1. Fourier Transform (Continuous time) A finite energy signal is a signal f(t) for which. f(t) 2 dt < Scalar product: f(t)g(t)dt

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1 1. Fourier Transform (Continuous time) 1.1. Signals with finite energy A finite energy signal is a signal f(t) for which Scalar product: f(t) 2 dt < f(t), g(t) = 1 2π f(t)g(t)dt The Hilbert space of all these functions is L 2 (IR) or L 2 of the... Page 1 of 100

2 1.2. The Fourier Transform Suppose f(t) L 2, then its Fourier-transform is This is: Inverse transform F (ω) = F{f(t)} = 1 2π f(t) = 1 2π Convolution product then: (f g)(t) = F (ω) = e iωt, f(t) f(t)e iωt dt F (ω)e iωt dω = e iωt, F (ω) f(u)g(t u)du = F{f g} = F{f(t)}F{g(t)} f(t u)g(u)du of the... Page 2 of 100

3 1.3. Parseval s identity The Fourier transform preserves energy: or: f(t) = F (ω) f(t) 2 dt = F (ω) 2 dω This is Parseval s identity. It can be stated more generally: f(t), g(t) = F (ω), G(ω) The norm-equivalence follows for g(t) = f(t). of the... Page 3 of 100

4 1.4. Heisenberg s uncertainty In (t, ω) plane, signal concentrated at (t 0, ω 0 ) f(t) 2 tdt F (ω) 2 ωdω IR t 0 = IR ; ω 0 = f(t) 2 dt F (ω) 2 dω IR with variances σt 2 = IR f(t) 2 (t t 0 ) 2 dt ; σω 2 = f(t) 2 dt then: IR IR IR F (ω) 2 (ω ω 0 ) 2 dω F (ω) 2 dω σ t σ ω 1 2 or σ2 t σ 2 ω 1 4 Higher resolution in the time domain Lower resolution in the frequency domain IR of the... Page 4 of 100

5 1.5. Time-frequency plane! "#%$'&)( *,+.-/! 1. basis functions δ(t t k ) give 1st picture 2. basis functions e iω kt give 2nd picture how to get pictures like 3 or 4? needs local basis in (t, ω)-plane 3. is more natural for human observation (sound/vision) = wavelets 4. corresponds to windowed or short time Fourier transform (WFT/STFT) WFT is older than wavelets of the... Page 5 of 100

6 1.6. Windowed Fourier Transform (WFT) Use a windowed instead of a pure sine/cosine: F ψ (f(t)) = 1 f(t)e iωt ψ(x b)dt 2π ψ is a (localizing) window function: for example a Gaussian: ψ(t) = e t2 /2σ 0 This leads to the Gabor transform real part psi(t)*sin(3*t) psi(t) -psi(t) imaginary part psi(t)*cos(3*t) psi(t) -psi(t) However, this approach has a drawback: the time window is constant for all frequencies: for high frequencies, we have more oscillations in the window. of the... Page 6 of 100

7 1.7. The continuous wavelet transform (CWT) (sneak preview) W ψ (a, b) = 1 a ( t b f(t)ψ a ) dt This is an overcomplete function representation. (from one dimension, we transform into a 2D space!!). the translation factor b shifts along t-axis the scaling factor a shifts along ω-axis shorter time support higher frequencies for example mexican hat of the... Page 7 of 100

8 Objective and Definition approximation of finite energy, continuous time signals. multi-resolution analysis: { } successive ( multi ) approximations ( resolutions ) of a function. Successive basis functions all have same shape: translations and dilations of one function. of the... Page 8 of 100

9 The precise mathematical definition is as follows: A multi-resolution analysis (MRA) is a sequence of closed subspaces V j (j ZZ) of L 2 (IR) with the following properties 1. V j V j+1 2. (a) (b) + j= + j= V j is dense in L 2 (IR) i.e.: + j= V j = {0} (0 is constant function) 3. (shift invariance) v(t) V 0 v(t k) V 0, k ZZ 4. (scale invariance) v(t) V j v(2t) V j+1, j ZZ 5. (shift invariant basis) ϕ(t) V 0 : {ϕ(t k)} k ZZ is a stable basis (Riesz-basis) for V 0 V j = L 2 (IR) of the... Page 9 of 100

10 2.2. The dilation equation (refinement equation, two-scale equation) The function ϕ(t) is called scaling function or father function. Dilation equation: Immediate consequence: ϕ(t)dt = k ZZ Hence, choosing we get {c k } k ZZ : ϕ(t) = k ZZ c k ϕ(2t k) c k ϕ(2t k)dt = k ZZ ϕ(t)dt = θ 0 Φ(0) = 1 2π c k = 2 k ZZ 1 c k ϕ(t)dt 2 ϕ(t)dt = θ 2π of the... Page 10 of 100

11 2.3. Example V 1 V 0 of the Terminology: V V -1-2 j is called the resolution level 2 j is called the scale Page 11 of 100

12 2.4. Solution of the dilation equation By iteration Numerical solution in time domain: Hence, start with (for instance) ϕ(t) = k ZZ c k ϕ(2t k) ϕ (0) (t) = χ [0,1] (t) (Box function on [0, 1]) and iterate ϕ (i) (t) = k ZZ c k ϕ (i 1) (2t k) until convergence. This algorithm is proposed in continuous time. While implementing it (e.g. in MATLAB), be careful with discretisation!! Theorem if c n = 0 for n < n and for n > n +, then supp(ϕ) [n, n + ] of the... Page 12 of 100

13 By recursion The iteration procedure starts with an initial value in all points t (in practice only in points of discretisation) Each iteration step computes a new function value in each point. Recursion starts from the exact function values in some points and uses the dilation equation to compute other points, exactly (no function value iteration, the iteration is now on the number of points) We have: ϕ( t 2 ) = k ZZ c k ϕ(t k) If we know ϕ(t) for all integer t, then we can compute it for all t = k/2. Repeating this yields ϕ(t) for all dyadic points t = k/2 j. We know ϕ(k) = 0 if k {n,... n + }. of the... Page 13 of 100

14 Example: n = 0, n + = 5. Then we have ϕ(0) c 0 ϕ(1) c 2 c 1 c 0 ϕ(2) ϕ(3) = c 4 c 3 c 2 c 1 c 0 c 5 c 4 c 3 c 2 c 1 ϕ(4) c 5 c 4 c 3 ϕ(5) c 5 ϕ(0) ϕ(1) ϕ(2) ϕ(3) ϕ(4) ϕ(5) The matrix must have eigenvalue 1 and the function values are the components of the corresponding eigenvector. We call this matrix C. From this, it follows: ϕ(n ) = 0 = ϕ(n + ) (unless c n = 1 or c n + = 1) of the... Page 14 of 100

15 Cascade algorithm Recall V j V j+1. Suppose f(t) = s j,k ϕ(2 j t k) = s j+1,l ϕ(2 j+1 t l) k l Use ϕ(x) = i c iϕ(2x i) to compute s j+1,k from s j,k Result: s j+1,l = k c l 2k s j,k Now, start the iteration with s 0,k = δ k. This corresponds to f(t) = ϕ(t). If j, the support of ϕ(2 j t k) gets smaller and smaller. At a certain moment it can only be represented by one pixel. The s j,k at this level are an approximation for ϕ(k/2 j ). of the... Page 15 of 100

16 Fourier domain solution Define the Fourier transform Φ(ω) = 1 2π ϕ(t)e iωt dt Then the dilation equation becomes in the Fourier domain Φ(ω) = C( ω 2 )Φ(ω 2 ) = = Φ(ω/2k ) with C(ω) = c k e iωk. k Since lim k ω/2 k = 0, and Φ(0) = k C(ω/2 j ) j=1 θ 2π, we have: of the... Φ(ω) = θ C(ω/2 j ) 2π j=1 Page 16 of 100

17 3. Properties of the father function Theorem: If ϕ(t) is Riemann-integrable, then ϕ(t k) = θ Theorem: We have k ZZ This is called partition of unity. k ZZ ϕ(t k) = θ n ZZ c 2n = n ZZ C(ω) = 1 c k e ikω 2 k ZZ c 2n+1 C(π) = 0 and since k ZZ ( 1)k c k = 0 this becomes: C(π) = 0 and since k ZZ c k = 2 also C(0) = 1 of the... Page 17 of 100

18 4. L 2 is a unitary space with dot/inner/scalar product: f(t), g(t) = 1 2π f(t)g(t)dt This allows to define the concept of orthogonality: f(t) g(t) f(t), g(t) = 0 If ϕ(t k) ϕ(t l), k l, we have an orthogonal basis. If f(t) = k ZZ α k ϕ(t k) we can easily find α m : multiply both sides by ϕ(t m) and integrate to get α m = ϕ(t m), f(t) ϕ(t m) 2 = ϕ(t m), f(t) ϕ(t) 2 since it is easy to see that ϕ(t m) = ϕ(t). of the... Page 18 of 100

19 5. theorem: ϕ(t m) is orthogonal basis then 2 ϕ(s)ds = ϕ(s) 2 ds If we choose ϕ(s)ds = θ, then ϕ(t) = θ If ϕ(t m) is an orthonormal basis for V 0 ϕ(2 j t m) is an orthogonal basis for V j ϕ j,k (t) = 2 j/2 ϕ(2 j t k) is an orthonormal basis for V j, j ZZ So h k = c k 2 is an normalised sequence: h k 2 = 1 and h k = 2 k ZZ k ZZ The dilation equation ϕ(t) = k ZZ c kϕ(2t k) becomes: ϕ j,m (t) = k ZZ h k ϕ j+1,2m+k (t). of the... Page 19 of 100

20 of the basis, can be expressed as c n 2k c n = 2δ k ( ) n ZZ The coefficients have double shift orthogonality. The inverse is also true: coefficients with double shift orthogonality lead to orthogonal scaling functions. In the frequency domain, this is: C(ω) 2 + C(ω + π) 2 = 1 Corollary: Orthogonal scaling functions with compact support must have an even number of non-zero coefficients in the dilation equation. Furthermore for Φ(ω) we have: Theorem: If {ϕ 0,k } is orthonormal k ZZ partition of unity property. Φ(ω + 2kπ) 2 = 1 2π of the... Page 20 of 100

21 6. Recall: V j V j+1 We now try to catch the information present in V j+1, but lost in V j : Suppose we have W j, such that: V j+1 = V j W j V j+1 = V j + W j and V j W j = {0} V j+1 is the direct sum of V j and W j. V j +W j is a linear sum: all elements of V j +W j can be written as a sum of an element from V j and an element from W j. If V j W j = {0}, this decomposition is unique. special case V j W j : orthogonal sum; W j is the orthogonal complement of V j in V j+1. W j V V j j+1 of the... Page 21 of 100

22 Theorem: If {φ(t k)} k ZZ constitute an orthogonal basis for V 0, then there exists one function ψ(t) such that {ψ(t k)} k ZZ forms an orthogonal basis for the orthogonal complement W 0 of V 0 in V 1. We call this function mother function or wavelet function. We know: A function J 1 V J = V J 1 W J 1 = V L f(t) = 2 J 1 k=0 can be written as: (suppose L 0) J 1 f(t) = j=l 2 j 1 k=0 j=l s k ϕ(2 J t k) w j,k ψ(2 j t k) + 2 L 1 k=0 W j s k ϕ(2 L t k) This decomposition is a discrete wavelet expansion of the... Page 22 of 100

23 7. Since W 0 V 1, we have ψ(x) = k ZZ d k ϕ(2x k) ψ(x)ϕ(x k)dx = 0 & implies: ψ(x)ψ(x k)dx = δ k d n 2k d n = 2δ k c n 2k d n = 0 n ZZ n ZZ A possible solution is: d k = ( 1) k c 1 k If we put both sequences c k and d k under each other c... c 3 c 2 c 1 c 0 c 1 c 2 c 3 c 4... d... c 4 c 3 c 2 c 1 c 0 c 1 c 2 c 3... we immediately see that their inner product is zero. of the... Page 23 of 100

24 8. The integral of the mother function Since ψ(x) = d k ϕ(2x k) k ZZ we have (in the orthogonal case): Since this is: ψ(x)dx = k ZZ ( 1) k c 1 k ϕ(2x k)dx ( 1) k c 1 k = 0 k ZZ ψ(x)dx = 0 of the... Page 24 of 100

25 9. of the wavelets Suppose the fully orthogonal case and call: ψ j,m (t) = 2 j/2 ψ(2 j t m) Then this is an orthonormal basis for L 2. Also: ψ j,m (t) = 2 j/2 ψ(2 j t m) = 2 j/2 d k ϕ(2 j+1 t 2m k) k ZZ of the... = k ZZ d k 2 2 (j+1)/2 ϕ(2 j+1 t 2m k) = k ZZ g k ϕ j+1,2m+k (t) where g k = d k 2 is a normalised sequence: g k 2 = h k 2 = 1 and k ZZ k ZZ g k = 0, k ZZ h k = 2 k ZZ Part 3 Page 25 of 100