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I've read (<> 'have understood') that for real symmetric signals the DCT and DFT are equivalent. That got me thinking about a particular set of pseudo noise signals and how information is encoded. I'm interested in things loosely associated with speech recognition. Wouldn't be reasonable to considered voiced phonemes to be nearly pink noise. Unvoiced phonemes are noise. The information is encoded by modulating pinkish &/or less-pink noise sources. Part of my justification is having spoken with a man who had had his voice box removed. Intelligibility was fine. Also, lossy compression can transmit intelligible speech. So, if I'm interested in how formant amplitudes vary in time, is there any reason to strongly prefer a DFT over a DCT for "spectral" analysis?______________________________
On Nov 23, 10:25 am, Richard Owlett <rowl...@atlascomm.net> wrote: > I've read (<> 'have understood') that for real symmetric signals the DCT > and DFT are equivalent. ... Maybe. There are two kinds if symmetry: "odd" and "even". The names describe the graphs of power series with only odd and only even terms respectively. Cosines have even symmetry, sines have odd. (Think of their power series) The DFT of an even function will have only cosine terms. Is that the same as a DCT? Why? Jerry -- The world is what you make of it. If you don't like it, make something else.______________________________
Richard Owlett wrote: > I've read (<> 'have understood') that for real symmetric signals the DCT > and DFT are equivalent. Symmetric is an important boundary condition. (snip) > So, if I'm interested in how formant amplitudes vary in time, is there > any reason to strongly prefer a DFT over a DCT for "spectral" analysis? As I understand it, DFT has periodic boundary conditions, DST has the function go to zero at the boundary, and DCT has the derivative go to zero at the boundary. A symmetric (around zero) function has its derivative zero at zero. If it is also periodic, the derivative will be zero periodically, and so the DCT can be used. The symmetric condition throws out all the antisymmetric sine terms. If your data wasn't symmetric, but was periodic you lose important terms. -- glen______________________________