由H(=)=(5+3=3)(1+=1+=2) 得 k H()=H(=)2x2=(5+32)1+e 即H(0)=24()=2-23H(2)=0 H(3)=2H(4)=0 H(5)=2+23 则H(2) H(0 24 1-0.9z H H(3) 1+0.9z( ) 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