[FFmpeg-devel] [PATCH] RDFT for Bink and QDM2
Alex Converse
alex.converse
Thu Jan 22 17:56:24 CET 2009
On Wed, Jan 21, 2009 at 6:14 PM, Michael Niedermayer <michaelni at gmx.at> wrote:
> On Wed, Jan 21, 2009 at 06:03:36PM -0500, Alex Converse wrote:
>> On Wed, Jan 21, 2009 at 5:53 PM, Michael Niedermayer <michaelni at gmx.at> wrote:
>> > On Wed, Jan 21, 2009 at 04:45:09PM -0500, Alex Converse wrote:
>> >> On Wed, Jan 21, 2009 at 4:25 PM, Michael Niedermayer <michaelni at gmx.at> wrote:
>> >> > On Wed, Jan 21, 2009 at 01:12:30PM -0500, Alex Converse wrote:
>> >> >> On Wed, Jan 21, 2009 at 8:36 AM, Michael Niedermayer <michaelni at gmx.at> wrote:
>> >> >> >
>> >> >> > On Tue, Jan 20, 2009 at 03:19:46PM -0500, Alex Converse wrote:
>> >> >> > > On Tue, Jan 20, 2009 at 1:50 PM, Michael Niedermayer <michaelni at gmx.at> wrote:
>> >> >> > > > On Tue, Jan 20, 2009 at 12:24:06PM -0500, Alex Converse wrote:
>> >> >> > > >> On Tue, Jan 20, 2009 at 11:52 AM, Michael Niedermayer <michaelni at gmx.at> wrote:
>> >> >> > > >> > On Mon, Jan 19, 2009 at 07:44:38PM -0500, Alex Converse wrote:
>> >> >> > > > [...]
>> >> >> > > >> >
>> >> >> > > >> > [...]
>> >> >> > > >> >
>> >> >> > > >> >> +/**
>> >> >> > > >> >> + * Sets up a real FFT.
>> >> >> > > >> >> + * @param nbits Log2 of the length of the input array
>> >> >> > > >> >
>> >> >> > > >> >> + * @param inverse If TRUE, perform the inverse of the transform
>> >> >> > > >> >
>> >> >> > > >> > i suggest if 0 perform the forward transform, if 1 perform the inverse
>> >> >> > > >>
>> >> >> > > >> I believe these are equivalent.
>> >> >> > > >
>> >> >> > > > I searched for TRUE in the iso C standard and found no match.
>> >> >> > > > But lets assume it where defined as 1, this doesnt document any
>> >> >> > > > other value, is inverse=2 or 0 invalid? the forward transform?
>> >> >> > > > 99% of the people will guess correct but why write something ambigous if
>> >> >> > > > it can be written unambigous ...
>> >> >> > > >
>> >> >> > > >
>> >> >> > > >>
>> >> >> > > >> >
>> >> >> > > >> >
>> >> >> > > >> >> + * @param sign_convention The sign of j of the forward FFT.
>> >> >> > > >> >
>> >> >> > > >> > i do not understand this, j is a variable that has no clear relation to a FFT
>> >> >> > > >> >
>> >> >> > > >>
>> >> >> > > >> j is the unit vector in the vertical direction on the complex plane. j
>> >> >> > > >> = sqrt(-1).
>> >> >> > > >
>> >> >> > > > wasnt that i, i for imaginary?
>> >> >> > > >
>> >> >> > > >
>> >> >> > > >>
>> >> >> > > >> The forward DFT can be defined as:
>> >> >> > > >> X_k = \sum_{n=0}^{N-1} x_n e^{-{2\pi j \over N} nk }
>> >> >> > > >>
>> >> >> > > >> It also can be defined as
>> >> >> > > >> X_k = \sum_{n=0}^{N-1} x_n e^{{2\pi j \over N} nk }
>> >> >> > > >
>> >> >> > > > you can, but why would we need it?
>> >> >> > > >
>> >> >> > >
>> >> >> > > Bink uses it. As per the comments, the inverse of this transform is
>> >> >> > > not simply the transform with the opposite sign convention.
>> >> >> >
>> >> >> > maybe iam too tired but i see just one transform
>> >> >> >
>> >> >> > X[ k] = sum(n=0..N-1) x[n] e^(-2pi*i*n*k / N)
>> >> >> > | k= N-m
>> >> >> > X[N-m] = sum(n=0..N-1) x[n] e^(-2pi*i*n*(N-m) / N)
>> >> >> > X[N-m] = sum(n=0..N-1) x[n] e^(-2pi*i*n*N / N - 2pi*i*n*-m / N)
>> >> >> > X[N-m] = sum(n=0..N-1) x[n] e^(-2pi*i*n + 2pi*i*n*m / N)
>> >> >> > X[N-m] = sum(n=0..N-1) x[n] e^( 2pi*i*n*m / N)e^(-2pi*i*n)
>> >> >> > X[N-m] = sum(n=0..N-1) x[n] e^( 2pi*i*n*m / N)
>> >> >>
>> >> >> The twiddle has to happen in the frequency domain so we have 4 cases:
>> >> >
>> >> > i have no clue what you talk about, but after a little testing i have
>> >> > some doubt that your code works at all.
>> >> > for example
>> >> >
>> >> > ff_rdft_init(&one, 7, 1, 0);
>> >> >
>> >>
>> >> sign_convention takes +/- 1. I'm sorry if that wasn't clear.
>> >
>> > followng shows empirically
>> > that the transforms are identical short of trivial changes
>> >
>> > ff_rdft_init(&one, bits, 1, -1);
>> > ff_rdft_init(&two, bits, 1, 1);
>> >
>> > for(i=0; i<N; i++)
>> > src[i] = dst1[i] = (i + 2*i*i - 7*i*i*i + 123) % 1000;
>> >
>> > for(i=0; i<N; i++){
>> > dst2[i] = dst1[i];
>> > if((i&1) && i!=1)
>> > dst2[i]*=-1;
>> > }
>> >
>> > ff_rdft_calc(&one, dst1);
>> > ff_rdft_calc(&two, dst2);
>> >
>>
>> Yes, it can be done as yet another pre/post processing loop. Or it can
>> be simply included as part of the transform.
>
> or the decoder can be fixed so it stores the values with the correct signs
>
There is no such thing as "the correct sign." It's simply a matter of
convention, and in this case, considering the code we already have,
the problem is easier to solve with the positive sign convention.
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