I have series of $ N =133 $ data sampled symmetrically around the origin in an actual space at $ x(n \tau) = x_n $ with $ n \in \{ -(N-1)/2, \, ..., \, 0, \, ... , \, (N-1)/2 \} $. Now, I'm a bit unsure of how a typical FFT implementation performed on my data will be correct as a result of $ N $ being odd. My data will be zero padded to a length of $ (2N-1)_2 = 512 $, which is even. Ordering the sample from $ 0 $ to $ N $ as is done in an implementation, $$ X_f = \frac{1}{N} \sum_{n=0}^{N-1} x_n e^{-2 \pi i f n / N} $$ I know that $ x_0 = 0 $, and $ X_0 \stackrel{!}{=} 0 $ needs to be my DC term, as $ f $ is sampled at intervals of $ m / N \tau, \, m \in \{0, \, ..., \, N-1 \}$. But now my data has one more element on one side of my zero-datum than the other since all the spatial frequencies above $ N/2 = 256 $ are negative frequencies due to the periodicity $ X_{-m} = X_{N-m} $: there are $ 256 $ points to the right of zero, and $ 255 $ points to the left (as $ X_{257} \rightarrow -255 $ cycles per length, ..., $ X_{511} \rightarrow -1 $ cycles per length).
So my real data is sample-symmetric about the origin, but a zero-padded data set isn't. It's hard to explain exactly what I'm having difficulty with, but it seems like my transform is going to have some kind of "fencepost" error in it due to some kind of offset by one in my data about the center. There will be one extra zero-pad data point on the right side than the left. In fact, if I had taken an even number of real samples from the start, there is no datum actually associated with the origin at all, so shouldn't an $ n = 0 $ term zero out some poor piece of data that doesn't deserve it? I know my fear is unjustified, but I can't seem to work through whatever imaginary problem it is I'm having.
