Signal-Space Analysis

ENSC 428 – Spring 2008

Reference: Lecture 10 of Gallager

Digital Communication System

Representation of Bandpass Signal

Bandpass real signal x(t) can be written as:

cos 2 cx t s t f t

22 Re where is complex envelopcj f tx t x t e x t

Note that I Qx t x t j x t

In-phase Quadrature-phase

Representation of Bandpass Signal

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

(1)

(2) Note that j tx t x t e

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

Relation between and

2

2

x t x t

fx

2 cj f te

fc-fcf fc

f f

x t x t

*1

2

( ), 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

Energy of s(t)

2

2

2

0

2

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

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

Key Ideas

Examples (1): BPSK

Examples (2): QPSK

Examples (3): QAM

Geometric Interpretation (I)

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

properties Constructing optimal receivers for a given modulation Analyzing the performance of a modulation.

View the set of signals as a vector space!

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.

Group: example

A set of non-singular n×n matrices of real numbers, with matrix multiplication

Note; the operation does not have to be commutative to be a Group.

Example of non-group: a set of non-negative integers, with +

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.

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 + .

Application? Later in channel coding (for error correction or

error detection).

Algebra: field

A field is a set of two or more elements F={,,..} closed under two operations, + (addition) and * (multiplication) with the following properties F is an Abelian group under addition The set F−{0} is an Abelian group under

multiplication, where 0 denotes the identity under addition.

The distributive law is satisfied: (++

Immediately following properties impliesor For any non-zero

therefore

For a non-zero its additive inverse is non-

zero.

Examples: the set of real numbers The set of complex numbers Later, finite fields (Galois fields) will be

studied for channel coding E.g., {0,1} with + (exclusive OR), * (AND)

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 in F, we have

(vv) (vvv v+w)=vw 1*v=v

Examples of vector space Rn over R Cn over C L2 over

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 V

S V

V

v w S av bw S

Linear independence of vectors

1 2

Def)

A set of vectors , , are linearly independent iffnv v v V

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 V

B 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 B

x a v a v

space do not permit us

to meaningfully speak about an infinite sum of vectors.

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 V

u V v v v

R

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.

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 V

n 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

Example: Rn and its Basis Vectors

Inner product space: for length and angle

Example: Rn

Orthonormal set and projection theorem

Def)

A non-empty subset of an inner product space is said to be

orthonormal iff

1) , , 1 and

2) If , and , then , 0.

S

x S x x

x y S x y x y

Projection onto a finite dimensional subspace

Gallager Thm 5.1

Corollary: norm bound

Corollary: Bessel’s inequality

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 V

V

span s s span

Gram-Schmidt Orthog. Procedure

Step 1 : Starting with s1(t)

Step 2 :

Step k :

Key Facts

Examples (1)

cont … (step 1)

cont … (step 2)

cont … (step 3)

cont … (step 4)

Example application of projection theorem

Linear estimation

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 k

TT

u t L T u u

2

.

Thm: Every orthonormal set in is contained in some

complete orthonormal set.

Note that the complete orthonormal set above is not unique.

L

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

On Hilbert space over C. For special folks (e.g., mathematicians) only

L2 is a separable Hilbert space. We have very useful results on

1) 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.

Hilbert spaceDef)

A complete inner product space.

Def) A space is complete if every Cauchy sequence converges to a point in the space.

Example: L2

Orthonormal set S in Hilbert space H is complete if

22

Equivalent definitions

1) 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

S

x 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:

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

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.

Only for mathematicians. (Separability is nice.)

Euivalent definitions

Def) 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 D

H D H

H D

x 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. .....

Signal Spaces: L2 of complex functions

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 t

f t f t f t

s t s t s t span f t f t f t

Examples (1)

2

T

2

T

Signal Constellation

cont …

cont …

cont …

QPSK

Examples (2)

Example: Use of orthonormal set and basis Two square functions

Signal Constellation

Geometric Interpretation (III)

Key Observations

Vector XTMR/RCVR Model

t t

t 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

t t

s(t)

n (t)

r(t) = s(t) + n (t)

0

Tz

0

Tz

0

Tz

}}

i 1

i 1

N

i i j t i = j

i t

s i

n i

s(t) =

n (t) =

A

.

.

.

.

.

.

.

.

....