modul 03 siskom2 signal space
DESCRIPTION
signal spaceTRANSCRIPT
Modul #03
TE3223 TE3223 SISTEM KOMUNIKASI SISTEM KOMUNIKASI 22
SIGNAL SPACE ANALYSIS
Program Studi S1 Teknik TelekomunikasiDepartemen Teknik Elektro - Sekolah Tinggi Teknologi Telkomp gg g
Bandung – 2008
Signal space?What is a signal space?
Vector representations of signals in an N-dimensional orthogonal spaceorthogonal space
Why do we need a signal space?It is a means to convert signals to vectors and vice versa.It is a means to calculate signals energy and Euclidean distances between signals.
Why are we interested in Euclidean distancesWhy are we interested in Euclidean distances between signals?
For detection purposes: The received signal is p p gtransformed to a received vectors. The signal which has the minimum distance to the received signal is estimated as the transmitted signal.
2
g
Modul 3 - Siskom 2 - Signal Space
Schematic example of a signal space)(2 tψ
),( 12111 aa=s
)(1 tψ),( 21 zz=z
)(s
),( 32313 aa=s
),( 21
)()()()(),()()()( 121112121111
aatatatsaatatats
⇔+=⇔+=
ss
ψψψψ
),( 22212 aa=s
Transmitted signal
),()()()(),()()()(),()()()(
212211
323132321313
222122221212
zztztztzaatatatsaatatats
=⇔+==⇔+==⇔+=
zss
ψψψψψψg
alternatives
Received signal at t h d filt t t 212211matched filter output
3Modul 3 - Siskom 2 - Signal Space
Signal spaceTo form a signal space, first we need to know the inner product between two signals (functions):
Inner (scalar) product:
∫∞
∞−
>=< dttytxtytx )()()(),( *
= cross correlation between x(t) and y(t)
Properties of inner product:= cross-correlation between x(t) and y(t)
><>=< )(),()(),( tytxatytax ><>< )(),()(),( tytxatytax
><>=< )(),()(),( * tytxataytx
)()()()()()()( ttttttt ><+>>=<+< )(),()(),()(),()( tztytztxtztytx
4Modul 3 - Siskom 2 - Signal Space
Signal space – cont’dThe distance in signal space is measure by calculating the norm.What is norm?What is norm?
Norm of a signal:
Edttxtxtxtx >< ∫∞ 2)()()()( xEdttxtxtxtx ==><= ∫ ∞−
)()(),()(
)()( txatax == “length” of x(t)
Norm between two signals:
)()(
)()( tytxd −=
We refer to the norm between two signals as the E clidean distance bet een t o signals
)()(, tytxd yx =
Euclidean distance between two signals.5Modul 3 - Siskom 2 - Signal Space
Example of distances in signal space
)(2 tψ),( 12111 aa=s
)(tψzsd ,1
1E
)(1 tψ),( 21 zz=z
dzsd ,3
3E
2E
),( 22212 aa=s
),( 32313 aa=s zsd ,2zs ,3 2
The Euclidean distance between signals z(t) and s(t):
)()()()( 222
211, −+−=−= zazatztsd iiizsi
3,2,1=i
6Modul 3 - Siskom 2 - Signal Space
Signal space - cont’dN di i l th l i l iN-dimensional orthogonal signal space is characterized by N linearly independent functions
called basis functions. The basis functions{ }Nj t)(ψ called basis functions. The basis functions
must satisfy the orthogonality condition{ }
jj t1
)(=
ψ
T Tt ≤≤0
h
jiijiji Kdttttt δψψψψ =>=< ∫ )()()(),( *
0
Tt ≤≤0Nij ,...,1, =
where
⎩⎨⎧
≠→=→
=jiji
ij 01
δ ∫Ψ==T
iii dttEK0
2 ).(
If all the signal space is orthonormal
⎩ ≠→ ji0 0
1=KIf all , the signal space is orthonormal.
7Modul 3 - Siskom 2 - Signal Space
1=iK
Example of an orthonormal basis functionsExample: 2-dimensional orthonormal signalExample: 2 dimensional orthonormal signal space
0)/2cos(2)(1⎪⎪⎧
<≤= TtTtT
t πψ )(2 tψ
0)/2sin(2)(2⎪⎪⎩
⎪⎨
<≤= TtTtT
t
T
πψ)(tψ
0)()()(),( 20
121 =>=< ∫ dtttttT
ψψψψ
)(1 tψ0
Example: 1-dimensional orthonornal signal space
1)()( 21 == tt ψψ
)(1 tψ
T1 1)(1 =tψ
)(1 tψ0
T t08Modul 3 - Siskom 2 - Signal Space
Signal space – cont’dAn arbitrar finite set of a eforms { }Mts )(Any arbitrary finite set of waveforms where each member of the set is of duration T, can be expressed as a linear combination of N
{ }ii ts 1)( =
can be expressed as a linear combination of N orthonogal waveforms where .{ }N
jj t1
)(=
ψ MN ≤N
where
∑=
=N
jjiji tats
1
)()( ψ Mi ,...,1=MN ≤
where
dtttsK
ttsK
aT
jij
jij
ij )()(1)(),(1
0
*∫>=<= ψψ Tt ≤≤0Mi ,...,1=Nj ,...,1=
jj 0
),...,,( 21 iNiii aaa=s2
1ij
N
jji aKE ∑
=
=Vector representation of waveformVector representation of waveform Waveform energy
9Modul 3 - Siskom 2 - Signal Space
Signal space - cont’dN
∑=
=N
jjiji tats
1
)()( ψ ),...,,( 21 iNiii aaa=s
)(1 tψ)(1 tψ
Waveform to vector conversion Vector to waveform conversion
⎥⎥⎤
⎢⎢⎡ iaM1
1ia
)(tsi
∫T
0⎥⎤
⎢⎡ iaM1
s)(tsi
1ia
ms
⎥⎥⎦⎢
⎢⎣ iNaM )(tNψ
a
)(i
∫T
)(tNψ
⎥⎥⎥
⎦⎢⎢⎢
⎣ iNaM ms=
a iNa∫0 iNa
10Modul 3 - Siskom 2 - Signal Space
Example of projecting signals to an orthonormalsignal space
)(2 tψ )(2 tψ),( 12111 aa=s
)(1 tψ
),( 32313 aa=s
),()()()( 121112121111 aatatats =⇔+= sψψ
),( 22212 aa=s
Transmitted signal
),()()()(),()()()(
),()()()(
323132321313
222122221212
121112121111
aatatatsaatatats
=⇔+==⇔+=
ss
ψψψψψψTransmitted signal
alternatives
T
dtttsaT
jiij )()(0∫= ψ Tt ≤≤0Mi ,...,1=Nj ,...,1=
11Modul 3 - Siskom 2 - Signal Space
Signal space – cont’dTo find an orthonormal basis f nctions for a gi enTo find an orthonormal basis functions for a given set of signals, Gram-Schmidt procedure can be used.Gram-Schmidt procedure:
Given a signal set , compute an orthonormalb i
{ }Mii ts 1)( =
{ }basis1. Define2. For compute
{ }Njj t
1)(
=ψ
)(/)(/)()( 11111 tstsEtst ==ψMi ,...,2= ∑
−
><−=1
)()(),()()(i
jjiii tttststd ψψIf letIf , do not assign any basis function.
3. Renumber the basis functions such that basis is
=1j0)( ≠tdi)(/)()( tdtdt iii =ψ
0)( =tdi
This is only necessary if for any i in step 2.
{ })(),...,(),( 21 ttt Nψψψ
0)( =tdiy y y pNote that
)(i
MN ≤
12Modul 3 - Siskom 2 - Signal Space
Example of Gram-Schmidt procedure
Find the basis functions and plot the signal space for the following transmitted signals:
)(1 ts )(2 ts
T t
)(1
T t
)(2 ts
TA
A−0
0
Using Gram-Schmidt procedure:
T t T0
)( )()()(
)()(
21
12
11
AAtAts
tAts
−==−=
=
ssψ
ψ)(1 tψ
T1
/)(/)()(
)(
1111
0
2211
==
== ∫AtsEtst
AdttsET
ψ
1
)()( 21
)(tψ1s2sT t0
0)()()()(
)()()(),(
)()()(
122
0 1212
1111
=−−=
−=>=< ∫tAtstd
AdtttsttsT
ψ
ψψ
ψ
2
)(1 tψ-A A0
0)()()()( 122 tAtstd ψ
13Modul 3 - Siskom 2 - Signal Space
Implementation of matched filter receiver
)(1 Tz
Bank of N matched filters
)(1 tz
)(tr
)(1 Tz)(1 tT −∗ψ Observation
vector
⎥⎥⎤
⎢⎢⎡z1
z= z
1
)( tTN −∗ψ)(TzN
⎥⎥⎦⎢
⎢⎣ Nz
)(tzN
∑=N
tats )()( ψ Mi 1
)(N
N1
),...,,( 21 Nzzz=z∑=
=j
jiji tats1
)()( ψ
MN ≤Mi ,...,1=
)()( tTtrz jj −∗= ψ Nj ,...,1=
14Modul 3 - Siskom 2 - Signal Space
Implementation of correlator receiver
T
)(1 tψz
Bank of N correlators
∫T
0
)( ⎥⎥⎤
⎢⎢⎡zM1
z=)(tr
1z
z Observationvector
∫T
0
)(tNψ
⎥⎥⎦⎢
⎢⎣ NzM
Nz
vector
∫0 N
∑=
=N
jjiji tats
1
)()( ψ Mi ,...,1=
),...,,( 21 Nzzz=z
Nj 1=dtttrzT
)()( ψ∫=
=j 1
MN ≤
Nj ,...,1=dtttrz jj )()(0
ψ∫=15Modul 3 - Siskom 2 - Signal Space
Example of matched filter receivers using basic functions
)(1 ts )(2 ts )(1 tψ
T1
TA
T tT t
T t0TA−0
0
1 matched filter)()( 11 ttT ψψ =−
1 )(1 Tz)(tz[ ]1z z=)(tr z
T t
T1
0
)(1z)(1 tz
Number of matched filters (or correlators) is reduced by 1 compared to i t h d filt ( l t ) t th t itt d i lusing matched filters (correlators) to the transmitted signal.
16Modul 3 - Siskom 2 - Signal Space
White noise in orthonormal signal space
AWGN n(t) can be expressed as)(~)(ˆ)( )(~)(ˆ)( tntntn +=
Noise projected on the signal space Noise outside on the signal spacep j g pwhich impacts the detection process.
Noise outside on the signal space
)()(ˆ tntnN
jj∑= ψ )(ˆ
>=< )(),( ttnn jj ψ
)()(~
)()(1
tntnj
jj∑=
ψ
Nj ,...,1=Nj 1
Vector representation of
),...,,( 21 Nnnn=n)(ˆ tn
{ }Nn0)(),(~ >=< ttn jψ Nj ,...,1= independent zero-mean
Gaussain random variables with
variance
{ }jjn
1=
2/)var( 0Nnj =variance 2/)var( 0Nnj
17Modul 3 - Siskom 2 - Signal Space