Slepian Functions Why, What, How?

Transcription

1 Slepian Functions Why, What, How? Lecture 1: Basics of Fourier and Fourier-Legendre Transforms Lecture : Spatiospectral Concentration Problem Cartesian & Spherical Cases Lecture 3: Slepian Functions and Sequences, and Applications

2 In general, one may think of the Slepian functions in connection with the tradeoff between concentrating a signal in the time (or, space) domain versus the frequency domain principal applications in communications theory (D. Slepian, Bell Labs, and others) geophysical applications (F. Simons, Princeton U., and others) A brief review of spectral methods prepares for the fundamental setup of the spatiospectral concentration problem Instead of the time variable, we consider a spatial variable, primarily in one dimension, but with generalization to two dimensions, in particular the sphere, on which many geodetic and geophysical signals reside 1.1

3 Fourier transform pair for continuous, integrable functions, g ( x) G( f ) i π f x g x = G f e df, < x < i π f x G f = g x e dx, < f < G(f) = F(g(x)) is the Fourier transform (spectrum) of g(x) g(x) = F 1 (G(f)) is the inverse Fourier transform of G(f) Discrete Fourier transform pair for finite sequences, gɶ l Gɶ k N 1 π i kl ɶ N k 1 gɶ = l G e, l N 1 N x k= N 1 π i kl ɶ N l Gɶ k = x g e, k N 1 l= 1.

4 Fourier transforms and their inverses have many properties similarity, or scaling, property (Fourier integral transform) 1 f a 1 a g ax G 1 a a is a constant Proof: iπ fax 1 f iπ f x g ( ax) = G( f ) e df = G e df, f = af a a Proof: an expansion of the spatial domain implies a contraction of the frequency domain, and vice versa Parseval s theorem: = g x dx G f df * i π f x * iπ f x = = = g x dx g x G f e df dx G f g x e dxdf G f df total energy of a function is the same in space domain as in frequency domain 1.3

5 y x = g x h x = g x h x x dx Convolution: * Fourier transform (convolution theorem): Y ( f ) = G( f ) H ( f ) Proof: iπ f x iπ f x iπ f ( x x ) Y ( f ) = g ( x ) h( x x ) dx e dx = g ( x ) e h( x x ) e dx dx useful also for numerical calculation of convolution: 1 ( ( ) ) y x =F F g x F h x Also, Fourier transform of a product is the convolution of Fourier transforms dual convolution theorem: w( x) = g ( x) h( x) W ( f ) = G( f )* H ( f ) 1.4

6 Example 1: Rectangle function b x 1, x < 1 =.5, x = 1, x > 1 B f ( π f ) sin = π f sinc ( f ) b(x) is extent-limited ; its spectrum is different from zero almost everywhere 1.5

7 MathCad Ch_rectangle_fncs.xmcd Digression: Dirac delta function δ ( x) 1 x = lim b T T T hence, δ ( x) =, x δ ( x) dx = 1 Fourier transform: ( f ) = 1, for all f Proof: F δ ( x ) B( Tf ) ( ( Tf ) ) T T = lim = lim sinc = 1 by convolution theorem g ( x ) δ ( x x ) dx = g ( x) 1.6

8 Example : Gaussian function Fourier transform: γ Γ β 1 π x β ( x) = e β >, a constant β ( β ) π ( β f ) ( f ) = e Figures.ppt Figures.ppt varying β demonstrates similarity (scaling) property 1.7

9 Extent-limited function: g ( x) =, x > T Band-limited function: G f =, f > f bandwidth = f Fundamental result: A (non-trivial) function cannot be both extent-limited and band-limited Proof (heuristic): If band-limited, then extended over the complex plane it is an entire function (all derivatives exist everywhere). Thus, if then also extent-limited, i.e., zero on some interval, then it must be zero everywhere. 1.8

10 Band-limited approximation of g(x): f Gibbs s Effect = = iπ fx iπ fx iπ fx g f x G f e df g x e dx e df f f f ( π f ( x x )) f iπ f ( x x) 1 sin = g x e df dx = g x dx π x x f MathCad: Ch3_Gibbs_phenom.xmcd b( x) b( x) g x b f = ( x) b f = 1 ( x) x 1.9

11 Questions: In the space of extent-limited functions (given T), which one has most of its energy concentrated in a given spectral band? In the space of band-limited functions (given f ), which one has most of its energy concentrated in a given spatial extent? Answers: Series of 5 papers, , by D. Slepian, H.O. Pollak, H.J. Landau (Bell Labs) Prolate Spheroidal Wave Functions Fourier Analysis and Uncertainty I, II, III, IV, V Sleptian, D. (1983), SIAM Review, 5(3): I am going to use this occasion to tell you in detail about a problem in Fourier analysis that arose in a quite natural manner in a corner of electrical engineering known as Communication Theory. The problem was first attacked more than years ago jointly by me and two colleagues at Bell Labs-Henry Pollak and Henry Landau. It differed from other problems I have worked on in two fundamental ways. First, we solved it--completely, easily and quickly. Second, the answer was interesting-even elegant and beautiful. [It] had so much unexpected structure that we soon saw that we had solved many other problems as well. 1.1

12 Spherical domain: Fourier-Legendre transform pair {,, } Ω = θ λ θ π λ π n (, ) (, ), (, ) g θ λ = G Y θ λ θ λ Ω n= m= n n, m n, m 1 Gn, m = g ( θ, λ ) Yn, m ( θ, λ ) dω, n m n, n 4π Ω Parseval s Theorem 1 4π Ω n ( θ, λ ) Ω n, m g d G = n= m= n g θ, λ =, θ, λ ΩR Ω Extent-limited function: Band-limited function: G, =, n > nmax n m Same questions may be posed First studied to some extent by (Gilbert and) Slepian in 197s, Albertella et al. (1999), and a few others in other fields; see references in (Simons et al. 6) Comprehensively developed by F. Simons, F. Dahlen, and colleagues in the s 1.11

13 Slepian Functions Why, What, How? Lecture 1: Basics of Fourier and Fourier-Legendre Transforms Lecture : Spatiospectral Concentration Problem Cartesian & Spherical Cases Lecture 3: Slepian Functions and Sequences, and Applications

14 Recall - Questions: In the space of extent-limited functions (given T), which one has most of its energy concentrated in a given spectral band, f < f? In the space of band-limited functions (given f ), which one has most of its energy concentrated in a given spatial extent, x < T/? Note: Gaussian is practically extent- and band-limited, but we seek functions of two parameters Figures.ppt γ β 1 π x β ( x) = e β Γ ( β ) π ( β f ) ( f ) = e β >, a constant.

15 Spectral concentration problem we seek an extent-limited function whose spectral energy is maximally concentrated in a finite bandwidth The finite extent and bandwidth are independently given finite-extent function: g ( x) T g ( x), x T =, x > T Fourier transform: define finite bandwidth: f f T =, < < iπ fx GT f gt x e dx f T Consider the ratio: λ ( f ) T = f f T T G f df G f df Find the function that maximizes this ratio.1

16 Numerator: substituting the Fourier transform Denominator: substitute Parseval s relationship λ T ( f ) f T T iπ ( x x ) f gt ( x) gt ( x ) e df dxdx gt ( x) gt ( x ) D( x x ; f ) dxdx f T T = = T where ( ; ) T T T g x dx g x dx Variational problem: Find g T (x) such that small variations produce no variation in λ T (f ) in the limit. Use to denote a small variation and consider ( ) sin π ( ) π ( x x ) iπ x x f D x x f = e df = f f T T T ( ; ) λ f g x dx = g x g x D x x f dxdx. T T T T T T T λ T ( x x f ) ( f ) = f f T T G f df G f df

17 After some manipulations, we can arrive at (details provided in supplemental notes) λ T T f g x dx T T T T = ( gt ( x ) D( x x ; f ) λt ( f ) gt ( x ) δ ( x x )) dx gt ( x) dx T T The left side goes to zero for arbitrarily small variations, g T (x), only if the parenthetical integral vanishes: T T ( ) ( ; ) = λ g x D x x f dx f g x T T T this is a homogeneous Fredholm integral equation of the second kind Before solving this equation, we consider the complementary spatial concentration problem.3

18 We seek a band-limited function whose spectral energy is maximally concentrated in a finite extent band-limited function: G ( f ) f G f, f f =, f > f inverse Fourier transform: define finite extent: x T Consider the ratio: f = < < iπ fx g f x G f f e df, x f µ f ( T ) T T g f x dx = = * ( ; ) f f f f f f G f G f D f f T dfdf f f f f g x dx G f df.4

19 The variational problem is solved, as before, by a solution to the integral equation f f ( ) ( ; ) = µ G f D f f T df T G f f f f A change in variable reduces each integral equation to a more convenient form or 1 1 ψ ( c ) y D y y dy = λ ψ y c c c c ( c) y = x T, ψ y = g Ty, λ = λ f ( c ) f T ( c) T y = f f, ψ y = G f y, λ = µ T f c = πtf where and ( ( )) ( y y ) ( c ) sin c y y c D ( y y ) = = sinc c y y π π ( π ).5

20 Repeating: 1 1 ψ y D y y dy = λ ψ y c c c c the left side is almost a convolution of ψ (c) (y) with the sinc function therefore, in the dual domain, this is almost a product of the transform of ψ (c) (y) and a rectangle function e.g., for the first (spectral) concentration problem, this almost-convolution, transformed to the frequency domain, is an almost-product of the transform of ψ (c) (y) and the rectangle function of frequency i.e., the right side is almost band-limited Finding the solution(s) to the integral equation is an eigenfunction/eigenvalue problem: Dψ = λψ This is explored in more detail in the next lecture the solution to the concentration problem is the eigenfunction with the maximum eigenvalue.6

21 Instead of a function on the continuous domain, the spatiospectral concentration problem can be formulated also for sequences, which addresses more practical applications in the case of spectral concentration extent-limited sequence ( g ) N l gl, N l N 1 =, l < N or l > N 1 Fourier transform: ɶ Concentration ratio λ ( f ) f N 1 iπ f x GN f = x l gn e, f f < f l N N l= N N 1 N 1 Gɶ ( f ) df x ( gn ) ( g N ) D( x( l l ); f l l ) f l= N l = N = = = N Gɶ ( f ) df ( gn ) l N f N 1 T where f N sin ( l l ) D ll, π ( l l ) l= N c c c N D = = g = g N l ( c) T g D g g g.7

22 Maximizing the concentration ratio is equivalent to find the eigenvector (the finiteextent sequence) with maximum eigenvalue for the vector-matrix equation: ( c) ( c) D g = λ g this is investigated in more detail in the next lecture An illustrative example rectangle sequence: Fourier transform: N N ( N ) 1, l 1 bl =, otherwise ( Nπ xf ) ( π xf ) N 1 N iπ x f sin iπ xf = l = sin l= N Bɶ f x e x e compare spectral energy distribution of this to that of solution to spectral concentration problem.8

23 Choose parameters: x = 1 N = 5 f = 1. N x = (first zero of Bɶ ( N ) ( f ) ) 1 MathCad: discrete_prolate_spheroidal_sequence.xmcd rectangle sequence dpss eigensequence* with the maximum eigenvalue..15 MathCad: discrete_prolate_spheroidal_sequence.xmcd frequency, f sequence index *called discrete prolate spheroidal sequence, dpss the energy of the rectangle sequence is not well concentrated in the band, f < f spectral energy of eigensequence is well concentrated in the spectral band, f < f.9

24 Finally, we can set up the spatiospectral concentration problem for the sphere consider only the second problem of finding the band-limited function whose energy is most concentrated on a given patch of the sphere this leads to the problem of how to represent most efficiently a given spherical harmonic model of the potential in a given area K n,, band-limited function: g ( θ λ ) G Y ( θ λ ) = K n, m n, m n= m= n limited extent: ΩR Ω Concentration ratio: K n K n K n, m n, m n, m n, m 1 1 g θ, λ dω G G Y ( θ, λ ) Y ( θ, λ ) dω 4π 4π Ω n= m= n n = m = n R K R = = K n 1 K ( θ, λ ) Ω µ Ω 4π Ω g d G n= m= n Ω R n, m.11

25 That is, where µ Ω K n K n n = m = n n = m = n K R K n G G D n= m= n ( Ω ) ( Ω ) n, m n, m n, m, n, m R T = = G n, m 1 Dn, m, n, m ( ΩR ) = Yn, m ( θ, λ ) Yn, m ( θ, λ ) dω 4π Ω R G D R T G G G Maximizing this ratio leads to the matrix equation, ( Ω ) ( Ω ) R R D G G = µ The difficult part is the integration of the product of spherical harmonics over Ω R if Ω R is a polar cap, {,, } Ω = Ω = θ λ λ π θ θ R s s then θ s ( m ) ε m Dn, n ( θs ) = P, ( cosθ ) P, ( cosθ ) sinθdθ n m n m θ = Solution is the eigenvector, G, with maximum eigenvalue; it contains the spectrum of the band-limited function with maximally concentrated energy in the polar cap.1

26 Compare the truncated (band-limited) Dirac delta function on the sphere with the solution to the spatial concentration problem band-limited Dirac delta function: δ ( θ ) ( n 1) P ( cosθ ) K K = + n= Legendre spectrum is rectangle in frequency domain maximally concentrated function on polar cap: g(θ), θ s = 1º both δ K (θ) and g(θ) are band-limited, K = 6 n Legendre spectrum of g(θ).13

27 Slepian Functions Why, What, How? Lecture 1: Basics of Fourier and Fourier-Legendre Transforms Lecture : Spatiospectral Concentration Problem Cartesian & Spherical Cases Lecture 3: Slepian Functions and Sequences, and Applications

28 Recall the equation to be solved for the spatiospectral concentration problem: 1 1 ψ y D y y dy = λ ψ y c c c c where the kernel function is ( ( )) ( y y ) ( c ) sin c y y D ( y y ) = π c = πtf From the theory of integral equations, it is known that for this particular kernel, the integral operator has eigenvalues, λ (c), that are real, distinct, and countably infinite ( c) ( c) 1 > λ > λ > 1 the eigenvalues, being concentration ratios, must all be less than unity the corresponding eigenfunctions are orthogonal on both [ 1,1], as well as (, ), and can be normalized so that they are orthonormal on [ 1,1] 1 ( c ) ( c ψ ) k y ψl ( y) dy = δk l 1 Interestingly, like the Gaussian function, the eigenfunction and its Fourier transform have the same form, one a scaled version of the other see the supplemental notes 3.1

29 The eigenfunction for the maximum eigenvalue solves the concentration problem The eigenfunction for the next largest eigenvalue is next best in concentrating its energy, and so on The total set of eigenfunctions is a basis for the space of extent- (or band-) limited functions however, not all eigenfunctions are needed in practice to represent an extent- (or band-) limited function Analogous to the eigenvalues of a matrix operator, the sum of the eigenvalues equals the trace of the kernel function of the integral equation 1 1 ( c) ( c λ ) k k= 1 1 c c = D y y dy = dy = = ft π π it can be shown that the values of the eigenvalues transition sharply from near unity to near zero therefore, the number of significant eigenvalues, and correspondingly significant eigenfunctions, is approximately c ft 1 = k= E f T λ ( c) k This is known as the Shannon number (also the extent-bandwidth product) c = πtf 3.

30 By a seemingly lucky accident (Slepian 1983), the eigenfunctions are also certain solutions to the Helmholz wave equation in prolate spheroidal coordinates 1 d u du + = ζ, χ are constants dy dy ( y ) y ( ζ χ y ) u ( y) thus, the eigenfunctions of the integral operator are also known as prolate spheroidal wave functions Indeed, writing the integral and differential equations in terms of operators, that are operating on an arbitrary function, h(y), 1 1 ( c ) ( c) D h y = D y y h y dy = λ h y d dh ( c) L h( y) = ( 1 y ) + c y h( y) = λ h y dy dy it can be shown that these operators commute see the supplemental notes ( h( y) ) = L D h( y) D L Hence they have the same eigenfunctions proved on next slide 3.3

31 To prove that D and L share eigenfunctions, let h (y) be an eigenfunction of D then = λ D h y h y ( h y ) = λ h ( y) ( h ( y) ) = λ h ( y) L D L D L L and, Lh (y) is an eigenfunction of D with the same eigenvalue this means Lh (y) and h (y) are linearly related, i.e., the latter is an eigenfunction of L it turns out that finding the eigenfunctions of L is easier than for D Once the eigenfunction, ψ k (y), is obtained, the eigenvalue with respect to D is simply λ k ψ = D ψ k k ( y) ( y) These ideas considerably simplify also the practical application with sequences, i.e., functions defined on a discrete domain 3.4

32 For sequences, recall that the spectral concentration problem is solved by finding the eigenvector with maximum eigenvalue for the matrix equation, ( c) ( c) D g = λ g ( c) ( c) sin c( l l ) N where D = D ll, = g = [ gl ] π ( l l ) Slepian (1978) found a difference equation whose operator commutes with D (c) n c N 1 1 N n gn 1 + cos n ζ gn + ( n + 1)( N 1 n) gn+ 1 = N c N 1 1 cos ( 1)( N 1) N 1 ( 1 )( N 1 ) g g g 1 g 1 c N 1 cos n = ζ N gn gn 1 ( N 1)( 1) g N 1 g N 1 1 c N 1 ( N 1)( 1) cos N + 1 N or, ( c) S g = ζ g and: ( c) ( c) ( c) ( c) S D = D S 3.5

33 Thus, S (c) and D (c) commute and have the same eigenvectors, but not the same eigenvalues. Having obtained the eigenvectors for S (c), hence D (c), the eigenvalues with respect to D (c) are ( c) ( c) D g ( c) k λ k =, k =,, N 1 ( c) g k For example, N = 6 (extent-limited sequence), f =.5 (Shannon number = N xf = 6) 3.6

34 The Fourier spectra of the eigensequences show good concentration of energy for k less than the Shannon number, k < 6 frequency G G G G ( c ) ( f ) 6 ( c ) ( f ) 4 ( c ) ( f ) ( c ) ( f ) Similar results are obtained for the spherical case instead, later, we consider the spatial concentration problem 3.7

35 Estimating the Power Spectral Density Definition of power spectral density (PSD) Fourier transform of the covariance function of a stationary stochastic process i ( ξ ) = ξ covariance function: cg, g ( ξ ) =E ( gxg x+ ξ ) πξ f Cg, g f cg, g e d Estimate of the covariance function: (, ) N 1 1 l g g l n n+ l N n= N c = g g, l =,, N 1 It can be shown that an estimator of the PSD at discrete frequencies is 1 1 C g, g fk Gk gn f k k k N x N x N x = ɶ = DFT, = ( ) this is known as the periodogram it is a biased estimator due to aliasing and spectral leakage 3.8

36 Aliasing: due to sampling of a function with spectral content beyond the spatial resolution defined by the sampling; i.e., content beyond the Nyquist frequency, f > f N = 1/( x) Spectral leakage: blurring and/or biasing of a spectrum due to truncation of the function. using finite extent of a function, i.e., windowing a function, is the same as multiplying by the rectangle function g x, T x T x gt ( x) = = b g x, otherwise T by the dual convolution theorem, the Fourier transform of g T (x) is the convolution of the Fourier transforms of b(x/t) and g(x) GT ( f ) = TB( ft )* G( f ) = T G( f ) sinc( ( f f ) T ) df 3.9

37 Example of Spectral Leakage polar motion MathCad Ch1_polar_motion.xmcd x p amplitude spectrum [arcsec/(cy/yr)] 5 years of data frequency [cy/yr] MathCad Ch1_polar_motion.xmcd MathCad Ch1_polar_motion.xmcd 6 years of data 8 years of data MathCad Ch1_polar_motion.xmcd 1 years of data 3.1

38 Window Function, Data Taper MathCad Ch3_windows_comparison.xmcd MathCad Ch3_windows_comparison.xmcd extent frequency MathCad Ch1_polar_motion.xmcd x p amplitude spectrum [arcsec/(cy/yr)] 1 years of data frequency [cy/yr] 3.11

39 Another Example PSD Determination MathCad: Magnetic data periodograms with tapers.xmcd MathCad: Magnetic data periodograms with tapers.xmcd MathCad: Magnetic data periodograms with tapers.xmcd 3.1

40 Recall the matrix equation for the spherical spatial concentration problem ( Ω ) ( Ω ) R R D G G where = µ n= ( K 1) ( Ω ) D ( Ω ) K n K n R 1 = Dn, m, n, m ( Ω ) R Yn, m ( θ, λ ) Yn, m ( θ, λ ) dω = 4π ΩR The eigenvalues, µ (Ω R), are real and distinct; the eigenvectors, G, are orthonormal G j = (( G ) ) T, GK, K The eigenfunctions, g ( θ λ ) ( G ) Y ( θ λ ) and form a basis for band-limited functions; they are also orthonormal on Ω The sum of eigenvalues is the trace of D (Ω R) and approximates the Shannon number ( Ω ) ( ) 1 R ΩR E µ n, m = tr D = Yn, m ( θ, λ ) Yn, m ( θ, λ ) dω 4π n= m= n n= m= n Ω K 1 = ( n + 1) Pn ( cos ) dω 4π = + j A j n, m j n, m n= m= n 4π R K n,, Ω R j G G =, are orthogonal on Ω R ; R = δ T j k j k where A(Ω R ) is the area of Ω R 3.13

41 For the special case of a polar cap for the concentration = = region, {,, } Ω Ω θ λ λ π θ θ R s s For m = K,,K, the matrix, D (m) (θ s ), commutes with T (m) (θ s ), defined by ( m ( θs )) T = n n + 1 cos θ, n = m,, K ( m ) ( m ( ( θ )) ) s ( θs ) ( m ( T ) ( θs )) n m + 1, n m + 1 n m + 1, n m + n m +, n m + 1 n, n ( K ( D ) ( θs )) 1 1 ( ( R ) 1 D Ω = D ( θ )) s K K ( m ) ( diag D ( θ ) ( ) s = D ( θ s )) K + 1 K + 1 ( 1 ( ) D ( θs )) K K ( K D ( θs )) 1 1 ( n + 1) m ( n + 1)( n + 3) T = T = n n + K K +, n = m,, K 1 =, otherwise s θ s ( m ) ε m Dn, n ( θs ) = P, ( cosθ ) P, ( cosθ ) sinθdθ n m n m θ = T (m) (θ s ) is tri-diagonal and symmetric 3.14

42 ( m ) m The eigenvectors, G j ( θ s ), j = 1,, K m + 1, of T θ s are eigenvectors of D ( m ) θ s eigenvalues are obtained from: ( m ) µ θ ( m ) ( θ ) = D G j s m G j ( θ ) ( m ) ( θ ) s j s s 1 s 1 s α s K + 1 s eigenvalues can be ranked: > µ ( θ ) > µ ( θ ) > > µ ( θ ) > > µ ( θ ) each α corresponds to a particular m and j Example: K = 6, θ s = 1º, E = 8; (K + 1) =

43 The corresponding eigenfunctions (basis functions for band-limited functions) are For λ = º g α (θ,λ = ) K cos mλ, m gα ( θ, λ ) = ( Gn, m ) P, ( cosθ n m ) j sin m λ, m < g g 1 n= m g 3 g6 g1 g 15 g 19 g 3 polar angle, θ [deg] g 3 µ α α m 3.16

44 For θ s = 1º, K = 1, E = 77 (L. Wang 1, OSU Report 498) 3.17

45 Given a band-limited representation of a function (signal) on the sphere, K n s, S Y, ( θ λ ) ( θ λ ) = n= m= n n, m n, m we wish to represent it with maximum efficiency in a local domain i.e., fewer basis functions, but with the full energy of the signal ( K + 1) α = 1 α = 1 s θ, λ S g θ, λ S g θ, λ where E is the Shannon number ( + 1) Thus, ( θ λ ) = K K n β = 1 n= m= n α α α α E n, m n, m ( θ λ ) S g, = S Y, β β with g ( θ λ ) ( G ) Y ( θ λ ) α K n,, = and orthonormality of g α, n= m= n n, m α n, m as well as Y n,m, on Ω K S S G α n = n= m= n ( n m ) n, m, α 3.18

46 Model-predicted co-seismic gravity changes for 1 Chile earthquake (Mw 8.8), band-limited to spherical harmonic degree and order 1; ±8 microgal (L. Wang 1; OSU Report 498) For θ s = 1º, K = 1, E = 77 full spherical harmonic series, 11 spherical harmonics Slepian series, 77 basis functions rms difference:.14% of signal in cap Sn, m S α 3.19