fftshift in MATLAB with even number of data points in double sided spectrum I have a question with reference to this Table.


With even N, the frequency axis extremes should be $\pm$Fs/2, where Fs is the sampling frequency. However in the array we have only one value corresponding to the Nyquist frequency. Following the notation of the table in the first column, N=10, we have

[$c_{0}$ $c_{1}$ $c_{2}$ $c_{3}$ $c_{4}$ $c_{-5}$ $c_{-4}$ $c_{-3}$ $c_{-2}$ $c_{-1}$] 

which after fftshift should become

[$c_{-5}$ $c_{-4}$ $c_{-3}$ $c_{-2}$ $c_{-1}$ $c_{0}$ $c_{1}$ $c_{2}$ $c_{3}$ $c_{4}$  ]

The $c_{-5}$  value corresponds to the Nyquist frequency. So how is our double sided frequency spectrum symmetric when N is even? For real valued function it will be zero, but is there a reason that MATLAB calculates the negative frequency first as shown in the Table?

Example:
A_even=[ 0 0 0 1 1 1 1  0 0 0],

N=10

B=fft(A_even)'

C=fftshift(B)

Then C is equal to

$$\begin{matrix} 0.0000 + 0.0000i\\
-0.3090 - 0.9511i\\
0.4271 + 0.5878i\\
0.8090 + 0.5878i\\
-2.9271 - 0.9511i\\
4.0000 + 0.0000i\\
-2.9271 + 0.9511i\\
0.8090 - 0.5878i\\
0.4271 - 0.5878i\\
-0.3090 + 0.9511i\\
\end{matrix}$$

This implies $c_{-5}$ =  0.0000 + 0.0000i  and $c_{0}$= 4.0000+0000i
Thanks.
If the size $N$ of the DFT is even, only one "extremal" point (after fftshift) is Nyquist.  If $N$ is even, you cannot have an arbitary $c_{-5}$ and $c_5$.  They must add to be whatever your $c_{-5}$ term is.

If the input to the DFT is purely real and if $N$ is even, consider the Nyquist point, at $c_{N/2}$ to be split in half.  One half is the negative-frequency component and the other half is the positive-frequency component.  Since the negative and positive frequency components must be complex conjugates of each other, and since $c_{-N/2}=c_{N/2}$ after splitting, that means that $c_{N/2}$ must be purely real just as $c_0$ must also be real.