State space modeling of LTI

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Muhammad Atif 2020-4-8

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https://ww2.mathworks.cn/matlabcentral/answers/516376-state-space-modeling-of-lti

评论： Ameer Hamza 2020-4-20

I have a LTI system is given in the attached picture. For which I have A=[0 1;1 2] B=[1;1], C=[3,4], D=0.1 and L=[0 1]. I need to find z(t) over an interval [0,20] and initial condition is assumed as x0=[1;-1] with an input w=0.5sin(2pi)t and v=t, how can I make a code for this??

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Answer 1

Ameer Hamza 2020-4-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/516376-state-space-modeling-of-lti#answer_424815

编辑：Ameer Hamza 2020-4-8

在 MATLAB Online 中打开

Such LTI system can be solved using lsim

C=[3,4]; 
D=0.1;
L=[0 1];
sys = ss(A,B,C,D);
t = linspace(0,20,1000)';
x0 = [1; -1];
[t,x] = ode45(@odeFun, t, x0);
v = t;
y = C*x' + D*v';
z = L*x';
plot(t,z)
function dxdt = odeFun(t, x)
    A=[0 1;1 2];
    B=[1;1];
    w = 0.5*sin(2*pi*t);
    
    dxdt = A*x+B*w;
end

However, the system is unstable and the output diverges to infinity.

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Muhammad Atif 2020-4-11

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Hi Sir, I modified the code that you created above, In this I added the input w(t) and noise signal v(t). The plot of (z=L*v' ) should start when input starts to take effects. I attached two images for references. This plot I obtained from this code, while I wanna start it from t=5. How can I change this?

%%function to get dxdt
function dxdt = odeFun(t, x)
ms = 973;
ks = 42720;
cs = 3000;
ku = 101115;
mu = 114;
G = 512*10^(-6)
q = 0.1;
v1 = 12.5;
A = [0 0 1 -1 ; 0 0 0 1 ; -ks/ms 0 -cs/ms  cs/ms; ks/mu -ku/mu cs/mu -cs/mu];
B = [0 0 -2*pi*q*(G*v1)^(1/2) 0]';
a = 0.2 ;
l = 5 ;
%%input w(t)=1 for 5<=t<=10 and w(t)=-1 for 15<=t<=20
w_int1 = t>=5 & t<=10; 
w_int2 = t>=15 & t<=20;
w = zeros(1,numel(t));%add this
rng(0);
w(w_int1)=1;
w(w_int2)=-1;
dxdt = A*x+B*w';
end
C = [0 0 0 1];
D = 0.1;
L = [0 0 1 0];
t = linspace(0,30,1000)';
x0 = [0.1; 0; -0.1; -0.2];
[t,x] = ode45(@odeFun, t, x0);
T = linspace(0,20,1000)';
%%noise v(t) distributed over an interval [0,20]
v_int = t>=0 & t<=20; 
v = zeros(1,numel(t)); 
rng(0);
v(v_int)= 0.2*rand(1,length(t(v_int)));
y = C*x' + D*v;
z = L*x';
plot(t,z)

Muhammad Atif 2020-4-18

在 MATLAB Online 中打开

Sorry to bother you again. Now I have no idea how to use the the state of xf0 for the other system. It's a sort of switching condition in which two subsystems. When 't' reaches the defined interval the system one will run, otherwise the 2nd system will run. But how can I use the state xf0 of the first system for system 2 when system 1 run for the last time??

clc
clear all
Af1 = [-2.8682 0.5240 0.1093 0.4576; 0.6676 -2.3337 -0.1285 -0.5333; 4.2978 1.3131 -13.1636 3.3939; -30.1511 55.6790 48.9362 -218.6780];
Bf1 = [0.0167; -0.0400; -0.1676; -3.6279];
Bf2 = [0;0;0;0];
Af2 = [-2.8682 0.5240 0.1093 0.4576; 0.6676 -2.3337 -0.1285 -0.5333; 4.2978 1.3131 -13.1636 3.3939; -30.1511 55.6790 48.9362 -218.6780];
Cf1 = [1.6042 1.1359 -0.7995 -0.1854];
Cf2 = [0.9791 0.7264 -0.6448 -0.0632];
Df1=0;
Df2=0;
C = [0 0 0 1];
D = 0.1;
L = [0 0 1 0];
t = linspace(0,30,1000);
x0 = [0; 0; -0.00; -0.00];
[t,x] = ode45(@odeFun12, t, x0);
%%noise v(t) distributed over an interval [0,20]
v = 0.05*rand(size(t));
v = v.*(t<20);
y = C*x' + D*v';
z = L*x';
%plot(t,y)
%grid on
%%system 1
if t<=1 | (t>=2 & t<=7) | (t>=12 & t<=17) | (t>=22 & t<=27)
sys = ss(Af1,Bf1,Cf1,Df1);
  xf0 = [0; 0; 0.00; 0.00];
  lsim(sys,y,t,xf0);
%%system 2 I need to use xf0 of the 1st system when program switches to system 2 
else
    sys1 = ss(Af2,Bf2,Cf2,Df2);
    y1=zeros(size(t));
    lsim(sys1,y1,t,xf0);
end

Ameer Hamza 2020-4-18

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I guess something like this will work

Cf1 = [1.6042 1.1359 -0.7995 -0.1854];
Cf2 = [0.9791 0.7264 -0.6448 -0.0632];
Df1=0;
Df2=0;
t = linspace(0,30,1000);
x0 = [0; 0; -0.00; -0.00];
[t,x] = ode45(@odeFun12, t, x0);
%%noise v(t) distributed over an interval [0,20]
mask = t<=1 | (t>=2 & t<=7) | (t>=12 & t<=17) | (t>=22 & t<=27);
y(mask) = Cf1*x(mask,:).';
y(~mask) = Cf2*x(~mask,:).';
function dxdt = odeFun12(t, x)
Af1 = [-2.8682 0.5240 0.1093 0.4576; 0.6676 -2.3337 -0.1285 -0.5333; 4.2978 1.3131 -13.1636 3.3939; -30.1511 55.6790 48.9362 -218.6780];
Bf1 = [0.0167; -0.0400; -0.1676; -3.6279];
Bf2 = [0;0;0;0];
Af2 = [-2.8682 0.5240 0.1093 0.4576; 0.6676 -2.3337 -0.1285 -0.5333; 4.2978 1.3131 -13.1636 3.3939; -30.1511 55.6790 48.9362 -218.6780];
ms = 973;
ks = 42720;
cs = 3000;
ku = 101115;
mu = 114;
G = 512*10^(-6);
q = 0.1;
v1 = 12.5;
a = 0.2 ;
l = 25;
%%input w = a*pi*v/l.*sin(2*pi*v/l.*t) for 5<=t<=l/v and zero otherwise, while 'v' is uniformely distributed noise between[0,0.1] for an interval 0 to 20
w = a*pi*v1/l.*sin(2*pi*v1*t/l);
if  t<=1 || (t>=2 && t<=7) || (t>=12 && t<=17) || (t>=22 && t<=27)
    dxdt = Af1*x+Bf1*w';
else
    dxdt = Af2*x+Bf2*w';
end
end

Ameer Hamza 2020-4-19

在 MATLAB Online 中打开

I think something like this will work. You can extend it to the full time duration.

clc
clear all
Af1 = [-2.8682 0.5240 0.1093 0.4576; 0.6676 -2.3337 -0.1285 -0.5333; 4.2978 1.3131 -13.1636 3.3939; -30.1511 55.6790 48.9362 -218.6780];
Bf1 = [0.0167; -0.0400; -0.1676; -3.6279];
Bf2 = [0;0;0;0];
Af2 = [-2.8682 0.5240 0.1093 0.4576; 0.6676 -2.3337 -0.1285 -0.5333; 4.2978 1.3131 -13.1636 3.3939; -30.1511 55.6790 48.9362 -218.6780];
Cf1 = [1.6042 1.1359 -0.7995 -0.1854];
Cf2 = [0.9791 0.7264 -0.6448 -0.0632];
Df1=0;
Df2=0;
C = [0 0 0 1];
D = 0.1;
L = [0 0 1 0];
t = linspace(0,30,1000);
x0 = [0; 0; -0.00; -0.00];
[t,x] = ode45(@odeFun12, t, x0);
%%noise v(t) distributed over an interval [0,20]
v = 0.05*rand(size(t));
v = v.*(t<20);
y = C*x' + D*v';
z = L*x';
%plot(t,y)
%grid on
%%system 1
% if t<=1 | (t>=2 & t<=7) | (t>=12 & t<=17) | (t>=22 & t<=27)
sys1 = ss(Af1,Bf1,Cf1,Df1);
sys2 = ss(Af2,Bf2,Cf2,Df2);
xf0 = [0; 0; 0.00; 0.00];
t1 = t(t<=1);
y1 = y(t<=1);
[~,Y1,X1] = lsim(sys1,y1,t1,xf0);
t2 = t(1<t & t<=2);
y2 = 0*y(1<t & t<=2);
[~,Y2,X2] = lsim(sys2,y2,t2,X1(end,:));
t3 = t(2<t & t<=7);
y3 = y(2<t & t<=7);
[~,Y3,X3] = lsim(sys1,y3,t3,X2(end,:));
t4 = t(7<t & t<=12);
y4 = 0*y(7<t & t<=12);
[~,Y4,X4] = lsim(sys2,y4,t4,X3(end,:));
Y = [Y1;Y2;Y3;Y4];
X = [X1;X2;X3;X4];
function dxdt = odeFun12(t, x)
ms = 973;
ks = 42720;
cs = 3000;
ku = 101115;
mu = 114;
G = 512*10^(-6);
q = 0.1;
v1 = 12.5;
A = [0 0 1 -1; 0 0 0 1; -ks/ms 0 -cs/ms  cs/ms; ks/mu -ku/mu cs/mu -cs/mu];
B = [0 -2*pi*q*(G*v1)^(1/2) 0 0]';
a = 0.2 ;
l = 25;
%%input w = a*pi*v/l.*sin(2*pi*v/l.*t) for 5<=t<=l/v and zero otherwise, while 'v' is uniformely distributed noise between[0,0.1] for an interval 0 to 20
w = zeros(size(t));
mask = t>=5 & t<=5+l/v1;
w = a*pi*v1/l.*sin(2*pi*v1*t/l).*mask;
dxdt = A*x+B*w';
end

Muhammad Atif 2020-4-20

That's it, finally I got the results. Thank you very much for your help

Ameer Hamza 2020-4-20

I am glad to be of help.

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State space modeling of LTI

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