(b)超前定理 ZIX(t+kT I=Z[X(2-2X(nToZI 证明:Xt+=∑X(nT+kT)z =X(kT)+X(k+1)Tz+X(k+2)T0z+……+X(nT+k)z”+ =z-[X(kT)Z+X(k+1)T01z#)+… =z{X(0)+X(T)Z+……+X(k-1)T2)+X(kT0)Z+X(K+1)T0Z+) X(0)-X(0)Z-1-……-X(k-1)TZ--)B IX(Z)-∑X(mT)z="1 k=1时 Z1X(t+2T0)=2X(Z)-ZX(0) k=2时 ZIX(t+2T=Z X(2-Z X(O)-ZX(To 当k=m时 ZIX(t+mTo)l=Z X(Z-Z X(0)-Z-XTo)-Z X(2T.o) XI(m-1)ToI....... [( 1) ] k m [ ( )] ( ) (0) ( ) (2 ) ......... 2 [ ( 2 )] ( ) (0) ( ) 1 [ ( 2 )] ( ) (0) [ ( ) ( ) ] (0) ( ) ...... [( 1) ] ]} { (0) ( ) ...... [( 1) ] ( ) [( 1) ] [ ( ) [( 1) ] ......] ( ) [( 1) ] [( 2) ] ....... ( ) ...... : Z[X(t kT )] ( ) 0 0 2 0 1 0 0 2 2 0 0 1 0 0 ( 1) 0 1 0 ( 1) ...... 0 0 ( 1) 0 1 0 ( ! ) 0 0 0 0 2 0 1 0 0 0 0 0 0 ZX m T Z X t mT Z X Z Z X Z X T Z X T k Z X t T Z X Z Z X ZX T k Z X t T ZX Z ZX Z X Z X nT Z X X T Z X k T Z Z X X T Z X k T Z X kT Z X K T Z Z X kT Z X k T Z X kT X k T Z X k T Z X nT kT Z X nT kT Z m m m m k n k n k k k k k k k k n n n − − − = + = − − − = + = − − = + = − = − − − − − − = + + + − + + + = + + + = + + + + + + + + + = + − − − = − − − − − − − − − + + − − − + − − − −  =   当 时 时 时 证 明 [ ( )] [X(Z) - X(nT ) ] 0 K-1 n 0 T0 Z Z k n Z X t k − = + =  (b)超前定理