Discrete Fourier Series - Time Index I have been trying to understand Discrete Fourier Time Series (NOT Transform).

It is defined as 

$$a_k = \sum_{n=0}^{N-1} x[n]e^{jk\frac{2\pi}{N}n}$$

where N is the Time period and an integer (by definition), and $n$ goes from $0 ... N-1$.

Fair enough. But when N = 0.2 second as the time period of the wave, then how we define $n$, as $N-1$ is negative. I think here we should take $N$ as 200 miliseconds. And then loop $n$ from 0 ... 199.

But then the terms $x[n]$ (the samples) can be defined over either 200 milisec OR $2*10^8$ nanoseconds and so on. So for 200 ms case, we need 200 samples. But lets say I have taken 100 samples distributed equally over 200 ms time period. Now how we define $x[n]$ in this case, should I do like this:  $x[0]=1, x[1]=0, x[2]=1,...$ That is, set alternate values to zero?

Because in the above equation, I need 200 values of $x$.

Or Better (I think):

Define $1 sec = 500ss$, where $ss$ is a hypothetical new unit of time, in that case time period is $N=0.2*500ss=100ss$ (with the condition that the final value is an integer). In this case we can loop over 100 samples easily without worrying over setting alternate values.

Can anyone kindly help me with this or on my thinking above.

Best Regards,
The integer $N$ is the number of samples of $x[n]$ in one period. There need not be any relation to absolute time whatsoever, simply because $x[n]$ are just discrete data that are not necessarily samples of a function of time.

However, if the data $x[n]$ are samples of some continuous-time signal, then the time difference between two adjacent samples is, by definition, the sampling interval $T_s$, which is the inverse of the sampling frequency $f_s=1/T_s$.

In that case, the time corresponding to one period of $x[n]$, i.e., the time difference between two samples $x[n]$ and $x[n+N]$, equals $N\cdot T_s=N/f_s$.