wrong outcomes when using lsim and step function

Question

Eliraz Nahum 2018-12-7

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/434412-wrong-outcomes-when-using-lsim-and-step-function

hello everyone,

I am trying to define a 2nd order system y''(t)+5y'(t)+6y(t)=15u(t) in the state space and get outcomes (y) for different incomes (u).

the initial conditions are y(0)=y'(0)=0.

u=const=2;

the analytical solution is y=5 -15*exp(-2t) +10*exp(-3t), so it's clear that yss, the steady state output, is 5.

but when using the lsim or even step functions I am getting strange results (yss=0.33). any idea what's wrong?

the code is:

clear all
close all
clc
syms t y(t) %creating symbolic variable
%defining the ODE that describes the physical behaviour of the system
a=[6,5,1]; %creating a vector that represents the Input function derivatives coefficients - [a0,a1,...,an-1,...,a0] - 6y+5y'+y''
b=[15,0,0]; %creating a vector that represents the Output function derivatives coefficients - [b0,b1,...,bn-1,...,b0] - 15u+0u'+0u''
ode_order=length(a)-1; %determines the system order
%creating the matrices of the space state y''=-(a1/a2)*y'-(a0/a2)*y+(b0/a0)*u
%X=[x1,x2]=[y,y'] and X'=[x1',x2']=[y',y'']=[x2,-(a0/a2)*x1-(a1/a2)*x2+(b0/a0)*u]
    A=zeros(ode_order,ode_order);
        for k=[1:1:(ode_order-1)]
        A(k,k+1)=1;
        end
        for k=[1:1:ode_order]
        A(ode_order,k)=-(a(k)/a(ode_order+1));
        end
    B=zeros(ode_order,1);
    B(ode_order,1)=1;
    C=zeros(1,ode_order);
    C(1,1)=1;
    D=zeros(1,1);
end
%DEFINING OUR SYSTEM IN TERMS OF STATE SPACE
sys=ss(A,B,C,D);
t2=[0:0.01:15];
u2=zeros(1,length(t2));
for counter=1:1:length(t2)
u2(counter)=2;
end
lsim(sys,u2,t2,[0,0])