Extended singular spectrum analysis for processing incomplete heterogeneous geodetic time series

Abstract

The singular spectrum analysis (SSA) is a powerful tool to de-noise the geodetic time series and extract geophysical signals of interest. However, when missing data exists in geodetic time series, the ordinary SSA cannot be directly used to process them. Moreover, the heterogeneous properties of the geodetic time series are usually not considered in ordinary SSA, though their formal errors are provided beforehand. In this contribution, we develop an extended singular spectrum analysis (ESSA) to directly process the incomplete and heterogeneous geodetic time series. To validate the proposed approach, we select the 27 vertical position time series of GNSS permanent stations located in the Chinese mainland for analysis, and the results are compared with that of the improved SSA (ISSA, Shen et al. in Nonlinear Process Geophys 22(4):371–376, 2015) which is an effective approach for analyzing the time series with missing data. The results show that (i) our ESSA method performs faster than ISSA, with mean reductions in computation time by 40.14% for 27 stations; moreover, the time consumption of ISSA is more sensitive to the window size and the length of observations than that of ESSA; (ii) the ESSA method can extract more signals than ISSA in terms of fitting errors and root-mean-square ratios of extracted signals over residuals, specifically for the consideration of formal errors. The power spectrum analysis shows that the power of annual oscillation extracted by ESSA is stronger than that by ISSA; (iii) the repeated simulations based on the real data further demonstrate that the signals extracted by ESSA are closer to the true signals than ISSA, and the accuracy improvement is portal to the percentage of data missing.

Appendices

Appendix A: Mapping original time series to reconstructed components

Taking Eq. (6) into account, we can rewrite the term $a_{k,i - j + 1} v_{j,k}$ in Eq. (7) as,

$$ a_{k,i - j + 1} v_{j,k} { = }\left( {\sum\limits_{h = 1}^{L} {x_{i - j + h} v_{h,k} } } \right)v_{j,k} = \sum\limits_{h = 1}^{L} {v_{h,k} v_{j,k} x_{i - j + h} } $$

(A1)

According to Eq. (7), the reconstructed component of any layer is composed of three segments. For the first segment (${1} \le i \le L - 1$), the element ${\text{RC}}_{i}^{k}$ reads as

$$ {\text{RC}}_{i}^{k} = \frac{1}{i}\sum\limits_{j = 1}^{i} {\sum\limits_{h = 1}^{L} {v_{h,k} v_{j,k} x_{i - j + h} } } $$

(A2)

Equation (A2) indicates that there may exist a strong connection between the reconstructed components and the original time series; however, it cannot build a linear mapping between the two items since the elements of the time series overlap in the right hand. To separate these like terms, we rewrite Eq. (A2) as the matrix form,

$$ {\text{RC}}_{i}^{k} = \frac{1}{i}{\varvec{H}}_{i}^{1} \odot {\varvec{G}}_{i}^{1} $$

(A3)

where the symbol ‘$\odot$’ is the Hadamard product operator (Marcus and Khan 1959), which takes two matrices of the same dimensions, and produces another matrix where each element is the product of elements of the original two matrices in the same position. The two matrices ${\varvec{H}}_{i}^{1}$ and ${\varvec{G}}_{i}^{1}$ with the size of $i \times L$ are defined as

$$ \begin{aligned} {\varvec{H}}_{i}^{1} & = \left[ {\begin{array}{*{20}c} {\quad x_{i} } & {\quad x_{{i + 1}} } & {\quad \cdots } & {\quad x_{{i - 1 + L}} } \\ {\quad x_{{i - 1}} } & {\quad x_{i} } & {\quad \cdots } & {\quad x_{{i - 2 + L}} } \\ {\quad \vdots } & {\quad \vdots } & {\quad \ddots } & {\quad \vdots } \\ {\quad x_{1} } & {\quad x_{2} } & {\quad \cdots } & {\quad x_{L} } \\ \end{array} } \right]{\text{ }} \hfill \\ {\varvec{G}}_{i}^{1} & = \left[ {\begin{array}{*{20}c} {\quad v_{{1,k}} v_{{1,k}} } & {\quad v_{{2,k}} v_{{1,k}} } & {\quad \cdots } & {\quad v_{{L,k}} v_{{1,k}} } \\ {\quad v_{{1,k}} v_{{2,k}} } & {\quad v_{{2,k}} v_{{2,k}} } & {\quad \cdots } & {\quad v_{{L,k}} v_{{2,k}} } \\ {\quad \vdots } & {\quad \vdots } & {\quad \cdots } & {\quad \vdots } \\ {\quad v_{{1,k}} v_{{i,k}} } & {\quad v_{{2,k}} v_{{i,k}} } & {\quad \cdots } & {\quad v_{{L,k}} v_{{i,k}} } \\ \end{array} } \right] \hfill \\ \end{aligned} $$

(A4)

Since ${\varvec{H}}_{i}^{1}$ is a Toeplitz matrix, we can reformulate Eq. (A3) by diagonal summation as,

$$ \begin{aligned}{\text{RC}}_{i}^{k} & = \frac{1}{i}\left( {\sum\limits_{j = 1}^{i} {\sum\limits_{h = 1}^{j} {v_{h,k} v_{i - j + h,k} } x_{j} + \sum\limits_{j = i + 1}^{L} {\sum\limits_{h = 1}^{i} {v_{h,k} v_{j - i + h,k} x_{j} } } } }\right. \\ & \quad \quad \quad \left.{+ \sum\limits_{j = L + 1}^{i - 1 + L} {\sum\limits_{h = 1}^{i + L - j} {v_{h,k} v_{j - i + h,k} x_{j} } } } \right)\end{aligned} $$

(A5)

It is now clear from Eq. (A5) that the first segment (${1} \le i \le L - 1$) of kth RC can be independently expressed by parts of the original time series. Similarly, for the second segment ($L \le i \le K$), the element ${\text{RC}}_{i}^{k}$ reads as,

$$ {\text{RC}}_{i}^{k} = \frac{1}{L}\sum\limits_{j = 1}^{L} {\sum\limits_{h = 1}^{L} {v_{h,k} v_{j,k} x_{i - j + h} } } = \frac{1}{L}{\varvec{H}}_{i}^{2} \odot {\varvec{G}}_{i}^{2} $$

(A6)

where ${\varvec{H}}_{i}^{2}$ and ${\varvec{G}}_{i}^{2}$ are $L \times L$ matrices and defined as,

$$ \begin{aligned} {\varvec{H}}_{i}^{2} & = \left[ {\begin{array}{*{20}c} {\quad x_{i} } & {\quad x_{{i + 1}} } & {\quad \cdots } & {\quad x_{{i - 1 + L}} } \\ {\quad x_{{i - 1}} } & {\quad x_{i} } & {\quad \cdots } & {\quad x_{{i - 2 + L}} } \\ {\quad \vdots } & {\quad \vdots } & {\quad \ddots } & {\quad \vdots } \\ {\quad x_{{i - L + 1}} } & {\quad x_{{i - L + 2}} } & {\quad \cdots } & {\quad x_{i} } \\ \end{array} } \right] \hfill \\ {\varvec{G}}_{i}^{2} & = \left[ {\begin{array}{*{20}c} {\quad v_{{1,k}} v_{{1,k}} } & {\quad v_{{2,k}} v_{{1,k}} } & {\quad \cdots } & {\quad v_{{L,k}} v_{{1,k}} } \\ {\quad v_{{1,k}} v_{{2,k}} } & {\quad v_{{2,k}} v_{{2,k}} } & {\quad \cdots } & {\quad v_{{L,k}} v_{{2,k}} } \\ {\quad \vdots } & {\quad \vdots } & {\quad \cdots } & {\quad \vdots } \\ {\quad v_{{1,k}} v_{{L,k}} } & {\quad v_{{2,k}} v_{{L,k}} } & {\quad \cdots } & {\quad v_{{L,k}} v_{{L,k}} } \\ \end{array} } \right] \hfill \\ \end{aligned} $$

(A7)

By summing the diagonal elements of ${\varvec{H}}_{i}^{2} \odot {\varvec{G}}_{i}^{2}$ and combining like terms, we can rewrite Eq. (A6) as,

$$\begin{aligned}{\text{RC}}_{i}^{k} &= \frac{1}{L}\left( {\sum\limits_{j = i - L + 1}^{i - 1} {\sum\limits_{h = 1}^{L - i + j} {v_{h,k} v_{i - j + h,k} } x_{j} }}\right. \\& \quad \quad \quad \left.{+ \sum\limits_{h = 1}^{L} {v_{h,k}^{2} } x_{i} +\sum\limits_{i + 1}^{i - 1 + L} {\sum\limits_{h = 1}^{i + L - j}{v_{h,k} } } v_{j - i + h,k} } \right)\end{aligned}$$

(A8)

For the last segment of the time series ($K + 1 \le i \le N$), the element ${\text{RC}}_{i}^{k}$ reads as

$$\begin{aligned} {\text{RC}}_{i}^{k} &= \frac{1}{N - i + 1}\sum\limits_{j = i - N + L}^{L} {\sum\limits_{h = 1}^{L} {v_{h,k} v_{j,k} x_{i - j + h} } } \\ &= \frac{1}{N - i + 1}{\varvec{H}}_{i}^{3} \odot {\varvec{G}}_{i}^{3} \end{aligned}$$

(A9)

where

$$ \begin{gathered} {\varvec{H}}_{i}^{3} = \left[ {\begin{array}{*{20}c} {\quad x_{{N - L + 1}} } & {\quad x_{{N - L + 2}} } & {\quad \cdots } & {\quad x_{N} } \\ {\quad x_{{N - L}} } & {\quad x_{{N - L + 1}} } & {\quad \cdots } & {\quad x_{{N - 1}} } \\ {\quad \vdots } & {\quad \vdots } & {\quad \ddots } & {\quad \vdots } \\ {\quad x_{{i - L + 1}} } & {\quad x_{{i - L + 2}} } & {\quad \cdots } & {\quad x_{i} } \\ \end{array} } \right]{\text{ }} \hfill \\ {\varvec{G}}_{i}^{3} = \left[ {\begin{array}{*{20}c} {\quad v_{{1,k}} v_{{i - N + L,k}} } & {\quad v_{{2,k}} v_{{i - N + L,k}} } & {\quad \cdots } & {\quad v_{{L,k}} v_{{i - N + L,k}} } \\ {\quad v_{{1,k}} v_{{i - N + L + 1,k}} } & {\quad v_{{2,k}} v_{{i - N + L + 1,k}} } & {\quad \cdots } & {\quad v_{{L,k}} v_{{i - N + L + 1,k}} } \\ {\quad \vdots } & {\quad \vdots } & {\quad \cdots } & {\quad \vdots } \\ {\quad v_{{1,k}} v_{{L,k}} } & {\quad v_{{2,k}} v_{{L,k}} } & {\quad \cdots } & {\quad v_{{L,k}} v_{{L,k}} } \\ \end{array} } \right] \hfill \\ \end{gathered} $$

(A10)

By following the previous diagonal summation technique, we can rewrite Eq. (A9) as follows:

$$ \begin{aligned}{\text{RC}}_{i}^{k} &= \frac{1}{N - i + 1}\left( {\sum\limits_{j = i - L + 1}^{N - L} {\sum\limits_{h = 1}^{L - i + j} {v_{h,k} v_{i - j + h,k} } x_{j} } }\right.\\ & \quad \left.{+ \sum\limits_{j = N - L + 1}^{i} {\sum\limits_{h = i}^{N} {v_{L + j - h,k} } v_{L + i - h,k} x_{j} } }\right.\\ & \quad \left.{+ \sum\limits_{i + 1}^{N} {\sum\limits_{h = j}^{N} {v_{L + j - h,k} } } v_{i + L - h,k} x_{j} } \right) \end{aligned}$$

(A11)

With Eqs. (A5), (A8), and (A11), we can map the original time series to the kth RC,

$$ {\varvec{RC}}^{k} = \left[ {\begin{array}{*{20}c} {{\text{RC}}_{1}^{k} } & {{\text{RC}}_{2}^{k} } & \cdots & {{\text{RC}}_{N}^{k} } \\ \end{array} } \right]^{T} = {\varvec{B}}^{k} {\varvec{x}} $$

(A12)

where ${\varvec{B}}^{k}$ is a $N \times N$ coefficient matrix with the entries of,

$$ \begin{aligned} {\varvec{B}}^{k} \left( {i,j} \right) &= \left\{ {\begin{array}{*{20}c} {b_{1}^{k} \left( {i,j} \right),} & {{1} \le i \le L - 1,{ 1} \le {{j}} \le {{N}}} \\ {b_{2}^{k} \left( {i,j} \right),} & {L \le i \le K,{ 1} \le {{j}} \le {{N}}} \\ {b_{3}^{k} \left( {i,j} \right),} & {K + 1 \le i \le N,{ 1} \le {{j}} \le {{N}}} \\ \end{array} } \right. \hfill \\ b_{1}^{k} \left( {i,j} \right) &= \left\{ \begin{gathered} \sum\limits_{h = 1}^{j} {v_{h,k} v_{i - j + h,k} } ,{ 1} \le {{j}} \le {{i}} \hfill \\ \sum\limits_{h = 1}^{i} {v_{h,k} v_{j - i + h,k} } ,{{ i + 1}} \le {{j}} \le L \hfill \\ \sum\limits_{h = 1}^{i + L - j} {v_{h,k} v_{j - i + h,k} } ,{{ L + 1}} \le {{j}} \le i - 1 + L \hfill \\ 0,{{ i + L}} \le {{j}} \le {{N}} \hfill \\ \end{gathered} \right.,\\ b_{2}^{k} \left( {i,j} \right) &= \left\{ \begin{gathered} 0,{ 1} \le {{j}} \le {{i}} - {{L}} \hfill \\ \sum\limits_{h = 1}^{L - i + j} {v_{h,k} v_{i - j + h,k} } ,{{ i}} - {{L + 1}} \le {{j}} \le {{i}} \hfill \\ \sum\limits_{h = 1}^{i + L - j} {v_{h,k} v_{j - i + h,k} } ,{{ i + 1}} \le {{j}} \le i - 1 + L \hfill \\ 0,{{ i + L}} \le {{j}} \le N \hfill \\ \end{gathered} \right. \hfill \\ b_{3}^{k} \left( {i,j} \right) &= \left\{ \begin{gathered} 0,{ 1} \le {{j}} \le {{i}} - {{L}} \hfill \\ \sum\limits_{h = 1}^{L - i + j} {v_{h,k} v_{i - j + h,k} } , \, i - L + 1 \le j \le N - L \hfill \\ \sum\limits_{h = i}^{N} {v_{L + j - h,k} } v_{L + i - h,k} ,{{ N}} - L + 1 \le j \le i \hfill \\ \sum\limits_{h = j}^{N} {v_{L + j - h,k} } v_{L + i - h,k} , \, i + 1 \le j \le N \hfill \\ \end{gathered} \right. \hfill \\ \end{aligned} $$

(A13)

Appendix B: Review of ISSA

The key procedure of ISSA is to calculate the PCs with the available observed data based on the minimum norm criteria. By considering the missing data, the Eq. (6) can be reformulated as

$$ a_{k,i} = \sum\limits_{{i + j - 1 \in S_{i} }} {x_{i + j - 1} v_{j,k} } + \sum\limits_{{i + j - 1 \in \overline{S}_{i} }} {x_{i + j - 1} v_{j,k} } $$

(B1)

where $1 \le i \le K$, $S_{{\text{i}}}$ and $\overline{S}_{{\text{i}}}$ are the index sets of sampling data and missing data, respectively, within the integer interval $\left[ {i, \, i + L - 1} \right]$, i.e., $S_{i} \cap \overline{S}_{i} = 0$ and $S_{i} \cup \overline{S}_{i} = \left[ {i, \, i + L - 1} \right]$. The missing data in Eq. (B1) can be computed with

$$ x_{i + j - 1} = \sum\limits_{m = 1}^{L} {a_{m,i} v_{j,m} } $$

(B2)

By substituting Eq. (B2) into Eq. (B1) and collecting all terms for $k=1, 2,\dots ,L$, we have

$$ {\varvec{G}}_{i} {{\varvec{\xi}}}_{i} = {\varvec{y}}_{i} $$

(B3)

where

$$ \begin{gathered} {\varvec{G}}_{i} = \left[ {\begin{array}{*{20}c} {\quad 1 - \sum\limits_{{i + j - 1 \in \bar{S}_{i} }} {v_{{j,1}}^{2} } } & {\quad - \sum\limits_{{i + j - 1 \in \bar{S}_{i} }} {v_{{j,1}} v_{{j,2}} } } & {\quad \cdots } & {\quad - \sum\limits_{{i + j - 1 \in \bar{S}_{i} }} {v_{{j,1}} v_{{j,L}} } } \\ {\quad - \sum\limits_{{i + j - 1 \in \bar{S}_{i} }} {v_{{j,2}} v_{{j,1}} } } & {\quad 1 - \sum\limits_{{i + j - 1 \in \bar{S}_{i} }} {v_{{j,2}}^{2} } } & {\quad \cdots } & {\quad - \sum\limits_{{i + j - 1 \in \bar{S}_{i} }} {v_{{j,2}} v_{{j,L}} } } \\ \vdots & \vdots & {\quad \ddots } & \vdots \\ {\quad - \sum\limits_{{i + j - 1 \in \bar{S}_{i} }} {v_{{j,L}} v_{{j,1}} } } & {\quad - \sum\limits_{{i + j - 1 \in \bar{S}_{i} }} {v_{{j,L}} v_{{j,2}} } } & {\quad \cdots } & {\quad 1 - \sum\limits_{{i + j - 1 \in \bar{S}_{i} }} {v_{{j,L}}^{2} } } \\ \end{array} } \right] \hfill \\ {\varvec{\xi }}_{i} = \left[ {\begin{array}{*{20}c} {\quad a_{{1,i}} } & {\quad a_{{2,i}} } & {\quad \cdots } & {\quad a_{{L,i}} } \\ \end{array} } \right]^{T} \hfill \\ {\varvec{y}}_{i} = \left[ {\begin{array}{*{20}c} {\quad \sum\limits_{{i + j - 1 \in S_{i} }} {x_{{i + j - 1}} v_{{j,1}} } } & {\quad \sum\limits_{{i + j - 1 \in S_{i} }} {x_{{i + j - 1}} v_{{j,2}} } } & {\quad \cdots } & {\quad \sum\limits_{{i + j - 1 \in S_{i} }} {x_{{i + j - 1}} v_{{j,L}} } } \\ \end{array} } \right]^{T} \hfill \\ \end{gathered} $$

(B4)

The coefficient matrix ${\varvec{G}}_{i}$ is symmetric and rank-deficient with the number of rank deficiencies equaling the number of missing data within the interval $\left[ {x_{i} , \, x_{i + L - 1} } \right]$. To obtain a unique solution, Shen et al. (2015) imposed an additional constraint condition,

$$ \min :{{\varvec{\xi}}}_{i}^{T} {{\varvec{\Lambda}}}^{ - 1} {{\varvec{\xi}}}_{i} $$

(B5)

where ${{\varvec{\Lambda}}}$ is a diagonal matrix, with the diagonal elements of $\lambda_{i} \left( {i = 1, \, 2, \, \ldots , \, L} \right)$. The solution to problem (B3) under the criteria (B5) is

$$ {{{\xi}}}_{i} = {\varvec{\Lambda G}}_{i}^{T} \left( {{\varvec{G}}_{i} {\varvec{\Lambda G}}_{i}^{T} } \right)^{ - } {\varvec{y}}_{i} $$

(B6)

The symbol ‘-’ denotes the pseudo-inverse of a matrix. Once all PCs are computed, we can derive the signals with the dominant PCs as the ordinary SSA. The equation (B6) can be equivalently rewritten as,

$$ {{\varvec{\xi}}}_{i} = {\varvec{\Lambda}} {\varvec{G}}_{i}^{T} \left( {{\varvec{G}}_{i} {\varvec{\Lambda}} {\varvec{G}}_{i}^{T} } \right)^{ - } {\varvec{Q}}_{i} {\varvec{x}}_{1} $$

(B7)

where ${\varvec{Q}}_{i}$ is a $L \times N_{S}$ matrix with the entries of

$$ {\varvec{Q}}_{i} \left( {j,k} \right) = \left\{ {\begin{array}{*{20}c} {v_{S(k) - i + 1,j} } & {S\left( k \right) \in S_{i} } \\ 0 & {{\text{others}}} \\ \end{array} } \right. $$

(B8)

By stacking all vectors ${{\varvec{\xi}}}_{i} \left( {i = 1,2, \ldots ,K} \right)$ in column order, we will immediately obtain the PC matrix as,

$$ {\varvec{A}} = \left( {\begin{array}{*{20}c} {{\varvec{J}}_{1} {\varvec{x}}_{1} } & {{\varvec{J}}_{2} {\varvec{x}}_{1} } & \cdots & {{\varvec{J}}_{K} {\varvec{x}}_{1} } \\ \end{array} } \right) $$

(B9)

where ${\varvec{J}}_{i} = {\varvec{\Lambda}} {\varvec{G}}_{i}^{T} \left( {{\varvec{G}}_{i}^{T} { {\varvec{\Lambda}} {\varvec{G}}}_{i} } \right)^{ - } {\varvec{Q}}_{i}$. Denoting ${\varvec{\varphi }}_{j}^{i}$ as the ith row vector of ${\varvec{J}}_{j}$, we can readily derive the signals  ${\widehat{{\varvec{s}}}}^{I}$ and the interpolation of missing data  ${\widehat{{\varvec{x}}}}_{2}^{I}$ as

$$ \begin{gathered} {\hat{\varvec{s}}}^{I} = {\varvec{F}}_{1} {\hat{\varvec{s}}} = {\varvec{F}}_{1} {\varvec{Mx}}_{1} \hfill \\ {\hat{\varvec{x}}}_{2}^{I} = {\varvec{F}}_{2} {\hat{\varvec{s}}} = {\varvec{F}}_{2} {\varvec{Mx}}_{1} \hfill \\ \end{gathered} $$

(B10)

Here, the superscript ‘I’ denotes the ISSA and ${\varvec{M}}$ is defined as,

$$ {\varvec{M}}_{i} = \left\{ \begin{gathered} \;\;\;\sum\limits_{{k = 1}}^{d} {\sum\limits_{{j = 1}}^{i} {\frac{1}{i}v_{{j,k}} {\varvec{\varphi }}_{{i - j + 1}}^{k} {\text{ }}\left( {\;\;\;1 \le i \le L - 1} \right)} } \hfill \\ \;\;\;\sum\limits_{{k = 1}}^{d} {\sum\limits_{{j = 1}}^{L} {\frac{1}{L}v_{{j,k}} {\varvec{\varphi }}_{{i - j + 1}}^{k} {\text{ }}\left( {\;\;\;L \le i \le K} \right)} } \hfill \\ \;\;\;\sum\limits_{{k = 1}}^{d} {\sum\limits_{{j = i - N + L}}^{L} {\frac{1}{{N - i + 1}}v_{{j,k}} {\varvec{\varphi }}_{{i - j + 1}}^{k} {\text{ }}\left( {\;\;\;K + 1 \le i \le N} \right)} } \hfill \\ \end{gathered} \right. $$

(B11)

where ${{\varvec{M}}}_{i}$ is the ith row of ${\varvec{M}}$. By comparing Eqs. (31), and (33), and Eq. (B10), we can find that the ESSA and ISSA share similar structures of solutions though they are derived via different strategies.