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7 Oct. 2009 TELE3113 1
TELE3113 Analogue and Digital Communications –
Detection Theory (2)Wei Zhang
School of Electrical Engineering and TelecommunicationsThe University of New South Wales
7 Oct. 2009 TELE3113 2
Integrate-and-Dump detectorIntegrate-and-Dump detector
r(t)=si(t)+n(t)
≤≤−=≤≤+=
=0for 0 )(1for 0 )(
)(2
1
TtAtsTtAts
tsi
[ ]
++
=+= ∫+
0for )(1for )(
)()()(2
10
0 o
oTt
ti nta
ntadttntstzOutput of the integrator:
where
( )
∫
∫
∫
+
+
+
=
−=−=
==
Tt
to
Tt
t
Tt
t
dttnn
ATdtAa
ATAdta
0
0
0
0
0
0
)(
2
1
7 Oct. 2009 TELE3113 3
Integrate-and-Dump detectorno is a zero-mean Gaussian random variable.
{ } { }∫∫++
==
=Tt
t
Tt
to dttnEdttnEnE
0
0
0
0
0)()(
{ } { }
{ }
( )
22
2
)()(
)(
0
0
0
0
0
0
0
0
0
0
2
22
Td
dtdt
dtdntnE
dttnEnEnVar
Tt
t
Tt
t
Tt
t
Tt
t
Tt
t
Tt
toon
o
o
o
ηεη
εεδη
εε
σ
==
−=
=
===
∫
∫ ∫
∫ ∫
∫
+
+ +
+ +
+
( ) )/()2/( 222 12
1 T
nn e
Tef on
o
o
ηασα
πησπα −− ==pdf of no:
7 Oct. 2009 TELE3113 4
Integrate-and-Dump detector
≤≤−=≤≤+=
=0for 1for
TtAtsTtAts
tsi 0)(0)(
)(0
1
AT
s
+
1
AT
s
−
0 As
0We choose the decision threshold to be 0.
Two cases of detection error:
(a) +A is transmitted but (AT+no)<0 no<-AT
(b) -A is transmitted but (-AT+no)>0 no>+AT
Error probability:
[ ]
Tudue
APAPdT
e
dT
eAPdT
eAP
APAATnPAPAATnPP
TA
u
AT
T
AT
TAT T
ooe
ηα
π
απη
απη
απη
η
ηα
ηαηα
2 2
)()(
)()(
)()|()()|(
2
2
2
22
2
2/
)/(
)/()/(
==
−+=
−+=
−>+−<=
∫
∫
∫∫
∞ −
∞ −
∞ −−
∞−
−
Q
( )
∫
∫
=
=
=
=
∞ −
T
bb
x
u
e
dtAEEQ
duexQTAQP
0
2
2/2
2
2 2
2
Q
Q
η
πηThus,
7 Oct. 2009 TELE3113 5
Integrate-and-Dump detector
2 s
signal symbol energy=si2
Consider two signal symbols s1 and s2.
1 sLet Ed be the energy of the difference signal (s1- s2),
i.e. [ ]∫ −=T
d dttstsE0
221 )()(
=
η2d
eEQP
For example: If
≤≤−=≤≤+=
=0for 0 )(1for 0 )(
)(2
1
TtAtsTtAts
tsi
[ ] [ ] TAdtAAdttstsETT
d2
0
2
0
221 4)()()( =−−=−= ∫∫
=
=⇒
ηηTAQTAQPe
22 22
4
Thus
7 Oct. 2009 TELE3113 6
Integrate-and-Dump detectorExample: In a binary system with bipolar binary signal which is a +A volt or –A volt pulse
during the interval (0,T), the sending of either +A or –A are equally probable. The value of A is 10mV. The noise power spectral density is 10-9 W/Hz. The transmission rate of data (bit rate) is 104 bit/s. An integrate-and-dump detector is used.
( )
49
4232
108.710102
)10()1010(22 −−
−−
×==
××
=
= QQTAQPe η
(a) With , P(+A)=P(-A)=0.5 , bit interval T=10-4 seconds, A=10mV910
2−=
η
(a) Find the probability of error, Pe.
(b) If the bit rate is increased to 105 bit/s what value of A is needed to attain the same Pe, as in part (a).
Solution:
(b) Bit interval T=10-5 seconds, for the same Pe, i.e. ( )
102 2
QTAQPe =
=
η
( )( )
mVATA 62.31102
102101025
92
=×
=→= −
−
η
7 Oct. 2009 TELE3113 7
Optimal Detection Threshold
[ ]∫
∫∫
∫∫
∞−
∞−∞−
∞
∞−
−+=
−+=
+=
+=
λ
λλ
λ
λ
drsrfsPsrfsPsP
drsrfsPdrsrfsP
drsrfsPdrsrfsP
sPssPsPssPPe
)|()()|()()(
)|(1)()|()(
)|()()|()(
)()|()()|(
22112
2211
2211
221112 detect detect
7 Oct. 2009 TELE3113 8
Optimal Detection ThresholdTo find a threshold λo which minimizes Pe, we set
gives
0=λd
dPe
)()(
)|()|(
)|()()|()(
1
2
2
1
2211
sPsP
sfsf
sfsPsfsP
o
o
oo
=
=
λλ
λλ
)()(
)|()|(
1
2)2/()(
)2/()(
2
122
2
221
sPsP
e
esfsf
ono
ono
s
s
o
o ==−−
−−
σλ
σλ
λλ
)()(
1
2)2/()(/)( 222
21
221
sPsPe onono ssss =−−− σσλ
)()(ln
2
)()(
ln2
)()(
ln2
)(
1
2
21
221
1
2
21
221
1
22
22
21
221
sPsP
ssss
sPsP
ssss
sPsPssss
o
o
oo
no
no
nn
o
−+
+=
−=
+−
=−
−−
σλ
σλ
σσλ
If2
)()( 2121
sssPsP o
+=⇒= λ
Taking ln(.) on both sides, then
7 Oct. 2009 TELE3113 9
Detected signal
symbol
Correlator ReceiverRecall ML decision criterion: minimize 2
isr rr−
[ ] ∫∫∫∫ −+=−=−T
i
l
Ti
TTii dttstrdttsdttrdttstrsr )()(2)()()()( 2222
signa th-i of energyconstant4342143421
rrWith
minimize 2isr rr
− minimize ∫∫ −T
iT
i dttstrdtts )()(2)(2
Let denotes the energy of si(t). iξ
ML decision criterion becomes
Find i to maximize
43421i
Ti
Ti dttsdttstr
ξ
∫∫ − )()()( 221
if all signal symbols have the same energy
Find i to maximize ∫T
i dttstr )()(Or simply:
Correlation Receiver
7 Oct. 2009 TELE3113 10
Detected signal
symbol
Matched FilterThe multiplying and integrating in correlation receiver can be reduced to a linear filtering.
Consider the received signal r(t) passes through a filter hi(t):
i.e.
Let
Then we sample the filter output at t=T,
thus
ML decision criterion:
Find i to maximize
becomes43421
i
Ti
Ti dttsdttstr
ξ
∫∫ − )()()( 221
43421i
Ti
TtTi dttsdthr
ξ
τττ ∫∫ −−=
)()()( 221
Find i to maximize
∫ −=∗T
ii dthrthtr τττ )()()()(
Ttfor dTtsrdthrTshT
iT
iii ≤≤+−=−⇒−= ∫∫ 0 )()()()( )()( ττττττττ
∫∫∫ =+−=−== T
i
TtTi
TtTi dsrdTtsrdthr τττττττττ )()()()()()(
Matched-Filter Receiver
(correlation)
7 Oct. 2009 TELE3113 11
≤ ≤
Matched Filter
0 T t
si(t)
-T 0 t
si(-t)
0 T t
hi(t)
)()( tTsth ii −=Consider the matched filtersi(-t)
ttt
Equivalence of matched filter and correlator:
Matched filter receiver
Correlator receiver
7 Oct. 2009 TELE3113 12
Detection of PAMDetection of M-ary PAM (one-dimension)
)12( and 21 MiA,...,M,iwhereAEs iii −−===
0
Let , energy of si is
amplitude EM )1( −
sM/2+1 sMsM-1sM/2sM/2-1s1 s2
… …
EE−
sM/2+2
E3E3−
symbol
For equally probable signals, i.e. ,...,M,for iMsP i 21 /1)( ==
average energy
2is
( ) ( ) EMMMMEMi
MEA
MEs
M
M
i
M
ii
M
iiav
−=
−=−−=== ∑∑∑
=== 31
31121 22
1
2
1
2
1
2ε
Received signal where nAEnsr ii +=+= 2 ,0 2 ησ == nn
( )
−=
−=
−=
>−−
=
∫
∫∞
−
∞−
ηπ
πη
η
η
EQM
MdyeM
M
dxeM
M
EsrPM
MP
E
y
E
x
ie
2)1(2 221
21
1 )error symbol(Prob Average
/2
22
2
η2let xy =
7 Oct. 2009 TELE3113 13
Detection of QAMDetection of M-ary QAM (two-dimension) : M=2k
… …
E2
……
E2
−=
−
ηEQ
MM
P DPAMMe
2)1(2
)error symbol(Prob Average )1,(
For one-dimension -ary PAM:
For two-dimension M-ary QAM:
( )2)1,()2,( 1 )decisioncorrect (Prob Average DPAMMeDQAMMc PP−− −=
( )2)1,(
)2,()2,(
11
1 )error symbol(Prob Average
DPAMMe
DQAMMcDQAMMe
P
PP
−
−−
−−=
−=
M
For M=4,
=
− ηEQP DPAMe
2)1,4(
( )
−
=
−−=
−−=−−
ηηηEQEQEQ
PP DPAMeDQAMe
222 211
11 )error symbol(Prob Average2
2)1,4()2,4(