你是无线通信方向的院士 我是大学教授，正在写一篇文章。下面是我们的一个推导的结果，现在我们需要一起对齐整理一下思路，现在的有点乱。我们想要的推导结果是\mathbf{Q}k^{\mathrm{sp}}和\mathbf{Q}k^{\mathrm{bulk}}的矩阵形式解析解，即最重要的结果是：
\begin{equation}\label{eq:Qsp_compact_engl}
\mathbf{Q}k^{\mathrm{sp}}
=\Big(\tfrac{1}{N^{3/2}}\mathbf{I}+\tfrac{\sqrt{N}}{2N}\mathbf{A}k\Big)\odot \mathbf{S}(\mathbf{s}),
\qquad \mathbf{s}=\mathbf{F},\tilde{\boldsymbol{\kappa}}.
\end{equation}
和
\begin{align}
\mathbf Q_k^{\mathrm{bulk}}
=\frac{1}{N^2}\mathbf D{-k}\mathbf K{\mathrm{bulk}}\mathbf D_k
=\frac{1}{N^2}(\mathbf d_k\mathbf d_k^H)\odot \mathbf K{\mathrm{bulk}}.
\end{align}
下面是我们的推导过程。需要注意的是，我们在推导中尽量少定义新变量，避免混乱。另外，下面的推到中没有加入
\begin{equation}
\mathbb{E}{|r_k|^2}\approx
\frac{1}{N^2}\sum{n,m}\tilde{g}{n,k}\tilde{g}{m,k},e^{j\frac{2\pi}{N}k(n-m)}
\sum_{p,q,p',q'}e^{j\tilde{\phi}{p,q,p',q'}}
e^{-j\frac{2\pi}{N}\left((n-1)(q-p)+(m-1)(p'-q')\right)}
\exp!\big[-\frac{\tilde{h}^2}{N^2} S(p,q,p',q')\big],
\label{eq:Ek_rk2_start}
\end{equation}
中的e^{j\tilde{\phi}{p,q,p',q'}}这一项。这一项是我们自己可设计的相位， $e^{j\tilde{\phi}_{p,q,p',q'}} \triangleq e^{j\phi_{p,q}}e^{-j\phi_{p',q'}}$，这一项最好在我们最终的解析解和其他部分独立出来，但是又不是仅由k点决定，而是由多个位置的点联合决定的。请用中文文本方式进行总结推导，用IEEE TSP的风格来写。下面是推导过程：
\subsection{Hyperplane Decomposition and the 1D Difference Weight $\kappa[d]$}
After removing the pairing set, the remaining non-pairing index set can be further decomposed into four structured hyperplanes and a residual bulk complement:
\begin{equation}
\text{(H1)}\ p'=q',\qquad
\text{(H2)}\ p=q,\qquad
\text{(H3)}\ q=q',\qquad
\text{(H4)}\ p=p',\qquad
\text{Bulk: the remaining index combinations}.
\end{equation}
On each hyperplane, the shift invariance of $\Psi(\cdot)$ reduces the four-argument quadratic form to a function of a single modular difference $d\in\mathbb{Z}_N$. To directly characterize this reduction, we introduce the one-dimensional weight:
\begin{equation}\label{eq:kappa_def_engl}
\kappa[d]\triangleq \exp!\left(-\alpha,\mathbf{b}^T\mathbf{K}_2(d,0,0,0)\mathbf{b}\right)
=\exp!\left(-2\alpha\big(N_d-\Psi(d)\big)\right),\qquad d\in\mathbb{Z}_N,
\end{equation}
which serves as a shorthand representation of the original kernel. Define $\tilde{\kappa}[0]=0$ and $\tilde{\kappa}[d]=\kappa[d]$ for $d=1,\ldots,N-1$, let $\tilde{\boldsymbol{\kappa}}=[\tilde{\kappa}[0],\ldots,\tilde{\kappa}[N-1]]^T$, and compute the single DFT vector:
\begin{equation}
\mathbf{s}\triangleq \mathbf{F},\tilde{\boldsymbol{\kappa}}.
\end{equation}
Based on the above notation, the total contribution of the hyperplanes can be expressed as a Hermitian quadratic form:
\begin{equation}\label{eq:Sk_sp_quadratic_engl}
\mathcal{S}k^{\mathrm{sp}}(\mathbf{g})=\mathbf{g}^H\mathbf{Q}k^{\mathrm{sp}}\mathbf{g},
\qquad
\mathbf{Q}k^{\mathrm{sp}}=\mathbf{Q}{\mathrm{diag}}+\mathbf{Q}{k,ut}+\mathbf{Q}{k,v0}.
\end{equation}
Introduce the shared lifting'' matrix $\mathbf{S}(\mathbf{s})\triangleq \mathbf{1}\mathbf{s}^H+\mathbf{s}\mathbf{1}^T$ and $\mathbf{A}_k\triangleq \mathbf{f}_k\mathbf{f}_{-k}^H+\mathbf{f}_{-k}\mathbf{f}_k^H$. The diagonal layer induced by (H3) and (H4) then simplifies to: \begin{equation} \mathbf{Q}_{\mathrm{diag}} =\frac{2}{N^{3/2}}\operatorname{diag}(\operatorname{Re}\{\mathbf{s}\}) =\frac{1}{N^{3/2}}\big(\mathbf{I}\odot \mathbf{S}(\mathbf{s})\big), \end{equation} while the two (Hermitianized) low-rank layers induced by (H1) and (H2) combine into a compact Hadamard form: \begin{equation} \mathbf{Q}_{k,ut}+\mathbf{Q}_{k,v0} =\frac{\sqrt{N}}{2N}\big(\mathbf{A}_k\odot \mathbf{S}(\mathbf{s})\big). \end{equation} In summary, we obtain: \begin{equation}\label{eq:Qsp_compact_engl} \mathbf{Q}_k^{\mathrm{sp}} =\Big(\tfrac{1}{N^{3/2}}\mathbf{I}+\tfrac{\sqrt{N}}{2N}\mathbf{A}_k\Big)\odot \mathbf{S}(\mathbf{s}), \qquad \mathbf{s}=\mathbf{F}\,\tilde{\boldsymbol{\kappa}}. \end{equation} This result indicates that the structured baseline'' background is fully governed by the one-dimensional difference profile $\kappa[d]$ via a single DFT, and its matrix form exhibits DFT-friendliness (diagonal-plus-low-rank structure).

\subsection{Bulk Complement Analysis via Circulant Subspace Projection}
The remaining bulk complement excludes both perfect pairings and single-hyperplane structures, and primarily captures the residual high-dimensional coupling effects. Let $\boldsymbol{\gamma}=\mathbf{F}^H\mathbf{g}$ and define:
\begin{equation}
[\mathbf{H}k]{p,q}\triangleq \gamma_{\langle k+p-q\rangle}.
\end{equation}
Since $\mathbf{g}$ is a real-valued vector, it satisfies $\gamma_{\langle-\ell\rangle}=\gamma_{\langle\ell\rangle}^*$, such that the bulk contribution can be written as a Hermitian quadratic form:
\begin{equation}\label{eq:Sk_bulk_vec_engl}
\mathcal{S}_k^{(\mathrm{bulk})}(\mathbf{g})
=\frac{1}{N^2}\operatorname{vec}(\mathbf{H}k)^H\mathbf{M}{\mathrm{bulk}}\operatorname{vec}(\mathbf{H}_k).
\end{equation}
A key property is that $\mathbf{H}_k$ lies in the circulant subspace, so there exists a vector $\mathbf{c}_k$ such that $\mathbf{H}_k=\operatorname{circ}(\mathbf{c}_k)$. Using the circulant synthesis matrix $\mathbf{B}=[\operatorname{vec}(\mathbf{J}_0),\ldots,\operatorname{vec}(\mathbf{J}_{N-1})]$, we have $\operatorname{vec}(\mathbf{H}_k)=\mathbf{B}\mathbf{c}_k$, which projects the $N^2\times N^2$-dimensional kernel matrix to an $N\times N$-dimensional matrix:
\begin{equation}
\mathcal{S}k^{(\mathrm{bulk})}(\mathbf{g})
=\frac{1}{N^2}\mathbf{c}k^H\mathbf{R}{\mathrm{bulk}}\mathbf{c}k,
\qquad
\mathbf{R}{\mathrm{bulk}}\triangleq \mathbf{B}^H\mathbf{M}{\mathrm{bulk}}\mathbf{B}\in\mathbb{C}^{N\times N}.
\end{equation}
Returning to the $\mathbf{g}$-domain and utilizing the modulation identity (which induces a Hadamard form in $k$), the bulk term can be expressed as:
\begin{equation}\label{eq:Sk_bulk_hadamard_engl}
\mathcal{S}_k^{(\mathrm{bulk})}(\mathbf{g})=\mathbf{g}^H\mathbf{Q}_k^{\mathrm{bulk}}\mathbf{g},
\qquad
\mathbf{Q}k^{\mathrm{bulk}}
=\frac{1}{N^2}(\mathbf{d}k\mathbf{d}k^H)\odot \mathbf{K}{\mathrm{bulk}},
\qquad
\mathbf{K}{\mathrm{bulk}}=\mathbf{F}\mathbf{R}{\mathrm{bulk}}\mathbf{F}^H,
\end{equation}
where $\mathbf{d}_k=[e^{j\omega k(n-1)}]_{n=1}^N$.

\subsection{Unified Matrix Form and Physical Interpretation}
By integrating the coherent pairing layer, the structured hyperplane layer, and the bulk complement, we finally obtain the closed-form solution in a unified matrix form:
\begin{equation}\label{eq:final_matrix_form_engl}
\mathbb{E}{|r_k|^2}
\approx
\left(2-\frac{1}{N}\right)|\mathbf{f}_k^H\mathbf{g}|^2
+\mathbf{g}^H\mathbf{Q}_k^{\mathrm{sp}}\mathbf{g}
+\mathbf{g}^H\mathbf{Q}_k^{\mathrm{bulk}}\mathbf{g}

\mathbf{g}^H\mathbf{Q}_k\mathbf{g},
\end{equation}
where:
\begin{equation}
\mathbf{Q}_k\triangleq
\left(2-\frac{1}{N}\right)\mathbf{f}_k\mathbf{f}_k^H+\mathbf{Q}_k^{\mathrm{sp}}+\mathbf{Q}_k^{\mathrm{bulk}}.
\end{equation}
The first term in \eqref{eq:final_matrix_form_engl} is a purely filter-dependent coherent peak, determined solely by the DFT bin projection $|\mathbf{f}_k^H\mathbf{g}|^2$; the remaining terms constitute the background ``baseline''. Specifically, the structured background is controlled by the one-dimensional attenuation profile $\kappa[d]=\exp(-2\alpha(N_d-\Psi(d)))$ via $\mathbf{s}=\mathbf{F}\tilde{\boldsymbol{\kappa}}$, and its matrix form is DFT-friendly (diagonal-plus-low-rank structure); the bulk term characterizes the residual fine-grained modulation through the Hadamard factor $(\mathbf{d}_k\mathbf{d}_k^H)\odot \mathbf{K}_{\mathrm{bulk}}$.

We partition the non-pairing set into structured hyperplanes and a residual bulk:

On each hyperplane, $\mathbf b^T\mathbf K_2(p,q,p',q')\mathbf b$ reduces (by shift-invariance of $\Psi(\cdot)$) to a function of a single modular difference $d\in\mathbb{Z}N$. We encode this \emph{using $K_2$ itself} via the 1D weight
\begin{equation}\label{eq:kappa_def}
\kappa[d]\triangleq \exp!\left(-\alpha,\mathbf b^T\mathbf K_2(d,0,0,0)\mathbf b\right),\qquad d\in\mathbb{Z}N ,
\end{equation}
which is merely a shorthand (no new model) since it is generated by the original four-argument kernel.

\begin{equation}
\kappa[d]\triangleq \exp!\left(-\alpha,\mathbf b^T\mathbf K_2(d,0,0,0)\mathbf b\right)
=\exp!\left(-2\alpha\big(N_d-\Psi(d)\big)\right).
\end{equation}

% (Optional) elementwise definition

\emph{Low-rank layers:} the hyperplanes (H1) and (H2) lead to two (Hermitianized) rank-$\le2$ quadratic forms

% ===== Key identity: elementwise product of DFT columns =====
For normalized DFT columns, the elementwise products satisfy

Hence,

% ===== Compact forms for Q_{k,ut}^{(0)} and Q_{k,v0}^{(0)} without D_k =====
Using $\mathbf f_{-k}^*=\mathbf f_k$ and $\mathbf f_k^*=\mathbf f_{-k}$,

% ===== Final low-rank sum (no D_k) =====

\emph{Diagonal layer:} the hyperplanes (H3) and (H4) collapse (via the DFT shift/modulation property) to a diagonal matrix

% ===== Key idea: represent the diagonal operator by its diagonal vector =====
Since $\mathbf Q_{\mathrm{diag}}$ is diagonal, define its diagonal vector

% ===== Compute q_diag explicitly =====
For each $n=1,\ldots,N$,
\begin{align*}
(\mathbf q_{\mathrm{diag}})n
&=\frac{1}{N^2}\sum{d=1}^{N-1}\kappa[d]\Big(e^{-j\omega d(n-1)}+e^{+j\omega d(n-1)}\Big)\
&=\frac{2}{N^2}\sum_{d=1}^{N-1}\kappa[d]\cos!\big(\omega d(n-1)\big).
\end{align*}

% ===== Pack kappa into a length-N vector and express via DFT columns =====
Define the length-$N$ vector $\tilde{\boldsymbol\kappa}\in\mathbb R^N$ by

Then

and

Therefore,

and the diagonal operator can be applied without any $\mathrm{diag}(\cdot)$ as

Thus the structured part is
\begin{equation}\label{eq:Sk_sp_quadratic}
\boxed{
\mathcal S_k^{\mathrm{sp}}(\mathbf g)=\mathbf g^H\mathbf Q_k^{\mathrm{sp}}\mathbf g,\qquad
\mathbf Q_k^{\mathrm{sp}}=\mathbf Q_{\mathrm{diag}}+\mathbf Q_{k,ut}+\mathbf Q_{k,v0}.
}
\end{equation}

% ===== Normalized DFT =====

and let $\mathbf f_d$ be the $d$-th column of $\mathbf F$ (indices mod $N$).

% ===== kappa vector and one FFT =====

% ===== s-factor matrix shared by all terms =====

% ===== Q_diag without diag(·): diagonal picked by Hadamard with I =====

% ===== Low-rank sum expressed with the SAME S(s) =====

% Assume normalized DFT F and its columns f_k.
% Let z = F \tilde{kappa}. To make the Hadamard product well-defined, lift z to an NxN matrix:

Note that $\mathrm{diag}(\tilde{\mathbf S})=2\mathbb{R}e\{\mathbf z\}=2\mathbb{R}e\{\mathbf F\tilde{\boldsymbol\kappa}\}$.

Define

If $\mathbf Q_k^{\mathrm{sp}}$ is Hermitian PSD, then

Let $\boldsymbol\gamma\triangleq \mathbf F^H\mathbf g$ so that $\gamma_\ell=\mathbf f_\ell^H\mathbf g$.
Define the two $N\times N$ matrices
\begin{align}
[\mathbf H_k]{p,q} &\triangleq \gamma{\langle k+p-q\rangle}, \
[\mathbf G_k]{p',q'} &\triangleq \gamma{\langle -k+(q'-p')\rangle}.
\end{align}
Let the bulk kernel (including masking) be
\begin{align}
\mathbf M_{\mathrm{bulk}} \triangleq \operatorname{Herm}{\mathbf K\odot \mathbf P^{\mathrm{bulk}}},
\qquad
[\mathbf K]{(p,q),(p',q')} \triangleq \exp!\big(-\alpha S_2(p,q,p',q')\big),
\qquad \alpha\triangleq \frac{\tilde h^2}{N^2}.
\end{align}
The bulk contribution is (cf. the original vec--matrix--vec form)
\begin{align}
\mathcal S_k^{(\mathrm{bulk})}(\mathbf g)
=\frac{1}{N^2},\operatorname{vec}(\mathbf H_k)^T,
\mathbf M{\mathrm{bulk}},
\operatorname{vec}(\mathbf G_k).
\label{eq:Sk_bulk_vec_start}
\end{align}

\paragraph*{Step 1: Hermitian quadratic form.}
Assume $\mathbf g\in\mathbb R^N$, hence $\gamma_{\langle-\ell\rangle}=\gamma_{\langle\ell\rangle}^*$.
Then for all $(p',q')$,
\begin{align}
[\mathbf G_k]{p',q'}
=\gamma{\langle -k-(p'-q')\rangle}
=\gamma_{\langle k+(p'-q')\rangle}^*
=[\mathbf H_k]{p',q'}^,
\end{align}
i.e.,
\begin{align}
\boxed{\mathbf G_k=\mathbf H_k^,\quad \operatorname{vec}(\mathbf G_k)=\operatorname{vec}(\mathbf H_k)^*.}
\label{eq:Gk_equals_Hk_conj}
\end{align}
Substituting \eqref{eq:Gk_equals_Hk_conj} into \eqref{eq:Sk_bulk_vec_start} yields
\begin{align}
\boxed{
\mathcal S_k^{(\mathrm{bulk})}(\mathbf g)
=\frac{1}{N^2},\operatorname{vec}(\mathbf H_k)^H,
\mathbf M{\mathrm{bulk}},
\operatorname{vec}(\mathbf H_k).
}
\label{eq:Sk_bulk_quadratic_N2}
\end{align}

\paragraph*{Step 2: Circulant subspace parameterization.}
Define the circulant shift basis
\begin{align}
[\mathbf J_d]{p,q}\triangleq \mathbb 1{p-q\equiv d\ (\mathrm{mod}\ N)},
\qquad d\in\mathbb Z_N.
\end{align}
Any circulant matrix can be written as
\begin{align}
\operatorname{circ}(\mathbf c)=\sum{d=0}^{N-1} c[d]\mathbf J_d.
\end{align}
Let $c_k[d]\triangleq \gamma_{\langle k+d\rangle}$ and $\mathbf c_k=[c_k[0],\ldots,c_k[N-1]]^T$.
Then
\begin{align}
\boxed{\mathbf H_k=\operatorname{circ}(\mathbf c_k).}
\label{eq:Hk_circ_ck}
\end{align}
Define the circulant synthesis matrix
\begin{align}
\mathbf B\triangleq \big[\operatorname{vec}(\mathbf J_0),\ldots,\operatorname{vec}(\mathbf J_{N-1})\big]\in\mathbb R^{N^2\times N}.
\end{align}
Using \eqref{eq:Hk_circ_ck},
\begin{align}
\boxed{\operatorname{vec}(\mathbf H_k)=\mathbf B,\mathbf c_k.}
\label{eq:vecHk_Bck}
\end{align}

\paragraph*{Step 3: Projection of the $N^2\times N^2$ kernel.}
Substitute \eqref{eq:vecHk_Bck} into \eqref{eq:Sk_bulk_quadratic_N2}:
\begin{align}
\mathcal S_k^{(\mathrm{bulk})}(\mathbf g)
&=\frac{1}{N^2},\mathbf c_k^H\underbrace{\big(\mathbf B^H\mathbf M_{\mathrm{bulk}}\mathbf B\big)}{\triangleq,\mathbf R{\mathrm{bulk}}\in\mathbb C^{N\times N}}\mathbf c_k.
\end{align}
Hence,
\begin{align}
\boxed{
\mathcal S_k^{(\mathrm{bulk})}(\mathbf g)
=\frac{1}{N^2},\mathbf c_k^H \mathbf R_{\mathrm{bulk}},\mathbf c_k,
\qquad
\mathbf R_{\mathrm{bulk}}\triangleq \mathbf B^H\mathbf M_{\mathrm{bulk}}\mathbf B.
}
\label{eq:Sk_bulk_ck_Rbulk}
\end{align}

\subsection{Explicit Difference-Domain Formula for $\mathbf R_{\mathrm{bulk}}$}

Let $[\mathbf R_{\mathrm{bulk}}]_{d,d'}=\operatorname{vec}(\mathbf J_d)^H\mathbf M_{\mathrm{bulk}}\operatorname{vec}(\mathbf J_{d'})$.
Expanding the vec inner product gives
\begin{align}
[\mathbf R_{\mathrm{bulk}}]{d,d'}
&=\sum{p,q}\sum_{p',q'} \mathbb 1{p-q=d}\mathbb 1{p'-q'=d'},
[\mathbf M_{\mathrm{bulk}}]{(p,q),(p',q')}.
\label{eq:Rbulk_expand_pq}
\end{align}
Using the constraints $p=q+d$ and $p'=q'+d'$ (all indices modulo $N$), we can rewrite
\begin{align}
[\mathbf R{\mathrm{bulk}}]{d,d'}
=\sum{q=0}^{N-1}\sum_{q'=0}^{N-1}
[\mathbf M_{\mathrm{bulk}}]_{(q+d,q),(q'+d',q')}.
\label{eq:Rbulk_qqprime}
\end{align}

\paragraph*{Shift-invariance reduction to a single modular difference $t=q-q'$.}
The term $S_2(p,q,p',q')$ and the bulk mask $\mathbf P^{\mathrm{bulk}}$ depend only on modular differences
among $(p,q,p',q')$, hence the summand in \eqref{eq:Rbulk_qqprime} depends on $(q,q')$ only through
$t\triangleq q-q'$.
For each fixed $t\in\mathbb Z_N$, there are exactly $N$ pairs $(q,q')$ such that $q-q'=t$.
Thus,
\begin{align}
\boxed{
[\mathbf R_{\mathrm{bulk}}]{d,d'}
= N\sum{t\in\mathbb Z_N}
\iota_{\mathrm{bulk}}(d,d',t),
\exp!\Big(-\alpha,S_2(t+d,t,d',0)\Big),
}
\label{eq:Rbulk_ddprime_tsum}
\end{align}
where $\iota_{\mathrm{bulk}}(d,d',t)\in\{0,1\}$ is the indicator that removes pairings and single-pair hyperplanes.

\paragraph*{Bulk indicator in $(d,d',t)$.}
With the parametrization
\begin{align}
(p,q,p',q')=(t+d,\ t,\ d',\ 0),
\end{align}
the excluded (structured) hyperplanes correspond to the following modular equalities:
\begin{align}
p=q \Leftrightarrow d=0,\qquad
p'=q' \Leftrightarrow d'=0,\qquad
q=q' \Leftrightarrow t=0,\qquad
p=q' \Leftrightarrow t=-d,\qquad
q=p' \Leftrightarrow t=d',\qquad
p=p' \Leftrightarrow t=d'-d.
\end{align}
Hence a convenient explicit choice is
\begin{align}
\boxed{
\iota_{\mathrm{bulk}}(d,d',t)
\triangleq
\mathbb 1{d\neq 0,\ d'\neq 0,\ t\notin{0,-d,d',d'-d}}.
}
\label{eq:iota_bulk}
\end{align}

\paragraph*{Closed form of $S_2(t+d,t,d',0)$.}
Recall the cosine kernel $\Psi(\Delta)\triangleq\sum_{r=1}^{N_d}\cos(\omega r\Delta)$ and the closed-form
expression for
\begin{align}
S_2(p,q,p',q')=(4N_d+2)\Big[
\Psi(p-q')+\Psi(q-p')
-\Psi(p-q)-\Psi(p-p')-\Psi(q'-q)-\Psi(q'-p')
\Big].
\label{eq:S2_general_Psi}
\end{align}
Substituting $(p,q,p',q')=(t+d,\ t,\ d',\ 0)$ into \eqref{eq:S2_general_Psi} yields
\begin{align}
\boxed{
S_2(t+d,t,d',0)
=(4N_d+2)\Big[
\Psi(t+d)+\Psi(t-d')-\Psi(d)-\Psi(d')-\Psi(t)-\Psi(t+d-d')
\Big].
}
\label{eq:S2_bulk_ddprime_t}
\end{align}
Combining \eqref{eq:Rbulk_ddprime_tsum}--\eqref{eq:S2_bulk_ddprime_t} gives a fully explicit
difference-domain construction of $\mathbf R_{\mathrm{bulk}}$.

\subsection{Back to $\mathbf g$: sp-like Hadamard-modulated quadratic / norm form}

Let $\boldsymbol\gamma=\mathbf F^H\mathbf g$ and define the circulant shift (permutation) matrix $\mathbf\Pi_k$
such that $\mathbf c_k=\mathbf\Pi_k\boldsymbol\gamma$ (i.e., $(\mathbf c_k)[d]=\gamma_{\langle k+d\rangle}$).
Then from \eqref{eq:Sk_bulk_ck_Rbulk},
\begin{align}
\mathcal S_k^{(\mathrm{bulk})}(\mathbf g)
&=\frac{1}{N^2},\boldsymbol\gamma^H\mathbf\Pi_k^H\mathbf R_{\mathrm{bulk}}\mathbf\Pi_k\boldsymbol\gamma
=\mathbf g^H\underbrace{\Big(\frac{1}{N^2}\mathbf F\mathbf\Pi_k^H\mathbf R_{\mathrm{bulk}}\mathbf\Pi_k\mathbf F^H\Big)}{\triangleq,\mathbf Q_k^{\mathrm{bulk}}}\mathbf g.
\end{align}
Therefore,
\begin{align}
\boxed{
\mathcal S_k^{(\mathrm{bulk})}(\mathbf g)=\mathbf g^H\mathbf Q_k^{\mathrm{bulk}}\mathbf g,
\qquad
\mathbf Q_k^{\mathrm{bulk}}\triangleq
\frac{1}{N^2}\mathbf F\mathbf\Pi_k^H\mathbf R{\mathrm{bulk}}\mathbf\Pi_k\mathbf F^H.
}
\label{eq:Sk_bulk_g_Qbulk}
\end{align}

\paragraph*{Hadamard modulation in $k$.}
Using $\mathbf\Pi_k=\mathbf F^H\mathbf D_k\mathbf F$ with
$\mathbf D_k\triangleq\operatorname{diag}(e^{-j\omega k(n-1)})_{n=1}^N$,
let $\mathbf K_{\mathrm{bulk}}\triangleq \mathbf F\mathbf R_{\mathrm{bulk}}\mathbf F^H$ and
$\mathbf d_k\triangleq[e^{j\omega k(n-1)}]_{n=1}^N$.
Then
\begin{align}
\mathbf Q_k^{\mathrm{bulk}}
=\frac{1}{N^2}\mathbf D_{-k}\mathbf K_{\mathrm{bulk}}\mathbf D_k
=\frac{1}{N^2}(\mathbf d_k\mathbf d_k^H)\odot \mathbf K_{\mathrm{bulk}}.
\end{align}
Hence,
\begin{align}
\boxed{
\mathcal S_k^{(\mathrm{bulk})}(\mathbf g)
=\mathbf g^H\Big(\frac{1}{N^2}(\mathbf d_k\mathbf d_k^H)\odot \mathbf K_{\mathrm{bulk}}\Big)\mathbf g,
\qquad
\mathbf K_{\mathrm{bulk}}=\mathbf F\mathbf R_{\mathrm{bulk}}\mathbf F^H,
}
\label{eq:Sk_bulk_hadamard}
\end{align}
where $\mathbf R_{\mathrm{bulk}}$ is given explicitly by \eqref{eq:Rbulk_ddprime_tsum}--\eqref{eq:S2_bulk_ddprime_t}.

\paragraph*{Norm form (PSD case).}
If $\mathbf Q_k^{\mathrm{bulk}}\succeq \mathbf 0$,
\begin{align}
\boxed{
\mathcal S_k^{(\mathrm{bulk})}(\mathbf g)
=\big|(\mathbf Q_k^{\mathrm{bulk}})^{1/2}\mathbf g\big|_2^2

\left|
\left(
\frac{1}{N^2}(\mathbf d_k\mathbf d_k^H)\odot \mathbf K_{\mathrm{bulk}}
\right)^{1/2}\mathbf g
\right|_2^2.
}
\label{eq:Sk_bulk_norm}
\end{align}
Otherwise one may use the standard stabilized surrogate
$\mathbf Q_{k,+}^{\mathrm{bulk}}\triangleq \Pi_{\succeq 0}(\operatorname{Herm}\{\mathbf Q_k^{\mathrm{bulk}}\})$,
yielding $\mathcal S_k^{(\mathrm{bulk})}(\mathbf g)\approx\|(\mathbf Q_{k,+}^{\mathrm{bulk}})^{1/2}\mathbf g\|_2^2$.

我们现在不用K_2的结构，用s\begin{align}
S(p,q,p',q')&\triangleq(4N_d+2)[\Psi(p-q')+\Psi(q-p')-\Psi(p-q)\notag\
&\quad-\Psi(p-p')-\Psi(q'-q)-\Psi(q'-p')].
\end{align}
然后我觉得推导过程太简略了，还是要用DFT矩阵的列这种给公式看起来高级且solid。重新更新一版。然后Phi从最开始的推导就扔进去，一起算