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Signal-Space Analysis ENSC 428 – Spring 2008 Reference: Lecture 10 of Gallager Digital Communication System Representation of Bandpass Signal x t s t cos 2 fct Bandpass real signal x(t) can be written as: x t 2 Re x t e j 2 fct where x t is complex envelop Note that x t xI t j xQ t In-phase Quadrature-phase Representation of Bandpass Signal (1) x t 2 Re x t e j 2 fct 2 Re xI t j xQ t cos 2 f ct j sin 2 f ct xI t 2 cos 2 f ct xQ t 2 sin 2 f ct (2) Note that x t x t e j t x t 2 Re x t e j 2 fct 2 Re x t e j t e j 2 fct x t 2 cos 2 f ct t Relation between x t and x t e j 2 f c t x t x t 2 x f 2 2 -fc fc f 1 X f f c X * f f c 2 X ( f ), f 0 X f , X f X f fc 0, f 0 Xf fc f f Energy of s(t) E s 2 t dt S f df 2 2 S f df 2 0 S f df 0 2 (Rayleigh's energy theorem) (Conjugate symmetry of real s(t ) ) Representation of bandpass LTI System s t h t r t s t h t r t r t s t h t R f S f H f S f H f f c because s(t ) is band-limited. H f H f f c H * f f c H ( f ), f 0 H f f 0 0, H f H f fc 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 nonnegative 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={a,b,..} 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: (a+bg ag+bg Immediately following properties ab0 implies a0 or b0 For any non-zero a, a0 ? a0 a a0 a 1 a0 1 a1a; therefore a0 0 00 ? For a non-zero a, its additive inverse is non-zero. 00a a 0 a0 a0 000 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 a,b in F, we have (abv abv) (abv avbv a v+w)=av a w 1*v=v Examples of vector space Rn over R Cn over C L2 over Subspace. Let V be a vector space. Let V be a vector space and S V . If S is also a vector space with the same operations as V , then S is called a subspace of V . S is a subspace if v, w S av bw S Linear independence of vectors Def) A set of vectors v1 , v2 , vn V are linearly independent iff Basis Consider vector space V over F (a field). We say that a set (finite or infinite) B V is a basis, if * every finite subset B0 B of vectors of linearly independent, and * for every x V , it is possible to choose a1 , ..., an F and v1 , ..., vn B such that x a1v1 ... an vn . The sums in the above definition are all finite because without additional structure the axioms of a vector space do not permit us to meaningfully speak about an infinite sum of vectors. Finite dimensional vector space A set of vectors v1 , v2 , vn V is said to span V if every vector u V is a linear combination of v1 , v2 , vn . Example: R n 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 Let V be a finite dimensional vector space. Then If v1 , v2 , vm are linearly independent but do not span V , then V has a basis with n vectors (n m) that include v1 , v2 , vm . If v1 , v2 , vm span V and but are linearly dependent, then a subset of v1 , v2 , vm is a basis for V with n vectors (n m) . Every basis of V contains the same number of vectors. Dimension of a finiate dimensional vector space. 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 S of an inner product space is said to be orthonormal iff 1) x S , x, x 1 and 2) If x, y S and x y, then x, y 0. Projection onto a finite dimensional subspace Gallager Thm 5.1 Corollary: norm bound Corollary: Bessel’s inequality Gram –Schmidt orthonormalization Consider linearly independent s1 , ..., sn V , and inner product space. We can construct an orthonormal set 1 , ..., n V so that span{s1 , ..., sn } span 1 , ..., n 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.) Consider an orthonormal set 1 2 kt exp j k t k 0, 1, 2,... . T T Any function u (t ) in L2 0, T is u k u , k k . Fourier series. For this reason, this orthonormal set is called complete. Thm: Every orthonormal set in L2 is contained in some complete orthonormal set. Note that the complete orthonormal set above is not unique. Significance? IQ-modulation and received signal in L2 r t , s t N t , L2 0, T s t span 2 T cos 2 f ct , 2 T sin 2 f ct Any signal in L2 can be represented as There exist a complete orthonormal set r (t ). i i i 2 cos 2 f c t , 2 sin 2 f ct , 3 (t ), 4 (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 Thm If 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 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 Orthonormal set S in Hilbert space H is complete if Equivalent definitions 1) There is no other orthonormal set strictly containing S . (maximal) 2) x H , x x, ei ei 3) x, e , e S implies x 0 4) x H , x 2 x, ei 2 Here, we do not need to assume H is separable. Summations in 2) and 4) make sense because we can prove the following: Only for mathematicians (We don’t need separability.) Let O be an orthonormal set in a Hilbert space H . For each vector x H , set S e O x, e 0 is either empty or countable. Proof: Let Sn e O x, e 2 x 2 n . Then, Sn n (finite) Also, any element e in S (however small x, e is) is in S n for some n (sufficiently large). Therefore, S n 1 Sn . Countable. 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) H is separable iff there exists a countable subset D which is dense in H , that is, D H . Def) H is separable iff there exists a countable subset D such that x H , there exists a sequence in D convergeing to x. Thm: If H has a countable complete orthonormal set, then H is separable. proof: set of linear combinations (loosely speaking) with ratioanl real and imaginary parts. This set is dense (show sequence) Thm: If H is separable, then every orthogonal set is countable. proof: normalize it. Distance between two orthonormal elements is 2. ..... Signal Spaces: L2 of complex functions Use of orthonormal set M-ary modulation {s1 (t ), s2 (t ),..., sM (t )} Find orthonormal functions f1 (t ), f 2 (t ),.., f K (t ) so that {s1 (t ), s2 (t ),..., sM (t )} span{ f1 (t ), f 2 (t ),.., f K (t )} Examples (1) T 2 T 2 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 N s(t) = s i i t , r(t) = s(t) + n(t) i 1 Waveform channel / Correlation Receiver 1 t Vector XTMR . . . sN i=j i 1 s2 j n(t) = ni i t n(t) s1 i A s(t) t . . . s(t) n(t) 1 t r(t) . . . z r1 = s 1 + n1 z r2 = s 2 + n2 T 0 T 0 . . . t t t z T 0 rN = sN + nN } Vector RCVR