Output FFT, iFFT

[vrpascuzzi]

Very nice work here.

Wondering, how does one extract a generated kernel and the (i)FFT output for comparisons? For the former, I've done:

printf("%s", sc->output);

just before return res; in vkFFT::shaderGenVkFFT(). However, calling this code directly within my own application does not give the expected result. For example, input x and -X 64 -Y 1 -Z 1 and no lut using OpenCL backend, I get:

x=[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63]
FFT=[(224, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (224, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (224, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (224, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (224, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (224, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (224, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (224, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0)]

while I believe I should get something akin to:

[2016.  +0.j          -32.+651.374964j    -32.+324.9014524j
  -32.+215.72647697j  -32.+160.87486375j  -32.+127.75116108j
  -32.+105.48986269j  -32. +89.43400872j  -32. +77.254834j
  -32. +67.65831544j  -32. +59.86778918j  -32. +53.38877458j
  -32. +47.89138441j  -32. +43.14700523j  -32. +38.99211282j
  -32. +35.30655922j  -32. +32.j          -32. +29.00310941j
  -32. +26.26172131j  -32. +23.73281748j  -32. +21.38171641j
  -32. +19.18006188j  -32. +17.10435635j  -32. +15.13487283j
  -32. +13.254834j    -32. +11.44978308j  -32.  +9.70709388j
  -32.  +8.01558273j  -32.  +6.36519576j  -32.  +4.7467516j
  -32.  +3.15172491j  -32.  +1.57205919j  -32.  +0.j
  -32.  -1.57205919j  -32.  -3.15172491j  -32.  -4.7467516j
  -32.  -6.36519576j  -32.  -8.01558273j  -32.  -9.70709388j
  -32. -11.44978308j  -32. -13.254834j    -32. -15.13487283j
  -32. -17.10435635j  -32. -19.18006188j  -32. -21.38171641j
  -32. -23.73281748j  -32. -26.26172131j  -32. -29.00310941j
  -32. -32.j          -32. -35.30655922j  -32. -38.99211282j
  -32. -43.14700523j  -32. -47.89138441j  -32. -53.38877458j
  -32. -59.86778918j  -32. -67.65831544j  -32. -77.254834j
  -32. -89.43400872j  -32.-105.48986269j  -32.-127.75116108j
  -32.-160.87486375j  -32.-215.72647697j  -32.-324.9014524j
  -32.-651.374964j  ]

Could you provide an example kernel, e.g.,

./Vulkan_FFT -benchmark_vkfft -X 64

[DTolm]

Hello,

There is an option called keepShaderCode that makes all VkFFTAppend calls print executed kernels. You can enable it in input configuration in user_benchmark_VkFFT.cpp

Best regards,
Dmitrii

[vrpascuzzi]

FFT problem turned out to be an issue with work group size. Fixed and now getting the correct result.

Peculiarity of the iFFT now is that each output c-number is multiplied by 64:

x=[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63]
FFT(x)=[(2016, 0), (-32, -651.375), (-32, -324.901), (-32, -215.726), (-32, -160.875), (-32, -127.751), (-32, -105.49), (-32, -89.434), (-32, -77.2548), (-32, -67.6583), (-32, -59.8678), (-32, -53.3888), (-32, -47.8914), (-32, -43.147), (-32, -38.9921), (-32, -35.3066), (-32, -32), (-32, -29.0031), (-32, -26.2617), (-32, -23.7328), (-32, -21.3817), (-32, -19.1801), (-32, -17.1044), (-32, -15.1349), (-32, -13.2548), (-32, -11.4498), (-32, -9.70709), (-32, -8.01558), (-32, -6.3652), (-32, -4.74676), (-32, -3.15172), (-32, -1.57208), (-32, 0), (-32, 1.57205), (-32, 3.15172), (-32, 4.74675), (-32, 6.3652), (-32, 8.01558), (-32, 9.70709), (-32, 11.4498), (-32, 13.2548), (-32, 15.1349), (-32, 17.1044), (-32, 19.1801), (-32, 21.3817), (-32, 23.7328), (-32, 26.2617), (-32, 29.0031), (-32, 32), (-32, 35.3066), (-32, 38.9921), (-32, 43.147), (-32, 47.8914), (-32, 53.3888), (-32, 59.8678), (-32, 67.6583), (-32, 77.2548), (-32, 89.434), (-32, 105.49), (-32, 127.751), (-32, 160.875), (-32, 215.727), (-32, 324.901), (-32.0001, 651.375)]
iFFT(FFT(x))=[(0, 0), (64, 0.0001297), (128, 9.15527e-05), (192, 9.15527e-05), (256, 7.98547e-05), (320, 0.000137329), (384, -0.000152588), (448, 9.91821e-05), (512, -1.52588e-05), (576, 1.52588e-05), (640, 0), (704, 6.10352e-05), (768, 2.69569e-05), (832, 3.05176e-05), (896, -0.000137329), (960, 0.000144958), (1024, 0), (1088, -2.28882e-05), (1152, -3.05176e-05), (1216, -0.00012207), (1280, -1.16981e-05), (1344, -1.52588e-05), (1408, -9.15527e-05), (1472, 5.34058e-05), (1536, 1.52588e-05), (1600, -0.000198364), (1664, -3.05176e-05), (1728, -0.000213623), (1792, -9.51135e-05), (1856, -0.000213623), (1920, -6.10352e-05), (1984, -2.28882e-05), (2048, 0), (2112, -3.8147e-05), (2176, 0), (2240, -9.15527e-05), (2304, 1.88195e-05), (2368, -0.000106812), (2432, 3.05176e-05), (2496, 9.91821e-05), (2560, -1.52588e-05), (2624, 1.52588e-05), (2688, 0), (2752, 0), (2816, 8.7992e-05), (2880, 3.05176e-05), (2944, 0.000137329), (3008, -3.8147e-05), (3072, 0), (3136, -6.86646e-05), (3200, -6.10352e-05), (3264, 0), (3328, -7.27332e-05), (3392, 4.57764e-05), (3456, 0.000152588), (3520, -0.0001297), (3584, 1.52588e-05), (3648, 0.000167847), (3712, 3.05176e-05), (3776, 0.000274658), (3840, -3.40783e-05), (3904, 9.15527e-05), (3968, 0.00012207), (4032, -0.000205994)]

Again, likely something I'm overlooking. Will also check sample_11_precision_VkFFT_single. Thanks.

[DTolm]

Almost all FFT libraries produce unnormalized IFFT results. That is just how they define IFFT. You can enable normalization in VkFFT, though, with normalize parameter.

[vrpascuzzi]

I see. Not being an FFT expert, I wouldn't have known.

So it's safe to say I've extracted your kernels properly and they are working---modulo normalization---as expected?

[DTolm]

I guess so. I don't know how you launch them so it may not work for more complex systems, but for simple 1D systems it should be pretty straightforward.

[vrpascuzzi]

Will be in touch. For now, I thank you for the help and discussion.