
ver a short ran
ge cab
le
No problem
of IS
I
Tran
smitted
pulse g
(t)for each
bit is u
naffected
by th
e tran
smissio
n excep
t for th
e additive w
hite n
ose a
t the receiver fro
nt
end
6/8/20
109
Basic p
roblem of d
etecting a p
ulse tran
smitted
over a ch
annel th
at is corru
pted
by ad
ditive w
hite
noise at th
e receiver front en
d
Match
ed Filter
Received
(or, in
put) sig
nal: x(t) =
g(t) +
w(t), 0
≤t≤T
T: an
arbitrary o
bservatio
n in
tervalg(t): rep
resents a b
inary sym
bol 1 o
r 0w(t): w
hite G
aussian
noise p
rocess o
f zero m
ean an
d power
spectru
m den
sity N0 /2
Output sig
nal: y(t) =
x(t) h(t) =
g0 (t) +
n(t)
6/8/20
1010
Detectio
n of R
eceived Sign
al
6/8/20
1011
s1 (t)
tT
0
s0 (t)
tT
0
tT
0
tT
0
1T
T
dtt
s0
1)
(
T
dtt
s0
0)
(
Optim
al detectio
n tim
e
Optim
al detectio
n tim
e
Match
ed Filter (co
ntd.)
Problem
Find h(t)
to m
aximize th
e peak
pulse sig
nal‐to
‐noise
ratio at th
e samplin
g in
stant t=
T:
If G
(f)
g(t), H
(f)
h(t), th
en g
0 (t)
H(f)G
(f). W
e can derive g
0 (t)by in
verse Fourier tran
sform
:
The in
stantan
eous sig
nal p
ower at t=
Tis:
6/8/20
1012
power
noise
output
Average
signaloutput
in thepow
er
ousInstantane
E
)(
2
2
0
(t)
n Tg
df
ftj
fG
fH
tg
)2
exp()
()
()
(0
22
0)
2exp(
)(
)(
)(
df
fTj
fG
fH
Tg
Instan
taneo
us
power
Why sq
uare?
Match
ed Filter (co
ntd.)
The averag
e noise p
ower
The p
ower sp
ectral den
sity of th
e output n
oise n
(t)is
The averag
e noise p
ower is
The p
eak pulse sig
nal‐to
‐noise ratio
is
6/8/20
1013
df
fH
Ndf
fS
tn
N
20
2)
(2
)(
)(
E
20
)(
2)
(f
HN
fS
N
dff
HN
dffT
jf
Gf
H
20
2
)(
2
)2
exp()
()
(
Find H
(f)
h(t) th
at m
axim
izes η
Match
ed Filter (co
ntd.)
Sch
warz’s in
equality
Therefo
re, we h
ave
6/8/20
1014
If an
d
The eq
uality h
olds if an
d only if:
when
Complex
conjugatio
n
Match
ed Filter (co
ntd.)
Excep
t for th
e factor k∙exp
(‐j2πfT), th
e transfer fu
nctio
n of th
e optim
al filter is the sam
e as the co
mplex co
njugate o
f the
spectru
m of th
e input sig
nal
k: scales th
e amplitu
de
exp
(‐j2πfT): tim
e shift
For real sig
nal g
(t), we h
ave G*(f)=
G(‐f): tim
e inversed
The o
ptim
al filter is found by in
verse Fourier tran
sform
Match
ed filter: m
atch
ed to th
e signal
A tim
e‐inversed
and delayed
version of th
e input sig
nal g
(t)
6/8/20
1015
Properties o
f Match
ed Filters
Match
ed filter:
Received
signal:
Noise p
ower:
6/8/20
1016
Properties (co
ntd.)
Maxim
um peak
pulse sig
nal‐to
‐noise ratio
Observatio
ns
Indep
enden
t of g
(t): removed
by th
e match
ed filter
Signal en
ergy (o
r, transm
it power) m
atters
For co
mbatin
g ad
ditive w
hite G
aussian
noise, all sig
nals th
at have th
e same en
ergy are eq
ually effective
Not tru
e for IS
I, where th
e signal w
ave form m
atters
E/N
0 : signal en
ergy‐to
‐noise sp
ectral den
sity ratio
6/8/20
1017
Example 8
.1 M
atched
Filter for
Rectan
gular P
ulse
Rectan
gular p
ulse fo
r g(t)
Match
ed filter
6/8/20
1018
)2 1
(rect
)(
T tA
tg
)2 1(
rect
)2 1
(rect
)(
)(
T tkA
T
tT
kA
tT
kgt
h
Example 8
.1 M
atched
Filter for
Rectan
gular P
ulse (co
ntd.)
Output g
o (t)
Max o
utput k
A2T
occu
rs at t=
T
Optim
al samplin
g
instan
ce
Im
plem
ented
usin
g th
e integ
rate‐and‐dump
circuit
6/8/20
1019
)(
)(
)(
th
tg
tg
o
Sam
plin
g tim
eInteg
rator is resto
re to
initial co
nditio
n
k=1/A
Effect of N
oise
6/8/20
1020
William
Stallin
gs, D
ata and Computer C
ommunicatio
ns, 8
/E, P
rentice H
all, 2007.
Bit erro
r rate (BER) =
2/15=13.3%
Line co
ding:
‐“1”: ‐5 vo
lts‐“0”: 5 vo
lts
A properly
chosen
decisio
n
thresh
old
Sam
plin
g
instan
ce
Probability o
f Error d
ue to
Noise
Assu
me p
olar n
onretu
rn‐to‐zero
(NRZ) sig
nalin
g1: p
ositive rectan
gular p
ulse, +
A
0: n
egative rectan
gular p
ulse, ‐A
Additive w
hite G
aussian
Noise w
(t)of zero
mean
and
power sp
ectral den
sity N0 /2
Received
signal is
The receiver h
as prio
r knowled
ge o
f the p
ulse sh
ape, n
eed
to decid
e 1 or 0 fo
r a received am
plitu
de in
each sig
nalin
g
interval 0
≤t≤tb
6/8/20
1021
Probability o
f Error d
ue to
Noise
(contd.)
Sam
pled valu
e y, with
thresh
old λ,
6/8/20
1022
received 0
sym
bol ,
if
received 1
symbol
,
if
y y
Two kinds o
f errors
01
01
1 ‐
1 ‐
‐How to quantify resid
ual B
ER?
‐How to ch
oose th
reshold to m
inim
ize BER?
Consid
er the C
ase When
Symbol 0
Was Tran
smitted
Receiver g
ets x(t) = ‐A + w(t), fo
r 0≤t≤Tb
The m
atched filter o
utput, sam
pled at t=
Tb , is th
e sam
pled valu
e of a ran
dom variab
le Y
Since w
(t)is w
hite an
d Gau
ssian, Y
is also Gau
ssian
with m
ean E[Y]=–A, an
d varian
ce
6/8/20
1023
See P
age 19
Gaussia
n
distrib
utio
n ca
n be
completely
determ
ined by it
mean and va
riance
When
0 W
as Transm
itted (co
ntd.)
Since w
(t)is w
hite G
aussian
,
The varian
ce is
Yis G
aussian
with m
ean µ
Y =–A
, variance σ
Y2=N
0 /(2Tb )
The co
nditio
nal p
robab
ility den
sity functio
n (P
DF) o
f Y,
conditio
ned on th
at symbol 0
was tran
smitted
, is
6/8/20
1024
T
N
Ay
TN
yy
fb
Y
Y
Y
Y/
)(
exp/
1
2
)(
exp2 1
)0|
(0
2
02
2
Stan
dard
PDF of G
aussian
r.v., with m
ean µ
Yan
d varian
ce σY2
When
0 W
as Tran
smitted
(co
ntd.)
W
hen no noise, Y
=‐A
W
ith noise, d
rifts away fro
m –
AIf less th
an λ, o
utput 0 (n
o bit
error)
If larg
er thanλ, o
utput 1 (b
it erro
r occu
rs)
6/8/20
1025
When
0 W
as Transm
itted (co
ntd.)
Assu
me sym
bol 1 an
d 0 are eq
ual lik
ely to be
transm
itted, w
e choose λ
=0, d
ue to
symmetry
Defin
e , we h
ave
Eb : th
e transm
itted sig
nal en
ergy p
er bit
Eb /(N
0 /2) : (signal p
ower p
er bit)/(n
oise p
ower p
er Hz)
6/8/20
1026
When
0 W
as Transm
itted (co
ntd.)
Defin
e Q‐Functio
n:
The co
nditio
nal b
it error
probab
ility when 0 w
as tran
smitted
is
6/8/20
1027
u
Q‐fu
nctio
n, see P
age 4
01
x=[‐3:0
.1:3];for i=
1:length(x)
Q(i)=
0.5*erfc(x(i)/sq
rt(2));en
dplot(x,Q
)
When
1 W
as Transm
itted
6/8/20
1028
Receives: x(t)=
A+w(t), 0
≤t≤Tb
Yis G
aussian
with µ
Y =A, σ
Y2=N
0 /(2Tb )
W
e have
The co
nditio
nal b
it error rate is
Choosin
g λ=0, an
d defin
ing , w
e have
Bit Erro
r Probability (o
r, Bit Erro
r Rate –
BER
)
6/8/20
1029
Bit E
rror R
ate (BER) is
Dep
ends o
nly o
n E
b /N0 , th
e ratio of th
e transm
itted sig
nal
energ
y per b
it to th
e noise sp
ectral den
sity
Noise is u
sually fixed
for a g
iven tem
peratu
re
Energ
y plays th
e crucial ro
le tran
smit p
ower
W
hat are th
e limitin
g facto
rs?Battery life, in
terference to
others, d
ata rate requirem
ent
00
01
10
0
10
22
2 12
2 1
}ed
transmitt
is {1
Pr
}ed
transmitt
is {0
Pr
N EQ
N EQ
N EQ
Pp
Pp
PP
P
bb
be
e
ee
e
Wider p
ulse
Higher am
plitu
de
BER
6/8/20
1030
Intersym
bol In
terference
The n
ext source o
f bit erro
rs to be ad
dressed
Hap
pen
s when th
e chan
nel is d
ispersive
The ch
annel h
as a frequen
cy‐dep
enden
t (or, freq
uen
cy‐selective) am
plitu
de sp
ectrum
e.g
., ban
d‐lim
ited ch
annel:
passes all freq
uen
cies |f|<W with
out d
istortio
n
Block
s all frequen
cies |f|>W
Use d
iscrete pulse‐am
plitu
de m
odulatio
n (P
AM) as
example
First exam
ine b
inary d
ata
Then co
nsid
er the m
ore g
eneral case o
f M‐ary d
ata
6/8/20
1031
Example 8
.2: Th
e Disp
ersive
Natu
re of a Telep
hone C
hannel
Ban
d‐lim
ited an
d disp
ersive
6/8/20
1032
Block high frequ
encies:
cut‐o
ff at 3.5 k
Hz
Block dc
Example 8
.2: Th
e Disp
ersive natu
re of a Telep
hone C
hannel
Conflictin
g req
uirem
ents fo
r line co
ding
High freq
uen
cies block
ed
need
a line co
de w
ith a n
arrow
spectru
m
polar N
RZ
But p
olar N
RZ has d
c
Low freq
uen
cies block
ed
need
a line co
de th
at has n
o dc
Man
chester co
de
But M
anch
ester code h
as high freq
uen
cy
6/8/20
1033
Example 8
.2: Th
e Disp
ersive natu
re of a Telep
hone C
hannel (co
ntd.)
Previo
us p
age: d
ata rate at 1600 bps
This p
age: d
ata rate at 3200 bps
6/8/20
1034
Eye Pattern
An operatio
nal to
ol fo
r evaluatin
g th
e effects of IS
I
Syn
chronized
superp
ositio
n of a
llpossib
le realizations o
f the sig
nal view
ed w
ithin a p
articular sig
nalin
g in
terval
6/8/20
1035
http
://mem
bers.ch
ello.nl/~
m.heijlig
ers/DAChtm
l/digco
m/d
igco
m.htm
l
Eye Pattern
(contd.)
Eye o
pen
ing: th
e interio
r region of th
e eye pattern
6/8/20
1036
Interp
reting Eye P
atternThe w
idth
of th
e eye open
ing
Defin
es the tim
e interval o
ver which th
e received sig
nal can
be
sampled w
ithout erro
r from IS
I
The b
est samplin
g tim
e: when th
e eye is open th
e widest
The slo
pe
The sen
sitivity of th
e system to tim
ing erro
rs
The rate o
f closu
re of th
e eye as the sam
plin
g tim
e is varied
The h
eightof th
e eye open
ing
Noise m
argin of th
e system
Under severe IS
I: the eye m
ay be co
mpletely clo
sed
Im
possib
le to avo
id erro
rs due to
ISI
6/8/20
1037
Example 8
.3
6/8/20
1038
The ch
annel h
as no ban
dwidth
limitatio
n: th
e eyes are open
Ban
d‐lim
ited ch
annel: b
lurred
reg
ion at sam
plin
g tim
e
Eye Pattern
on Oscillo
scope
6/8/20
1039
http
://www.m
ypriu
s.co.za/p
cmx2_
processo
r1.ht
m
Baseb
and Binary P
AM System
Input seq
uen
ce: {bk | b
k = 0 or 1}
Tran
smitted
signal:
The receiver filter o
utput:
where:
Assu
me p
(t)is n
orm
alized, p
(o)=1, u
sing μ
as a scaling facto
r to
account fo
r amplitu
de ch
ange d
urin
g tran
smissio
n
6/8/20
1040
k
bk
kTt
ga
ts
)(
)(
)(
)(
)(
tn
kTt
pa
ty
kb
k
)
()
()
()
(t
ct
ht
gt
p
)(
)(
)(
)(
fC
fH
fG
fP
Baseb
and Binary P
AM System
(co
ntd.)
For th
e i‐th received
symbol, sam
ple th
e output y(t)
at ti =iT
b , yielding
6/8/20
1041
)(
])
[(
)(
])
[()
(
i
ik k
bk
i
ik
bk
i
tn
Tk
ip
aa
tn
Tk
ip
at
y
Contrib
utio
n of th
e i‐th
transm
itted bit
Resid
ual effect d
ue to
the
occu
rrence o
f pulses
befo
re and after th
e sam
plin
g in
stant t
i : the IS
I
Effect o
f noise,
taken care o
f by
match
ed filter
In th
e absen
ce of b
oth IS
I and noise:
ii
at
y
)(
Nyquist’s C
riterion fo
r Disto
rtionless Tran
smissio
n
Recall th
at:
The p
roblem
: Usu
ally the tran
sfer functio
n of th
e chan
nel h
(t)an
d th
e tran
smitted
pulse sh
ape are sp
ecified (e.g
., p32, telep
hone ch
annel)
to determ
ine th
e transfer fu
nctio
ns o
f the tran
smit an
d receive
filters so as to
reconstru
ct the in
put b
inary {b
k }
The receiver p
erform
sExtractio
n: sam
plin
g y(t)
at time t=
iTb
Deco
ding:
req
uires th
e ISI to
be zero
at the sam
plin
ginstan
ce:
6/8/20
1042
)(
])
[()
(i
ik k
bk
ii
tn
Tk
ip
aa
ty
)(
)(
)(
)(
tc
th
tg
tp
.,0 ,1
])
[(k
i
ki
Tk
ip
b
Nyquist’s C
riterion fo
r Disto
rtionless
Transm
ission (co
ntd.)
Sam
plin
g p(t)
at nTb , n
=0, ±
1, ±2, …
{p(nTb )}
Sam
plin
g in th
e time d
omain produces p
eriodicity in
the freq
uen
cy domain
, we h
ave
On th
e other h
and, th
e sampled sig
nal is
Its Fourier tran
sform is
If the co
nditio
n is satisfied
6/8/20
1043
nb
bnR
fP
Rf
P)
()
(
)(
)(
)(
bn
bnT
tnT
pt
p
dt
ftnT
tnT
pf
Pb
nb
)2
exp()]
()
([
)(
1)0
()
2exp(
)(
)0(
)(
p
dtft
tp
fP
.,0 ,1
])
[(k
i
ki
Tk
ip
b
Nyquist’s C
riterion fo
r Disto
rtionless
Transm
ission (co
ntd.)
Finally w
e have
The N
yquist C
riterion fo
r disto
rtionless b
aseban
d
transm
ission in th
e absen
ce of n
oise: b
bn
b
nb
b
TR
nRf
P
nRf
PR
fP
/1
)(
1
)(
)(
The freq
uen
cy functio
n P(f)
eliminates IS
I for sam
ples
taken at in
tervals Tbprovid
ed th
at it satisfies
bn
bT
nRf
P
)(
P(f): fo
r the o
verall system,
inclu
ding th
e transm
it filter, th
e chan
nel, an
d th
e receive filter
Ideal N
yquist C
hannel
The sim
ples w
ay of satisfyin
g The N
yquist
Criterio
n is
Where W
=Rb /2=
1/(2Tb ).
The sig
nal th
at produces zero
ISI is th
e sinc
functio
n
Nyq
uist b
andwidth: W
Nyq
uist rate: R
b =2W
6/8/20
1045
Wf
Wf
WW
W f
Wf
P|
|,0
,2 1
2rect
2 1)
(
)2(
sinc2
)2
sin()
(W
tW
t Wt
tp
Ideal N
yquist C
hannel (co
ntd.)
6/8/20
1046
Raised
Cosin
e Spectru
mP(f)
is physically u
nrealizab
leNo filter can
have th
e abrupt tran
sitions at f=
±W
p(t)
decays at rate 1/|t|: to
o slo
w, n
o m
argin fo
r samplin
g tim
e erro
r (see Fig. 8
.16)
Use raised
cosin
e spectru
m: a flat to
p + a ro
lloff p
ortio
n
6/8/20
1047
Raised
Cosin
e Sp
ectrum (co
ntd.)
6/8/20
1048
Rollo
ff factor: α
=1‐f1 /W
Indicates th
e excess ban
dwidth
over th
e ideal so
lutio
n, W
)1(
)1
1()
2(1
1
WW f
Wf
WB
T
W=Rb /2=
1/(2Tb ).
Sam
e property as sin
c(2Wt), b
ut
now it is p
ractical
How to
Design
the Tran
sceiverNyq
uist C
riterion
P(f)
= raised
cosin
e spectru
m
Study th
e chan
nel
find h(t)
Match
ed filter (to
cope w
ith noise)
c(t)an
d g(t)
are sym
metric
solve fo
r c(t)an
d g(t)
6/8/20
1049
)(
)(
)(
)(
tc
th
tg
tp
)
()
()
()
(f
Cf
Hf
Gf
P
)2
exp()
()
(fT
jf
kGf
C
Example 8
.4 Bandwidth
Req
uirem
ent o
f the T1
SystemT1 system
: multip
lexing 24 vo
ice calls, each 4 kHz,
based
on 8‐bit P
CM w
ord, T
b =0.647μs
Assu
ming an id
eal Nyq
uist ch
annel, th
e minim
um
required
ban
dwidth is
In practice, a fu
ll‐cosin
e rollo
ff spectru
m is u
sed w
ith
α=1. T
he m
inim
um tran
smissio
n ban
dwidth is
In Chap
ter 3, if use S
SB an
d FDM, th
e ban
dwidth is
6/8/20
1050
kHz
773
)2
/(1
2/
b
bT
TR
WB
MH
z
544.1
2)
1(
W
WB
T
kHz
96
424
T
B
Digital tran
smissio
n is n
ot b
andwidth efficien
t
Baseb
and M
‐ary PAM Tran
smissio
nBaseb
and M‐ary P
AM system
M
possib
le amplitu
de levels, w
ith M
>2
Log2 (M
) bits are m
apped to one o
f the levels
6/8/20
1051
Baseb
and M
‐ary PAM Tran
smissio
n
(contd.)
Sym
bol d
uratio
n: T
Signalin
g rate: R
=1/T
, in sym
bols p
er second, o
r bau
ds
Binary sym
bol d
uratio
n: T
b
Binary d
ata rate: Rb =1/T
b
Sim
ilar proced
ure u
sed fo
r the d
esign of th
e filters as in th
e binary d
ata case
6/8/20
1052
MT
Tb
2log
M
RR
b2
log
Eye Pattern
for M
‐ary Data
Contain
s (M‐1) eye o
pen
ings stack
ed up vertically
6/8/20
1053
Tapped
‐Delay‐Lin
e Equalizatio
nISI is th
e majo
r cause o
f bit erro
r in baseb
and
transm
issions
If ch
annel h
(t)or H
(f) is known precisely, o
ne can
desig
n
transm
it and receiver to
mak
e ISI arb
itrarily small
Find P(f)
find G(f)
find C(f)
However in
practice, h
(t)may n
ot b
e known, o
r be k
nown
with erro
rs (i.e., time‐varyin
g ch
annels)
Cau
se residual d
istortio
nA lim
iting facto
r for d
ata rates
Use a p
rocess, eq
ualizatio
n, to
compen
sate for th
e intrin
sic residual d
istortio
nEqualizer: th
e filter used
for su
ch process
6/8/20
1054
Tapped
‐Delay‐Lin
e FilterTotally (2N
+1) tap
s, with w
eights w
‐N , …w
1 , w0 , w
1 , …w
N
6/8/20
1055
T=Tb : sym
bol d
uratio
n
Tapped
‐Delay‐Lin
e Filter (co
ntd.)
W
e have
and
6/8/20
1056
Tapped
‐Delay‐Lin
e Filter (co
ntd.)
The N
yquist criterio
n m
ust b
e satisfied. W
e have
Den
ote c
n =c(n
T), w
e have
6/8/20
1057
Tapped
‐Delay‐Lin
e Filter (co
ntd.)
Rem
arks
Referred
to as a zero
‐forcin
g eq
ualizer
Optim
um in th
e sense th
at it minim
izes peak
disto
rtion
(ISI)
Sim
ple to
implem
ent
The lo
nger, th
e better, i.e., th
e closer to
the id
eal co
nditio
n as sp
ecified by th
e Nyq
uist criterio
n
For tim
e‐varying ch
annels
Train
ing
Adap
tive equalizatio
n: ad
justs th
e weig
hts
6/8/20
1058
Them
e Example –
100Base‐TX
–Tran
smissio
n of 1
00 M
bps o
ver Tw
isted Pair
Fast E
thern
et: 100BASE‐TX
Up to 10
0Mbps
Usin
g tw
o pairs o
f twisted
copper w
ires Categ
ory 5
cable
One p
air for each
directio
n
Maxim
um distan
ce: 100 m
eters
First stag
e: NRZ 4B5B
provid
e clock
ing in
form
ation
Seco
nd stag
e: NRZI
Third stag
e: three‐level sig
nalin
g M
LT‐3
W
ith tap
ped‐delay‐lin
e equalizatio
n
6/8/20
1059
Summary
Two tran
smissio
n im
pairm
ents
Noise
ISI
Im
pact o
f the tw
o tran
smissio
n im
pairm
ents
How to m
itigate th
e effects of tran
smissio
n im
pairm
ents
Match
ed filter
Evalu
ating th
e BER
Eye p
attern
Nyq
uist criterio
ns fo
r disto
rtionless criterio
n
Tap
ped‐delay‐lin
e equalizatio
n
Binary tran
smissio
ns an
d M‐ary tran
smissio
ns
6/8/20
1060