chapter 5 signal-space analysisshannon.cm.nctu.edu.tw/comtheory/chap5.pdf5.2 geometric...
TRANSCRIPT
Chapter 5 Signal-Space Analysis
Signal space analysis provides a mathematically elegant and highly insightful tool for the study of data transmission.
© Po-Ning [email protected] Chapter 5-2
5.1 Introduction
o Statistical model for a genetic digital communication systemn Message source: A priori probabilities for information
source
n Transmitter: The transmitter takes the message source output mi and (en-)codes it into a distinct signal si(t) suitable for transmission over the channel. So:
MimPp ii ,...,2,1for )( ==
MitsPmPp iii ,...,2,1for ))(()( ===
© Po-Ning [email protected] Chapter 5-3
5.1 Introduction
o si(t) must be a real-valued energy signal (i.e., a signal with finite energy) with duration T.
n Channel: The channel is assumed linear and with a bandwidth wide enough to pass si(t) with no distortion.o A zero-mean additive white Gaussian noise (AWGN)
is also assumed to facilitate the analysis.
.)(0
2 ¥<= òT
ii dttsE
o We can simplify the previous system block diagram to:
o Upon the reception of x(t) with a duration of T, the receiver makes the best estimate of mi. (We haven’t defined what “the best” means.)
© Po-Ning [email protected] Chapter 5-4
5.1 A Mathematical Model
© Po-Ning [email protected] Chapter 5-5
5.1 Criterion for the “Best” Decision
o Best = Minimization of the average probability of symbol error.
n It is optimum in the minimum-probability-of-error sense.n Based on this criterion, we begin to design the receiver
that can give the best decision.
å=
¹×=M
iiiie mmmPpP
1
)|ˆ(
© Po-Ning [email protected] Chapter 5-6
5.2 Geometric Representation of Signals
o Signal space conceptn Vectorization of the (discrete or continuous) signals
removes the redundancy in signals, and provides a compact representation for them.
n Determination of the vectorization basiso Gram-Schmidt orthogonalization procedure
© Po-Ning [email protected] Chapter 5-7
5.2 Gram-Schmidt Orthogonalization Procedure
Given !v1,!v2,…, !vk, how to find an orthonormal basis for them?
(step i) Let !u1 =!v1
|| !v1 ||.
(step ii) !u2' =!v2 − (!v2 ⋅
!u1)!u1. Set !u2 =!u2
'
|| !u2' ||
.
(step iii) For i = 3, 4,..., Let !ui
' =!vi − (!vi ⋅
!ui−1)!ui−1 − (!vi ⋅!ui−2 )!ui−2 −"− (!vi ⋅
!u1)!u1.
Set !ui =!ui
'
|| !ui' ||
.
(step iv) Then !u1,!u2,", !uk( ) forms an orthonormal basis.
© Po-Ning [email protected] Chapter 5-8
© Po-Ning [email protected] Chapter 5-9
(viii) Cauchy Schwartz inequality : |!v1 ⋅
!v2 |≤|| !v1 || ⋅ || !v2 || with equality holding if !v1 = a
!v2.(xi) norm square : || !v1 +
!v2 ||2=|| !v1 ||2 + || !v2 ||2 +2!v1 ⋅!v2.
(x) Pythagorean property : If orthogonal, || !v1 +
!v2 ||2=|| !v1 ||2 + || !v2 ||2 .(xi) Matrix transformation w.r.t. matrix A : !v1 = A
!v2.(xii) eigenvalues w.r.t. matrix A: solution λ of det A-λI[ ] = 0.(xiii) eigenvectors w.r.t. eigenvalue λ : solution !v of A!v = λ!v.
© Po-Ning [email protected] Chapter 5-10
5.2 Signal Space Concept for Continuous Functions
Properties for continuous functions(i) (complex valued) signal : z(t)
(ii) inner product : < z(t), z(t)>= z(t)z*(t)dta
b∫ .
(iii) orthogonal, if inner product = 0.
(iv) norm : z(t) = | z(t) |2 dta
b∫
(v) orthonormal, if inner product = 0, and indivual norm =1.(vi) linearly independent vectors, if none can be represented as a linear combination of others.(vii) triangle inequality : z(t)+ z(t) ≤|| z(t) ||+ || z(t) || .
© Po-Ning [email protected] Chapter 5-11
© Po-Ning [email protected] Chapter 5-12
© Po-Ning [email protected] Chapter 5-13
o Problem : How to minimize the “energy” of ? )(ˆ)()( tstste -=
Q.E.D.
© Po-Ning [email protected] Chapter 5-14
)(ts
)(1 ty
)(2 ty
)(ˆ ts)(),( 22 ttsa y=
)(),( 11 ttsa y=)(te
o Interpretationn aj is the projection of s(t) onto the Yj(t)-axis.n |aj|2 is the energy-projection of s(t) onto the Yj(t)-axis.
© Po-Ning [email protected] Chapter 5-15
)(ts
)(1 ty
)(2 ty
)(ˆ ts)(),( 22 ttsa y=
)(),( 11 ttsa y=)(te
Slide 5-13 also yields:
© Po-Ning [email protected] Chapter 5-16
5.2 Signal Space Concept for Continuous Functions
o For simplicity, we now focus on real-valued functions.o Completeness
n If every finite energy signal s(t) satisfies
n Example. Fourier series
is a complete orthonormal set.
complete for signals defined over [0,T).
© Po-Ning [email protected] Chapter 5-17
5.2 Geometric Representation of Signals
MiTttfstsN
jjiji ,...,2,1 ,0 ,)()(
1
=<£=å=
NjMi
dttftssT
jiij
,...,2,1,,...2,1
,)()(0
==
= ò
lorthonorma }{ 1Niif =
© Po-Ning [email protected] Chapter 5-18
5.2 Geometric Representation of Signals
o Through the signal space concept, si(t) (where 1 £ i £ M) can be unambiguously represented by an N-dimensional signal vector (si1, si2,…, siN) over an N-dimensional signal space.
o The design of transmitters becomes the selection of Mpoints over the signal space, and the receivers make a guess about which of the M points was transmitted.
o In the N-dimensional signal space,n length square of the vector = energy of the signaln angle between vectors = energy correlation between
signalsåò=
==N
jij
T
ii sdttsts1
2
0
22 )(||)(||||)(||||)(||
)(),()cos(
tstststs
ki
kiik ×=q
© Po-Ning [email protected] Chapter 5-19
5.2 Geometric Representation of Signals
n The length square of a vector and the angle between vectors are independent of the basis used (Note that no translation of the origin is allowed).
o From this view, n the transmitter may be viewed as a synthesizer, which synthesizes the transmitted signal by a bank of Nmultipliers.
n the receiver may be viewed as an analyzer, which correlates (product-integrate) the common input into individual informational signal.
© Po-Ning [email protected] Chapter 5-20
5.2 Geometric Representation of Energy Signals
o Illustration the geometric representation of signals for the case of N = 2 and M = 3
© Po-Ning [email protected] Chapter 5-21
5.2 Euclidean Distance
o After vectorization, we can calculate the Euclidean distancebetween two signals, which is the squared root of:
åò=
-=-=-N
jkjijki
T
ki sststsdttsts1
22
0
2 )(||)()(||))()((
© Po-Ning [email protected] Chapter 5-22
Example 5.1 Schwarz Inequality
o Cauchy-Schwarz inequality and angle between signalsn Cauchy-Schwarz inequality said that
n Also, the angle between signals gives that
n Hence, Cauchy-Schwarz inequality can be equivalently stated as:
s1(t), s2 (t)2≤|| s1(t) ||2 ⋅ || s2 (t) ||2 with equality holding if s1(t) = cs2 (t).
||)(||||)(||)(),(
)cos(21
2112 tsts
tsts×
=q
| cos(θ12 ) |2≤1 with equality holding if θ12 = 0 or π
© Po-Ning [email protected] Chapter 5-23
5.2 Basis
o The (complete) orthonormal basis for a signal space is not unique!n So, the synthesizer and the analyzer for the transmission
of the same informational messages are not unique!o One way to determine a set of orthonormal basis is the
Gram-Schmidt orthogonalization procedure.
o Try and practice Example 5.2 yourself!
© Po-Ning [email protected] Chapter 5-24
5.3 Conversion of the Continuous AWGN Channel to a Vector Channel
o Influence of the AWGN noise to the signal space concept
where w(t) is zero-mean AWGN with PSD N0/2.o After the correlator at the receiver, we obtain:
n Equivalently,n Notably, there is no “information loss”
by the signal space representation.
)()()( twtstx i +=
.jijj wsx +=
)(),()(),()(),( tftwtftstftx jjij += )(tx
)(tfN
)(1 tf
Nx
××,
××,
1x
© Po-Ning [email protected] Chapter 5-25
5.3 Conversion of the Continuous AWGN Channel to a Vector Channel
o Statistics of {wj}
o Since {sij} is deterministic, the distribution of x is a mean-shift of the distribution of w.
o Observe that w is Gaussian distributed because w(t) is AWGN. The distribution of w can, therefore, be determined by its mean vector and covariance matrix.
úúú
û
ù
êêê
ë
é+úúú
û
ù
êêê
ë
é=
úúú
û
ù
êêê
ë
é
NiN
i
N w
w
s
s
x
x!!!111
© Po-Ning [email protected] Chapter 5-26
5.3 Conversion of the Continuous AWGN Channel to a Vector Channel
o Mean
o Covariance
[ ] 0)()]([)()(][00
=== òòT
j
T
jj dttftwEdttftwEwE
( )( )[ ]
ij
T
ji
T T
ji
T T
ji
T
j
T
iji
NdttftfN
dsdttfsftsN
dsdttfsftwswE
dttftwdssfswEwwE
d
d
2)()(
2
)()()(2
)()()]()([
)()()()(][
0
0
0
0 0
0
0 0
00
==
-=
=
=
ò
ò ò
ò òòò
© Po-Ning [email protected] Chapter 5-27
5.3 Conversion of the Continuous AWGN Channel to a Vector Channel
o As a result, [w1, w2, …, wN] are zero-mean i.i.d. Gaussian distributed with variance N0/2.
o This shows that x is independent Gaussian distributed with common variance N0/2 and mean vector si = [si1, si2, …, siN]. Equivalently,
Õ=
úû
ùêë
é--=
N
jijji sx
NNf
1
2
00
)(1
exp1
)|(p
sx
© Po-Ning [email protected] Chapter 5-28
5.3 Conversion of the Continuous AWGN Channel to a Vector Channel
o Remainding term in noisen It is possible that
n However, it can be shown that (as an error term) w’(t) is orthogonal to si(t) for 1 £ i £ M. Hence, w’(t) will not affect the decision error rate on message i.
0)()()('1
¹×-= å=
N
iii tfwtwtw
1.y probabilit with0)(),(' =tstw i
o An equivalent signal-space channel model
o The best decision function d( ) that minimizes the decision error is:
n This is the maximum a posteriori probability (MAP) decision rule.
© Po-Ning [email protected] Chapter 5-29
5.4 Likelihood Functions
}|{maxarg1 allfor }|{}|{ if ,)(
},...,{ 1
xxxx
mPMkmPmPmd
Mmmm
kii
Î=
££³=
© Po-Ning [email protected] Chapter 5-30
5.4 Likelihood Functionso With equal prior probabilities, d(x) = arg max
m∈{m1,...,mM }P{m | x}
= argmax P{m1 | x},P{m2 | x},...,P{mM | x}{ }
= argmax P{m1 | x} f (x)P(m1)
, P{m2 | x} f (x)P(m2 )
,..., P{mM | x} f (x)P(mM )
⎧⎨⎩
⎫⎬⎭
= argmax P{m1 | x} f (x)1 /M
, P{m2 | x} f (x)1 /M
,..., P{mM | x} f (x)1 /M
⎧⎨⎩
⎫⎬⎭
= argmax f (x |m1), f (x |m2 ),..., f (x |mM ){ }f(x|mi) is named the likelihood function given mi is transmitted. Hence, the above rule is named the maximum-likelihood (ML) decision rule.
© Po-Ning [email protected] Chapter 5-31
5.4 Likelihood Functions
o MAP rule = ML rule, if equal prior probability is assumed.o In practice, it is more convenient to work on the log-
likelihood functions, defined by
o Why log-likelihood functions are more convenient? The decision function becomes “sum of (squared) Euclidean distances” in AWGN channel.
{ }{ })|(log),...,|(log),|(logmaxarg
)|(),...,|(),|(maxarg)(
21
21
M
M
mfmfmfmfmfmfd
xxxxxxx
==
d(x) = argmax1≤i≤M
log f (x |mi ) = argmax1≤i≤M
log f (x | si )
= argmax1≤i≤M
log 1πN0
exp −1N0
(x j − sij )2
⎡
⎣⎢
⎤
⎦⎥
j=1
N
∏
= argmax1≤i≤M
−12
logπN0 −1N0
(x j − sij )2
⎛
⎝⎜
⎞
⎠⎟
j=1
N
∑
= argmin1≤i≤M
(x j − sij )2
j=1
N
∑
= argmin1≤i≤M
|| x − si ||2 ( = argmin1≤i≤M
|| x − si ||)
© Po-Ning [email protected] Chapter 5-32
Upon the reception of received signal point x, find the signal point si that is closest in Euclidean distance to x.
© Po-Ning [email protected] Chapter 5-33
5.5 Coherent Detection of Signals in Noise: Maximum Likelihood Decoding
decision region for s1
decision region for s2
decision region for s3
decision region for s4
o Signal constellationn The set of M signal points in the signal space
o Example. Signal constellation for 2B1Q code
© Po-Ning [email protected] Chapter 5-34
5.5 Coherent Detection of Signals in Noise: Maximum Likelihood Decoding
o Decision regions for N = 2 and M = 4
© Po-Ning [email protected] Chapter 5-35
5.5 Coherent Detection of Signals in Noise: Maximum Likelihood Decodingo Usually, s1, s2, …, sM are named the message points.o The received signal point x wanders about the transmitted
message point in a Gaussian-distributed random fashion.
© Po-Ning [email protected] Chapter 5-36
5.5 Coherent Detection of Signals in Noise: Maximum Likelihood Decoding
o Constant-energy signal constellationn In this case, the ML decision rule can be reduced to an
inner-product.
( )( )
constant. is if ,,maxarg
,2minarg
||||,2||||minarg
||||minarg)(
1
1
22
1
2
1
iiMi
iiMi
iiMi
iMi
E
E
d
sx
sx
ssxx
sxx
££
££
££
££
=
+-=
+-=
-=
© Po-Ning [email protected] Chapter 5-37
5.6 Correlation Receiver
o If signals do not have equal energy, we can use
to implement the ML rule.n The receiver is coherent because the receiver requires to
be in perfect synchronization with the transmitter (more specifically, the integration must begin at exactly the right time instance).
.21,maxarg)(
1÷øö
çèæ -=
££ iiMiEd sxx
demodulator or detector
decision maker
© Po-Ning [email protected] Chapter 5-38
N product-integrators or
correlators
Correlation receiver
1x
2x
Nx
matched filter decision maker
© Po-Ning [email protected] Chapter 5-39
N product-integrators or
correlators
1x
2x
Nx
© Po-Ning [email protected] Chapter 5-40
5.6 Equivalence of Correlation and Matched Filter Receivers
o The correlator and matched filter can be made equivalent.o Specifically,
© Po-Ning [email protected] Chapter 5-41
5.7 Probability of Symbol Error
o Average probability of symbol error
{ }åò
å
å
å
=
=¹££
=
=
-=
-£--=
=-=
=-=-=
M
i Zi
M
iijijMji
M
iii
M
iiiice
i
dfM
mM
mmdPM
mmdPmPPP
1
1
2
,1
2
1
1
)|(11
ed transmitt||||min|||| Pr11
ed) transmitt|)((11
ed) transmitt|)(()(11
xsx
sxsx
x
x
{ }.||||min||:|| where 2
,1
2jijMji
NiZ sxsxx -£-ÂÎ=
¹££
© Po-Ning [email protected] Chapter 5-42
5.7 Invariance of Probability of Symbol Error
o Probability of symbol error is invariant with respect to basis change (i.e., rotation and translation of the signal space).
o Specifically, the symbol error rate (SER) only depends on the relative “Euclidean distances” between the message points.
{ }å=
¹££-£--=
M
iijijMjie m
MP
1
2
,1
2 ed transmitt||||min|||| Pr11 sxsx
© Po-Ning [email protected] Chapter 5-43
5.7 Invariance of Probability of Symbol Error
o Specifically, if Q is a reversible transform (matrix), such as rotation, then
{ }{ }2
,1
2
2
,1
2
||||min||:||
||||min||:||
jijMjiN
jijMjiN
sxsxx
sxsxx
QQQQ -£-ÂÎ=
-£-ÂÎ
¹££
¹££
© Po-Ning [email protected] Chapter 5-44
5.7 Invariance of Probability of Symbol Error
o A pair of signal constellation for illustrating the principle of rotational invariance.
© Po-Ning [email protected] Chapter 5-45
5.7 Invariance of Probability of Symbol Error
o The invariance in SER for translation can be likewise proved.
n Is the transmission power the same for both constellations?
© Po-Ning [email protected] Chapter 5-46
5.7 Minimum Energy Signals
o Since SER is invariant to rotation and translation, we may rotate and translate the signal constellation to minimize the transmission power without affecting SER.
n Since Q does not change the norm (i.e., transmission power), we only need to determine the right a.
å=
=M
iiig pE
1
2|||| s
minimized. is ||)(||),( that such and Find1
2å=
-=M
iiig pE asaa QQQ
© Po-Ning [email protected] Chapter 5-47
5.7 Minimum Energy Signals
o Determine the optimal a.
( )
2
11
2
1
22
1
2
||||2||||
||||2||||
||||)(
asas
asas
asa
+÷øö
çèæ-=
+-=
-=
åå
å
å
==
=
=
M
iii
TM
iii
M
ii
Tii
M
iiig
pp
p
pE
å=
=ÞM
iiip
1optimal sa
2
11
2optimal ||||)( and åå
==
-=M
iii
M
iiig ppE ssa
© Po-Ning [email protected] Chapter 5-48
5.7 Minimum Energy Signals
o So, subfigure (a) below has minimum average energy.
© Po-Ning [email protected] Chapter 5-49
5.7 Union Bound on Probability of Error
o Union bound
{ }( )! ( )
( )! ( )
{ }å å
å
åå
å
= ¹=
=¹
=¹
=
=¹££
->-£
ïþ
ïýü
ïî
ïíì
->-Ú
Ú->-=
ïþ
ïýü
ïî
ïíì
-£-Ù
Ù-£--÷øö
çèæ=
-£--=
M
i
M
ijjiji
M
ii
Miij
i
M
ii
Miij
iM
i
M
iijijMjie
mM
mM
mMM
mM
P
1 ,1
22
1
22
21
2
1
22
21
2
1
1
2
,1
2
ed transmitt |||||||| Pr1
ed transmitt ||||||||||||||||
Pr1
ed transmitt ||||||||||||||||
Pr111
ed transmitt||||min|||| Pr11
sxsx
sxsx
sxsx
sxsx
sxsx
sxsx
"
"
"
"
)()()( BPAPBAP +£!
© Po-Ning [email protected] Chapter 5-50
5.7 Union Bound on Probability of Error
( )( )( )ïþ
ïý
ü
ïî
ïí
ì
->-Ú
->-Ú
->-
24
21
23
21
22
21
||||||||||||||||||||||||
sxsx
sxsx
sxsx
{ }222
1 |||||||| sxsx ->- { }232
1 |||||||| sxsx ->- { }242
1 |||||||| sxsx ->-
© Po-Ning [email protected] Chapter 5-51
5.7 Union Bound on Probability of Error
å å= ¹=
£M
i
M
ijjjie P
MP
1 ,12 ),(1 ss
{ }. mean withddistribute Gaussian is ed, transmitt given Notably,
.ed transmitt |||||||| Pr),( where 222
ii
ijiji
m
mP
sx
sxsxss ->-=
is js w
ji ss -21
© Po-Ning [email protected] Chapter 5-52
5.7 Union Bound on Probability of Error
Since x = si + w when si was transmitted, we have
Pr��x � si�2 > �x � sj�2
��si transmitted�
= Pr��(si + w) � si�2 > �(si + w) � sj�2
��si transmitted�
= Pr��w�2 > �w + (si � sj)�2
��si transmitted�
= Pr��w�2 > �w�2 + �si � sj�2 + 2(si � sj)
T w��si transmitted
�
= Pr
�(si � sj)
T w < �1
2�si � sj�2
����si transmitted
�
= Pr
�n < �1
2�si � sj�2
����si transmitted
�
where n � (si � sj)T w.
Chapter 5-53
5.7 Union Bound on Probability of ErrorObserve that w is zero-mean Gaussian distributedwith covariance matrix E[wwT ] = N0
2 I, where I is the identity matrix.
Hence, n � (si � sj)T w is Gaussian distributed with
E[n] = E�(si � sj)
T w�
= (si � sj)T E[w] = (si � sj)
T 0 = 0
This implies that w � n/�si � sj� is Gaussian distributedwith mean zero and variance N0/2.
E[n2] = E�(si � sj)
T w · ((si � sj)T w)T
�
= E�(si � sj)
T wwT (si � sj)�
= (si � sj)T E
�wwT
�(si � sj)
=N0
2(si � sj)
T I(si � sj)
=N0
2�si � sj�2.
© Po-Ning [email protected] Chapter 5-54
5.7 Union Bound on Probability of ErrorAs a result,
Pr
�n < �1
2�si � sj�2
����si transmitted
�
= Pr
��si � sj�w < �1
2�si � sj�2
����si transmitted
�
= Pr
�w < �1
2�si � sj�
����si transmitted
�
= Pr
�w >
1
2�si � sj�
����si transmitted
�,
where the last equality is valid because the probability density function of azero-mean Gaussian random variable is symmetric with respect to w = 0 (hence,Pr[w > a] = Pr[w < �a] for any a > 0).
© Po-Ning [email protected] Chapter 5-55
5.7 Union Bound on Probability of Error
o Hence,n when N = 1,
{ }
.)exp(2erfc(u) where,2
erfc21
|| where,exp1
||21
Pr
ed transmitt |||||||| Pr),(
2
0
2/0
2
0
222
ò
ò
¥
¥
-=÷÷ø
öççè
æ=
-=÷÷ø
öççè
æ-=
þýü
îíì ->=
->-=
u
ij
jiijd
ji
ijiji
dzzNd
ssddvNv
N
ssw
mP
ij
p
p
sxsxss
n For N = 2,
n The same formula is valid for any N.
© Po-Ning [email protected] Chapter 5-56
5.7 Union Bound on Probability of Error
w
ji ss -21
P2 (si, s j ) = Pr || x − si ||2>|| x − s j ||2 mi transmitted{ }
= Pr w1 >12dij and w2 = don't care
⎧⎨⎩
⎫⎬⎭
, where dij =|| si − s j ||
=1πN0
exp −v2
N0
⎛
⎝⎜
⎞
⎠⎟dv
dij /2
∞
∫
=12
erfcdij
2 N0
⎛
⎝⎜⎜
⎞
⎠⎟⎟, where erfc(u) = 2
πexp(−z2 )dz
u
∞
∫ .
© Po-Ning [email protected] Chapter 5-57
5.7 Union Bound on Probability of Error
o Consequently, the union bound for symbol error rate is:
o The above bound can be further simplified when additional condition is given.n For example, if the signal constellation is circularly
symmetric in the sense that “{di1, di2, …, diM} is a permutation of {dk1, dk2, …, dkM} for i ¹ k,” then
å åå å= ¹== ¹=
÷÷ø
öççè
æ=£
M
i
M
ijj
ijM
i
M
ijjjie N
dM
PM
P1 ,1 01 ,1
2 2erfc211),(1 ss
å¹=
÷÷ø
öççè
æ£
M
ijj
ije N
dP
,1 02erfc21
© Po-Ning [email protected] Chapter 5-58
5.7 Union Bound on Probability of Error
o Another simplification of union boundn Define the minimum distance of a signal constellation
as:
Then, by the strict decreasing property of erfc function,
ijjiMjMidd
¹££££=
,1,1min min
÷÷ø
öççè
æ£÷
÷ø
öççè
æ
0
min
0 2erfc
2erfc
Nd
Ndij
÷÷ø
öççè
æ-=÷
÷ø
öççè
æ£÷
÷ø
öççè
æ£Þ å åå å
= ¹== ¹= 0
min
1 ,1 0
min
1 ,1 0 2erfc
21
2erfc211
2erfc211
NdM
Nd
MNd
MP
M
i
M
ijj
M
i
M
ijj
ije
© Po-Ning [email protected] Chapter 5-59
5.7 Union Bound on Probability of Error
o We may use the bound for erfc function to realize the relation between SER and dmin.
n Conclusion: SER decreases exponentially as the squared minimum distance grows.
608131.0for )exp()(erfc2
>-
£ uuup
.47929.1 if ,4
exp2
12
erfc2
10
2min
0
2min
0
min NdNdM
NdMPe >÷÷
ø
öççè
æ-
-£÷
÷ø
öççè
æ-£Þ
p
© Po-Ning [email protected] Chapter 5-60
5.7 Relation between BER and SER
o The information bits are transmitted in group of log2M bits to form an M-ary symbol.
o This gives the result that a large symbol error rate (SER) may not cause a large bit error rate (BER).n For example, a symbol error (for large M) may be due to
only 1 bit error.n Optimistically, if every symbol error is due to a single bit
error, then (assuming that n symbols are transmitted)
.)(log
SER)(log
SER BER22 MMn
n=
××
= ÷÷ø
öççè
æ³ .
)(logSER
BERgeneral, In2 M
© Po-Ning [email protected] Chapter 5-61
5.7 Relation between BER and SER
n Pessimistically, if every symbol error causes log2M bit errors, then (assuming that n symbols are transmitted)
n Summary:
.SERlog
SERlog BER2
2 =×
××=
MnMn ( ).SER BERgeneral, In £
SER BERlogSER
2
££M
© Po-Ning [email protected] Chapter 5-62
5.7 Relation between BER and SER
o If the statistics for “number of bit error patterns that causes one symbol error” is known, we can determine the exact relation between BER and SER.
Mn
PnM
jjj
2
1
1
log
)()(#SER BER
×
×××=
å-
=
bb
where #(bj ) = number of 1's in bj, and bj represents a binary permutation of log2M bit pattern.
Here, a 1’s in bj means a bit error occurs in the corresponding position; hence, the all-zero pattern is excluded because it represents no symbol error.
© Po-Ning [email protected] Chapter 5-63
5.7 Relation between BER and SER
o Example. If all bit error patterns (including no error pattern) are equally likely, then