Capacity of the AWGN channel in bits per channel use (bpcu) is $$\mathsf{C}(\text{SNR})=\frac{1}{2}\log_2(1+\text{SNR})$$ where $\text{SNR}$ is the Signal-to-Noise-Ratio.

We consider equidistant bi-polar amplitude shift keying (ASK) constellations $\mathcal{X}$ with $2^m$ signal points:$$\mathcal{X}= \{\pm 1,\pm 3,\dotsc,\pm(2^m-1)\}.$$

The input-output-relation is:

$$Y=\Delta\cdot X+Z.$$
• Input $X$ with distribution $P_X$ on $\mathcal{X}$.
• $\Delta$ scales the constellation.
• Average signal power $P=\text{E}[(\Delta X)^2]$.
• Noise term $Z$ is zero mean Gaussian with power one: $Z \sim \mathcal{N}(0,1)$.
• Signal-to-Noise-Ratio is $\text{SNR}=P/1=\text{E}[(\Delta X)^2]$.

A good choice for $P_X$ is the Maxwell-Boltzmann (MB) distribution with parameter $\nu$, which resembles a sampled Gaussian distribution: $$P_{X_\nu}(x) = \frac{e^{-\nu x^2}}{\sum_{\tilde{x}\in\mathcal{X}}e^{-\nu \tilde{x}^2}}$$.

In order to find the optimal $\nu$ and $\Delta$, we solve the optimization problem:

$$\mathsf{R}_\text{MB}(\text{SNR})=\max_{\nu,\Delta\colon\text{E}[(\Delta X_\nu)^2]\leq P}I(X_\nu;\Delta X_\nu+Z),$$ where $I(X_\nu;\Delta X_\nu+Z)$ is the mutual information of input $X_\nu$ and output $Y=\Delta X_\nu+Z$.