Time - Evolving Signal Analysis Roza ACESKA Ball State University - PowerPoint PPT Presentation

Time - Evolving Signal Analysis Roza ACESKA Ball State University Midwestern Workshop on Asymptotic Analysis Indiana U. Purdue U. Fort Wayne, 2016 Roza ACESKA Ball State University Time - Evolving Signal Analysis

Introduction The problem of recovering an evolving signal from a set of samples taken at different time instances is motivated by research questions emerging in dynamical systems. Lu, Veterli. Spatial super-resolution of a diffusion fieldby temporal oversampling in sensing networks. IEEE Int. Conf. Acoustics, Speech and Signal Proc. 2009. Sampling problems in dynamical systems: eg. sampling of air pollution, wireless networks, temperature distribution over a metropolitan area etc. We state the problem of spatio-temporal sampling for different classes of functions (signals), and provide specific reconstruction results. Roza ACESKA Time - Evolving Signal Analysis

What is Dynamical Sampling? Let f describe the initial state of a physical system on domain D . Over time, the system evolves to the state f t = A t f , where { A t } t ≥ 0 is a family of evolution operators. at time instances t i : sampling sets X i ⊂ D S X i - corresponding downsampling operator (obtains insufficiently many samples for ’successful’ recovery) The fundamental question in dynamical sampling Is the recovery of the initial state f possible from the coarsely under-sampled initial state and its altered states f t = A t f at time instances { t i : i = 1 , · · · , L } ? Goal : Reconstruct f = f 0 from S X 0 ( f ) , S X 1 ( A t 1 f ) , ..., S X L ( A t L f ) . (1) Roza ACESKA Time - Evolving Signal Analysis

Conditions ... ... on evolution operators A i = A t i , sampling sets X i , number L of repeated subsampling harvests, and sampling time instances t i Recovery of the initial state f = f 0 is possible, if we have: IS Invertibility sampling property . Within a class of signals, any signal h is associated with a samples data set { S X i ( A i h i ) } which uniquely determines h . SS Stability sampling property. Within a class of signals, given any two signals h , ˜ h , the following two norms, L � h − ˜ h � 2 � � S X i A i ( h − ˜ h ) � 2 p and ℓ p are equivalent . i =0 Roza ACESKA Time - Evolving Signal Analysis

Convolution operators in Shift-Invariant Spaces (I) f ∈ [ − 1 / 2 , 1 / 2] and f ∈ L 1 ∩ L 2 ( R ), then Example: If supp ˆ � (Shannon Sampling Thm) f ( x ) = f ( k ) sinc ( x − k ) . k Let ϕ ∈ { f | � k ∈ Z sup 0 ≤ x ≤ 1 | f ( x + k ) | < ∞} ∩ C . Shift-Invariant space � c k ϕ ( · − k ) | c = ( c k ) ∈ ℓ 2 ( Z ) } V ( ϕ ) := { c ∗ ϕ = (2) k ∈ Z ϕ ( ξ + k ) | 2 ≤ M 2 < ∞ , then � If 0 < M 1 ≤ | ˆ k ∈ Z every f ∈ V ( ϕ ) can be recovered from the samples f ( Z ). Question : Let f = f 0 ∈ V ( ϕ ) and f j = a ∗ · · · ∗ a ∗ f ; can we recover f from subsamples { f ( m Z ) , f 1 ( m Z ) , f 2 ( m Z ) , . . . } ? Answer: Yes. Roza ACESKA Time - Evolving Signal Analysis

Convolution operators in Shift-Invariant Spaces (II) Fact: Let Φ j = ϕ j | Z , where ϕ j = a j ∗ ϕ , 0 ≤ j ≤ L . Then ˆ Φ j ∈ C ([0 , 1]). Theorem Every f ∈ V ( ϕ ) can be recovered in a stable way from the measurements y j = ( a j ∗ f ) | m Z , 0 ≤ j ≤ m − 1, if and only if det A m ( ξ ) � = 0 for all ξ ∈ [0 , 1], where Φ 0 ( ξ ˆ Φ 0 ( ξ +1 ˆ Φ 0 ( ξ + m − 1 ˆ  m ) m ) . . . )  m ˆ ˆ ˆ Φ 1 ( ξ Φ 1 ( ξ +1 Φ 1 ( ξ + m − 1 m ) m ) . . . )   m A m ( ξ ) =  .  . . . .  . . . .   . . . .  Φ m − 1 ( ξ ˆ Φ m − 1 ( ξ +1 ˆ Φ m − 1 ( ξ + m − 1 ˆ m ) m ) . . . ) m Proof: m − 1 Due to F ( f j | m Z )( ξ ) = 1 c ( ξ + l Φ j ( ξ + l � m )ˆ ˆ m ) . (3) m l =0 Roza ACESKA Time - Evolving Signal Analysis

Special case: A i = A in a separable Hilbert space I Roza ACESKA Time - Evolving Signal Analysis

Special case: A i = A in a separable Hilbert space II Roza ACESKA Time - Evolving Signal Analysis

Special case: A i = A in a separable Hilbert space III Roza ACESKA Time - Evolving Signal Analysis

Special case: A i = A in a separable Hilbert space IV Roza ACESKA Time - Evolving Signal Analysis

Frames = generalization of orthonormal bases Frame definition A sequence F = { f i } i ∈ I ( I -count. index set) of H \ { 0 } is a frame for H , if there exist 0 < C ≤ D < ∞ such that C � f � 2 ≤ |� f , f i �| 2 ≤ D � f � 2 for all f ∈ H . � (4) i ∈ I The frame operator S f := � i ∈ I � f , f i � f i is invertible For each frame F of H there exists at least one dual frame G = { g i } i ∈ I , satisfying � � f = � f , f i � g i = � f , g i � f i for all f ∈ H . (5) i ∈ I i ∈ I The set { g i = S − 1 f i } i ∈ I is called the canonical dual frame. The frame F is C - tight , if C = D in (4), and f = 1 � f , f i � f i = 1 � C S f for all f ∈ H . (6) C i ∈ I Roza ACESKA Time - Evolving Signal Analysis

Dynamical Frames and Canonical Duals in Hilbert spaces Consider G = { f i } i ∈ I ⊆ H , and a bounded operator A : H → H . If F L G ( A ) := ∪ i ∈ I { A j f i | f ∈ G} L i j =0 is a frame for H , (7) then we call (7) a dynamical frame , gen. by A and f i . Theorem 1 Let F L G ( A ) be a dynamical frame for H , with frame operator S . The canonical dual frame of F L G ( A ) is also dynamical, generated by B := S − 1 AS and g i = ( S − 1 f i ) i.e. L i � � � f , A j f i � B j g i f = for every f ∈ H . (8) i ∈ I j =0 Roza ACESKA Time - Evolving Signal Analysis

Dynamical Frames and Scalability (joint work with Y. Kim, arxiv: 1608.05622v1) If the frame F is tight then S − 1 ≡ I . In general, computing S − 1 can be a challenging problem. If there exist scaling coefficients w i ≥ 0, i ∈ I , such that F w := { w i f i } i ∈ I is a tight frame, then we call F a scalable frame. Property G ( A ) = ∪ i ∈ I { A j f i | f ∈ G} L i Say F L j =0 is a scalable frame; then L i � � w 2 ij � f , A j f i � A j f i for every f ∈ H . f = (9) i ∈ I j =0 � 1 � cos(2 π/ 3) � � − sin(2 π/ 3) Example: e 1 = , A = . 0 sin(2 π/ 3) cos(2 π/ 3) { e 1 , A e 1 , A 2 e 1 } is a scalable frame for R 2 (mercedes). Roza ACESKA Time - Evolving Signal Analysis

Dynamical Frames and Scalability in Finite Dimensions Let H = R n , and let { e j } n j =1 be the std orthonormal basis of H . Theorem * Let G = { e k 1 , . . . , e k p } , p < n , and L = ( L 1 , . . . , L p ) ∈ Z p + . Let A be an operator on H . If B is a unitary operator on H , then TFAE ∪ p s =1 { A j e k s } L s j =0 is a (scalable) frame ∪ p s =1 { C j g s } L s j =0 is a (scalable) frame, where C := B − 1 AB , and g s := B − 1 e k s , s = 1 , . . . , p . Note: if A = URU T is a Schur decomposition of A , with a unitary matrix U and a matrix of Schur form R , then TFAE (i) ∪ p s =1 { A j e k s } L s j =0 is a (scalable) frame for H , (ii) ∪ p s =1 { R j v s } L s j =0 is a (scalable) frame for H . Roza ACESKA Time - Evolving Signal Analysis

Scalable Dynamical Frames: Normal Operators in R n (I) Let A = UDU T be a normal operator on H , D = diag ( a 1 , . . . , a n ), U -orthogonal; let e l k , k = 1 , . . . , s , s ≤ n , be standard basis vectors for H . TFAE: F G ( A ) = ∪ p s =1 { A j e l s | j = 0 , 1 , . . . , L s } is a scalable frame ∪ p s =1 { D j v s | j = 0 , 1 , . . . , L s } is a scalable frame for R n where v s = U T e l s = ( x s (1) , . . . , x s ( n )) T , 1 ≤ s ≤ p . The scaling coefficients w s , t of F G ( A ) are solutions to (*): p � x s ( i ) � 2 � s , 1 � a i � 2 + · · · + w 2 s , L s � a n � 2 L s � � w 2 s , 0 + w 2 = 1 , s =1 p � a j ) L s � � w 2 s , 0 + w 2 a j + · · · + w 2 x s ( i ) ¯ x s ( j ) s , 1 a i ¯ s , L s ( a i ¯ = 0 , s =1 for all i , j = 1 , . . . , n , i � = j . Roza ACESKA Time - Evolving Signal Analysis

Example 1: Let  1 0 0   1   1   , ; v 1 =  , v 2 =  . D = 0 1 0 − 1 1    0 0 − 1 1 − 1 The frame { v 1 , D v 1 , v 2 , D v 2 } is a scalable frame in R 3 , with weights w k , i = 0 . 5, 0 ≤ i ≤ 1, 1 ≤ k ≤ 2. Example 2: Let v 1 = ( x 1 , x 2 , x 3 ) T , v 2 = ( y 1 , y 2 , y 3 ) T and D = diag ( a , b , 0), where ab � = 0, 1 + ab < 0. Set 1 1 x 1 = ± , x 2 = ± , b 2 ( a 2 + 1) − a 2 (1 + ab ) a 2 ( b 2 + 1) − b 2 (1 + ab ) � � a b � − (1 + ab ) 3 , y 1 = − x 1 x 3 , y 2 = − x 2 x 3 � 1 − x 2 x 3 = ± , and y 3 = ± . ab y 3 y 3 Then { v 1 , A v 1 , v 2 , A v 2 , A 2 v 2 } is a tight frame of R 3 . Roza ACESKA Time - Evolving Signal Analysis

Scalable Dynamical Frames: Normal Operators in R n (II) Theorem 2 Let D = diag ( a 1 , · · · , a n ), where a 1 , . . . , a n ∈ R , and let v s = ( x s (1) , . . . , x s ( n )) T ∈ H , s ∈ { 1 , · · · , p } , p ≥ 1. The sequence ∪ p s =1 { D j v s | j = 0 , 1 , . . . , L s } is a scalable frame for H if and only if there exist scaling coefficients w s , 0 , w s , 1 , . . . , w s , L s , s = 1 , . . . , p , which satisfy conditions (*). j =0 is never a scalable frame for R n ( we need p ≥ 2) { D j v } L Roza ACESKA Time - Evolving Signal Analysis