topic Re: IPP v3.0 DCT transform in Intel® Integrated Performance Primitives

IPP v3.0 DCT transform

olivier_prat — Wed, 28 Apr 2004 23:40:23 GMT

Hello,

I'm a bit puzzled with the general DCT transform in IPP v3.0 (not the 8x8 versions).

After forward transforming an image, the inverse transform of this transformed image doesn't yield the original image. Is this normal? I've called both initialisation functions (forward and inverse) and I'm working with the 32f_C1R version of the functions.

By the way, are the low frequencies of the transform at the begining of the image (nearest to the original image pointer)?

Hope someone can help!

Re: IPP v3.0 DCT transform

olivier_prat — Thu, 29 Apr 2004 19:02:08 GMT

ok, sorry everyone. My fault. I wasn't comparing the same images... silly me.

Re: IPP v3.0 DCT transform

Intel_C_Intel — Sat, 01 May 2004 02:04:32 GMT

Hi Oliver

I'm glad you solved your problem. Feel free to ask questions if youhave something to ask.

Regards,

Vladimir

Re: IPP v3.0 DCT transform

olivier_prat — Wed, 05 May 2004 16:48:52 GMT

Allright :-). Here I go!

I'm using the DCT transform functions to solve a 2D partial differential equation. I suppose the DCT transform used in the IPP is the classical:

F(i,j) = sum(k: 0 to N) sum(l: 0 to M) f(k,l).cos[Pi.k.(i+1/2)/N].cos[Pi.l.(j+1/2)/N]

and inverse:

f(i,j) = (4/NM) sum(k: 0 to N) sum(l: 0 to M) F(k,l).cos[Pi.k.(i+1/2)/N].cos[Pi.l.(j+1/2)/N]

I haven't got access to the book referenced as [Rao90] in the documentation so I'm not absolutely sure of that. Can someone confirm this?

Also, is the data in the transformed image packed the straightforward way :F(i,j) stored at pixel location (i,j)?

Are there any constraints involved when using these transform functions not cited in the documentation?

Thanks for any answers.

Re: IPP v3.0 DCT transform

Intel_C_Intel — Thu, 06 May 2004 23:48:10 GMT

According developer in ippiDCT the following equation are used:

F(k,l) = (1/sqrt(N*M))*C(k)*C(l) * sum(i: 0 to N-1) sum(j: 0 to M-1) f(i,j)*cos[Pi*k*(i+1/2)/N]*cos(Pi*l*(j+1/2)/M]

f(i,j) = (1/sqrt(N*M)) * sum(k: 0 to N-1) sum(l: 0 to M-1) C(k)*C(l) * F(k,l)*cos[Pi*k*(i+1/2)/N]*cos(Pi*l*(j+1/2)/M]

where?(?)=1for ?=0and ?(k)=sqrt(2)for k>0

Doesit helps you?

Vladimir

Re: IPP v3.0 DCT transform

olivier_prat — Fri, 07 May 2004 14:14:27 GMT

Thanks, that should do the trick. I'll check my code against that...