Sparse Sampling. Pier Luigi Dragotti 1. investigator award Nr (RecoSamp). December 17, 2012

Transcription

1 1 December 17, is supported by the European Research Council (ERC) starting investigator award Nr (RecoSamp).

2 Outline Problem Statement and Motivation Classical Sampling Formulation Sampling using expansion-based sparsity Sparsity in Complete and Over-complete Dictionaries Compressed Sensing Applications Sampling using sparsity in parametric spaces Signals with Finite Rate of Innovation (FRI) Sampling Kernels: E-splines and B-splines Sampling FRI Signals: the Basic Set-up and Extensions Applications New Domains of Applications of the Sparsity and Sampling Paradigm Diffusion Fields and Neuroscience Conclusions and Outlook

3 Problem Statement You are given a class of functions. You have a sampling device. Given the measurements y n = x(t), ϕ(t/t n), you want to reconstruct x(t). x(t) h(t)=!(!t/t) y(t) T y n =<x(t),!(t/t!n)> Acquisition Device Natural questions: When is there a one-to-one mapping between x(t) and y n? What signals can be sampled and what kernels ϕ(t) can be used? What reconstruction algorithm?

4 Problem Statement Observed scene Lens Acquisition System CCD Array Samples The low-quality lens blurs the images. The images are sampled by the CCD array.

5 Outline Problem Statement and Motivation Classical Sampling Formulation Sampling using expansion-based sparsity Sparsity in Complete and Over-complete Dictionaries Compressed Sensing Applications Sampling using sparsity in parametric spaces Signals with Finite Rate of Innovation (FRI) Sampling Kernels: E-splines and B-splines Sampling FRI Signals: the Basic Set-up and Extensions Applications New Domains of Applications of the Sparsity and Sampling Paradigm Diffusion Fields and Neuroscience Conclusions and Outlook

6 Motivation: Sparsity and Sampling Everywhere In 2005, the U.S. spent 16% of its GDP on health care. It is projected that this will reach 20% by Goal: Individualized treatments based on low-cost and effective medical devices. Image Formation End!Users Image Processing

7 Motivation: Sparsity and Sampling Everywhere Wide-Band Communications: t t TX RX Current A-to-D converters in UWB communications operate at several gigaherz. This is a sparse parametric estimation problem, only the location and amplitude of the pulses need to be estimated.

8 Motivation: Sparsity and Sampling Everywhere Sensor networks The source (phenomenon) is distributed in space and time. The phenomenon is sampled in space (finite number of sensors) and time. When the sources are localized the problem is sparse.

9 8 Motivation: Sparsity and Sampling Everywhere Neural acquisi)on and sor)ng Applications in Neuroscience Neuroprosthesis ADC Spike sor)ng Processing unit The energy budget of implanted devices is limited. Implanted High neuronal sampling prostheses rates impose require wired data low-processing transmission and low-sampling limits the rate. quality and diversity of experiments. Spike sorting is based on a sparse description of the action potentials.

10 Motivation: Free Viewpoint Video Multiple cameras are used to record a scene or an event. Users can freely choose an arbitrary viewpoint for 3D viewing. This is a multi-dimensional sampling and interpolation problem.

11 Classical Sampling Formulation Sampling of x(t) is equivalent to projecting x(t) into the shift-invariant subspace V = span{ϕ(t/t n)} n Z. If x(t) V, perfect reconstruction is possible. Reconstruction process is linear: ˆx(t) = P n ynϕ(t/t n). For bandlimited signals ϕ(t) = sinc(t). x(t) ~ φ(t) ~ T y n φ(t) x(t) ^

12 Sampling as Projecting into Shift-Invariant Sub-Spaces 1! !4!3!2! !1 0 1!4!3!2!

13 Classical Sampling Formulation The Shannon sampling theorem provides sufficient but not necessary conditions for perfect reconstruction. Moreover: How many real signals are bandlimited? How many realizable filters are ideal low-pass filters? By the way, who discovered the sampling theorem? The list is long ;-) Whittaker 1915, 1935 Kotelnikov 1933 Nyquist 1928 Raabe 1938 Gabor 1946 Shannon 1948 Someya 1948

14 Key elements in the novel sampling approaches Classical Sampling Formulation: In classical sampling formulation, the reconstruction process is linear. Innovation is uniform. New formulation: The reconstruction process can be non-linear. Innovation can be non-uniform.

15 Sparse Representations in a Basis Wavelets provide sparse representations of images. In matrix/vector form α = W 1 Y is sparse. Here the matrix W has size N N and models the discrete-time wavelet transform of finite dimensional signals. Figure: Cameraman is reconstructed using only 8% of the wavelet coefficients.

16 Notation The l 0 norm of a N-dimensional vector X is X 0 = the number of i such that x i 0 The l 1 norm of a N-dimensional vector X is: X 1 = P N i=1 x i The Mutual Coherence of a given N M matrix A is the largest absolute normalized inner product between different columns of A: a T k µ(a) = max a j 1 k,j M;k j a k 2 a j 2

17 Sparsity in Redundant Dictionaries Y The above signal, Y, is a combination of two spikes and two complex exponentials of different frequency (real part of Y plotted). In matrix vector form: Y = (I N F N ) α = Dα, where I N is the N N identity matrix and F N is the N N Fourier transform. The matrix D models an over-complete dictionary and has size N M with M > N, α has only K non-zero coefficients (in the example K = 4, N = 128, M = 2N).

18 Sparsity in Redundant Dictionaries You are given Y and want to find its sparse representation. Ideally, you want to solve (P 0 ) : min α 0 s.t. Y = Dα. This is a combinatorial problem which requires N chooses K operations. You may instead solve the convex problem: (P 1 ) : min α 1 s.t. Y = Dα. Key result due to Donoho et al.: (P 0 ) is unique when K < 1/µ(D) = N. (P 0 ) and (P 1 ) are equivalent when K < ( 2 0.5)/µ(D) 0.9 N.

19 Sparsity ibounds n Pairs of Bases (P 0 ) Bound: 1/µ Tight (P 1 ) bound Simplified (P 1 ) bound: 0.91/µ 7 6 K q K p Uniqueness of (P 0 ) and the two l 1 bounds for the case of two orthogonal bases and µ(d) = 0.1. See [Elad 2010, page 59] for more details.

20 Sparsity in Redundant Dictionaries Sketch on the proof of unicity of (P 0 ). (P 0 ) is unique when K is such that, given Y 1 = Dα 1 and Y 2 = Dα 2, then Y 1 Y 2 for any possible K-sparse α 1, α 2. Consider α n = α 1 α 2, this new vector has sparsity 2K and unicity is lost when Y = Dα n = 0. «ˆX α n is in the null space of D = (I N F N ) when α n =, where X ˆX = F N X. In fact: Y = Dα n = (I N F N ) ˆX X «= 0, X is an N dimensional vector and cannot be simultaneously sparse in both the time and the frequency domain. Donoho uncertainty principle says that the number of non-zero entries in α n must be 2K 2/µ(D) = 2 N. Thus, (P 0 ) can be solved when K < N.

21 Sparsity in Redundant Dictionaries (cont d) Extensions [Tropp-04, GribonvalN:03, Elad-10] For a generic over-complete dictionary D, (P 1 ) is equivalent to (P 0 ) when 2 K < «. µ When D is a concatenation of J orthonormal dictionaries (P 1 ) is equivalent to (P 0 ) when» 2 1 K < 1 + µ 1 2(J 1) 2 Proof in Appendix.

22 Compressed Sensing The fat matrix D now plays the role of the acquisition device and we denote it with Φ. The entries of Y = Φα are the samples. Based on the previous analysis, we want to reconstruct the signal α from the samples Y using l 1 minimization. We want maximum incoherence of the columns of Φ. We consider large M, N. Key insight: Relax the condition of a deterministic perfect reconstruction and accept that, with an extremely small probability, there might be an error in the reconstruction.

23 The power of randomness Key theorem due to Candès et al.[candes:06-08]: if Φ is a proper random matrix (e.g., a matrix with normalized Gaussian entries), then with overwhelming probability the signal can be reconstructed from the samples Y when N C K log(m/k) for some constant C. Assume that the measured signal X is not sparse but has a sparse representation: X = Dα. We have that Y = ΦX = ΦDα. The new matrix ΦD is essentially as random as the original one. Therefore the theorem is still valid. Thus random matrices provides universality. However, very redundant dictionaries implies larger M and therefore larger N.

24 Restricted Isometry Property (RIP) In order to have perfect reconstruction, Φ must satisfy the so called Restricted Isometry Property: (1 δ S ) x 2 2 Φx 2 2 (1 + δ S ) x 2 2 for some 0 < δ S < 1 and for any S-sparse vector x. Candes et al.: If x is K-sparse and δ 2K + δ 3K < 1 then the l 1 minimization finds x exactly. if Φ is a random Gaussian matrix, the above condition is satisfied with probability 1 O(e γm ) for some γ > 0, when N C K log(m/k). if Φ is obtained by extracting at random N rows from the Fourier matrix, then perfect reconstruction is satisfied with high probability when: N C K(log M) 4. NB: When the signal x is not exactly sparse, solve: y Φˆx 2 + λ ˆx 1 It is proved that linear programming achieve the best solution up to a constant factor.

25 Compressed Sensing. Simulation Results Image Boat. (a) Recovered from random projections using Compressed Sensing. PSNR=31.8dB. (b) Optimal 7207-approximation using the wavelet transform with the same PSNR as (a). (c) Zoom of (a). (d) Zoom of (b). Images courtesy of Prof. J. Romberg.

26 Application in MRI Image taken from Lustig, Donoho, Santos, Pauly-08.

27 Problem Statement You are given a class of functions. You have a sampling device. Given the measurements y n = x(t), ϕ(t/t n), you want to retrieve the degrees of freedom of x(t). x(t) h(t)=!(!t/t) y(t) T y n =<x(t),!(t/t!n)> Acquisition Device Natural questions: When is there a one-to-one mapping between x(t) and y n? What kernels ϕ(t) can be used? What reconstruction algorithm?

28 Sparsity in Parametric Spaces Consider a continuous-time stream of pulses or a piecewise sinusoidal signal These signals are not bandlimited. are not sparse in a basis or a frame. However: they are completely determine by a finite number of free parameters.

29 Signals with Finite Rate of Innovation Consider a signal of the form: x(t) = X k Z γ k g(t t k ). (1) The rate of innovation of x(t) is then defined as 1 ρ = lim τ τ τ Cx 2, τ, (2) 2 where C x( τ/2, τ/2) is a function counting the number of free parameters in the interval τ. Definition A signal with a finite rate of innovation is a signal whose parametric representation is given in (1) and with a finite ρ as defined in (2).

30 Examples of Signals with Finite Rate of Innovation Filtered Streams of Diracs Piecewise Polynomial Signals Piecewise Sinusoidal Signals Mondrian paintings ;-)

31 The Sampling Kernel x(t) h(t)=!(!t/t) y(t) T y n =<x(t),!(t/t!n)> Acquisition Device Given by nature Diffusion equation, Green function. Ex: sensor networks. Given by the set-up Designed by somebody else. Ex: Hubble telescope, digital cameras. Given by design Pick the best kernel. Ex: engineered systems.

32 The Sampling Kernel 1 Line spread function Pixels (perpendicular) (a)original ( ) (b) Point Spread function

33 Sampling Kernels Any kernel ϕ(t) that can reproduce exponentials: X c m,nϕ(t n) = e αmt, α m = α 0 + mλ and m = 0, 1,..., L. n This includes any composite kernel of the form γ(t) β α (t) where β α (t) = β α0 (t) β α1 (t)... β αl (t) and β αi (t) is an Exponential Spline of first order [UnserB:05]. E Spline β α (t) e α t Notice: β α(t) ˆβ(ω) = 1 eα jω jω α α can be complex. E-Spline is of compact support. E-Spline reduces to the classical polynomial spline when α = 0.

34 Kernels Reproducing Exponentials Exponential e 0.06t Reproduction Scaled and shifted E splines Exponential e 0.5t Reproduction Scaled and shifted E splines t t t Here the E-spline is of second order and reproduces the exponential e α 0t, e α 1t : with α 0 = 0.06 and α 1 = 0.5.

35 Examples of E-Splines Kernels

36 Examples of Best Kernels

37 The Sampling Kernel 1 Line spread function Pixels (perpendicular) (a)original ( ) (b) Point Spread function

38 Kernel Reproducing Exponential Any functions with rational Fourier transform: Q i ˆϕ(ω) = (jω b i ) Qm (jω am) m = 0, 1,..., L. is a generalized E-splines. This includes practical devices as common as an RC circuit: + R + x(t) C y(t) - -

39 : Basic Set-up Assume the sampling period T = 1. Consider any x(t) with t [0, N). Assume the sampling kernel ϕ(t) is any function that can reproduce exponentials of the form X c m,nϕ(t n) = e αmt m = 0, 1,..., L, n We want to retrieve x(t), from the samples y n = x(t), ϕ(t n), n = 0, 1,..., N 1.

40 : Basic Set-up We have that s m = N 1 n=0 c m,ny n = x(t), N 1 n=0 c m,nϕ(t n) = x(t)eαmt dt, m = 0, 1,..., L. s m is the bilateral Laplace transform of x(t) evaluated at α m. When α m = jω m then s m = ˆx(ω m ) where ˆx(ω) is the Fourier transform of x(t). When α m = 0, the s m s are the polynomial moments of x(t).

41 Sampling Streams of Diracs Assume x(t) is a stream of K Diracs on the interval of size N: x(t) = K 1 k=0 x kδ(t t k ), t k [0, N). We restrict α m = α 0 + mλ m = 0, 1,..., L and L 2K 1. We have N samples: y n = x(t), ϕ(t n), n = 0, 1,...N 1: We obtain s m = N 1 n=0 c m,ny n = x(t)eαmt dt, = K 1 k=0 x ke αmt k = K 1 k=0 ˆx ke λmt k = K 1 k=0 ˆx kuk m, m = 0, 1,..., L.

42 The quantity The Annihilating Filter Method s m = K 1 k=0 ˆx k u m k, m = 0, 1,..., L is a sum of exponentials. We can retrieve the locations u k and the amplitudes ˆx k with the annihilating filter method (also known as Prony s method since it was discovered by Gaspard de Prony in 1795). Given the pairs {u k, ˆx k }, then t k = (ln u k )/λ and x k = ˆx k /e α0t k.

43 The Annihilating Filter Method 1. Call h m the filter with z-transform H(z) = P K i=0 h iz i = Q K 1 k=0 (1 u kz 1 ). We have that KX KX K 1 X X KX h m s m = i=0 h i s m i = i=0 k=0 K 1 ˆx k h i u m i k = k=0 ˆx k u m k i=0 h i u i k {z } 0 = 0. This filter is thus called the annihilating filter. In matrix/vector form, we have that SH = 0 and using the fact that h 0 = 1, we obtain s K 1 s K 2 s 0 h 1 s K s K s K 1 s 1 h 2 s K+1 = B. C B. C 4 A s L 1 s L 2 s L K h K s L Solve the above system to find the coefficients of the annihilating filter.

44 The Annihilating Filter Method 2. Given the coefficients {1, h 1, h 2,..., h k }, we get the locations u k by finding the roots of H(z). 3. Solve the first K equations in s m = P K 1 k=0 ˆx kuk m to find the amplitudes ˆx k. In matrix/vector form u 0 u 1 u K u K 1 0 u K 1 1 u K 1 K B ˆx 0 ˆx 1. ˆx K = C B Classic Vandermonde system. Unique solution for distinct u k s. s 0 s 1. s K 1 1. (3) C A

45 Sampling Streams of Diracs: Numerical Example t (a) Original Signal t (b) Sampling Kernel (β 7 (t)) t (c) Samples t (d) Reconstructed Signal

46 Sampling Streams of Diracs: Sequential Reconstruction (a) Original Signal (b) Sampling Kernel (β 7(t)) (c) Samples (d) Reconstructed Signal

47 Sampling Streams of Diracs: Sequential Reconstruction Noiseless samples Noise Time (s) (a) y n samples (b) Reconstructed stream stream of ion of the Fig. 6: Noisy samples with a SNR 10 db and reconstructed In this N 50, stream example: from10kthesamples, peaks of1000 the histogram Diracs, SNR of the = 15dB, retrieved Execution locations. time: one minute, The Success temporal ratelocations 100%, one are very falseaccurately positive. estimated Original Diracs Estimated Diracs Time (s) led with shows one realisation of the procedure explained before.

48 Note on the proof Linear vs Non-linear Problem is Non-linear in t k, but linear in x k given t k The key to the solution is the separability of the non-linear from the linear problem using the annihilating filter. The proof is based on a constructive algorithm: 1. Given the N samples y n, compute the moments s m using the exponential reproduction formula. In matrix vector form S = CY. 2. Solve a K K Toeplitz system to find H(z) 3. Find the roots of H(z) 4. Solve a K K Vandermonde system to find the a k Complexity 1. O(KN) 2. O(K 2 ) 3. O(K 3 ) 4. O(K 2 ) Thus, the algorithm complexity is polynomial with the signal innovation.

49 Sampling Piecewise Sinusoidal Signals: Numerical Example [sec] (a) [sec] (b) [sec] (c)

50 plane χ ds here the d on (5) he Sampling 2-D domains Algorithm (a) Original (c) 1: Difference Annihilation indicator image plane of indicator (b)(d) Reconstructed Samples function of indicator plane lane χ 1. Compute Fourier transform of the indicator plane χ evaluated ( at uniformly sampled frequency grids 2πk Fourier transd exactly. the FRI 2. Create annihilation filter according to (6) where the M, ) 2πl N ; ere either streams elements are rearranged lexicographically; whose Fourier on in (5) the case of 3. Solve the annihilation coefficients c k,l based on (5) of the Fourier under the Hermitian symmetry constraint on the n not use DFT coefficients in (2). (c) Difference image (d) Samples of indicator plane e χ Γ has infirs from server Fig. 2. Annihilation results for noiseless case sform ˆχ Γ with4.1. Fourier Transform of Indicator Plane rm ourier exactly. transexactly. As Notice that the annihilation algorithm presented here is noniterative to apply and the is extremely annihilation fast. Algorithm Except for1, the Fourier DFT step trans- which In order can prove FRI that form of imited ither streams discrete is O(n the annihilable log n) withcurve the FFT Γ has implementation, to be evaluatedexecution exactly. FRI time for signals allinvestigated the other steps previously are linear inwith 1-DKL, cases, theare degree either ofstreams freedom of hose Fourier of Diracs the annihilable [7] [2] or curve. piecewise Thus polynomials the algorithm[1] can whose easilyfourier cope with the case of transform largecan scalebeproblems. computed exactly. However, in the case of the Fourier annihilable curves here, no explicit expression of the Fourier, not the use annihilatwhich χ Γ haswe infi- canof χ Γ directlyfig. either. 2. Annihilation Because theresults indicator for noiseless plane χ Γ case has infi- DFT transform ˆχ Γ is available in general. And we can not use DFT 5. SIMULATION RESULTS AND DISCUSSIONS lter from Pier method, Luigi server Dragotti nite a bandwith, Several experiments results obtained are set from up to DFT verify suffers the exact from annihilation server ormsparse I signals. ˆχ Γ with Sampling Sup- aliasing for effect. Notice noiseless But that the cases. we can annihilation In relate all the the algorithm subsequent Fourier transform presented simulations, ˆχ Γ with here is wenon- use Fig. 2. Annihilation results for noiseles Notice that the annihilation algorithm presen (a) Original indicator plane (b) Reconstructed (c) Difference image (d) Samples of in

51 Robust x(t) h(t)=!(!t/t) y(t) T + y =<x(t), (t/t!n)>+" n! n " n Acquisition Device The measurements are noisy The noise is additive and i.i.d. Gaussian

52 Robust In the presence of noise, the annihilation equation SH = 0 is only approximately satisfied. Minimize: SH 2 under the constraint H 2 = 1. This is achieved by performing an SVD of S: S = UλV T. Then H is the last column of V. Notice: this is similar to Pisarenko s method in spectral estimation.

53 Robust : Cadzow s algorithm For small SNR use Cadzow s method to denoise S before applying TLS. The basic intuition behind this method is that, in the noiseless case, S is rank deficient (rank K) and Toeplitz, while in the noisy case S is full rank. Algorithm: SVD of S = UλV T. Keep the K largest diagonal coefficients of λ and set the others to zero. Reconstruct S = Uλ V T. This matrix is not Toeplitz, make it so by averaging along the diagonals. Iterate.

54 Robust Samples are corrupted by additive noise. This is a parametric estimation problem. Unbiased algorithms have a covariance matrix lower bounded by CRB. The proposed algorithm reaches CRB down to SNR of 5dB.

55 Robust Noiseless Samples (N=128) Noisy Samples (N=128) Original and estimated Diracs : SNR = 5 db; original Diracs estimated Diracs t

56 Page 23 of 108 Piecewise sinusoidal signal location [seconds] Robust 1 IEEE TRANSACTIONS ON SIGNAL PROCESSING Observed Standard Deviation 1 6 Cramer!Rao Bound ! ! ! ! Input SNR [db] Input SNR [db] (a) (b) Fig. 8. Retrieval of the switching point of a step sine (ω1 = 12.23π [1/sec] and t1 = [sec]) in 128 noisy samples. (a) Scatter plot of the estimated location. (b) Standard deviation (averages over 1000 iterations) of the location retrieval compared 27 to the Cramér-Rao bound. 28 For Re Standard deviation for location t1

57 Robust Page 24 of IEEE TRANSACTIONS ON SIGNAL PROCESSING !1 9! [sec] (a) Ground Truth Reconstructed Signal !0.5 16! [sec] (b) Fig. 9. Recovery of a truncated sine wave at SNR = 8 [db]. (a) The observed noisy samples. (b) The reconstructed signal 21 SNR= 8dB, N= along with the ground truth signal (dashed) Pier 27Luigi Dragottitime piecewise sinusoidal signal (with t 1 = [sec], t 2 = [sec] and ω 1 = 12.23π [rad/sec]) 28 given 128 noisy samples at an SNR of 8 [db]. Note that despite the small error in the estimation of the For Re

58 Application: Image Super-Resolution Super-Resolution is a multichannel sampling problem with unknown shifts. Use moments to retrieve the shifts or the geometric transformation between images. (a)original ( ) (b) Low-res. (64 64) (c) Super-res ( PSNR=24.2dB) Forty low-resolution and shifted versions of the original. The disparity between images has a finite rate of innovation and can be retrieved. Accurate registration is achieved by retrieving the continuous moments of the Tiger from the samples. The registered images are interpolated to achieve super-resolution.

59 Application: Image Super-Resolution Image super-resolution basic building blocks Restoration Super-resolved image LR image 0... LR image k Set of low-resolution images Image Registration HR grid estimation

60 Application: Image Super-Resolution For each blurred image I (x, y): A pixel Pm,n in the blurred image is given by P m,n = I (x, y), ϕ(x/t n, y/t m), where ϕ(t) represents the point spread function of the lens. We assume ϕ(t) is a spline that can reproduce polynomials: X X c m,n (l,j) ϕ(x n, y m) = x l y j l = 0, 1,..., N; j = 0, 1,..., N. n m We retrieve the exact moments of I (x, y) from Pm,n: τ l,j = X X Z Z c m,n (l,j) P m,n = I (x, y)x l y j dxdy. n m Given the moments from two or more images, we estimate the geometrical transformation and register them. Notice that moments of up to order three along the x and y coordinates allows the estimation of an affine transformation.

61 Application: Image Super-Resolution 1 Line spread function Pixels (perpendicular) (a)original ( ) (b) Point Spread function

62 Application: Image Super-Resolution (a)original ( ) (b) Super-res ( )

63 Application: Image Super-Resolution (a)original (48 48) (b) Super-res ( )

64 Application in Neuroscience Applications Neural in Neuroscience acquisi)on and sor)ng 8 Neuroprosthesis ADC Spike sor)ng Processing unit The energy budget of implanted devices is limited. High sampling rates impose wired data transmission and limits the quality and diversity of experiments.

65 Objec)ve Application in Neuroscience Insight: Sample at lower rate and reconstruct the signal outside the implant Sample at a lower rate and reconstruct the signal outside the implant. 6 Neuroprosthesis ADC Classical Processing unit Spike sor)ng Neuroprosthesis ADC Sparse Reconstruc)on Processing unit Spike sor)ng

66 Application in Neuroscience Spike sor)ng valida)on (iii) Classical Sampling (C) f s = 24KHz Sparse Classical Sampling sampling (F) f s = at 5.8KHz. Sparse sampling at (x4 undersampling) % 90% 80% 70% 60% 50% 40% 30% 20% 10% 0% C F C F C F C F C F σ = 0.05 σ = 0.1 σ = 0.15 σ = 0.2 Average ClassificaFon Error DetecFon Error Correctly Sorted

67 Calcium Transient Detection Figure 6: Double consistency spike search. (i) and (ii) show the detected locations in red and the locations of the original spikes in green for two different window sizes. In (i) the algorithm runs estimating the number of spikes within the sliding window. In (ii) the algorithm runs assuming a fixed number of spikes equal to one for each position of the sliding window. (iii) shows the joint histogram of the detected locations. (iv) shows the fluorescence signal in blue with the original spikes in green and the detected spikes in red.

68 Application in Sensor Networks Localizing Point Sources in Diffusion Fields Localized and instantaneous sources: KX f t = 4f + c k (x x k) (t t k) Diffusion field: k=1 KX c k f(x,t)= 4 (t t e kx xkk 2 4(t t k) U(t t k) k=1 k) finite degrees of freedom p 4 p 3 x 2 x 1 p 5 p p 2 1 Goal: Es-mate the unknown parameters {c k} k, {t k} k, {x k} k from the spa-otemporal samples taken by distributed sensors. Source Localiza-on in Diffusive Fields IMA 2012 Tuesday, 11 September 12 8

69 Sampling signals using sparsity models: Conclusions and Outlook New framework that allows the sampling and reconstruction of signals at a rate smaller than Nyquist rate. It is a non-linear problem Different possible algorithms with various degrees of efficiency and robustness Applications: Many actual and potential applications: But you need to fit the right model! Carve the right algorithm for your problem: continuous/discrete, fast/ complex, redundant/ not-redundant Still many open questions from theory to practice!

70 On sparsity in over-complete dictionaries References D. L. Donoho and P.B. Starck, Uncertainty principles and signal recovery, SIAM J. Appl. Math., 1989, pp D.L. Donoho and X. Huo, Uncertainty principles and ideal atomic decomposition, IEEE Trans. on Info.. Theory, vol.47(7, pp , November M. Elad, Sparse and Redundant Representations, Springer, On Compressed Sensing and its applications E. J. Candès, J. Romberg, and T. Tao, Robust Uncertainty Principle: Exact signal reconstruction from highly incomplete frequency information, IEEE Trans. Info. Theory, vol. 52(2), pp , February E.J. Candés and M.B. Wakin, An introduction to compressive sampling, IEEE Signal Processing Magazine, vol. 25(2), pp , March M. Lustig, D.L. Donoho, J.M. Santos and J.M. Pauly, Compressed Sensing MRI, IEEE Signal Processing Magazine, vol. 25(2), pp , March 2008.

71 References On sampling FRI Signals M. Vetterli, P. Marziliano and T.Blu, Sampling Signals with Finite Rate of Innovation, IEEE Trans. on Signal Processing, 50(6): , June T. Blu, P.L. Dragotti, M. Vetterli, P. Marziliano and L. Coulot of Signal Innovations: Theory, Algorithms and Performance Bounds, IEEE Signal Processing Magazine, vol. 25(2), pp , March 2008 P.L. Dragotti, M. Vetterli and T. Blu, Sampling Moments and Reconstructing Signals of Finite Rate of Innovation: Shannon meets Strang-Fix, IEEE Trans. on Signal Processing, vol.55 (5), pp , May J.Berent and P.L. Dragotti, and T. Blu, Sampling Piecewise Sinusoidal Signals with Finite Rate of Innovation Methods, IEEE Transactions on Signal Processing, Vol. 58(2),pp , February J. Uriguen, P.L. Dragotti and T. Blu, On the Exponential Reproducing Kernels for Sampling Signals with Finite Rate of Innovation in Proc. of Sampling Theory and Application Conference, Singapore, May P.L. Dragotti, M. Vetterli and T. Blu, Exact Sampling Results for signals with finite rate of innovation using Strang-Fix conditions and local kernels in Proc. of ICASSP, Philadelphia, March 2005.

72 On Image Super-Resolution References (cont d) L. Baboulaz and P.L. Dragotti, Exact Feature Extraction using Finite Rate of Innovation Principles with an Application to Image Super-Resolution, IEEE Trans. on Image Processing, vol.18(2), pp , February On Diffusion Fields Y. Lu, P.L. Dragotti and M. Vetterli, Localization of diffusive sources using spatio-temporal measurements, 49th Allerton Conference, Allerton D. Malioutov, M. Cetin and A. Willsky, A sparse signal reconstruction perspective for source localization with sensor arrays, IEEE Trans, on Signal Processing, vol. 53(8) pp , August On Neuroscience: J. Onativia, S. Schultz and P.L. Dragotti, A Finite Rate of Innovation algorithm for fast and accurate spike detection from two-photon calcium imaging, submitted to Journal of Neural Engineering, Nov J. Caballero, J.A. Uriguen, S. Schultz and P.L. Dragotti, Spike Sorting at Sub-Nyquist Rates, in Proc. of IEEE (ICASSP), Kyoto, Japan, April 2012.

73 Appendix Orthogonal matching pursuit (OMP) finds the correct sparse representation when K < «. (4) 2 µ Sketch of the Proof (Elad 2010, pages 65-67): Assume the K non-zero entries are at the beginning of the vector in descending order with y = Dx. Thus KX y = x l D l (5) First iteration of OMP work properly if D1 T y > Di T y for any i > K. Using (5) KX KX x l D1 T D l > x l Di T D l l=1 l=1 l=1

74 Sketch of the Proof (cont d): But l=1 Appendix (cont d) KX KX KX x l D1 T D l x 1 x l D1 T D l x 1 x l µ x 1 (1 µ)(k 1). Moreover, l=2 l=2 KX x l Di T l=1 D l KX x l Di T D l l=1 KX x l µ x 1 µk l=1 Using these two bounds, we conclude that D1 T y > Di T condition (4) is met. y is satisfied when