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The question of how one can calculate the Fourier series coefficients of a continuous signal from the discrete Fourier transform of samples of thesignal is similar to asking how one calculates the discrete wavelet transform from samples of the signal. And the answer is similar. Thesamples must be “dense" enough. For the Fourier series, if a frequency can be found above which there is very little energy in the signal (abovewhich the Fourier coefficients are very small), that determines the Nyquist frequency and the necessary sampling rate. For the waveletexpansion, a scale must be found above which there is negligible detail or energy. If this scale is $j={J}_{1}$ , the signal can be written
or, in terms of wavelets, [link] becomes
This assumes that approximately $f\in {\mathcal{V}}_{{J}_{1}}$ or equivalently, $\parallel f-{P}_{{J}_{1}}f\parallel \approx 0$ , where ${P}_{{J}_{1}}$ denotes the orthogonal projection of $f$ onto ${\mathcal{V}}_{{J}_{1}}$ .
Given $f\left(t\right)\in {\mathcal{V}}_{{J}_{1}}$ so that the expansion in [link] is exact, one computes the DWT coefficients in two steps.
For ${J}_{1}$ large enough, ${\phi}_{{J}_{1},k}\left(t\right)$ can be approximated by a Dirac impulse at its center of mass since $\int \phi \left(t\right)\phantom{\rule{0.166667em}{0ex}}dt=1$ . For large $j$ this gives
where ${m}_{0}=\int t\phantom{\rule{0.166667em}{0ex}}\phi \left(t\right)\phantom{\rule{0.166667em}{0ex}}dt$ is the first moment of $\phi \left(t\right)$ . Therefore the scaling function coefficients at the $j={J}_{1}$ scale are
which are approximately
For all 2-regular wavelets (i.e., wavelets with two vanishing moments, regular wavelets other than the Haar wavelets—even in the $M$ -band case where one replaces 2 by $M$ in the above equations, ${m}_{0}=0$ ), one can show that the samples of the functions themselves form a third-orderapproximation to the scaling function coefficients of the signal [link] . That is, if $f\left(t\right)$ is a quadratic polynomial, then
Thus, in practice, the finest scale ${J}_{1}$ is determined by the sampling rate. By rescaling the function and amplifying it appropriately, onecan assume the samples of $f\left(t\right)$ are equal to the scaling function coefficients. These approximations are made better by setting some ofthe scaling function moments to zero as in the coiflets. These are discussed in Section: Approximation of Scaling Coefficients by Samples of the Signal .
Finally there is one other aspect to consider. If the signal has finite support and $L$ samples are given, then we have $L$ nonzero coefficients $\u27e8f,{\phi}_{{J}_{1},k}\u27e9$ . However, the DWT will typically have more than $L$ coefficients since the wavelet and scaling functions are obtained by convolution and downsampling. In other words,the DWT of a $L$ -point signal will have more than $L$ points. Considered as a finite discrete transform of one vector into another, this situationis undesirable. The reason this “expansion" in dimension occurs is that one is using a basis for ${L}^{2}$ to represent a signal that is of finite duration, say in ${L}^{2}[0,P]$ .
When calculating the DWT of a long signal, ${J}_{0}$ is usually chosen to give the wavelet description of the slowly changing or longer duration featuresof the signal. When the signal has finite support or is periodic, ${J}_{0}$ is generally chosen so there is a single scaling coefficient for the entire signal or for one period of the signal. To reconcile thedifference in length of the samples of a finite support signal and the number of DWT coefficients, zeros can be appended to the samples of $f\left(t\right)$ or the signal can be made periodic as is done for the DFT.
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