I want to plot the control signal (u) when we solve the equation with ode45. how I can plot control signal (u) in the following code.
tspan = 0:1:10000;
y0 = [0.2,0.3];
[t,y] = ode45(@(t,y) odefcn(t,y), tspan, y0);
%
function dx = odefcn(t,x)
dx = zeros(2,1);
u=[2 -2]*x(1:2)+1;
dx(1)=x(1)-2*x(2)
dx(2)=-x(2)+u
end

 Réponse acceptée

Create a second output for ‘u’ (ode45 will ignore it during the integration) and then return it using a for loop after the integration finishes.
Try this —
tspan = 0:0.001:15;
y0 = [0.2,0.3];
[t,y] = ode45(@(t,y) odefcn(t,y), tspan, y0);
for k = 1:numel(t)
[dx,u(k)] = odefcn(t(k),y(k,:).');
end
figure
plot(t,y(:,1), 'DisplayName','y_1(t)')
hold on
plot(t,y(:,2), 'DisplayName','y_2(t)')
plot(t,u, 'DisplayName','u(t)')
hold off
grid
legend('Location','best')
% set(gca,'XScale','log')
function [dx,u] = odefcn(t,x)
dx = zeros(2,1);
u=[2 -2]*x(1:2)+1;
dx(1)=x(1)-2*x(2);
dx(2)=-x(2)+u;
end
I changed ‘tspan’ because with a long vector, the details of the initial transient disappear from the plot.
.

1 commentaire

Dyuman Joshi
Dyuman Joshi le 12 Oct 2023
Nice answer, @Star Strider.
Just a small suggestion to preallocate u and negate the output dx if it is needed.

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Plus de réponses (1)

Your initial query has already been addressed. To achieve step profile tracking using your designed gain matrix, you should multiply the reference input (xref) by a scaling factor (sf). In doing so, the blue curve will consistently attain its steady-state position .
tspan = 0:0.001:10;
x0 = [0.2, 0.3];
[t, x] = ode45(@odefcn, tspan, x0);
% Computing the control signal u from the ode solution
u = [2 -2]*x.' + (-0.5)*(1); % sf = -0.5; xref = 1;
plot(t, x(:,1), 'DisplayName', 'x_1(t)'), hold on
plot(t, x(:,2), 'DisplayName', 'x_2(t)')
plot(t, u, 'DisplayName', 'u(t)'), hold off, grid on
xlabel('Time, (seconds)')
title('Step Response')
legend('location', 'SE', 'fontsize', 12)
% Dynamics
function [dx, u] = odefcn(t, x)
dx = zeros(2,1);
xref = 1; % reference input
sf = -0.5; % scaling factor
u = [2 -2]*x + sf*xref; % control signal
dx(1) = x(1) - 2*x(2); % x'
dx(2) = - x(2) + u; % x"
end

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