你是无线通信方向的院士,正在写一篇IEEE顶级学术论文。下面想请你用IEEE TSP的风格润色一下下...
创建于:2025年12月23日
使用 GPT-5.2 Thinking 回答 Chat01
创建于:2025年12月23日
使用 GPT-5.2 Thinking 回答 Chat01
你是无线通信方向的院士,正在写一篇IEEE顶级学术论文。下面想请你用IEEE TSP的风格润色一下下面的表达,下面的内容写的不够深刻。我们想说的是a^Hv这部分是inid的分布,所以可以用lindberg CLT对期望进行展开到后面的解析解。先给我润色后的中文文本版本的:
We further simplify the inner product by exploiting the Hermitian symmetry of the random vector and the conjugate property of DFT columns ( for ), which implies . This symmetry reduces the inner product to a real-valued sum of the real parts of complex terms:
\begin{equation}
\mathbf a^{\mathsf H}\mathbf v=\sum_{r=1}^{N_d} (a[r]^* s_r + a[r] s_r^)=2\sum_{r=1}^{N_d}\mathbb{R}e{a[r]^ s_r},
\label{eq:inner_prod_sym}
\end{equation}
where denotes the -th component of and is the number of distinct frequency bins after symmetry reduction. Note that the random variables comprising are independent but not identically distributed (i.n.i.d.), which necessitates a Lyapunov-style convergence argument for the characteristic function approximation.
Under the Lindeberg condition for i.n.i.d. random variables,
\begin{equation}
\max_{r}\frac{|a[r]|^2}{\sum_{r=1}^{N_d}|a[r]|^2}\to 0, \qquad \mu_4<\infty,
\label{eq:lindeberg}
\end{equation}
where denotes the fourth-order moment (assumed finite), a cumulant expansion truncated to the fourth order yields the Gaussian approximation for the characteristic function:
\begin{equation}\label{eq:charfun_approx_new}
\mathbb{E}!\left[\exp!\big(j\mathbf a^{\mathsf H}\mathbf v\big)\right]
\approx \exp!\big(-\sum_{r=1}^{N_d}|a[r]|^2\big).
\end{equation}
下面给出一版更贴近 IEEE TSP 论文写法、逻辑更“深”、更强调 i.n.i.d. → Lindeberg CLT → 特征函数/累积量展开 → 解析近似 的中文润色稿(保留你原有符号与结构,但把“为什么能这么做/条件是什么/结论如何得到”写得更学术):
我们进一步简化内积 。利用随机向量 的厄米对称性以及 DFT 列向量的共轭对称性质(对 ,有 ,从而 ),可将原本复值的内积等价地规约为一组共轭成对项之和,最终写成仅由复随机项实部构成的实值和:
其中 表示 的第 个分量, 为对称规约后的独立频点数。由此可见, 本质上是由一组相互独立的加权随机变量的线性组合所构成;然而由于权系数 随 变化,各加和项呈现独立但非同分布(i.n.i.d.)结构。这一点排除了直接套用 i.i.d. 情形下的经典 CLT,而需要在 Lindeberg(或 Lyapunov)条件下建立其渐近高斯性,并据此对特征函数进行可控的近似。
具体而言,在 i.n.i.d. 情形下若满足 Lindeberg 条件
其中 为四阶矩(假设有限),则加权和中任一单项对总方差的贡献渐近可忽略,从而可由 Lindeberg–Feller 中心极限定理推出 的归一化和在分布意义下收敛到高斯变量。等价地,在特征函数层面,可对 进行以累积量(cumulant)为核心的展开,并在四阶截断后证明高阶项在上述条件下渐近消失,从而得到高斯型指数闭式近似:
该结果表明,在 Lindeberg 条件保证的“无主导项”加权机制下, 的特征函数可由其二阶统计量主导,进而为后续期望项的解析化推导提供了可操作的闭式近似基础。
如果你愿意,我也可以再给你一个更“硬核”的版本:把 写成 后,明确写出 ,并把二阶/四阶累积量项与 的阶数量级联系起来(例如 的收敛速度),这样会更像 TSP 的证明风格。
因为我们的调制是在QAM或者PSK下的,都是亚高斯星座,所以mu_4<2。把这个内容加入到前面的分析中,然后你建议的求和的期望那部分的解析解的推导也放进去
下面给出把“QAM/PSK 为亚高斯星座 有界(你给定 )”以及“把求和期望写成乘积并用 Lindeberg/累积量展开得到解析近似”的内容完整融入前述推导后的中文版本(偏 IEEE TSP 叙事与证明口吻,符号尽量沿用你原文)。
我们进一步简化内积 。利用随机向量 的厄米对称结构以及 DFT 列向量的共轭对称性质(,从而 ),可将复值内积规约为由共轭成对项组成的实值和:
其中 为 的第 个分量,,并定义
由于 相互独立,故 亦相互独立;但权系数 随 变化,使得 呈现**独立但非同分布(i.n.i.d.)**结构,因此需要在 Lindeberg(或 Lyapunov)框架下刻画其渐近高斯性并构造可控的特征函数近似。
在本文的调制假设下, 取自 PSK 或(能量归一化的)QAM 星座。此类有限字母表(bounded-alphabet)信号天然满足亚高斯性质,从而其高阶矩一致有界。特别地,设
并利用星座关于原点与 I/Q 对称性(PSK 显然成立;对称 QAM 亦满足 ),则四阶矩满足统一上界
其中不等式常数与你的星座归一化约定一致(这里直接采用你给定的 )。该界将用于控制特征函数展开中的四阶余项,保证“二阶项主导”的高斯近似在 i.n.i.d. 加权求和下仍然成立。
考虑加权和 ,在 i.n.i.d. 情形下我们采用 Lindeberg 条件(等价地也可用 Lyapunov 条件):
该条件刻画了“单个频点的能量权重不支配总体”的渐近机制,是保证中心极限定理适用、并使高阶累积量项整体可忽略的关键。
由独立性可将目标期望写成特征函数的乘积形式:
进一步取对数得到
下面分别给出二阶主项与四阶余项的可控界。
(i) 二阶主项(方差项):由于 为实值且零均值(由 得 ),并利用 与 ,有
因此 。
(ii) 四阶余项(由亚高斯四阶矩控制):对 在 处作泰勒/累积量展开并截断到四阶,可写成
其中余项 可由四阶矩界进行控制,典型地满足 (常数 与截断形式有关但与 无关)。又由
结合 得到一致的四阶控制,从而
在 Lindeberg “无主导项”条件 \eqref{eq:lindeberg_cn} 下, 相对 渐近可忽略(等价于 Lyapunov 条件的一种表达),因此总体余项 在渐近意义下收敛为 ,保证高斯主项的有效性。
将 \eqref{eq:cumulant_cn}–\eqref{eq:var_cn} 代回 \eqref{eq:logsum_cn},得到
指数化即得所需的闭式近似:
该推导清晰表明:在 PSK/QAM 亚高斯星座保证的四阶矩一致有界()以及 Lindeberg “无主导项”加权条件下, 的特征函数由二阶统计量主导,因而可用高斯型指数项给出可操作的解析近似,为后续期望项的闭式化计算奠定基础。
如果你接下来要把它写成“Lemma + Proof”的 TSP 形式,我也可以把上面 的关键一步用更标准的 Lyapunov 条件写成:
并直接由 推到成立(只需把 、 代入即可),这样整体更“证明化”。
把这部分的推导写成proposition+proof的形式。proof直接就写成appendix就好。然后proposition里面只放\begin{equation}\label{eq:charfun_approx_new}
\mathbb{E}!\left[\exp!\big(j\mathbf a^{\mathsf H}\mathbf v\big)\right]
\approx \exp!\big(-\sum_{r=1}^{N_d}|a[r]|^2\big).
\end{equation},剩下上面的推导都放到附录中。按照TSP的语言风格,分别给出中文文本版本和英文latex代码的版本。appendix中不要分section,直接在一个推导中来写就行
下面按 TSP 常用写法把该段落改成 Proposition + Proof(附录)。我分别给出:** 中文文本版本**(便于你直接替换正文/附录措辞)和 英文 LaTeX 代码版本(可直接粘到论文里编译)。
latex\begin{proposition}\label{prop:charfun_approx} \begin{equation}\label{eq:charfun_approx_new} \mathbb{E}\!\left[\exp\!\big(j\mathbf a^{\mathsf H}\mathbf v\big)\right] \approx \exp\!\big(-\sum_{r=1}^{N_d}|a[r]|^2\big). \end{equation} \end{proposition}
正文紧随其后可加一句(不放进 proposition 环境里):
latex命题~\ref{prop:charfun_approx} 的证明见附录。
latex\begin{proof}[命题~\ref{prop:charfun_approx} 的证明] 利用 $\mathbf v$ 的厄米对称结构以及 DFT 列向量的共轭性质 $\mathbf f_{N-r}=\mathbf f_r^*$(从而 $a[N-r]=a^*[r]$),可将内积规约为实值和 $$ \mathbf a^{\mathsf H}\mathbf v =\sum_{r=1}^{N_d}\big(a[r]^*s_r+a[r]s_r^*\big) =2\sum_{r=1}^{N_d}\Re\{a[r]^*s_r\} \triangleq \sum_{r=1}^{N_d}X_r, $$ 其中 $X_r\triangleq 2\Re\{a[r]^*s_r\}\in\mathbb R$,$s_r$ 为 $\mathbf v$ 的第 $r$ 个分量,$N_d=\lfloor N/2\rfloor$。由于 $\{s_r\}$ 相互独立,$\{X_r\}$ 亦相互独立,但一般为 i.n.i.d.。因此特征函数可因子分解为 $$ \mathbb E\!\left[e^{j\mathbf a^{\mathsf H}\mathbf v}\right] =\mathbb E\!\left[e^{j\sum_{r=1}^{N_d}X_r}\right] =\prod_{r=1}^{N_d}\mathbb E\!\left[e^{jX_r}\right], \qquad \log \mathbb E\!\left[e^{j\mathbf a^{\mathsf H}\mathbf v}\right] =\sum_{r=1}^{N_d}\log \mathbb E\!\left[e^{jX_r}\right]. $$ 在 PSK/QAM 调制下,$s_r$ 为亚高斯(有限字母表)星座点;在能量归一化且星座 I/Q 对称的常用设置下有 $\mathbb E\{s_r\}=0$、$\mathbb E\{|s_r|^2\}=1$、$\mathbb E\{s_r^2\}=0$,并且四阶矩一致有界 $\mu_4\triangleq \mathbb E\{|s_r|^4\}<2$。 由此可得 $$ \mathbb E\{X_r^2\} =\mathbb E\{a[r]^{*2}s_r^2+a[r]^2s_r^{*2}+2|a[r]|^2|s_r|^2\} =2|a[r]|^2. $$ 另一方面,对每一项 $\log\mathbb E[e^{jX_r}]$ 进行四阶截断的累积量(或等价的泰勒)展开,可写成 $$ \log \mathbb E[e^{jX_r}] =-\frac{\mathbb E\{X_r^2\}}{2}+R_r, $$ 其中余项满足 $|R_r|\le C\,\mathbb E\{|X_r|^4\}$(常数 $C$ 与截断形式有关但与 $r$ 无关)。注意到 $|X_r|=|a[r]^*s_r+a[r]s_r^*|\le 2|a[r]||s_r|$, 从而 $$ \mathbb E\{|X_r|^4\}\le 16|a[r]|^4\,\mu_4 \le 32|a[r]|^4, \qquad \sum_{r=1}^{N_d}|R_r| \le 32C\sum_{r=1}^{N_d}|a[r]|^4. $$ 在 i.n.i.d. 的 Lindeberg“无主导项”条件 $\max_r \frac{|a[r]|^2}{\sum_{k=1}^{N_d}|a[k]|^2}\to 0$ 下,有 $\sum_{r=1}^{N_d}|a[r]|^4=o\!\left(\left(\sum_{r=1}^{N_d}|a[r]|^2\right)^2\right)$, 从而 $\sum_{r=1}^{N_d}R_r=o(1)$(在相应的渐近尺度下)。代入并合并主项可得 $$ \log \mathbb E\!\left[e^{j\mathbf a^{\mathsf H}\mathbf v}\right] = -\frac{1}{2}\sum_{r=1}^{N_d}\mathbb E\{X_r^2\}+o(1) = -\sum_{r=1}^{N_d}|a[r]|^2+o(1), $$ 两边取指数即得到 $$ \mathbb{E}\!\left[\exp\!\big(j\mathbf a^{\mathsf H}\mathbf v\big)\right] \approx \exp\!\big(-\sum_{r=1}^{N_d}|a[r]|^2\big), $$ 从而命题得证。 \end{proof}
latex\begin{proposition}\label{prop:charfun_approx} \begin{equation}\label{eq:charfun_approx_new} \mathbb{E}\!\left[\exp\!\big(j\mathbf a^{\mathsf H}\mathbf v\big)\right] \approx \exp\!\big(-\sum_{r=1}^{N_d}|a[r]|^2\big). \end{equation} \end{proposition}
Right after it in the main text (outside the proposition):
latexThe proof of Proposition~\ref{prop:charfun_approx} is provided in the Appendix.
latex\begin{proof}[Proof of Proposition~\ref{prop:charfun_approx}] Exploiting the Hermitian symmetry of $\mathbf v$ and the conjugate pairing of DFT columns $\mathbf f_{N-r}=\mathbf f_r^*$ (hence $a[N-r]=a^*[r]$), the inner product can be reduced to a real-valued sum $$ \mathbf a^{\mathsf H}\mathbf v =\sum_{r=1}^{N_d}\big(a[r]^*s_r+a[r]s_r^*\big) =2\sum_{r=1}^{N_d}\Re\{a[r]^*s_r\} \triangleq \sum_{r=1}^{N_d}X_r, $$ where $X_r\triangleq 2\Re\{a[r]^*s_r\}\in\mathbb R$, $s_r$ is the $r$th entry of $\mathbf v$, and $N_d=\lfloor N/2\rfloor$. Since $\{s_r\}$ are independent, so are $\{X_r\}$, albeit generally i.n.i.d. Hence the characteristic function factorizes as $$ \mathbb E\!\left[e^{j\mathbf a^{\mathsf H}\mathbf v}\right] =\mathbb E\!\left[e^{j\sum_{r=1}^{N_d}X_r}\right] =\prod_{r=1}^{N_d}\mathbb E\!\left[e^{jX_r}\right], \qquad \log \mathbb E\!\left[e^{j\mathbf a^{\mathsf H}\mathbf v}\right] =\sum_{r=1}^{N_d}\log \mathbb E\!\left[e^{jX_r}\right]. $$ Under PSK/QAM signaling, $\{s_r\}$ are sub-Gaussian constellation symbols; with the standard energy normalization and I/Q symmetry one has $\mathbb E\{s_r\}=0$, $\mathbb E\{|s_r|^2\}=1$, $\mathbb E\{s_r^2\}=0$, and a uniform fourth-moment bound $\mu_4\triangleq \mathbb E\{|s_r|^4\}<2$. Consequently, $$ \mathbb E\{X_r^2\} =\mathbb E\{a[r]^{*2}s_r^2+a[r]^2s_r^{*2}+2|a[r]|^2|s_r|^2\} =2|a[r]|^2. $$ Moreover, a fourth-order cumulant (equivalently, Taylor) truncation yields $$ \log \mathbb E[e^{jX_r}] =-\frac{\mathbb E\{X_r^2\}}{2}+R_r, $$ where the remainder satisfies $|R_r|\le C\,\mathbb E\{|X_r|^4\}$ for some absolute constant $C$. Using $|X_r|=|a[r]^*s_r+a[r]s_r^*|\le 2|a[r]||s_r|$, we obtain $$ \mathbb E\{|X_r|^4\}\le 16|a[r]|^4\,\mu_4 \le 32|a[r]|^4, \qquad \sum_{r=1}^{N_d}|R_r| \le 32C\sum_{r=1}^{N_d}|a[r]|^4. $$ Under the i.n.i.d. Lindeberg ``no-dominant-term'' condition $\max_r \frac{|a[r]|^2}{\sum_{k=1}^{N_d}|a[k]|^2}\to 0$, it holds that $\sum_{r=1}^{N_d}|a[r]|^4=o\!\left(\left(\sum_{r=1}^{N_d}|a[r]|^2\right)^2\right)$, which implies $\sum_{r=1}^{N_d}R_r=o(1)$ at the relevant asymptotic scale. Therefore, $$ \log \mathbb E\!\left[e^{j\mathbf a^{\mathsf H}\mathbf v}\right] = -\frac{1}{2}\sum_{r=1}^{N_d}\mathbb E\{X_r^2\}+o(1) = -\sum_{r=1}^{N_d}|a[r]|^2+o(1). $$ Exponentiating both sides gives $$ \mathbb{E}\!\left[\exp\!\big(j\mathbf a^{\mathsf H}\mathbf v\big)\right] \approx \exp\!\big(-\sum_{r=1}^{N_d}|a[r]|^2\big), $$ which concludes the proof. \end{proof}
如果你希望把“”替换为更严格的 asymptotic equivalence(例如写成 或者在指数里写 ),我也可以按你全文的渐近参数(、SNR、过采样率等)把小 的量纲写得更精确、更像 TSP 的证明口径。
proposition的结构和风格按照下面这种:
\begin{proposition}\label{prop:Ik_closed_form}
Consider the transmission of random CE-OFDM symbols, the contribution to the squared ACF arising from the vanishing-kernel index configurations could be expressed as:
\begin{align}\label{eq:Ik_def_engll}
\mathcal{I}k(\mathbf{g})=(2N-1)\big|\mathbf{f}{k+1}^{H}\mathbf{\tilde{g}}_k\big|^2,
\end{align}
where .
\end{proposition}。然后只给出符合IEEE TSP语言风格的英文版本latex代码即可
latex\begin{proposition}\label{prop:charfun_approx_new} Consider the inner product $\mathbf a^{\mathsf H}\mathbf v$ formed by an i.n.i.d.\ sum after Hermitian-symmetry reduction, where $N_d=\lfloor N/2\rfloor$ denotes the number of distinct frequency bins and $\{a[r]\}_{r=1}^{N_d}$ are the associated (generally nonuniform) weighting coefficients. Assume $\{s_r\}_{r=1}^{N_d}$ are independent sub-Gaussian constellation symbols generated by PSK/QAM modulation with $\mathbb{E}\{|s_r|^4\}=\mu_4<2$, and the Lindeberg no-dominant-term condition holds, i.e., $\max_r \frac{|a[r]|^2}{\sum_{k=1}^{N_d}|a[k]|^2}\to 0$. Then the characteristic function of $\mathbf a^{\mathsf H}\mathbf v$ admits the Gaussian-form approximation \begin{equation}\label{eq:charfun_approx_new} \mathbb{E}\!\left[\exp\!\big(j\mathbf a^{\mathsf H}\mathbf v\big)\right] \approx \exp\!\big(-\sum_{r=1}^{N_d}|a[r]|^2\big). \end{equation} \end{proposition}
用高级、科学、符合IEEE顶级期刊的语言润色翻译下面的内容:对于具有i.n.i.d的,其中的随机发射符号满足假设1和2时,其特征函数可采用高斯近似:
对于呈 独立非同分布(i.n.i.d.) 结构的内积项 ,当其中的随机发射符号满足假设 1 与假设 2 时,其特征函数可在渐近意义下由二阶统计量主导,从而可用高斯型特征函数进行精确且可控的近似:
翻译成英文
For the i.n.i.d.-structured inner product , when the underlying random transmit symbols satisfy Assumptions 1 and 2, its characteristic function is asymptotically dominated by the second-order statistics and can therefore be accurately approximated by a Gaussian-form characteristic function:
给我latex代码的形式
latexFor the i.n.i.d.-structured inner product $\mathbf a^{\mathsf H}\mathbf v$, when the underlying random transmit symbols satisfy Assumptions~1 and~2, its characteristic function is asymptotically dominated by the second-order statistics and can therefore be accurately approximated by a Gaussian-form characteristic function:
用latex英文代码,精简,深化一下下面的推导过程:
Recall that the inner product is an independent but not identically distributed (i.n.i.d.) sum of random variables, i.e., , where , with denoting the -th component of . By the factorization property of characteristic functions for independent random variables, we have
\begin{equation}
\mathbb E[e^{j\mathbf a^{H}\mathbf v}] = \mathbb E[e^{j\sum_{r=1}^{N_d} z_r}] = \prod_{r=1}^{N_d}\mathbb E[e^{j z_r}],
\label{eq:charfun_factor}
\end{equation}
Taking the natural logarithm of both sides converts the product to a sum, simplifying subsequent analysis:
\begin{equation}
\log \mathbb E[e^{j\mathbf a^{\mathsf H}\mathbf v}] = \sum_{r=1}^{N_d}\log \mathbb E[e^{j z_r}.
\label{eq:log_charfun_sum}
\end{equation}
With the assumptions of unit power and rotational symmetry, the second moment of could be computed as:
\begin{align}
\mathbb E\left{z_r^2\right}
&=\mathbb E\left{\left(a[r]^{}s_r + a[r]s_r^\right)^2\right} \notag\
&=\mathbb E\left{a[r]^{*2}s_r^2 + a[r]^2s_r^{*2} + 2|a[r]|^2|s_r|^2\right} \notag\
&= 2|a[r]|^2,
\label{eq:zr_second_moment}
\end{align}
Next, we truncate the Taylor series expansion of at the second order (equivalent to a fourth-order cumulant truncation for sub-Gaussian random variables). This yields
\begin{equation}
\log \mathbb E\left[e^{j z_r}\right] = -\frac{1}{2}\mathbb E\left{z_r^2\right} + R_r,
\label{eq:taylor_truncation}
\end{equation}
where the remainder term satisfies the bound for some absolute constant (cf. \cite{vershynin2018high}). To estimate the fourth moment , we use the triangle inequality , leading to
\begin{align}
\mathbb E\left{|z_r|^4\right} &\le 16|a[r]|^4\mathbb E\left{|s_r|^4\right} = 16\mu_4|a[r]|^4,
\label{eq:zr_fourth_moment_bound}
\end{align}
where (a given constraint on the sub-Gaussian symbols). Summing the remainder bounds over to gives
\begin{equation}
\sum_{r=1}^{N_d}|R_r| \le 16C\mu_4\sum_{r=1}^{N_d}|a[r]|^4 \le 32C\sum_{r=1}^{N_d}|a[r]|^4,
\label{eq:remainder_sum_bound}
\end{equation}
with the second inequality following from .
We now invoke the Lindeberg no-dominant-term condition for i.n.i.d. sums:
\begin{equation}
\max_{r} \frac{|a[r]|^2}{\sum_{k=1}^{N_d}|a[k]|^2} \to 0 \quad (N_d \to \infty),
\label{eq:lindeberg_condition}
\end{equation}
which ensures no single term dominates the sum. A key consequence of this condition is the asymptotic relation
\begin{equation}
\sum_{r=1}^{N_d}|a[r]|^4 = o!\left(\left(\sum_{r=1}^{N_d}|a[r]|^2\right)^2\right),
\label{eq:asymptotic_4th_moment}
\end{equation}
implying the remainder sum satisfies (vanishes asymptotically as ).
Substituting \eqref{eq:zr_second_moment} and the vanishing remainder into \eqref{eq:log_charfun_sum}, we obtain
\begin{align}
\log \mathbb E!\left[e^{j\mathbf a^{\mathsf H}\mathbf v}\right]
&= -\frac{1}{2}\sum_{r=1}^{N_d}\mathbb E\left{z_r^2\right} + o(1) \notag\
&= -\sum_{r=1}^{N_d}|a[r]|^2 + o(1).
\label{eq:log_charfun_final}
\end{align}
Exponentiating both sides and neglecting the asymptotically negligible term yields the Gaussian-form approximation:
\begin{equation}
\mathbb{E}!\left[\exp!\big(j\mathbf a^{\mathsf H}\mathbf v\big)\right]
\approx \exp!\big(-\sum_{r=1}^{N_d}|a[r]|^2\big),
\label{eq:charfun_approx_final}
\end{equation}
which completes the proof.
latexRecall that $\mathbf a^{\mathsf H}\mathbf v$ can be written as an i.n.i.d.\ sum $$ \mathbf a^{\mathsf H}\mathbf v=\sum_{r=1}^{N_d} z_r,\qquad z_r \triangleq a^*[r]\,s_r+a[r]\,s_r^* \in \mathbb R, $$ where $\{s_r\}$ are independent. Hence, by independence, \begin{equation}\label{eq:charfun_factor} \mathbb E\!\left[e^{j\mathbf a^{\mathsf H}\mathbf v}\right] =\prod_{r=1}^{N_d}\mathbb E\!\left[e^{j z_r}\right], \qquad \log \mathbb E\!\left[e^{j\mathbf a^{\mathsf H}\mathbf v}\right] =\sum_{r=1}^{N_d}\log \mathbb E\!\left[e^{j z_r}\right]. \end{equation} Under the unit-power and rotational-symmetry assumptions (i.e., $\mathbb E\{|s_r|^2\}=1$ and $\mathbb E\{s_r^2\}=0$), we have \begin{equation}\label{eq:zr_second_moment} \mathbb E\{z_r^2\} =\mathbb E\!\left\{a[r]^{*2}s_r^2+a[r]^2 s_r^{*2}+2|a[r]|^2|s_r|^2\right\} =2|a[r]|^2 . \end{equation} Moreover, for sub-Gaussian PSK/QAM symbols with $\mu_4\triangleq \mathbb E\{|s_r|^4\}<2$, a second-order expansion of $\log \mathbb E[e^{j z_r}]$ yields \begin{equation}\label{eq:taylor_truncation_refined} \log \mathbb E\!\left[e^{j z_r}\right] =-\frac{1}{2}\mathbb E\{z_r^2\}+R_r, \qquad |R_r|\le C\,\mathbb E\{|z_r|^4\}, \end{equation} for some absolute constant $C>0$ (e.g., via a cumulant/Taylor remainder bound). Using $|z_r|=|a^*[r]s_r+a[r]s_r^*|\le 2|a[r]||s_r|$, we obtain \begin{equation}\label{eq:zr_fourth_moment_bound_refined} \mathbb E\{|z_r|^4\}\le 16|a[r]|^4 \mu_4 \le 32|a[r]|^4, \qquad \sum_{r=1}^{N_d}|R_r| \le 32C\sum_{r=1}^{N_d}|a[r]|^4. \end{equation} Invoking the i.n.i.d.\ Lindeberg no-dominant-term condition \begin{equation}\label{eq:lindeberg_condition_refined} \max_{r}\frac{|a[r]|^2}{\sum_{k=1}^{N_d}|a[k]|^2}\to 0, \end{equation} implies $\sum_{r=1}^{N_d}|a[r]|^4=o\!\left(\left(\sum_{r=1}^{N_d}|a[r]|^2\right)^2\right)$, and thus $\sum_{r=1}^{N_d}R_r=o(1)$ at the corresponding asymptotic scale. Substituting \eqref{eq:zr_second_moment}--\eqref{eq:taylor_truncation_refined} into \eqref{eq:charfun_factor} gives \begin{equation}\label{eq:log_charfun_final_refined} \log \mathbb E\!\left[e^{j\mathbf a^{\mathsf H}\mathbf v}\right] = -\frac{1}{2}\sum_{r=1}^{N_d}\mathbb E\{z_r^2\}+o(1) = -\sum_{r=1}^{N_d}|a[r]|^2+o(1). \end{equation} Exponentiating \eqref{eq:log_charfun_final_refined} yields \begin{equation}\label{eq:charfun_approx_final} \mathbb{E}\!\left[\exp\!\big(j\mathbf a^{\mathsf H}\mathbf v\big)\right] = \exp\!\Big(-\sum_{r=1}^{N_d}|a[r]|^2+o(1)\Big) \approx \exp\!\Big(-\sum_{r=1}^{N_d}|a[r]|^2\Big), \end{equation} which is the desired Gaussian-form approximation.