Frequency Axis of Discrete Fourier Transform (DFT) with Odd Number of Data Points I am trying to understand the logic behind making a frequency axis in DFT. I am using for time based light absorbance. When we have even number of data points (N= even integer), collected over a length of time A, the lowest frequency $\omega_1$ that can be resolved is 1/A, the next step is 2/A, the $k^{th}$ step is k/A, where k =1,2,3..., N/2. The highest frequency $\omega_{max}$ = N/2A based on Shannon's or Nyquist criterion. This the maximum value on the frequency axis.

How do we scale the frequency axis of FFT when we have odd number of data points (N+1) collected over a length of time A, where N is a even number? I cannot find similar reasoning for scaling when we have odd number of data points except in the book excerpt below.




The step size of frequency in case of odd number of data points:  k/A whether the data points are even or odd.


Please see the excerpt from a book called DFT: An Ownwer's Manual by Briggs. What is the author trying to say when he says that the highest frequency $$\frac{N}{(2A)}$$ does not quite coincide with the endpoint of the frequency domain which have the values $\Omega$/2?Please note that the author is using N as an even number.

Is he indicating that there is a slight error when we have a DFT of an odd number of datapoints?
This can be answered simply by considering the definition of the $N$-point DFT:

$$X_N[n] = \sum_{k=0}^{N-1} x[k]e^{-j2\pi \frac {n}{N}k }$$

where it's easy to see that the DFT just compares your $N$-point input signal $x[k]$ to a sinusoid of frequency $\frac nN$. 

Thus, the lowest frequency is always $0$, and the resolution is always $\frac{f_\text{sample}}{N}$, no matter whether $N$ is even or odd. That also means that the lowest non-DC frequency is also $\frac{f_\text{sample}}{N}$.