I've written my own FFT but wanted to know if there was perhaps a way to combine the Cooley and Turkey methods. So far I've got N/2 log N real multiplies for transforming a real signal.

My current implementation has N log N multiplies (well a little more because I've not optimised out the case of the complex sinusoid at 0 radians), but I have an optimisation that reduces the number of multiplies by 50%.

Note: this is absolutely not homework, I've written this algorithm because I found the explanations of the Cooley/Turkey method far too convoluted, and at the same time I wanted the ability to select which bins I processed in both the FFT and IFFT as well as being able to output the IFFT to multiple channels. So I just decided to figure out my own radix-2 while at the same time understanding how and why the transform works (which I have).

ORIGINAL QUESTION (edited for clarity) ---

Example code: kfft.c

I wanted to know whether there is a similar FFT algorithm to mine.

I basically perform FFT by factorising calculations by recursively subtracting the second half of a signal from the first, meaning x[n] only needs to be multiplied by the complex sinusoid for only half of the duration, resulting in N/2 log N real multiplies (shown below). The summed signal is for multiplying with complex sinusoids that are multiples of the sampling period. For example groups {2, 6} are multiples of 2, and therefore have a periodicity of 4 samples.

// For bins { 1, 3, 5 and 7 }

x1[0] = x0[0] - x0[4]
x1[1] = x0[1] - x0[5]
x1[2] = x0[2] - x0[6]
x1[3] = x0[3] - x0[7]
// Calculate sum (required for the next level of bins)
x0[0] = x0[0] + x0[4]
x0[1] = x0[1] + x0[5]
x0[2] = x0[2] + x0[6]
x0[3] = x0[3] + x0[7]

// For bins { 2 and 6 }

x2[0] = x0[0] - x0[2]
x2[1] = x0[1] - x0[3]
// Calculate sum (required for the next level of bins)
x0[2] = x0[0] + x0[2]
x0[3] = x0[1] + x0[3]

// For bin { 4 }
x3[0] = x0[0] - x0[1]
x3[1] = x0[0] + x0[1] // <-- This is the 0hz bin (sum)
  • 1
    $\begingroup$ Your scheme is hardly any different from what the FFT does in the case when $n = 2^m$. $\endgroup$ Sep 15, 2014 at 14:19
  • $\begingroup$ Well that's what I mean. It is FFT, but not quite radix 2, or is it? I'm really not sure as I haven't quite wrapped my head around radix-2 completely (instead I just wrote this). Also does what I said make much sense, such as considering e^ (-i * 2pi * k * n) as a resonator? $\endgroup$ Sep 15, 2014 at 14:27
  • $\begingroup$ Why the down vote with no helpful comment or anything? Some strange people??? The only algorithm I've seen is radix-2/4, so surely it's understandable that i might not know this same process is the more general form when I've not seen it described in this way anywhere. $\endgroup$ Sep 15, 2014 at 15:02
  • $\begingroup$ most of us, during or since our college daze, have written a simple Tukey-Cooley radix-2 FFT in some programming language. i have done so in C (can send you the code if i can find it) and in MC68000 asm and have modified the DSP56000 asm FFT to be a either unitary ($\frac{1}{\sqrt{N}}$) gain or a decent block-floating-point (mot's code sucked). the only thing i can think of is to write the code at the beginning of each group so that multiplication by $1$ or by $j$ is avoided in the butterfly. $\endgroup$ Sep 18, 2014 at 14:32
  • $\begingroup$ Yes that is one of my planned optimisations, but with an upcoming optimisation I'm now down to N/2 log N real multiplies, so I'm wondering whether there are any further optimisation left by perhaps borrowing from optimisations used in other algorithms $\endgroup$ Sep 18, 2014 at 19:50

1 Answer 1


Indeed, this is more or less the usual butterfly used in radix-2 FFT (can be extended to any prime number): http://en.wikipedia.org/wiki/Butterfly_diagram

  • $\begingroup$ Thank you Matthieu and Dilip. And I can see clearly how and why the butterfly works. $\endgroup$ Sep 15, 2014 at 15:16

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