# Where did the length of time’s period disappear in Periodic Fourier Series Discrete Time

In continuous time Periodic Fourier Series has smallest n as possible, since it is an integral and a length of the repeating time (period time) which is T0. In discrete time however we don’t have units of time at all. Explain in your own words the intuition of avoiding time units in Discrete Time Fourier Series of a signal periodic in time.

Time T0 should be devided by N and the time intervals between N0 N1 and N2 should all be T/N. But, that T0 could be more than 2pi! Could you please explain the 2pi/N? What are the units of 2pi/N? The units of 2pi/T in contniuous time were 1/second. N is an integer, it shall have no units, it shall be the numbers of parts we devide some time T0 by. No? Yes? I'm confused. Could you help? Thanks ahead.

## 2 Answers

A DFT (Discrete Fourier Transform) is applied to an interval of a signal with uniformly sampled values on the underlying dimension, which is often time. $$N$$ is a variable that is conventionally used to define the number of sample points. The set of points in the interval is called the DFT frame. Thus $$N$$ has units of $$\frac{samples}{frame}$$.

The DFT does not care about the sampling rate. The sampling rate is used to convert frequencies in the DFT of cycles per frame to cycles per the underlying dimension's unit.

$$2\pi$$ has units of $$\frac{radians}{cycle}$$. The $$kth$$ bin in a DFT represents a frequency of $$k$$ $$\frac{cycles}{frame}$$.

Therefore, the expression

$$\frac{2\pi}{N}k$$ has units of

$$\frac{\frac{radians}{cycle}}{\frac{samples}{frame}} \cdot \frac{cycles}{frame} = \frac{radians}{sample}$$

Now, look at the unnormalized definition of the DFT:

$$X[k] = \sum_{n=0}^{N-1} x[n] e^{-i \left(\frac{2\pi}{N}k\right)n }$$

The units of $$n$$ is samples, so the units of the exponent is:

$$\frac{radians}{sample} \cdot sample = radians$$

Which is what it should be.

$$F_s$$ is the usual symbol for the sampling rate, also called the sampling frequency and sometimes the sampling density. With seconds as the unit of time, it has units of $$\frac{samples}{second}$$. I do not like the use of $$Hz$$ as a unit here, it causes confusion with $$Hz$$ being a unit of the frequency domain.

$$T = \frac{N}{F_s}$$

$$\frac{seconds}{frame} = \frac{ \frac{samples}{frame} }{ \frac{samples}{second} }$$

The frequency of the $$k$$th bin.

$$f_k = \frac{k}{T} = k \frac{F_s}{N} = \frac{k}{N} F_s$$

\begin{aligned} \frac{cycles}{second} &= \frac{ \frac{cycles}{frame} }{ \frac{seconds}{frame} } \\ &= \frac{cycles}{frame} \frac{ \frac{samples}{second} }{ \frac{samples}{frame} } \\ &= \frac{ \frac{cycles}{frame} }{ \frac{samples}{frame} } \frac{samples}{second} \\ \end{aligned}

Frequency spacing on the DFT is also known as the bin width. The difference in frequency between any two adjacent bins (except DC and Nyquist) is going to be $$\frac{F_s}{N}$$ in units of $$\frac{cycles}{second}$$ also called $$Hz$$.

We could define with T0 and the relative time units to restore original time, indeed. But apparently we can relate just to N indices and that’s what definition did.