第22回 ロボット聴覚特集 - 奥乃研究室 - 京都大学

k lo (l), k c (l), k hi (l) l k c (l) =k hi (l − 1) = k lo (l + 1) (4)k c (l) k c (l) Mel kc(l) {Mel kc(l) = 2595 log 10 1+ k }c(l)f s(5)1400(N − 1) f s N FFT 3.3 (6) Y (k) Z(k) (7) m(l) =k hi (l)∑k=k lo (l)W (k; l){|Y (k)| 2 − α(l) · β·|Z(k)| 2} 1 2(6)(if |Y(k)| 2 − α(l) · β·|Z(k)| 2 ≥ 0)k hi (l)∑m(l) = W (k; l) { γ ·|Y (k)| } ( otherwise ) (7)k=k l o (l)m(l) l β L γ α(l) 0 ( ) 2 mY (l)α(l) =m Z (l)⎛⎞2k hi (l)∑W (k; l)|Y (k)|k=k=l o (l) k⎜ hi (l)(8)∑⎟⎝ W (k; l)|Z(k)| ⎠k=k l o (l)m Y (l), m Z (l) SSA m(l) 24 SSA m(l) GJ 1( k l)W ;l −1 l l + 1・・・・ ・・・k c( l −1)k hi( l −1)k hi() lk lo() l k c() l k c( l +1)( l +1)k loFigure 3: SSA SSA MFCC (6) (7) m(l) √2L∑{ }{ } (l 1)iπMFCC(i) = logL e m(l) cos − (9)2 Ll=1i MFCC 3.4 [5]SSA NBF SSA NBF NBF SSA X(k) [x 1 (t), ,x J (t)] T NBF Y NBF (k) (10) 31Y NBF (k) =A T NBF(k)X(k) (10)A NBF (k) NBF A NBF (k) =[A NBF1 (k), ···,A NBFJ (k)] T= { [1, 0] · [e NBF (k, θ N ), e NBF (k, θ U )] +} T(11)θ N θ U e NBF (k, θ) (12) e NBF (k, θ) = [ exp(j2π k K f sd 1 sinθ/c)···, exp(j2π k K f sd J sinθ/c) ] T(12)c d 1 d J k