Recursive Hilbert Transform Method: Algorithm and Convergence Analysis

Abstract

The Hilbert transform (HT) is an important method for signal demodulation and instantaneous frequency (IF) estimation. The modulus of the analytic signal constructed by the HT is considered as the amplitude, and the derivative of the instantaneous phase of the extracted pure frequency modulation signal is the IF. When the spectra of the amplitude function and the oscillation term overlap each other and fail to satisfy the Bedrosian condition, the instantaneous amplitude and frequency calculated by the HT will contain errors. The recursive Hilbert transform (RHT) is an effective method to overcome this problem. The RHT considers the pure frequency modulation signal obtained by the previous HT as a new signal and recursively computes its HT until convergence. The final pure frequency modulation signal of the recursive procedure has the same zero-crossing points as the original signal, and its corresponding quadrature error signal vanishes. We emphasize the convergence analysis of the algorithm and study the convergent tendency of the quadrature error signal in each recursive step. The key to the proof is that the discrete Fourier transform value of the quadrature error signal is regarded as a vector, and the length/norm of the vector decreases with the recursion process. Finally, we used three examples to demonstrate the effectiveness of this method in signal demodulation, IF identification and damped vibrating signal analysis, which indicate the application potential of the RHT method in mono-component signal processing.

Appendix: Singular Values of Matrix \({\varvec{A}}\)

Let us consider the characteristic of matrix ${\varvec{A}}$:

$$ {\varvec{B}} = {\varvec{AA}}^{{\text{T}}} . $$

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According to matrix theory, the singular values of matrix ${\varvec{A}}$ are the square roots of the eigenvalues of matrix ${\varvec{B}}.$ Matrix ${\varvec{B}}$ is a positive-definite matrix, whose elements are given by

$$ {\mathbf{B}}_{ij} = \left\{ {\begin{array}{*{20}l} {0.5} \hfill & {{\text{if }}i = j \le M - n_{2\omega } , } \hfill \\ {0.25} \hfill & {{\text{if }}i = j > M - n_{2\omega } ,} \hfill \\ { - 0.25} \hfill & {{\text{if }}\left| {{ }i - j} \right| = n_{2\omega } ,} \hfill \\ 0 \hfill & {{\text{else}}.} \hfill \\ \end{array} } \right. $$

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Fig. 16

Example matrices presenting the relationship of matrix ${\varvec{A}}$, ${\varvec{B}}$, ${{\varvec{B}}}_{d}$ and ${\varvec{J}}$. (${n}_{2\omega }$=4, M = 10)

The first $M-{n}_{2\omega }$ diagonal elements equal to 0.5, and the rest ${n}_{2\omega }$ diagonal elements equal to 0.25. In addition, at positions ($i, j)$, if $\left|i-j\right|={n}_{2\omega }$, the elements are − 0.25. A typical example of matrix ${\varvec{B}}$ is shown in Fig. 16, wherein ${n}_{2\omega }$=4 and M = 10. With a permutation matrix ${\varvec{P}}$, matrix ${\varvec{B}}$ is transformed into ${{\varvec{B}}}_{d}$ as

$$ {\varvec{B}}_{d} = {\varvec{PBP}}^{T} , $$

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where matrix ${{\varvec{B}}}_{d}$ is a block diagonal matrix written as

$$ {\varvec{B}}_{d} = \left[ {\begin{array}{*{20}c} {{\mathbf{J}}_{1} } & {\mathbf{0}} & {\mathbf{0}} & {\mathbf{0}} \\ {\mathbf{0}} & {{\mathbf{J}}_{2} } & {\mathbf{0}} & {\mathbf{0}} \\ {\mathbf{0}} & {\mathbf{0}} & \ddots & {\mathbf{0}} \\ {\mathbf{0}} & {\mathbf{0}} & {\mathbf{0}} & {{\mathbf{J}}_{{n_{2\omega } }} } \\ \end{array} } \right]. $$

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Each block sub-matrix ${\mathbf{J}}_{i}$ in ${{\varvec{B}}}_{d}$ is a tridiagonal Toeplitz matrix, which has the form

$$ {\mathbf{J}}_{i} = \left[ {\begin{array}{*{20}c} a & b & \cdots & 0 & 0 \\ b & a & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & a & b \\ 0 & 0 & \cdots & b & c \\ \end{array} } \right]. $$

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The elements in matrix ${\mathbf{J}}_{i}$ are $a=0.5$, $b=-0.25$ and $c=0.25$. Except for the last diagonal element $c=0.25$, the other diagonal elements are all 0.5.

The permutation matrix can be obtained by rearranging the rows of the identity matrix ${\varvec{P}}={\varvec{I}}({I}_{p},:)$, where ${\varvec{I}}$ is an ${\text{M}}\times {\text{M}}$ identity matrix and ${I}_{p}$ is the index number of the row. Figure 16 depicts the rearrangement of matrix ${\varvec{B}}$ and the index number is ${I}_{p}=[\mathrm{1,5},\mathrm{9,2},\mathrm{6,10,3},\mathrm{7,4},8]$. Matrix ${\varvec{B}}$ is multiplied by ${\varvec{P}}$ to the left and then multiplied by ${{\varvec{P}}}^{T}$ to the right, the effect is the same as ${{\varvec{B}}}_{d}={\varvec{B}}({I}_{p},{I}_{p})$. Therefore, we can also directly use the method of changing the matrix index to calculate ${{\varvec{B}}}_{d}$. In fact, in the RHT algorithm, there is no need for a computing matrix ${{\varvec{B}}}_{d}$. Still, the equation for computing the index number is

$$ {\text{Ind}}\left( k \right) = \left\{ {\begin{array}{*{20}l} {\left\lceil\frac{k}{r + 1}\right\rceil + n_{\omega } \cdot {\text{mod}}\left( {k - 1,{ }r + 1} \right)} \hfill & {k \le m_{d} \left( {r + 1} \right),} \hfill \\ {\left\lceil\frac{{k - m_{d} \left( {r + 1} \right)}}{r}\right\rceil + m_{d} + n_{\omega } \cdot {\text{mod}}\left( {k - m_{d} \left( {r + 1} \right) - 1,{ }r} \right)} \hfill & {k > m_{d} \left( {r + 1} \right),} \hfill \\ \end{array} } \right. $$

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where the intermediate variable ${m}_{d}$ and $r$ are given by

$$ m_{d} = {\text{mod}}\left( {N,n_{\omega } } \right), $$

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$$ r = \left\lfloor {\frac{N}{{n_{\omega } }}} \right\rfloor . $$

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The command ${\text{mod}}\left(a, b\right)$ returns the remainder after division of $a$ by $b$, $\left\lceil x \right\rceil$ and $\left\lfloor x \right\rfloor$ round $x$ toward positive and negative infinity, respectively.

Because the permutation matrix is an orthogonal matrix, the eigenvalues of matrix ${\varvec{B}}$ are the same as matrix ${{\varvec{B}}}_{d}$. Furthermore, according to the mathematical result in Refs. [4, 29], for an $M\times M$ tridiagonal Toeplitz matrix, we write

$$ {\mathbf{J}}_{i} = \left[ {\begin{array}{*{20}c} a & b & \cdots & 0 & 0 \\ b & a & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & a & b \\ 0 & 0 & \cdots & b & {a - \beta } \\ \end{array} } \right] \in R^{N \times N} . $$

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Suppose $\beta =-b$, then the eigenvalues ${\lambda }_{k}$ are given by

$$ \lambda_{k} = a + 2b\cos \frac{2k\pi }{{2M + 1}},{ }k = 1,{ }2,{ },{ }M. $$

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In our problem, nonzero elements of matrix ${\mathbf{J}}_{i}$ are $a=0.5$, $b=-0.25$ and $\beta =0.25$, then $a-\beta =0.25$. The eigenvalues are

$$ \lambda_{k} = 0.5 + 0.5\cos \frac{2k\pi }{{2M + 1}} < 1,{\text{ for }}k = 1,{ }2,{ },{ }M. $$

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As shown in the above equation, each sub-block matrix ${\mathbf{J}}_{i}$ in the block diagonal matrix ${{\varvec{B}}}_{d}$ has eigenvalues less than 1, so all the eigenvalues of matrix ${{\varvec{B}}}_{d}$ are less than 1. Thus, the eigenvalues of matrix ${\varvec{B}}$ are all less than 1. Since the singular values of matrix ${\varvec{A}}$ are the square roots of the eigenvalues of matrix ${\varvec{B}}$, the singular values of matrix ${\varvec{A}}$ are all less than 1.