A Correction for the Paper "Symplectic geometry mode decomposition and its application to rotating machinery compound fault diagnosis"

Hong-Yan Zhang; Haoting Liu; Rui-Jia Lin; Yu Zhou

arXiv:2508.20990·eess.SP·September 1, 2025

A Correction for the Paper "Symplectic geometry mode decomposition and its application to rotating machinery compound fault diagnosis"

Hong-Yan Zhang, Haoting Liu, Rui-Jia Lin, Yu Zhou

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TL;DR

This paper identifies and corrects limitations in the symplectic geometry mode decomposition (SGMD) method by fixing bugs related to the diagonal averaging principle, enhancing its accuracy for time series analysis.

Contribution

The authors provide a correction to SGMD by applying the pulling back theorem, improving the method's reliability for decomposing time series data.

Findings

01

Corrected the SGMD method with the pulling back theorem

02

Enhanced accuracy in time series decomposition

03

Identified limitations in the original SGMD approach

Abstract

The symplectic geometry mode decomposition (SGMD) is a powerful method for decomposing time series, which is based on the diagonal averaging principle (DAP) inherited from the singular spectrum analysis (SSA). Although the authors of SGMD method generalized the form of the trajectory matrix in SSA, the DAP is not updated simultaneously. In this work, we pointed out the limitations of the SGMD method and fixed the bugs with the pulling back theorem for computing the given component of time series from the corresponding component of trajectory matrix.

Tables2

Table 1. Table 1: Structure of G 0 ( n , 3 , 0 , 6 , 0 , 8 ) = { ( x n − 3 x ) : x min 0 ≤ x ≤ x max 0 } G_{0}(n,3,0,6,0,8)=\left\{\binom{x}{n-3x}:{x}^{\mathrm{0}}_{\mathrm{min}}\leq x\leq{x}^{\mathrm{0}}_{\mathrm{max}}\right\} for x min 0 = max ⁡ ( 0 , ⌈ n − 8 3 ⌉ ) {x}^{\mathrm{0}}_{\mathrm{min}}=\max\left(0,\left\lceil{\frac{n-8}{3}}\right\rceil\right) and x max 0 = min ⁡ ( 6 , ⌊ n 3 ⌋ ) {x}^{\mathrm{0}}_{\mathrm{max}}=\min\left(6,\left\lfloor{\frac{n}{3}}\right\rfloor\right)

$n$	$x_{\min}^{0}$	$x_{\max}^{0}$	$G_{0} (n, 3, 0, 6, 0, 8)$	$\| G_{0} (n, 3, 0, 6, 0, 8) \|$
$0$	$0$	$0$	${(\frac{0}{0})}$	$1$
$1$	$0$	$0$	${(\frac{0}{1})}$	$1$
$2$	$0$	$0$	${(\frac{0}{2})}$	$1$
$3$	$0$	$1$	${(\frac{0}{3}), (\frac{1}{0})}$	$2$
$4$	$0$	$1$	${(\frac{0}{4}), (\frac{1}{1})}$	$2$
$5$	$0$	$1$	${(\frac{0}{5}), (\frac{1}{2})}$	$2$
$6$	$0$	$2$	${(\frac{0}{5}), (\frac{1}{2}), (\frac{2}{0})}$	$3$
$7$	$0$	$2$	${(\frac{0}{7}), (\frac{1}{4}), (\frac{2}{1})}$	$3$
$8$	$0$	$2$	${(\frac{0}{8}), (\frac{1}{5}), (\frac{2}{2})}$	$3$
$9$	$1$	$3$	${(\frac{1}{6}), (\frac{2}{3}), (\frac{3}{0})}$	$3$
$10$	$1$	$3$	${(\frac{1}{7}), (\frac{2}{4}), (\frac{3}{1})}$	$3$
$11$	$1$	$3$	${(\frac{1}{8}), (\frac{2}{5}), (\frac{3}{2})}$	$3$
$12$	$2$	$4$	${(\frac{2}{6}), (\frac{3}{3}), (\frac{4}{0})}$	$3$
$13$	$2$	$4$	${(\frac{2}{7}), (\frac{3}{4}), (\frac{4}{1})}$	$3$
$14$	$2$	$4$	${(\frac{2}{8}), (\frac{3}{5}), (\frac{4}{2})}$	$3$
$15$	$3$	$5$	${(\frac{3}{6}), (\frac{4}{3}), (\frac{5}{0})}$	$3$
$⋮$	$⋮$	$⋮$	$⋮$	$⋮$
$26$	$6$	$6$	${(\frac{6}{8})}$	$1$

Table 2. Table 2: Structure of G 1 ( n , 3 , 1 , 7 , 1 , 9 ) = { ( x n + 3 − 3 x ) : x min 1 ≤ x ≤ x max 1 } G_{1}(n,3,1,7,1,9)=\tiny{\left\{\binom{x}{n+3-3x}:\tiny{{x}^{\mathrm{1}}_{\mathrm{min}}\leq x\leq{x}^{\mathrm{1}}_{\mathrm{max}}}\right\}} for x min 1 = max ⁡ ( 1 , ⌈ n − 6 3 ⌉ ) {x}^{\mathrm{1}}_{\mathrm{min}}=\max\left(1,\left\lceil{\frac{n-6}{3}}\right\rceil\right) and x max 1 = min ⁡ ( 7 , ⌊ n + 2 3 ⌋ ) {x}^{\mathrm{1}}_{\mathrm{max}}=\min\left(7,\left\lfloor{\frac{n+2}{3}}\right\rfloor\right) .

$n$	$x_{\min}^{1}$	$x_{\max}^{1}$	$G_{1} (n, 3, 1, 7, 1, 9)$	$\| G_{1} (n, 3, 1, 7, 1, 9) \|$
$1$	$1$	$1$	${(\frac{1}{1})}$	$1$
$2$	$1$	$1$	${(\frac{1}{2})}$	$1$
$3$	$1$	$1$	${(\frac{1}{3})}$	$1$
$4$	$1$	$2$	${(\frac{1}{4}), (\frac{2}{1})}$	$2$
$5$	$1$	$2$	${(\frac{1}{5}), (\frac{2}{2})}$	$2$
$6$	$1$	$2$	${(\frac{1}{6}), (\frac{2}{3})}$	$2$
$7$	$1$	$3$	${(\frac{1}{7}), (\frac{2}{4}), (\frac{3}{1})}$	$3$
$8$	$1$	$3$	${(\frac{1}{8}), (\frac{2}{5}), (\frac{3}{2})}$	$3$
$9$	$1$	$3$	${(\frac{1}{9}), (\frac{2}{6}), (\frac{3}{3})}$	$3$
$10$	$2$	$4$	${(\frac{2}{7}), (\frac{3}{4}), (\frac{4}{1})}$	$3$
$11$	$2$	$4$	${(\frac{2}{8}), (\frac{3}{5}), (\frac{4}{2})}$	$3$
$12$	$2$	$4$	${(\frac{2}{9}), (\frac{3}{6}), (\frac{4}{3})}$	$3$
$13$	$3$	$5$	${(\frac{3}{7}), (\frac{4}{4}), (\frac{5}{1})}$	$3$
$14$	$3$	$5$	${(\frac{3}{8}), (\frac{4}{5}), (\frac{5}{2})}$	$3$
$15$	$3$	$5$	${(\frac{3}{9}), (\frac{4}{6}), (\frac{5}{3})}$	$3$
$16$	$4$	$6$	${(\frac{4}{7}), (\frac{5}{4}), (\frac{6}{1})}$	$3$
$⋮$	$⋮$	$⋮$	$⋮$	$⋮$
$27$	$7$	$7$	${(\frac{7}{9})}$	$1$

Equations124

X = {x [n] : 1 \leq n \leq N} = {x [1], x [2], \dots, x [N]}

X = {x [n] : 1 \leq n \leq N} = {x [1], x [2], \dots, x [N]}

X = {x [n] : 0 \leq n \leq N - 1} = {x [0], x [1], \dots, x [N - 1]}

X = {x [n] : 0 \leq n \leq N - 1} = {x [0], x [1], \dots, x [N - 1]}

X = {x [n] : s \leq n \leq N - 1 + s} = {x [n]}_{n = s}^{N - 1 + s}

X = {x [n] : s \leq n \leq N - 1 + s} = {x [n]}_{n = s}^{N - 1 + s}

X = {X = {x [n]}_{n = s}^{N - 1 + s} : N \in N, x [n] \in R},

X = {X = {x [n]}_{n = s}^{N - 1 + s} : N \in N, x [n] \in R},

m τ \in N = N - (d - 1) τ

m τ \in N = N - (d - 1) τ

X_{ssa}

X_{ssa}

m τ = 1 N - (d - 1) τ = N - d + 1

m τ = 1 N - (d - 1) τ = N - d + 1

X_{sgmd}

X_{sgmd}

Φ_{s, τ} : X

Φ_{s, τ} : X

X

M

M

= x [s] x [τ + s] ⋮ x [(d - 1) τ + s] x [s + 1] x [τ + s + 1] ⋮ x [(d - 1) τ + s + 1] \dots \dots ⋱ \dots x [s + m - 1] x [τ + s + m - 1] ⋮ x [(d - 1) τ + s + m - 1]

Φ_{s, τ}^{- 1} : R^{d \times m}

Φ_{s, τ}^{- 1} : R^{d \times m}

M

M = Φ_{0, τ} (X) = [x_{0}, x_{1}, \dots, x_{m - 1}] = (M_{j}^{i})_{d \times m} = (x [i τ + j])_{d \times m} = x [0] x [τ] ⋮ x [(d - 1) τ] x [1] x [τ + 1] ⋮ x [(d - 1) τ + 1] \dots \dots ⋱ \dots x [m - 1] x [τ + m - 1] ⋮ x [(d - 1) τ + m - 1]

M = Φ_{0, τ} (X) = [x_{0}, x_{1}, \dots, x_{m - 1}] = (M_{j}^{i})_{d \times m} = (x [i τ + j])_{d \times m} = x [0] x [τ] ⋮ x [(d - 1) τ] x [1] x [τ + 1] ⋮ x [(d - 1) τ + 1] \dots \dots ⋱ \dots x [m - 1] x [τ + m - 1] ⋮ x [(d - 1) τ + m - 1]

M = Φ_{1, τ} (X) = [x_{1}, x_{2}, \dots, x_{m}] = (M_{j}^{i})_{d \times m} = (x [(i - 1) τ + j])_{d \times m} = x [1] x [τ + 1] ⋮ x [(d - 1) τ + 1] x [2] x [τ + 2] ⋮ x [(d - 1) τ + 2] \dots \dots ⋱ \dots x [m] x [τ + m] ⋮ x [(d - 1) τ + m]

M = Φ_{1, τ} (X) = [x_{1}, x_{2}, \dots, x_{m}] = (M_{j}^{i})_{d \times m} = (x [(i - 1) τ + j])_{d \times m} = x [1] x [τ + 1] ⋮ x [(d - 1) τ + 1] x [2] x [τ + 2] ⋮ x [(d - 1) τ + 2] \dots \dots ⋱ \dots x [m] x [τ + m] ⋮ x [(d - 1) τ + m]

x_{j} = [x [j], x [τ + j], \dots, x [(d - 1) τ + j]]^{T}

x_{j} = [x [j], x [τ + j], \dots, x [(d - 1) τ + j]]^{T}

X_{ssa} = M ∣_{s = 1, τ = 1}, X_{sgmd} = M^{T}_{s = 1} .

X_{ssa} = M ∣_{s = 1, τ = 1}, X_{sgmd} = M^{T}_{s = 1} .

M = k = 1 \sum r Z_{k},

M = k = 1 \sum r Z_{k},

X^{(k)} = {x^{(k)} [n]}_{n = s}^{N - 1 + s} \in X, 1 \leq k \leq r

X^{(k)} = {x^{(k)} [n]}_{n = s}^{N - 1 + s} \in X, 1 \leq k \leq r

Ω = Ω (α_{1}, α_{2}, β_{1}, β_{2}) = {(y x) \in Z^{2} : α_{1} \leq x \leq α_{2}, β_{1} \leq y \leq β_{2}}

Ω = Ω (α_{1}, α_{2}, β_{1}, β_{2}) = {(y x) \in Z^{2} : α_{1} \leq x \leq α_{2}, β_{1} \leq y \leq β_{2}}

τ x + y = n + s τ, \forall n \in Z^{+}

τ x + y = n + s τ, \forall n \in Z^{+}

G_{s} (n, τ, α_{1}, α_{2}, β_{1}, β_{2}) = {(y x) \in Ω : xτ + y = n + s τ}

G_{s} (n, τ, α_{1}, α_{2}, β_{1}, β_{2}) = {(y x) \in Ω : xτ + y = n + s τ}

\frac{n + s τ - β _{2}}{τ} \leq x \leq \frac{n + s τ - β _{1}}{τ}

\frac{n + s τ - β _{2}}{τ} \leq x \leq \frac{n + s τ - β _{1}}{τ}

max (α_{1}, \frac{n + s τ - β _{2}}{τ}) \leq x \leq min (α_{2}, \frac{n + s τ - β _{1}}{τ})

max (α_{1}, \frac{n + s τ - β _{2}}{τ}) \leq x \leq min (α_{2}, \frac{n + s τ - β _{1}}{τ})

⎩ ⎨ ⎧ x_{min}^{s} (n, τ, α_{1}, β_{2}) x_{max}^{s} (n, τ, α_{2}, β_{1}) = max (α_{1}, ⌈ \frac{n + s τ - β _{2}}{τ} ⌉) = min (α_{2}, ⌊ \frac{n + s τ - β _{1}}{τ} ⌋)

⎩ ⎨ ⎧ x_{min}^{s} (n, τ, α_{1}, β_{2}) x_{max}^{s} (n, τ, α_{2}, β_{1}) = max (α_{1}, ⌈ \frac{n + s τ - β _{2}}{τ} ⌉) = min (α_{2}, ⌊ \frac{n + s τ - β _{1}}{τ} ⌋)

τ x + y = n + s τ, (y x) \in Ω, s \in {0, 1}

τ x + y = n + s τ, (y x) \in Ω, s \in {0, 1}

G_{s} (n, τ, α_{1}, α_{2}, β_{1}, β_{2}) = {(n + s τ - xτ x) : x_{min}^{s} \leq x \leq x_{max}^{s}}

G_{s} (n, τ, α_{1}, α_{2}, β_{1}, β_{2}) = {(n + s τ - xτ x) : x_{min}^{s} \leq x \leq x_{max}^{s}}

∣ G_{s} (n, τ, α_{1}, α_{2}, β_{1}, β_{2}) ∣ = x_{max}^{s} (n, τ, α_{2}, β_{1}) - x_{min}^{s} (n, τ, α_{1}, β_{2}) + 1.

∣ G_{s} (n, τ, α_{1}, α_{2}, β_{1}, β_{2}) ∣ = x_{max}^{s} (n, τ, α_{2}, β_{1}) - x_{min}^{s} (n, τ, α_{1}, β_{2}) + 1.

\tilde{Z}_{j}^{i} (k) = {Z_{j}^{i} (k), Z_{i}^{j} (k), m < d; m \geq d .

\tilde{Z}_{j}^{i} (k) = {Z_{j}^{i} (k), Z_{i}^{j} (k), m < d; m \geq d .

x^{(k)} [n]

x^{(k)} [n]

d^{*} = min (d, m), m^{*} = max (d, m) .

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Full text

A Correction for the Paper \saySymplectic geometry mode decomposition and its application to rotating machinery

compound fault diagnosis††thanks: Corresponding author: Hong-Yan Zhang, e-mail: [email protected]

Hong-Yan Zhang1,2, Haoting Liu2,1, Rui-Jia Lin3, Yu Zhou1

$1$ . School of Information Science and Technology, Hainan Normal University, Haikou 571158, China

$2$ . School of Automation, University of Science and Technology Beijing, Beijing 100083, China

$3$ . Information Network and Data Center, Hainan Normal University, Haikou 571158, China

0000-0002-4400-9133

Abstract

The symplectic geometry mode decomposition (SGMD) is a powerful method for decomposing time series, which is based on the diagonal averaging principle (DAP) inherited from the singular spectrum analysis (SSA). Although the authors of SGMD method generalized the form of the trajectory matrix in SSA, the DAP is not updated simultaneously. In this work, we pointed out the limitations of the SGMD method and fixed the bugs with the pulling back theorem for computing the given component of time series from the corresponding component of trajectory matrix.

Keywords: Time series; Symplectic geometry mode decomposition (SGMD); Singular spectrum analysis (SSA); Diagonal averaging principle (DAP); Pulling back theorem

1 Introduction

The symplectic geometry mode decomposition (SGMD) is originally proposed by Pan et al. in the paper \saySymplectic geometry mode decomposition and its application to rotating machinery compound fault diagnosis published in Mechanical System and Signal Processing [1] for decomposing time series111The time series is also named with time sequence.. The SGMD depends on the diagonal averaging principle (DAP) proposed by Vautard et al. [2] in the classic paper about the singular spectrum analysis (SSA).

In recent five years, there are active researches about the SGMD, such as Jin et al. [3], Zhang [4], Guo et al. [5, 6], Chen et al. [7], Liu et al. [8], Hao et al. [9], Zhan et al [10], and Xin et al. [11]. Although the researches which demonstrate the merits of the SGMD method are still increasing, there is no doubt about the key principle behind the SGMD. It should be noted that there are two limitations for the DAP in SSA

•

it just holds for the time delay $\tau=1$ in the step of constructing the trajectory matrix for embedding;

•

it just works for the time series denoted by $X=\left\{x[n]\right\}^{N}_{n=1}$ but fails for the time series denoted by $X=\left\{x[n]\right\}^{N-1}_{n=0}$ due to different structures of the trajectory matrix.

Although the trajectory matrix is generalized by Pan et al. in SGMD, the DAP is not generalized. Consequently, the reconstruction principle is not correct if the time delay $\tau>1$ and time series is denoted by $X=\left\{x[n]\right\}^{N-1}_{n=0}$ . In this paper, our purpose is to fix theses bugs with a general formula in which both the time delay $\tau$ and type flag $s$ are involved explicitly.

2 Preliminaries

2.1 Notations

The time series $X$ of length $N\in\mathbb{N}$ can be denoted by $x[n]$ for simplicity. There are two types of concrete representations for different computer programming languages:

•

type-1 for the Fortran/MATLAB/Octave/…

[TABLE]

•

type-0 for the C/C++/Python/Java/Rust/…

[TABLE]

It is trivial to find that the unified formula for these two types can be expressed by

[TABLE]

for $s\in\left\{0,1\right\}$ . The set of time sequence with length $N$ is denoted by

[TABLE]

which is called the sequence space. Suppose $d$ is the dimension of immersion space and

[TABLE]

is the number of $d$ -dim signals in the immersion space for the time series $X$ . In the work of SSA [2], the trajectory matrix of the time series is expressed by

[TABLE]

where $m$ is a positive integer such that

[TABLE]

In the work of SGMD [1], Pan et al. generalized the form of the trajectory matrix with a time delay $\tau\in\mathbb{N}$ , which is written by

[TABLE]

Obviously, if $\tau=1$ , then ${\bm{X}}_{\mathrm{sgmd}}$ is the transpose of ${\bm{X}}_{\mathrm{ssa}}$ . In order to unify the notations, the embedding mapping parameterized by the type $s$ and time delay $\tau$ is denoted by

[TABLE]

such that

[TABLE]

for $0\leq i\leq d-1$ and $0\leq j\leq m-1$ for the time series $X$ of type- $s$ with length $N$ . Intuitively, the inverse of the embedding mapping

[TABLE]

can be regarded as the pulling back mapping since it pull back the trajectory matrix to the original time series.

Please note there are two types of representation methods for the time sequece, which leads to different forms of the trajectory matrix. Particularly, for $s=0$ , we have the following trajectory matrix of type-0

[TABLE]

for the row index $0\leq i\leq d-1$ and column index $0\leq j\leq m-1$ where $\mathbf{x}_{j}={\begin{bmatrix}x[j],x[\tau+j],\cdots,x[(d-1)\tau+j]\end{bmatrix}}^{\textsf{T}}$ is a $d$ -dim vector. It is easy to find that

•

$M^{0}_{0}=x[0]$ and $M^{d-1}_{m-1}=x[(d-1)\tau+m-1]=x[N-1]$ ;

•

each of the $x[n]$ for $0\leq n\leq N-1$ has been embedded in the trajectory matrix;

•

for the given $n$ , the $x[n]$ appears at the position of $\binom{x}{y}$ such that $\tau x+y=n$ for $0\leq x\leq d-1$ and $0\leq y\leq m-1$ .

Similarly, for $s=1$ , we have the trajectory matrix of type-1 as follows

[TABLE]

for $1\leq i\leq d$ and $1\leq j\leq m$ where

[TABLE]

is a $d$ -dim signal vector. It is easy to find that

•

$M^{1}_{1}=x[1]$ and $X^{d}_{m}=x[(d-1)\tau+m]=x[N]$ ;

•

each of the $x[n]$ for $1\leq n\leq N$ has been embedded into the trajectory matrix;

•

for given $n$ , the $x[n]$ appears at the position of $\binom{i}{j}$ such that $\tau(i-1)+j=n$ for $1\leq i\leq d$ and $1\leq j\leq m$ .

It is easy to check that

[TABLE]

Suppose the trajectory matrix can be decomposed into $r$ components $\left\{\bm{Z}_{1},\cdots,\bm{Z}_{k},\cdots,\bm{Z}_{r}\right\}$ such that

[TABLE]

then the $k$ -th component $\bm{Z}_{k}\in\mathbb{R}^{{d}\times{m}}$ can be used to compute the $k$ -th component

[TABLE]

of the time series $X=\left\{x[n]\right\}^{N-1+s}_{n=s}$ . According to Pan et al [1], the matrices $\left\{\bm{Z}_{1},\cdots,\bm{Z}_{r}\right\}$ are computed based on symplectic orthogonal transform. By comparison, they are computed with singular value decomposition in the SSA method.

2.2 Binary Diophantine Equation $\tau x+y=n+s\tau$ with Rectangular Constraint

Assume that $N,d,\tau\in\mathbb{N}$ are positive integers. Suppose that $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}\in\mathbb{Z}^{+}$ such that $0\leq\alpha_{1}<\alpha_{2}$ and $0\leq\beta_{1}<\beta_{2}$ . Let

[TABLE]

For the given $n\in\mathbb{Z}^{+}$ and $\tau\in\mathbb{N}$ , we have the greatest common divisor $\operatorname{GCD}(\tau,1)=1$ and $n>\tau\cdot 1-\tau-1=-1$ , thus the set of solutions to the Diophantine equation

[TABLE]

must be non-empty according to the Theorem 7 and Theorem 8 in Appendix A. For the type flag $s\in\left\{0,1\right\}$ and $\forall n\in\mathbb{Z}_{+}$ , let

[TABLE]

be the set of the solutions to the Diophantine equation $x\tau+y=n+s\tau$ with rectangular constraint $\binom{x}{y}\in\Omega$ . The cardinality of $G_{s}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})$ is denoted by $\left|G_{s}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})\right|$ . The geometric interpretation of the set $G_{s}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})$ is all of the 2-dim points which satisfy (16) with interger coordinates on the line $x\tau+y=n+s\tau$ and in the rectangle domain $\Omega$ . Figure 1 illustrates the scenario intuitively.

The equation $x\tau+y=n+s\tau$ implies that $y=n+s\tau-x\tau$ . Thus $\beta_{1}\leq y\leq\beta_{2}$ implies that

[TABLE]

It is obvious that

[TABLE]

For the purpose of finding integer solutions, the $\dfrac{n+s\tau-\beta_{2}}{\tau}$ should be replaced by $\left\lceil{\dfrac{n+s\tau-\beta_{2}}{\tau}}\right\rceil$ and the $\dfrac{n+s\tau-\beta_{1}}{\tau}$ should be replaced by $\left\lfloor{\dfrac{n+s\tau-\beta_{1}}{\tau}}\right\rfloor$ , where $\left\lceil{x}\right\rceil$ and $\left\lfloor{x}\right\rfloor$ denote the ceiling and floor of $x\in\mathbb{R}$ respectively. Let

[TABLE]

for $s\in\left\{0,1\right\}$ , then it is easy to prove the following lemma according to (19) and (20).

Lemma 1.

For the constrained Diophantine equation

[TABLE]

the set of its solutions can be written by

[TABLE]

and the number of solutions is

[TABLE]

Some special cases for the set $G_{s}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})$ is discussed in Appendix B.

3 Bugs and Correction for the Key Formula in SGMD

3.1 Bugs in the SGMD Method

The key formula of Pan et al. can be stated by the following result [1]:

Result 2.

Suppose that $d$ is the embedding dimension, $m=N-(d-1)\tau$ , and $\bm{Z}_{k}=(Z^{i}_{j}(k))_{d\times m}\in\mathbb{R}^{{d}\times{m}}$ is the $k$ -th component of the trajectory matrix $\bm{M}=\Phi_{1,1}(X)$ in the immersion space $\mathbb{R}^{{d}\times{m}}$ for the time series $X=\left\{x[n]\right\}^{N}_{n=1}$ . Let

[TABLE]

for $1\leq i\leq d$ and $1\leq j\leq m$ , the $k$ -th component of time series $x^{(k)}[n]$ rebuilt from the $\bm{Z}_{k}$ can be computed by

[TABLE]

where

[TABLE]

We remark that the rebuilding process based on (25) does not depend on the time delay $\tau$ , which is conflict with the definition of the trajectory matrix ${\bm{X}}_{\mathrm{sgmd}}$ in (6) or its equivalent form $\bm{M}$ for $s=1$ in (8). Consequently, the equation (25) just holds for $\tau=1$ . Moreover, the MATLAB code released by Pan et al. can not be applied if $\tau>1$ or $s=0$ .

3.2 Correction for the SGMD Method

Our study shows that the bug in SGMD can be fixed if the DAP is generalized correctly. We now propose the following pulling back theorem about the generalized diagonal averaging principle (GDAP) for the SGMD and SSA:

Theorem 3 (Pulling Back Theorem).

For the type $s\in\left\{0,1\right\}$ , embedding dimension $d\in\mathbb{N}$ and time series $X=\left\{x[n]\right\}^{N-1+s}_{n=s}$ of length $N$ , suppose $\bm{M}=\Phi_{s,\tau}(X)\in\mathbb{R}^{{d}\times{m}}$ is the trajectory matrix of $X$ with $r$ components $\bm{Z}_{1},\cdots\bm{Z}_{r}$ such that $\displaystyle\bm{M}=\sum^{r}_{k=1}\bm{Z}_{k}$ . Let

[TABLE]

and

[TABLE]

be the data set specified by the matrix $\bm{Z}_{k}=(Z^{i}_{j}(k))_{d\times m}\in\mathbb{R}^{{d}\times{m}}$ , then the $k$ -th component of time sequence $X^{(k)}=\left\{x^{(k)}[n]\right\}^{N-1+s}_{n=s}\in\mathscr{X}$ corresponding to the component $\bm{Z}_{k}$ can be computed by

[TABLE]

where

[TABLE]

for $1\leq k\leq r$ .

Proof:

In order to find the inverse $X=\Phi^{-1}_{s,\tau}(\bm{M})$ such that the decomposition $\bm{M}=\sum^{r}_{k=1}\bm{Z}_{k}$ , what we should do is to find the position of $x^{(k)}[n]$ appearing in the $k$ -th component matrix $\bm{Z}_{k}$ .

According to the definition of the trajectory matrix $\bm{M}=\sum^{r}_{k=1}\bm{Z}_{k}$ of type-0 in (11) or of type-1 in (10), it is equivalent to find the solution $\binom{i}{j}$ to the Diophantine equation $\tau i+j=n$ or $\tau i+j=n+\tau$ for the given $n$ . Obviously, the key issue lies in solving the binary Diophantine equation $\tau x+y=n+s\tau$ for $s\in\left\{0,1\right\}$ . According to the Lemma 1, the pair of row and column indices $\binom{i}{j}$ are the elements of $G_{s}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})$ for $(s,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(0,0,d-1,0,m-1)$ or $(s,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(1,1,d,1,m)$ with the form $\binom{i}{j}=\binom{q}{n+s\tau-q\tau}$ for ${q}_{\mathrm{min}}\leq q\leq{q}_{\mathrm{max}}$ .

On the other hand, the data set

[TABLE]

for $s=0$ or

[TABLE]

for $s=1$ contains all of the candidates or copies of $x^{(k)}[n]$ appearing in the matrix $\bm{Z}_{k}$ . Consequently, the $x^{(k)}[n]$ can be rebuilt by averaging all of the elements in the data set $Q_{k}(n,\tau,s)$ , which is computed by (29). This completes the proof. $\blacksquare$

The Theorem 3 essentially states the method for computing the pulling back mapping $\Phi^{-1}_{s,\tau}:\mathbb{R}^{{d}\times{m}}\to\mathscr{X}$ for computing the time sequence $X^{(k)}=\Phi^{-1}_{s,\tau}(\bm{Z}_{k})$ , thus it can be called the pulling back theorem intuitively.

3.3 Corollaries

There are some trivial corollaries for $s\in\left\{0,1\right\}$ which can be applied for solving the components of the time sequence with concrete programming languages.

3.3.1 Type-0 Time Series in C/C++/Pyhton/Rust/…

For $s=0$ and $\tau\in\mathbb{N}$ , we immediately have the following corollary by Theorem 3:

Corollary 4.

For the $n\in\left\{0,1,2,\cdots,N-1\right\}$ and the $k$ -th component $\bm{Z}_{k}=(Z^{i}_{j}(k))_{d\times m}\in\mathbb{R}^{{d}\times{m}}$ of the trajectory matrix $\bm{Z}$ constructed from the time series $X=\left\{x[n]\right\}^{N-1}_{n=0}$ such that $\bm{Z}=\sum^{r}_{k=1}\bm{Z}_{k}=\Phi_{0,\tau}(X)$ , let

[TABLE]

then we can convert the $\bm{Z}_{k}$ to the time series $X^{(k)}=\left\{x^{(k)}[n]\right\}^{N-1}_{n=0}=\Phi^{-1}_{0,\tau}(\bm{Z}_{k})$ by

[TABLE]

3.3.2 Type-1 Time Series in MATLAB/Octave/Fortran

For $s=1$ and $\tau\in\mathbb{N}$ , we immediately have the following corollary by Theorem 3:

Corollary 5.

For the $n\in\left\{1,2,\cdots,N\right\}$ and the $k$ -th component $\bm{Z}_{k}=(Z^{i}_{j}(k))_{d\times m}\in\mathbb{R}^{{d}\times{m}}$ of the trajectory matrix $\bm{Z}$ constructed from the time series $X=\left\{x[n]\right\}^{N}_{n=1}$ such that $\displaystyle\bm{Z}=\sum^{r}_{k=1}\bm{Z}_{k}=\Phi_{1,\tau}(X)$ , let

[TABLE]

then we can convert the $\bm{Z}_{k}$ to the time series $X^{(k)}=\left\{x^{(k)}[n]\right\}^{N}_{n=1}=\Phi^{-1}_{1,\tau}(\bm{Z}_{k})$ by

[TABLE]

The Corollary 5 is the correct version for rebuilding the components of time series from the components of trajectory matrix, which depends on the time delay $\tau$ explicitly and can be applied for the type-1 time series.

3.3.3 Equivalent Form of DAP in SGMD and SSA subjecto to $(s,\tau)=(1,1)$

For $s=1$ and $\tau=1$ , we immediately have the following corollary by Theorem 3:

Corollary 6.

For the $n\in\left\{1,2,\cdots,N\right\}$ and the $k$ -th component $\bm{Z}_{k}=(Z^{i}_{j}(k))_{d\times m}\in\mathbb{R}^{{d}\times{m}}$ of the trajectory matrix $\bm{Z}$ constructed from the time series $X=\left\{x[n]\right\}^{N}_{n=1}$ such that $\bm{Z}=\sum^{r}_{k=1}\bm{Z}_{k}=\Phi_{1,1}(X)$ , let

[TABLE]

then we can convert the $\bm{Z}_{k}$ to the time series $X^{(k)}=\left\{x^{(k)}[n]\right\}^{N}_{n=1}=\Phi^{-1}_{1,1}(\bm{Z}_{k})$ by

[TABLE]

We remark that Corollary 6 is equivalent to the Result 2 according to the equivalent form of $G_{1}(n,1,1,d,1,m)$ proposed in Appendix B.3, where an intuitive geometric interpretation is illustrated in Figure 2.

4 Conclusion

The original SGMD method should be used carefully since the DAP are limited by two rules: the time delay $\tau$ must be set with $\tau=1$ ; and the time series must be of type-1 with the form $X=\left\{x[n]\right\}^{N}_{n=1}$ .

For the time delay $\tau>1$ and time series $X=\left\{x[n]\right\}^{N-1+s}_{n=s}$ of type- $s$ and length $N$ , the components of time series should be computed according to (29) for $s\in\left\{0,1\right\}$ . Our new formula holds for time series denoted by $X=\left\{x[n]\right\}^{N-1+s}_{n=s}$ and any time delay $\tau\geq 1$ , which can be applied not only for the SGMD method but also for the SSA method.

Acknowledgments

This work was supported in part by the National Natural Science Foundation of China under grant numbers 62167003 and 62373042, in part by the Hainan Provincial Natural Science Foundation of China under grant number 720RC616, in part by the Research Project on Education and Teaching Reform in Higher Education System of Hainan Province under grant number Hnjg2025ZD-28, in part by the Specific Research Fund of the Innovation Platform for Academicians of Hainan Province, in part by the Guangdong Basic and Applied Basic Research Foundation under grant number 2023A1515010275, and in part by the Foundation of National Key Laboratory of Human Factors Engineering under grant number HFNKL2023WW11, in part by the Hainan Province Key R & D Program Project under grant number WSJK2024MS234.

Data Availability

Not applicable

Code Availability

Not applicable

Declaration of interests

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

Appendix A Diophantine Equation $\tau x+y=n+s\tau$

For the Diophantine equation $ax+by=n$ , we have the following important properties [12, 13]:

Theorem 7.

For $a,b\in\mathbb{N}=\left\{1,2,3,\cdots\right\}$ and $n\in\mathbb{Z}=\left\{0,\pm 1,\pm 2,\cdots\right\}$ , the equation $ax+by=n$ has solution $\binom{x}{y}\in\mathbb{Z}^{2}$ if and only if the greatest common divisor of $a$ and $b$ is a factor of $n$ , i.e., $\operatorname{GCD}(a,b)\mid n$ .

Theorem 8.

Suppose that $a,b\in\mathbb{N}$ such that $\operatorname{GCD}(a,b)=1$ . For any $n\in\mathbb{Z}$ such that $n>ab-a-b$ , there must exist $x,y\in\mathbb{Z}^{+}=\left\{0,1,2,\cdots\right\}$ such that $n=ax+by$ .

If we take the assignments for the parameters $a$ , $b$ and $s$ as follows

[TABLE]

for $s\in\left\{0,1\right\}$ and $\tau\in\mathbb{N}$ , then the solution of the Diophantine equation $\tau x+y=n+s\tau$ have non-negative solution $\binom{x}{y}\in(\mathbb{Z}^{+})^{2}$ for any $n\in\mathbb{Z}^{+}$ .

Appendix B Special Cases for $G_{s}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})$

There are some typical special cases about the structure of $G_{s}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})$ for $s\in\left\{0,1\right\}$ , $(\alpha_{1},\alpha_{2})\in\left\{(0,d-1),(1,d)\right\}$ and $(\beta_{1},\beta_{2})\in\left\{(0,m-1),(1,m)\right\}$ because of the way of representation of time sequences and coding with concrete computer programming languages.

B.1 Case of $(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(0,d-1,0,m-1)$

Particularly, for $d,m\in\mathbb{N}$ and $(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(0,d-1,0,m-1)$ , we have

[TABLE]

and

[TABLE]

B.2 Case of $(\alpha_{2},\alpha_{2},\beta_{1},\beta_{2})=(1,d,1,m)$

Particularly, for $d,m\in\mathbb{N}$ and $(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(1,d,1,m)$ , we can find that

[TABLE]

and

[TABLE]

B.3 Case of $(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(1,d,1,m)$ and $\tau=1$

A special case for the $G_{1}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})$ such that $(\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(1,1,d,1,m)$ is of interest in the references. Figure 2 illustrates such a scenario. Let

[TABLE]

then $d+m=\min(d,m)+\max(d,m)=d^{*}+m^{*}$ . Thus $N=m+(d-1)\tau=m+d-1=d^{*}+m^{*}-1$ or equivalently $d^{*}+m^{*}=N+1$ . For the set

[TABLE]

we can obtain some interesting observations from Figure 2:

①

For $1\leq n<d^{*}$ , there are $n$ solutions to the Diophantine equation $\tau x+y=n+\tau$ for $\binom{x}{y}\in\Omega(1,d,1,m)$ and $s=1$ , thus

[TABLE]

Geometrically, the $G_{1}^{*}(n)$ corresponds to the red points on the red hypotenuse of the equilateral right angled triangle shown in the Figure 2.

②

For $d^{*}\leq n\leq m^{*}$ , there are always $d^{*}$ solutions to the Diophantine equation $\tau x+y=n+s\tau$ for $\binom{x}{y}\in\Omega(1,d,1,m)$ , thus

[TABLE]

Geometrically, the $G_{1}^{*}(n)$ corresponds to the blue points on the blue edge of the parallelogram shown in the Figure 2.

③

For $m^{*}<n\leq N$ , there are $N-n+1$ solutions to the Diophantine equation $\tau x+y=n+s\tau$ for $\binom{x}{y}\in\Omega(1,d,1,m)$ , thus

[TABLE]

for $m^{*}<n\leq N$ . Geometrically, the $G_{1}^{*}(n)$ corresponds to the black points on the black edge of the parallelogram shown in the Figure 2.

Appendix C Example of Embedding and Pulling Back

For the purpose of illustrating the scenarios of embedding and pulling back mapping, we set $(N,d,m,\tau)=(27,7,9,3)$ for the time series $X$ and observe the the trajectory matrix and the set $G_{s}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})$ for $s\in\left\{0,1\right\}$ .

C.1 Type-0 Time Series

For $s=0$ and $(N,d,m,\tau)=(27,7,9,3)$ , the time series must be

[TABLE]

Its trajectory matrix is shown in Figure 3.

On the other hand, we can set $(\alpha_{2},\alpha_{2},\beta_{1},\beta_{2})=(0,6,0,8)$ for the domain $\Omega$ . Equation (39) implies that

[TABLE]

and equation (40) implies that

[TABLE]

The structure of $G_{0}(n,\tau,0,d-1,0,m-1)=G_{0}(n,3,0,6,0,8)$ is listed in Table 1.

Table 1 describes the position of $x[n]$ in the trajectory matrix $\bm{M}$ , which can be used to estimate the $x[n]$ with generalized diagonal averaging principle. For example, when $n=9$ , we have

[TABLE]

which means $x[9]$ appears as the entries $M^{1}_{6},M^{2}_{3}$ and $M^{3}_{0}$ in the trajectory matrix $\bm{M}$ shown in Figure 3. In other words, we can take the pulling back theorem about the GDAP to rebuild the $x[9]$ by $x[9]=\frac{1}{3}(M^{1}_{6}+M^{2}_{3}+M^{3}_{0})$ .

C.2 Type-1 Time Series

For $s=1$ and $(N,d,m,\tau)=(27,7,9,3)$ , the time series must be

[TABLE]

Its trajectory matrix is shown in Figure 4.

On the other hand, we can set $(\alpha_{2},\alpha_{2},\beta_{1},\beta_{2})=(1,7,1,9)$ for the domain $\Omega$ . With the help of (41) and (42), we immediately have

[TABLE]

The structure of $G_{1}(n,\tau,1,d,1,m)=G_{1}(n,3,1,7,1,9)$ is listed in Table 2.

Similarly, Table 2 describes the position of $x[n]\in X=\left\{x[1],\cdots,x[27]\right\}$ in the trajectory matrix $\bm{M}$ . For example, when $n=9$ , we have

[TABLE]

which means $x[9]$ appears as the entries $M^{1}_{9},M^{2}_{6}$ and $M^{3}_{3}$ in the trajectory matrix $\bm{M}$ shown in Figure 4. The pulling back theorem about the GDAP implies that the $x[9]$ can be estimated by $x[9]=\frac{1}{3}(M^{1}_{9}+M^{2}_{6}+M^{3}_{3})$ .

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TL;DR

Contribution

Findings

Abstract

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A Correction for the Paper \saySymplectic geometry mode decomposition and its application to rotating machinery

Abstract

Contents

1 Introduction

2 Preliminaries

2.1 Notations

2.2 Binary Diophantine Equation τx+y=n+sτ\tau x+y=n+s\tauτx+y=n+sτ with Rectangular Constraint

Lemma 1**.**

3 Bugs and Correction for the Key Formula in SGMD

3.1 Bugs in the SGMD Method

Result 2**.**

3.2 Correction for the SGMD Method

Theorem 3** (Pulling Back Theorem).**

3.3 Corollaries

3.3.1 Type-0 Time Series in C/C++/Pyhton/Rust/…

Corollary 4**.**

3.3.2 Type-1 Time Series in MATLAB/Octave/Fortran

Corollary 5**.**

3.3.3 Equivalent Form of DAP in SGMD and SSA subjecto to (s,τ)=(1,1)(s,\tau)=(1,1)(s,τ)=(1,1)

Corollary 6**.**

4 Conclusion

Acknowledgments

Data Availability

Code Availability

Declaration of interests

Appendix A Diophantine Equation τx+y=n+sτ\tau x+y=n+s\tauτx+y=n+sτ

Theorem 7**.**

Theorem 8**.**

Appendix B Special Cases for Gs(n,τ,α1,α2,β1,β2)G_{s}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})Gs​(n,τ,α1​,α2​,β1​,β2​)

B.1 Case of (α1,α2,β1,β2)=(0,d−1,0,m−1)(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(0,d-1,0,m-1)(α1​,α2​,β1​,β2​)=(0,d−1,0,m−1)

B.2 Case of (α2,α2,β1,β2)=(1,d,1,m)(\alpha_{2},\alpha_{2},\beta_{1},\beta_{2})=(1,d,1,m)(α2​,α2​,β1​,β2​)=(1,d,1,m)

B.3 Case of (α1,α2,β1,β2)=(1,d,1,m)(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(1,d,1,m)(α1​,α2​,β1​,β2​)=(1,d,1,m) and τ=1\tau=1τ=1

Appendix C Example of Embedding and Pulling Back

C.1 Type-0 Time Series

C.2 Type-1 Time Series

2.2 Binary Diophantine Equation $\tau x+y=n+s\tau$ with Rectangular Constraint

Lemma 1.

Result 2.

Theorem 3 (Pulling Back Theorem).

Corollary 4.

Corollary 5.

3.3.3 Equivalent Form of DAP in SGMD and SSA subjecto to $(s,\tau)=(1,1)$

Corollary 6.

Appendix A Diophantine Equation $\tau x+y=n+s\tau$

Theorem 7.

Theorem 8.

Appendix B Special Cases for $G_{s}(n,\tau,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})$

B.1 Case of $(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(0,d-1,0,m-1)$

B.2 Case of $(\alpha_{2},\alpha_{2},\beta_{1},\beta_{2})=(1,d,1,m)$

B.3 Case of $(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})=(1,d,1,m)$ and $\tau=1$