由1()=(5+3z)1+1+2） 得 =He-5+30+。- 即H(0)=24H(①=2-2N5j H(2)=0 H(3)=2H(4)=0 H(5)=2+237 则H(a)= H(0) 24 1-2可 1-0.9z1 H,(e) H(3) 2 1+ 1+0.9z1 2024/10/21( ) 2 3 3 5 3 1 j k j k j k e e e    − − −   = + + +     即 0 24 H ( ) = H j (1 2 2 3 ) = − H (2 0 ) = H (3 2 ) = H (4 0 ) = H j (5 2 2 3 ) = + 则 ( ) ( ) 0 1 1 0 24 1 1 0.9 H H z rz z − − = = − − ( ) ( ) 3 1 1 3 2 1 1 0.9 H H z rz z − − = = + + ( ) ( )( ) 3 1 2 5 3 1 H z z z z − − − 由 = + + + 得 ( ) ( ) H k H z = z k N =2 2024/10/21 11