low-${{\text{\footnotesize{SNR}}}}$ regime. We first introduce the notation $\zeta=\frac{1}{B}$. Note that as $B \to \infty$, we have $\zeta \to 0$. Moreover, with this notation, the normalized effective capacity can be expressed as[^6] $$\label{eq:spececb}
\hspace{-0.03cm}{{\sf{R}}}_E({{\text{\footnotesize{SNR}}}})=-\frac{\zeta}{\theta
T}\log_e\Big(1-P\{z>{\alpha_{\text{opt}}}\}\big(1-e^{-\theta T{r_{\text{opt}}}}\big)\Big).$$ Note that ${\alpha_{\text{opt}}}$ and ${r_{\text{opt}}}$ are also in general dependent on $B$ and hence $\zeta$. The following result provides the expressions for the bit energy at zero spectral efficiency (i.e., as $B \to \infty$) and the wideband slope, and characterize the spectral efficiency-bit energy tradeoff in the wideband regime.

\[theo:wideband\] In the wideband regime, the bit energy at zero spectral efficiency, and wideband slope are given by $$\begin{gathered}
\frac{E_b}{N_0}\bigg|_{{{\sf{R}}}= 0}=\frac{-\delta\log_e2}{\log_e\xi} \quad \text{and} \label{eq:ebminwb}\\
\mathcal{S}_0=\frac{2\xi\log_e^2\xi}{(\delta{{\alpha^{\ast}_{\text{opt}}}})^2
P\{z>{{\alpha^{\ast}_{\text{opt}}}}\}e^{-\delta{{\alpha^{\ast}_{\text{opt}}}}}},\label{eq:s0wb}\end{gathered}$$ respectively, where $\delta=\frac{\theta T{\bar{P}}}{N_0\log_e2}$ and $\xi=1-P\{z>{{\alpha^{\ast}_{\text{opt}}}}\}(1-e^{-\delta{{\alpha^{\ast}_{\text{opt}}}}})$. ${{\alpha^{\ast}_{\text{opt}}}}$ is defined as ${{\alpha^{\ast}_{\text{opt}}}}=\lim_{\zeta\rightarrow0}{\alpha_{\text{opt}}}$ and ${{\alpha^{\ast}_{\text{opt}}}}$ satisfies $$\label{eq:stacondwideband}
\delta{{\alpha^{\ast}_{\text{opt}}}}=\log_e\left(1+\delta\frac{P\{z>{{\alpha^{\ast}_{\text{opt}}}}\}}{p_z({{\alpha^{\ast}_{\text{opt}}}})}\right).$$

*Proof:* Assume that the Taylor series expansion of ${r_{\text{opt}}}$ with respect to small $\zeta$ is $$\label{eq:substRb}
{r_{\text{opt}}}={{{r^{\ast}_{\text{opt}}}}}+\dot{r}_{\text{opt}}(0)\zeta+o(\zeta)$$ where ${{{r^{\ast}_{\text{opt}}}}}=\lim_{\zeta\rightarrow0}{r_{\text{opt}}}$ and $\dot{r}_{\text{opt}}(0)$ is the first derivative with respect to $\zeta$ of ${r_{\text{opt}}}$ evaluated at $\zeta=0$. From (\[eq:thresh\]), we can find that $$\begin{aligned}
{\alpha_{\text{opt}}}=\frac{2^{{r_{\text{opt}}}\zeta}-1}{\frac{{\bar{P}}\zeta}{N_0}}=\frac{{{{r^{\ast}_{\text{opt}}}}}\log_e2}{\frac{{\bar{P}}}{N_0}}+\frac{\dot{r}_{\text{opt}}(0)\log_e2+\frac{({{{r^{\ast}_{\text{opt}}}}}\log_e2)^2}{2}}{\frac{{\bar{P}}}{N_0}}\,\zeta+o(\zeta)\end{aligned}$$ from which we have as $\zeta\rightarrow0$ that $$\label{eq:stab}
{{\alpha^{\ast}_{\text{opt}}}}=\frac{{{{r^{\ast}_{\text{opt}}}}}\log_e2}{\frac{{\bar{P}}}{N_0}}$$ and that $$\label{eq:dotalpha}
{\dot{\alpha}_{\text{opt}}}(0)=\frac{{\dot{r}_{\text{opt}}}(0)\log_e2+\frac{({{{r^{\ast}_{\text{opt}}}}}\log_e2)^2}{2}}{\frac{{\bar{P}}}{N_0}}$$ where ${\dot{\alpha}_{\text{opt}}}(0)$ is the first derivative with respect to $\zeta$ of ${\alpha_{\text{opt}}}$ evaluated at $\zeta=0$. According to (\[eq:stab\]), ${{{r^{\ast}_{\text{opt}}}}}=\frac{{\bar{P}}{{\alpha^{\ast}_{\text{opt}}}}}{N_0\log_e2}$. 