CSR-DCF視頻目標跟蹤論文筆記（2）——關於濾波器Learning的推導（Augmented Lagrangian方法）

1. 論文基本信息

• 論文標題：Discriminative Correlation Filter with Channel and Spatial Reliability
• 作者：Alan Lukezic等
• 出處：CVPR，2017
• 文章鏈接：https://arxiv.org/abs/1611.08461
• 補充材料：https://www.semanticscholar.org/paper/Discriminative-Correlation-Filter-with-Channel-and-Luke%C5%BEi%C4%8D-Voj%C4%B1-%C5%99/7b485979c75b46d8c194868c0e70890f4a0f0ede
• 源碼鏈接：https://github.com/alanlukezic/csr-dcf

這篇筆記主要針對濾波器求解的推導過程進行分析（拉格朗日乘子法），主要參考內容是原文的補充材料，關於論文其他部分創新點及其整體思路會在後續文章中進行分析。（筆記1的鏈接：http://blog.csdn.net/discoverer100/article/details/78182306）

2. 濾波器求解目標函數的構建

在多通道情況下，目標函數爲

argminh∑d=1Nd(∥fd⊙hd−g∥2+λ∥hd∥2)=argminh∑d=1Nd(∥∥h^Hddiag(f^d)−g^d∥∥2+λ∥∥h^d∥∥2)(1)$$\begin{array}{}\text{(1)}& \begin{array}{c}\underset{\mathbf{h}}{\arg min}\sum _{d=1}^{N_d}\left({\Vert {\mathbf{f}}_d\odot {\mathbf{h}}_d-\mathbf{g}\Vert }^2+\lambda {\Vert {\mathbf{h}}_d\Vert }^2\right)\\ =\underset{\mathbf{h}}{\arg min}\sum _{d=1}^{N_d}\left({\Vert {\hat{\mathbf{h}}}_d^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left({\hat{\mathbf{f}}}_d\right)-{\hat{g}}_d\Vert }^2+\lambda {\Vert {\hat{\mathbf{h}}}_d\Vert }^2\right)\end{array}\end{array}$$

其中，h$\mathbf{h}$ 表示濾波器，d=1toNd$d=1to{N}_d$ 表示Nd$N_d$ 個通道，g$\mathbf{g}$ 表示期望的響應輸出，λ$\lambda$ 表示正則項用於防止過擬合（關於正則項爲什麼可以防止過擬合可以參考：http://www.cnblogs.com/alexanderkun/p/6922428.html）

根據上述(1)式，爲簡化推導過程，將多通道情況改爲單通道情況模式，則目標函數爲

argminh∥f⊙h−g∥2+λ∥h∥2=argminh∥∥h^Hdiag(f^)−g^∥∥2+λ∥∥h^∥∥2(2)$$\begin{array}{}\text{(2)}& \begin{array}{c}\underset{\mathbf{h}}{\arg min}{\Vert \mathbf{f}\odot \mathbf{h}-\mathbf{g}\Vert }^2+\lambda {\Vert \mathbf{h}\Vert }^2\\ =\underset{\mathbf{h}}{\arg min}{\Vert {\hat{\mathbf{h}}}^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2+\lambda {\Vert \hat{\mathbf{h}}\Vert }^2\end{array}\end{array}$$

引入變量hc${\mathbf{h}}_{\mathbf{c}}$ 並定義約束條件

hc−hm=0(3)$$\begin{array}{}\text{(3)}& {\mathbf{h}}_c-{\mathbf{h}}_m=0\end{array}$$

其中，hm≡m⊙h${\mathbf{h}}_m\equiv \mathbf{m}\odot \mathbf{h}$ ，而m$\mathbf{m}$ 表示論文中的空間置信圖（spatial reliability map），也可以理解爲一個mask，具體概念可以參考前面的一篇文章：http://blog.csdn.net/discoverer100/article/details/78182306，上述(3)式中引入的變量hc${\mathbf{h}}_{\mathbf{c}}$ 可以先不理會其物理意義，它的主要作用是讓算法能夠收斂（論文原文表述：prohibits a closed-form solution），個人猜測：這裏的下標命名爲c，可能就是取constrained的第一個字母。

對(2)式引入上述約束條件，並進一步調整，得到最終的目標函數

argminhc,hm∥∥h^Hcdiag(f^)−g^∥∥2+λ2∥∥h^m∥∥2s.t.hc−hm=0(4)$$\begin{array}{}\text{(4)}& \begin{array}{l}\underset{{\mathbf{h}}_c,{\mathbf{h}}_m}{\arg min}{\Vert {\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2+\frac{\lambda }{2}{\Vert {\hat{\mathbf{h}}}_m\Vert }^2\\ s.t.{\mathbf{h}}_c-{\mathbf{h}}_m=0\end{array}\end{array}$$

上述的正則項前面多出了一個係數1/2$1/2$ ，其主要意圖是求導數後係數可以變爲1$1$ ，便於公式書寫。

這樣，公式(4)就是我們推導的起始表達式。

3. 構建Lagrange表達式

根據上述目標函數，以及Augmented Lagrangian方法（參考Distributed optimization and statistical learning via the alternating direction method of multipliers），構建Lagrang表達式，如下

L(h^c,h,I^|m)=∥∥h^Hcdiag(f^)−g^∥∥2+λ2∥hm∥2+[I^H(h^c−h^m)+I^H(h^c−h^m)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]+μ∥∥h^c−h^m∥∥2(5)$$\begin{array}{}\text{(5)}& \mathcal{L}\left({\hat{\mathbf{h}}}_c,\mathbf{h},\hat{\mathbf{I}}|\mathbf{m}\right)={\Vert {\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2+\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2+\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-{\hat{\mathbf{h}}}_m\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-{\hat{\mathbf{h}}}_m\right)}\right]+\mu {\Vert {\hat{\mathbf{h}}}_c-{\hat{\mathbf{h}}}_m\Vert }^2\end{array}$$

其中，字母I$\mathbf{I}$ 表示Lagrange乘數，字母上面的橫槓表示共軛矩陣，字母右上方的H$H$ 表示共軛轉置矩陣，因此有規律：A¯T=AH${\overline{\mathbf{A}}}^T={\mathbf{A}}^H$ （後面的推導中可能同時存在兩種表示，需要留意）。將上述(5)式進行向量化表示，可得

L(h^c,h,I^|m)=∥∥h^Hcdiag(f^)−g^∥∥2+λ2∥hm∥2+[I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]+μ∥∥h^c−D−−√FMh∥∥2(6)$$\begin{array}{}\text{(6)}& \mathcal{L}\left({\hat{\mathbf{h}}}_c,\mathbf{h},\hat{\mathbf{I}}|\mathbf{m}\right)={\Vert {\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2+\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2+\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\right]+\mu {\Vert {\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\Vert }^2\end{array}$$

不難看出，上述(5)式到(6)式，主要變化就是將變量h^m${\hat{\mathbf{h}}}_m$ 的表達式替換爲D−−√FMh$\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}$ ，其中F$F$ 表示離散傅里葉變換矩陣，它相當於一個常量，D$D$ 是F$F$ 的大小（F$F$ 是一個D×D$D\times D$ 的方陣），M=diag(m)$\mathbf{M}\mathbf{=}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\mathbf{m}\right)$

將上述(6)式簡單表述爲四個項的和，爲

L(h^c,h,I^)=L1+L2+L3+L4(7)$$\begin{array}{}\text{(7)}& \mathcal{L}\left({\hat{\mathbf{h}}}_c,\mathbf{h},\hat{\mathbf{I}}\right)={\mathcal{L}}_1+{\mathcal{L}}_2+{\mathcal{L}}_3+{\mathcal{L}}_4\end{array}$$

其中，

⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪L1=∥∥h^Hcdiag(f^)−g^∥∥2=(h^Hcdiag(f^)−g^)(h^Hcdiag(f^)−g^)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯TL2=λ2∥hm∥2L3=I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯L4=μ∥∥h^c−D−−√FMh∥∥2=μ(h^c−D−−√FMh)(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T(8)$$\begin{array}{}\text{(8)}& \{\begin{array}{l}{\mathcal{L}}_1={\Vert {\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2=\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right){\overline{\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right)}}^T\\ {\mathcal{L}}_2=\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2\\ {\mathcal{L}}_3={\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\\ {\mathcal{L}}_4=\mu {\Vert {\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\Vert }^2=\mu \left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right){\overline{\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}}^T\end{array}\end{array}$$

4. 開始優化，首先對h_c求偏導數

對上述公式(4)的優化可以表述爲下面的迭代過程

h^optc=argminhcL(h^c,h,I^)hopt=argminhL(h^optc,h,I^)(9)$$\begin{array}{}\text{(9)}& \begin{array}{l}{\hat{\mathbf{h}}}_c^{\mathrm{o}\mathrm{p}\mathrm{t}}=\underset{{\mathbf{h}}_c}{\arg min}L\left({\hat{\mathbf{h}}}_c,\mathbf{h},\hat{\mathbf{I}}\right)\\ {\mathbf{h}}^{\mathrm{o}\mathrm{p}\mathrm{t}}=\underset{\mathbf{h}}{\arg min}L\left({\hat{\mathbf{h}}}_c^{\mathrm{o}\mathrm{p}\mathrm{t}},\mathbf{h},\hat{\mathbf{I}}\right)\end{array}\end{array}$$

現在看關於變量h^c${\hat{\mathbf{h}}}_c$ 的優化，需要令滿足∇h^c¯¯¯¯¯¯L≡0${\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}\mathcal{L}\equiv 0$ ，也就是

∇h^c¯¯¯¯¯¯L1+∇h^c¯¯¯¯¯¯L2+∇h^c¯¯¯¯¯¯L3+∇h^c¯¯¯¯¯¯L4≡0(10)$$\begin{array}{}\text{(10)}& {\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_1+{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_2+{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_3+{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_4\equiv 0\end{array}$$

對各個分量求偏導數，有

∇h^c¯¯¯¯¯¯L1=∂∂h^c¯¯¯¯¯[(h^Hcdiag(f^)−g^)(h^Hcdiag(f^)−g^)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T]=∂∂h^c¯¯¯¯¯[h^Hcdiag(f^)diag(f^)Hh^c−h^Hcdiag(f^)g^H−g^diag(f^)Hh^c+g^g^H]=diag(f^)diag(f^)Hh^c−diag(f^)g^H−0+0=diag(f^)diag(f^)Hh^c−diag(f^)g^H(11)$$\begin{array}{}\text{(11)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_1& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right){\overline{\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right)}}^T\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[{\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H-\hat{\mathbf{g}}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c+\hat{\mathbf{g}}{\hat{\mathbf{g}}}^H\right]\\ & =\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H-0+0\\ & =\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H\end{array}\end{array}$$

∇h^c¯¯¯¯¯¯L2=∇h^c¯¯¯¯¯¯[λ2∥hm∥2]=0(12)$$\begin{array}{}\text{(12)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_2& ={\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}\left[\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2\right]\\ & =0\end{array}\end{array}$$

∇h^c¯¯¯¯¯¯L3=∂∂h^c¯¯¯¯¯[I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h^c¯¯¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^TD−−√FMh¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=0−0+I^T−0=I^(13)$$\begin{array}{}\text{(13)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_3& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-{\hat{\mathbf{I}}}^T\overline{\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =0-0+{\hat{\mathbf{I}}}^T-0\\ & =\hat{\mathbf{I}}\end{array}\end{array}$$

∇h^c¯¯¯¯¯¯L4=∂∂h^c¯¯¯¯¯[μ(h^c−D−−√FMh)(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T]=∂∂h^c¯¯¯¯¯[μ(h^c⋅h^Hc−h^c⋅D−−√hHMFH−D−−√FMhh^Hc+D−−√FMh⋅D−−√hHMFH)]=∂∂h^c¯¯¯¯¯[μ(h^c⋅h^Hc−h^c⋅D−−√hHMFH−D−−√FMhh^Hc+DFMhhHMFH)]=μ(h^c−0−D−−√FMh+0)=μh^c−μD−−√FMh(14)$$\begin{array}{}\text{(14)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_4& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[\mu \left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right){\overline{\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}}^T\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[\mu \left({\hat{\mathbf{h}}}_c\cdot {\hat{\mathbf{h}}}_c^H-{\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H-\sqrt{D}{\mathbf{F}\mathbf{M}\mathbf{h}\hat{\mathbf{h}}}_c^H+\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H\right)\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[\mu \left({\hat{\mathbf{h}}}_c\cdot {\hat{\mathbf{h}}}_c^H-{\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H-\sqrt{D}{\mathbf{F}\mathbf{M}\mathbf{h}\hat{\mathbf{h}}}_c^H+D\mathbf{F}\mathbf{M}\mathbf{h}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H\right)\right]\\ & =\mu \left({\hat{\mathbf{h}}}_c-0-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+0\right)\\ & =\mu {\hat{\mathbf{h}}}_c-\mu \sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\end{array}\end{array}$$

於是，上述公式(10)可以寫成

diag(f^)diag(f^)Hh^c−diag(f^)g^H+0+I^+μ(h^c−0−D−−√FMh+0)≡0diag(f^)diag(f^)Hh^c−diag(f^)g^H+I^+μ(h^c−0−D−−√FMh+0)≡0(15)$$\begin{array}{}\text{(15)}& \begin{array}{c}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H+0+\hat{\mathbf{I}}+\mu \left({\hat{\mathbf{h}}}_c-0-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+0\right)\equiv 0\\ \mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H+\hat{\mathbf{I}}+\mu \left({\hat{\mathbf{h}}}_c-0-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+0\right)\equiv 0\end{array}\end{array}$$

回顧公式(6)，我們曾將變量h^m${\hat{\mathbf{h}}}_m$ 的表達式替換爲D−−√FMh$\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}$ ，現在我們將它替換回來，得

diag(f^)diag(f^)Hh^c−diag(f^)g^H+I^+μ(h^c−0−h^m+0)≡0(16)$$\begin{array}{}\text{(16)}& \mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H+\hat{\mathbf{I}}+\mu \left({\hat{\mathbf{h}}}_c-0-{\hat{\mathbf{h}}}_m+0\right)\equiv 0\end{array}$$

針對h^c${\hat{\mathbf{h}}}_c$ 合併同類項，得

h^c⋅[diag(f^)diag(f^)H+μ]=μh^m+diag(f^)g^H−I^(17)$$\begin{array}{}\text{(17)}& {\hat{\mathbf{h}}}_c\cdot \left[\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H+\mu \right]=\mu {\hat{\mathbf{h}}}_m+\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H-\hat{\mathbf{I}}\end{array}$$

於是，

h^c=diag(f^)g^H+μh^m−I^diag(f^)diag(f^)H+μ(18)$$\begin{array}{}\text{(18)}& {\hat{\mathbf{h}}}_c=\frac{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H+\mu {\hat{\mathbf{h}}}_m-\hat{\mathbf{I}}}{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H+\mu }\end{array}$$

根據對角矩陣的乘法性質以及相關濾波基礎概念，我們可以將上述(18)式的表達進行簡化，得

h^c=f^⊙g^∗+μh^m−I^f^⊙f^∗+μ(19)$$\begin{array}{}\text{(19)}& {\hat{\mathbf{h}}}_c=\frac{\hat{\mathbf{f}}\odot {\hat{\mathbf{g}}}^{\ast }+\mu {\hat{\mathbf{h}}}_m-\hat{\mathbf{I}}}{\hat{\mathbf{f}}\odot {\hat{\mathbf{f}}}^{\ast }+\mu }\end{array}$$

由於右上角得星號表示共軛矩陣，爲與作者原文表述一致，也可用頂部的橫槓表示，有

h^c=f^⊙g^¯¯¯+μh^m−I^f^⊙f^¯¯¯+μ(20)$$\begin{array}{}\text{(20)}& {\hat{\mathbf{h}}}_c=\frac{\hat{\mathbf{f}}\odot \overline{\hat{\mathbf{g}}}+\mu {\hat{\mathbf{h}}}_m-\hat{\mathbf{I}}}{\hat{\mathbf{f}}\odot \overline{\hat{\mathbf{f}}}+\mu }\end{array}$$

這樣就完成了對變量h^c${\hat{\mathbf{h}}}_c$ 的最優化求解，它對應論文官方源碼中的變量G（位於create_csr_filter.m中）

G = (Sxy + mu*H - L) ./ (Sxx + mu);

5. 對變量h求解偏導數

前面的公式(10)-(20)都是針對變量h^c${\hat{\mathbf{h}}}_c$ 求解偏導數，由於論文提出的Lagrange表達式中含有兩個變量，現在還需要針對變量h¯¯¯$\overline{\mathbf{h}}$ 求解偏導數，也就是

∇h¯¯¯¯L1+∇h¯¯¯¯L2+∇h¯¯¯¯L3+∇h¯¯¯¯L4≡0(21)$$\begin{array}{}\text{(21)}& {\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_1+{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_2+{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_3+{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_4\equiv 0\end{array}$$

首先回顧(8)式，也就是L1、L2、L3和L4這四個優化項

⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪L1=∥∥h^Hcdiag(f^)−g^∥∥2=(h^Hcdiag(f^)−g^)(h^Hcdiag(f^)−g^)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯TL2=λ2∥hm∥2L3=I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯L4=μ∥∥h^c−D−−√FMh∥∥2=μ(h^c−D−−√FMh)(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T(8)$$\begin{array}{}\text{(8)}& \{\begin{array}{l}{\mathcal{L}}_1={\Vert {\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2=\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right){\overline{\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right)}}^T\\ {\mathcal{L}}_2=\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2\\ {\mathcal{L}}_3={\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\\ {\mathcal{L}}_4=\mu {\Vert {\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\Vert }^2=\mu \left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right){\overline{\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}}^T\end{array}\end{array}$$

其關於變量h¯¯¯$\overline{\mathbf{h}}$ 的偏導數爲

∇h¯¯¯¯L1=∂∂h¯¯¯[(h^Hcdiag(f^)−g^)(h^Hcdiag(f^)−g^)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T]=0(22)$$\begin{array}{}\text{(22)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_1& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right){\overline{\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right)}}^T\right]\\ & =0\end{array}\end{array}$$

∇h¯¯¯¯L2=∂∂h¯¯¯[λ2∥hm∥2]=∂∂h¯¯¯[λ2∥m⊙h∥2]=∂∂h¯¯¯[λ2(Mh)¯¯¯¯¯¯¯¯¯¯¯¯T(Mh)]=∂∂h¯¯¯[λ2hHMHMh](23)$$\begin{array}{}\text{(23)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_2& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\Vert \mathbf{m}\odot \mathbf{h}\Vert }^2\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\overline{\left(\mathbf{M}\mathbf{h}\right)}}^T\left(\mathbf{M}\mathbf{h}\right)\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\mathbf{h}}^H{\mathbf{M}}^H\mathbf{M}\mathbf{h}\right]\end{array}\end{array}$$

觀察(23)式，由於其中的M$\mathbf{M}$ 表示論文定義的mask（也就是Spatial reliability map）的對角線元素，這個矩陣中的所有元素僅爲0或者1，它們都是實數，因此(23)式中的MHM${\mathbf{M}}^H\mathbf{M}$ 可以簡化表示爲M$\mathbf{M}$ ，於是，

∇h¯¯¯¯L2=∂∂h¯¯¯[λ2hHMh]=∂∂h¯¯¯[λ2((hHMh)T)T]=∂∂h¯¯¯[λ2(hTMTh¯¯¯)T]=(λ2(hTMT))T=λ2Mh(24)$$\begin{array}{}\text{(24)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_2& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\mathbf{h}}^H\mathbf{M}\mathbf{h}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\left({\left({\mathbf{h}}^H\mathbf{M}\mathbf{h}\right)}^T\right)}^T\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\left({\mathbf{h}}^T{\mathbf{M}}^T\overline{\mathbf{h}}\right)}^T\right]\\ & ={\left(\frac{\lambda }{2}\left({\mathbf{h}}^T{\mathbf{M}}^T\right)\right)}^T\\ & =\frac{\lambda }{2}\mathbf{M}\mathbf{h}\end{array}\end{array}$$

2018年8月更新：特別注意，上述公式(24)推導過程有誤，應改爲如下（此處特別感謝CSDN網友weixin_41660496）：

c∇h¯¯¯¯L2=∂∂h¯¯¯[λ2hHMh]=∂∂h¯¯¯[λ2(h¯¯¯)TMh]=λ2Mh(24)$$\begin{array}{}\text{(24)}& \begin{array}{rl}c{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{L}_2& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\mathbf{h}}^H\mathbf{M}\mathbf{h}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\left(\overline{\mathbf{h}}\right)}^T\mathbf{M}\mathbf{h}\right]\\ & =\frac{\lambda }{2}\mathbf{M}\mathbf{h}\end{array}\end{array}$$

∇h¯¯¯¯L3=∂∂h¯¯¯[I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^TD−−√FMh¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=0−0+0−I^TD−−√FM¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=−D−−√⋅I^T⋅F¯¯¯¯⋅M¯¯¯¯¯¯=−D−−√⋅(I^)T⋅(FH)T⋅(MH)T=−D−−√(MHFHI^)T=−D−−√(MFHI^)T(25)$$\begin{array}{}\text{(25)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_3& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-{\hat{\mathbf{I}}}^T\overline{\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =0-0+0-{\hat{\mathbf{I}}}^T\overline{\sqrt{D}\mathbf{F}\mathbf{M}}\\ & =-\sqrt{D}\cdot {\hat{\mathbf{I}}}^T\cdot \overline{\mathbf{F}}\cdot \overline{\mathbf{M}}\\ & =-\sqrt{D}\cdot {\left(\hat{\mathbf{I}}\right)}^T\cdot {\left({\mathbf{F}}^H\right)}^T\cdot {\left({\mathbf{M}}^H\right)}^T\\ & =-\sqrt{D}{\left({\mathbf{M}}^H{\mathbf{F}}^H\hat{\mathbf{I}}\right)}^T\\ & =-\sqrt{D}{\left(\mathbf{M}{\mathbf{F}}^H\hat{\mathbf{I}}\right)}^T\end{array}\end{array}$$

2018年8月更新：特別注意，上述公式(25)推導過程有誤，應改爲如下（此處特別感謝CSDN網友weixin_41660496）：

∇h¯¯¯¯L3=∂∂h¯¯¯[I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^H(h^c−D−−√FMh)+I^Hh^c¯¯¯¯¯¯¯¯¯¯¯−I^HD−−√FMh¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^HD−−√FMh¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^HD−−√FMh¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^H¯¯¯¯¯¯⋅D−−√¯¯¯¯¯¯¯¯¯⋅FMh¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^H¯¯¯¯¯¯⋅D−−√⋅FMh¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^H¯¯¯¯¯¯⋅D−−√⋅F¯¯¯¯⋅M¯¯¯¯¯¯⋅h¯¯¯]=0−0+0−(I^H¯¯¯¯¯¯⋅D−−√)⋅(F¯¯¯¯⋅M¯¯¯¯¯¯)T=0−0+0−(I^H¯¯¯¯¯¯⋅D−−√)⋅M¯¯¯¯¯¯T⋅F¯¯¯¯T=0−0+0−(I^⋅D−−√)⋅MT⋅FH=−D−−√MTFHI^=−D−−√MFHI^$$\begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{L}_3& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H}\cdot \overline{\sqrt{D}}\cdot \overline{\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H}\cdot \sqrt{D}\cdot \overline{\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H}\cdot \sqrt{D}\cdot \overline{\mathbf{F}}\cdot \overline{\mathbf{M}}\cdot \overline{\mathbf{h}}\right]\\ & =0-0+0-\left(\overline{{\hat{\mathbf{I}}}^H}\cdot \sqrt{D}\right)\cdot {\left(\overline{\mathbf{F}}\cdot \overline{\mathbf{M}}\right)}^T\\ & =0-0+0-\left(\overline{{\hat{\mathbf{I}}}^H}\cdot \sqrt{D}\right)\cdot {\overline{\mathbf{M}}}^T\cdot {\overline{\mathbf{F}}}^T\\ & =0-0+0-\left(\hat{\mathbf{I}}\cdot \sqrt{D}\right)\cdot {\mathbf{M}}^T\cdot {\mathbf{F}}^H\\ & =-\sqrt{D}{\mathbf{M}}^T{\mathbf{F}}^H\hat{\mathbf{I}}\\ & =-\sqrt{D}\mathbf{M}{\mathbf{F}}^H\hat{\mathbf{I}}\end{array}$$

上述(25)式中，利用了矩陣D−−√,M$\sqrt{D},\mathbf{M}$ 爲實數矩陣的性質，所以有M¯¯¯¯¯¯=M$\overline{\mathbf{M}}=\mathbf{M}$ , D¯¯¯¯¯=D$\overline{\mathbf{D}}=\mathbf{D}$ ，另外M$\mathbf{M}$ 也是一個對角矩陣，因此MT=M${\mathbf{M}}^T=\mathbf{M}$ 。

最後是L4項

∇h¯¯¯¯L4=∂∂h¯¯¯[μ(h^c−D−−√FMh)(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T]=∂∂h¯¯¯[μ(h^c⋅h^Hc−h^c⋅D−−√(F¯¯¯¯⋅M¯¯¯¯¯¯⋅h¯¯¯)T−D−−√FMhh^Hc+DFMh(F¯¯¯¯⋅M¯¯¯¯¯¯⋅h¯¯¯)T)]=∂∂h¯¯¯[μ(h^c⋅h^Hc−h^c⋅D−−√hHMFH−D−−√FMhh^Hc+DFMhhHMHFH)]=0−∂∂h¯¯¯[μ⋅h^c⋅D−−√hHMFH]−0+∂∂h¯¯¯[μ⋅DFMhhHMHFH]=−[∂∂hH[μ⋅h^c⋅D−−√hHMFH]]T+[∂∂hH[μ⋅DFMhhHMHFH]]T=−[∂∂hH[μ⋅D−−√⋅h^c⋅hHMFH]]T+[∂∂hH[μ⋅DFMh⋅hH⋅MHFH]]T=−μ⋅D−−√⋅[(h^c)T⋅(MFH)T]T+μ⋅D⋅[(FMh)T⋅(MHFH)T]T=−μ⋅D−−√⋅[(MFH)T]T⋅[(h^c)T]T+μ⋅D⋅[(MHFH)T]T⋅[(FMh)T]T=−μ⋅D−−√⋅MFH⋅h^c+μ⋅D⋅MHFH⋅FMh=−μ⋅D−−√⋅MFH⋅h^c+μ⋅D⋅MHIMh=−μ⋅D−−√⋅MFH⋅h^c+μ⋅D⋅(M⋅M)h=−μD−−√MFHh^c+μDMh(26)$$\begin{array}{}\text{(26)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_4& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\mu \left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right){\overline{\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}}^T\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\mu \left({\hat{\mathbf{h}}}_c\cdot {\hat{\mathbf{h}}}_c^H-{\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\left(\overline{\mathbf{F}}\cdot \overline{\mathbf{M}}\cdot \overline{\mathbf{h}}\right)}^T-\sqrt{D}{\mathbf{F}\mathbf{M}\mathbf{h}\hat{\mathbf{h}}}_c^H+D\mathbf{F}\mathbf{M}\mathbf{h}{\left(\overline{\mathbf{F}}\cdot \overline{\mathbf{M}}\cdot \overline{\mathbf{h}}\right)}^T\right)\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\mu \left({\hat{\mathbf{h}}}_c\cdot {\hat{\mathbf{h}}}_c^H-{\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H-\sqrt{D}{\mathbf{F}\mathbf{M}\mathbf{h}\hat{\mathbf{h}}}_c^H+D\mathbf{F}\mathbf{M}\mathbf{h}{\mathbf{h}}^H{\mathbf{M}}^H{\mathbf{F}}^H\right)\right]\\ & =0-\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\mu \cdot {\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H\right]-0+\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\mu \cdot D\mathbf{F}\mathbf{M}\mathbf{h}{\mathbf{h}}^H{\mathbf{M}}^H{\mathbf{F}}^H\right]\\ & =-{\left[\frac{\mathrm{\partial }}{\mathrm{\partial }{\mathbf{h}}^H}\left[\mu \cdot {\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H\right]\right]}^T+{\left[\frac{\mathrm{\partial }}{\mathrm{\partial }{\mathbf{h}}^H}\left[\mu \cdot D\mathbf{F}\mathbf{M}\mathbf{h}{\mathbf{h}}^H{\mathbf{M}}^H{\mathbf{F}}^H\right]\right]}^T\\ & =-{\left[\frac{\mathrm{\partial }}{\mathrm{\partial }{\mathbf{h}}^H}\left[\mu \cdot \sqrt{D}\cdot {\hat{\mathbf{h}}}_c\cdot {\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H\right]\right]}^T+{\left[\frac{\mathrm{\partial }}{\mathrm{\partial }{\mathbf{h}}^H}\left[\mu \cdot D\mathbf{F}\mathbf{M}\mathbf{h}\cdot {\mathbf{h}}^H\cdot {\mathbf{M}}^H{\mathbf{F}}^H\right]\right]}^T\\ & =-\mu \cdot \sqrt{D}\cdot {\left[{\left({\hat{\mathbf{h}}}_c\right)}^T\cdot {\left(\mathbf{M}{\mathbf{F}}^H\right)}^T\right]}^T+\mu \cdot D\cdot {\left[{\left(\mathbf{F}\mathbf{M}\mathbf{h}\right)}^T\cdot {\left({\mathbf{M}}^H{\mathbf{F}}^H\right)}^T\right]}^T\\ & =-\mu \cdot \sqrt{D}\cdot {\left[{\left(\mathbf{M}{\mathbf{F}}^H\right)}^T\right]}^T\cdot {\left[{\left({\hat{\mathbf{h}}}_c\right)}^T\right]}^T+\mu \cdot D\cdot {\left[{\left({\mathbf{M}}^H{\mathbf{F}}^H\right)}^T\right]}^T\cdot {\left[{\left(\mathbf{F}\mathbf{M}\mathbf{h}\right)}^T\right]}^T\\ & =-\mu \cdot \sqrt{D}\cdot \mathbf{M}{\mathbf{F}}^H\cdot {\hat{\mathbf{h}}}_c+\mu \cdot D\cdot {\mathbf{M}}^H{\mathbf{F}}^H\cdot \mathbf{F}\mathbf{M}\mathbf{h}\\ & =-\mu \cdot \sqrt{D}\cdot \mathbf{M}{\mathbf{F}}^H\cdot {\hat{\mathbf{h}}}_c+\mu \cdot D\cdot {\mathbf{M}}^H\mathbf{I}\mathbf{M}\mathbf{h}\\ & =-\mu \cdot \sqrt{D}\cdot \mathbf{M}{\mathbf{F}}^H\cdot {\hat{\mathbf{h}}}_c+\mu \cdot D\cdot \left(\mathbf{M}\cdot \mathbf{M}\right)\mathbf{h}\\ & =-\mu \sqrt{D}\mathbf{M}{\mathbf{F}}^H{\hat{\mathbf{h}}}_c+\mu D\mathbf{M}\mathbf{h}\end{array}\end{array}$$

上述(26)式末尾利用了MM=M$\mathbf{M}\mathbf{M}=\mathbf{M}$ ，這是因爲M$\mathbf{M}$ 表示空間置信圖（Spatial reliability map），根據論文中的定義，矩陣M$\mathbf{M}$ 中的元素要麼爲1，要麼爲0，因此MM=M$\mathbf{M}\mathbf{M}=\mathbf{M}$

將上述(22)-(26)式代入到(21)式中，有

λ2Mh−D−−√MFHI^−μD−−√MFHh^c+μDMh=0(27)$$\begin{array}{}\text{(27)}& \frac{\lambda }{2}\mathbf{M}\mathbf{h}-\sqrt{D}\mathbf{M}{\mathbf{F}}^H\hat{\mathbf{I}}-\mu \sqrt{D}\mathbf{M}{\mathbf{F}}^H{\hat{\mathbf{h}}}_c+\mu D\mathbf{M}\mathbf{h}=0\end{array}$$

對上式中的Mh$\mathbf{M}\mathbf{h}$ 進行合併同類項，有

(λ2+μD)MhMhMhMhMh=D−−√MFHI^+μD−−√MFHh^c=D−−√MFHI^+μD−−√MFHh^cλ2+μD=M⋅D−−√FH(I^+μh^c)λ2+μD=M⋅D√DFH(I^+μh^c)1D(λ2+μD)=M⋅1D√FH(I^+μh^c)λ2D+μ(28)$$\begin{array}{}\text{(28)}& \begin{array}{rl}\left(\frac{\lambda }{2}+\mu D\right)\mathbf{M}\mathbf{h}& =\sqrt{D}\mathbf{M}{\mathbf{F}}^H\hat{\mathbf{I}}+\mu \sqrt{D}\mathbf{M}{\mathbf{F}}^H{\hat{\mathbf{h}}}_c\\ \mathbf{M}\mathbf{h}& =\frac{\sqrt{D}\mathbf{M}{\mathbf{F}}^H\hat{\mathbf{I}}+\mu \sqrt{D}\mathbf{M}{\mathbf{F}}^H{\hat{\mathbf{h}}}_c}{\frac{\lambda }{2}+\mu D}\\ \mathbf{M}\mathbf{h}& =\mathbf{M}\cdot \frac{\sqrt{D}{\mathbf{F}}^H\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2}+\mu D}\\ \mathbf{M}\mathbf{h}& =\mathbf{M}\cdot \frac{\frac{\sqrt{D}}{D}{\mathbf{F}}^H\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{1}{D}\left(\frac{\lambda }{2}+\mu D\right)}\\ \mathbf{M}\mathbf{h}& =\mathbf{M}\cdot \frac{\frac{1}{\sqrt{D}}{\mathbf{F}}^H\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2D}+\mu }\end{array}\end{array}$$

現在簡單回顧一下傅里葉逆變換的定義

F−1(x^)=1D−−√FHx^(29)$$\begin{array}{}\text{(29)}& {\mathcal{F}}^{-1}\left(\hat{\mathbf{x}}\right)=\frac{1}{\sqrt{D}}{\mathbf{F}}^H\hat{\mathbf{x}}\end{array}$$

據此，可以將上述(28)式表示爲

Mh=M⋅F−1(I^+μh^c)λ2D+μdiag(m)h=diag(m)⋅F−1(I^+μh^c)λ2D+μm⊙h=m⊙F−1(I^+μh^c)λ2D+μ(30)$$\begin{array}{}\text{(30)}& \begin{array}{r}\mathbf{M}\mathbf{h}=\mathbf{M}\cdot \frac{{\mathcal{F}}^{-1}\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2D}+\mu }\\ \mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\mathbf{m}\right)\mathbf{h}=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\mathbf{m}\right)\cdot \frac{{\mathcal{F}}^{-1}\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2D}+\mu }\\ \mathbf{m}\odot \mathbf{h}=\mathbf{m}\odot \frac{{\mathcal{F}}^{-1}\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2D}+\mu }\end{array}\end{array}$$

由於論文中最優化表達式的約束爲m≡m⊙h$\mathbf{m}\equiv \mathbf{m}\odot \mathbf{h}$ （這裏左邊的h$\mathbf{h}$ 實際上也就是hm${\mathbf{h}}_m$ ），因此上述(30)式也可以表示爲

h=m⊙F−1(I^+μh^c)λ2D+μ(31)$$\begin{array}{}\text{(31)}& \mathbf{h}=\mathbf{m}\odot \frac{{\mathcal{F}}^{-1}\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2D}+\mu }\end{array}$$

上述(31)式就對應論文原文中的公式(10)，相應的代碼位於create_csr_filter.m中的變量H

H = fft2(real((1/(lambda + mu)) * bsxfun(@times, P, ifft2(mu*G + L))));

6. 小結

論文爲求解最優的濾波器值，在原始目標函數的基礎上，引入約束條件，利用Augmented Lagrangian方法構造最優化表達式，最後利用偏導數值爲0的情況進行求解，這種利用拉格朗日最優化方法進行目標跟蹤算法的建模思想值得我們學習，關於(25)式的轉置問題，也歡迎大家共同討論交流。

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