[FFmpeg-devel] [PATCH] RDFT for Bink and QDM2
Alex Converse
alex.converse
Thu Jan 22 00:03:36 CET 2009
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.
--Alex
More information about the ffmpeg-devel
mailing list