ACM 126a Solutions for Homework Set 4

Transcription

1 ACM 26a Solutions for Homewor Set 4 Laurent Demanet March 2, 25 Problem. Problem 7.7 page 36 We need to recall a few standard facts about Fourier series. Convolution: Subsampling (see p. 26): Zero insertion (see p. 26): ĥ g(ω) = ĥ(ω)ĝ(ω) x[2n]e inω = 2 (ˆx(ω 2 ) + ˆx(ω 2 + π)). ˇx[n]e inω = ˆx(2ω). (a) For part (a), apply these identities repeatedly to obtain ã. â (ω) = 2 (â h( ω 2 ) + â h( ω 2 + π)) = 2 (â ( ω 2 )ĥ(ω 2 ) + â ( ω 2 + π)ĥ(ω 2 + π)). ˆã (ω) = ˆǎ (ω)ˆ h(ω) + ˆď (ω)ˆ g(ω) = â (2ω)ˆ h(ω) + ˆd (2ω)ˆ g(ω) = 2 (â (ω)ĥ(ω) + â (ω + π)ĥ(ω + π))ˆ h(ω) + 2 (â (ω)ĝ(ω) + â (ω + π)ĝ(ω + π))ˆ g(ω) = â (ω) 2 (ĥ(ω)ˆ h(ω) + ĝ(ω)ˆ g(ω)) + â (ω + π) 2 (ĥ(ω + π)ˆ h(ω) + ĝ(ω + π)ˆ g(ω)). On the other hand, we wish to recover the sequence a up to a shift l, ˆã (ω) = a [n l]e inω = â (ω)e ilω. If this is to be satisfied for every sequence a, the filters h, h, g and g must obey the perfect reconstruction conditions ĥ(ω)ˆ h(ω) + ĝ(ω)ˆ g(ω) = 2e ilω, ()

2 ĥ(ω + π)ˆ h(ω) + ĝ(ω + π)ˆ g(ω) =. (2) It is now straightforward to see that the choice of filters suggested in the wording, coupled with the QMF condition, offer one possible way to satisfy equations () and (2). (b) Let us evaluate the QMF condition at ω + π, By -periodicity, this is also ĥ 2 (ω + π) ĥ2 (ω + ) = 2e ilω e ilπ. (ĥ2 (ω) ĥ2 (ω + π)) = 2e ilω e ilπ. If we compare with the original QMF relation, we see that this expression should also be equal to 2e ilω, which in turn implies = e ilπ. This is only true if l is an odd integer. (c) The Haar transfer function is ĥ(ω) = 2 ( + e iω ). So we obtain ĥ 2 (ω) = 2 ( + 2e iω + e 2iω ), ĥ 2 (ω + π) = 2 ( 2e iω + e 2iω ). Subtracting these two equation we get ĥ 2 (ω) ĥ2 (ω + π) = 2e iω and we conclude that l = for the Haar filter. Problem Problem 7. page 36 The following argument is strongly inspired from Daubechies proof of frame bounds for wavelet frames (Ten Lectures, p.67). We use the same convention as in class for the Fourier transform and series. Positive indexes small scales. Let f, g be two unspecified functions at this point. 2

3 f, g = Z f, ψ g, ψ = 4π 2 ˆf(ω)2 /2 ˆψ(2 ω)e inω2 dω = 4π 2 2 dω e inω2 m Z ĝ(ω )2 /2 ˆψ(2 ω )e inω 2 dω ˆf(ω + 2 m) ˆψ(2 ω + m) dω e inω 2 Z ĝ(ω + 2 ) ˆψ(2 ω + ) = 2 dθ e inθ ˆf(2 (θ + m)) ˆψ(θ + m) m Z dθ e inθ ĝ(2 (θ + )) ˆψ(θ + ) Z The tric is to recognize that each dθ e ±inθ ( ) is a Fourier series coefficient or its conugate (see equation 3.24 in Mallat), and the sum over n hides an inner product in l 2 between two sequences of coefficients. By the Parseval formula for Fourier series, we get f, g = = = 2 dθ ˆf(2 (θ + m)) ˆψ(θ + m) ĝ(2 (θ + )) ˆψ(θ + ) m Z Z dω ˆf(ω + 2 m)ĝ(ω + 2 ) ˆψ(2 ω + m) ˆψ(2 ω + ) m dω ˆf(ω)ĝ(ω + 2 ) ˆψ(2 ω) ˆψ(2 ω + ). (To obtain the last line we made the change of variables ω ω 2 m.) Let us now choose ˆf(ω) = δ(ω ω ) and g continuous so that the pairing f, g is well defined. The left-hand side becomes by Parseval ĝ(ω ), hence ĝ(ω ) = ĝ(ω + 2 ) ˆψ(2 ω ) ˆψ(2 ω + ). Since the values of ĝ can be specified arbitrarily, only the term = is nonzero and yields bac the left-hand side. As a result, ˆψ(2 ω ) ˆψ(2 ω + ) = δ. The conclusion follows from letting =. We have left out a few technicalities due to the use of distributions. Also, please note that the wording of the problem is imprecise in the following sense. The equality ˆψ(2 ω) 2 = cannot in general be expected to hold for all ω R, but instead for almost every ω R. The origin has a special role because it always holds that ψ(t) dt = ˆψ() =. Here are examples of functions which satisfy ˆψ(2 ω) 2 = but which do not form orthobases: Tae ψ(t) the Haar wavelet and compress is by a factor of 2, ψ(t) = 2ψ(2t). Then the integer translates remain orthogonal but don t span the whole L 2, there are gaps at every other dyadic interval. 3

4 Alternatively, dilate ψ(t) by a factor 2, ψ(t) = 2 ψ( t 2 ). The integer translates now overlap and are not orthogonal to each other. For a more interesting counter-example, tae a Meyer wavelet and twiddle the linear phase factor in Fourier to destroy orthogonality. Problem Problem 7.6 page 38 (a) We recognize φ as the Haar scaling function, { when t <, φ (t) = otherwise. It is not hard to chec that φ 2 (t) is an affine function on the same interval [, ], { 2t when t <, φ 2 (t) = otherwise. Note that φ, φ 2 = as needed, but φ 2 is not normalized. This is a typo in the wording. The coefficients of the second scaling equation can be changed to accomodate the normalization or, alternatively, the normalization can be performed subsequently. Either way, we obtain instead { 3(2t ) when t <, φ 2 (t) = otherwise. The space V spanned by the integer translates of φ and φ 2 consists of all the piecewise affine functions, not necessarily continuous, with nodes at integer asbcissae. (b) By dilation invariance, we expect V to be the space of piecewise affine functions, not necessarily continuous, with nodes /2 for Z. Hence the two wavelets must be themselves piecewise affine on each subinterval [, /2] and [/2, ]. We chec that the following two functions obey all the constraints, namely that they are orthogonal to each other, to φ and φ 2, and they are normalized. ψ (t) = ψ 2 (t) = 3(4t ) when t < 2, 3( 4t + 3) when 2 t <, otherwise. 6t when t < 2, 6t 5 when 2 t <, otherwise. Because the two wavelets are supported on [, ], orthogonality to φ and φ 2 is equivalent to having two vanishing moments. Note that we could not have obtained these wavelets from the scaling functions taen individually. The general theory of chapter 7 does not apply in the multiwavelet case. Define the spaces V as usual. It is easy to chec that they form a multiresolution approximation. In particular, Z V = {} because there is no L 2 function other than zero which is affine over arbitrarily large intervals. 4

5 Z V = L 2 (R). This follows from the fact that V Haar V (piecewise constants are special cases of piecewise affine functions) and that the union property holds for the Haar wavelets. As a result, we claim that we have completeness, namely, W = L 2 (R), Z where V + = V W. Indeed, let ɛ > and assume there is a function f which proects to zero in all the W s. By completeness of the V it must be approximated arbitrarily well by f in some V, i.e. f f < ɛ. The proection of f onto all the coarser spaces V, <, is f itself by assumption. But only the zero function belongs to all the V. Therefore f = and, by letting ɛ, f =. This proves completeness. We had orthogonality by construction. We therefore have an orthobasis of L 2 (R). 5