Confused about the period of a discretised time-series I have managed to confuse myself about the periods of discretised time series. It is all rather embarrassing. 

This is one period extracted from my continuous signal (radians on y, time on x):



To use it in my calculations I use a discretised vector of N points. The above signal has a period of $T=0.95s$. The discretised version A has $N=100$ and the angle, over time, is indexed from 1 to 100.

Q1: is the period of my angle vector $[\theta_1,\ldots,\theta_{100}]$; $T=100$? 

Now suppose I need to standardise my data as such, incl the index:

% standardise training inputs
m = mean(A);
s = std(A);
x(:,1) = 1/s(2) * (A(:,2) - m(2));  % index
x(:,2) = 1/s(1) * (A(:,1) - m(1));  % velocity


Q2: Is the new period then simply max(x(:,1))?
On the continuous setting, the period is the minimum $T\ne 0$ such that $\theta(t+T)=\theta(t)$. Once you decide to have $N$ regularly sampled points inside a period, you introduce a sampling period $T_N$ that is directly proportional to the period: $T_N=T/N$. Hence, you can describe your period as a quantity with a time unit (second) or as a unitless number of samples, equivalently. 

Now, if for some reason you normalize the time axis with an affine function, the mean will only shift the whole axis (no change for the period), and the scaling only will dilate or shrink the interval between samples.

So here, the effect of  $s(1)$ will turn $T_N$ into $T_N/s(1)$, and also  $T$ into $T/s(1)$. 

In your case, you can directly compute $s(1)=T_N\sqrt{\frac{N(N+1)}{12}}$. In this formula, the scale factor depends on $N$ only, and not on the start and end indices.
For $N=100$, the new period should be now a unitless quantity of 
$$T/s(1) = \sqrt{\frac{12N}{N+1}}\,.$$