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Signal-Space Analysis
ENSC 428 – Spring 2008Reference: Lecture 10 of Gallager
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Digital Communication System
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Representation of BandpassSignal
Bandpass real signal x(t) can be written as:
( ) ( ) ( )cos 2 cx t s t f tπ=
( ) ( ) ( )22 Re where is complex envelopcj f tx t x t e x tπ =
Note that ( ) ( ) ( )I Qx t x t j x t= + ⋅
In-phase Quadrature-phase
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Representation of BandpassSignal
( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
22 Re
2 Re cos 2 sin 2
2 cos 2 2 sin 2
cj f t
I Q c c
I c Q c
x t x t e
x t j x t f t j f t
x t f t x t f t
π
π π
π π
= = + ⋅ +
= + −
(1)
(2) Note that ( ) ( ) ( )j tx t x t e θ=
( ) ( ) ( ) ( )
( ) ( )( )
2 22 Re 2 Re
2 cos 2
c cj tj f t j f t
c
x t x t e x t e e
x t f t t
θπ π
π θ
= = ⋅
= +
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Relation between and
2
2
( )x t ( )x t
fx
2 cj f te π−
fc-fc f fc f f
( )x t ( )x t
( ) ( ) ( )( )
( ) ( ) ( )
*12
( ), 0,
0, 0
c c
c
X f X f f X f f
X f fX f X f X f f
f+ +
= − + − +
>= = + <
2
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Energy of s(t)
( )
( )
( )
( )
2
2
2
02
0
(Rayleigh's energy theorem)
2 (Conjugate symmetry of real ( ) )
E s t dt
S f df
S f df s t
S f df
∞
−∞
∞
−∞
∞
∞
=
=
=
=
∫∫∫
∫
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Representation of bandpass LTI System
( )h t
( )h t
( )s t
( )s t
( )r t
( )r t
( ) ( ) ( )( ) ( ) ( )
( ) ( ) because ( ) is band-limited.c
r t s t h t
R f S f H f
S f H f f s t
= ∗
=
= +
( ) ( ) ( )( )
( )
( ) ( )
*
( ), 0
0, 0
c c
c
H f H f f H f f
H f fH f
f
H f H f f
+
+
= − + − + >
= <= +
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Key Ideas
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Examples (1): BPSK
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Examples (2): QPSK
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Examples (3): QAM
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Geometric Interpretation (I)
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Geometric Interpretation (II)I/Q representation is very convenient for some modulation types.We will examine an even more general way of looking at modulations, using signal space concept, which facilitates
Designing a modulation scheme with certain desired propertiesConstructing optimal receivers for a given modulationAnalyzing the performance of a modulation.
View the set of signals as a vector space!
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Basic Algebra: Group
A group is defined as a set of elements G and a binary operation, denoted by · for which the following properties are satisfied
For any element a, b, in the set, a·b is in the set.The associative law is satisfied; that is for a,b,c in the set (a·b)·c= a·(b·c)There is an identity element, e, in the set such that a·e= e·a=a for all a in the set.For each element a in the set, there is an inverse element a-1 in the set satisfying a· a-1 = a-1 ·a=e.
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Group: example
A set of non-singular n×n matrices of real numbers, with matrix multiplicationNote; the operation does not have to be commutative to be a Group.Example of non-group: a set of non-negative integers, with +
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Unique identity? Unique inverse fro each element?
a·x=a. Then, a-1·a·x=a-1·a=e, so x=e.x·a=a
a·x=e. Then, a-1·a·x=a-1·e=a-1, so x=a-1.
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Abelian group
If the operation is commutative, the group is an Abelian group.
The set of m×n real matrices, with + .The set of integers, with + .
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Application?
Later in channel coding (for error correction or error detection).
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Algebra: field
A field is a set of two or more elements F={α,β,..} closed under two operations, + (addition) and * (multiplication) with the following properties
F is an Abelian group under additionThe set F−{0} is an Abelian group under multiplication, where 0 denotes the identity under addition.The distributive law is satisfied:
(α+β)∗γ = α∗γ+β∗γ
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Immediately following properties
α∗β=0 implies α=0 or β=0For any non-zero α, α∗0= ?
α∗0 + α = α∗0 + α ∗1= α∗(0 +1)= α∗1=α; therefore α∗0 =0
0∗0 =?For a non-zero α, its additive inverse is non-zero.
0∗0=(α+(− α) )∗0 = α∗0+(− α)∗0 =0+0=0
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Examples:
the set of real numbersThe set of complex numbersLater, finite fields (Galois fields) will be studied for channel coding
E.g., {0,1} with + (exclusive OR), * (AND)
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Vector space
A vector space V over a given field F is a set of elements (called vectors) closed under and operation + called vector addition. There is also an operation * called scalar multiplication, which operates on an element of F (called scalar) and an element of V to produce an element of V. The following properties are satisfied:
V is an Abelian group under +. Let 0 denote the additive identity.For every v,w in V and every α,β in F, we have
(α∗β)∗v= α∗(β∗v)(α+β)∗v= α∗v+β∗vα∗( v+w)=α∗v+ α ∗w1*v=v
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Examples of vector space
Rn over RCn over CL2 over
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Subspace.
Let V be a vector space. Let be a vector space and . If is also a vector space with the same operations as ,then S is called a subspace of .
S is a subspace if ,
V S VS V
V
v w S av bw S
⊂
∈ ⇒ + ∈
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Linear independence of vectors
1 2
Def)A set of vectors , , are linearly independent iffnv v v V∈
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Basis
0
Consider vector space V over F (a field).We say that a set (finite or infinite) is a basis, if * every finite subset of vectors of linearly independent, and * for every , it
B VB B
x V
⊂⊂
∈
1 1
1 1
is possible to choose , ..., and , ..., such that ... .
The sums in the above definition are all finite because without additional structure the axioms of a vector
n n
n n
a a F v v Bx a v a v
∈ ∈= + +
space do not permit us to meaningfully speak about an infinite sum of vectors.
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Finite dimensional vector space
1 2
1 2
A set of vectors , , is said to span if every vector is a linear combination of , , .
Example:
n
n
n
v v v V Vu V v v v
R
∈∈
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Finite dimensional vector space
A vector space V is finite dimensional if there is a finite set of vectors u1, u2, …, un that span V.
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Finite dimensional vector space
1 2
1 2
1 2
Let V be a finite dimensional vector space. Then
If , , are linearly independent but do not span , then has a basis with vectors ( ) that include , , .
If , , span and but ar
m
m
m
v v v V Vn n m v v v
v v v V
•>
•
1 2
e linearly dependent, then a subset of , , is a basis for with vectors ( ) .
Every basis of contains the same number of vectors.
Dimension of a finiate dimensional vector space.
mv v v V n n m
V
<
•
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Example: Rn and its Basis Vectors
•••
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Inner product space: for length and angle
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Example: Rn
•••
•••
•••
•••
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Orthonormal set and projection theorem
Def)A non-empty subset of an inner product space is said to beorthonormal iff1) , , 1 and2) If , and , then , 0.
S
x S x xx y S x y x y
∀ ∈ < >=∈ ≠ < >=
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Projection onto a finite dimensional subspace
Gallager Thm 5.1
Corollary: norm bound
Corollary: Bessel’s inequality
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Gram –Schmidt orthonormalization
{ }{ }
1
1
1 1
Consider linearly independent , ..., , and inner product space.We can construct an orthonormal set , ..., so that
{ , ..., } , ...,
n
n
n n
s s VV
span s s span
φ φ
φ φ
∈
∈
=
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Gram-Schmidt Orthog. Procedure
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Step 1 : Starting with s1(t)
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Step 2 :
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Step k :
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Key Facts
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Examples (1)
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cont … (step 1)
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cont … (step 2)
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cont … (step 3)
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cont … (step 4)
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Example application of projection theorem
Linear estimation
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L2([0,T])(is an inner product space.)
( )
[ ]( )2
Consider an orthonormal set
1 2 exp 0, 1, 2,... .
Any function ( ) in 0, is , . Fourier series.
For this reason, this orthonormal set is called complete
k
k kk
ktt j kTT
u t L T u u
πφ
φ φ∞
=−∞
= = ± ±
= ∑
2
.
Thm: Every orthonormal set in is contained in some complete orthonormal set.
Note that the complete orthonormal set above is not unique.
L
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Significance? IQ-modulation and received signal in L2
( ) ( ) ( ) [ ]( )( ) { }
{ }
2
2
3 4
, , 0,
span 2 cos 2 , 2 sin 2
Any signal in can be represented as ( ).
There exist a complete orthonormal set
2 cos 2 , 2 sin 2 , ( ), ( ),...
c c
i ii
c c
r t s t N t L T
s t T f t T f t
L r t
f t f t t t
ξ ξ
π π
φ
π π φ φ
= + ∈
∈ −
−
∑
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On Hilbert space over C. For special folks (e.g., mathematicians) only
L2 is a separable Hilbert space. We have very useful results on1) isomorphism 2)countable complete orthonormal set
ThmIf H is separable and infinite dimensional, then it is
isomorphic to l2 (the set of square summable sequence of complex numbers)
If H is n-dimensional, then it is isomorphic to Cn.The same story with Hilbert space over R. In some sense there is only one real and one
complex infinite dimensional separable Hilbert space.L. Debnath and P. Mikusinski, Hilbert Spaces with Applications, 3rd Ed., Elsevier, 2005.
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Hilbert space
Def)A complete inner product space.
Def) A space is complete if every Cauchy sequence converges to a point in the space.
Example: L2
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Orthonormal set S in Hilbert space H is complete if
22
Equivalent definitions1) There is no other orthonormal set strictly containing . (maximal)2) , ,
3) , , implies 0
4) , ,
Here, we do not need to assume H is separable.
i i
i
Sx H x x e e
x e e S x
x H x x e
∀ ∈ =
∀ ∈ =
∀ ∈ =
∑
∑
Summations in 2) and 4) make sense because we can prove the following:
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Only for mathematicians (We don’t need separability.)
{ }
{ }2 2
Let be an orthonormal set in a Hilbert space .
For each vector x , set , 0 is
either empty or countable.
Proof: Let , .
Then, (finite)
Also, any element in (however small
n
n
O H
H S e O x e
S e O x e x n
S n
e S
∈ = ∈ ≠
= ∈ >
<
1
, is)
is in for some (sufficiently large).
Therefore, . Countable.n
nn
x e
S n
S S∞
==∪
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Theorem
Every orothonormal set in a Hilbert space is contained in some complete orthonormal set. Every non-zero Hilbert space contains a complete orthonormal set.
(Trivially follows from the above.)
( “non-zero” Hilbert space means that the space has a non-zero element.We do not have to assume separable Hilbert space.)
Reference: D. Somasundaram, A first course in functional analysis, Oxford, U.K.: Alpha Science, 2006.
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Only for mathematicians. (Separability is nice.)
Euivalent definitionsDef) is separable iff there exists a countable subset which is dense in , that is, .Def) is separable iff there exists a countable subset such that ,
H DH D H
H Dx H
=
∀ ∈ there exists a sequence in convergeing to .
Thm: If has a countable complete orthonormal set, then is separable. proof: set of linear combinations (loosely speaking)
D x
H H
with ratioanl real and imaginary parts. This set is dense (show sequence)Thm: If is separable, then every orthogonal set is countable.
proof: normalize it. Distance between two orthonorma
H
l elements is 2. .....
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Signal Spaces:L2 of complex functions
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Use of orthonormal set
1 2
1 2
1 2 1 2
M-ary modulation { ( ), ( ),..., ( )}Find orthonormal functions ( ), ( ),.., ( ) so that { ( ), ( ),..., ( )} { ( ), ( ),.., ( )}
M
K
M K
s t s t s tf t f t f t
s t s t s t span f t f t f t⊂
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Examples (1)
2T
2T
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Signal Constellation
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cont …
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cont …
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cont …
QPSK
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Examples (2)
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Example: Use of orthonormal set and basis
Two square functions
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Signal Constellation
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Geometric Interpretation (III)
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Key Observations
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Vector XTMR/RCVR Model
⊗ ⊗
⊗ ⊗
⊗ ⊗
⊕φ ( )1 t φ ( )1 t
φ ( )2 t φ ( )2 t
r = s + n1 1 1
s1
VectorRCVR
VectorXTMR
Waveform channel / CorrelationReceiver
s(t)
n(t)
r(t)s2
sN
r = s + n2 2 2
r = s + nN N N
φ ( )Ν t φ ( )Ν t
∑
⊕s(t)
n(t)
r(t) = s(t) + n(t)
0
Tz
0
Tz
0
Tz
i 1=
∞
∑i 1
N
=∑ φ ( ) , φ φi i j t i = j
φ ( )i t
si
ni
s(t) =
n(t) =
A
.
.
.
.
.
.
..
....