Can someone please help me with the code. Speech recognition system.
이전 댓글 표시
Hi there,
whenever I am trying to run the code, it first throws an error at "ms" in "mfcc.m" and then if cont. the running process, it stuck at the "x=disteu(......)" (line 25, "vector.m"). Can any one pls help with the code.
Phase one is main training code from where mfcc and vector code will be called.
댓글 수: 7
Walter Roberson
2018년 11월 24일
what is the error message at ms and what are the sizes of the variables on that line ?
Gaurav Sharma
2018년 11월 24일
편집: Gaurav Sharma
2018년 11월 24일
Walter Roberson
2018년 11월 24일
if (((distortion - t)/t) < e)
appears to be attempting to calculate a proportional error, but looks to be failing to take into account absolute value, that the range -e to +e should probably be accepted. Still, though, that would tend to break more often rather than less often.
Gaurav Sharma
2018년 11월 24일
Walter Roberson
2018년 11월 24일
You have
ms = MelFrequencyVector*abs(freqDomMat(1:nby2)).^2; % mel spectrum
c = dct(log(ms)); % mel-frequency cepstrum coefficients
c(1,:) = [] ; % exclude 0'th order cepstral coefficient
Your ms is a row vector, 1 x 501. dct of a row vector gives you a row vector. You then ask to delete row 1 of the row vector, which is all that exists in the row vector, so your c is going to come out empty.
Gaurav Sharma
2018년 11월 24일
Walter Roberson
2018년 11월 24일
In my test, if the data you pass as the first parameter of vector() has 1 or 2 columns, then the way you handle the means makes it certain that t will become 0. When it does then
(distortion - t)/t
has a division by 0 and becomes infinite, which is not less than e. This leads to 0 being assigned to distortion, and t becomes 0 again, leading to (0-0)/0 as the next test, which is NaN, which again is not less than e. You are then stuck in an infinite loop.
In the case of 3 or more columns then this problem is not certain to happen, and vector can potentially exit.
답변 (0개)
카테고리
도움말 센터 및 File Exchange에서 Deep Learning Toolbox에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!