Interpreting N in DFT as the Number of Points vs. Number of Intervals The &quot;N&quot; is DFT is understood to be the number of data points in a given sequence or in other words the length of the sequence. We recently have had discussions here Indexing in DFT (from an old paper) and someone's old question How do I measure the time duration of a finite-length discrete sequence?. One of the popular symbolic versions of DFT is
$$
X(j)=\sum_{k=0}^{N-1} x(k) \exp \left(-i 2 \pi\left(\frac{j}{N}\right) k\right)
$$
Suppose someone gives us the sequence consisting of N=11 points without telling us the total time or the sampling rate. If we apply DFT on it in MATLAB, the output is 11 points

Notational problems start when we wish to determine the time interval $\Delta$t and the frequency step $\frac{1}{N\Delta t}$ when the $\Delta$t is revealed.
a) If we wish to determine the sampling frequency it is (N-1) points collected in 1 seconds. The last point belongs to the next cycle. The correct sampling rate is 10 Hz not 11.
b) The frequency resolution is shown to be $\frac{1}{N\Delta t}$. In order to get the correct  frequency step, we to have to put $\frac{1}{10(0.1)}$  NOT $\frac{1}{11(0.1)}$.
Therefore, the frequency resolution, if we accept that N=11,
$$\frac{1}{(N-1)\Delta t}$$ but nobody shows this formula.
It seems that we are using N in two ways

N as the length of the sequence
N when we have to determine the frequency step where actually it is one less than N to get the correct result.

A respected colleague says that N should be interpreted as the no of intervals not the number of points. This is inconsistent with defining N as the number of points in the sequence. Unfortunately, I cannot find any reference which says that N is the number of intervals.
How can we make this consistent?
Thanks.
The key is to understand what the DFT says, vs what we seek. Consider a cosine, where we change $f \text[Hz]$, $N$, and $t$ and observe the effect on DFT:


[1]: DFT &quot;sees&quot; 1 cycle in the &quot;analysis frame&quot; (i.e. what we feed it), so nonzero at $k=1$, as expected.
[1] to [2]: we double our time duration without changing $f$ or $N$; DFT sees this as two cycles spanning the analysis frame, so $k=2$.
[1] to [3]: we double the physical frequency without changing $N$ or $t$; DFT sees this as two cycles spanning the frame.
[1] to [4]: we double the number of samples, $N$, without changing $f$ or $t$; DFT sees this as still 1 cycle spanning the frame; bin location, $k$, remains unchanged, but (unnormalized) correlation strength doubles (not relevant here).
[4] to [5]: now we also double duration, yielding two cycles in analysis frame.

You should be seeing a pattern. Without reading further, try to establish a relationship between $k$, $N$, $t$, and $f$. Hint: units.

Here's the deal: DFT has no idea what Hz, or physical frequency, is. All it knows is samples and cycles spanning the analysis frame. In [1], the &quot;DFT frequency&quot; is
$$
f_{\text{DFT}} = \frac{k}{N} = \frac{1 \text{ cycle}}{10 \text{ samples}} = .1 \left[ \frac{\text{cycles}}{\text{samples}} \right]
$$
[2] = [3] = 2 cycles / 10 samples, [4] = 1 cycle / 20 samples, [5]  = 2 cycles / 20 samples. Now let's take what we know about physical frequency, $f_p$ and DFT frequency, and relate them. In [2], DFT says $k=2$, but we know $f_p = 1$. It could also be (not shown in any [1]-[5]) $f_p=2$ and $k=1$. How to convert?
The unifying relation is:

 \begin{align} f_p \left[ \frac{\text{cycles}}{\text{second}} \right] &amp; = \left( f_{\text{DFT}} \left[ \frac{\text{cycles}}{\text{samples}} \right] \right) \cdot \left( f_s \left[ \frac{\text{samples}}{\text{second}} \right] \right) \end{align}

So, for [2]:
$$
f_{\text{DFT}} \cdot f_s =  \left( \frac{2 \text{ cycles}}{10 \text{ samples}} \right) \cdot \left( \frac{10 \text{ samples}}{2 \text{ seconds}} \right) = 1 \left[ \frac{\text{cycles}}{\text{second}} \right] = 1\ \text{Hz} = f_p
$$

But how to determine $f_s$?
By a simple definition, it's the reciprocal of sampling period, $\Delta t$, making everything above consistent. However, one must ask, if &quot;sampling frequency&quot; is defined as &quot;# of samples / total duration&quot;, and &quot;total duration&quot; of
$$
[0, .1, .2, .3, .4, .5, .6, .7, .8, .9]\ \text{sec}
$$
is clearly $0.9\ \text{sec}$, then isn't $f_s$ actually $.9 / 10 = 0.9\ \text{Hz}$? No; the duration is actually 1 sec. Here's why: $0.9\ \text{sec}$ here is actually the duration of something else entirely. Namely, &quot;what is the duration of the signal?&quot; can ask two things:

For how long have we been sampling?
How much time's worth of information is contained in our signal?

The answer to former is $0.9\ \text{sec}$, but to latter is $1\ \text{sec}$. Former computed via $(N-1)\Delta t$, latter via $N \Delta $, and if we insist on 0.9 for #2, we're saying that one sample holds no time-representative information, which implies all signals have duration of zero.
I clarify this with examples here. In a nutshell, the end-goal is description of information, not of process used to obtain it.

So what is the frequency resolution (rather DFT bin spacing)${}^{1}$?
It's defined as spacing between DFT bins, $df$; the answer depends on units of choice. For Hertz, per all of the above,
$$
df_p = \frac{1}{N \Delta t} \tag{1}
$$
meaning, for [2], $k=1$ corresponds to $f_p = 0.5\ \text{Hz}$, $k=2$ corresponds to $f_p = 1\ \text{Hz}$, and so on. Alternatively, if you insist on defining duration via $(N-1)$, then it'll be per $(N-1)$ in Hertz, but not in DFT frequencies; latter is unambiguous:
$$
k=1 \rightarrow \frac{1\ \text{cycle}}{N\ \text{samples}} = \frac{1}{N} \left[ \frac{\text{cycles}}{\text{sample}} \right]
$$
You can again convert between DFT frequency resolution and physical; taking [2], $\text{Duration} / N = 2 \text{ sec} / 10 = .2 \text{ sec}$, so bin spacing is
$$
df_p = \frac{1}{N \Delta t} = .5\ \text{Hz}
$$
The spacing can change, but it's by redefining $\Delta t$ rather than changing $N$ to $(N - 1)$ in $(1)$. Suppose we say $\text{Duration} = 1.8\ \text{sec}$; then, $\Delta t = 0.18\ \text{sec}$, and
$$
df_p = \frac{1}{N \Delta t} = 0.\bar{5}\ \text{Hz}
$$
So in [2], $k=1$ corresponds to $0.56\ \text{Hz}$, and $k=2$ corresponds to $1.1\ \text{Hz}$, which agrees with completing 2 cycles in 1.8 secs = 1.1 Hz.

Your specific example: $N=11$, $\Delta t = 0.1\ \text{sec}$:
$$
df_p = \frac{1}{N \Delta t} = \frac{1}{11 \cdot 0.1\ \text{sec}} = 0.909\ \text{Hz}
$$
So $k=1$ corresponds to $0.909\ \text{Hz}$, not to 1Hz, since you've included a sample from the next cycle in the analysis frame.

1: NOTE: $df$ is DFT bin spacing, not &quot;frequency resolution&quot;. DFT has perfect frequency resolution and no time resolution. But if you define it as discrimination of continuous-time frequencies, then the resolution and bin spacing are inversely-related (lesser spacing -&gt; more bins -&gt; more granular resolution). This is its own topic so I'll avoid clarifying in detail, feel free to open new q.