Re: FFT VS DCT
- From: glen herrmannsfeldt <gah@xxxxxxxxxxxxxxxx>
- Date: Fri, 27 Apr 2007 12:34:19 -0800
cincydsp@xxxxxxxxx wrote:
(I wrote)
The difference is boundary conditions. FFT (or DFT) has periodic
boundary conditions. DST has f(0)=f(L)=0, DCT has f'(0)=f'(L)=0.
For image processing the DCT boundary conditions are less noticeable.
I've read that the DCT is preferable for data compression applications
because it has good energy-compaction properties; that is, when
applying the DCT to a signal, a higher ratio of the energy is
concentrated in a small number of coefficients relative to the DFT or
other similar transforms. Therefore, it's possible to throw away more
frequency bins at the DCT output, resulting in better compression. I'm
not sure if this is something that is mathematically proven, or if
it's just an empirical observation.
I don't think you can separate the questions. The usual use is many
little DCTs, such as 8 by 8. You then ask about the quality of the
compression as you through away some bins. The DCT boundary condition
will be less noticeable at block boundaries even with many bins gone.
With FFT or FST, many bins will be needed to make up for the boundary
mismatch, and so the compression won't be as good.
If instead you do the whole image as one large two dimensional FFT,
DCT, or DST, I would again not be surprised that you need more bins
to make up for the boundary conditions, unless your source data
naturally has FFT (periodic) or DST (zero at the edge) boundary
conditions. Then you have to determine the quality as bins
are discarded to compare them.
-- glen
.
- References:
- FFT VS DCT
- From: senthil_lsk
- Re: FFT VS DCT
- From: glen herrmannsfeldt
- Re: FFT VS DCT
- From: cincydsp
- FFT VS DCT
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