# Ambiguity in the IFFT process in OFDM

I am still trying to iron out some ambiguities in my understanding of the IFFT component of OFDM modulation schemes.

So here we have a QAM symbol $$s_0$$ being multiplied with the subcarrier for that frequency bin, $$s_1$$ being multiplied with carrier 1, $$s_2$$ carrier 2, and it shows clearly that the sum of these products produces the OFDM symbol in the time domain. This would logically follow the equation:

$$c\left(t\right) = \sum^{N-1}_{n=0}{x_n e^{-i\frac{2\pi n}{N}}}$$

Where c(t) is the output symbol, x is the list of complex QAM symbols and n is the subcarrier index, N is the number of subcarriers. (Note there is no 1/N at the start and i'm not sure why that needs to be introduced)

Why is it that IDFT is stated on Wikipedia to be

$$x_n = \frac{1}{N} \sum^{N-1}_{k=0}{X_k . e^{i 2 \pi k n / N}}$$

Where x is the output list and xn is merely a single output sampling point in the time domain.

Here is an example of application of the DFT/IDFT formula :

Let : $$N = 4$$ and : $$x = \left(\begin{array}{c}x_0\\ x_1\\ x_2\\ x_3\\ \end{array} \right) = \left(\begin{array}{c}1\\ 2-i\\ -i\\ -1+2i\\ \end{array} \right)$$

then

$$X_0 = e^{-i 2 \pi 0.0/4} + e^{-i 2 \pi 0.1/4}.\left(2-i\right) + e^{-i 2 \pi 0.2/4}.\left(-i\right)+ e^{-i 2 \pi 0.3/4}.\left(-1+2i\right) = 2$$ $$X_1 = e^{-i 2 \pi 1.0/4} + e^{-i 2 \pi 1.1/4}.\left(2-i\right) + e^{-i 2 \pi 1.2/4}.\left(-i\right)+ e^{-i 2 \pi 1.3/4}.\left(-1+2i\right) = -2-2i$$ $$X_2 = e^{-i 2 \pi 2.0/4} + e^{-i 2 \pi 2.1/4}.\left(2-i\right) + e^{-i 2 \pi 2.2/4}.\left(-i\right)+ e^{-i 2 \pi 2.3/4}.\left(-1+2i\right) = -2i$$ $$X_3 = e^{-i 2 \pi 3.0/4} + e^{-i 2 \pi 3.1/4}.\left(2-i\right) + e^{-i 2 \pi 3.2/4}.\left(-i\right)+ e^{-i 2 \pi 3.3/4}.\left(-1+2i\right) = 4+4i$$

$$X = \left(\begin{array}{c}X_0\\ X_1\\ X_2\\ X_3\\ \end{array} \right) = \left(\begin{array}{c}2\\ -2-2i\\ -2i\\ 4+4i\\ \end{array} \right)$$

If we pretend this worked example is IDFT and not DFT, it clearly shows that for each output sampling point, Xk, in the time domain, every point in the constellation was multiplied by a single frequency bin and then summated. This is directly backed up by the fact that IDFT is stated to have a complexity of O(n^2) (as far as I'm aware IFDT and IFFT are the same equation but implemented differently algorithmically).

Why is this? This directly contradicts many OFDM examples and diagrams I've seen and it fundamentally makes sense to me that a single frequency would be phase and amplitude shifted by a single complex input from the constellation, and not by all of them.

• I have worked it out. In essence Commented Nov 12, 2018 at 13:20
• FFT (respectively IFFT) are algorithms which calculate efficiently the DFT (resp. IDFT) by taking advantages of arithmetical symmetries. Commented Nov 12, 2018 at 13:21
• @MaximGi I'm about to answer my own question, because I have cleared it up Commented Nov 12, 2018 at 13:22
• Ok, very good. I was just adding info on your subquestion about FFT/DFT difference without having fully understood your question Commented Nov 12, 2018 at 13:26