KLT & DCT

Hi 
We know that DCT is almost as good as KLT for a 1st order markov process.
But under what cases ( if any) will they both KLT and DCT be the same.
regards
Abraham


		
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abraham.dsp skrev:
> Hi
> We know that DCT is almost as good as KLT for a 1st order markov process.
> But under what cases ( if any) will they both KLT and DCT be the same.

Homework questions don't get answered here. The very, very least
you could do, is to try and reformulate the text from the assignment.

It takes about as much work as answering the question yourself.

Rune

"abraham.dsp" <abraham.dsp@gmail.com> writes:

> Hi 
> We know that DCT is almost as good as KLT for a 1st order markov process.
> But under what cases ( if any) will they both KLT and DCT be the same.
> regards
> Abraham

Perhaps it says in your textbook. Have you read it?
-- 
%  Randy Yates                  % "Remember the good old 1980's, when 
%% Fuquay-Varina, NC            %  things were so uncomplicated?"
%%% 919-577-9882                % 'Ticket To The Moon' 
%%%% <yates@ieee.org>           % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr

"abraham.dsp" <abraham.dsp@gmail.com> writes:

> We know that DCT is almost as good as KLT for a 1st order markov process.
> But under what cases ( if any) will they both KLT and DCT be the same.

OK, so we know:

        p(DCT(x)) < p(KLT(x)) if x is markov(1)
       
Assuming that the performance of the KLT [p(KLT(.))] remains the same
for all processes (it "adapts" to the structure of different
situations), then changing something about the markov process will
either improve or disimprove the performance of the DCT.

You can change the order of the process, or you could make it
non-markov.

What do you think will happen in either case?

Ciao,

Peter K.

in article 64u9tzu1.fsf@ieee.org, Randy Yates at yates@ieee.org wrote on
08/13/2005 15:13:

> "abraham.dsp" <abraham.dsp@gmail.com> writes:
> 
>> We know that DCT is almost as good as KLT for a 1st order markov process.
>> But under what cases ( if any) will they both KLT and DCT be the same.
>> regards
> 
> Perhaps it says in your textbook. Have you read it?

i don't even know what-the-fuck a KLT is, other than the plaid skirt that a
Scot wears.

-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

robert bristow-johnson skrev:
> in article 64u9tzu1.fsf@ieee.org, Randy Yates at yates@ieee.org wrote on
> 08/13/2005 15:13:
>
> > "abraham.dsp" <abraham.dsp@gmail.com> writes:
> >
> >> We know that DCT is almost as good as KLT for a 1st order markov process.
> >> But under what cases ( if any) will they both KLT and DCT be the same.
> >> regards
> >
> > Perhaps it says in your textbook. Have you read it?
>
> i don't even know what-the-fuck a KLT is, other than the plaid skirt that a
> Scot wears.

KLT - Karhunen-Loeve Transform.

It's a signal decomposition that works in terms of the eigenvectors
of the data covariance matrix, as opposed to a pre-determined set of
orthogonal vectors.

Has little to do with clothing, really.

Rune

Peter K. skrev:
> "abraham.dsp" <abraham.dsp@gmail.com> writes:
>
> > We know that DCT is almost as good as KLT for a 1st order markov process.
> > But under what cases ( if any) will they both KLT and DCT be the same.
>
> OK, so we know:
>
>         p(DCT(x)) < p(KLT(x)) if x is markov(1)
>
> Assuming that the performance of the KLT [p(KLT(.))] remains the same
> for all processes (it "adapts" to the structure of different
> situations), then changing something about the markov process will
> either improve or disimprove the performance of the DCT.
>
> You can change the order of the process, or you could make it
> non-markov.

"Make the process non-Markov"... how do you achieve that?

Rune

"Rune Allnor" <allnor@tele.ntnu.no> writes:

> 
> "Make the process non-Markov"... how do you achieve that?
> 

:-) Your homework, too? :-)

Acutally, when I wrote it, I was thinking that a "zeroth order" markov
process would be non-markov...  but that's a little bit of an
oxymoron, no?

Ciao,

Peter K.

Peter K. skrev:
> "Rune Allnor" <allnor@tele.ntnu.no> writes:
>
> >
> > "Make the process non-Markov"... how do you achieve that?
> >
>
> :-) Your homework, too? :-)
>
> Acutally, when I wrote it, I was thinking that a "zeroth order" markov
> process would be non-markov...  but that's a little bit of an
> oxymoron, no?

You know me, I am a bit too concerned about practical details.
My point was that in a practical situation where the data are
measurements, you usually can't do much to change the process
that generates the data. What you *can* do, is to change the
*model* of the process.

Most people don't see the difference, though.

Rune