这里有一个持续时间传递函数(G(s))
,形式为:
G(s) = N(s)/D(s);
G(s) = (s^3+4s^2-s+1)/(s^5+2s^4+32s^3+14s^2-4s+50) (1)
和(s = j*w)
其中w = frequency symbol.
现在,如何分解分子和分母
方程的多项式(1)进入偶数和奇数部分并得到G(jw)
为(使用Matlab):
答案 0 :(得分:6)
您可以在用s=j*w
替换后使用实部和虚部。但是,您实际上可以选择多项式的偶数和奇数部分:
% G(s) = N(s)/D(s);
syms s;
N = s^3+4*s^2-s+1;
p = sym2poly(N);
%// do this in fewer lines:
%{
/*
if mod(length(p),2)==0 %// then first index is odd
imin_o = 1; %// for odd part
imin_e = 2; %// for even part
else
imin_o = 2; %// for odd part
imin_e = 1; %// for even part
end
*/
%}
imin_o = mod(length(p),2) + 1;
imin_e = 2 - mod(length(p),2);
% odd part of numerator
p_o = zeros(size(p));
p_o(imin_o:2:end) = p(imin_o:2:end);
% even part of numerator
p_e = zeros(size(p));
p_e(imin_e:2:end) = p(imin_e:2:end);
% restore
N_o = poly2sym(p_o,s);
N_e = poly2sym(p_e,s);
和分母一样。