第4意快速傅里叶夜换(H 用同样的方法可计算出 X2(k)=X5(k)+WN2X6(k) k=0,1,…N/4-1(42.1) X2(k+N/4)=X(k 其中 N/4-1 Xs()=2x(OWN/4=DFTLxs(D N/4-1 X6(k) 0 xpk=DFT[() N/4 x5()=x2(2) 1=0.1.…N/4-1 x6()=x2(2/+1)第4章 快速傅里叶变换(FFT) 用同样的方法可计算出 2 5 / 2 6 2 5 / 2 6 ( ) ( ) ( ) , 0,1, / 4 1 ( / 4) ( ) k N k N X k X k W X k k N X k N X k W X k = +   =  − + = −  (4.2.11) 其中 / 4 1 5 5 / 4 5 0 / 4 1 6 6 / 4 6 0 5 2 6 2 ( ) ( ) [ ( )] ( ) ( ) [ ( )] ( ) (2 ) , 0,1, / 4 1 ( ) (2 1) N kl N i N kl N i X k x l W DFT x l X k x l W DFT x l x l x l l N x l x l − = − = = = = = =   =  − = +   