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The Fourier transform itself is an exact mathematical transformation with an exact inverse transform. There's nothing lossy about it.
by Bou 15y ago
The Fourier transform itself is an exact mathematical transformation with an exact inverse transform. There's nothing lossy about it.
- bwarp 15y agoMathematically, you are correct. Practically you are not if you consider the transform source.
- psykotic 15y agoWhat do you mean? If the original source is analog and sampled below its Nyquist rate in the analog-to-digital conversion, the process is indeed irreversible. But that all happens before any transforms from the time domain to the frequency domain are in play, so it's a separate issue. Beyond that, discrete Fourier and cosine transforms as usually implemented are not fully reversible because of loss in precision. A colleague of mine blogged about the issue in the context of Haar transforms a few years ago: http://cbloomrants.blogspot.com/2008/09/09-08-08-1.html http://cbloomrants.blogspot.com/2008/09/09-08-08-1.html. By decomposing an orthogonal transform into shears as explained by Charles, you can design reversible fixed-precision variants of the DCT like binDCT: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.41.8531 http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.41.8...
- bwarp 15y agoSorry it is a separate issue - I should have been more clear. The original point the OP made was a bad example. Unfortunately I made a poor attempt at explaining that.