F28377D , VCU2 and FFT

Hi Vishal,
Thanks for reply so fast.

Vishal_Coelho said:

marco zanardi

1.) as I understand, array input to CFFT and ICFFT must be {imag0,real0, imag1,real1 .... imag(N-1),real(N-1)} ?

If you are using VCU2 code, its [real, imag, real, imag ....]

honestly, now I'm a little confused.

looking at the source vcu2_cfft_xx.asm , algorithm works with 32-bit and CPACK=1 [Lo:Hi] => [Real:Imag]

in my case     {imag0,real0, (imag0 => 16-bit ,real0=> 16-bit)

                         [ HI   , LO]

in your case   [real   , imag, real, imag ....]

mistake?

No, sorry i misunderstood your earlier post. you have it right:

low word is real, high word is imaginary

[LO = real : HI = imag]

Hi Vishal,

thanks for example.I really appreciate your work.

I analyzed the code and for me something is wrong.

if I want to do RFFT -> switched pInBuffer and pOutBuffer -> RIFFT

I expect to have the same signal, but not so.

code: real data

CFFT.pInBuffer  = input;
CFFT.pOutBuffer = output;
CFFT.init = (void (*)(void *))CFFT_init64Pt;
CFFT.run  = (void (*)(void *))CFFT_run64Pt;
CFFT.init(handleCFFT);
CFFT.run(handleCFFT);
CFFT_unpack(handleCFFT);

pTemp = CFFT.pInBuffer;
CFFT.pInBuffer = CFFT.pOutBuffer;
CFFT.pOutBuffer = pTemp;

CFFT_pack(handleCFFT);
CFFT.run(handleCFFT);
CFFT_conjugate(handleCFFT->pOutBuffer, handleCFFT->nSamples);

output:

how do you see the input and output does not coincide.

1943 cycles(pack+run+conjugate) for unfinished data, are so many.

the amplitude is 240 times smaller, does not fit, I expect 4 times smaller and no more.

I did not stop, I did go through with it.

the problem is that they are not well documented input and output structure.

1. the output of function CFFT_unpack has the structure =>

{real0,imag0  ... real(N-1),imag(N-1)} (-f/2 .. 0)  [real0,imag0] -> 0,  [real(N-1),imag(N-1)] -> -f/2 , [HI,LO]

2. CFFT_pack(handleCFFT)

the input of function  has the structure => {imag0,real0...imag(N-1),real(N-1)} (0 .. f/2)

[imag0,real0] -> 0, [imag(N-1),real(N-1)] -> f/2 , [HI,LO]

the output of function gives Z(k) = Fe(k) + jFo(k) k = 0..N/2-1 ,

     for information http://www.engineeringproductivitytools.com/stuff/T0001/PT10.HTM

 That said, just do

a.) pack

b.) runs inverse complex FFT

3. the output structure of RIFFT => data are inverted, CFFT.pOutBuffer[0] -> N,

    [CFFT.pOutBuffer[1]..CFFT.pOutBuffer[N-1]] -> [0 .. N-1]

control code:

//RFFT
CFFT.pInBuffer  = input;
CFFT.pOutBuffer = output;
CFFT.init = (void (*)(void *))CFFT_init64Pt;
CFFT.run  = (void (*)(void *))CFFT_run64Pt;
CFFT.init(handleCFFT);
CFFT.run(handleCFFT);
CFFT_unpack(handleCFFT);

//adjust CFFT.pInBuffer
/*for(i = 0; i < handleCFFT->nSamples; i+=2){
	CFFT.pInBuffer[2*i] = CFFT.pOutBuffer[2*i+1];
	CFFT.pInBuffer[2*i+1] = CFFT.pOutBuffer[2*i];
}
*/
for(i = 0; i < handleCFFT->nSamples*2; i+=2){
        CFFT.pInBuffer[i+1] = CFFT.pOutBuffer[i];
        CFFT.pInBuffer[i] = CFFT.pOutBuffer[i+1];
}
CFFT_conjugate(handleCFFT->pInBuffer, handleCFFT->nSamples);

//RIFFT
CFFT.run  = (void (*)(void *))ICFFT_run64Pt;
CFFT_pack(handleCFFT);
CFFT.run(handleCFFT);

output :

how do you see, 1669 clk for finite data and the amplitude is 4 times smaller,as must be. that a good result.

//adjust RIFFT output, CFFT.pOutBuffer -> CFFT.pInBuffer
CFFT.pInBuffer[handleCFFT->nSamples*2-1] = CFFT.pOutBuffer[0]*-1;
for(i=0;i<handleCFFT->nSamples*2-1; i++)
	CFFT.pInBuffer[i] = CFFT.pOutBuffer[i+1]*-1;

adjust:

this is my conclusion.

as you think? this procedure is right?

thanks.

update,

the analysis made above is valid only if real signal has data on even bins (odd bins without data).
seems algorithm of function CFFT_pack(handleCFFT) contains bug.

some help.

update,

sorry, my fault.
error in my code:

//adjust CFFT.pInBuffer
for(i = 0; i < handleCFFT->nSamples; i+=2){
    CFFT.pInBuffer[2*i] = CFFT.pOutBuffer[2*i+1];
    CFFT.pInBuffer[2*i+1] = CFFT.pOutBuffer[2*i];
}
CFFT_conjugate(handleCFFT->pInBuffer, handleCFFT->nSamples);

must be

//adjust CFFT.pInBuffer
for(i = 0; i < handleCFFT->nSamples*2; i+=2){
	CFFT.pInBuffer[i+1] = CFFT.pOutBuffer[i];
	CFFT.pInBuffer[i] = CFFT.pOutBuffer[i+1];
}
CFFT_conjugate(handleCFFT->pInBuffer, handleCFFT->nSamples);

now seems fully functional.

can someone test?

thanks.

Hi Vishal,

thanks for new library. I tried it.
work, but not completely correct, the amplitude remain 240 times smaller.
must be 4 times smaller or mistake?

with the old library goes, try the code posted above(post 6)with old library,output is 4 times smaller.
I think it was a misunderstanding among you regarding structure input and output buffers.

Hi Vishal,

in the new library, input buffer to CFFT_pack(handleCFFT),
will have the structure:
low word is imaginary, high word is real

[LO = imag : HI = real] ?

thanks.

Marco - No, it should be [LO = real: HI = imag] (i.e. CPACK = 1). Ill try your approach (post 6) and get back to you this week itself...

Marco,

Sorry for the rather late reply. I finally got the example working - the output buffer needed to be aligned to a 2N word boundary as well as the input, since we switch them for the RIFFT, i.e. the output becomes the input for the RIFFT process. Anyway ive attached the example (which will also go out in v2.10.00.00).

The pack/unpack and FFT scale down the input, so the output will be <0.25x the original - I will try to quantify exactly what this factor is, but i cannot get rid of the divisions in each of the assembly routine as they are there to prevent overflows - so you might have to scale up the output of the unpack to get as close to the original - i tried scaling up by 4*sqrt(N) and it was 0.5x the original - scaling up further would saturate some values after the pack.

.2837x_vcu2_rfft_128_150219.zip