Time-varying Sea Surface Temperature Reconstruction Leveraging Low Rank and Joint Smoothness Constraints
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摘要: 海表面温度对于海洋动力过程及海气相互作用等具有重要意义,是海洋环境关键要素之一。浮标是海表面温度观测的常用手段,但由于浮标在空间的分布不规则,浮标采集的海表面温度数据也呈现非规则性。另外,浮标在实际工作中难免存在故障,致使采集的海表面温度数据存在缺失。因此对存在缺失的非规则海表面温度数据进行重构具有重要意义。该文通过将海表面温度数据建立为时变图信号,利用图信号处理方法解决海表面温度缺失数据重构问题。首先,利用数据的低秩性和时域-图域联合变差特性构建海表面温度重构模型;其次,基于交替方向乘子法框架提出一种求解该优化模型的基于低秩和联合平滑性(LRJS)的时变图信号重构方法,并分析该方法的计算复杂度和估计误差的理论极限;最后,采用南海和太平洋海域海表温度数据对方法的有效性进行了评估,结果表明,与现有缺失数据重构方法相比,该文所提LRJS方法有更高的重建精度。Abstract: Sea surface temperature is one of the key elements of the marine environment, which is of great significance to the marine dynamic process and air-sea interaction. Buoy is a commonly used method of sea surface temperature observation. However, due to the irregular distribution of buoys in space, the sea surface temperature data collected by buoys also show irregularity. In addition, it is inevitable that sometimes the buoy is out of order, so that the sea surface temperature data collected is incomplete. Therefore, it is of great significance to reconstruct the incomplete irregular sea surface temperature data. In this paper, the sea surface temperature data is established as a time-varying graph signal, and the graph signal processing method is used to solve the problem of missing data reconstruction of sea surface temperature. Firstly, the sea surface temperature reconstruction model is constructed by using the low rank data and the joint variation characteristics of time-domain and graph-domain. Secondly, a time-varying graph signal reconstruction method based on Low Rank and Joint Smoothness (LRJS) constraints is proposed to solve the optimization problem by using the framework of alternating direction multiplier method, and the computational complexity and the theoretical limit of the estimation error of the method are analyzed. Finally, the sea surface temperature data of the South China Sea and the Pacific Ocean are used to evaluate the effectiveness of the method. The results show that the LRJS method proposed in this paper can improve the reconstruction accuracy compared with the existing missing data reconstruction methods.
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1 共轭梯度法
输入:$ {\boldsymbol{Y}},{\boldsymbol{J}},{{\boldsymbol{L}}_{\rm{G}}},{{\boldsymbol{L}}_T},{{\boldsymbol{Z}}^k},{{\boldsymbol{P}}^k},{\boldsymbol{D}},\alpha ,\beta ,\rho $,终止迭代阈值$ \varepsilon $,最
大迭代次数$ K $输出:重构信号时变图信号$ {{\boldsymbol{X}}_{^i}} $ 初始化设置:$ {{\boldsymbol{X}}_i} = 0,\Delta {{\boldsymbol{X}}_i} = 0,i = 0 $ 迭代: (1) 确定步长:
$ \tau = - \dfrac{{\left\langle {\Delta {{\boldsymbol{X}}_i},\nabla f({{\boldsymbol{X}}_i})} \right\rangle }}{{\left\langle {\Delta {{\boldsymbol{X}}_i},\nabla f({{\boldsymbol{X}}_i}) + {\boldsymbol{Y}} + \rho {{\boldsymbol{Z}}_i} - {{\boldsymbol{P}}_i}} \right\rangle }} $其中,$ \Delta {{\boldsymbol{X}}_i} $为第$ i $步的搜索方向,$ \tau $为第$ i $步的最优步长,其由线性最小化步长准则$ \mathop {\min }\limits_\tau f({{\boldsymbol{X}}_i} + \tau \Delta {{\boldsymbol{X}}_i}) $决定。 确定步长: (2) 更新搜索方向: $ \begin{aligned} & {{\boldsymbol{X}}_{i + 1}} = {{\boldsymbol{X}}_i} + \tau \Delta {{\boldsymbol{X}}_i}; \xi {\text{ = }}\frac{{||\nabla f({{\boldsymbol{X}}_{i + 1}})||_{{\mathrm{F}}} ^2}}{{||\nabla f({{\boldsymbol{X}}_{i + 1}})||_{{\mathrm{F}}} ^2}};\\& \Delta {{\boldsymbol{X}}_{i + 1}} = - \nabla f({{\boldsymbol{X}}_{i + 1}}) + \xi \Delta {{\boldsymbol{X}}_i}\end{aligned}$ 终止条件:如果$ i = K $或者$ {\text{||}}\Delta {{\boldsymbol{X}}_i}|{|_{\mathrm{F}}} \le \varepsilon $,则停止迭代;否则令$ i = i + 1 $,继续迭代,直至满足终止条件。 2 LRJS算法求解步骤
输入:$ {\boldsymbol{Y}},{\boldsymbol{J}},{{\boldsymbol{L}}_{{\mathrm{G}}} },{{\boldsymbol{L}}_{{{T}}} },{{\boldsymbol{Z}}^k},{{\boldsymbol{P}}^k},{\boldsymbol{D}},\alpha ,\beta ,\rho ,\mu $,终止迭代阈值$ \varepsilon $,
最大迭代次数$ K $输出:重构海表面温度数据$ {{\boldsymbol{X}}^k} $ 初始化设置:$ {{\boldsymbol{X}}^0} = {{\boldsymbol{Z}}^0} = {\boldsymbol{Y}},{{\boldsymbol{P}}^0} = 0,k = 0 $ 迭代: (1) 更新${{\boldsymbol{X}}^{k + 1}}$:利用算法1的共轭梯度法; (2) 更新${{\boldsymbol{Z}}^{k + 1}}$:利用式(22); (3) 更新${{\boldsymbol{P}}^{k + 1}}$:利用式(17); 终止条件:如果$ k = K $或者${\text{||}}\Delta {{\boldsymbol{X}}^k}|{|_{{\mathrm{F}}} } \le \varepsilon $,则停止迭代;否则
令$k = k + 1$,继续迭代,直至满足终止条件。 -
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