Thanks to my sum_discrete_fourier.pdf paper, to me it seems that if we have a finite signal on a finite time scale with graininess set \(M =\{\mu_1, \mu_2,...,\mu_n\},\), with \(p:=\sum_{k=1}^{n} \mu_k\) then we are really just taking a sum of Fourier Transforms of *period-one* sequences (the sequence \(\{f_m^{j-1}\}\) in the original paper would be constant on the time scale \(\mathbb{Z}\) (we could also move to \(p \mathbb{Z}\))). So, we are going to get a (kind of) complicated sum of just \(n\) things. Those \(n\) things will be weighted by the hilger real part and hilger imaginary part. So, in the forward direction we’ll get sums of multiples of 1/(1-e^(-i \omega)) for various values of \omega and \delta functions. We should get similar things in the backwards direction.

If we have a finite signal over a periodic time scale where the signal length is longer than the period of the time scale, then we’ll have a sum of DTFs

Useful slides: https://spinlab.wpi.edu/courses/ece503_2014/11-3dtft_of_periodic_sequence.pdf

Let’s do an example

\mathbb{T}*{1,2} = {0,1,3,4,6,7,9,…}
f(t) = \sin( \pi/32 t); 0 \leq t \leq 63
x(n) = {\sin(0), \sin(3 \pi/32), \sin(6 \pi/32),…., \sin(63 \pi/32)} = {\sin(3n \pi/32)}*{n=0}^{21}
y(n) = {\sin(\pi/32), \sin(4 \pi/32),\sin(7 \pi/32),… \sin(64 \pi/32)} = {\sin((3n+1) \pi/32)}_{n=0}^{21}

According the the slides, the DTFT of x(n) is

{\cal F}*{\mathbb{Z}}{x}(e^{i \omega}) = frac{2 \pi}{22} \sum*{t=\infty}^{\infty} \sum_{n=0}^{21} x(n) e^{-2 \pi i r n/22} \delta(\omega-\frac{2 \pi r}{N})