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To construct local trigonometric bases we have to choose: (a) the window functions ${w}_{k}\left(t\right)$ ; and (b) the trigonometric functions (i.e., $\alpha $ , $\beta $ and $\gamma $ in Eq. [link] ). If we use the rectangular window (which we know is a bad choice),then it suffices to find a trigonometric basis for the interval that the window spans. Without loss of generality, we couldconsider the unit interval $(0,1)$ and hence we are interested in trigonometric bases for ${L}^{2}\left((0,1)\right)$ . It is easy to see that the following four sets of functions satisfy this requirement.
Indeed, these orthonormal bases are obtained from the Fourier series on $(-2,2)$ (the first two) and on $(-1,1)$ (the last two) by appropriately imposing symmetries and hence are readily verified to becomplete and orthonormal on $(0,1)$ . If we choose a set of nonoverlapping rectangular window functions ${w}_{k}\left(t\right)$ such that ${\sum}_{k}{w}_{k}\left(t\right)=1$ for all $t\in \text{\mathbb{R}}$ , and define ${\chi}_{k,n}\left(t\right)={w}_{k}\left(t\right){\Phi}_{n}\left(t\right)$ , then, $\left\{{\chi}_{k,n},\left(t\right)\right\}$ is a local trigonometric basis for ${L}^{2}\left(\text{\mathbb{R}}\right)$ , for each of the four choices of $ph{i}_{n}\left(t\right)$ above.
We know how to construct orthonormal trigonometric bases for disjoint temporal bins or intervals. Now we need to construct smooth windows ${w}_{k}\left(t\right)$ that when applied to cosines and sines retain orthonormality. An outline of the process is as follows:A unitary operation is applied that “unfolds” the discontinuities of all the local basis functions at the boundaries ofeach temporal bin. Unfolding leads to overlapping (unfolded) basis functions. However,since unfolding is unitary, the resulting functions still form an orthonormal basis. The unfolding operator is parameterizedby a function $r\left(t\right)$ that satisfies an algebraic constraint (which makes the operator unitary). The smoothness of the resulting basisfunctions depends on the smoothness of this underlying function $r\left(t\right)$ .
The function $r\left(t\right)$ , referred to as a rising cutoff function, satisfies the following conditions (see [link] ) :
$r\left(t\right)$ is called a rising cutoff function because it rises from 0 to 1 in the interval $[-1,1]$ (note: it does not necessarily have to be monotone increasing). Multiplying a function by $r\left(t\right)$ would localize it to $[-1,\infty ]$ . Every real-valued function $r\left(t\right)$ satisfying [link] is of the form $r\left(t\right)=sin\left(\theta \right(t\left)\right)$ where
This ensures that $r(-t)=sin\left(\theta (-t)\right)=sin(\frac{\pi}{2}-\theta \left(t\right))=cos\left(\theta \left(t\right)\right)$ and therefore ${r}^{2}\left(t\right)+{r}^{2}(-t)=1$ . One can easily construct arbitrarily smooth risingcutoff functions. We give one such recipe from [link] (p.105) . Start with a function
It is readily verified to be a rising cutoff function. Now recursively define ${r}_{\left[1\right]}\left(t\right),{r}_{\left[2\right]}\left(t\right),...$ as follows:
Notice that ${r}_{\left[n\right]}\left(t\right)$ is a rising cutoff function for every $n$ . Moreover, by induction on $n$ it is easy to show that ${r}_{\left[n\right]}\left(t\right)\in {C}^{{2}^{n}-1}$ (it suffices to show that derivatives at $t=-1$ and $t=1$ exist and are zero up to order ${2}^{n}-1$ ).
Using a rising cutoff function $r\left(t\right)$ one can define the folding operator, $U$ , and its inverse, the unfolding operator ${U}^{\u2606}$ as follows:
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