【正文】 )???? () 5 ??(??) = 1(2??)2∫???????????(??)???? () wave atom is noted as ????(??), with subscript ?? = (??,??,??) = (??,??1,??2,??1,??2), j is scale or level, ?? is the frequency index, ?? is the time index. All five quantities ??,??1,??2,??1,??2 are intervalued and index a point (????,????) in phasespace. They are related together as follows [1]. ???? = 2?????, ???? = ??2????, 121 , 22 m a x 2jjiiC m C??? () where 1C and 2C are two parameters which are all for implementation of puting boundary. x? is the center coordinate of ()x?? in phasespace. At the same time, ???? presents each bump of ?????(??) as 177。 s n r fo r 0 d Bp a r a m e t e r amp。 s n r fo r 5 d Bp a r a m e t e r amp。 s n r fo r 5 d B0 1 2 3 4 5 6 7 8 9 1020246810p a r a m e t e rSNRd e t e r m i n t i o n o f p a r a m e t e r o f w a v e a t o m t a n s fo r m fo r w o m a n 2 p a r a m e t e r amp。 s n r fo r 5 d Bp a r a m e t e r amp。 s n r fo r 0 d Bp a r a m e t e r amp。 s n r fo r 5 d B0 1 2 3 4 5 6 7 8 9 1050510p a r a m e t e rSNRd e t e r m i n a t i o n o f p a r a m e t e r o f w a v e a t o t r a n s fo r m fo r w o m a n 2 p a r a m e t e r amp。????. So wave atom should obey the relation from phasespace to frequency domain as the point (????,????). Under the condition above, x? and ???? in Eq. with the parameter 1C and 2C to keep the boundary should build a frame of wave packets {}?? which supports the equations below. 6 |?????(??) | ≤ ???? ? 2???(1+2???|??? ????|)??? +?????2???(1+2???|??+????|)??? ( ) 2 (1 2 )j j MMx C x x??? ?? ? ? ? for all 0M? () The parabolic scaling is applied in the definition of wave atom. When the scale is 22j? or |????|~22?? , both bumps are of the size of ~2j in frequency domain and in phasespace the size of each wave atom is ~2j? . Some detailed mathematical proofs are given well in [2]. 1D Characteristics in Wave Atom Transform Before introducing the 1D characteristics of wave atom transform, it is helpful to understand 2D conditions. The origin of phasespace is from lots of plex mathematic derivation. The main purpose of the origin of phasespace is to extend the ordinary space domain. In order to depose the signal with the feature of texture, more information of direction is needed in phasespace domain. So, around the concept of the “direction”, more bound conditions are defined to build the domain of phasespace [1]. From the mathematical 7 derivation based on ordinary space, we can see that the definition of phasespace is stricter than ordinary space because of the parameter of the “direction”. Fig. conveys the relationship between the phasespace and frequency domain. From Fig. , it is clearly shown that each singularity in phasespace is corresponding to a pair of polar bump in the frequency domain. The left side of is the singularities in phasespace, and the right side is the frequency domain. Here, bumps are distributed as a surround status from outsidecircle to insidecircle. Of course, no matter where the bump is, a polar bump also exists as its symmetry. In Fig. , we should note that there are two parameters named( , )?? to describe the size of the bump. The parameter ? describes the multiscale nature of the transform, from 0 (uniform) to 1 (dyadic). And the parameter?indexes the wave packet?s directional selectivity, from 0 (best selectivity) to 1 (poor selectivity) [2]. Wave atom?s parameter for( , )?? is 11( , )22 . Parameters for other conventional transforms are shown in Fig. . Here in , the sign “~” presents a relationship of size other than a puting method of plex numbers. Meanwhile, it?s used for the definition of the size of each bump in the frequency domain. 8 Fig. Relationship of phasespace and frequency domain Fig. Various transforms as ( , )?? families of wave packets 9 For puting the coefficients of wave atom, we must pay attention to the Fig. and understand the symmetrical conditions. As a matter of fact, wave atom coefficients are symmetric to the origin in the frequency domain. The point of the center of the circle in Fig. is zero point in frequency axis. Suppose that the bumps of upper 180 degree belong to the positive part of frequency axis, on the contrary, the other bumps of 180 degrees will be in the negative part of the frequency axis. So, the result of rebuilding the bumps in frequency axis is shown in Fig. . Where each bump in frequency of length is 2??2?? and the center frequency of the positive frequency bump is ??2????. Fig. Result of rebuilding the bumps in 1D frequency axis 10 The Structure of 1D filterbank in Wave Atom Wave atom has a special parameter ( , )?? , and ? , ? are equal to 12 , respectively. Therefore, there exists a g function given in Eq. [6] which can make symmetry of the bumps in the negative part of the frequency axis to the positive part. ( 2 ) ( )22gg????? ? ? ? () Thus, based on wave atom?s special definition and the parameters, coefficients of wave atom can be calculated along each bump by using and Eq. . 0 / 2 111? ( ) ( ( ( ) ) ) ( ( ( ) ) )22mmiiim m me e g m e g m???? ? ? ? ? ? ? ??? ???? ? ? ? ? ????? () / 2 0, ( ) ( 2 ) 2 ( 2 )j j j j jm n m mx x n x n? ? ??? ? ? ? () where ???? = (?1)?? and ???? = ??2(?? + 12). Next, the structure of filterbank for wave atom transform will be explained. In wavelet transform, filterbank can be represented like the form given in Fig. . 11 (a) (b) Fig. (a) Filterbank representation of wavelet transform (b) wavelet tree representation Like the same way, a wavelet packet tree representing