Hi!

I’v been doing some FFT:s on a 4800 FX this morning and the results are:

The FFT is done using CUFFT with toolkit 2.3 for complex single precision, i.e. 8 bytes per element.

The measurement of time is the time taken for a transform followed by its inverse. The number of floating point operations are m * 2 * (5 * n * ln(n)) where m is the number of batches, n is the number of elements per batch, and the “2” comes from their being 2 transforms.

As can be seen for batches of 128 elements a piece or more, the number of GFLOPS attained are mostly a function of the number of elements, not the relation between the size of the batch or the length of the input vector.

Any comments? Do you think the results reasonable? Do you think I’m uncool seeing as I use Excel and not say Matlab?

Cheers!

Ian