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   -> Volume 6, Issue 8


Answer: Construction almost tight frames (WD 6.7 #23)
 
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Sergio Schuler (ss4@iprg.cise.ufl.edu)
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PostPosted: Tue Jul 15, 1997 4:41 pm    
Subject: Answer: Construction almost tight frames (WD 6.7 #23)
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#15 Answer: Construction almost tight frames (WD 6.7 #23)

Response:

>I want to construct a tight frame or a frame where the upper and lower
>bound are not far apart. I would prefer the frame to have an analytic
>expression in the time domain.

The paper by Daubechies, "The wavelet transform, time-frequency
localization and signal analysis", in IEEE Transactions on Information
Theory, vol 36, no 5, Sept. 1990, has a good section on frame bounds
for arbitrary dilation and translation steps. She also shows the frame
bounds for several continuous functions (with analytic expressions in
the time domain). Her book also has details about the frame bounds and
more examples.

Question: What kind of wavelet do you need? Are there any requirements you
would like to satisfy?

>I only want one scale however. Any help
>in doing this would be greatly appreciated.

I guess that you don't want to have voices, is that correct? Voices
usually make the frame tighter but add more redundancy. The tightness
of the frame basically depends on four things the dilation step ao,
the translation step bo, the number of voices and the function itself.

>Also information on
>computing frame bounds would be helpful. Specifically, computations
>such as the following: the Haar basis forms a tight frame when they
>are orthogonal and when they are used in a continuous wavelet
>transform, what happens for the case where the Haar functions are more
>redundant than the orthogonal case, how does one compute frame bounds
>for different redundancies?

The Haar basis are orthogonal when the dilation step ao=2 and the
translation step bo=1. If you change those values the Haar wavelet is
no longer orthogonal. However, the formulas by Daubechies will give
you the frame bounds. A word of caution, if the frame bounds are not
tight and not equal to one that does not mean that the wavelet is not
orthogonal. Confused? The following is true: if the bounds are equal
and equal to one then the wavelet is orthogonal. The opposite however
is not true. The bounds are simply bounds. They are not necessarily
the best bounds. So in some cases you might get bounds that do not
reflect how tight the frame really is. A very good example of this is
in the paper by Daubechies mentioned above (page 984, right column
after Theorem 2.9). The wavelet by Meyer mentioned there is
orthogonal, however, if one uses the wrong bounds it is impossible to
recognize that it is a frame. However, the modified bounds of Theorem
2.9 do show that the wavelet is indeed orthogonal.

With regards to computing the frame bounds, one needs to implement the
formulas given by Daubechies numerically. Although they look very
complicated the implementation is not too complicated. However, since
you are going to compute the bounds numerically, you must first prove
that the function you are using is indeed a frame. Again Daubechies
has a Theorem that shows what needs to be satisfied. Going back to the
numerical computation, since the sums are infinite one needs to
truncate them at some point. The approach that I use is to compute the
terms until they fall below the epsilon of the computer and then sum
them from smallest to biggest. This reduces the numerical error. Sup
and inf are of course replaced by max and min. Also you only need to
compute the sums in the interval 1 to ao since the expressions are
"exponentially" periodic (i.e., the expressions for A and B as a
function of omega have the same behavior in 1 to ao, ao to ao^2, ao^2
to ao^3, ...). In the dyadic case 1 to 2, 2 to 4, 4 to 8, etc.

If you have any questions let me know.

Sergio
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