c n -M 0 M 2M X.e)立eaa啡立2m,a 00 M-1 n=-00 n=-o0k=0 22r*) M ∑X(zW) k=0 n=-00 =0 Using Eq.(10.8)to Eq.(10.11),we get1 int 0 1 () [ ][ ] ( )[ ] M n kn n M n n k X z cnxnz W xnz M ∞ ∞ − − − − =−∞ =−∞ = = ∑ ∑∑ = 1 1 0 0 1 1 ( [] ) ( ) M M kn n k M M k n k x n W z X zW M M − ∞ − − − = =−∞ = = ∑∑ ∑ = Using Eq.(10.8) to Eq. (10.11), we get -M 0 M 2M 1 c[n] n