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Design of Time–Frequency Optimal Three-Band Wavelet Filter Banks with Unit Sobolev Regularity Using Frequency Domain Sampling

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Abstract

In this paper, we design three-band time–frequency-localized orthogonal wavelet filter banks having single vanishing moment. We propose new expressions to compute mean and variances in time and frequency from the samples of the Fourier transform of the asymmetric band-pass compactly supported wavelet functions. We determine discrete-time filter of length eight that generates the time–frequency optimal time-limited scaling and wavelet functions using cascade algorithm. Time–frequency product (TFP) of a function is defined as the product of its time variance and frequency variance. The TFP of the designed functions is close to 0.25 with unit Sobolev regularity. Three-band filter banks are designed by minimizing a weighted combination of TFPs of wavelets and scaling functions. Interestingly, empirical results show that time–frequency optimal, filter banks of length nine, designed with the proposed methodology, have unit Sobolev regularity, which is maximum achievable with single vanishing moment. Design examples for length six and length nine filter banks are given to demonstrate the effectiveness of the proposed design methodology.

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Acknowledgments

The authors acknowledge the support received from the Bharti Center for Communication, Department of Electrical Engineering, Indian Institute of Technology (IIT) Bombay; from the ‘Knowledge Incubation under TEQIP’ Initiative of the Ministry for Human Resource Development (MHRD) at IIT Bombay; from the research group associated with Dr. Ram Bilas Pachori at IIT Indore and from Acropolis Institute of Technology and Research, Indore, toward the research work carried out and reported in this manuscript.

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Correspondence to Dinesh Bhati.

Appendix

Appendix

1.1 Mean in Time from Samples of Band-Limited Function

In this appendix, we derive the expression for mean in time \(\mu _t\) for real-valued band-limited function \(f(t) \in L_2(\mathfrak {R})\) in terms of samples \(f(nT_s)\) sampled at \(T_s < \frac{\pi }{\sigma }\). Substituting f(t) from (16) in (1), we get

$$\begin{aligned} \mu _t= \frac{1}{E}\int _{t=-\infty }^{\infty } \bigg [\left( t\sum f(mT_s)g(t-mT_s)\right) \left( \sum f(nT_s)g(t-nT_s)\right) \hbox {d}t \bigg ] \end{aligned}$$

where \(m,n \in Z\). or,

$$\begin{aligned} \mu _t= \frac{1}{E} \sum \sum f(mT_s)f(nT_s) \bigg [ \int _{t=-\infty }^{\infty } tg(t-mT_s)g(t-nT_s)\hbox {d}t \bigg ] \end{aligned}$$

Let \(H_1({\varOmega })\) and \(H_2({\varOmega })\) are Fourier transforms of \(tg(t-nT_s)\) and \(g(t-mT_s)\), respectively. Then, using generalized Parseval’s theorem,

$$\begin{aligned} \int _{t=-\infty }^{\infty } tg(t-mT_s)g(t-nT_s)\hbox {d}t =\frac{1}{2\pi }\int _{{\varOmega }=-\infty }^{\infty } H_1({\varOmega })H_2^{*}({\varOmega })\hbox {d}{\varOmega }\end{aligned}$$

Thus,

$$\begin{aligned} \mu _t= \frac{1}{2\pi E} \sum \sum f(mT_s) f(nT_s) \int _{-\infty }^{\infty } H_1({\varOmega })H_2^{*}({\varOmega }) \hbox {d}{\varOmega }\end{aligned}$$
(33)

Since \(G({\varOmega })\) is differentiable, the Fourier transform \(H_1({\varOmega })\) and \(H_2({\varOmega })\) is given by

$$\begin{aligned} H_1({\varOmega })= & {} j\frac{\hbox {d}}{\hbox {d}{\varOmega }}\left[ G({\varOmega })\hbox {e}^{-j{\varOmega }nT_s}\right] =j\frac{\hbox {d}G({\varOmega })}{\hbox {d}{\varOmega }}\hbox {e}^{-j{\varOmega }n T_s} +nT_sG({\varOmega })\hbox {e}^{-j{\varOmega }n T_s}\\ H_2({\varOmega })= & {} G({\varOmega })\hbox {e}^{-j{\varOmega }mT_s} \end{aligned}$$

Since \(G({\varOmega })\) is a real function,

$$\begin{aligned} H_2^{*}({\varOmega })=G({\varOmega })\hbox {e}^{j{\varOmega }mT_s} \end{aligned}$$

Then,

$$\begin{aligned} H_1({\varOmega })H_2^{*}({\varOmega })=\bigg [ jG({\varOmega })\frac{\hbox {d}G({\varOmega })}{\hbox {d}{\varOmega }} +(nT_s)|G({\varOmega })|^2 \bigg ] \hbox {e}^{j{\varOmega }(m-n)T_s} \end{aligned}$$

To simplify the integral in (33), define the integrals \(I_1\) and \(I_2\) as,

$$\begin{aligned} I_1(m,n)=\int _{{\varOmega }=-\infty }^{\infty } jG({\varOmega })\frac{\hbox {d}G({\varOmega })}{\hbox {d}{\varOmega }} \hbox {e}^{j{\varOmega }(m-n)T_s}\hbox {d}{\varOmega }\end{aligned}$$

and,

$$\begin{aligned} I_2(m,n)=\int _{{\varOmega }=-\infty }^{\infty } (nT_s)|G({\varOmega })|^2\hbox {e}^{j{\varOmega }(m-n)T_s} \hbox {d}{\varOmega }\end{aligned}$$

Differentiating \(G({\varOmega })\) in (9), we get,

$$\begin{aligned} \frac{\hbox {d}G({\varOmega })}{\hbox {d}{\varOmega }}={\left\{ \begin{array}{ll} \frac{-T_s\pi }{2\epsilon }\sin \left( \frac{\pi ({\varOmega }+\sigma )}{\epsilon }\right) &{} ~~:~~ -\sigma - \epsilon \le {\varOmega }< -\sigma \\ \frac{-T_s\pi }{2\epsilon }\sin \left( \frac{\pi ({\varOmega }-\sigma )}{\epsilon }\right) &{} ~~:~~ \sigma < {\varOmega }\le \sigma + \epsilon \\ 0 &{} ~~:~~ \text {elsewhere}. \end{array}\right. } \end{aligned}$$

Split the integral \(I_1\) into \(C_1\) and \(C_2\) corresponding to frequency range \(-\sigma - \epsilon \le {\varOmega }< -\sigma \) and \(\sigma < {\varOmega }\le \sigma + \epsilon \). Let \(I_1(m,n)=C_1(m,n)+C_2(m,n)\), where

$$\begin{aligned}&C_1(m,n)\\&\,=\int _{-\sigma -\epsilon }^{-\sigma } j\bigg [\frac{T_s}{2}\left( \cos \left( \frac{\pi ({\varOmega }+ \sigma )}{\epsilon }\right) +1\right) \left( \frac{-T_s\pi }{2\epsilon }\sin \left( \frac{\pi ({\varOmega }+\sigma )}{\epsilon }\right) \right) \hbox {e}^{j{\varOmega }(m-n)T_s}\hbox {d}{\varOmega }\bigg ]\\&C_2(m,n)\\&\,=\int _{\sigma }^{\sigma +\epsilon } j \bigg [ \frac{T_s}{2}\left( \cos \left( \frac{\pi ({\varOmega }- \sigma )}{\epsilon }\right) +1\right) \left( \frac{-T_s\pi }{2\epsilon }\sin \left( \frac{\pi ({\varOmega }-\sigma )}{\epsilon }\right) \right) \hbox {e}^{j{\varOmega }(m-n)T_s}\hbox {d}{\varOmega }\bigg ] \end{aligned}$$

Solving the above integrals and substituting \(\sigma =\frac{\pi }{T_s}-\epsilon \), we get

$$\begin{aligned} C_1= & {} \frac{j\pi ^2T_s^2\hbox {e}^{j(-\pi +T_s\epsilon )(m-n)}\left( 3\pi ^2\hbox {e}^{-jT_s(m-n)\epsilon } + 5\pi ^2 - 2T_s^2{(m-n)}^2\epsilon ^2\right) }{4\left( T_s^4(m-n)^4\epsilon ^4 - 5\pi ^2T_s^2(m-n)^2\epsilon ^2 + 4\pi ^4\right) }\\ C_2= & {} \frac{-j}{4}\frac{\pi ^2T_s^2\hbox {e}^{-j(-\pi +T_s\epsilon )(m-n)}\left( 3\pi ^2\hbox {e}^{jT_s(m-n)\epsilon } + 5\pi ^2 - 2T_s^2(m-n)^2\epsilon ^2\right) }{T_s^4(m-n)^4\epsilon ^4 - 5\pi ^2T_s^2(m-n)^2\epsilon ^2 + 4\pi ^4} \end{aligned}$$

Let

$$\begin{aligned} I_2(m,n)=\int _{{\varOmega }=-\infty }^{\infty } (nT_s)|G({\varOmega })|^2\hbox {e}^{j{\varOmega }(m-n)T_s} \hbox {e}{\varOmega }=nT_sA \end{aligned}$$

then A is given by [50],

$$\begin{aligned} A(m,n)= & {} 2\sigma T_s^2\delta _{mn}+ \frac{T_s^2(-1)^{(m-n)}}{2}\bigg (\epsilon \frac{\sin ((m-n)T_s\epsilon )}{(m-n)T_s\epsilon } \\&+\,\frac{\sin ((m-n)T_s\epsilon )}{2}\frac{2(m-n)T_s}{{(\frac{\pi }{\epsilon })}^2-(m-n)^2T_s^2}\bigg )\\&+\,\frac{T_s^2(-1)^{(m-n)}}{2}\left( \epsilon \frac{\sin ((m-n)T_s\epsilon )}{(m-n)T_s\epsilon }\left( 1-\frac{\frac{1}{2}}{\left[ 1-{\left( \frac{(m-n)T_s\epsilon }{2\pi }\right) }^2\right] }\right) \right) \end{aligned}$$

Thus,

$$\begin{aligned} \mu _t= \frac{1}{2\pi E}\sum _{m \in \mathbb {Z}} \sum _{n \in \mathbb {Z}} f(mT_s)f(nT_s)(C_1(m,n)+C_2(m,n)+nT_sA(m,n)) \end{aligned}$$

or,

$$\begin{aligned} \mu _t=\frac{1}{2\pi E}(\mathbf{f}^H \mathbf{S}_\mu \mathbf{f}) \end{aligned}$$

where

$$\begin{aligned} S_\mu (m,n)=C_1(m,n)+C_2(m,n)+nT_sA(m,n) \end{aligned}$$

1.2 Mean in Frequency from Samples of Band-Limited Function

Mean frequency of band-pass function \(f(t) \in L_2(\mathfrak {R})\) can be obtained from its spectrum for positive frequencies. It is defined by (4). Since we have to compute \(\mu _{{\varOmega }}\) from the samples of band-limited function f(t), representing Fourier transform \(F({\varOmega })\) in terms of DTFT \(F(\hbox {e}^{j\omega })\), and using the relation \({\varOmega }=\frac{\omega }{T_s}\), we have

$$\begin{aligned} F({\varOmega }) ={\left\{ \begin{array}{ll} T_sF(\hbox {e}^{j{\varOmega }T_s}) &{} ~~:~~ {\forall |{\varOmega }| \le \frac{\pi }{T_s}}\\ 0 &{} ~~:~~ \text {elsewhere}. \end{array}\right. } \end{aligned}$$
(34)

Substituting for \(F({\varOmega })\) from (34) in the expression for \(\mu ({\varOmega })\), we get

$$\begin{aligned} \mu _{{\varOmega }}=\frac{1}{\pi E} \int _{\omega =0}^{\pi } \omega {|F(\hbox {e}^{j\omega )}|}^2 \hbox {d}\omega \end{aligned}$$
(35)

Since f[n] is a real-valued sequence, we have

$$\begin{aligned} |F(\hbox {e}^{j\omega })|^2=F(\hbox {e}^{j\omega })F(\hbox {e}^{j\omega })^* =\left( \sum _{n \in Z} f[n]\hbox {e}^{-jn\omega }\right) \left( \sum _{m \in Z} f[m]\hbox {e}^{jm\omega }\right) \end{aligned}$$
(36)

Substituting \(|F(\hbox {e}^{j\omega })|^2\) from (36) in (35), we get

$$\begin{aligned} \mu _{{\varOmega }}= & {} \frac{1}{\pi E} \int _{\omega =0}^{\pi } \omega \left( \sum _{m \in Z} f[m]\hbox {e}^{jm\omega }\right) \left( \sum _{n \in Z} f[n]\hbox {e}^{-jn\omega }\right) \hbox {d}\omega \\= & {} \frac{1}{\pi E} \sum \sum f[m]f[n] \int _{\omega =0}^{\pi } \omega \hbox {e}^{j\omega (m-n)} \hbox {d}\omega \end{aligned}$$

or,

$$\begin{aligned} \mu _{{\varOmega }}=\frac{1}{\pi E}(\mathbf{f}^H \mathbf{B}_v \mathbf{f}) \end{aligned}$$

where

$$\begin{aligned} {B}_v(m,n)=\int _{\omega =0}^{\pi } \omega {\hbox {e}^{j\omega (m-n)}} \hbox {d}\omega =-\frac{[(-1)^{(m-n)}(-1+\pi j (m-n))+1]}{{(m-n)}^2} \end{aligned}$$

Note that for \(m=n\),

$$\begin{aligned} {B}_\mu (m,n)=\pi ^2 /2 \end{aligned}$$

1.3 Mean of Time-Limited Functions from the Samples of Fourier Transform

For a real-valued function \(f(t) \in L_2(\mathfrak {R})\), mean in time \(\mu _t\) is defined as

$$\begin{aligned} \mu _t= \frac{1}{E}\int _{t=-\infty }^{\infty } (t{f(t)})({f(t)})^* \hbox {d}t \end{aligned}$$

The Dual of (34) is given by,

$$\begin{aligned} f(t) ={\left\{ \begin{array}{ll} \frac{{\varOmega }_s}{2\pi }\sum _{k=-\infty }^{\infty } F(k{\varOmega }_s)\hbox {e}^{jk{\varOmega }_s t} &{} ~~:~~ {\forall |{\varOmega }| \le \frac{\pi }{T_s}}\\ 0 &{} ~~:~~ \text {elsewhere}. \end{array}\right. } \end{aligned}$$
(37)

Then,

$$\begin{aligned} \mu _t=\frac{1}{E} \int _{t=-\pi /{\varOmega }_s}^{\pi /{\varOmega }_s} \bigg [t{\bigg (\frac{{\varOmega }_s}{2\pi }\sum _m F(m{\varOmega }_s)\hbox {e}^{jm{\varOmega }_s t}\bigg )}\bigg ] \bigg [{\bigg (\frac{{\varOmega }_s}{2\pi }\sum _n F(n{\varOmega }_s)\hbox {e}^{jn{\varOmega }_s t}\bigg )}\bigg ]^* \hbox {d}t \end{aligned}$$

or,

$$\begin{aligned} \mu _t= \frac{{\varOmega }_s^2}{4\pi ^2 E} \sum _m \sum _n F(m{\varOmega }_s) F(n{\varOmega }_s)^* \int _{t=-\pi /{\varOmega }_s}^{\pi /{\varOmega }_s} t{ \hbox {d}^{j{\varOmega }_s(m-n) t}} \hbox {d}t \end{aligned}$$

Let

$$\begin{aligned}&\int _{t=-\pi /{\varOmega }_s}^{\pi /{\varOmega }_s} [t{ \hbox {e}^{j{\varOmega }_s(m-n) t}}] \hbox {d}t=2j\bigg [\frac{\sin (\pi (m-n)) - \pi (m-n)\cos (\pi (m-n))}{((m-n){\varOmega }_s)^2}\bigg ]\\&\quad =R(m,n) \end{aligned}$$

Then,

$$\begin{aligned} \mu _t=\frac{{\varOmega }_s^2}{4\pi ^2 E}({\hat{\mathbf{f}}}^H \mathbf{R} {\hat{\mathbf{f}}}) \end{aligned}$$

Note that for \(m=n\),

$$\begin{aligned} R(m,n)=0 \end{aligned}$$

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Bhati, D., Sharma, M., Pachori, R.B. et al. Design of Time–Frequency Optimal Three-Band Wavelet Filter Banks with Unit Sobolev Regularity Using Frequency Domain Sampling. Circuits Syst Signal Process 35, 4501–4531 (2016). https://doi.org/10.1007/s00034-016-0286-7

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