pulse is that it is both a Nyquist and a root-Nyquist pulse.

In contrast to Nyquist pulses, from which nonnegative Nyquist pulses can be generated by squaring the original pulse (see Sec. \[sec:Pulse-shaping\]), the square of a root-Nyquist pulse is not root-Nyquist anymore. Moreover, [@4132995] has proven that there is no nonnegative root-Nyquist pulse with strictly limited bandwidth.

Required DC Bias \[sec:Required-DC-bias\]
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Our goal is to find the lowest $\mu$ that guarantees the nonnegativity of $x(t)$. From (\[eq:TxSignal\]) and $x(t)\geq0$, the smallest required DC bias is $$\begin{aligned}
\mu & =-\min_{\forall a,-\infty< t < \infty}\sum_{k=-\infty}^{\infty}a_{k}q(t-k{T_\mathrm{s}}) \label{eq:DC1}\\
 &=-\min_{\forall a,-\infty< t < \infty}\sum_{k=-\infty}^{\infty}\left[\left(a_{k}-L\right)q(t-k{T_\mathrm{s}})+Lq(t-k{T_\mathrm{s}})\right] \label{eq:DC2}\end{aligned}$$ where $L=(\hat{a}+\check{a})/2$, $\hat{a}=\max_{a\in \mathcal{C}}a$, and $\check{a}=\min_{a\in \mathcal{C}}a$. The notation $\forall a$ in and means that the minimization should be over all $a_k \in \mathcal{C}$ where $k = \ldots,-1,0,1,2,\ldots$ Going from (\[eq:DC1\]) to (\[eq:DC2\]), we created a factor ($a_{k}-L$) which is a function of $a_{k}$ and symmetric with respect to zero. As a result, the minimum of the first term in (\[eq:DC2\]) occurs if, for all $k$, either $a_{k}=\hat{a}$ and $q(t-k{T_\mathrm{s}})<0$ or $a_{k}=\check{a}$ and $q(t-k{T_\mathrm{s}})>0$. In both cases, due to the fact that the factor $\hat{a}-L=-(\check{a}-L)$, $$\begin{aligned}
 \mu &=\max_{0\leq t< {T_\mathrm{s}}}\left[\left(\hat{a}-L\right)\sum_{k=-\infty}^{\infty}\left|q(t-k{T_\mathrm{s}})\right|-L\sum_{k=-\infty}^{\infty}q(t-k{T_\mathrm{s}})\right].\label{eq:DC}\end{aligned}$$ The reason why (\[eq:DC\]) is minimized over $0\leq 