22.08.2015 Views

Jian-Jiun Ding, Wei-Lun Chao, Jiun-De Huang, and Cheng-Jin Kuo

Jian-Jiun Ding, Wei-Lun Chao, Jiun-De Huang, and Cheng-Jin Kuo

Jian-Jiun Ding, Wei-Lun Chao, Jiun-De Huang, and Cheng-Jin Kuo

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REPLACE THIS LINE WITH YOUR PAPER IDENTIFICATION NUMBER (DOUBLE-CLICK HERE TO EDIT) < 4snsW ns nWEOriginalsegmentLinear offsetAnti-symmetricextensionDFTresponseReconstructedsegment(2.5% compression rate)Smaller highfrequencycoefficientsFig. 5. Illustrations of the segment shapes (Row 1), DFT responses (Row 2), <strong>and</strong> reconstruction shapes with 2.5% coefficients (Row 3) afterapplying linear offset (Column 2) <strong>and</strong> anti-symmetric extension (Column 3). The original segment is 321-point long <strong>and</strong> its DFT has 321coefficients. The anti-symmetric extended segment is 640-point long then with 640 coefficients; however, only part of them are essential. Forclearness, in Row 2 only 20% of the coefficients around the zero frequency are shown; the magnitude responses are plotted in the log scale. Row3 shows the reconstructed segments with only 8 complex values (about 2.5% of the original 321 coefficients) based on each input signatures. Thered boxes in Row 2 mark the coefficients preserved, <strong>and</strong> for s W(n) <strong>and</strong> s W+E(n), 2 of the 8 remained complex values are the end point locations.an anti-symmetric segment of s W (n) behind s W (n), as illustratedin Fig. 4(b). The new (2N-2)-point segment s W+E (n), withWarping <strong>and</strong> Extension, is defined as follows:sW( n) , for n 0,1,..., N 1sWE( n) , (4) sW(2N 2 n) , for n N,...,2N 3which is closed <strong>and</strong> has continuous first-order <strong>and</strong> zero secondorderdifferences at the two end points (hence much smoother).Note that adding the anti-symmetric segment in s W+E (n) doesnot increase the amount of coefficients to preserve, since theDFT frequency response of an anti-symmetric signal is alsoanti- symmetric. Therefore, only half of the Fourier coefficientsof s W+E (n) are required to keep for lossless reconstruction.As illustrated in Fig. 5, s W+E (n) has even smaller highfrequencycomponents than s W (n); besides, the reconstructedsegment from s W+E (n) with 2.5% compression rate has betterquality — with the exact end point locations — than that froms(n) <strong>and</strong> s W (n).D. DFT <strong>and</strong> Boundary CompressionAfter linear offset <strong>and</strong> anti-symmetric extension, the resultingsequence s W+E (n) is then fed into the DFT to get the frequencycoefficients S(k):2N31j2 nk/(2N2)Sk ( ) sWE( ke ) . (5)2N 2k 0Since s W+E (n) is of zero mean <strong>and</strong> S(k) is anti-symmetric withS(k) = -S(2N-2-k), both S(0) <strong>and</strong> S(N-1) are 0. Thus to achievelossless reconstruction, only S(k) at k = 1, 2, ……, N-2 arerequired.For boundary compression, the first P coefficients (S(k) at k= 1, 2, ……, P; 0≦ P ≦ N-2) are remained, <strong>and</strong> totally P+2complex values, including the two end point locations, arerecorded for reconstruction. The compression rate (CR) utilizedin this paper is defined as:# recorded complex valuesCR 100%. (6)# original segment lengthFor the proposed scheme, CR is computed as:P 2CR 100%.(7)NE. Boundary Reconstruction from the Proposed SchemeThe reconstruction process from the preserved P coefficients<strong>and</strong> the two end point locations is listed as follows:• (Step1): Regenerate the (2N-2)-long frequency responseS*(k) from the P preserved coefficients by zero padding <strong>and</strong>anti-symmetric extension.• (Step 2): Perform the inverse DFT to get s W+E *(n).• (Step 3): Eliminate the anti-symmetric part in s W+E *(n) to gets W *(n) <strong>and</strong> perform the inverse operations of (2) to achievethe reconstructed segment s*(n).In Fig. 6, both the boundary compression <strong>and</strong> reconstructionprocesses are summarized.F. Proof of the Exact End Point Preserving PropertyIn this subsection, we give the proof of the exact end point

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