Re: DCT by convolution and convolution theorem ?
From: Bernie (a.buisson_at_nextamp.com)
Date: 08/04/04
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Date: Wed, 4 Aug 2004 10:42:27 +0200
"Bernie" <a.buisson@nextamp.com> a écrit dans le message de
news:ceobp4$a1v$1@s5.feed.news.oleane.net...
> Cp(u,S,p) u-eme coefficient of a n point DCT on the signal S with pixel
> origin at p as (1):
>
> __n-1
> \
> Cp(u,S,p) = alpha(u)* | S(x+p)cos(pi*u*(2*x+1)/(2*n)
> /__x=0
>
> with S signal of m real data and p in [0..n-1] (n=8 and m =16)
>
> to write Cp as a convolution product (2):
> __m-1
> \
> Cp(u,S,p) = | S(x)*G(u-x,u,p)
> /__x=0
>
> we introduce G(x,u,p) define by :
> if p<=u-x<p+n then alpha(u)cos(pi*u*(2*(u-x-p)+1))
> else 0
>
> we can also write the following relation (3):
>
>
> __m-1 __m-1 __
m-1
> \ (-2*Pi*I*u*t/m) \ \
> (-2*Pi*I*u*t/m)
> |Cp(u,S,p) *e = | [ |
> S(x)*G(u-x,u,p) ] * e
> /__u=0 /__u=0 /__x=0
>
>
> with :
> (-2*Pi*I*u*t/m) (-2*Pi*I*x*t/m) (-2*Pi*I*(u-x)*t/m)
> e =e *e
>
> (3) can be rewrite as (4) :
> __m-1 __m-1
__
> m-1
> \ (-2*Pi*I*u*t/m) \ (-2*Pi*I*x*t/m) \
> (-2*Pi*I*(u-x)*t/m)
> |Cp(u,S,p) *e = | S(x)*e [
> |G(u-x,u,p)*e ]
> /__u=0 /__x=0
> /__u=0
>
>
> we can see appear the DFT of S on m points in 0..m-1
>
> but my question is how transform (4) to find in the second part of the
> equation a DFT of G and obtain a relation as describe in lots of book by
> "Convolution Theorem" ?
>
> -1
> g*h = F [F(g).F(h)]
>
>
> with * is a convolution product and . a multiplication
>
> i know the conclusion, but if i cannot understand why and proof that this
is
> ok with my G function...
> so if someone could help me, any advice would be welcome.
> thanks
>
>
ok equation formulation can be destroy by news reader so in only-text
notations
Cp(u,S,p) u-eme coefficient of a n point DCT on the signal S with pixel
origin at p as (1):
Cp(u,S,p) = alpha(u)*Sum( S(x+p)cos(pi*u*(2*x+1)/(2*n), x=0 to n-1)
with S signal of m real data and p in [0..n-1] (n=8 and m =16)
to write Cp as a convolution product (2):
Cp(u,S,p) = Sum( S(x)*G(u-x,u,p), x=0 to m-1)
we introduce G(x,u,p) define by :
if p<=u-x<p+n then alpha(u)cos(pi*u*(2*(u-x-p)+1))
else 0
we can also write the following relation (3):
Sum(Cp(u,S,p)*exp(-2*I*Pi*u*t/m), u=0 to m-1) = Sum(Sum( S(x)*G(u-x,u,p),
x=0 to m-1)*exp(-2*I*Pi*u*t/m), u=0 to m-1)
with :
exp(-2*Pi*I*u*t/m)=exp(-2*Pi*I*x*t/m)*exp(-2*Pi*I*(u-x)*t/m)
(3) can be rewrite as (4) :
Sum(Cp(u,S,p)*exp(-2*I*Pi*u*t/m), u=0 to m-1) =
Sum(S(x)*exp(-2*I*Pi*x*t/m)*Sum( G(u-x,u,p)*exp(-2*I*Pi*(u-x)*t/m), u=0 to
m-1), x=0 to m-1)
we can see appear the DFT of S on m points in 0..m-1
but my question is how transform (4) to find in the second part of the
equation a DFT of G and obtain a relation as describe in lots of book by
"Convolution Theorem" ?
g*h = IF[F(g).F(h)]
with * is a convolution product and . a multiplication
i know the conclusion, but if i cannot understand why and proof that this is
ok with my G function...i think that i need to transform G by Gp (periodic,
without piecewise) * Rp a window (1 or 0)
in fact i am in the darkness ...
so if someone could help me, any advice would be welcome.
thanks
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