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| Re: Question: Storage scheme of 2D-wavelet-coefficients [message #5149 is a reply to message #5054] |
Fri, 29 September 1995 00:00  |
agraps
Messages: 35
Registered: September 1994
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In article <44gheq$a69@news.tuwien.ac.at>, "Christian Oehreneder
(PHOTO-Dissertand)" <coehrene> wrote:
> I'm testing idl's 2D-wavelet-transform, but I could not find out how the
> wavelet-coefficienmts with different dilation level and displacement are
> stored in the result-matrix. Can anyone help?
>
> Thanks
> Christian
> (coehrene@fbgeo1.tuwien.ac.at)
I haven't used the NR_WTN routine, but if it works like the usual Discrete
Wavelet Transforms in the literature and textbooks, the output decomposition
matrix, which contains the wavelet coefficients, works something like the
following.
The algorithm goes like this.
- Coefficient matrix size is same is input size of matrix.
(square, length 2^n on each side, so a 64x64, or 128x128, or 256x256
matrix etc.)
- Choose a scale
- For the first row:
Perform a down-sample, high-pass smoothing operation on 1/2 of the row.
Perform a down-sample, low-pass operation (throw away every 2nd data
point) of the other half of the row.
Increment row by one, and repeat until you've gone through all the rows.
-For the first column:
Perform a down-sample, low-pass operation (throw away every 2nd data
point) of one half of the column.
Perform a down-sample, high-pass smoothing operation on the other 1/2
of the column.
Increment column by one, and repeat until you've gone through all the columns.
-Increment the scale by one (next 2^n) and repeat until you've covered
all of your desired scales.
So the result is a square with squares-within-squares of low-pass (L)
operations and high-pass operations (H). The filter operations always
work in pairs ("Quadrature Mirror Filters"), the filters being the
wavelets you've chosen (say the Daubechies wavelet).
Here is an ASCII sketch of some output filter operations. The sketch
shows 3 scale levels of smoothing, downsampled operations. The square
labelled all "HH" will be the most "bland", smoothed result.
The square with all "LL" will be the tiniest representation of your
original image (just downsampled, with every other point thrown
away several times). The others will show you different things.
Some combinations of LH will show the fine detail in your image.
Other combinations will show the sharp contrast portions of your
image (those are the combinations that make wavelets good "edge
detectors").
-----------------------------------
| | |
| | |
| | |
| | |
| | |
| L,H | H,H |
| | |
| | |
| | |
| | |
-----------------------------------
| | | |
| LL,LH |LH,LH | |
| | | |
| | | |
| | | |
|-----------------| H,L |
|LLL|LLH,| | |
|LLH|LLH | | |
|--------| LH,LL | |
|LLL|LLH,| | |
|LLL|LLL | | |
-----------------------------------
Hope this makes sense!! Your output coefficient matrix may have the
squares flipped to have the smallest scale be at the top left corner,
but this explanation should still apply.
Amara
--
************************************************************ *********
Amara Graps email: agraps@netcom.com
Computational Physics vita: finger agraps@sunshine.arc.nasa.gov
Multiplex Answers URL: http://www.amara.com/
************************************************************ *********
"Sometimes I think I understand everything... Then I regain
consciousness." --Ashleigh Brilliant
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