Week6 PDF
Week6 PDF
Week6 PDF
EE4601
Communication Systems
Week 6
Orthogonal Expansions
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2011, Georgia Institute of Technology (lect6 1)
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Basic Problem
Problem:
Suppose that we have a set of M finite energy signals S = {s1 (t), s2(t), . . . , sM (t)},
where each signal has a duration T seconds.
Every T seconds one of the waveforms from the set S is selected for transmission
over an AWGN channel. The transmitted waveform is
X
x(t) = sn (t nT )
n
By observing r(t) we wish to determine the time sequence of waveforms {sn (t)}
that was transmitted. That is, in each T second interval, we must determine
which si (t) S was transmitted.
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2011, Georgia Institute of Technology (lect6 2)
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Orthogonal Expansions
Consider a real valued signal s(t) with finite energy Es,
Z
Es = s2 (t)dt
0
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2011, Georgia Institute of Technology (lect6 3)
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Orthogonal Expansions
To minimize the mean square error, we take the partial derivative with respect
to each of the sk and set equal to zero, i.e., for the nth term we solve
Z XN
=2 s(t) sk fk (t) fn (t)dt = 0.
sn k=1
R
Using the orthonormal property of the basis functions, sn = s(t)fn (t)dt and
2
Z N
X
= s(t) sk fk (t) dt
k=1
Z Z N N
Z X N
2 X X
= s (t)dt 2 s(t) sk fk (t)dt + sk fk (t) s f (t)dt
k=1 k=1 =1
Z N Z N N Z
s2 (t)dt 2
X X X
= sk s(t)fk (t)dt + sk s fk (t)f(t)dt
k=1 k=1 =1
N
s2k
X
= Es
k=1
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2011, Georgia Institute of Technology (lect6 4)
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Gram-Schmidt Orthonormalization
Suppose that we have a set of finite energy real signals {si (t)}, i = 1, . . . , M}.
We wish to obtain a complete set of orthonormal basis functions for the signal
set. This can be done in 2 steps.
Step1: Determine if the set of waveforms is linearly independent. If they are
linearly dependent, then there exists a set of coefficients a1 , a2 . . . , aM , not all
zero, such that
a1 s1 (t) + a2 s2 (t) + + aM sM (t) = 0.
Suppose, without loss of generality, that aM 6= 0. If aM = 0, then the signal set
can be permuted so that aM 6= 0. Then
a1 a2 aM 1
!
sM (t) = s1 (t) + s2 (t) + + sM (t) .
aM aM aM
Next consider the reduced signal set {si (t)}M 1
i=1 . If this set of waveforms is
linearly dependent, then there exists another set of co-efficients {bi}M 1
i=1 , not all
zero, such that
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2011, Georgia Institute of Technology (lect6 5)
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Gram-Schmidt Orthonormalization
We continue until a set {si(t)}Ni=1 of linearly independent waveforms is obtained.
Note that N M with equality if and only if the set of waveforms {si(t)}M i=1 is
linearly independent.
If N < M, then the set of linearly independent waveforms {si (t)}N i=1 is not
unique, but any one will do.
Step 2: From the set {si (t)}N i=1 construct the set of N orthonormal basis func-
N
tions {fi(t)}i=1 as follows. First, let
s1 (t)
f1(t) =
E1
where E1 is the energy in the waveform s1 (t), given by
Z T
E1 = s21 (t)dt
0
Then
s1(t) = E1f1 (t) = s11 f1(t)
where s11 = E1 .
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Gram-Schmidt Orthonormalization
Next, by using the waveform s2(t) we obtain
Z T
s21 = s2(t)f1(t)dt
0
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Gram-Schmidt Orthonormalization
Continuing in the above fashion, we define the ith intermediate function
i1
X
gi(t) = si (t) sij fj (t)
j=1
where Z T
sij = si(t)fj (t)dt
0
The set of functions
gi (t)
fi (t) = qR
T
i = 1, 2, , . . . , N
2
0 (gi (t))
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2011, Georgia Institute of Technology (lect6 8)
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Gram-Schmidt Orthonormalization
We can now write the signals as weighted linear combinations of the basis func-
tions, i.e.,
s1(t) = s11f1(t)
s2(t) = s21f1(t) + s22f2 (t)
s3(t) = s31f1(t) + s32f2 (t) + f33f3(t)
.. .
. = ..
sN (t) = sN 1f1 (t) + + sN N fN (t)
where Z T
sik = si(t)fk (t)dt
0
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Signal Vectors
It follows that the signal set si (t), i = 1, . . . , M can be expressed in terms of a
set of signal vertors si , i = 1, . . . , M in an N -dimensional signal space, i.e.,
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Example
s 1(t) s (t)
2
1 1
0 T/3 T 0 2T/3 T
s (t) s (t)
3 4
1 1
0 T/3 T 0 T
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2011, Georgia Institute of Technology (lect6 10)
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Example
Step 1: This signal set is not linearly independent because
Therefore, we will use s1 (t), s2 (t), and s3(t) to obtain the complete orthonormal
set of basis functions.
Step 2:
a)
Z T
E1 = s21 (t)dt = T /3
0
q
s1 (t) 3/T , 0 t T /3
f1 (t) = =
E1 0 , else
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Example
b)
Z T
s21 = s2(t)f1(t)dt
0
Z T /3 q q
= 3/T dt = T /3
0
Z T
E2 = s22 (t)dt = 2T /3
0
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2011, Georgia Institute of Technology (lect6 12)
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Example
c)
Z T
s31 = s3(t)f1(t)dt = 0
0
Z T
s32 = s3(t)f2(t)dt
0
Z 2T /3 q q
= 3/T dt = T /3
T /3
g3(t)
f3(t) = qR
T 2
0 g3 (t)dt
q
3/T , 2T /3 t T
=
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2011, Georgia Institute of Technology (lect6 13)
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Example
f 1(t) f (t)
2
3/T 3/T
f (t)
3
3/T
0 T/3 2T/3 T
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Example
q
s1(t) s1 = ( T /3, 0, 0)
q q
s2(t) s2 = ( T /3, T /3, 0)
q q
s3(t) s3 = (0, T /3, T /3)
q q q
s4(t) s4 = ( T /3, T /3, T /3)
f (t)
2
T/3 s
2
s s
3 4
s
1
T/3 f (t)
1
T/3
f (t)
3
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2011, Georgia Institute of Technology (lect6 15)
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2011, Georgia Institute of Technology (lect6 16)
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1 N
X
= q sjn skn
Ej Ek n=1
sj sk
=
ksj k ksk k
Note that
0 , if sj (t) and sk (t) are orthogonal
=
1 , if sj (t) = sk (t)
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(Z )1/2
T 2
djk = (sj (t) sk (t)) dt
0
2 1/2
Z T
N
X N
X
= sjn fn (t) skm fm (t) dt
0
n=1 m=1
1/2
N
(sjn skn )2
X
=
n=1
o1/2
= ksj sk k2
n
= ksj sk k
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2011, Georgia Institute of Technology (lect6 18)
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Example
Consider the earlier example where
q
s1 = ( T /3, 0, 0)
q q
s2 = ( T /3, T /3, 0)
q q
s3 = (0, T /3, T /3)
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2011, Georgia Institute of Technology (lect6 19)