Re: Hartley transforms

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2 total messages Started by Thomas M. Breuel Sun, 05 Aug 1990 07:17

#903

Author: Thomas M. Breuel
Date: Sun, 05 Aug 1990 07:17

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Oh, well, the answers turn out to be simple (I thought
about it while waiting for the subway--"think--then post")... 
sorry to have bothered the list with the first set of questions, but
I would still be interested in the answer to this question:

|Also, does anyone have code for the FHT (C/Fortran)? Currently, I am
|using a REALFFT routine and re-arrange the output, which is probably
|not very inefficient, but not quite as fast as it could be.  Have the
|patent issues involving the FHT been resolved?

					Thanks, Thomas.
					tmb@ai.mit.edu

PS: the other questions were:

|Does anyone know or have pointers to the literature or code about how
|to do multidimensional Hartley transforms (i.e., is it obvious that a
|2D HT is the same as 2 1D HT's, or do I need some correction), 

The answer is obvious if you write f(x,y)=(cas px)(cas qy) as the
set of basis functions, where cas x=(sin x)+(cos x).

|and what
|the correlation and convolution theorems look like for multidimensional
|Hartley transforms? [and: can we get the same number of multiplies
|as for the FT case]

It turns out that it isn't too hard to show that if you apply
the HT 1D correlation/convolution theorem first in the X and
then in the Y direction, you get the correct answer and in
the same number of multiplies as the complex (Fourier) version. 
You get the 1D versions from the corresponding Fourier theorems
and the relation 2F=(H+HR)+i(-H+HR), where HR(x)=H(-x).

Re: Hartley transforms

#904

Author: H.V. Sorensen
Date: Sun, 05 Aug 1990 14:58

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>tmb@wheaties.ai.mit.edu (Thomas M. Breuel) writes

>Also, does anyone have code for the FHT (C/Fortran)? Currently, I am
>using a REALFFT routine and re-arrange the output, which is probably
>not very inefficient, but not quite as fast as it could be.  Have the
>patent issues involving the FHT been resolved?

I am glad to see you solved your own problem, but will have to
dissapooint you. If you are interested in speed, you should stick with
the real-input FFTs. It has been shown several places, that the Hartley
transform requires more arithmetic operations than the real-input FFT.
(The number of multiplications are the same, while the number of
additions are N-2 more for the Hartley transform, where N is the
transform length).

Hope this helps you,   Henrik Sorensen

******************************************************************************
*   Henrik Sorensen               Department of Electrical Enginering        *
*   Internet: hvs@ee.upenn.edu    University of Pennsylvania                 *
******************************************************************************

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