1
Optimum Receiver
Signal Alphabet
Transmitter ReceiverChannel
MMMMM mpmpmptststs ,,,,...,, 2121 P
• Set of signals used by transmitter and receiver is a set of deterministic energy signals, all the signals have duration T seconds, apriori probabilities
• This set is called the SIGNAL ALPHABET• The receiver will get messages made out of signals from the signal alphabet• The channel adds white noise, therefore it is an Additive White Gaussian Noise (AWGN) channel
Example
• Assume that you transmit using Manchester code, then your signal alphabet is
0 T
A
-A
ts1
t 0 T
A
-A
ts2
t
“1” “0”
2
Matched Filter
• The matched filter is the system that maximizes signal-to-noise ratio at the output.
• Given a signal alphabet, the impulse response of the matched filter for the i-th signal is given by
• K is any constant different from zero• T>0 can take any value that makes the filter causal, e.g., T
could be the maximum signal duration of the alphabet
,tTKsth ii
Example: Design of Receiver with Matched Filters
ts1
1 2 3 t
1
th1
1 2 3 t
1
SignalsMatched filter’s
Impulse Response
3T ts2
1 2 3 t
1
ts3
2 3t
-1
1
th2
1 2 3 t
1
th3
2 3t
-1
1
Example: Design of Receiver with Matched Filters
Assume signal was transmitted ts2
th1
1 2 3 t
1
* Convolution3T
ts2
1 2 3 t
1
1 2 3 4 5
1
6
3
Example: Design of Receiver with Matched Filters
Assume signal was transmitted ts2
* Convolution3T
ts2
1 2 3 t
1
th2
1 2 3 t
1
1 2 3 4 5
123
6
Example: Design of Receiver with Matched Filters
Assume signal was transmitted ts2
* Convolution3T
ts2
1 2 3 t
1
th3
2 3t
-1
1
1 2 3 4 5-1 6
-2
Example: Design of Receiver with Matched Filters
thts 12
thts 22
thts 32
3T
1 2 3 4 5
1
6
1 2 3 4 5
123
6
1 2 3 4 5-1 6
-2
4
Example: Design of Receiver with Matched Filters
th1
th2
th3
ts2
,3,2, TTT
,3,2, TTT
,3,2, TTT
choosemaximum
Example: Design of Receiver with Matched Filters
22
2
0
22
20
2
20
2
20
222
s
E
ds
dss
dtTss
dthsthts
T
TTt
T
Tt
T
Tt
Result of the convolution of the signal with its corresponding matched filter sampled at T seconds is the energy of the signal
Example: Design of Receiver with Matched Filters
ts2
,3,2, TTT
,3,2, TTT
,3,2, TTT
choosemaximum
T
dt0
T
dt0
T
dt0
ts2
ts1
ts3
Correlators dttsts
T
10
2 dttsT
0
22
dttstsT
30
2Output is energy of
signal in the branch of the corresponding
match filter dss
t
30
2
5
Signal Representation
Signal Representation
Gram Schmidt Orthogonalization, i.e., geometric representation of signals
• Any set of M energy signals can be represented as linear combinations of N orthonormal basis functions where
• In other words
Mii ts 1
MN
.,0,,1
,,
.,2,1,,2,1,
.,2,1,0,
0
11
0
1
jiji
dttt
ttt
NjMidtttss
MiTttsts
j
T
i
NNjj
j
T
iij
N
jjiji
where satisfy (orthonormal)
Signal Representation
tsi
t1
t2
tN
1is
2is
iNs
Synthesis
tsi
T
dt0
T
dt0
T
dt0
1is
2is
iNs
t1
t2
tN
Analysis
6
Gram-Schmidt
tststs M,...,, 21
Energy or norm of a signal
dttstsET
iii 0
22
Step 1
1
1
1
11 E
tststst
Gram-Schmidt
Step 2
Step 3
tgtgt
2
22
,
,
10
122
11222
tdtttsts
tttststgT
tgtgt
3
33
,
,,
20
2310
133
22311333
tdtttstdtttsts
tttstttststgTT
Gram-Schmidt
Step k
tgtgt
k
kk
,
,
1
1 03
1
1
k
jj
T
jk
k
jjjkkk
tdtttsts
tttststg
Stop until ALL signals have been used
7
Gram-Schmidt
From Step 1
From Step 2
1
11 E
tst tttEts N 00 2111 0,,0,0,11 Es
tgtgt
2
22 ,, 11222 tttststg
,, 1122222 tttststtgtg
,, 221122 ttgtttsts 0,,0,,, 2122 tgttss
Gram-Schmidt
From Step k
,,1
1
k
jjjkkk tttststg
tgtgt
k
kk
,,1
111 ttgtttsts kk
k
jkk
0,,0,,,,,,,, 121 tgttsttsttss kkkkkk
Example: Orthogonalization
ts1
1 2 3 t
1
Signals
3T ts2
1 2 3 t
1
ts3
2 3t
-1
1
8
Example: Orthogonalization
ts1
1 2 3 t
1
Signals
3T ts2
1 2 3 t
1
ts3
2 3t
-1
1
t1
1 2 3 t
1
t2
1 2 3 t
21
Orthonormal Basis
0,11 s
2,12 s
2,03 s
Constellation
0,11 s
2,12 s
2,03 s 1
2
1s
3s
2s
Decision Regions
1
2
1s
3s
2sDistance from
any signal point to the origin is the square root of the energy of
such signal
9
Example: Design of Receiver with Matched Filters
ts2
,3,2, TTT
,3,2, TTT
,3,2, TTT
choosemaximum
T
dt0
T
dt0
T
dt0
ts2
ts1
ts3
Correlators dttsts
T
10
2 dttsT
0
22
dttstsT
30
2
RECALL
Output is energy of signal in the branch of
the corresponding match filter
Receiver: Design with Constellation
tsi
T
dt0
T
dt0
1is
2is t1
t2
Compute Euclidian Distance
and compare
Output is coordinates of the signal within the
constellation space
Optimal Receiver with AWGN Channel
10
Received Signal
.,,2,1
,Mi
tsi
.,,2,1,
MitNtstR ii
tNAWGN
Deterministic signal
Received signal
CHANNEL
2,0 02 Nm NN
fSN
20N
f
2
0NRN
Autocorrelation
22 2/22
1)( Nx
NN exf
Receiver: Design with Constellation
tNtstR ii T
dt0
T
dt0
t1
t2
Output is coordinates of the signal within the
constellation space plus noise
Compute Euclidian Distance
and compare
Distribution of Received Signal
tsrFtsrtNPrtNtsP
rtRPrF
MitNtstR
itN
i
i
itR
ii
i
.,,2,1,
• cdf of received signal
• Received signal has the same distribution as that of the noise N, i.e., Gaussian
• Received signal
MMMMM mpmpmptststs ,,,,...,, 2121 P
11
Optimal Receiver: Generalization
bT
b
dtT 0
1t=iTb R1
bT
b
dtT 0
1t=iTb R2
bT
b
dtT 0
1t=iTb RM
DecisionRi(t)
Matched filter bank
t1
t2
tM
,,,, 21 MRRRR
T
jj dtttRR0
,NsR i
,,,, 21 MNNNN
,,,, 21 iMiii ssss
T
jj dtttNN0
T
jiij dtttss0
,jijj NsR
Output is coordinates of the signal within the constellation
space plus noise
Recall: Distribution of Received Signal
tsrFtsrtNPrtNtsP
rtRPrF
MitNtstR
itN
i
i
itR
ii
i
.,,2,1,
• cdf of received signal
• Received signal has the same distribution as that of the noise N, i.e., Gaussian
• Received signal
Distribution of Received Signal in Constellation
iN
iMMMii
iiMMMii
iMMiMii
iMMimR
srFsrNsrNsrNP
mmsrNsrNsrNPmmrNsrNsrNsP
mmrRrRrRPmrF
,,,
|,,,|,,,
|,,,|
222111
222111
222111
2211|
tNmk oft independen,
T
jj dtttNN0
Nj is obtained from a linear operation of a Gaussian process, hence N1, N2, ..., NM are jointly Gaussian, then we need mean and variance of Nj
iNimR srfmrf ||
12
Distribution of Received Signal in Constellation
T
j
T
jj
dtttNE
dtttNENE
0
0
0
jijiji NENENNENN ,cov
If independence of vector components, then we have
0 0
but we do not know that.
Distribution of Received Signal in Constellation
.,0,,2/
2
2
2
,cov
0
0
0
0 0
0
0 0
0
0 0
0 0
jijiN
dtttN
dssstdttN
dtdsststN
dtdsstsNtNE
dsssNdtttNE
NNENN
T
ji
T
j
T
i
T
ji
T
T
ji
T
T T
ji
jiji
Autocorrelation
.0
.,0,,0
,0
0
tdst
tstts
sst
fdttft
j
T
j
jj
a
a
This implies uncorrelation, and because they are
Gaussian, then they are independent
Distribution of Received Signal in Constellation
Hence
0
1
2
02
/
2/0
1
/
0
02
1
1
,2
N
M
M
iiNN
NiN
N
M
ii
i
i
i
i
eN
ff
eN
f
N
And
01
2 /
2/0
|1|
Nsr
MiNimR
M
jijj
eN
srfmrf
Euclidian distance of signal received and signal from
alphabet
13
Optimal Receiver: Generalization
bT
b
dtT 0
1t=iTb R1
bT
b
dtT 0
1t=iTb R2
bT
b
dtT 0
1t=iTb RM
Decision
g(r)=mi
Ri(t)
Matched filter bank
t1
t2
tM
,,,, 21 MRRRR
T
jj dtttRR0
,NsR i
,,,, 21 MNNNN
,,,, 21 iMiii ssss
T
jj dtttNN0
T
jiij dtttss0
,jijj NsR
MAP: Maximum Aposteriori Decision Rule
.11
,IFF,
01
20
1
2 /
2/0
/
2/0
Nsr
MkM
Nsr
MiM
ikiM
jkjj
M
jijj
eN
mpeN
mp
mmmrg
apriori probability
Taking logarithm, and comparing, we have
.ln
smallest. isln
or
largest, islnln
IFF,
02
0
2
0
2
0
1
2
iMii
iMi
iiM
M
jijj
iM
i
mpNsrU
mpN
sr
Nsr
mpN
srmp
mrg
If apriori probabilities are all equal, then we have
Maximum Likelihood (ML) rule
Optimal Receiver: Generalization
bT
b
dtT 0
1t=iTb R1
bT
b
dtT 0
1t=iTb R2
bT
b
dtT 0
1t=iTb RM
Compute
Ui
Ri(t)
Matched filter bank
t1
t2
tM
is 0N
U1
U2
UM
Compare, choose smallest
14
Probability of Error and Modulation Techniques
Modulation Techniques: BASK
UNIPOLAR INFORMATION
Signal alphabet
ts2 ts1
Each signal “carries” one bit
Modulation Techniques: BPSK
POLAR INFORMATION
Signal alphabet
ts2 ts1
Each signal “carries” one bit
15
Example: Energy Signal (time-limited)
.0
,2cos)(Tt
tfAtx c
Signal
Phase
Amplitude A
Signal with A=1, and = 0 and
T=2.
some constant
Signal duration
T
-2 -1 0 1 2 3 4
RECALL: Time Averages for Energy Signals
• Mean value, i.e., DC component
• Mean squared value
• RMS value
.0T
dttxtx
.0
22 T
dttxtx
.2/12 txxRMS
Example:
• Mean value, i.e., DC component
• Mean squared value
• RMS value
.0tx
.2
22
bss ETPETAtx
.2
2/12bRMS ETAtxx
Average bit
Energy
Energy = Power * Duration
TEA b2
.0,2cos TttfAtx c
Each signal “carries” one bit
16
BPSK Constellation
• Signal set:
• Orthonormal basis (after applying Gram-Schmidt):
• Signal representation:
• Signal constellation:
,2cos21 tf
TEts c
b
.0,2cos212 Tttstf
TEts c
b
.0,2cos21 Tttf
Tt c
tEtstEts bb 1211 ,
10
1s2s
bEbE
Optimal Receiver
T
dt0
0 2,1
,
i
tNtstR ii
t1
T
AWGN
tsi transports one information bit
Received signal coordinates in the constellation, i.e.,
MRRRR ,,, 21
T
jjj dtttRTRR0
Received signal or process, i.e.,
t
jj dRtR0
)(
Optimal Receiver
T
dt0
0 tNtstR ii
t1
T
Assume is transmitted
tsi
tN
t
ts
t
i
t
i dNdsdRtR
i 11
0 10 10 11
Correlation of noise
and signal
111 RNETR b
TNN
T
Es
T
i
T
i
dNds
dR
TRR
bi 111
0 10 1
0 1
11
17
BPSK BER
QPSK
Received Signal Characterization
18
• Assume a system that transmits bits {0, 1}• Assume a signal alphabet given by
• Assume an antipodal signal set, i.e., , for example
• We can obtain the Energy of each signal
Scenario
tsts 21 , tsts 21
T0
A
ts1
t
T0
-A
ts2
t
b
TTEEdttsTAdttsE 20
22
2
0
211
Since each signal transports one bit, the energy is the same as the bit energy
Matched Filters
T0
A
ts1
t
T0
-A
ts2
t
T0
A
th1
t
T0
-A
th2
t
Design of Receiver with Matched Filters
.2,1
,
itNtstr ii
,3,2, TTT
,3,2, TTT
choosemaximum
T
dt0
T
dt0
ts2
ts1
Correlators dttsts
T
10
2
dttstsT
30
2
AWGN
2,0 02 Nm NN
tN
19
Convolutions: Outputs of Matched Filters
T0
A
ts1
t
T0
-A
ts2
t
T0
A
th1
t
T0
-A
th2
t
T0
A
th1
t
T0
-A
th2
t
T0
thts 11
t
TA2
2T
T0
thts 21
t
TA2
2T
T0 t
TA2
2T
T0t
TA2
2T
thts 12
thts 22
Orthogonalization
• Since we have antipodal signals, (i.e., one of the signals is a linear combination of the other,) there is only one dimension in the vector space
• Recall that the signal energy is• The orthonormal function is
• The linear combinations and constellation are
• Bit error rate
T0
1
11 E
tst
t
bEETAE 22
1
T1
tTAtsts 112
tTAts 11 11 ETAs
12 ETAs
10 TATA
1s2s
bEbE
Constellation
2dQP
Optimal Receiver
T
dt0
0 2,1
,
itNtstr ii
ts1
T
AWGN
tsi transports one information bit
20
Optimal Receiver
T
dt0
0 tNtstr 11
ts1
T
Assume is transmitted ts1
TTT
dttstNdttsdttstrTyy0 10
210 1111
Signal energy
Correlation of noise
and signal
111 WETy
ttt
dsNdsdsrty0 10
210 111
Optimal Receiver
T
dt0
0 tNtstr 22
ts1
T
Assume is transmitted tsts 12
TTT
dttstNdttsdttstrTy0 10
210 122
Signal energy
Correlation of noise
and signal
112 WETy
ttt
dsNdsdsrty0 10
210 122
Optimal Receiver
T
dt0
0
tNtstr ii
ts1
T
012 bbb1mm bb
mbm
b
m
WEbWEWETy
1
11
bits being transmitted, can be a plus one or a minus one
We look the m-th bit which is transmitted using si(t)
Since the m-th bit can be a plus one or a minus one, we have plus or minus the signal energy E1
1mb
Decision variable for m-th bit
21
In General
• Consider that the m-th information bit is transmitted,• The received signal after sampling, i.e., the decision variable, is
• Recall that is AWGN and that it is WSS, i.e., Gaussian Random process, its distribution is determined by
• Its Power Spectral Density and autocorrelation are
1mb
mmmbmm WTbAWbEWbEy 2111
tN
,0~ΝtN
fSN
20N
f
2
0NRN
Autocorrelation
Distribution of Received Signal
T
dt0
0
tNtstr ii
ts1
T
012 bbb1mm bb
mbm
b
m
WEbWEWETy
1
11
TYEb
dttstNdttsb
dttstrTY
bm
TT
m
T
im
0 10
21
0 1
What is the distribution of
Y(t) ?
tYdsb
dsNdsb
dsrtY
t
m
tt
m
t
im
0
21
0 10
21
0 1
Distribution of Y(t)
• Y(t) can be seen as the output of a stable linear filter with input N(t) , and by property 1 (page 56) of Haykin’s book, and since N(t) is Gaussian, then Y(t) is also Gaussian.
• We need only to find the mean value of Y(t) and its variance
0
0 1
0 1
dttNEts
dttNtsEtYE
T
T
Mean value
2
2
22var
Y
tYEtYEtYEtY
Variance
0
0
22
Distribution of Y(t)
dzdtzNtNEzsts
dzdtzNtNzstsE
dzzNzsdttNtsE
tYE
T T
T T
TT
Y
0 0 11
0 0 11
0 10 1
22
Autocorrelation of N(t)
ztNzNtNEztRN 2
, 0
Distribution of Y(t)
dzdtztNzsts
dzdtztRzsts
dzdtzNtNEzsts
T T
T T
N
T T
Y
0 00
11
0 0 11
0 0 112
2
Then we have
The last integral exists when t=z due to the term , therefore zt
TAN
dttsN T
Y
20
0
21
02
2
2
2,0~ 0TNAtY YΝ
Hence
Distribution of Received Signal
T
dt0
0
tNtstr ii
ts1
T
012 bbb1mm bb
mbm
b
m
WEbWEWETy
1
11
TYEb
dttstNdttsb
dttstrtY
bm
TT
m
T
im
0 10
21
0 1
What is the distribution of Wm ?
Recall
2,0~ 0TNAtY YΝ
23
Distribution of Received Signal
• Wm is the sampled version of Y(t), i.e.,
• Since Y(t) is a Gaussian random process, with zero mean and , Y(T) is a Gaussian random variable with zero mean and , i.e.,
TYWm
YY
2,0~ 0TNATYW Ym Ν
Optimal Receiver
T
dt0
0
tNtstr ii
ts1
T
012 bbb1mm bb
mbm
b
m
WEbWEWETy
1
11
bits being transmitted, can be a plus one or a minus one
We look the m-th bit which is transmitted using si(t)
Since the m-th bit can be a plus one or a minus one, we have plus or minus the signal energy E1
1mb
Decision variable for m-th bit
In Conclusion
mbm
b
mm
WEbWEWE
Tyr
1
11
Y,0ΝDeterministic
Gaussian
What is the mean value and variance of rm =ym(T) ?
24
Distribution of ym(t)
0
mb
mmb
mmbm
bEWEbEE
WbEErE
Mean value
222
22
2 var
mbmmb
mm
mr
bEWbEE
rErEr
Variance
0
0
is always 12mb
22
22
2222
2
2
Yb
mmmbb
mmmbmbmmb
EWEWEbEE
WWbEbEEWbEE
Therefore ,22222YbYbr EE and YmbbE ,Ν~rm
Distribution of Y(t)
dzdtzNtNEzsts
dzdtzNtNzstsE
dzzNzsdttNtsE
tYE
T T
T T
TT
Y
0 0 11
0 0 11
0 10 1
22
Autocorrelation of N(t)
ztNzNtNEztRN 2
, 0
Bit Error Rate
0
0
2
22
2
2
NEQ
NE
Q
dQP
b
b
25
Bit Error Rate
Y
b
b
Y
b
Y
b
EQ
E
EQ
TA
EQdQP
2
22
,2
, 0
TATNAbE Ymb Ν~rm
Recall
bEETAE 22
1
2,0~ 0NtN Ν
Spread Spectrum: A Discrete-Time Approach
Dr. Cesar Vargas RosalesCenter for Electronics and Telecommunications
Digital Communication System
.1, mm bb
Transmitter ReceiverChannel
Transmitted sequence,mmm wbr
Received sequence
• E>0, is the energy of the pulse representing each bit
• wm is zero-mean additive white Gaussian noise (AWGN) with autocovariance
sent" was1",0sent" was1",0
.2
m
m
kmm
rr
kwwE
26
Correlation Receiver
T
dt0
)()()( twtvtr
)(tv
mm yr Decision making
Decision variable
,mmm wbr
Gaussian random variable with mean
Ebm and variance 2
System Modeling in Discrete-Time
• Let the transmitted pulse during signaling interval m be v(t-mT) with duration T.
• Consider the case m=0, i.e., the bit interval [0,T], the received signal is
• w(t) is continuous-time additive white Gaussian noise (AWGN) with power spectral density N0/2
• Using correlation receiver, we get
),()()( twtvtr
T T
m
T
m wdttwtvdttvdttvtrr0 0 0
2 .E)()()()()(
System Modeling
• If the signal is a rectangular pulse
then
and
and it can be shown that the BER is
hence, performance is determined by
A
-A
T2T
0 t
v(t)
,E 2TA
,2
202
2 TNAwE m
,E2E
0
NQQPe
E/
27
• For each information bit, instead of one pulse, a sequence of Npulses is transmitted, i.e., if bm=i is transmitted, a sequence {i,i,…,i} is used instead, where i can be +1 or -1
• Consider the first transmitted bit, so that, the subindex m could be dropped
• These sub-bits are known as chips
Generalization
1 2 3 N
Information bit b
transmitted signal
.1,,0, NnbEs cn NEc /EPulse energy
Generalization
• The received sequence representing the bit is
where the noise now has variance• We employ a discrete correlation receiver, where the
decision variable is
thus, the decision variable y has mean and variance hence performance is determined by
.1,,0, NnwbEr ncn ./22 NwE n
,1
0
1
0
1
0
N
nnc
N
nnc
N
nn wbNEwbEry
bbNEc E2
2
NN
E/
Spread Spectrum Concept
• The transmitted signal with chips can be written as
where is the sequence of chips • The spread-spectrum property arises from the fact that the chips,
rather than being identically valued, are drawn from a known (i.e., deterministic) binary source, thus
will be sent
,1,,0, NnbcEs ncn ,11
0
Nnnc
110110 ,,,or,,,, NN cccccc
28
Spreading sequence
• Assume that the sequence is defined for all the period, i.e.,
• Mean value
• Autocorrelation (orthonormal sequence)
.,1
0kNcc
N
nkNnn
.01 1
0
N
nnc
N
.0,0
,0,11 1
0 Nii
ccN
N
ninn
In practice
• Spreading sequences in practice are based on the maximum-length, or m-sequences, hence the discrete-time autocorrelation function for the m-sequences is
• These sequences are known as Pseudo-noise (PN) sequences for the noise-like properties.
.0,/1
,0,11 1
0 NiNi
ccN
N
ninn
Correlation Receiver
1
0
N
nnxnr
nc
Decision making
nxmb̂
,1
0
1
0
1
0
N
nnnc
N
nnnnc
N
nnn cwbNEcwbcEcry
The decision variable is again Gaussian with mean Eb and variance 2 hence the spreading yields no improvement in the ideal AWGN channel because the independent and uncorrelated noise process contributes the same power in both cases
29
Spread Spectrum
• The real power of spreading comes from its effect on narrowband or correlated signals, these include– Interference suppression– Multipath mitigation– Multiuser interference: Signals from other transmitters in the network
employing spreading signals (Multiple access)• We require that the receiver spreading sequence be synchronized
with the received version, we assume perfect synchronization
Interference Suppression
• Suppose that the channel contains an interferer, i.e., an unknown constant I, is added to the received signal, i.e.,
• The decision variable for the nonspread system would have a mean of
which will render the system unusable for I sufficiently large
IbEN c
,1
0
1
0
1
0NIwbNEIwbEry
N
nnc
N
nnc
N
nn
Interference Suppression
• For the spread system, the received sequence is
where in=I.• The decision variable produced by the correlation receiver is
• The interference is suppressed by the despreading
.1,,1,0, NnwibcEr nnncn
.01
0
1
0
1
0
1
0
N
nnnc
N
nnn
N
nnc
N
nnnnnc
cwbNE
cwcIbNE
cwibcEy
30
Multipath Mitigation
• Mitigation of time-dispersive effects of multipath channels• Consider the sequence• We incorporate the chip energy into the bits by setting
• Consider the multipath channel 0mmb
.cm Eb
Signal received with strength
Signal received with strength and a
delay l which is less than one bit duration
Multipath Mitigation
• The received chips during the mth bit interval are
• The decision variable for the mth bit interval is
• The multipath signal is suppressed by the despreading
.1,,,,1,,0,1
Nlncbcblncbcb
rlnmnm
nlNmnmn
.001
0
1
0
11
01
N
nnnm
N
nnn
N
lnnlnm
l
nnnlNmmm
cwbN
cwccbccbbNy
Multipath Mitigation
• We can see the effect of multipath on the unspread system by letting cn=1 for all n. Then we obtain
• The energy from adjacent bits causes severe intersymbol interference (ISI), resulting in a performance loss that depends upon the delay and amplitude of the reflected components, as well as equalization
.1
01
N
nnnmmmm cwblNlbbNy
31
Multiple Access
• Consider that there are K transmitting users, where the kthtransmitter modulates its data with a unique spreading sequence
• These spreading sequences have the cross-correlation property
• Thus we have a set of sequences with zero cross-correlations and impulse-valued autocorrelations
)(knc
.,0,0,,0
,0,,11 1
0
)()(
jkNijk
ijkcc
N
N
n
jin
kn
Multiple Access
• Assume that the K signals share a channel simultaneously, and we are interested only in signal k=1
• Assume time synchronization among signals, the received signal is
• The correlation receiver for signal k=1 generates the decision variable
n
K
k
kn
kmn wcbr
1
)()(
1
0
)1()1(
1
0
)1(
2
1
0
)1()()(1
0
2)1()1()1(
0N
nnnm
N
nnn
K
k
N
nn
kn
km
N
nnmm
cwNb
cwccbcby
Multiuser interference (MUI) or Multiple Access Interference (MAI)
Top Related