使用residuez的时候,b和a是按照z^-1升幂排列的多项式,
F(z)=z(2.5z-0.9)/[(z-0.6)(z-0.3)]=(2.5-0.9z^-1)/(1-0.9z^-1+0.18z^-2)中,b=[2.5,-0.9]代表2.5-0.9*z^-1,后面加几个0都无所谓滴啦。
F(z)=(2.5z-0.9)/[(z-0.6)(z-0.3)]中,b=[0,2.5,-0.9]; 计算如下:
>> a=[1,-0.9,0.18];b=[2.5,-0.9];[r,p,c]=residuez(b,a)
r =
2.0000
0.5000
p =
0.6000
0.3000
c =
[]
>> a=[1,-0.9,0.18];b2=[0,2.5,-0.9];[r,p,c]=residuez(b2,a)
r =
3.3333
1.6667
p =
0.6000
0.3000
c =
-5
结果并不相同。
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