Performance analysis of the maximum likelihood sequence estimator for known channel impulse
responses
by Harold Fred Fisher
A thesis submitted to the Graduate Faculty in partial fulfillment of the requirements for the degree of
MASTER OF SCIENCE in Electrical Engineering
Montana State University
© Copyright by Harold Fred Fisher (1973)
Abstract:
In a pulse amplitude modulation system, intersymbol interference is the primary detriment to reliable
high rate digital transmission over narrow bandwidth high signal to noise ratio channels. The maximum
likelihood sequence estimator is a receiver structure designed for use under intersymbol interference
conditions. The objective of this thesis is to develop and use an analysis technique for evaluating the
performance of the maximum likelihood sequence estimator for linear channels with known impulse
responses. The performance criterion used is the fractional difference in minimum euclidean weight
due to intersymbol interference. Discrete time signal models and flow graph theory are used to develop
the analysis technique. PERMISSION TO COPY
In p r e s e n tin g t h i s th e s is in p a r t i a l
f u l f i l l m e n t o f th e requirem ents
f o r an advanced degree a t Montana S ta te U n i v e r s i t y ,
L i b r a r y s h a ll make i t
I agree t h a t the
f r e e l y a v a ila b le f o r in s p e c tio n .
I f u r t h e r agree
t h a t perm ission f o r e x t e n s iv e copying o f t h i s th e s is f o r s c h o l a r l y purposes
may be granted by my m a jor p r o f e s s o r , o r ,
o f L ib ra rie s .
It
in h is absence, by t h e . D i r e c t o r
i s understood t h a t any copying o r p u b l i c a t i o n o f t h i s
th e s is f o r f in a n c ia l
g a in s h a l l not be a llo w e d my w r i t t e n p e rm is s io n .
S ig n a tu r e
Date
v (V v if.
/R
/ 9
~7 3
PERFORMANCE ANALYSIS OF THE MAXIMUM LIKELIHOOD SEQUENCE
ESTIMATOR FOR KNOWN CHANNEL IMPULSE RESPONSES
by
HAROLD FRED FISHER
A th e s is subm itted to th e Graduate F a c u lt y in p a r t i a l
f u l f i l l m e n t o f th e requirem ents f o r the degree
of
MASTER OF SCIENCE
in
E le c tric a l
E ngineering
Approved:
Head, M ajor Department
MONTANA STATE UNIVERSITY
Bozeman, Montana
August, 1973
iii
ACKNOWLEDGMENT
The a u th o r wishes t o thank h is committee C . K . R u s h fo r t h 9 D .A . Rudberg
and M.A .F a u lk n e r f o r t h e i r guidance in h e lp in g p re p are t h i s t h e s i s .
A ls o , th e a u th o r expresses h is a p p r e c i a t i o n to R.E.Leo h is re se a rch
a s s i s t a n t s h i p a d v is o r f o r h is p a tie n c e s and understand ing showed
d u rin g th e e n t i r e tim e spent p r e p a r in g th e t h e s i s .
iv
TABLE OF CONTENTS
Page
ii
V I T A ........................
.
iii
TABLE OF CONTENTS
iv
LIST OF TABLES
vi
ACKNOWLEDGMENT
.
v ii
LIST OF FIGURES
ABSTRACT
v iii
< . . .
Chapter
I.
IL
I
IN T R O D U C T IO N ............................................«
CHANNEL MODELS AND THE OPTIMUM RECEIVER
■ ... DISCRETE TIME A N A L Y S I S ............................
6
. . .
8
........................
Tl
’ A RUNNING EXAMPLE
......................................
14
' PERFORMANCE ANALYSIS
....................................
21
D-TRANSFORM
................................. ....
CHIP D-TRANSFORM
III.
6
ERROR EVENTS
. . . .
. .
21
..........................................
23
. . . .
EUCLIDEAN WEIGHT
FINDING:MINIMUM WEIGHT
* ............... ...
............................
24
V
Chapter
IV .
. Page
ANALYSIS USING FLOW GRAPHS
ERROR STATE
. . . .
.
. .
...............................................
27
......................................................................
28
MINIMUM PATH THROUGH FLOW GRAPH
............................
29
MINIMUM EUCLIDEAN DISTANCE WITH NO INTERSYMBOL
INTE RFE RE NCE ................................................................. ....
FRACTIONAL DIFFERENCE IN MINIMUM WEIGHT
MINIMUM DISTANCE PATH ALGORITHM
V.
V I.
SEVERAL EXAMPLES
..........................................
REFERENCES................................. ...
32
■.
33
.........................................
.
35
. .-.
43
.
52
.......................................
.
. .
.
....................................... ; ................... ....
CONCLUSION AND SUMMARY
APPENDIX
.
.
.
.
.
.
. ................................................ • • • •
............................................................. ....
54
65
vi
LIST OF TABLES
T a b le
Page
1.
Program O utput f o r Examples I through 5
.
.
.
.
.
.
2.
Program Output f o r Examples .6 through 8
............................
49
50
v ii
LIST OF FIGURES
F ig u re
I*
2.
■ 3.
Page
P.A.M. Channel w i t h White Gaussion Noise and
th e Maximum L i k e lih o o d Sequence R e c eiv er
s tru c tu re
. ........................................................
D i s c r e t e Time Model o f P.A.M.
System
3
. . . . . . . .
P.A.M. Channel w i t h White Gaussian N o is e, F i l t e r
Matched t o h ( t ) , and T ra n s v e rs a l f i l t e r
w i t h D i s c r e t e Time Response 1 / f (D- ^ )
. . . . . .
11
.
14
4.
Channel Response t o ( a ) x Q = I , .(b) x-j = I , ( c )
X3 = I , (d ) X zl = I and ( e ) Xn = I , X1 = I ,
x 2 = Oi x3 =
and x^ = I
.................................. 15
5.
C onvolutional Encoder E q u iv a le n t t o th e F l a t
Top Pulse Channel
............................ ................... ...
16
6.
Chip Function f o r i = 0,1 and
17
7.
Flow Graph o f F l a t Top Pulse ..............................
8.
Closed Loop Paths Adding Zero Weight
9.
Flow Graph w i t h Minimum D is ta n c e Paths Only . . . . . .
I
31
Flow Graph o f F l a t Top Pulse Example
38
10.
"
.
"
'
2
■ ■
. . . . . . . . . . . .
. . . . . .
30
.................................
31
. ............................
v
‘
■
viii
ABSTRACT
In a pulse a m p litu d e m odulation system, intersym bol i n t e r f e r e n c e is
th e p rim a ry d e t r i m e n t to r e l i a b l e high r a t e d i g i t a l tr a n s m is s io n over
narrow bandwidth high s ig n a l to noise r a t i o c h a n n els .
The maximum l i k e ­
lih o o d sequence e s t im a t o r i s a r e c e i v e r s t r u c t u r e designed f o r use under
intersym bol i n t e r f e r e n c e c o n d i t i o n s .
The o b j e c t i v e o f t h i s th e s is i s to
develop and use an a n a l y s is te c h n iq u e f o r e v a l u a t i n g th e performance o f
th e maximum l i k e l i h o o d sequence e s t im a t o r f o r l i n e a r channels w ith known
impulse responses.
The performance c r i t e r i o n used i s th e f r a c t i o n a l
d i f f e r e n c e i n minimum e u c lid e a n w e ig h t due t o intersym bol i n t e r f e r e n c e .
D is c r e t e tim e s ig n a l models and f l o w graph th e o r y a r e used to develop
th e a n a l y s is te c h n iq u e .
I.
INTRODUCTION
In a p ulse a m p litu d e m odulation system whenever th e in fo r m a tio n
r a t e , measured in u n i t s o f pulses per second (b a u d ) , i s l a r g e r than th e
bandwidth o f th e tra n s m is s io n c h a n n e l, intersym bol
s u lt [ I ] .
i n t e f e r e n c e may r e ­
In o t h e r words, th e pulses o v e r la p i n t o a d j a c e n t tim e
and may cause an erroneous d e c is io n a t th e r e c e i v e r .
i s th e p rim a ry d e t e r r e n t t o r e l i a b l e ,
s lo ts
T h is phenomenon
high r a t e d i g i t a l
tra n s m is s io n
over narrow bandw idth, high s i g n a l - t o - n o i s e r a t i o c h a n n e ls .
te le p h o n e channels used f o r d a ta tr a n s m is s io n a r e t y p i c a l
V o ic e -g ra d e
examples.
The o b j e c t i v e o f t h i s th e s is i s to develop and use a te c h n iq u e f o r
a n a ly z in g th e performance o f th e maximum l i k e l i h o o d sequence e s t im a to r
[ 2] f o r l i n e a r channels whose im pulse response i s lo n g e r than th e source
symbol s e p a r a tio n .
Many r e c e i v e r s t r u c t u r e s have been discussed and developed [ 3 , 4 ,
5 ].
P robably th e most t a l k e d about s t r u c t u r e has been th e optimum r e ­
c e i v e r s t r u c t u r e - - t h e one t h a t makes symbol d e c is io n s based bh the
e n t i r e r e c e iv e d sequence [ 5 ] .
This s t r u c t u r e was not pursued, however,
s in c e maximum l i k e l i h o o d c a l c u l a t i o n s in c re a se d e x p o n e n t ia l I y w ith s e ­
quence le n g th making hardware development too complex and d i f f i c u l t .
As
an a l t e r n a t i v e to t h i s overwhelming c o m p le x ity problem , s t r u c t u r e s were
developed t h a t made sim ple symbol-by-symbol d e c is io n s [ 4 ] . I The best
forms o f these r e c e i v e r s t r u c t u r e s wefe d e riv e d using c e r t a i n c r i t e r i o n
o f o p t i m a l i t y , such as minimum p r o b a b i l i t y o f e r r o r , minimum p r o b a b l i t y o f
e r r o r w i t h intersym bol
i n t e r f e r e n c e fo r c e d to z e r o , and minimum mean-square
2
erro r.
filte r
In e v e ry case these r e c e i v e r s t r u c t u r e s tu rn e d o u t to be a matched
in cascade w it h a tapped d e la y l i n e
(tran sversal f i l t e r s )
N o n lin e a r r e c e i v e r s have a ls o been looked a t [ 6] .
[3 ].
S everal optimum
n o n l i n e a r s t r u c t u r e s have been d e v e lo p e d , but these tu rn e d o u t t o be v e r y
complex.
T h is caused th e e x p e r ts t o lo o k a t suboptimaI n o n lin e a r r e c e i ­
v ers such as d e c is io n fe edback.
These were f a r too complex and d i f f i c u l t
to a n a ly z e t o j u s t i f y t h e i r use.
1
.
^
i
The most r e c e n t r e c e i v e r s t r u c t u r e devised f o r channels in tr o d u c in g
gaussian n oise and intersym bol
i n t e r f e r e n c e i s a maximum l i k e l i h o o d s e ­
quence e s t im a t o r o f th e e n t i r e r e c e iv e d sequence [ I ] .
U n lik e p a s t form u­
l a t i o n s o f l i k e l i h o o d sequence e s t im a to r s i t s c o m p le x ity does not in c re a s e
w i t h sequence le n g th ,' but i s p r o p o r tio n a l
to
where M i s th e s iz e o f
th e i n p u t a lp h a b e t and L i s th e le n g th o f channel impulse response in
u n i t s o f symbol s e p a r a t i o n .
In f a c t t h i s r e c e i v e r can e a s i l y be im p le ­
mented and a n a ly z e d .
F ig u re I
i s a b lock diagram o f a P .A.M . channel w i t h a d d i t i v e w h ite
gaussian n oise cascaded w i t h th e maximum l i k e l i h o o d sequence r e c e i v e r .
The r e c e i v e r c o n s is ts o f a l i n e a r f i l t e r c a l l e d th e whitened matched f i l ­
t e r , a sampler ta k in g samples once e v e r y T seconds (symbol
s e p a ra tio n ),
and a n o n l i n e a r r e c u r s i v e a lg o r it h m c a l l e d the V i t e r b i A lg o r ith m .
F ig u re I
From
,
i=N
s ( t ) =;
I
1=0
6
xi h (t-iT )
(I)
;
3
n ( t ) White Gaussian
9
Noise
x ,x
O
Sample Once Every
T Seconds
Channel
V ite rb i
s (t)
i=N
x (t)
=
I
A lg o rith m
Whitened Matched
F ilte r
x . 6( t - i T )
i =0 1
F ig u re I .
r(t
z0,z l
P.A.M. channel w it h w h ite gaussian noise and th e maximum
l i k e l i h o o d sequence r e c e i v e r s t r u c t u r e .
where
s ( t ) = o u tp u t o f channel whose impulse response i s h ( t )
x ( t ) = in p u t sequence
I = in p u t symbol s e p a r a tio n
Xi = ( i
+ l ) th symbol o f th e i n p u t sequence
The im plem entation o f the whitened matched f i l t e r
In s te a d o f having a bank o f matched f i l t e r s ,
s e n t , o n ly one f i l t e r
s t a n t i s necessary.
one f o r e v e r y in p u t symbol
[ w ( - t ) ] , w ith samples taken once e v e r y symbol
This lea v es the o u tp u t samples w it h a l l
i n fo r m a tio n making them a s e t o f s u f f i c i e n t s t a t i s t i c s
o f th e t r a n s m it t e d sequence.
" ' ’ zN + L - I
i s q u i t e sim ple.
the necessary
f o r th e e s t im a tio n
In a d d i t i o n , the o u tp u t sequence, Zq ,
a sequence o f s t a t i s t i c a l l y
d i s t r i b u t e d gaussian random v a r i a b l e s .
in ­
z ^,
in dependent, i d e n t i c a l l y
This i s necessary to insure the
simple and r e c u r s i v e p r o p e r ty o f th e V i t e r b i a lg o r it h m .
The V i t e r b i a lg o r it h m was o r i g i n a l l y used to decode c o n v o lu tio n a l
4
codes.
Because o f th e s i m i l a r i t y between a d a ta tra n s m is s io n channel
causing intersym bol
i n t e r f e r e n c e and a c o n v o lu t io n a l
encoder the V i t e r b i
a lg o r it h m can be used f o r maximum l i k e l i h o o d sequence e s t im a t i o n [ I ] .
mentioned e a r l i e r ,
th e a lg o r it h m g iv es l i k e l i h o o d c a l c u l a t i o n w ith com­
p l e x i t y p r o p o r tio n a l
isons and
As
t o M*- .
To implement th e a lg o r it h m o n ly
a d d i t i o n s per r e c e iv e d symbol a r e r e q u i r e d .
L -I
i s t e r s f o r remembering th e s u r v i v i n g path and M
compar­
A ls o ,
reg-
r e g i s t e r s f o r remem­
be rin g , th e a p p r o p r i a t e m e tr ic s a re needed [ 7 ] .
Much o f th e l i t e r a t u r e d e a lin g w i t h d e t e c t io n th e o r y c l a s s i f i e s a
r e c e i v e r optimum when th e p r o b a b i l i t y o f e r r o r i s m inim ized [ 2 , 3 , 8 , 9 ,
10].
For th e maximum l i k e l i h o o d sequence e s t i m a t o r , a bound on the minimum
p r o b a b i l i t y o f e r r o r , can be determ ined by knowing th e minimum e u c lid e a n
d is ta n c e o r w eig h t o f th e o u tp u t s ig n a l
space.
Minimum eucludean d i s ­
tance i s th e d is ta n c e between th e two c l o s e s t s ig n a ls o r sequences o f the
o u tp u t s ig n a l
space.
I f th e channel causes intersym bol
th e minimum d is ta n c e may become s m a l l e r .
in te rfe re n c e ,
Some channels cause a l a r g e r
r e d u c t io n in d is ta n c e then o t h e r s , hence, causing a l a r g e r increase, in
p ro b a b ility o f e rro r.
th e system.
If
it
This decrease d i r e c t l y reduces th e r e l i a b i l i t y o f
i s known which channels y i e l d th e s m a lle s t d i f f e r e n c e s
(p e r c e n t r e d u c t io n ) then communication systems can be de vis e d t h a t p ro v id e
maximum r e l i a b i l i t y
in th e presence o f intersym bol
i n t e r f e r e n c e and
a d d i t i v e w h ite gaussi an n o is e .
The performance c r i t e r i a
used i s th e f r a c t i o n a l
d i f f e r e n c e in. minimum
5
e u c lid e a n w e ig h t.
To de te rm in e t h i s f o r each c h a n n e l, minimum e u c lid e a n
w e ig h t w i t h and w it h o u t inte rsy m b o l i n t e r f e r e n c e must be found.
A s sociated w it h each channel
impulse response th e r e i s a s e t o f
e r r o r s t a t e s t h a t form nodes o f a flo w graph.
These nodes a re connected
i n such a way t h a t a t r a n s i t i o n from one node to a n o th e r determ ines a
s in g le r e c e i v e r o u tp u t e r r o r e v e n t.
F u r t h e r , e u c lid e a n w e ig h t f o r each
t r a n s i t i o n can be shown t o be th e square o f a s in g le o u tp u t e r r o r e v e n t.
T h is m ig h t suggest using the f lo w graph to s o lv e f o r minimum e u c lid e a n
w e ig h t by f i n d i n g the s h o r t e s t path from some i n i t i a l
node t o a f i n a l
node.
Indeed t h i s can be done and i s developed and used here in th e form o f an
.
XDS Extended F o r tr a n IV program as our performance a n a l y s is te c h n iq u e .
As a f i r s t step toward th e flo w gra p h , a more com plete a n a ly s is o f
th e w h ite n e d , matched f i l t e r w i l l
be g iv e n .
A ls o , an e q u i v a l e n t d i s c r e t e ­
tim e model o f th e P.A.M. channel and maximum l i k e l i h o o d sequence r e c e i v e r
w ill
be p r e s e n te d .
T h is w i l l
le a d t o d e f i n i t i o n s o f in p u t and outp u t
e r r o r e v e n ts , s t a t e e r r o r e v e n ts , e u c lid e a n w e ig h t , and then f i n a l l y to
■
minimum e u c lid e a n w e ig h t.
1.
II.
CHANNEL MODELS AND THE OPTIMUM RECEIVER
Reference to F ig u re T shows th e o u tp u t r ( t )
o f th e n o is y channel
to
be a sum o f two s i g n a l s , w h ite gaussian noise n ( t ) and s i g n a l ; s ( t ) given
by ( I ) .
It
i s w e ll
known t h a t d e t e c t i o n o f s ig n a ls t h a t a r e l i n e a r com-
b in a t io n s o f some s e t o f b a s is s ig n a ls i s accomplished using a bank o f
filte rs
each matched to a ba sis s ig n a l
a r e tim e t r a n s l a t i o n s o f h ( t )
[8] ,
(th e channel
Since th e s ig n a ls o f ( I )
impulse r e s p o n s e ), o n ly one
f i l t e r matched to th e channel w it h samples taken once e v e ry T seconds i s
n e cessary.
its
L e t t i n g a ( t ) r e p r e s e n t th e o u tp u t o f h ( - t )
samples form a sequence o f numbers a ( 0) , a ( l ) ,
...,
(matched f i l t e r ) ,
a (k ),
....
a(N + L - I ) .
.+ O O
a (k ) =
k = 0, I ,
r(t)h (t-k T )d t
—oo
i+OO
i=N
i
Now, l e t each i n t e g r a l
h (t-iT )h (t-k T )d t +
—oo
...,
N + L - I
I+Oo
n (t)h (t-k T )d t
—OO
(2)
term o f ( 2 ) be re p re s e n te d as
»+oo
R
k -i
h (t-k T )h (t-iT )d t
—00
(3 )
and
.+ o o
n (t)h (t-k T )d t
\
(4 )
Equation ( 2 ) now has a d i s c r e t e r e p r e s e n t a t i o n given by
i=N
3 (k )
^ l0
(5 )
Xi Rk - i ' + nk
7
What can be s a id about the c o r r e l a t i o n o f the n o is e samples given by
(4 )?
To begin w i t h , th e i n p u t n oise n ( t )
c o r r e l a t i o n fu n c t i o n o <s ( t ) where a
u n i t bandwidth.
s e p a ra tio n , i . e . „
i s w h ite gaussian w it h a u t o -
i s th e v a lu e o f th e power over a
I f th e d u r a t i o n o f h ( t )
i s l a r g e r than th e in p u t symbol
i f L > I where L i s th e s m a lle s t i n t e g e r such t h a t :
h ( t ) - 0 f o r t > L I , then a t l e a s t a d ja c e n t samples a r e c o r r e l a t e d .
To
.
g e t an e x a c t measure o f th e c o r r e l a t i o n o f the n oise samples, t h e i r a u to ­
c o r r e l a t i o n c o e f f i c i e n t must be d e te rm in e d .
Taking th e expected v a lu e
o f th e p ro d u c t nj^nl, 0 <_ i and k <. N + L - I , gives
EK ni) ■ "2rR-I
R e f e r r in g to e q u a tio n ( 3 ) ,
because o f th e o v e r l a p .
it
is e v id e n t th a t
A ls o , Rk_t
<6>
i s nonzero f o r k - l < L
i s s y m m e tr ic a l, i . e . , R ^ j =. R ._ k
I t was mentioned e a r l i e r t h a t th e w h ite n e d , matched f i l t e r
p r o p e r t y t h a t i t s o u tp u t samples a r e s t a t i s t i c a l l y
independent,
has th e
This i s
necessary t o in s u r e th e sim ple and r e c u r s i v e n a tu re o f th e V i t e r b i a l g o r ­
B u t, e q u a tio n s ( 6 ) and ( 3 ) show t h a t th e noise sampled nk a r e c o r r ­
ith m .
e l a t e d whenever L > I *, t h e r e f o r e , th e sampled outpu ts o f h ( - t ) a re n o t
s t a t i s t i c a l l y independent.
formed.
Some w h ite n in g t r a n s fo r m a t io n must be p e r ­
From random v a r i a b l e th e o r y ] t
i s known t h a t a l i n e a r t r a n s f o r -
m ation on a sequence o f gaussian random v a r i a b l e s i s a ls o gaussian.
th e r,
if
F u r­
th e sequence i s s t a t i s t i c a l l y dependent, then th e r e i s a l i n e a r
tr a n s f o r m a t io n t h a t w i l l
produce a s t a t i s t i c a l l y independent sequence.
8
T h is i s th e f i n a l
process performed by th e w h ite n e d , matched f i l t e r and
i s accomplished, by a tr a n s v e r s a l
o f th e tr a n s v e r s a l
filte r
filte r
(tapped d e la y l i n e ) .
i s postponed u n t i l
Discussion
th e d i s c r e t e tim e model i s
in tr o d u c e d .
R e c all from F ig u re I t h a t th e in p u t to th e channel
i s a sequence
o f numbers, each separated from i t s a d ja c e n t neighbors by T seconds. As­
s o c ia te d w it h t h i s sequence ( o r any tim e sequence o f numbers) i s a formal
power s e r i e s i n D.
T h is s e r i e s i s c a l l e d th e D -tra n s fo rm o f x ( t ) and i s
g iven by
i=N
x (D ) =
I
.
x .D 1
i =0
1
(7 )
whehe x . i s a symbol chosen a t th e t r a n s m i t t e r ?rpm som§ predeterm ined
I
^
v.-
A
a lp h a b e t ; e . g . , in th e b in a r y c a s e , Xi e q u a lSr IroK O ". - D
can be thought
o f as a d e la y o p e r a t o r r e p r e s e n t in g I T " u n i t s o f d e la y .
Since th e o u tp u t samples o f h ( - t )
a ra tio n T, i t
sample a ( k )
R ..
I
too has a D -tr a n s fo r m .
form a sequence w it h symbol sep­
E quation ( 5 ) i n d i c a t e d t h a t each
i s a f u n c t i o n o f n|J, and th e f i r s t k + I v a lu e s f o r Xi and
T h is looks as i f th e D -tra n s fo rm o f th e sequence a ( 0 ) , a ( l ) ,
could be r e p re s e n te d in terms o f the D -tra n s fo rm s p f Xq , x-j,
...,
...,
x^ ;
Hq , n ^ ........... n^, and th e sequence whose c o e f f i c i e n t s a re given by ( 3 ) .
Now e q u a tio n ( 5 ) has terms r e p r e s e n tin g the c o n v o lu tio n o f th e c o e f f i ­
c i e n t s Xi and Ri .
Whenever two p o lynom ials a r e m u l t i p l i e d , each term o f
th e product i s th e c o n v o lu tio n o f terms from each.
Hence, th e D -tra n s fo rm
9
o f th e samples a ( k ) can be expressed as
a (D ) = X (D )R (D ) + n ' ( D )
(8 )
where x (D ) i s th e i n p u t sequence t r a n s fo r m , n ' (D) is th e c o r r e l a t e d n o is e
sample tr a n s fo r m and R(D) i s the tr a n s fo r m o f th e sequence whose c o e f f i ­
c i e n t s a r e g iven by ( 3 ) .
N o t ic e , e q u a tio n ( 3 ) i s a f u n c t i o n o f th e d i f f e r e n c e k - L
For e x ­
am ple,
R
2
•+CO
h (t-9 T )h (t-7 T )d t
9 -7
-CO
i+OO
= R
3-1 ~
A ls o , R ^ equals R
a bout th e k - i o r i g i n .
^
h (t-3 T )h (t-T )d t
making the c o rres ponding sequence symmetrical
The number o f c o e f f i c i e n t s in th e sequence depends
on th e amount o f o v e r l a p , o r th e v a lu e o f L .
A ll
t o t a l e d , t h e r e a re
2 ( L - 1 ) + I nonzero terms having a D -t r a n s fo r m .
i= L -l
I
R(D) =
i= l-L
The D -tra n s fo rm R(D) w i l l
(9 )
1
be c a l l e d th e a u t o c o r r e l a t i o n fu n c t io n o r
th e pulse a u t o c o r r e l a t i o n f u n c t i o n .
te r m s , i t
_•
R1D1
has 2 L -2 complex r o o t s .
Since i t c o n ta in s 2L-1 nonzero
Since R(D) i s sym m etrical
R(D™ ) ] an in v e r s e o f a r o o t i s a ls o a root,..
T h is makes i t
[R(D) =
p o s s ib le to
■'
I
‘
f a c t o r R(D) i n t o a p roduc t o f p o ly n o m ia ls , f ( D ) and f ( D
).
We a r e now in a p o s i t i o n t o t a l k about th e tr a n s v e r s a l
filte r,
filte r.
T h is
remember, must tr a n s fo r m th e c o r r e l a t e d noise sequence n ' ( D ) i n t o
10
But th e c o e f f i c i e n t s o f ( 6 ) a r e j u s t
p
those o f ( 3 ) tim es th e n o is e power pe r u n i t bandwidth (o ) ; t h e r e f o r e ,
a w h ite gaussian noise sequence.
th e p ulse a u t o c o r r e l a t i o n f u n c t i o n o f n ' ( D )
is
RJ1(D ) = O2 R(D) = o 2f ( D ) f ( D - 1 )
(1 0 )
The w h ite n o is e sequence must have a p u ls e a u t o c o r r e l a t i o n f u n c t i o n t h a t
i s a c o n s ta n t.
Let
n ' (D) = n(D )f(D ™ 1 )
(1 1 )
where n(D ) i s our d e s ir e d w h ite n o is e sequence.
E quation ( 8 ) now can be
w r i t t e n as
a (D ) = x ( D ) f ( D ) f (D- 1 ) + n(D )f(D ™ 1 )
(12)
Passing th e sequence a (D ) through a f i l t e r w i t h response f " 1 (D™1 ) ( t r a n s ­
versal f i l t e r )
giv es
z (D ) = a ( D ) / f ( D " ^ ) = x ( D ) f (D) + n(D)
(1 3 )
w it h p ulse a u t o c o r r e l a t i o n fu n c t i o n f o r n(D ) equal to
EWD,n(D-l)] , ^'(D)l
P1(Dl )I , 2 m
, ,2
(14)
f ( D ) f (D ' )
where o2 i s a c o n s ta n t.
Equation (1 3 ) t e l l s
us t h a t th e P.A.M. channel
and w h ite n e d , matched f i l t e r can be r e p re s e n te d as a d i s c r e t e time f i l t e r
w it h impulse response f (D) plus an a d d i t i v e w h ite gaussian noise sequence
(see F ig u r e 2 ) .
n
n(D )
o
White Noise
Sequence
si
In p u t
Sequence
V
o—
f (D)
■>
x(D )
e
O
- ^
y (D ) = x ( D ) f (D)
z(D )
Output Sequence
F ig u re 2.
D is c r e t e tim e model o f P.A.M. system.
Nothing a t t h i s p o i n t has been s aid about the o v e r a l l
w (t).
What kind o f waveform i s i t ?
w (t-k T ), k = 0, I ,
...»
the samples z k , k = 0 ,
f o r e s t im a t i n g x (D )?
Does indeed the s e t o f fu n c tio n s
N + L - I ,fo r m an orthonormal
I,
....
tim e response,
N + L - l
b a s i s , in s u r in g
are a set o f s u f f i c i e n t s t a t i s t i c s
These q ue stions can be answered by lo o k in g a t a n­
o th e r d i s c r e t e tim e t r a n s f o r m a t i o n , th e c h ip D -tra n s fo rm .
Think o f the response h ( t ) as a sum o f L m u tu a lly e x c l u s iv e chips
H1- ( t ) , 0 < i <_ L - I , where h ^ ( t ) = h ( t + i T ) f o r 0 £ t <_ T and zero el s e w here,
i= L -l
I
h (t) =
h .(t-iT )
i =0
Now ta k e th e D -tr a n s fo r m o f the c h ip sequence, i = 0 , I ,
i = L -I
h (D ,t) =
I
1=0
The fu n c t i o n h ( D , t )
(15)
1
...,
•
h .(t)D 1
1
i s c a l l e d the c h ip D -tra n s fo rm o f h ( t ) .
L - I ,
(16)
12
The f u n c t i o n f (D~^) i s a polynom ial
d iv is io n
I /
f(D
tim e sequence.
)
in D*^ o f degree L - I .
i s c a r r i e d o u t , what r e s u l t s i s p o s s ib ly an i n f i n i t e
Remember,
(F ig u r e I )
th e in p u t sequence Xq ,
i=N
continuous tim e r e p r e s e n t a t i o n x ( t ) = Y x . 6 ( t - i T ) .
i t 1
v e rs a ! f i l t e r
I f th e
x -j ,
had
L ik e w is e , th e t r a n s ­
sequences can be g e n e r a l l y r e p re s e n te d in terms o f a
w eighted sqm o f d i r a c d e l t a fu n c t i o n tim e t r a n s l a t i o n s .
I =oo
g ( t ) = I g 1S ( M T )
i =0
where g ( t )
is i t s
Since g ( t )
Each c h ip g . ( t )
.
(1 7 )
impulse response.
i s a tim e f u n c t i o n , i t
has a c h ip D -t r a n s fo r m .
i s a w eighted d e l t a f u n c t i o n , g . 6 ( t ) .
I
I
g (D ,t)
Therefore
•
*
= g , 6 ( I ) D i = g ( D ) 6( t )
1
(1 8 )
.
and a ls o .
x (D ,t)
= x ( D ) 6( t ) .
Now th e w h ite n e d , matched f i l t e r
filte r
h (-t)
i s re p re s e n te d in terms o f a matched
[ c h i p D -tra n s fo rm h(D \ t ) ]
t e r g(D ) = f™^(D- i ) .
(1 9 )
in cascade w it h a tr a n s v e r s a l
What i s th e c h ip D -tra n s fo rm o f w ( - t ) ?
fil
Note, t h a t
h ( D , t ) and g ( D , t ) a r e fu n c tio n s o f both D and T, t h e r e f o r e ,
W(D- 1 S t) = h(D- 1 , t ) * g ( D , t ) = h(D- 1 , t ) * 6 ( t ) g ( D )
= h(D- 1 , t ) g ( D ) = h(D- 1 , t ) / f ( D - 1 )
■
( 20)
13
A s i m i l a r development shows t h a t f o r th e o u tp u t s ig n a l
s ( t ) o f the chan­
n e l,
s (D ,t) = h (D ,t)x (D )
(2 1 )
E quation ( 9 ) g iv e s th e pulse a u t o c o r r e l a t i o n fu n c t io n o f h ( t ) where the
c o e f f i c i e n t s a re given by ( 3 ) .
R(D) has a n o th e r r e p r e s e n t a t i o n in terms
o f th e c h ip D -tr a n s fo r m h ( D , t ) :
■+00
R(D) =
h (D ,t)h (p ™
( 22)
,t)d t
How about th e o r t h o n o r m a l i t y o f w ( t ) o r w ( D , t ) ?
k = 0,- . . . ,
N +. L - I , r e p r e s e n t an orthonormal
p ulse a u t o c o r r e l a t i o n f u n c t i o n o f w ( t ) .
Does th e s e t w ( t - k T ) ,
basis?
From (2 2 ) and ( 2 0 ) , ,
-I
,” h ( D , t ) h ( D ~ 1 , t ) d t
w (D ,t)w (D ~ , t ) d t =
J O f ( D ) f ( D “V )
jO
Rw (D )
=
L e t ' s look a t th e
= I
(2 3 )
(2 3 )
f(D )f(D " ')
Rw(D) = I
i n d i c a t i n g th e s e t w ( t - k T ) ,
k = 0,
...,
N + L - I , i s o r th o n o r ­
m a l, t h e r e f o r e t h a t th e samples z k , k. = 0 , I ............ N + L - I , a re a s e t o f
s u f f i c i e n t s t a t i s t i c s f o r th e e s t im a t io n o f x ( D ) .
The w h ite n e d , matched f i l t e r
now c o n s is ts o f a matched f i l t e r , h ( - t ) ,
;■
I
H
t' ! ■ '
in cascade w i t h a t r a n s v e r s a l f i l t e r , f " (D“ ) ( s e e 'F i g u r e 3 ) w ith a c h ip
D -tr a n s fo r m w ( D , t ) g iv en by ( 2 0 ) .
14
n ( t ) White Gaussian
Noise
s (t) r ( t )
Channel
F ig u re 3.
Matched
F ilte r
a (0 ),a (l)
T ra n s v e rs a l
F ilte r
P.A.M. channel w i t h w h ite gaussian n o is e , f i l t e r matched to
h ( t ) , and t r a n s v e r s a l f i l t e r w it h d i s c r e t e tim e response
l / f ( D " 1 ).
To g a in deeper i n s i g h t i n t o the communication system i l l u s t r a t e d
by
F ig u re 3 , l e t ' s
work through a running example.
m ission channel
has an impulse response g iv en by th e f l a t pulse o f F ig u re
4 (a ).
L e t ' s assume the t r a n s ­
A ls o , l e t ' s c o n s id e r the case where the symbols to be tr a n s m it te d
are b in a ry .
S p e c i f i c a l l y , c o n s id e r an i n p u t sequence
X2 = 0 , X3 = I ,
and X4 = I .
xq
= 1 » xI = 1 •
T h e r e f o r e , from (1 5 )
s (t) = h (t) + h ( t - l ) + h (t-3 ) + h (t-4 )
The channel can be thought o f as a t h r e e - s t a g e s h i f t r e g i s t e r and an adder
( F ig u r e 5 ) .
15
h (t)
I
O
ro
__ I____ I____
1
4
5
6
7
8
6
7
8
(a)
h (t-l) 2 f
0
I
2
3
1
5
h(t-3 ) 2 41 -
O
I
4
(c)
^ (t-4 )
s (t)
21 -■
(e)
Fig u re 4 .
Channel response to ( a ) x n = I , (b ) x-, = 1 , ( c ) X3 = I ,
(d ) X, = I and ( e ) x n = I ; X 1 = I , x , = 0 , x . = I and
rI .
16
1 1 0 11
1 2 2 2 2 2 1
F ig u re 5.
C o n v o lu tio n a l
encoder e q u i v a l e n t to the f l a t top pulse c h an n el.
White gaussian noise i s added to the s i g n a l , r ( t )
want t o e s tim a te th e tr a n s m it t e d sequence.
a (k ) =
= s (t) + n ( t ) .
From e q u a tio n s ( 2 ) and ( 4 )
+ nj.
where
i= 4
'k
\L
h (t-iT )h (t-k T )d t
i =0
i= 4
h (t)h (t)d t = 3 + 2 = 5
5O
%
i= 4
h(t)h (t-l)d t = 2 + 3 + I = 6
si M o X1
i= 4
h ( t ) h ( t - 2)d t = 1
+ 2 + 2+1
'2 M o * 1
i= 4
h (t)h (t-3 )d t = 1 + 3
S3
We now
+ 2 = 6
=6
17
I
x.
s (t)h (t-4 )d t = 2 + 3 = 5
I = O
— 0°
Now, ta k e th e D -tra n s fo rm acc o rd in g to
(7 ),
s ' (D) = 5 + 6D + 6D2 + 6D3 + 5D4
But from ( 8 )
s ' ( D ) = X (D )R (D )
L e t ' s t r y f i n d i n g R(D) by using th e above e xpression x ( D ) R ( D ) .
sequence i s I I O l
I,
so i t s
The i n p u t
D -tr a n s fo r m i s
x (D ) = I + D + D3 + D4
We can solve f o r R(D) using one o f two methods, e i t h e r by (2 2 ) or ( 9 ) .
L e t ' s t r y both.
h (D ,t).
To use (2 2 ) we must s o lv e f o r th e c h ip D -tra n s fo rm
From (1 5 )
h( t) = hg(t) + Ii1U - I ) + h2( t - 2 )
where h ^ ( t )
i s given by F ig u re 6 .
h,(t)'
I
--
O
O
F ig u re 6 .
2
Chip f u n c t i o n f o r i = 0 , I ,
3
and 2.
t
Now, a p p ly in g th e d e f i n i t i o n o f th e c h ip D -tra n s fo rm (1 6 )
h ( D , t ) = hQ( t ) + h -|( t)D + h2 ( t ) D
= h .(t)
I
(I
2
+ D + D^)
L e t ' s solve f o r R(D) using both methods.
I.
From (2 2 )
h (D ,t)h (D " \t)d t
R(D)
0
(I
+ D + D2 ) ( I + D" 1 + D '2 ) h f ( t ) d t
R(D) = D" 2 + 2D™
1 + 3 + 2D + D+2
(I
2.
+ D1 + O f ) ( I
I
-I
+ D™' + D"^) = f ( D ) f ( D ™ ' )
From ( 3 ) and ( 9 )
fh 2 ( t ) d t
Jn
R1 ■ R_i
Ro = R o =
'2
hence
'- 2
h (t)h (t+ l)d t = 2
0
rl ■
h ( t ) h ( t + 2) d t = I
Jo
.
R(D) = I D ™
2 + 2D™1 + 3D + 2D + I D2
Now s' (D) i s
19
S1(D) = X (D )R (D ) = D“ 2 + 3D" 1 + 5 + 6D - 6D2 + GD3 + GD9 + 3D5 + D6
N o tic e th e e x t r a terms in s'(D).
M u l t i p l i c a t i o n o f R(D) and x (D ) im p lie s
c o n v o lu tio n o f th e tim e sequences.
T h is assumes sampling s t a r t e d a t
minus i n f i n i t y and te r m in a te d a t plus i n f i n i t y .
B u t, th e in p u t sequence
S1(D) s t a t e d a t tim e k = 0 and ended a t tim e k = 4 , so, o n ly those terms
w here, k = 0 ,
...»
Now, l e t ' s
4 o f s'(D) need to be remembered.
lo o k a t th e waveform
w (D ,t)
h .(t)(l
h (D ,t)
+ D + D2 )
hi ( t )
I + D + D2
f (D)
T h is im p lie s t h a t w ( t ) = h ^ ( t ) .
N e x t, l e t ' s
f i n d th e samples
fo r k = 0,
....
6.
/1+00
r(T )w (T -k )d t
-CO
hoo
[s (t) + n (t)]w (t-k )d t
-OO
The s ig n a l p o r tio n s o f z ^ , k = 0 ,
...,
6 , a re
s (t)h .. ( t - k ) d t
I,
Z^ = 2 ,
z'0
= 2, z ' = 2 , z l = 2, z ' = 2 ,
zl
Now, r e f e r t o th e d i s c r e t e tim e model o f F ig u re 2 where
= I
20
y (D ) = x (D )f(D )
= (I
=1
+ D + D3 + D4 ) ( l
+ D + D2 )
+ 2D + 2D2 + 2D3 + 2D4 + 2D5 + I D 6
N o t ic e , t h i s i s th e D -tra n s fo rm o f th e s ig n a l
p o r tio n s o f z ^ , k = 0 ,
The V i t e r b i a lg o r it h m r e q u ir e s th e samples z ^ , k = 0 , I ,
be s t a t i s t i c a l l y independent gaussian random v a r i a b l e s .
noise p o r tio n s o f th e samples a r e random, l e t ' s
I+OO » + 0 0
n (t)n (s )w (t-k )w (s -j)d td s
+00I +00,
& (t-s )w (t-k )w (s " j)d td s
J
6,
Since o n ly the
look a t t h e i r a u t o c o r r e l a
tio n c o e ffic ie n t.
E (n kn , )
...»
E (nkn . ) = 0; k M
0O
O
h j ( t - j ) d t = 'a ; -k = j
Hence, th e n oise sequence is independent ( w h i t e ) .
III.
PERFORMANCE ANALYSIS
Comparison o f th e o u tp u t sequence y ( D ) and in p u t sequence x (D ) sug­
gests t h a t th e f l a t - t o p
in te rfe re n c e .
pulse w it h L = 3T c r e a t e s severe intersym bol
How severe i s th e intersym bol
in te rfe re n c e .
What f r a c t i o n ­
a l r e d u c t io n in th e minimum e u c lid e a n w e ig h t,d o e s th e f l a t - t o p pulse p r o ­
duce?
This q u e s tio n lea d s d i r e c t l y to th e d e f i n i t i o n o f an e r r o r e vent
E and i t s e u c lid e a n w eig h t d ( E ) .
L e t ' s assume t h a t th e P.A.M. t r a n s m i t t e r sends th e sequence
Xg = I ,
X1 = 1 ,
X2 = 0 , Xg = I ,
X4 = I
b u t , th e r e c e i v e r e s tim a te s th e t r a n s m it t e d sequence as
xO = I, X1 ^ 0, X2 = !, x3 = I, X4 = I
These sequences have D -tran s fo rm s
x (D ) = I
+ D + D3 + D4
and
x ( D ) = I + D2 + D3 + D4 9 .
re s p e c tiv e ly .
Now, t a k in g th e d i f f e r e n c e between th e a c t u a l tr a n s m it t e d
sequence and th e e s tim a te d sequence g iv e s th e i n p u t e r r o r sequence
xO " x O = °» x i
" x i = + T » x2 ~ x2 =
1 x 3 " x3 = 0 ’ x4 ™ x4 = 0
w it h D -tra n s fo rm
Ex (D) = ( x 0 - x 0 )D ° + (X 1 - X1 ID 1 + .(x 2 - x 2 ) D2 + ( x 3 - x 3 )D 3 .+' (x 4 - %4 )D4
= 0 + D - D2 + OD3 + OD4
= D(1 - D ) =
x (D ) - x (D )
22
I f th e r e c e i v e r made a wrong e s t im a te o f x ( D ) , what about i t s
y ( D ) C y (D )]?
e s tim a te o f
From F ig u re 2
y ( D ) = f (D )x (D )
(2 5 )
y ( D ) = f (D )x (D )
(2 6 )
hence
The d i f f e r e n c e between y ( D ) and y (D )
Ey (D) = y (D ) - J (D )
is
.
.
-
A
= f ( D ) [ x ( D ) - x ( D ) ] = f (D)E (D)
.
(2 7 )
■
L e t ' s lo o k a t E^(D) f o r th e f l a t to p pulse [ f (D) = I + D + D^] and in p u t
e r r o r sequence D - D
2
:
Ey (D) = ( I
+ D + D2 H l
- D)D
= D - D^ = y (D ) - y (D )
Consider th e two sequences whose D -tran s fo rm s a re y (D ) and y ( D ) .
i s a . p o s s i b l e ouput v e c t o r c onta ined in th e o u tp u t v e c t o r space.
tance between th e two v e c to r s i s j u s t th e e u c lid e a n d is ta n c e
The d i s ­
(th e square
r o o t o f th e sum o f th e squares o f th e d i f f e r e n c e in each component).
e u c lid e a n w e ig h t i s j u s t th e square o f t h i s .
.
Each
The
B u t, Ew(D) i s a polynomial
y ...
in D whose c o e f f i c i e n t s a r e th e d i f f e r e n c e s o f the components o f y (D ) and
.•
A
••
-
.
y ( D ) , t h e r e f o r e , th e e u c lid e a n w e ig h t i s j u s t th e sum o f th e squares o f
23
th e components o f E y ( D ) , and g iven as
d 2 (E ) = •
j
(2 8 )
Equation (2 8 ) can be o b ta in e d by a n o th e r method.
Consider the p o l y ­
nomial
Ey (D)Ey ( D - I ) = Ex ( D ) f ( D ) f ( D ' 1 )Ex (D‘ 1 )
= E (D )R (D )E (D"1 )
X
*
The D^ c o e f f i c i e n t o f t h i s polynomial
(2 9 )
i s given by ( 2 8 ) , hence
d 2 (E ) = [E y (D )E y (D' 1) ] 0
(3 0 )
L e t ' s de te rm in e th e e u c lid e a n w e i g h t . o f our running example.
A pply­
in g ( 2 8 ) , we have
d 2 (E) = 1 + 1 = 2
I s t h i s th e s m a lle s t e u c lid e a n w e ig h t p o s s ib le w ith th e f l a t top pulse?
L e t ' s lo o k a t th e e u c lid e a n w e ig h t o f th e f l a t top p u ls e f o r several
o t h e r i n p u t e r r o r sequences and see.
L e t ' s f i r s t t r y s in g le e r r o r s
I .
A s in g le e r r o r
: .,
Ex (D) = ±Dk , 0 < k < N
Ex (D) = +Dk ( l + D + D2 )
'
.
•
24
d 2 (E ) = [ ( I
+ D + D2 ) ( + D " k ) ( + p “ k ) ( l . +
D' 1 + D" 2) ] 0
= [R (D )]q = 3
N o t ic e , sin ce (+ p k )(jjD™k ) = + 1 , Id2 (E ) i s independent o f o v e r a l l d e la y and
s ig n .
From example ( I )
e u c lid e a n w e ig h ts .
2.
above, we know s i n g l e e r r o r s do n o t g iv e s m a lle r
So, l e t ' s
t r y some more double e r r o r p a t t e r n s .
Ex (D) = I - D
E (D) = ( I
y
- D ) ( I + D + D2 ) = I
-D 3
d 2 (E ) = I + I = 2
3.
Ex (D) = D2 - D 3
Ey (D) = D2 ( I
d 2 (E ) = [ ( I
- D3 )
- D3 )D 2 D" 2 ( I
- D“ 3) ] 0 = 2
2
As f o r th e s in g le e r r o r s , f o r double e r r o r s d ( E ) , i s independent o f o v e r ­
a ll
d e la y .
4.
Ex (D) = I + Dk
Ew(D) = ( I
y
+ Dk ) ( I
I
<_ k <_ N
+ D + D2 )
= I + D + D2 + Dk + Dk+1 + Dk+2
.
25
k = I
d? (E ) = 1 + 22 + 2 2 + I = 1 0
k = 2
d2 (E ) = 1
+ I + 2 2+ 1 + 1 = 8
k > 3
d2 (E)
5,
1 + 1+1
Ex (D) = I - Dr
Ex (D) = ( I
+ 1 + 1 + 1=6
2 < k < N
- Dk ) ( l
+ D + D2 )
= I + D + D2 . - Dk - Dk+1 - Dk+2
k = 2
d2 (E ) = 1
k>
+1
+1
+1
=4
3
d2 (E ) = 1 + 1 + 1 + 1 + 1 + 1 = 6
L e t ' s t r y some t r i p l e e r r o r s .
6.
Ev (D) = I - D + D2
A
Ey (D) = I + D2 + D4
d2 (E ) = I + I + I = 3
26
7-
Ev (D) = I + D + D2
.
8.
2
d (E ) = 1 +
Ex (D) = 1
.
.
d 2 (E ) = 7
4 + 9 + 4 + 1=19
- D - D 2
IV .
ANALYSIS USING FLOW GRAPHS
P
From th e examples g iv e n , we can guess t h a t maybe d (E) - 2 , where
Ex (D) = Dk ( l
- D ),
is th e minimum e u c lid e a n w e ig h t f o r th e f l a t top p u ls e .
B u t, we need a more r e l i a b l e method t h a t w i l l
always g iv e th e minimum
e u c lid e a n w e ig h t f o r an a r b i t r a r y h ( t ) d r f ( D ) .
One such method i s to
f i n d the s h o r t e s t path through a f l o w graph r e p r e s e n tin g th e e u c lid e a n
w e ig h t d i s t r i b u t i o n o f th e o u tp u t e r r o r sequences.
Ey (D) = Ex ( D ) f (D)
L e t ' s ta k e a lo o k a t the i t h c o e f f i c i e n t o f Ev ( D ) , E , i = 0 , I ,
x
y^
...,
N + L - I.
Each such c o e f f i c i e n t o f th e product o f Ex (D) i s given
k=m
I <
"y i . k=0 ~x i - k
(3 1 )
k
where
L - I
i f L - I. < i
if L - I > i
What we need i s a flo w graph such t h a t a p a r t i c u l a r path through th e
graph w i l l
g iv e us th e c o e f f i c i e n t s
o f a p a r t i c u l a r o u tp u t e r r o r se­
quence.
From ( 3 1 ) , th e s t a t e e r r o r ,
E
si
can be d e f i n e d .
(3 2 )
i-L
Knowing two successive s t a t e s Ec
and Ec
i
th e o u tp u t c o e f f i c i e n t E , .
yi
, we can f i n d
i+1
I f we l e t each node o f th e graph be a
28
r
p o s s ib le s t a t e
node E
0
, then passing through i t am ounts.to s t a r t i n g a t
and p ro g re ss in g s e q u e n t i a l l y through to node E
.
SN+L-1
Each
O
l i n k between th e s t a t e E
and Ec
has assigned to i t
si
si+ l
summing th e E 's f o r each path w i l l
^i
E
yi
, th e re fo re ,
g iv e th e e u c lid e a n w e ig h t o f t h a t
p a th .
For a b in a r y i n p u t a lp h a b e t , which i s assumed thro u g h o u t th e e n t i r e
p a p e r, each e lem ent o f a s t a t e ta k e s one o f t h r e e p o s s ib le values
E
x
T h is means th e r e a r e 3 ^
a ll
(3 3 )
k
d i f f e r e n t , s t a t e s o f th e graph.
To in s u re t h a t
nonzero c o e f f i c i e n t s o f E (D) a r e d e te r m in e d , th e f i n a l
th e a l l
z e r o s t a t e . T h is makes a graph o f 3
and f i n a l
nodes being th e a l l
L- I
node must be
'
+ I nodes, th e i n i t i a l
zero s t a t e .
S in c e , t r a n s i t i o n through th e graph is done s e q u e n t i a l l y , l i n k s
from nodes w it h s t a t e
E
s
B u t, s t a t e
(3 2 ).
( 3 2 ) must go o n ly t o nodes w ith s t a t e
(3 4 )
1+1
M -L + l
( 3 4 ) c o n ta in s L -2 elements t h a t a r e a ls o c o n ta in e d in s t a t e
The o n ly e lem ent o f (3 4 ) t h a t i s not c onta ined in
»• - .
Since i t
'-I ■''VV -
• >•
(3 2 )
•
1•
i s Ex .
'
^
ta k e s on o n ly t h r e e v a l u e s , t h e r e a r e th r e e l i n k s l e a v i n g each
node and t h r e e l i n k s going t o each node.
Now since we a r e i n t e r e s t e d
29
2
o n ly in nonzero d (E) v a l u e s , th e r e is no l i n k from the i n i t i a l
th e f i n a l
node to
node; hence, t h e r e a r e o n ly two l i n k s le a v in g the i n i t i a l
node and two l i n k s going to th e f i n a l
node.
To see what t h i s flo w graph looks l i k e ,
l e t ' s c o n s t r u c t i t f o r the
f l a t top pulse w it h d i s c r e t e impulse response f (D) = I + D + D2 .
L = 3 , th e r e a r e 10 nodes composing th e graph (see F ig u re 7 ) .
s p e c tio n we see t h a t two s h o r t e s t paths through the graph a re
d
min = I + 0 + 0 + I
In p u t e r r o r 0 0 , - I
I 0 0
sequence
and
( o j ) — ^—
(T o )— >
d 2min = 1
—
+ 0 + 0 + 1
(<h
= 2
In p u t e r r o r 0 0 1 , - 1 0 0
sequence
)—
Since
By i n ­
30
F ig u re 7 .
Flow graph o f f l a t top p ulse example.
31
Look a t th e two loops shown below in F ig u re 8 .
Loop I
F ig u re 8 .
Closed loop paths adding zero w e ig h t.
We can t r a v e l
A ll
Loop 2
around these paths any number o f times and no t add w e ig h t.
o
paths c o n ta in in g these loops have d (E ) = 2 , corresponding to m i n i ­
mum d is ta n c e paths.
The flo w graph i s redrawn w it h a l l
d e le te d
e xc e p t minimum d is ta n c e routes
(F ig u r e 9 ) .
F ig u re 9.
Flow graph w ith minimum d is ta n c e paths o n ly .
32
Using a f l o w graph, we a r e a b le t o f i n d th e minimum d is ta n c e f o r a
g iven d i s c r e t e impulse response f ( D ) .
how severe th e intersym bol
th e r e i s no intersym bol
B u t, t h i s s t i l l
in te rfe re n c e is *
in te rfe re n c e .
is tra n s m itte d .
We need t o f i n d d
Then th e f r a c t i o n a l
minimum e u c lid e a n w e ig h t can be found .
s in g le symbol
does not t e l l
us
min when
d i f f e r e n c e in
Consider th e case.w here o n ly a
The w h ite n e d , matched f i l t e r would have
to process o n ly one p u ls e , so t h e r e would be no intersym bol
in te rfe re n c e .
Since we a r e assuming a b in a r y a lp h a b e t , t h e r e a r e o n ly two in p u t sequences p o s s ib le
X1 (D) = Dk
0 < k < N
I
H
O
O
■
I
I
rxT
and
I f th e t r a n s m i t t e r sends x-| ( D ) , but th e r e c e i v e r e s tim a te s x (D ) to be
X(D) = X2 (D) i n s t e a d , then
Ex (D) = x (D ) - x (D ) = X1 (D) 7 X1 (D) = Dk
th e re fo re ,
Ey (D) = Dkf (D)
and
d2 (E ) = [ f ( D ) D kD"kf (D- 1 J l 0 = [ f ( D ) f ( D - 1 ) ] 0
i= L -l
o
(3 5 )
33
Equation (3 5 ) would n o t change i f
in s te a d th e t r a n s m i t t e r s e n t a 0 and
th e r e c e i v e r made a wrong d e c i s i o n .
nonzero Ex ( D ) , i t
Since (3 5 ) i s t r u e f o r a l l o f th e
i s th e minimum e u c lid e a n w e ig h t w it h no intersym bol
in te rfe re n c e .
p
L e t ' s f i n d d min w ith no intersym bol
top p u l s e , f (D) = I + D + D2 .
'
i n t e r f e r e n c e f o r the f l a t
Using (3 5 )
d2min = [ ( I
= [I
+ D + D2 ) ( I
+ D™1 + D"2 ) ] 0
+ 2D"1 + 3 + 2D + 1 ] Q
= 3
T h is i s th e same r e s u l t o b ta in e d w i t h a s in g le in p u t e r r o r sequence w ith
inte rsy m b o l
in te rfe re n c e
(Example I ) .
What i s the f r a c t i o n a l
d i f f e r e n c e i n d min due to intersym bol
t e r f e r e n c e w i t h th e f l a t to p pulse?
Using (3 5 ) and d2min 1 = 2
in ­
(in te r-
symbol i n t e r f e r e n c e ) ,
(d 2min " d2min I ) / d 2min =
\ •
= .333 ...
, '
T h is says t h a t th e f l a t to p pulse reduces the minimum e u c lid e a n w eig h t
by o n e - t h i r d .
i•
•
G e n e r a liz in g th e above f o r an a r b i t r a r y f ( D ) g iv e s th e f r a c t i o n a l
d iffe re n c e
34
f d = (d 2min - d2min I ) / d 2min
(36)
where
d^min =. minimum e u c lid e a n w e ig h t w i t h o u t intersym bol
in te rfe re n c e .'
d m i n , I = minimum e u c lid e a n w eig h t w it h in te rs y m b o l i n t e r f e r e n c e .
Expression (3 6 ) i s th e c r i t e r i o n t h a t w i l l
be used t o compare the p e r ­
formance o f our maximum l i k e l i h o o d sequence r e c e i v e r f o r s e v e ra l channel
im pulse responses.
Its
performance can a ls o be e v a lu a te d by comparing
th e upper bounds on th e minimum p r o b a b i l i t y o f e r r o r .
The i n t e r e s t e d
re a d e r i s r e f e r r e d to r e fe r e n c e [ 2 ] .
The impulse response given as a running example was f o r the case
‘i
L = 2.
What about when L i s g r e a t e r than 3?
W i l T a minimum d is ta n c e
path through the; f l o w graph be as easy to f i n d ?
I f L = 4 , t h a t means
I
■
th e number o f nodes in th e graph i s t w e n t y - e i g h t .
For. L = 5 , th e r e a re
.
82 nodes.
3
L -I
+1,
•
.
•
As L ^becomes l a r g e r , th e graph grows e x p o n e n t i a l l y as
making; th e d e te r m in a tio n o f a minimum d is ta n c e path i n c r e a s i n g ­
ly d iffic u lt.
T h is makes i t necessary to use computer programming to
f i n d th e minimum d is ta n c e path .
I t was mentioned a t th e b eginnin g o f t h i s paper t h a t i t s o b j e c t i v e
'I
‘
:
i s to develop a te c h n iq u e f o r a n a ly z in g th e performance o f th e maximum
.
'
'
l i k e l i h o o d sequence e s t im a t o r f o r known c h a n n e ls .
■
i
\
■
:
f
.
-
■:
•;
■
■
.
Thd f l o w graph a n a l y s is
.u'
.
,
•
i
•
s____________________ =____________ •
r
I
35
was in tro d u c e d as a p a r t o f t h i s o b j e c t i v e .
y s i s us efu l
To make th e f l o w graph a n a l -
f o r a r b i t r a r y value s o f L, an XDS Extended F o r tr a n IV computer
*
program has been w r i t t e n t h a t s e ts up th e graph and f i n d s a minimum d i s ­
ta n c e path through i t f o r a r b i t r a r y L and f ( D ) p a i r s
(channel
impulse
re s p o n s e s ).
The minimum d is ta n c e path f i n d e r p o r t i o n o f th e program uses an a l ­
g o rith m in tro d u c e d by W. E. D i j k s t r a f o r d e te rm in in g s h o r t e s t paths
through a graph [ 1 1 , 1 2 ] .
L e t ' s ta k e a lo o k a t t h i s a lg o r it h m and use
i t t o f i n d th e minimum d is ta n c e f o r our ru n n in g example.
This p a r t i c u l a r s h o r t e s t path a lg o r it h m f i n d s th e s h o r t e s t r o u te
from a given s t a r t i n g node, Ng . to some o t h e r given node
in the graph*
I t accomplishes t h i s by f i n d i n g th e s h o r t e s t path from t h i s node to a l l
o t h e r nodes N. o f th e graph.
The reason f o r doing t h i s
i s t h a t any node
may be an i n t e r m e d i a t e node on th e s h o r t e s t path from Ng to Nf .
t h e r e i s more than one s h o r t e s t path from N^ to N^, i
d e le te s a l l
but one*
i s a t r e e w it h N - I
L (i,j)
p a th ).
(L (i,j)
The r e s u l t i n g gra p h , a f t e r a l l
f
If
s a lg o r it h m
d e le tio n s
(N i s the number o f nodes in th e f l o w graph) l i n k s
i s th e path from N^ to Nj. w i t h no o t h e r nodes on t h a t
Each l i n k L ( i , j )
i s th e s h o r t e s t l i n k between N^ and Nj..
As mentioned above, th e a lg o r it h m b u ild s a t r e e which c o n ta in s
s h o r t e s t paths from th e s t a r t i n g node"N
to a l l
g in n in g , t h e r e a r e no l i n k s o f th e t r e e so a l l
o th e r nodes.
At the be­
l i n k s a r e no n tre e l i n k s ,
The a lg o r it h m then a tte m p ts to in c re a s e th e number o f t r e e l i n k s u n t i l
•
36
th e re are N - I
in number.
To b e g in , l e t
be th e a c tu a l
s h o r t e s t d is ta n c e from Ng to
and
l e t LgJc be th e s h o r t e s t d is ta n c e from Ng to N^ using t r e e l i n k s and at,
l e a s t one no n tre e l i n k .
no n tre e l i n k then Lg^ =
I f th e path from Ng to N^ needs more than one
Now, i f
L (i,k )
Njc where N^ i s c o n ta in e d i n th e t r e e ,
If
i s the l i n k between N^ and
then i t
th e r e i s no l i n k between N. and N^ then d ^
has a d is ta n c e 0 I d ^
=
<_
A ls o , i f node N^
i s a n e ig h b o rin g node o f th e e x i s t i n g t r e e , whose nodes a r e designated
N^, then 0 <. d^^ <_
The a lg o r it h m then examines a l l
n e ig h b o rin g nodes
o f th e t r e e , and f o r each d e term ines th e minimum Lg^ f o r a l l
N .:(3 7 )
Lsk d m| n ( Ls i + dik>
I t then de te rm in e s th e minimum Lg^ f o r a l l
Lsk ■ 7
N^,
'
(3 8 )
' Lsk;
and makes t h i s node N^ a t r e e node w i t h d is ta n c e
L
(39)
sp
L in k L ( i , p ) now. becomes a n e w ,tr e e l i n k w it h d is ta n c e d ^ .
Each tim e e q u atio n s
added t o th e t r e e .
(3 7 ) through (3 9 ) a r e e x e c u te d , a new l i n k i s
T h is means t h a t we must R e c a lc u la te Lg^ f o r a l l
n e ig h b o rin g nodes o f th e new t r e e .
To do t h i s , we lo o k a t a l l
n e ig h b o rin g N^ and compare Lg^ determ ined e a r l i e r w i t h
i s l a r g e r th a n Lgp + dpj ,
th e new Lgk becomes Lgp
+
nodes
+ dp... I f
dp i .
I f Lgk i s
Lg^
37
s m a lle r no change i s made.
This o p e r a t io n is i n d i c a t e d by
LSk = " " ' " f - s k -
l SP + dPkj
(4 0 )
The above a lg o r it h m i s summarized using th e f o l l o w i n g s te p s .
I.
To s t a r t , Ng i s th e o n ly node o f th e t r e e so Lg^ = dg^ and Lgg
= 0.
II.
The minimum o f a l l
Lgk i s fo u n d , Lgp = min L 1g k , and L ( i , p )
is
made a new t r e e l i n k , Lcn = Lcn.
sp ■ sp
III.
The number o f l i n k s o f th e t r e e i s compared w i t h N - I ...
If
it
becomes equal t o t h i s v a lu e th e a lg o r it h m i s te r m in a t e d .
IV .
Re-examine each Lgk found.
L e t Lgk = min ( L g k , Lgp +
cW V.
Return to Step I I .
To c a r r y o u t th e above a l g o r i t h m , each node, N i s given a la b e l
(L ,i).
L i s e i t h e r Lgk o r Lgk depending on whether Nk i s a t r e e node o r
a no n tre e node.
The l a s t node on th e s h o r t e s t path from Ng l t o Nk is
d e s ig n a te d as i .
p
L e t ' s use th e above a lg o r it h m to f i n d d min w ith th e flo w graph o f
th e f l a t top p u ls e .
I t s graph i s redrawn here as F ig u re 10.
has been given a d e s ig n a tio n N1- , i = 1 , 2 , . . . ,
•
=
Each node
10.
"
'
To b e g in , th e t r e e c o n ta in s o n ly one node N^ and no t r e e l i n k s .
s i n g l e node N-j has two neighbors N^ and Ny.
The
We g iv e them temporary l a b e l s
/
I
38
F ig u r e 10.
Flow graph o f f l a t top p u ls e example.
39
( L j 4 J ) = ( d ^ . l ) = ( I 9I )
and ( L j 79I ) = ( 1 , 1 ) .
Now9 min ( I 9I )
= I,
but
both have d is t a n c e one; t h e r e f o r e , a r b i t r a r i l y chose e i t h e r , say N7 .
The t r e e now c o n s is ts o f th e l i n k L ( 1 , 7 ) o n ly .
The n e ig h b o rin g nodes o f the t r e e a r e N39 N59 Ng 9 Ng9 hence;
L j3 =
min ( L j 39 L17 + Ci73) = min
(“ , I +
L j4 =
min ( L j 49 L17 + Ci74) = min
( I , I + ») = I
L j6 =
min ( L j g 9L17 + d ^ , )
= min
(» , I +
0) = I
L jg =
min ( L j g9 L17 + d7 g ) = min
(<», I +
4) = 5
I) = 2
Since L j 4 and L j 6 a r e e q u a l , a r b i t r a r i l y chose one, say L j 4 .
now has th e permanent l a b e l
L ( 1 , 4 ) and L ( 1 » 7 ) .
(1 ,1 ).
Node N4
The t r e e c o n s is ts o f th e two l i n k s
C o n t in u in g 9 th e n e ig h b o rin g nodes a r e Ng 9 N59 Ng 9 N39
Ng 9 and Ng9 t h e r e f o r e ;
L j2 =
min ( L j g 9L^4 + dg4 ) = min
(°°, I +
I ) =2
L j3 =
min ( L j 39 L14 + d3 4 )
= min
(2 , I +
=°) = 2
L j5 =
min ( L j 59 L14 + d4 5 ) = min
(°°9 I +
4) = 5
L j 6 = min ( L j g 9 L14 + d4 g ) = min ( I 9 I + 00) . = !
Hs =m
1r^ (Hs' Lu + ^s 5 =m
1n, ("■ Ji+
*;’ .
- ' '''' (.'.y,
.
L j 9 = min ( L j g9 L 14 + d4 g ) = min ( 5 , 1 + » ) = 5
-
' -
(
Since L j g i s one o f th e s m a l l e s t i t
L ( 4 , 8 ) becomes a t r e e l i n k .
becomes a permanent l a b e l
( 1 , 4 ) and
P ro ce e d in g , th e n e ig h b o rin g nodes a re now
/
40
N2 , N3 , Ng, Ng and Ng w i t h d is ta n c e s
L j2 =
min ( L j 2 ,
L j3 =
min ( L j 3 , L18 + dg3 ) = min
(2 , I
+ 0)
= I
L j5 =
min ( L j 5 , L18 + d8 5 ) = min
(5 , I
+ ~)
= 5
L j6 =
min ( L j 6 , L18 + d8 6 ) = min
(I, I
+ I)
= 1
. L jg =
min ( L j g , L18 + d8 g ) = min
(5 , I
+ I)
= 2
+ dg2 )
Now choose L j 6 as a permenant l a b e l
=min ( 2 , I
(1 ,7 ).
+ «=)
L in k L ( 6 , 7 )
l i n k and N31 N2 , N5 and Ng a r e th e n e ig h b o rin g nodes.
= 2
;
becomes a t r e e
Pushing on,
L j 2 = min ( L j 2 , L16 + dg2 ) = min ( 2 , I + 0) = I
L j 3 = min ( L j 3 , L16 + dg3 ) = min ( 2 , I +<*>) = 2
L j5 =
min ( L j 5 , L16 + dg5 ) = min
(5 , I
+ I )
= 2
L j g = min ( L j g , L16 + d6 g) = min
(2, I
+ ~)
= 2
The s m a lle s t i s L j 2 and i t
becomes a permanent la b e l
f i v e l i n k s o f th e t r e e w i t h fo u r l e f t t o go.
(1 ,6 ).
There a re
C o n tin u in g ,
L j 3 = min ( L j 3 , L12 + d2 3 ) = min
(2 , I
+ °°)
= 2
L j 5 = min ( L j 5 , L 12 + d2 5 ) = min
(2 , I
+ °°)
= 2
L j g «? min ( L j g , L12 + d 2g) = min
(2 , I
+ °?)
= 2
L j L 10 = min ( L j 9 l 0 , Lj2 + d 2 1 0 ) = min ( » , I + I ) = 2
41
Now, p ic k a t random one o f th e f o u r as a permanent l a b e l , say L j 39 ( 2 , 8 ) .
There a r e
th r e e
nodes l e f t ,
N3 , Ng and N ^ ,
L j 5 = min
( L j 6 , L ^3 + d3 5 )
L j g = min
( L j g , L13
hence;
= min ( 2 , 2 + »
+ d3 g ) = min ( 2 , 2 + ~ )
=2
=2
Ljjlo = mi" (Lj.io. L13 + (I3 j l 0 ) = mf" (2. 2 + I ) = 2
Once a g a i n , randomly choose one, say L j g .
th e re a re
seven
t r e e l i n k s w ith two t o go.
' L j 5 = min
( L j 5 , Li g
I t has a l a b e l
(2 ,8 ).
So f a r
Proceeding as b e fo re
+ dg 5 ) = min ( 2 , 2 + 4 )
=2
H .IO = m,n <Ll ', 1 0 \ L19 + d9,lo ) * m' n (2i 2 + ” ) = 2
L e t ' s p ic k node N15 t h i s tim e .
I t s permanent la b e l
have e i g h t t r e e l i n k s so we must proceed on u n t i l
is
(2 ,2 ).
We now
t h e r e a r e n in e .
L1 5 = m1n 1M s 1 Ll , 1 0 + dI O 1S 1 = min (2> 2 I " ” ) = 2
Node N5 becomes a node o f th e t r e e w i t h l a b e l
nine l i n k s
(N - I )
( 2, 6) .
B u t, t h e r e a r e now
o f th e t r e e , so th e a lg o r it h m t e r m i n a t e s .
Remember, we wanted t o f i n d th e minimum d is ta n c e from th e s t a r t i n g
node N1 to th e f i n a l
node Ni q .
Node 10 has permanent l a b e l
(2 ,2 ),
th e re ­
fo re
min I
Ll , 1 0
Using th e a lg o r it h m j u s t i l l u s t r a t e d w it h th e f l a t top pulse example
we now have a te c h n iq u e f o r f i n d i n g th e minimum e u c lid e a n w e ig h t f o r an
a r b i t r a r y L and f ( D ) p a i r .
T h is te c h n iq u e along w i t h e q u a tio n s (3 5 ) and
42
( 3 6 ) a r e implemented by an Extended F o r tr a n IV .program and used to
a n a ly z e th e performance o f th e maximum l i k e l i h o o d sequence r e c e i v e r f o r
s e v e ra l channel
impulse response examples.
r
V.
SEVERAL EXAMPLES
To use the program in i t s e n t i r e t y ,
i t w ith v a lu e s f o r L and f ( D ) .
it
i s necessary o n ly to supply
For exam ple, the f l a t top pulse f o r
L = 3 has a d i s c r e t e tim e response f (D) = I + D + D^.
needs as d a ta i s L =
Al I th e program
3 and th e c o e f f i c i e n t s o f f ( D ) , f 0 = I , f , = I and
f2 = I .
Having developed a program t h a t f i n d s th e minimum d is ta n c e through
th e e r r o r s t a t e f lo w g ra p h , l e t ' s
use i t t o a n a ly z e the performance o f
o u r r e c e i v e r s t r u c t u r e f o r s ev e ral examples o f channel
Example Two:
Q uarter-W ave Cosine Impulse Response
cos
h (t)
h (t) = {
Tit
T
0
L = 3
T = I
I.
Pulse A u t o c o r r e l a t io n Function
S o lv in g f o r C o e f f i c i e n t s :
R0
=
R1 - R-1 *
impulse response.
hi ( t ) d t =
0
cos ( ^ ) d t = 1 .5
0
' c o s ( ^ ) cos ^ ^ d t
= .867
0 < T < 3
elsewhere
**
I
R2 = R- 2 = /
tTtT
cos^
0
COS(^ Z J l) d t = .2 5 0
R(D) = .2 5 D "2 +. .8 6 7 D "1 + 1 .5 + 867D1 + ..25D"2
2.
D i s c r e t e Impulse Response
L e t t i n g f (D) = a + bD + cD2
Then:
R(D), = acD " 2 + (ab + be) D" 1 + ( a 2 + b2 + c 2 ) + (ab + be) D+1 + acD2
Equating C o e f f i c i e n t s :
.
.
Rri = a 2 + b2 + c 2 + 1 .5
0
R-| = R_-j = ab + be = .867
R2 = R-2 = ac = '2 5 0
I
.
a 8 - a 6 + .12 6 7a 4 -
A p o s itiv e real
.0 6 2 5 a 2 + .00 3 9 = 0
r o o t is
a = .9 6 5 0
Hence
: f (D) = .9 6 5 0 + .7083D1 + . 2 5 9 1 D2
45
Example T h re e:
S t r a i g h t Line Impulse Response
-T /3 + I
0 <_ t <_ 3
h (t)= {
h (t)
elsewhere
L = 3
T = I
S o lv in g as in Example One:
1.
Pulse A u t o c o r r e l a t i o n Function
R(D) = 1 .333D " 2 + 4 .6 7 D " 1 + 8 . 9 9 9 + 4.67D + 1 .3 3 3 D 2
2.
D i s c r e t e Impulse Response
f (D) = 2 .5 2 8 + 1.528D + .528D2
Example Four:
Decaying E xponential
Function
!.S e ' 1
h (t)
h(t) = {
0 < t < 3
46
I.
Pulse A u t o c o r r e l a t i o n Function
hQ( t )
= h(t)
0 I t I
= 0
elsew here
h (D ,t) = (I
R(D)
(I
(I
1
+ B- 1 D + e " 2D2 )h Q( t )
+ e - 1 D + e ” 2 D2 ) ( I
+ t f V 1 + e ' 2D ' 2 )h 2 ( t ) d t
+ e ' V 1 + e " 1D"2 ) ( l + C- 1 D + e " 2D2 )
f ( - D ) X f (D)
2.
D is c r e t e Impulse Response
f (D) = I . +
Example F i v e :
e " 1 D + e " 2 D2
Suppressed C a r r i e r A.M. M odula tion
0 I t I 3
elsewhere
I.
Pulse A u t o c o r r e l a t io n Function
R(D) = .02 7 D "2 -
.0 8 2 D '1 + .4 7 6 -
.082D1 + .027D2
47
2.
D is c r e t e Impulse Response
f (D) = .6 8 0 -
Example S i x :
1.
.1 1 4 0 + .040D2
Voice-G rade Telephone Channel
Pulse A u t o c o r r e l a t io n Function
R(D) = -.194D"2 + .157D "12 + 1 .5 1 8 + .1570 -
2.
[ 1 3 , 14]
.194D 2
D is c r e te Impulse Response
f (D) = 1 .2 1 2 5 + .1492D1 Examples Seven and E ig h t:
.1599D2
E x p o n e n tia ls f o r I > 3
48
3 /2 . e
I.
h(t)
-3 /4 t.
= {
O < t < 4
elsewhere
O
L = 4
I.
R(D)
1 .1 6 4 (1 .+ e ' 3 / 4 D + e ' 3/ 2D2 + e " 9/ 4D3 ) ( l
e ' 3/ 2D" 2 + e " 9/ V
2.
+ e ^ V 1..+
3)
f (D) = 1 .0 7 9 + 509D1 + .242D2 + .114D3
3 /2 e “ 3 t/5
II.
h ( t ) ={ 0
0 < t < 5
elsewhere
L = 5
1.
R(D) = 1 .3 1 3 (1
+ e ” 2/ 5D1 + e " 6/ 5D2 + e " 9/ 5D3 + e “ 12/ 5D4 ) ( l
. e ' 2/ 5D' 1 + e ' 6/ 5D' 2 + e " 9/ 5 D™
3 + e " 12/f5D ' 4 )
2.
f (D) = 1 .1 4 6 + .6 2 8 0 + .345D 2 + .189D3 + .104D 4
+
I MPULSE
KLSPONSE
-V- -V-
CLRAlI UN
L
Y
Y
WI TH
IMERSYMfiCL
Y
Y
interference
Y
interference
Y
Y
F(D)
Y
V
V-
3
3
vF O=
YF I =
YF2 =
V
Y
Y
Y
Y
Y
*
V
3
Y
Y
Y
Y
3
B
«9558
I -1537
.
Y
Y
I • 266C
Y
Y
Y
8.6665
Y
Y
Y
Y
»4 7 7 0
Y
Dy y 2
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
EX(D)
V
Y
»3332
Y
.1560
.0370
•
.CCCC
Y
'4770
Y
Y
Y
Y
Y
Y C 1-1
Y
C
Y
Y
I .1537
V
V
YK O=
* 6 8 COY
v F I = " • 11 A C Y
Y f 2
= »C4C0v
2 .CCCC
Y
Y
YFO= 1 » 0 0 CCY
Y Fl = »3 6 7 9 v
Y F 2 = »I 3 5 3 Y
Y
Y
Y
;v
Y
YFO= 2 » b 2 7 5 v
Y F l =1 * 5 2 7 5 Y
v F 2 «= . 5 2 7 5 V
Y
Y
Y Y Y Y V Y V V V V V V Y V V V V Y V V Y V V- V V V V V V Y V V V Y V V V1 V V Y V V V V
Y
1*5000
Y SEQUENCE
Dy v 2 * D I y y 2 Y
• OCGC
Y
Y C 1-1 C
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y CM C
Y
Y
O
<-«
Y
3 * OCOO
ERROR
Y
H
X-*
3
C I YY2
Y
Y
»5650Y
• 7 O8 3 Y
• 2591v
Y
difference
O
2* *
V K O =
Y
Y
Y
V
Y
Y
Dw 2
Y
Y
I * OOOOv
v F l =1 • OOCOv
YFS =1 ♦ OCOOv
Y
Y
S.
RESPONSE
NO
I M F R SY MBOL
Y
I NPUT
Y
FRACTI ONAL Y
Y
v -V -V -V- V >;■ V -Y- v- V- V V -Y V V- V V- V V V V V -V- V V V V Y Y W
Y
4«
Y
Y
Y
2•
I MPULSE
Y
•V-
I •
Y
Y
Y
- V -
ELCi ICF.AN WEI GHT
MI NI MUM
discrete
-V*
O
E
X Y
A *
r V
F- Y
i Y
*
Program Output fo r Examples I through 5
O
#
*-*
Table I .
•
V
-V
-V
V
4
V
-V
8 .
•V
V
-V
V
V
5
I NTERSYMBOL
V
interference
V
interference
V
with
>
V
D v v 2.
V
V V V V V V V Y Y Y V V V V
V V V V V V V V
V
V
1*5180
V V V v- V
e
.OCCC
1.4581
.OCCO
V
V
V
V
V
V
V
V
V
V
V
V
V
I «8679
V
V
V
V
V
I
»
7 C2 1
ERROR
V
V V V V V V V V V V V V V V V V V V -V V V V
V
V
V
V SEQUENCE
V Dv v S - C I v v E V
V
V
EX(D)
V
D v v2
V
I . Sl SC
V
1*4981
FRACTI ONAL
DI FFERENCE
V
V
V
VFO= I • I 4 5 0 v
v-F I = » 6 2 0 0 v
vF2 = * 3 4 5 0 v
vF3 = . 1 8 7 0 v
vF 4 = * I 0 5 0v
D I vv2 ^
V
V
V-FO= I . 0 8 0 0 v
v F I = .SlOOv
v F 2
=
.2420V
v F3 * « 1140v
*
V
V
V
V
V
V
CTl
O
V
V
V
V
V
V C l C C
V
O
V
V
3
V
NC
I NTERSYMBOL
V
vFO* 1 . 2 1 2 5 v
v FI = . 1492v
vF2 * “ • 1599v
V
V
V
• 0 8 8 8
V
V
V
V
V
V
V
O
6•
I NPUT
weight
O
V-
euclidean
V
A
v
RESPONSE v I MPULSE
V
Viv V
P 'Y- DURATI ON V RESPONSE v
V
V
L V
-v
F(D)
v
E •V
I
v
v
// *
-V- V- V -V- -V- -V -V
- -Y- V V V -V- -V- V V V V V V V V V V V
-Y-
7
MI NI MUM
discrete
V
K
b*
V
I MPULSE
O
L
X *
Program Output fo r Examples 6 through 8
»->
8
M-
Table 2.
51
Tables I and 2 d i s p l a y th e program r e s u l t s f o r the e i g h t channel
impulse response examples.
The f i r s t s i x examples a r e f o r impulse responses w i t h L = 3.
these s i x th e l a s t t h r e e g iv e ze ro f r a c t i o n a l
e u c lid e a n w e ig h t.
opes.
N o tic e , a l l
Of
d i f f e r e n c e in minimum
th r e e have e x p o n e n t ia l I y decaying e n v e l ­
Of th e s i x examples we would e x p e c t these l a s t t h r e e to g ive the
b e s t performance in terms o f s m a lle s t in c re a s e in e r r o r r a t e .
Since those impulse responses having decaying e x p o n e n tia l envelopes
gave the best r e s u l t s , where L = 3 ,
v a lu e s f o r L.
th e y were e v a lu a te d using l a r g e r
For L = 4 (example seven) the decaying e x p o n e n tia l giv es
ze ro f r a c t i o n a l d i f f e r e n c e , but f o r L = 5 th e minimum e u c lid e a n w eig h t
i s reduced by a lm ost 9 p e r c e n t due t o intersym bol
in te rfe re n c e .
V I.
CONCLUSION AND SUMMARY ,
A performance c r i t e r i o n and a s s o c ia te d a n a l y s is te c h n iq u es have been
in tro d u c e d f o r e v a l u a t i n g th e performance o f the maximum l i k e l i h o o d se­
quence e s t im a t o r o v er l i n e a r channels w it h known channel
se.
impulse respon.-.
Since th e minimum e u c lid e a n w eig h t o r d is ta n c e o f th e o u tp u t s ig n a l
space g iv e s th e r e c e i v e r s t r u c t u r e s margin a g a in s t e r r o r when th e r e is
a d d i t i v e w h ite gaussian n oise on th e c h a n n e l, the f r a c t i o n a l
d iffe re n c e :
in th e minimum w e ig h t g iv e s us a way o f comparing th e performance o f th e
s tru c tu re fo r d iffe re n t.c h a n n e ls .
When a p a r t i c u l a r channel giv es the
s m a lle s t f r a c t i o n a l d i f f e r e n c e we know t h a t i t g iv es th e s m a lle s t i n c r ­
ease in e r r o r r a t e .
For th e l i n e a r channels a n a ly z e d , t h i s channel
would p ro b a b ly produce th e most r e l i a b l e communication, system.
In a P .A.M . system, when d a ta i s t r a n s m it t e d over narrow bandwidth
high s ig n a l t o n oise r a t i o channel w i t h an
i n fo r m a tio n r a t e
l a r g e r than th e channel bandw idth, inte rsy m bol
(bauds)
i n t e r f e r e n c e may r e s u l t .
The maximum l i k e l i h o o d sequence e s t im a t o r was designed t o help a l l e v i a t e
th is
in te rfe re n c e .
The o b j e c t i v e o f t h i s t h e s is i s to develop a c r i t e r i o n and a s s o c ia te d
a n a l y s is te c hniques t o e v a l u a t e th e performance o f th e maximum l i k e l i h o o d
sequence e s t im a t o r f o r l i n e a r channels w it h known impulse responses. .
T h is r e c e i v e r s t r u c t u r e i s optimum in th e sense t h a t i t m inim izes the
p ro b a b ility o f e r r o r .
Since bounds on th e p r o b a b i l i t y o f e r r o r can be
found by knowing th e minimum e u c lid e a n w e ig h t o f th e o u tp u t s ig n a l space.
53
th e performance c r i t e r i o n used i s th e f r a c t i o n a l d i f f e r e n c e in minimum
e u c lid e a n w e ig h t o f th e o u tp u t s ig n a l
space.
D i s c r e t e t i m e , D -tr a n s fo r m and c h ip D -tra n s fo rm models o f th e r e c e i v e r
s t r u c t u r e a re developed.
From these d i s c r e t e tim e m odels, d e f i n i t i o n s
o f i n p u t , o u tp u t and s t a t e e r r o r eve nts a r e g iv e n .
These a r e used to
c o n s t r u c t a flo w graph r e p r e s e n tin g th e e u c lid e a n w eig h t d i s t r i b u t i o n
o f th e o u tp u t e r r o r e v e n ts .
T h is f l o w graph along w it h a s h o r te s t path
I
f i n d i n g a lg o r it h m i s implemented by an XDS Extended F o r tr a n IV program
and used t o f i n d th e minimum e u c lid e a n w e ig h t in th e presence o f
intersym bol
in te rfe re n c e .
The program uses t h i s minimum e u c lid e a n w e ig h t
along w i t h th e minimum w e ig h t w ith no intersym bol
c a l c u l a t e th e f r a c t i o n a l
d iffe re n c e .
analy ze d using th e program.
i n t e r f e r e n c e to
S everal channel examples are
APPENDIX
1'
?•
2•
4«
5.
5.
7.
E.
5.
IC .
11 •
12 •
12 •
14.
IS*
16 *
17 *
IS ­
IS*
SC •
SI *
22*
25 *
2H*
25*
26 *
27 *
2S*
22 *
3C *
31 *
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
f
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
¥
tf
* * * *
T H I S PROGRAM CALCULATES THE f I M r U M E u C L l C C A\ wE I C h T Cr THE OUTPUT
Si OLENCE CF THE v A X l r U r u I K E L l H C C C SEQUENCE ESTI MATOR EOF CHANNEL
I H P l LSE RESPONSES CAUSI NG I M E c SYr SOL I NT E RF E RE NCE *
I T ACCOMPLI SHES
T H ' S SY F I N D I N G THE SHCRTEST PATH THROUGH A FLCw CRARH REPRESENTI NG
THE EUCLI DEAN WEI GHT D I S T R I B U T I O N CF JhE OUTFLT ERROR EVENTS*
THE
PROGRAM ALSL CALCULATES Tr-E FRACTI ONAL DI FFERENCE I N Ml N l ML H EuCLl CEAN
U t I G H T DUE TC I N T E r SY^BOL I NTERFERENCE*
I n p u t s tu t h e p r o g r a m are
l the
input
s y m b o l SEPARATI ON AND T h E C O E F F I C I E N T S OF F ( C ) /
THE DI SCRETE TI ME
MODEL I MPULSE RESPONSE•
SEVERAL L a n d F ( D ) PAI RS CA n be i n p u t t e d a t
one T I ME *
EACH p a i r T a k e s T wo CARDS U l T h T h E FOLLOWI NG FORMAT:
I
:
l
CARD 2
:
F(J)
card
>
format(
12)
in
,
FORMAT(LF6*4)
DI ME N S I O N F ( CU I O ) / E X ( O : 41> > L 1 ( 8 5 ) / N O D E ( 8 5 )
DI ME N S I O N L l N K T ( 8 5 ) , L I N K P ( 8 5 I , O I S T I 8 5 , 3 )
DI ME NS I ON X ( I O ) , I ASTNODE I 8 5 ) , P T R N l 8 5 )
I NTEGER E X , X , u , w , P E R M , V , T E v P , M, PT R N , U l / Wl
REAL L I , L l R
L2 = 0
CNT = O
C
Jf
2?
C
* * *
RFAD L
*
*
*
25
24
35
SCO
I
36
27
38
35
4C
41
860
C
C
*
2
OOO
42
42
44
84C
48
45
SC
51
52
55
54
SE
56
57
L
MAKES NC SENSE
RFADF
*
*
IN
T HI S
PROGRAM, t !
*
READ I 1 1'5> 2 ) L / ( F I J * 1 ) / J = I > L )
FORMAT I N F G * 4)
in it ia l iz e
variables
*
*
*
NUN * 3 9 9 V ♦ I
I F I L ‘ FO • I I
GOTO 8 5 0
LE=3*9(L-2)
’ NASKl =C
N A SK 3 3
CO 3 I = I z V - I
N A S K= I SC I NASKz 2 J
NASKl = I CR(MASKzNASKl )
3
55
6C
61
62
62
CONTI NUE
" ASK=ISC(MASKz-29( V - I ) )
P ERN=I
TEMP=N=O
CO 5 I =I z NUMl
PTRNt I ) = L I N K T t I I = L I N K P t D = C
LltI 1=1000000
64
66
? *
9 9 *
56
65
R E A D t l C S z I z En D = I I O )
FORMAT( 1 2 )
I F ( L - I ) S40z S6C# 8 6 0
OUTPUT 1LCOK B U O z L K l
GOTO H O
V= L - I
6
5
CU 6 J = I Z 3
D I S T t I z J ) = 1 OOOOOO
CONTI NUE
CONTI NUE
in
Ct *
*
'67
65
65
7C
71
7c
73
L I ( I ) - L A S T N d O E l l >*0
IP (L •EG •L2 ) GOTO 503
2 CO
h O D E (J)=G
CONTINUE
o o o o o n
4
DC 2 0 0 A = 1 , V
X(A)=O
CO 4 J = I f N U r
f
CONSTRUCTION CF FLOW GRAPH
building
the
nodes
* * v
*
CXj
Il
I
sc.
s i.
Sc .
SG*
Si •
Sb .
86 •
S7 •
Sg .
85 •
9C •
SI •
Sr •
SG •
94 .
95.
sc.
37 .
SE •
SS .
* * $
15
11
7
8
5
IC
12
13
5 C3
C
I=I
I F ( X( I )- I I 7 ,8,9
X(I)=I
g o t o 10
X < I >=2
COTOl C
X(I)=O
I F l x ( I ) . NE * O ) GOTO 12
1 = 1+1
I F ( I . L E . V ) GOTOl l
DO 1 3 I = 1 > V
X ( I I = I S C I X I I I f E jM I - I ) )
NODE( J) = I O R I X I I J f N C D E ( J ) )
Xl I J = I S C I X I I ) f - 2 * ( I - I ) )
CONTI NUE
J=J+l _
I F ( j • L E • NUM ) GOTO 15
CONTI NUE
Ul
"4
I OC
ICl
I OE
I 03
104
I CS
C
C
*
2C
NODE I = NODE( I )
XODEI = I AND ( NODE I ^ nI ASKI )
NODEK = NODEt K I
NODEK=I AND( I S C t N C D E K , 2 ) , M A S K l )
I F ( \ C D E K • n E • NODE I ) K = K t l J G G T C E C
PTRNt I I =K
CO 16 J = C , V - l
MASK=I S C ( M A S K , 2 * J )
E X t J ) = I ANDt MASKi NODE( I ) )
MASK=I SC I v A S K , - E f J I
E X ( J ) = IS C ( E X t J ) , =2*J)
IF tE X (J ).£0.2) EX(J)=-I
CONTI NUE
MASK= I SCt MAS K, 2 * ( V - l ) )
E X t V I = I A N O t v ASK, NOCE( K) )
MASK=I SC( MASK, - E v ( V - I ) )
E X ( V l = I S C t E X ( V ) , - 2 v ( V - l ))
I F t E X ( V ) . E C . 2) E X ( V ) = - I
J =V
EYJ = O
DC I S H = O , J
EYJ = E Y J f E X I I I I v F I J - I I I
CONTI NUE
F = ( KfLE-PTRNl I ) ) / L E
DISTI!,PI=EYJvEYJ
I F ( I • EQ » I ) NNN = EJ GOTO 19
ICg
ICS
17
111
112
1 12
IG
121
122
1 22
124
1 25
126
127
I2S
I2 S
13
I 3C
131
132
132
¥
I=I
K= 2
ICG
114
HS
HG
117
HS
HS
I2C
THE NCtES AND SOL v In G F:s
21
107
lie
linking
N N N = 3
IS
I F ( P . L I * NNN) K = Kt L E J GOTO 17
I =It l
I F ( I * L E • N U M - I ) GOTO 21
(TI
03
134
136
I 36
137
135
135
I4 C
C
C
C
C
•C
C
V V V
v
I 4I
142
143
22
MI Sl MUM
Ft-Tu
F I NDE R
INITIAL STEP CF ALGORITHM,
v
v = P T RN ( I )
TEMP- M •
F=I
LI ( M ) = D I S T ( I i P )
1 4 4
L aSTnCDC(M)=I
1 45
M=MVLE
P = ( Mv l E - P T R nM I ) I / L E
I F I P i L E i ? ) L I N K T t M - L E ) * M< GGTC2 2
LINKT(M-LE)=O
FLAC=O
GOTO 65
146
147
145
145
I SC
151
152
155
154
155
156
157
152
155
I6C
161
162
163
164
165
166
167
162
ALGCRI ThM
Ul
v
ALL NEI GHBORI NG
NODES
ARE LABELED
(LtI)
v
- = n T R N ( PERM)
P= (MtLE-PTRNtPERMe) )/LE
IE <P •G T • 3) GOTO 45
C
^=TEMP
IF(AiEGiC)
IF(w.EG.M)
55
36
GOTO 30
G=PERM
IF(QiEGiO)
IFI-.EQiQ)
3
G0 T 0 3 5
COT 0 3 7
W=LINKT( U )
C=LINKP(Q)
37
GOTO 36
M=MvLE
GOTO 40
LI NKT(M)-TEMPiTEMF = MiLASTNCDE(M)=PERMj
GOTO 37
GOTO 37
VD
165.I 7C*
171.
172.
* 172.
174.
175.
1 76 .
177.
178.
1/5.
I SC •
ISl .
162 .
185 •
1 84 .
185.
186 •
187.
188.
185.
I 5 C•
191 .
195.
193.
1 94 .
195 .
1 96 .
1 97 .
196.
139 .
20C •
C
C
C
45
.SC
51
52
6C
55
*
*
* =TEMP
IF(WEO-O)
Ll =W
LSK = L S K '
FLAC = I
i
71
L=LINKTIUl)
I F ( L . F G . O ) GOTO 75
I F I L l ( U ) - L I (w)) 7 1 ,7 2 ,7 2
W= U
Al =Ll
cn
o
= MI N I
GOTO 85
Wl = W
7C
<
+ DPK)
Is= TEPP
I F ( k . E Q . C ) 3 0 TO 65
P = Wt L E - P T P N I PERN I
CT = LE
F=P/ DT
I F ( P * E G • I * I GOTO 51
I F l p . E G . 2 . ) GOTO 51
I F I P . EG • 3 . ) GOTO Si
CT= 1COOOOC•
GOTO 52
CT = C I S T I P E R t f , P )
L l R = L I I PERM) +DT
I F l L l I w) - L 1 R ) 5 5 # 5 5 , 6 0
Ll ( W) =LlR
LASTNCOE( W) =PERM
W= L I N K T I w )
GOTO SC
C
C
C
65
LIK «
• ■
» P
r IM
N (LLI IKK ' , L S P
LSK'
¥
72
75
85
u l= L l\K l(U l )
GOTO 70
I F ( W l . E G . W) T E r P = L I N K T ( h )
L l N K T ( Wl J = L I N K T U ' )
L I n k T( W) = O
L l N K f I W) =PERM
FERr =W
N= N + 1
I F l N . L T . N U M - I ) GCTC 80
CONTI NUE
C
C
C
*
W = L a S T n CDE
*
I
*
calculation
cf
Ex(D)
*
*
*
NUM)
O
Il
' )
201 .
202.
202 .
2 04 .
205 .
206 •
207 .
208 .
202.
2 1 C•
211 •
212.
212 •
214.
215.
216 .
217 .
218 •
212 .
220 .
221 .
222.
222 •
224 .
225 .
226.
227 .
228 .
222 .
230 .
2 31 •
232 .
620
600
610
"ASK=I S C ( MASK, 2 » ( V - I ) )
NOOE I = NGDE ( W)
NODEI = I SC ( I A N D t N O D E U MA S K ) i - 2 * ( V - D )
IF IN G D E l-I1610,610,600
NuCEI=-I
E X ( J J ) =NODEI
W=L
a
S T
n
ODC(W)
IF (W .NE.CI
C
C
C
*
850
800
JJ = J J t l ; GOTO 6 2 0
*
*
CALCULATI ON
CF FRACTI ONAL
DI FFERENCE
D M I n =C
DO RCC I H = O , V
DMl N = D M l N t F ( I H U F ( I H )
I F ( L - E O - I ) L I ( NUM) = DMI N
D I F F = ( D M I N - L l ( NUM) ) / D M l N
CALL WRI TESUS I F H < L I , N L M , DM I N , D I F F / J J , CNT / E X )
223 •
224 .
225 .
2 36 .
237 .
2 3%.
lit
I COc
• L2 = L
GOTO 5 0 0
WRITE!1C8.1000)
FORMAT ( ' I « )
STOP
END
SUBRCLT I NE . ' RI TESUB ( F , L > L I > NLiB , DM I N # D I FF # L J# C NT # E X )
DI MENSI ON L l i 4 C 5 ) > F ( C : i C ) / E X ( C : 5 C )
i nteger ex
real LI
DATA L I N E / ' * ' /
I F I C N T • E G * ! I CCTD 4 0 0
.RITE 1108/323)
323
FORMAT I ' I ' )
CLTPLT • ' / •
«> • i / '
I, • •>• I , I •, I •
.R ITE (108,203)
30C
FORM AT I 1 4 X , ' E ' , 3 X , ' I M P U L S E ' , 4 X , ' D I S C R E T E ' , 1 2 X , ' MI N I p l m e u c l i d e a n W
CEI Gh T ' , 1 3 X , ' I N P U T ' )
301
. R l TE I 1 0 8 , 3 0 2 )
302
FORMAT I 1 4 X , ' X ' , 1 X / ' * ' , 1 C X , ' * ' , 1 C X , ' * ' , 4 4 X , ' * ' )
* . R I TE I 1 0 2 , 3 0 3 )
m
303
FORMAT I 1 4 X , ' A S l X , ' * ' , I X , ' RE S P ON S E ' , I X , ' * ' , I X , ' I M P U L S E ' , 2 X , ' * ' , 7 ^
C, ' N O ' , I 3 X , ' W I T H ' , 7 X , ' F R A C T I O N A L ' , I X , ' * ' , 2 X , ' ERROR • )
.HITE I 108,304)
304
FORMAT I l 4 X , 'Ml S I X , ' * ' , 1 C X , ' * ' , 1 0 X , * * ' , 2 X , • I n TERS y M B 0 L ' , 2 x , » * S 2 X
C , ' I NTERSYMBOL S 2 X , ' * S I X , ' D I F F E R E N C E ' , I X , ' * • )
WRI TE 11 0 8 , 3 0 5 )
305
FORMAT I I 4 X , ' P S I X , « * S I X, «CUR A T I CN S I X , ' * S I X , ' RESPONSE S I X , • * S EX
c , ' i nterference • , i x , • * s ? x , ' i nterference •, i x , •* •, i s x , • * s i x , ' secuen
CCE' )
WRITE!1 0 8 , 3 0 6 )
306
307
4 CI
4 0*
FORMAT! 1 4 X , ' L S
0*2',IX ,'*')
WRI TE I 1 0 8 , 3 0 7 )
FORMAT! 1 4 X , ' E S
C, ' 0 * * 2 S 5X, ' * S
CI
WRI TE! 1 0 8 , 4 0 4 )
FORMAT 11 4 X , ' F S
C ,'*')
I X , ' * S 1 0 X , ' * S l C X , ' * S 1 5 X , ' * S i 5 x , ' * ' , I X , 'C**2-CI*
I X , ' * S 4 X , ' L S 5 X , ' * ' , 3 X , 'FID ) S 3 X , S S 6 X
5 X , ' DI * * 2 S 5 X , ' * S I X,
S I X , ' * S 2 X , • FX I C ) '
I X , ' * S I C X , ' * S I C X , ' * S 1 5 X, ' * S 1 5 X , ' * S 4 X , ' C * * 2 S 4 X
35
3c
37
3S
35
.
•
.
•
•
2 C5
4C•
41 •
4 CC
402
42«
42 .
4 C3
44 .
45 .
46 •
47 .
42 .
45.
bC •
4C5
4 Cb
LP- SC
WHI TE I ! O S , 3 0 9 ) I P , ( L I N E
F u R y A T ( I 3 X , NAl )
CNT-I
NI = 6
wRITE( 1 0 2 , 4 0 2 I
FOKyAT I I 6 X, ' * ' , 1 CX, • * ' ,
w R I T E ( 1 0 S , 4 0 3 ) M , L , F ( O)
FCRy ATl I P X , I P , ' . ' , I X , ' *
C, ' * ' , 4 X , F 6 * 4 , 5 X , ' * ' , PX,
CG 4 0 5 J - l , L - I
LRI TE ( l f ; £ , 406 ) J , F ( J )
F O R y A T d O X , ' * ' , I OX , ' C F '
NI * M + 1
RETURN
END
>LP )
• )
- J ) ; J=C,JJI
4X,F6,4,5X
',IPX,'*')
CTl
4^
/■» * -
.HtiTtiaa.
--it •
65
REFERENCES
[1 ]
Mischa S c h w a rtz, In fo rm a tio n T ra n s m is s io n , M o d u la tio n and N o is e .
New Y ork:
M c G ra w -H ill, 1970* Ch. 2-3.
[2 ]
G. David Forn ey, "M axim um -Likelihood Sequence E s tim a tio n o f D i g i t a l
Sequences in th e Presences o f In tersym b o l In t e r f e r e n c e ,"
IEEE
T ra n s. In fo rm . T h e o ry, Vol I T - 1 8 , pp. 3 6 3 -3 7 8 , May 1972.
[3 ]
R. W. Lucky, J . S a lz , and E. J . Weldon, O r . , P r in c ip le s o f Data
Communication.
New Y ork: M c G ra w -H ill, 1 9 6 8 , Ch 5 .
[4 ]
Donald W. T u f t s , " N y q u is t's Problem — The J o in t O p tim iz a tio n o f Trans
m i t t e r and R e c e iv e r in Pulse A m plitude M o d u la tio n ," P ro c . IEEE,
V o l. 5 3 , pp. 2 4 8 -2 5 9 , Mar. 1965.
[5 ]
C. W. H e !s tro m , S t a t i s t i c a l
Pergamon, 1960, S e c t. I V . 5.
[ 6]
R. W. Chang and R. C. Hancock, "On R e c e iv e r S tru c tu re s f o r Channels
Having Memroy," IEEE T ra n s . In fo rm . T h e o ry, V o l. I T - 1 2 , pp. 4 6 3 4 6 8 , O ct. 1966.
[7 ]
Andrew 0 . V i t e r b i , "C o n v o lu tio n a l Codes and t h e j r Performance in
Communication System s," IEEE T ra n s . Commun. T e ch n o !. , V o l. COM-19,
O ct. 1 9 7 1 ^ pp. 7 4 1 -7 7 1 .
[ 8}
0 . M. W ozG ncraft and I . M. Jacobs, P r in c ip le s o f Communication
E n g in e e rin g .
New Y ork: W ile y , 1965.
[9 ]
D. J . S a k fis o n , Communication T h e o ry: Transm ission o f Waveforms
and D i g i t a l In fo r m a tio n .
New Y o rk:
John W ile y and Sons, I n c . ,
1 9 6 8 , Ch. 7 .
Theory o f S ignal
D e te c tio n . New Y ork:
[ 1 0 ] Jim K. Omuira, "O ptim al R e c e iv e r Design f o r C o n v o lu tio n a l Codes and
Channels w ith Memory V ia C ontrol T h e o re tic a l C oncepts," In fo rm a tio n
S c ie n c e s , V o l. 3 , 1 9 7 1 , pp. 2 4 3 -2 6 6 .
[1 1 ] T . C. Hu, In te g e r Programming and Network Flow s.,
Add 1son-W esley P u b lis h in g Company, 1 9 6 9 , Ch. 10.
R eading , Ma s s . :
[1 2 ] E. W. D i j k s t r a , "A Note on Two Problems in Connection w ith G raphs,"
Num. M a th ., V o l. I , 1 9 5 9 , pp. 2 6 9 -2 7 1 .
[ 1 3 ] J . R. Davey, "Modems," P ro c . o f th e IE E E , V o l. 6 0 , No. 1 1 , Nov. 1 97 2 ,
pp. 1 2 8 9 -1 2 9 0 .
66
[ M ] . F. P. D u ffy and I . .W. T h a tc h e r, J r . , "Analog T ransm ission P e r fo r ­
mance on th e Sw itched Telecom m unications N e tw o rk ,". B e ll System
T e ch n ic a l J o u r n a l, V o i. 5 0 , No. 4 , A p r i l , 1971, pp. .1 3 1 1 -1 3 9 7 ,
MONTANA STATE UNIVERSITY LIBRARIES
0013707 2
1762
I
•
- F533
cop.2
•
Fisher, Harold F
Performance analysis
of the maximum likeli­
hood sequence
estimator
mams and aooaksk
^
A
UM
M
vQrd.-t<.v ^vw ,
/I
^
7 /
/- _>
/-Xvfi.
College
Floce
Iindery