>>1. Why the manual example does not show the phantom
>>frequencies?
>
>Ah, I understand. They dropped the right part of the
>positive interval because it would show the phantom
>frequencies inherited from the negative domain.
>
Yes, although these frequencies are not always redundant. If the original function is complex rather than purely real, then half the information content of the transform is in the negative frequencies. Of course, in most practical cases, the function is some physical signal that is a function of time, and so it is real. In that case, the negative amplitudes are the conjugates of the positive ones, and so they can be constructed (if wanted) from the positive ones. The manual seems to assume that this will always be the case, and that therefore the negative frequencies will be redundant. The manual should not make these assumptions, of course.

>>2. You wrote "However, because of the way MatLab is set up,
>>it likes to use vectors that are indexed by nonnegative
>>integers, so the period that is chosen is from 0 to N-1." If
>>it's not an idiosyncrasy then what is?

Well, it is and it isn't. Since the transform is periodic, one must choose some particular period to display, and the choice is essentially arbitrary. I would prefer to see an interval that is symmetric around the origin, as that makes the interpretation easier, but that's just a matter of preference. It is I think equally reasonable to take a period that starts at zero. Where the manual fails is in not documenting that there is a choice being made at all.

By the way, I did not mean to say that there were no idiosyncracies in MatLab; it is full of them, which steepens its learning curve unnecessarily. What I meant was that there was nothing unusual about its implementation of the FFT algorithm itself. How to display the results of the calculation is a separate issue, and here I agree with you -- I would have preferred a symmetric interval.

--Stuart Anderson