例3-5
num=5*[1 5 6];den=[1 6 10 8];
%函数阶跃响应
sys0=tf(num,den)
step(sys0)
%零极点形式的传递函数
[z p k]=tf2zp(num,den);
sys=zpk(z,p,k)
%部分分式展开
[r p k]=residue(num,[den 0]);
%拉普拉斯反变换
syms s
f=ilaplace(sum(r./(s-p)));
temp=simple(f);
disp('时域方程')
temp
sys0 =
5 s^2 + 25 s + 30
----------------------
s^3 + 6 s^2 + 10 s + 8
Continuous-time transfer function.
sys =
5 (s+3) (s+2)
--------------------
(s+4) (s^2 + 2s + 2)
Continuous-time zero/pole/gain model.
时域方程
temp =
(exp(-t)*sin(t))/2 - (7*exp(-t)*cos(t))/2 - exp(-4*t)/4 + 15/4
