cosinus transform vs. real part of FFT
to...@tonyrobinsonnospam.com
> I was wondering why in speech recognition the cosinus transform
> is applied to the mfcc coefficients and not the real part of the FFT?
Primarily because it saves computing some zeros and so is "the right
thing to do".
Now here's another question - Numerical Recipes quite sensibly suggests
that the cosine transform is well suited to non-zero boundary conditions
and the sine transform is well suited to zero boundary conditions -
makes sense doesn't it. Since we don't care what the minimum (D.C.) and
maximum (lost through sampling) frequencies are - why do we use the
cosine and not sine transform?
Tony Robinson
Speaking for myself.
#include stdLegalDisclaimer
Robert Myers
> "Stevie Wonder" <x@x.x> writes:
>
>
>>I was wondering why in speech recognition the cosinus transform
>>is applied to the mfcc coefficients and not the real part of the FFT?
>>
>
> Primarily because it saves computing some zeros and so is "the right
> thing to do".
>
> Now here's another question - Numerical Recipes quite sensibly suggests
> that the cosine transform is well suited to non-zero boundary conditions
> and the sine transform is well suited to zero boundary conditions -
> makes sense doesn't it.
If that's really what Numerical recipes says, it makes no sense at all,
unless the data are truly periodic. The cosine transform (as I know it)
is commonly used to deal with an expansion in Chebyshev polynomials,
which is truly suited to dealing with non-periodic data.
> Since we don't care what the minimum (D.C.) and
> maximum (lost through sampling) frequencies are - why do we use the
> cosine and not sine transform?
>
Good question, to someone not invovled in speech recognition.
Stan Pawlukiewicz
This papaer by Gilbert Strang may explain some, particularly
about boundary conditions.
Jerry Avins
>
...
>
> This papaer by Gilbert Strang may explain some, particularly
> about boundary conditions.
>
> http://www-math.mit.edu/~gs/papers/dct.ps
Where is JJ when we need him?
Jerry
Robert Malek
> > that the cosine transform is well suited to non-zero boundary conditions
> > and the sine transform is well suited to zero boundary conditions -
> > makes sense doesn't it.
>
>
> If that's really what Numerical recipes says, it makes no sense at all,
> unless the data are truly periodic. The cosine transform (as I know it)
> is commonly used to deal with an expansion in Chebyshev polynomials,
> which is truly suited to dealing with non-periodic data.
>
>
> > Since we don't care what the minimum (D.C.) and
> > maximum (lost through sampling) frequencies are - why do we use the
> > cosine and not sine transform?
aren't our boundary conditions in time?
on the other hand, when we apply windowing anyway, our boundaries in
the time domain are 0, too.
but with: (sorry for thinking on "virtual paper")
from signal to spectrum
F_cos(signal) - i * F_sin(signal) = FT(signal)
back:
F_cos(spectrum) + i * F_sin(spectrum) = iFT(spectrum)
if we had a real signal, we can at least on the way back
rely on the sine_transformation part being (virtually) zero.
So calculation of cosine_transform has reduced computational cost.
and gives the result of an FT.
maybe confusion increased.
Robert Malek
> on the other hand, when we apply windowing anyway, our boundaries in
> the time domain are 0, too.
By the way: unfortunately boundary conditions means
"what we have" and not "what we are interested in"
.. unfortunately ;o(