Stabilization and variations to the adaptive local iterative filtering algorithm: the fast resampled iterative filtering method

Abstract

Non-stationary signals are ubiquitous in real life. Many techniques have been proposed in the last decades which allow decomposing multi-component signals into simple oscillatory mono-components, like the groundbreaking Empirical Mode Decomposition technique and the Iterative Filtering method. When a signal contains mono-components that have rapid varying instantaneous frequencies like chirps or whistles, it becomes particularly hard for most techniques to properly factor out these components. The Adaptive Local Iterative Filtering technique has recently gained interest in many applied fields of research for being able to deal with non-stationary signals presenting amplitude and frequency modulation. In this work, we address the open question of how to guarantee a priori convergence of this technique, and propose two new algorithms. The first method, called Stable Adaptive Local Iterative Filtering, is a stabilized version of the Adaptive Local Iterative Filtering that we prove to be always convergent. The stability, however, comes at the cost of higher complexity in the calculations. The second technique, called Resampled Iterative Filtering, is a new generalization of the Iterative Filtering method. We prove that Resampled Iterative Filtering is guaranteed to converge a priori for any kind of signal. Furthermore, we show that in the discrete setting its calculations can be drastically accelerated by leveraging on the mathematical properties of the matrices involved. Finally, we present some artificial and real-life examples to show the power and performance of the proposed methods.Kindly check and confirm that the Article note is correctly identified.

A Error estimation for non-stationary components

Here we report the proof for Theorem 17. First, let us show a more general result.

Proposition 19

Suppose $\alpha :{\mathbb {R}}\rightarrow {\mathbb {R}}$ is a $C^1$ function with $\alpha '(x)\in [a,b]$ where $a<b$ and $ab\ge 0$. Call $R:=b-a$ and suppose that $\alpha (1) = \alpha (0) + 2k\pi$ for some integer k. If now $d_j$ is the j-th Fourier coefficient of $e^{\textrm{i} \alpha (x)}$ as in

$$\begin{aligned} d_j:= \int _0^1 e^{\textrm{i} \alpha (x)} e^{-\textrm{i} 2\pi jx} dx, \end{aligned}$$

and if $2\pi j>b$, then

$$\begin{aligned} |d_j|\le \frac{1}{\pi }\frac{R}{2\pi j -b}. \end{aligned}$$

Proof

From the relation $\alpha (1) = \alpha (0) + 2k\pi$ we find that $2k\pi = \alpha (1) -\alpha (0) =\int _0^1 \alpha '(x) \in [a,b]$. Call $\psi (x):= 2\pi jx - \alpha (x)$ and notice that

$$\begin{aligned} \psi '(x) = 2\pi j - \alpha '(x) \in [q,p], \quad p:= 2\pi j -a\ge q:= 2\pi j-b>0, \end{aligned}$$

so $\psi$ is invertible with $\psi ^{-1}$ in $C^1$. We can then define the function $\varphi (y)$ as $\varphi (y):= (\psi ^{-1})'(y) = 1/\psi '(\psi ^{-1}(y) )\in [p^{-1},q^{-1}]$, and by the Fourier formula,

$$\begin{aligned} d_j= \int _0^1 e^{\textrm{i} [\alpha (x)- 2\pi jx]} dx = \int _0^1 e^{-\textrm{i} \psi (x)} dx = \int _{\psi (0)}^{\psi (1)} e^{-\textrm{i} y}\varphi (y) dy, \end{aligned}$$

where $\psi (1) - \psi (0) = 2\pi j - (\alpha (1)-\alpha (0)) = 2\pi j -2\pi k$. Call $s:= j-k$ and notice that $2\pi s = \int _0^1\psi '(x)\in [q,p]$ and in particular $p\ge 2\pi s\ge q>0$. Now

$$\begin{aligned} |d_j|= \left|\int _{-\alpha (0)}^{2\pi j-\alpha (1)} e^{-\textrm{i} y} \varphi (y) dy\right|&= \left|\int _{0}^{2\pi } e^{-\textrm{i} sz} e^{\textrm{i} \alpha (0)}\varphi (sz-\alpha (0))s dz\right|\\&= s \left|\int _{0}^{2\pi } e^{-\textrm{i} sz}\varphi (sz-\alpha (0)) dz\right|\end{aligned}$$

and the integral of the exponential is zero over $[0,2\pi ]$, therefore we can add to $\varphi (y)$ any constant without changing the result. As a consequence,

$$\begin{aligned} |d_j|=s \left|\int _{0}^{2\pi } e^{-\textrm{i} sz} \left( \varphi (sz-\alpha (0)) -\frac{q^{-1}+p^{-1}}{2}\right) dz\right|= s \left|\int _{0}^{2\pi } e^{-\textrm{i} sz} \phi (z) dz\right|\end{aligned}$$

where $\phi (z):= \varphi (sz-\alpha (0))-\frac{q^{-1}+p^{-1}}{2}$ is a real function bounded in absolute value by $\frac{q^{-1}-p^{-1}}{2}$. Suppose $z_0$ is the argument of $\int _0^{2\pi } e^{-\textrm{i} sz}\phi (z) dz$ so that

$$\begin{aligned} |d_j|=s e^{-\textrm{i} z_0} \int _{0}^{2\pi } e^{-\textrm{i} sz}\phi (z) dz \in {\mathbb {R}} \end{aligned}$$

and its imaginary part is zero, leading to

$$\begin{aligned} |d_j|&=s \Re \left( e^{-\textrm{i} z_0} \int _{0}^{2\pi } e^{-\textrm{i} sz}\phi (z) dz \right) \\&=s \int _{0}^{2\pi } \cos (sz+z_0)\phi (z) dz \\&\le s \frac{q^{-1}-p^{-1}}{2} \int _{0}^{2\pi } |\cos (sz+z_0)|dz = 2s (q^{-1}-p^{-1}). \end{aligned}$$

Using that $p\ge 2\pi s$ and $p-q = R$ we conclude that

$$\begin{aligned} |d_j|\le \frac{R}{q\pi }. \end{aligned}$$

$\square$

Corollary 20

Suppose $\alpha :{\mathbb {R}}\rightarrow {\mathbb {R}}$ is a $C^1$ function with $\alpha '(x)\in [a,b]$ where $0<a<b$. Call $R:=b-a$ and suppose that $\alpha (1) = \alpha (0) + 2k\pi$ for some integer k. If now $d_j$ is the j-th Fourier coefficient of $\cos (\alpha (x))$ as in

$$\begin{aligned} d_j:= \int _0^1 \cos (\alpha (x)) e^{-\textrm{i} 2\pi jx} dx, \end{aligned}$$

and $2\pi j - b>0$, then

$$\begin{aligned} |d_j|\le \frac{1}{\pi }\frac{R}{2\pi j - b}. \end{aligned}$$

Proof

Since $2\cos (\alpha (x)) = e^{\textrm{i} \alpha (x)}+e^{-\textrm{i} \alpha (x)}$ and both $\alpha (x)$ and $-\alpha (x)$ satisfy the hypotheses of Proposition 19, we can estimate $d_j$ through the mean of the Fourier coefficients $d_{j,1}$ and $d_{j,2}$ respectively of $e^{\textrm{i} \alpha (x)}$ and $e^{-\textrm{i} \alpha (x)}$

$$\begin{aligned} |d_j|\le \frac{|d_{j,1}|+ |d_{j,2}|}{2}\le \frac{1}{2\pi }\left( \frac{R}{2\pi j -b} +\frac{R}{2\pi j + b}\right) \le \frac{1}{\pi }\frac{R}{2\pi j - b}. \end{aligned}$$

$\square$

Going back to Theorem 17, notice that f(x) is a $C^1$ periodic function with continuous derivative $f'(x)$ whose Fourier coefficients are $2\pi \textrm{i}nd_n$ and $|f'(x)|= |\beta '(x)\sin (\beta (x))|\le b$. By Parseval Identity,

$$\begin{aligned} \Vert f'(x)\Vert _2^2 = \sum _{n=-\infty }^{+\infty } (2\pi n)^2 |d_n|^2 =\int _0^1 f'(x)^2 dx \le b^2. \end{aligned}$$

The series of $n^2|d_n|^2$ thus converges, and

$$\begin{aligned} \Vert f-f_N\Vert _2^2 =\sum _{|n|>N} |d_n|^2 =\sum _{|n|>N} \frac{(2\pi n)^2 |d_n|^2}{(2\pi n)^2} \le \left( \frac{b}{2\pi (N+1)}\right) ^2 = \left( \frac{b}{G+b+2\pi }\right) ^2. \end{aligned}$$

From Corollary 20, we already have a bound on $d_j$ leading to

$$\begin{aligned} \Vert f-f_N\Vert _2^2= & {} \sum _{|n|>N} |d_n|^2\le \frac{R^2}{\pi ^2} \sum _{|n|>N} \frac{1}{(2\pi n - b)^2}\\\le & {} \frac{R^2}{\pi ^2} \int _N^{\infty } \frac{1}{(2\pi x + b)^2} +\frac{1}{(2\pi x - b)^2}\\= & {} \frac{R^2}{2\pi ^3}\left[ \frac{1}{2\pi N + b} +\frac{1}{2\pi N - b}\right] \le \frac{R^2}{\pi ^3 G}. \end{aligned}$$