Multi-resolution Fourier analysis: time-frequency resolution in excess of Gabor–Heisenberg limit

Abstract

In this paper, multi-resolution Fourier analysis, labeled NY-MFA, suitable for time-varying spectra is proposed. It is shown that increasing time resolution has no detrimental effects on frequency resolution and achieved resolution bounds are in excess of Gabor–Heisenberg limit. Observation results with their increased time resolution, discriminated and precise frequency contents justify thoroughly the assumption of stationarity and consequently approach more precisely the concept of frequency and its changes than wavelets and short-term Fourier transform can do.

Appendices

Appendix A

Here, we use the only observed stationary signal $x_T(t)$ in the time interval of length $T$ to construct a double resolution signal for which $T\Delta B=1/2$ is satisfied. We propose “one-point” interpolation followed by zeros insertion in the frequency domain [24]. Notice that we propose here a simpler formulation of original computations appearing in [24].

1.1 One-point interpolation

Given $x_T(t)$, let us construct the overall signal defined in $[0, 2T]$ as follows,

$$\begin{aligned} \widehat{x}_{2T}(t)=\Pi _{2T}(t)[x_T(t)+z_T(t-T)], \end{aligned}$$

(26)

where $z_T(t)=0$, $\forall t\in [0, T]$ and $\Pi _{2T}(t)$ is the rectangular window of length $2T$. By writing $\widehat{x}_{2T}(t)$ as a function of the true signal $x_{2T}(t)$, we obtain,

$$\begin{aligned} \widehat{x}_{2T}(t)=\Pi _{2T}(t)x_{2T}(t)\Pi _T(t). \end{aligned}$$

(27)

One can see easily that (27) can be put under the form,

$$\begin{aligned} \widehat{x}_{2T}(t)=x_{2T}(t)\times w_{2T}(t), \end{aligned}$$

(28)

where the window $w_{2T}(t)$ is given by,

$$\begin{aligned} w_{2T}(t)=[\Pi _{2T}(t)+\xi _{2T}(t)]/2, \end{aligned}$$

(29)

and $\xi _{2T}(t)=\Pi _T(t)-\Pi _T(t-T)$.

1.2 Zeros insertion in the frequency domain

In order to insert zeros on the frequency axis able to define the bounded support, $1/(2T)$, we introduce a local period by substituting $4T$ for $2T$ in the subscripts of (28). This gives,

$$\begin{aligned} \widehat{x}_{4T}(t)=w_{4T}(t)\sum _{p=0}^{1}{x}_{2T}(t-2p T). \end{aligned}$$

(30)

It is easy to see that $w_{4T}(t)$ has the following Fourier transform,

$$\begin{aligned} W_{4}(f)=(H_ {\Pi _4}(f)+H_{\xi _ {4}}(f))/2 , \end{aligned}$$

(31)

where $H_{\Pi _4}(f)$ and $H_{\xi _{4}}(f)$ are, respectively, Fourier transforms of $\Pi _{4T}(t)$ and $\xi _{4T}(t)$, i.e.,

$$\begin{aligned} H_{\Pi _4}(f)&= 4T\phi _{\Pi _4}(f)\,S(4\pi f T),\nonumber \\ H_{\xi _{4}}(f)&= -\imath 4T S(\pi f T)\sin (\pi f T)\cos (2\pi f T) , \end{aligned}$$

(32)

where $S(x)=\sin (x)/x$. The complex exponential $\phi _{\Pi _4}(f)$ represents phase induced by the absolute position of the time interval on the timescale. Notice that,

$$\begin{aligned} \forall \,f,\;\;\vert H_{\Pi _{4}}(f)\vert >\vert H_{\xi _{4}}(f)\vert . \end{aligned}$$

(33)

1.3 Bandwidth

Here, our aim is to derive the bandwidth of $W_4(f)$ as given by (31). Notice that $W_4(f)$ consists of two terms. Let us consider the first term $H_{\Pi _4}(f)$. One can see from (32) that the interval for which $S(4\pi f T)=0$ for $f=k/(4T)$ where $k\pm 1$ is given by $1/(2T)$. Now, the bandwidth, defined as the main lobe width between 3-dB or $1/\sqrt{2}$ of $H_{\Pi _4}(f)$, is so that,

$$\begin{aligned} \mathrm{Bw}\left[ H_{\Pi _4}(f)\right] =1/(4T) . \end{aligned}$$

(34)

On the other hand,

$$\begin{aligned}&\!\!\! \left. { H_{\Pi _{4}}(f) }\right| _{\,f=\pm 1/(4T)}\nonumber \\&= \left. {H_{\xi _{4}}(f) }\right| _{\,f=\pm 1/(4T)}=\left. {H_{\xi _{4}}(f) }\right| _{\,f=\,0}=0, \end{aligned}$$

(35)

and particularly for points defining the bandwidth of $H_{\Pi _4}(f)$, we have,

$$\begin{aligned} \left. {\vert H_{\xi _{4}}(f)\vert }\right| _{\,f=\,1/(8T)}\ll \left. {\vert H_{\Pi _{4}}(f)\vert }\right| _{\,f=\,1/(8T)}. \end{aligned}$$

(36)

The bandwidth remains unchanged by the additional term $H_{\xi _{4}}(f)$. It follows from (34) and (36) that the bandwidth of $W_{4}(f)$, as given by (31), is so that,

$$\begin{aligned} \mathrm{Bw}\left[ W_{4}(f)\right] \simeq \mathrm{Bw}\left[ H_{\Pi _4}(f)\right] =1/(4T) . \end{aligned}$$

(37)

1.4 Double resolution properties

By considering (30) in the frequency domain, we can write,

$$\begin{aligned} \mathrm{FT}[\widehat{x}_{4T}(t)]&= \mathrm{FT}\left[ \sum _{p=0}^{1}{x}_{2T}(t-2p T)\right] \star W_{4}(f),\nonumber \\&= X(f)\star \theta (2fT)H_{\Pi _{2}}(f)\star W_{4}(f) , \end{aligned}$$

(38)

where FT$[x]$ is the Fourier transform of $x$. Here $X(f)$ and $H_{\Pi _{2}}(f)$ are, respectively, Fourier transforms of the signal $x(t)$ and the rectangular window $\Pi _{2T}(t)$ of length $2T$. Notice that $X(f)$, the true amplitude spectrum, is assumed bandlimited. Here the term $\theta (2fT)$ resulting from time translations is given by,

$$\begin{aligned} \theta (2fT)=\left[ 1+e^{-\imath 2\pi (2fT)}\right] . \end{aligned}$$

(39)

Notice that $\theta (2fT)$, as given by (39), defines the amplitude spectrum at the frequency locations $f=k/(4T)$. We have,

$$\begin{aligned}&\forall \,f=k/(4T),\;\;\;\theta (2fT)=(1+(-1)^k)\nonumber \\&\quad \quad \, =\left\{ \begin{array}{l} 2,\;\;\;\;k=2p\\ 0,\;\;\;\;k=2p+1, \end{array}\right. \end{aligned}$$

(40)

where $p=0,\pm 1,\pm 2\ldots$.

Now, convolution between the windows $H_{\Pi _{2}}(f)$ and $W_{4}(f)$ whose bandwidths are, respectively, given by $1/(2T)$ and $1/(4T)$ yields a window with the broadest of the two bandwidths. According to (37), this gives

$$\begin{aligned} {H}(f,1/(2T))&= \theta (2fT)H_{\Pi _{2}}(f)\star W_{4}(f)\nonumber \\&= \left\{ \begin{array}{l@{\quad }l@{\quad }l} 2(H_{\Pi _{2}}(f)\star W_{4}(f)),&{}k=2p\\ 0,&{}k=2p+1, \end{array}\right. \nonumber \\ \end{aligned}$$

(41)

where the bandwidth of ${H}(f,1/(2T))$ is specified by its second argument $1/(2T)$.

By setting (41), (38) yields,

$$\begin{aligned} \mathrm{FT}[\widehat{x}_{4T}(t)]&= X(f)\star {H}(f,1/(2T))\nonumber \\&= \widehat{X}(f,1/(2T)) . \end{aligned}$$

(42)

As shown by (42), the true spectrum $X(f)$ is convolved with the filter ${H}(f,1/(2T))$ whose bandwidth is $1/(2T)$. This yields the twofold resolution spectrum $\widehat{X}(f)$ for which,

$$\begin{aligned} T\Delta B=1/2, \end{aligned}$$

where $T$ and $\Delta B$ are widths in time and frequency domains.

Extraction of the twofold amplitude spectrum $X(f,1/(2T))$ of the signal is detailed in appendix B.

1.5 Expression of double resolution signals

By considering (30), the double resolution signal is given by,

$$\begin{aligned} \widehat{x}_{4T}(t)=w_{4T}(t)\sum _{p=0}^{1}{x}_{2T}(t-2p T). \end{aligned}$$

(43)

Here (43) can be written as a function of $x_T(t)$ as follows,

$$\begin{aligned} \widehat{x}_{4T}(t)=\Pi _{4T}(t)\sum _{p=0}^{1}x_T(t-2p T), \end{aligned}$$

(44)

which is given by $\mathfrak {R}_{(s)}[x_T(t)]$ for $s=2$. Derivation of remaining multi-resolution signals for $s=3,4,5$ is detailed in [24].

Appendix B

1.1 General case

Here, we extract the amplitude spectrum $X(f)$ of the signal $x(t)$ up to the resolution $1/(sT)$. We start by generalizing (42) to any level of resolution. This gives,

$$\begin{aligned} \mathrm{FT}[\mathfrak {R}_{(s)}[x_{T}(t)]]&= \widehat{X}(f,1/(sT))\nonumber \\&= X(f)\star H(f,1/(sT)), \end{aligned}$$

(45)

where,

$$\begin{aligned} H(f,1/(sT))=\theta (sfT)H_{\Pi {_{s}}}(f)\star W_{2s}(f), \end{aligned}$$

(46)

and $W_{2s}(f)$ is given by,

$$\begin{aligned} W_{2s}(f)=(H_{\Pi _{2s}}(f)+H_{\xi _{2s}}(f))/2. \end{aligned}$$

(47)

By setting (47), (45) yields the two terms,

$$\begin{aligned} \widehat{X}(f,1/(sT))&= X(f)\star \theta (sfT)H_{\Pi {_{s}}}(f)\star H_{\Pi _{2s}}(f)/2\nonumber \\&\quad +X(f)\star \theta (sfT)H_{\Pi {_{s}}}(f)\star H_{\xi _{2s}}(f)/2 \nonumber \\ \end{aligned}$$

(48)

For the sake of simplicity, let

$$\begin{aligned} X_{\Pi }(f,1/(sT))&= X(f)\star \theta (sfT)H_{\Pi {_{s}}}(f)\star H_{\Pi _{2s}}(f)/2\nonumber \\ X_{\xi }(f,1/(sT))&= X(f)\star \theta (sfT)H_{\Pi {_{s}}}(f)\star H_{\xi _{2s}}(f)/2 .\nonumber \\ \end{aligned}$$

(49)

By using (49), we can rewrite (48) under the form,

$$\begin{aligned} \widehat{X}(f,1/(sT))=X_{\Pi }(f,1/(sT)) +X_{\xi }(f,1/(sT)), \end{aligned}$$

(50)

Our aim, now, is to extract the spectrum $X(f)$ of the signal up to the resolution $1/(sT)$.

1.2 The spectrum of the signal

According to (49), $X(f)$ is involved in the expressions of $X_{\Pi }(f,1/(sT))$ and $X_{\xi }(f,1/(sT))$. Generalization of the important relation given by (33) to any level of resolution “$s$” yields,

$$\begin{aligned} \forall \,f,\;\;\vert H_{\Pi _{2s}}(f)\vert >\vert H_{\xi _{2s}}(f)\vert . \end{aligned}$$

(51)

Here (51) helps to find the convenient filtering able to extract $X(f)$ up to the resolution $1/(sT)$. Clearly, inequality (51) shows that spectral lines of $X_{\Pi }(f,1/(sT))$ are much more powerful than those of $X_{\xi }(f,1/(sT))$. Hence, the second right-hand side of (50) can be discarded and the first right-hand side preserved. Convenient filtering is therefore thresholding. We have shown in [25, 26] that thresholding (hard or soft) applied to simulated and experimental signals works for extraction of the amplitude spectrum of the signal. This is also observed in this work.

1.3 Frequency resolution

Consider the first right-hand side of (50). Convolution between transforms $H_{\Pi _{s}}(f)$ and $H_{\Pi _{2s}}(f)$ yields a transform with the broadest $1/(sT)$ of the two bandwidths. This gives,

$$\begin{aligned} X_{\Pi }(f,1/(sT))=X(f)\star \theta (sfT)H_{\Pi {_{s}}}(f)/2. \end{aligned}$$

(52)

Since $\forall \,f=k/(2sT)$, we have,

$$\begin{aligned} \theta (sfT)=\left\{ \begin{array}{l} 2,\;\;\;\;k=2p\\ 0,\;\;\;\;k=2p+1, \end{array}\right. \end{aligned}$$

(53)

where $p=0,\pm 1,\pm 2,\ldots$, then,

$$\begin{aligned} X_{\Pi }(f,1/(sT))=\left\{ \begin{array}{l@{\quad }l@{\quad }l} X(f)\star H_{\Pi {_{s}}}(f),&{} k=2p\\ 0,&{}k=2p+1. \end{array}\right. \nonumber \\ \end{aligned}$$

(54)

One can see immediately that since $H_{\Pi {_{s}}}(f)$ is the transform of the rectangular window $\Pi _{sT}(t)$, the true amplitude spectrum of the signal defined for the resolution $1/(sT)$ is represented by (54), i.e.,

$$\begin{aligned} \forall \,f&= k/(2sT)\;\;\mathrm{and}\;\;k=2p,\;\; \nonumber \\&\quad X_{\Pi }(f,1/(sT))=X(f)\star H_{\Pi {_{s}}}(f)\nonumber \\&= X(f,1/(sT)), \end{aligned}$$

(55)

where $p=0,\pm 1,\pm 2\ldots$.