0.5 The scaling function and scaling coefficients, wavelet and  (Page 10/13)

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Parameterization of the scaling coefficients

The case where $\phi \left(t\right)$ and $h\left(n\right)$ have compact support is very important. It aids in the time localization properties of the DWT andoften reduces the computational requirements of calculating the DWT. If $h\left(n\right)$ has compact support, then the filters described in  Chapter: Filter Banks and the Discrete Wavelet Transform are simple FIR filters. We have stated that $N$ , the length of the sequence $h\left(n\right)$ , must be even and $h\left(n\right)$ must satisfy the linear constraint of [link] and the $\frac{N}{2}$ bilinear constraints of [link] . This leaves $\frac{N}{2}-1$ degrees of freedom in choosing $h\left(n\right)$ that will still guarantee the existence of $\phi \left(t\right)$ and a set of essentially orthogonal basis functions generated from $\phi \left(t\right)$ .

Length-2 scaling coefficient vector

For a length-2 $h\left(n\right)$ , there are no degrees of freedom left after satisfying the required conditions in [link] and [link] . These requirements are

$h\left(0\right)\phantom{\rule{0.166667em}{0ex}}+\phantom{\rule{0.166667em}{0ex}}h\left(1\right)\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}\sqrt{2}$

and

${h}^{2}\left(0\right)+{h}^{2}\left(1\right)\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}1$

which are uniquely satisfied by

${h}_{D2}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\left\{h,\left(,0,\right),,,h,\left(,1,\right)\right\}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\left\{\frac{1}{\sqrt{2}},\phantom{\rule{0.166667em}{0ex}},,,\phantom{\rule{0.166667em}{0ex}},\frac{1}{\sqrt{2}}\right\}.$

These are the Haar scaling functions coefficients which are also the length-2 Daubechies coefficients [link] used as an example in Chapter: A multiresolution formulation of Wavelet Systems and discussed later in this book.

Length-4 scaling coefficient vector

For the length-4 coefficient sequence, there is one degree of freedom or one parameter that gives all the coefficients that satisfy the requiredconditions:

$h\left(0\right)\phantom{\rule{0.166667em}{0ex}}+h\left(1\right)\phantom{\rule{0.166667em}{0ex}}+h\left(2\right)\phantom{\rule{0.166667em}{0ex}}+h\left(3\right)\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}\sqrt{2},$
${h}^{2}\left(0\right)+{h}^{2}\left(1\right)+{h}^{2}\left(2\right)+{h}^{2}\left(3\right)=1$

and

$h\left(0\right)\phantom{\rule{0.166667em}{0ex}}h\left(2\right)\phantom{\rule{0.166667em}{0ex}}+\phantom{\rule{0.166667em}{0ex}}h\left(1\right)\phantom{\rule{0.166667em}{0ex}}h\left(3\right)\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}0$

Letting the parameter be the angle $\alpha$ , the coefficients become

$\begin{array}{cc}h\left(0\right)=\hfill & \left(1-cos\left(\alpha \right)+sin\left(\alpha \right)\right)/\left(2\sqrt{2}\right)\hfill \\ h\left(1\right)=\hfill & \left(1+cos\left(\alpha \right)+sin\left(\alpha \right)\right)/\left(2\sqrt{2}\right)\hfill \\ h\left(2\right)=\hfill & \left(1+cos\left(\alpha \right)-sin\left(\alpha \right)\right)/\left(2\sqrt{2}\right)\hfill \\ h\left(3\right)=\hfill & \left(1-cos\left(\alpha \right)-sin\left(\alpha \right)\right)/\left(2\sqrt{2}\right).\hfill \end{array}$

These equations also give the length-2 Haar coefficients [link] for $\alpha =0,\pi /2,3\pi /2$ and a degenerate condition for $\alpha =\pi$ . We get the Daubechies coefficients (discussed later in this book) for $\alpha =\pi /3$ . These Daubechies-4 coefficients have a particularly clean form,

${h}_{D4}=\left\{\frac{1+\sqrt{3}}{4\sqrt{2}},\phantom{\rule{0.166667em}{0ex}},,,\phantom{\rule{0.166667em}{0ex}},\frac{3+\sqrt{3}}{4\sqrt{2}},\phantom{\rule{0.166667em}{0ex}},,,\phantom{\rule{0.166667em}{0ex}},\frac{3-\sqrt{3}}{4\sqrt{2}},\phantom{\rule{0.166667em}{0ex}},,,\phantom{\rule{0.166667em}{0ex}},\frac{1-\sqrt{3}}{4\sqrt{2}}\right\}$

Length-6 scaling coefficient vector

For a length-6 coefficient sequence $h\left(n\right)$ , the two parameters are defined as $\alpha$ and $\beta$ and the resulting coefficients are

$\begin{array}{cc}h\left(0\right)=\hfill & \left[\left(1+cos\left(\alpha \right)+sin\left(\alpha \right)\right)\left(1-cos\left(\beta \right)-sin\left(\beta \right)\right)+2sin\left(\beta \right)\phantom{\rule{0.166667em}{0ex}}cos\left(\alpha \right)\right]/\left(4\sqrt{2}\right)\hfill \\ h\left(1\right)=\hfill & \left[\left(1-cos\left(\alpha \right)+sin\left(\alpha \right)\right)\left(1+cos\left(\beta \right)-sin\left(\beta \right)\right)-2sin\left(\beta \right)\phantom{\rule{0.166667em}{0ex}}cos\left(\alpha \right)\right]/\left(4\sqrt{2}\right)\hfill \\ h\left(2\right)=\hfill & \left[1+cos\left(\alpha -\beta \right)+sin\left(\alpha -\beta \right)\right]/\left(2\sqrt{2}\right)\hfill \\ h\left(3\right)=\hfill & \left[1+cos\left(\alpha -\beta \right)-sin\left(\alpha -\beta \right)\right]/\left(2\sqrt{2}\right)\hfill \\ h\left(4\right)=\hfill & 1/\sqrt{2}-h\left(0\right)-h\left(2\right)\hfill \\ h\left(5\right)=\hfill & 1/\sqrt{2}-h\left(1\right)-h\left(3\right)\hfill \end{array}$

Here the Haar coefficients are generated for any $\alpha =\beta$ and the length-4 coefficients [link] result if $\beta =0$ with $\alpha$ being the free parameter. The length-4 Daubechies coefficients are calculatedfor $\alpha =\pi /3$ and $\beta =0$ . The length-6 Daubechies coefficients result from $\alpha =1.35980373244182\phantom{\rule{4pt}{0ex}}\text{and}\phantom{\rule{4pt}{0ex}}\beta =-0.78210638474440$ .

The inverse of these formulas which will give $\alpha$ and $\beta$ from the allowed $h\left(n\right)$ are

$\alpha =arctan\left(\frac{2\left(h{\left(0\right)}^{2}+h{\left(1\right)}^{2}\right)-1+\left(h\left(2\right)+h\left(3\right)\right)/\sqrt{2}}{2\phantom{\rule{0.166667em}{0ex}}\left(h\left(1\right)\phantom{\rule{0.166667em}{0ex}}h\left(2\right)-h\left(0\right)\phantom{\rule{0.166667em}{0ex}}h\left(3\right)\right)+\sqrt{2}\left(h\left(0\right)-h\left(1\right)\right)}\right)$
$\beta =\alpha -arctan\left(\frac{h\left(2\right)-h\left(3\right)}{h\left(2\right)+h\left(3\right)-1/\sqrt{2}}\right)$

As $\alpha$ and $\beta$ range over $-\pi$ to $\pi$ all possible $h\left(n\right)$ are generated. This allows informative experimentation to better see whatthese compactly supported wavelets look like. This parameterization is implemented in the Matlab programs in Appendix C and in the Aware, Inc. software, UltraWave [link] .

Since the scaling functions and wavelets are used with integer translations, the location of their support is not important, only thesize of the support. Some authors shift $h\left(n\right)$ , ${h}_{1}\left(n\right)$ , $\phi \left(t\right)$ , and $\psi \left(t\right)$ to be approximately centered around the origin. This is achieved by having the initial nonzero scaling coefficient start at $n=-\frac{N}{2}+1$ rather than zero. We prefer to have the origin at $n=t=0$ .

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