In paper Convolutional neural network for speech recognition, they say
I didn't understand the highlighted sentence.
$\begingroup$
$\endgroup$
-
$\begingroup$ I don't understand it either; does it actually say consine? $\endgroup$ – Marcus Müller Jul 4 '16 at 8:03
-
$\begingroup$ @MarcusMüller Yes $\endgroup$ – Asuka Jul 4 '16 at 11:04
-
1$\begingroup$ @Asuka er, no, it doesn't. The transform is the discrete cosine transform. $\endgroup$ – Peter K.♦ Jul 4 '16 at 15:13
-
$\begingroup$ Your question has beeen answered. Do not hesitate to vote for the useful ones and accept the most suitable $\endgroup$ – Laurent Duval Feb 9 '17 at 19:44
$\begingroup$
$\endgroup$
In a frequency-like transformation (Fourier, discrete cosine, Walsh), one generally ends up with a sequence of coefficients that account for each frequency component over the support of the basis vector (a segment in time, a 2D patch in images). If you have a frequency component located on one half of the support, and zero the other half, or the converse, you often get the same absolute-valued coefficients. MFCCs fall in that context with:
the short-term power spectrum of a sound, based on a linear cosine transform.
Information about locality can usually be traced in the phase or the sign of the coefficients, but they are difficult to read in complex situations, and are not preserved with the absolute value or the power.
answered Jul 4 '16 at 9:30
-
$\begingroup$ I don't understand yes, does it mean that "DCT lose some locality feature"? If "Yes", I want to know what feature DCT lose? I'm student of CS and i'm unfamiliar with signal processing. $\endgroup$ – Asuka Jul 4 '16 at 11:02
-
2$\begingroup$ @Asuka well, in any case, you'll need to understand what locality means, if you want to ask questions regarding to its preservation, so you can't waltz in without any signal processing knowledge and hope to get sensible answers. Generally, as Laurent explained, DCT does preserve the information in form of a phase, but MFCCs throw that away. If you read the math behind DCT and MFCCs, you'll instantly understand what Laurent says – it doesn't make sense to explain the basics of DCT or MFCC, as there's plenty of good literature out there. $\endgroup$ – Marcus Müller Jul 4 '16 at 15:15
$\begingroup$
$\endgroup$
First you have a misconception in your question which I suggest you to edit first: The paper https://www.microsoft.com/en-us/research/wp-content/uploads/2016/02/TASLP2339736-proof.pdf is not authored by Hinton at all.
Second, their statement
As for frequency, the conventional use of MFCCs does present a major problem because the discrete consine transform projects the spectral energies into a new basis that may not maintain locality
They just failed to express themselves properly. The problem is not DCT but that in MFCC scheme you usually drop some coefficients in DCT result. From 40 transformed coefficients after DCT you just take 13 and drop the rest. They take all 40 instead and do not apply DCT transform taking more information. But they failed to explain it, their "locality" is completely senseless.
-
$\begingroup$ But I see something about "why CNN don't use MFCC" for many times and I saw "The standard MFCC features are not suitable because of DCT-based decorrelation transform". $\endgroup$ – Asuka Jul 4 '16 at 14:36
-
3$\begingroup$ There is a lot of misinformation and fairy tales in the field unfortunately. $\endgroup$ – Nikolay Shmyrev Jul 4 '16 at 14:40
-
1$\begingroup$ I just have to stress what Nikolay said: the sentence is simply wrong the way it's written. Simple as that. DCT is a reversible operation, which means that it does not lose any information (including locality). What loses information is discarding of DCT coefficients and calculating their magnitude. $\endgroup$ – Marcus Müller Jul 4 '16 at 15:20
-
1$\begingroup$ @Nikolay Shmyrev Indeed, a lot of misinformation and fairy tales, many people using NN (and in computer science) do not have a strong background in local harmonic transformations. The converse is valid as well, at least in my case $\endgroup$ – Laurent Duval Jul 4 '16 at 15:36
