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:
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?
