理论

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$\text{Lemma 1. ( S- Procedure[ 34] ) : Define the quadratic func- }$
tions w.r.t. $\mathbf{x}\in {\mathbb{C}}^M\times 1$ as
$$f_m\left(\mathbf{x}\right)={\mathbf{x}}^H{\mathbf{A}}_m\mathbf{x}+2Re\left\{{\mathbf{b}}_m^H\mathbf{x}\right\}+c_m,m=1,2,$$
where ${\mathbf{A}}_m\in {\mathbb{C}}^{M\times M},{\mathbf{b}}_m\in {\mathbb{C}}^{M\times 1},and$ $c_m\in \mathbb{R}.\text{The}$

$condition{f}_1\le 0\Rightarrow f_2\le 0holdsifandonlyifthereexists$

$a\text{ variable }\omega \ge 0\text{ such that}$

(19)
$$\omega \left[\begin{array}{cc}{\mathbf{A}}_1& {\mathbf{b}}_1\\ {\mathbf{b}}_1^H& c_1\end{array}\right]-\left[\begin{array}{cc}{\mathbf{A}}_2& {\mathbf{b}}_2\\ {\mathbf{b}}_2^H& c_2\end{array}\right]⪰0_{M+1}.$$

理论重述

Let $f(x)$ and $g(x)$ be two quadratic forms defined as:
$$f(x)=x^{H}Ax+2\Re (b^{H}x)+c$$
and
$$g(x)=x^{H}Dx+2\Re (e^{H}x)+f$$
where $A,D\in {\mathbb{C}}^{n\times n}$ are Hermitian matrices, $b,e\in {\mathbb{C}}^n$ are complex vectors, and $c,f\in \mathbb{R}$ are real constants. The superscript $H$ denotes the Hermitian (conjugate transpose) of the matrix or vector.

The implication
$$f(x)\le 0⟹g(x)\le 0$$
holds if and only if there exists a scalar $\lambda \ge 0$ such that:
$$f(x)+\lambda g(x)\le 0$$
or equivalently:
$$x^H(A+\lambda D)x+2\Re \left((b+\lambda e{)}^{H}x\right)+(c+\lambda f)\le 0\text{for all }x.$$

This condition can be rewritten as the following matrix inequality:
$$\left(\begin{array}{cc}A+\lambda D& b+\lambda e\\ (b+\lambda e{)}^H& c+\lambda f\end{array}\right)⪰0$$
where $⪰0$ denotes that the matrix is positive semidefinite (PSD).

Thus, the S-Procedure states that if such a non-negative $\lambda$ exists, then the implication $f(x)\le 0⟹g(x)\le 0$ holds.

实际案例

已知
$$\Delta {\mathbf{h}}^H\Delta \mathbf{h}\le a$$

如何根据S-Procedure 理论把下列形式转化成线性矩阵不等式呢
$$g(\Delta \mathbf{h})=\Delta {\mathbf{h}}^H\mathbf{D}\Delta \mathbf{h}+2\Re ({\mathbf{h}}^H\mathbf{D}\Delta \mathbf{h})+{\mathbf{h}}^H\mathbf{Dh}-z\ge 0$$

实际案例详细说明

\section*{S-Procedure 推导}

\textbf{已知条件}

1. 不等式 $f_1(\Delta \mathbf{h})$：
   $$f_1(\Delta \mathbf{h})=\Delta {\mathbf{h}}^H\Delta \mathbf{h}-a\le 0$$
   这表示：
   $$\Delta {\mathbf{h}}^H\Delta \mathbf{h}\le a$$

2. 需要证明的不等式 $f_2(\Delta \mathbf{h})$：
   $$f_2(\Delta \mathbf{h})=-\left(\Delta {\mathbf{h}}^H\mathbf{D}\Delta \mathbf{h}+2\Re ({\mathbf{h}}^H\mathbf{D}\Delta \mathbf{h})+{\mathbf{h}}^H\mathbf{Dh}-z\right)\le 0$$
   等价于：
   $$\Delta {\mathbf{h}}^H\mathbf{D}\Delta \mathbf{h}+2\Re ({\mathbf{h}}^H\mathbf{D}\Delta \mathbf{h})+{\mathbf{h}}^H\mathbf{Dh}-z\ge 0$$

\textbf{应用 S-Procedure}

为了应用 S-Procedure，我们需要构造两个二次型 $f_1$ 和 $f_2$ 的矩阵形式，并构造相应的线性矩阵不等式 (LMI)。

1. 构造 $f_1(\Delta \mathbf{h})$ 的矩阵形式：
   $$f_1(\Delta \mathbf{h})=\Delta {\mathbf{h}}^H\Delta \mathbf{h}-a$$
   其矩阵形式为：
   $$\left[\begin{array}{cc}\mathbf{I}& 0\\ 0& -a\end{array}\right]$$

2. 构造 $f_2(\Delta \mathbf{h})$ 的矩阵形式：
   $$f_2(\Delta \mathbf{h})=-\left(\Delta {\mathbf{h}}^H\mathbf{D}\Delta \mathbf{h}+2\Re ({\mathbf{h}}^H\mathbf{D}\Delta \mathbf{h})+{\mathbf{h}}^H\mathbf{Dh}-z\right)$$
   可以简化为：
   $$f_2(\Delta \mathbf{h})=-\Delta {\mathbf{h}}^H\mathbf{D}\Delta \mathbf{h}-2\Re ({\mathbf{h}}^H\mathbf{D}\Delta \mathbf{h})-({\mathbf{h}}^H\mathbf{Dh}-z)$$
   其矩阵形式为：
   $$\left[\begin{array}{cc}-\mathbf{D}& -\mathbf{Dh}\\ -{\mathbf{h}}^H\mathbf{D}& -({\mathbf{h}}^H\mathbf{Dh}-z)\end{array}\right]$$

3. 构造 S-Procedure 矩阵：

根据 S-Procedure，存在 $\mu \ge 0$ 使得：
$$\mu \left[\begin{array}{cc}\mathbf{I}& 0\\ 0& -a\end{array}\right]-\left[\begin{array}{cc}-\mathbf{D}& -\mathbf{Dh}\\ -{\mathbf{h}}^H\mathbf{D}& -({\mathbf{h}}^H\mathbf{Dh}-z)\end{array}\right]⪰0$$
进一步简化为：
$$\left[\begin{array}{cc}\mu \mathbf{I}+\mathbf{D}& \mathbf{Dh}\\ {\mathbf{h}}^H\mathbf{D}& -\mu a+({\mathbf{h}}^H\mathbf{Dh}-z)\end{array}\right]⪰0$$