Why is it that the input to an IFFT is first reordered in an OFDM Modulator

Question

I am back with my questions! I recently came across a paper titled "Low Latency IFFT Design for OFDM Systems Supporting Full-Duplex FDD" where they claim a new method of IFFT input mapping to reduce the output latency. However, they have mentioned how in a conventional OFDM, the null subcarriers are mapped in the middle and the data subcarriers to the side. I have attached a picture of this. But my question is why is it done in the first place? Is it something to do with the efficient use of butterfly operation?

Also, please mention a reference where it is cited that conventional OFDM systems use this method of reordering before performing an IFFT operation.

Thank you

Answer 1

Since Discrete Fourier Basis Vectors are $2\pi$-Periodic, hence, negative frequencies $-\frac{2\pi k}{N}$ are conventionally represented as $(2\pi - \frac{2\pi k}{N} = \frac{2\pi}{N} (N-k))$. It has nothing to do with Butterfly Structure of FFT Algorithm.

The above statement basically means that $-k^{th}$ tone is represented as $(N-k)^{th}$ tone in DFT/FFT. That is why you are seeing that DC and positive tones are placed as it is, but negative tone $-k$ is placed at $N-k$ before taking IFFT.

It is precisely the consequence of the convention that DFT coefficients are $X[k]$ where $k = 0,1,2,...,(N-1)$. $k$ goes from $0$ to $(N-1)$ and not from $(-\frac{N}{2})$ to $(\frac{N}{2} - 1)$.

So, DC + Positive $\frac{N_d}{2}$ tones are placed as first $\frac{N_d}{2}+1$ values, and then Negative $\frac{N_d}{2}$ tones are placed in reverse order from the end, i.e. $-1^{th}$ tone becomes $(N-1)^{th}$ value, $-2^{nd}$ tone becomes $(N-2)^{nd}$ value and so on to $-\frac{N_d}{2}^{th}$ tone becomes $(N-\frac{N_d}{2})^{th}$ value. The rest of the $(N - (N_d +1))$ tones are zero padded. This array of values becomes the IFFT input.