Lower Bound on the Capacity of the Continuous-Space SSFM Model of Optical Fiber

The capacity of a discrete-time model of optical fiber described by the split-step Fourier method (SSFM) as a function of the signal-to-noise ratio SNR and the number of segments in distance
$K$
is considered. It is shown that if
$K\geq \text {SNR} ^{2/3}$
and
$\text {SNR} \rightarrow \infty $
, the capacity of the resulting continuous-space lossless model is lower bounded by
$\frac {1}{2}\log _{2}(1+ \text {SNR}) - \frac {1}{2}+ o(1)$
, where
$o(1)$
tends to zero with SNR. As
$K \rightarrow \infty $
, the inter-symbol interference (ISI) averages out to zero due to the law of large numbers and the SSFM model tends to a diagonal phase noise model. It follows that, in contrast to the discrete-space model where there is only one signal degree-of-freedom (DoF) at high powers, the number of DoFs in the continuous-space model is at least half of the input dimension
$n$
. Intensity-modulation and direct detection achieves this rate. The pre-log in the lower bound when
$K= \sqrt [\delta]{ \text {SNR}}$
is generally characterized in terms of
$\delta $
. It is shown that if the nonlinearity parameter
$\gamma \rightarrow \infty $
, the capacity of the continuous-space model is
$\frac {1}{2}\log _{2}(1+ \text {SNR})+ o(1)$
. The SSFM model when the dispersion matrix does not depend on
$K$
is considered. It is shown that the capacity of this model when
$K= \sqrt [\delta]{ \text {SNR}}$
,
$\delta >3$
, and
$\text {SNR} \rightarrow \infty $
is
$\frac {1}{2n}\log _{2}(1+ \text {SNR})+ O(1)$
. Thus, there is only one DoF in this model. Finally, it is found that the maximum achievable information rates (AIRs) of the SSFM model with back-propagation equalization obtained using numerical simulation follows a double-ascent curve. The AIR characteristically increases with SNR, reaching a peak at a certain optimal power, and then decreases as SNR is further increased. The peak is attributed to a balance between noise and stochastic ISI. However, if the power is further increased, the AIR will increase again, approaching the lower bound
$\frac {1}{2}\log (1+ \text {SNR})- \frac {1}{2} + o(1)$
. The second ascent is because the ISI averages out to zero with
$K \rightarrow \infty $
sufficiently fast.