# Discrete state space find and plot

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Jack Reacher on 3 Dec 2016
Edited: bio lim on 3 Dec 2016
A = [0 1 0 0 0; -0.4 -1.3 0 0 0; 0 0 2 0 0; 0 0 0 0 -0.4; 0 0 0 1 -1.3];
B = [0; 1; 0; 0; 0];
C = [0 0 0 0 1];
x0 = transpose([0 0 0 0 0]); % initial condition
u = 1(k) % unit step
Find and plot:
x(k + 1) = A*x(k) + B*u(k)
How do I do this without using ss() and lsim(), and instead by using a for loop for 100 time units?
Thanks

bio lim on 3 Dec 2016
Edited: bio lim on 3 Dec 2016
Well, the nice thing about discrete time system is you can solve the discrete time equation with a loop.
clc;
clear all;
close all;
A = [0 1 0 0 0; -0.4 -1.3 0 0 0; 0 0 2 0 0; 0 0 0 0 -0.4; 0 0 0 1 -1.3];
B = [0; 1; 0; 0; 0];
C = [0 0 0 0 1];
x0 = transpose([0 0 0 0 0]); % initial condition
u = 1; % unit step
x(:,1) = A*x0 + B.*u;
% x(k + 1) = A*x(k) + B*u(k)
for k = 2:100
x(:,k) = A*x(:,k-1) + B*u;
end
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Jack Reacher on 3 Dec 2016
This is what I did to plot the 5 states. But, I'm not sure I got the right plots. -Thanks
clc;
clear;
close;
A = [0 1 0 0 0; -0.4 -1.3 0 0 0; 0 0 2 0 0; 0 0 0 0 -0.4; 0 0 0 1 -1.3];
B = [0; 1; 0; 0; 0];
C = [0 0 0 0 1];
x0 = transpose([0 0 0 0 0]);
u = 1;
x(:, 1) = A*x0 + B*u;
for k = 2: 100
x(:, k) = A*x(:, k - 1) + B*u;
end
t = 1: 100;
subplot(2, 3, 1);
plot(t, x(1, :))
subplot(2, 3, 2);
plot(t, x(2, :))
subplot(2, 3, 3);
plot(t, x(3, :))
subplot(2, 3, 4);
plot(t, x(4, :))
subplot(2, 3, 5);
plot(t, x(5, :))
bio lim on 3 Dec 2016
Edited: bio lim on 3 Dec 2016
You are plotting them right. I think, the last three states are not excited enough either due to initial condition, or your system dynamic configuration. Try changing the initial conditions, and see how it changes.