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Okay, so I am aware of the math behind perfect reconstruction in filterbank and wavelet theory when there is half-band symmetry. But I don't understand, when we have the computational power to do decent DSP to modulate the binary signal into the baseband IQ signal, why anyone puts up with any crossover from one FDM channel into another.

Consider this figure from Zhao 2014: DFT-based offset-QAM OFDM for optical communications:

frequency division multiplexing

Why use any other model than the Nyquist FDM model? Make maximum use of your channel's bandwidth and keep them other channels from fucking with the channel that your receiver will be demodulating.

Conceptually, the math is simple enough:

The discrete-time bipolar binary signal is

$$\begin{align} x[n] &\triangleq -1 + 2 a[n] \quad \in \{-1, 1 \} \\ &= -(-1)^{a[n]} \\ \end{align}$$

where $a[n] \in \{$0, 1$\}$ is our serial bit stream. And the baseband quadrature signals are:

$$\begin{align} i[n] &= \sum\limits_{m=-\infty}^{\infty} x[2m] \, p[n-2m] \\ \\ q[n] &= \sum\limits_{m=-\infty}^{\infty} x[2m+1] \, p[n-(2m+1)] \\ \end{align}$$

where

$$ p[n] = \operatorname{sinc}\left( \tfrac{n}{2} \right) w[n] $$

and

$$ \operatorname{sinc}(u) \triangleq \begin{cases} \frac{\sin(\pi u)}{\pi u} \qquad & u \ne 0 \\ 1 & u = 0 \\ \end{cases}$$

and you have a decent window function $w[n]$ such as a Kaiser:

$$ w[n] \triangleq \begin{cases} \frac{1}{J_0(\beta)} J_0\left(\beta \sqrt{1 - \left(\frac{n}{1+M/2}\right)^2} \right) \quad \quad & |n| \le \tfrac{M}{2} \\ \\ 0 & |n| > \tfrac{M}{2} \\ \end{cases} $$

$J_0(\cdot)$ is the 0th-order modified Bessel function of the first kind.

$$ J_0(u) = 1 \ + \ \sum\limits_{k=1}^{\infty} \frac{1}{(k!)^2} \left(-\frac{u^2}{4}\right)^{k} $$

$M+1$ is the number of non-zero samples or FIR taps, which gets you a finite number of terms in the summation:

$$\begin{align} i[n] &= \sum\limits_{m=\lfloor n/2-M/4 \rfloor}^{\lfloor n/2+M/4 \rfloor} x[2m] \, \operatorname{sinc}\left( \tfrac{n-2m}{2} \right) w[n-2m] \\ \\ q[n] &= \sum\limits_{m=\lfloor (n-1)/2-M/4 \rfloor}^{\lfloor (n-1)/2+M/4 \rfloor} x[2m+1] \, \operatorname{sinc}\left( \tfrac{n-(2m+1)}{2} \right) w[n-(2m+1)] \\ \end{align}$$

$i[n]$ would be sampled at every even $n$ and $q[n]$ would be sampled at every odd $n$.

This puts in a little constant delay of $\frac{M}{2}$ samples to do a nice sharply defined band for each OQPSK channel. Why would that be a problem, considering all of the other latency you will have in the signal chain? Why allow any significant leakages from adjacent channels? (Of course there should be a little bit of guard band between adjacent channels.) Why put up with any overlap?

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    $\begingroup$ Work out the bandwidth efficiency in bits/Hz for both cases and therein lies your answer (going away for a week starting tomorrow but if not done by the time I get back I will work through it analytically) $\endgroup$ Jul 17 at 3:20
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    $\begingroup$ The Kaiser window that you proposed, particularly for large values of $\beta$, can indeed yield very low sidelobes in the FFT per-bin frequency response. As with any window function, however, low sidelobes come at a cost of a wider main lobe. So, you actually get more overlapping with adjacent subcarriers if you were to window the transmitted signal. The sidelobes fall down much more quickly, so there will be less interference between far-away subcarriers, but much more interference between subcarriers near each other. $\endgroup$
    – Jason R
    Jul 17 at 15:38
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    $\begingroup$ Kaiser window will increase the main lobe at the benefit of low sidelobes but nothing beats the resolution of a non-windowed (rectangular window) function. It’s the fact that the signals are uncorrelated if the bin spacing is 1/T that makes it work so well. When the signal is considered over the duration T, there is no “overlap”— the overlap you see is what would occur if there was a frequency offset $\endgroup$ Jul 17 at 16:43
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    $\begingroup$ In the rectangular window DFT, each bin is an independent symbol with no cross symbol interference. As soon as you apply any other window, you get leakage from one bin to the next and therefore one symbol to the next. Hence the decreased spectral efficiency. $\endgroup$ Jul 17 at 16:57
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    $\begingroup$ Yes agreed - Kaiser better and DPSS based such as this even better (barely). dsp.stackexchange.com/questions/70516/… So if the window is much wider than the Sinc main lobe the frequency spearing of that would be minimized which is perhaps what you are doing— I am looking forward to looking at it closer. $\endgroup$ Jul 17 at 18:22

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