Passive, active, and digital filters (3ed., CRC, 2009) - tiera.ru

24-12 Passive, Active, and Digital Filters24.6 Good FIR Filters and Wavelets24.6.1 Perfect Reconstruction ConditionOne of the most important requirements for most filter banks is to be able to reconstruct perfectly theinput signal X(z) at the reconstruction filter bank output ^X(z) (see Figure 24.1).We now wish to find the constraints on arbitrary filters, fH 0 , H 1 , G 0 , G 1 g, such that perfect reconstruction(PR) occurs.Repeating the result from Equation 24.11, the input–output relationship for the pair of filter banks inFigure 24.1 is^X(z) ¼ 1 2 X(z)[G 0(z)H 0 (z) þ G 1 (z)H 1 (z)]þ 1 2 X( z)[G 0(z)H 0 ( z) þ G 1 (z)H 1 ( z)] (24:39)If we require ^X(z) X(z)—the PR condition—thenG 0 (z)H 0 (z) þ G 1 (z)H 1 (z) 2 (24:40)andG 0 (z)H 0 ( z) þ G 1 (z)H 1 ( z) 0 (24:41)Equation 24.41 is known as the antialiasing condition because the term in X( z) in Equation 24.39 is theunwanted aliasing term caused by the 2 : 1 downsampling of y 0 and y 1 .It is straightforward to show that the expressions for fH 0 , H 1 , G 0 , G 1 g, given in Equations 24.2 and 24.6for the filters based on the Haar wavelet basis, satisfy Equations 24.40 and 24.41. They are the simplest setof filters which do.24.6.2 Good Filters=WaveletsOur main aim now is to search for better filters which result in wavelets and scaling functions that aresmoother than the Haar functions (i.e., which avoid the discontinuities evident in the waveforms ofFigure 24.5).We start our search with the two PR identities, Equations 24.40 and 24.41.The usual way of satisfying the antialiasing condition Equation 24.41, while permitting H 0 and G 0 tohave low-pass responses (passband where Re[z] > 0) and H 1 and G 1 to have high-pass responses(passband where Re[z] < 0), is with the following relations in which k must be an odd integer:Hence:H 1 (z) ¼ z k G 0 ( z) and G 1 (z) ¼ z k H 0 ( z) (24:42)G 0 (z)H 0 ( z) þ G 1 (z)H 1 ( z) ¼ G 0 (z)H 0 ( z) þ z k H 0 ( z)( z) k G 0 (z)¼ G 0 (z)H 0 ( z) þ ( 1) k H 0 ( z)G 0 (z) 0 if k is odd (24:43)and so Equation 24.41 is satisfied, as required.