[Tale]

Originally Posted by clepsydrae

If so, is it the case that the JS function that reaper provides -- fft() -- takes n real samples and produces n/2 pairs of result values in the same buffer, where the first is the real component and the second is the imaginary?

JSFX's fft(buf, n) takes n complex pairs as input, and also produces n complex pairs. If you want to input a real signal, then you need to convert it to complex first by setting the imaginary part to 0:

i = 0;
loop(n,
  complex_buf[2*i] = real_buf[i];
  complex_buf[2*i+1] = 0;
  i += 1;
);

fft(complex_buf, n);
fft_permute(complex_buf, n);

Originally Posted by clepsydrae

Thus after calling fft_permute, the 0'th pair (indices 0 and 1) is <dc_component, 0> and thereafter the i'th pair (indices 2*i and 2*i+1) where i goes from 1 to windowsize/2 - 1 is the <real, imaginary> FFT value of the signal at frequency i/windowsize*srate ? No fencepost errors there?

Yeah, that sounds about right.

Originally Posted by clepsydrae

And magnitude would be sqrt(results[2*i]^2 + results[2*i+1]^2), and would range from 0 to 1?

Yes, except I think its range is 0 to +inf.

Originally Posted by clepsydrae

maybe fft() returns the values as <imaginary, real> ?

Nope, re=buf[2*i], im=buf[2*i+1].