# 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 – Lewis Kelsey Nov 12 '18 at 13:20
• FFT (respectively IFFT) are algorithms which calculate efficiently the DFT (resp. IDFT) by taking advantages of arithmetical symmetries. – MaximGi Nov 12 '18 at 13:21
• @MaximGi I'm about to answer my own question, because I have cleared it up – Lewis Kelsey Nov 12 '18 at 13:22
• Ok, very good. I was just adding info on your subquestion about FFT/DFT difference without having fully understood your question – MaximGi Nov 12 '18 at 13:26

I have worked it out.

For 802.11a, the usable subcarriers have frequency $$\frac{k}{T}$$ where k is the subcarrier number $$k \in [-26,-1] \lor k \in [1,26]$$ and $$T$$ is the symbol period being 3.2µs. $$\frac{1}{T}$$ is the minimum frequency separation in order for all 64 subcarrier sinusoids with uncertain phase to be orthogonal to eachother, $$\frac{1}{T}$$ = $$\frac{1}{3.2µs}$$ = 0.3125cpµs = 312.5KHz. $$\frac{64}{3.2µs}$$ = 20MHz

In the time domain, the samples are {$$x_0(t)$$ ... $$x_N(t)$$} where $$N$$ is the amount of samples = the amount of subcarriers (including pilots).

The general formula is

$$x_n(t) = \sum_{k=-\frac{N}{2}}^{-1} (A+Bj)_k e^{\frac{j2\pi kf_st}{N}} + \sum_{k=1}^{\frac{N}{2}-1} (A+Bj)_k e^{\frac{j2\pi kf_st}{N}}$$

Compute $$x_0(t)$$:

$$x_0(t) = \sum_{k=-\frac{N}{2}}^{-1} (A+Bj)_k e^{\frac{j2\pi kf_st}{N}} + \sum_{k=1}^{\frac{N}{2}-1} (A+Bj)_k e^{\frac{j2\pi kf_st}{N}}$$

Where $$n$$ is the time sample number, $$t$$ is the sampling point $$\frac{n}{f_s}$$, $$N$$ above the summation is the number of subcarriers, $$N$$ denominator of the fraction is the total number of time domain samples (they don't have to be equal), $$(A+Bj)_k$$ is complex constellation point $$X_k$$ at the $$k_{th}$$ frequency bin and $$f_s$$ is the sampling frequency.

Now, the frequency of each subcarrier about 0 is going to be given by multiplying $$k$$ by $$\frac{1}{T}$$ (the frequency granularity of the 64 point IFFT in a 20MHz channel). The subcarrier frequency $$\frac{kf_s}{N}$$ in the equation is the same thing as $$\frac{k}{T}$$ because sampling frequency*sample period = number of samples, hence sampling frequency / number of samples = 1 / sample period. ($$\frac{f_s}{N}$$ = $$\frac{1}{T}$$). From this we also realise that $$t = \frac{n}{f_s} = \frac{Tn}{N}$$ and hence the $$n_{th}$$ granularity of the symbol period aka the $$n_{th}$$ sample. We also observe that $$\frac{n}{N} = \frac{t}{T}$$ where $$N$$ here is the total number of time domain samples.

Simplify: $$x_0(\frac{n}{f_s}) = \sum_{k=-\frac{N}{2}}^{-1} X_k e^{\frac{j2\pi k*0}{N}} + \sum_{k=1}^{\frac{N}{2}-1} X_k e^{\frac{j2\pi k*0}{N}}$$

$$\small = \sum_{k=-\frac{N}{2}}^{-1} X_k ( sin (\frac{2\pi k*0}{N}) + j cos (\frac{2\pi k*0}{N})) + \sum_{k=1}^{\frac{N}{2}-1} X_k ( sin (\frac{2\pi k*0}{N}) + j cos (\frac{2\pi k*0}{N}))$$

Which will equal a complex value $$(C+Dj)$$, hence $$x_0(t) = (C+Dj)$$ and we have our first sample mapping from the frequency to the time domain. For $$x_1(t)$$, n = 1, naturally.

After the set of complex samples has been ascertained, the DC and guard bands are added, I assume $$(0,0j)$$ values. The sample is upconverted with a carrier sinusoid with frequency of the channel, i.e. 2.412MHz for channel 1, meaning the final real sample, $$x_n(\frac{n}{f_s})$$, is:

$$x_n(\frac{n}{f_s}) = Re \Bigg\{ p(\frac{n}{f_s}) e^{\frac{j2\pi f_c n}{f_s}} \Bigg[ \sum_{k=-\frac{N}{2}}^{-1} X_k e^{\frac{j2\pi kn}{N}} + \sum_{k=1}^{\frac{N}{2}-1} X_k e^{\frac{j2\pi kn}{N}} \Bigg] \Bigg\}$$

Where $$e^{\frac{j2\pi f_c n}{f_s}}$$ is a sinusoid at the carrier frequency sampled at $$\frac{n}{f_s}$$ time and $$p(t)$$ is the is the pulse-shaping function that gives the frequency domaim a sinc shape. The samples are clocked out at 20 Msps to create a 3.2 µs (20Msps/64) duration OFDM waveform. (Sample frequency = number of samples in the symbol duration = $$\frac{N}{T} = \frac{64}{3.2µs}$$ = 20MHz = $$2f_{max}$$ (which is 10MHz as the spectrum is about 0), Nyquist). To complete the OFDM symbol, a 0.8 µs duration Guard Interval (GI) / Cyclic prefix is then added to the beginning of the OFDM waveform (+16 samples), to eliminate ISI. This produces an OFDM symbol with a time duration of 4µs in length, (3.2 µs + 0.8 µs). This then goes through a DAC and it is then modulated with the carrier sinusoid as stated above. The imaginary part is also clocked out at 20MHz I'd assume at the same time, it goes through a separate DAC and is modulated by the same carrier but 90 degree phase-shifted which will be a cos wave due to periodicity at exactly the same frequency. These signals are then combined additively and transmitted.

Please do correct me if anyone has anything to add.

It is also worth noting that on the other end when it does an FFT on the time samples at a certain frequency it will produce a complex number which is then correlated to the closest point on the constellation.