(d.x4(回)=xn+2)+xn-2) 解， (a.X,(2=z3Xz+z3X2=z31+2z+z31+2z=2z+z4+23+2z4 d d X()=1+2z-2(-2)z2-2 2-=12z+2+2z=1+6e @X()=X3)=1+22)=1+: (d.X,(e)=z2X(e)+22X(e)=(e2+221+2=2+2:+2+2z 题5-6.用MATLAB计算下列多项式运算的结果 (a).X121-2z+3z2.4z(4+3z-2z2+z (b).X2(z=(z- +Z)(ZZ) (⊙.X2H1+z+z (d).X(z)=Xi(z)X-(z)+X(z) (e).X,(2(z.3z3+2z545z7z9(亿+32+2z3+4z 解：(a.X1=conv([1-2,3414,3-2,1月4，-5,4-2，-20,11，-4 即X4-5z 202 (b).X2=conv1,-232,1]1,0,0,0,0,0-1F1,-2,3,2,1,0-1,2,-3,-2,-1] X:(2)=222+32+22+zz+2z23z2z4z5 (c.X3=conv([1.1.11.conv1.1.11.1.1.1)=1.3.6.7.6.3.1l 即X321+3z+6z2+7z3+6z4+325+26 (d).X4-polyadd(cc nv(X2.X3).Izeros(1.5).X3])- =,136,17,27,32,25,121,-15,-24-27-19,-9,2,0,其首项为z (e).X5=conv(1.0.-30,20.5.0.-11.I3.1.0.0.0.2.4]= =3.1,-9-36,4,19,1,-15,3,810,20,-2,-4小，其首项为z 题5-9.用部分分式展开法和长除法分别求下列z反变换的前6个样本，设序列是右序列 @Xe)=1--42+4 1+--+2-2+2 b).X、(=)==-3+2=2+1.25:+025) (e).X(e)=z2-0.25)}2 解：(a).xl=deconv([l,-l,4,4,zeros(1,5l,1,2,1F[l,-2,-4,1l,-l,-l7,8 先化为负琴形式X,(e)=:+2+125：+02)=1+2+1252+0252 -2 x2=deconv(0,0,1,zeros(1,8.l2,125,0.25] =[0,0,1.0000,-2.0000,2.7500,-3.2500,3.5625,-3.7500] (d ). x 4 (n) = x(n+2) +x(n-2) 解： (a). X1(z)=z 3 X(z-1)+z -3X(z)==z 3 (1+2 z)+z -3(1+2 z -1)=2z4 +z3 ++z -3+2z -4 (b). 2 ( ) ( ) ( ) ( ) dX z d dX z X z X z z z z dz dz dz   = − −   −   -1 2 1 -1 -1 -1 1 2 ( ) 1+2 z ( 2) 2 1+2 z 2z 2z 1 6 d X z z z z z dz z − − − = − ⋅ − − = + + = + (c). ( ) 1 1 3 ( ) ( ) 1 2 2 1 0.5 z X z X z z − − = = + = + (d). 2 2 2 2 1 2 2 4 ( ) ( ) ( ) ( )(1 2 ) 2 2 3 X z z X z z X z z z z z z z z − − − − = + = + + = + + + − 题5-6. 用MATLAB计算下列多项式运算的结果 (a). X1(z)=(1-2z-1+3z-2-4z-3) (4+3z-1-2z-2+z-3) (b). X2(z)=( z2 -2z+3+2z -1+z-2) (z 3-z -3) (c). X3(z)=(1+z -1+z-2) 3 (d). X4(z)= X1(z) X2(z) + X3(z) (e). X5(z)=( z -1-3z -3+2z -5+5z -7-z -9) (z +3z2 +2z-3 + 4z-4) 解：(a). X1=conv([1,-2,3,-4],[4,3,-2,1])=[ 4, -5, 4, -2, -20, 11, -4 ] 即X1(z)=4 –5z -1+ 4 z -2 -2 z -3 -20 z -4 11 z -5 -4 z -6 (b). X2=conv([1,-2,3,2,1],[1,0,0,0,0,0,-1])=[1,-2, 3, 2, 1, 0, -1, 2, -3, -2, -1] 即X2(z)=z5 -2 z4 +3 z3 + 2 z2 + z- z -1+2 z -2-3 z -3-2 z -4- z -5 (c). X3=conv([1,1,1],conv([1,1,1],[1,1,1]))= [1, 3, 6, 7, 6, 3, 1] 即X3(z)=1+3 z -1+ 6 z -2+ 7 z -3+ 6 z -4+3 z -5+ z –6 (d). X4=polyadd(conv(X2,X3),[zeros(1,5),X3])= =[1,1,3 ,6,17,27,32 ,25,12 ,-1,-15 ,-24,-27,-19, -9 ,-2, 0]，其首项为z5 (e). X5=conv([1,0,-3,0,2,0,5,0,-1],[3,1,0,0,0,2,4])= =[3, 1, -9, -3, 6, 4, 19, -1, -15, 3, 8, 10, 20, -2, -4]，其首项为z 题5-9. 用部分分式展开法和长除法分别求下列z反变换的前6个样本，设序列是右序列。 (a). 1 2 3 1 1 2 3 1 4 4 ( ) , 1 2 z z z X z z z z − − − − − − − − + = + + + (b). 3 2 2 X z( ) = + z/(z 2z +1.25z + 0.25) (c). 2 2 3 X z( ) = − z z /( 0.25) , 解：(a). x1=deconv([1,-1,-4,4,zeros(1,5)],[1,1,2,1])= [1, -2, -4, 11, -1, -17, 8] (b). 先化为负幂形式 2 3 2 2 1 2 ( ) /( 2 1.25 0.25) 1 2 1.25 0.25 z X z z z z z z z − 3 z − − − = + ++= + + + x2=deconv([0,0,1,zeros(1,8)],[1,2,1.25,0.25])= =[0, 0, 1.0000, -2.0000, 2.7500, -3.2500, 3.5625, -3.7500]