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How can the fourier series / discrete fourier transform equation be modified in such a manner that they allow the existence of negative time units in addition to the positive ones?

The idea is to have the transforms behave more like the Fourier Transfor/DTFT which have negative as well as positive time and have the starting point of the exponentials at the "0" time unit, rather than at the start of the window. The phase is thus thrown around based on the ratio of negative and positive time compared to the regular Fourier series.

As an illustration, a discrete 5-sample length signal {0,0,1,0,0}, but now say the sample(time) values for the samples are {-2,-1,0,1,2}, instead of 0...4.

enter image description here

Taking a regular DFT of the signal produces {1,1,1,1,1} for the amplitudes and { 0., -2.51327412, 1.25663706, -1.25663706, 2.51327412} for the phases. Clearly the phases should be zero, however. A picture to illustrate the difference (wrapped phase):

enter image description here

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    It is unclear what you wan to ask. Can you elaborate or take an example? – 2017-01-06
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    Sure, for period $N$ you can take for example any interval $[-k, N-k)$ for $0 \leq k < N$. The DFT works exactly in the same way regardless of this choice of interval. – 2017-01-06
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    It is important to understand that the DFT is only applicable to periodic signals $x(t) = x(t + t_{\rm p})$. In that sense, negative time units are already included. We typically extract $N$ samples at the time points $0, t_0, \ldots, (N-1)t_0$ but since the function is periodic, its value at and $n t_0$ for $n<0$ is included in these samples at $\bar{n}t_0, \; \bar{n} = n +k\cdot N$ where $k \in \mathbb{N}$ such that $0\leq \bar{n} \leq N-1$. – 2017-01-06
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    @Florian Yes, but in the DFT the starting point of the exponentials is always at the start of the window and not say, middle, where the t=0 could reside. This produces some weirdness when approximating DTFT... – 2017-01-06
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    @Tony: And usually this make sense, since practically existing signals should be causal, so we look at them for $t\geq 0$. You could do the same for the DTFT. But if you want it to be symmetric you can tweak your DFT kernel. I tried to answer below. – 2017-01-06
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    Show/Use the shifting theorems – 2017-01-06

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The DFT is defined for periodic signals. The example you drew shows an excerpt of the signal around $t=0$, i.e., the values $[0,0,1,0,0]$ correspond to time points $-2t_0, -t_0, 0, t_0, 2t_0$. The standard approach to handle this case would be to periodify further and look at the 5-point interval $[0,4t_0]$ where the values are $[1,0,0,0,0]$ giving rise to the DFT you expected ($[1,1,1,1,1]$).

If you want to define a DFT that gives the same result but starts with the interval that is symmetric around zero, i.e., $[-\frac{N-1}{2}t_0, \frac{N-1}{2}t_0]$ instead of $[0,(N-1)t_0]$, all you need is to incorporate the shift. Similar to the shift theorem of the Fourier transform which reads as $${\mathcal F}(x(t-t_0)) = X(f) \cdot {\rm e}^{-\jmath 2\pi f t_0},$$ for the DFT you can write $${\mathcal D}(d[n-n_0]) = D[\mu]\cdot {\rm e}^{-\jmath 2\pi \frac{\mu n_0}{N}}.$$

This can be incorporated into your DFT kernel. It changes from $w_{n,\mu} = {\rm e}^{-\jmath 2\pi \frac{n \mu}{N}}$ to $\bar{w}_{n,\mu} = {\rm e}^{-\jmath 2\pi \frac{(n-n_0) \mu}{N}}$, where in your case $n_0 = \frac{N-1}{2}$. As before you use it via $D[\mu] = \sum_n \bar{w}_{n,\mu} d[n]$. You can see how this does nothing but shift the time axis $n$ by $n_0$ samples.