Math 2.0

Harmonic analysis on Euclidean spaces Wavelets Example

Shannon wavelet

Builds on Fourier transformMultiresolution analysis

Example

Write sinc⁡(t)=sin⁡(πt)/(πt)\operatorname{sinc}(t) = \sin(\pi t)/(\pi t) for t≠0t \ne 0 and sinc⁡(0)=1\operatorname{sinc}(0) = 1. Use the Fourier transform f^(ω)=∫ℝf(t)e−iωt dt\hat f(\omega) = \int_{\R} f(t) e^{-i\omega t} \, dt. The real Shannon wavelet is ψ(t)=2sinc⁡(2t)−sinc⁡(t)=sinc⁡(t2)cos⁡(3πt2).\psi(t) = 2\operatorname{sinc}(2t) - \operatorname{sinc}(t) = \operatorname{sinc}\left(\frac{t}{2}\right) \cos\left(\frac{3\pi t}{2}\right). Its Fourier transform is the indicator function of the set {ω:π<∣ω∣≤2π}\set{\omega : \pi < \abs{\omega} \le 2\pi}, up to a set of measure zero.

Check. The transform of sinc⁡\operatorname{sinc} is the indicator of [−π,π][-\pi, \pi]. The transform of 2sinc⁡(2t)2\operatorname{sinc}(2t) is the indicator of [−2π,2π][-2\pi, 2\pi]. Their difference is the indicator of the two bands. The second formula follows from sin⁡(2πt)−sin⁡(πt)=2cos⁡(3πt/2)sin⁡(πt/2)\sin(2\pi t) - \sin(\pi t) = 2\cos(3\pi t/2)\sin(\pi t/2).

Remarks

The functions ψj,k(t)=2j/2ψ(2jt−k)\psi_{j,k}(t) = 2^{j/2}\psi(2^j t - k), j,k∈ℤj, k \in \Z, form an orthonormal basis of L2(ℝ)L^2(\R). The reason is that the two bands, shifted by multiples of 2π2\pi, tile the line, and their dyadic dilates tile ℝ∖{0}\R \setminus \set{0}. The scaling function of the underlying multiresolution analysis is φ=sinc⁡\varphi = \operatorname{sinc}, and its approximation spaces are spaces of band-limited functions. The Shannon wavelet is perfectly localized in frequency but poorly localized in time: ∣ψ(t)∣\abs{\psi(t)} decays only like 1/∣t∣1/\abs{t}, so ψ\psi is not integrable. The Meyer wavelet smooths the sharp edges of ψ^\hat\psi to gain fast decay in time.

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  1. Fourier transform concept
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Further reading

  1. Web https://en.wikipedia.org/wiki/Shannon_wavelet
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Kindexample
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MSC42C40, 94A12
Instance ofWavelet
UsesFourier transform, Multiresolution analysis
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