Re: Aliasing in MDCT

casper.ptrsn@xxxxxxxxx wrote:
(snip on symmetry and transforms)

So what you're saying with the symmetry condition is that if I input X
samples, the symmetry condition applies only to the X/2 samples, and
thereby the X/2 + 1 ... X samples can be either symmetric or
asymmetric or zero or... whatever?

If you are used to the symmetry of the DFT (exponential
transform) then the symmetry of the DST and DCT can be
confusing. In all cases, N input points should result
in N output points.

For the exponential transform, the basis functions
have frequencies that are integer multiples
of the fundamental, where the fundamental
has one cycle over the length of the transform.

For DST and DCT the fundamental has one half cycle
over the length of the transform. The DST with
boundary conditions of zero at the ends does not
have sample points at the ends. Some forms of
DCT also don't have sample points at the ends.

If that's correct, giving the alternating even and odd boundary
conditions used in DCT Type-IV, then I should be able to see:
a) The other half of the input exactly mirrored / exactly symmetric to
the first half ?
and,

I am not sure what you mean by alternating even/odd boundary
conditions. For DCT the derivative goes to zero at the
boundary. (Which is not at a sample point for Type-IV.)

Consider the modes of an closed end tube or open end
transmission line. They either reflect with inversion
(DST) or without inversion (DCT).

b) The other half of the input exactly mirrored BUT with its sign
changed (i.e. if X is my inputvector then X(1) could be 4, and X/2 + 1
would be -4) ?

I hope that made more sense.

-- glen
.

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