Integration of a vector inside function for ode45
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I am trying to solve a set of differential equations using ode45 command. The problem is I want to integrate one of the variables of the differential equations. For that I need to save the values/solution of that variable in a vector inside the function. I am unable to store x(1) as a vector. In the code, I need to integrate x(1) for which I have used cumtrapz but its not helping . My differential equations arre x1dot=x2 ;x2dot=-9.81*sin(x1)+u where dot means diffeentiation with respect to time. For u, e, ei and de are as defined in the code. Kindly help.
function dxdt=slmc(t,x)
dx1dt=x(2);
e=-x(1);
de=-x(2);
ei=0.01*cumtrapz(e);
u=75*e+5*ei+25*de;
dx2dt=-9.81*sin(x(1,:))+u;
dxdt=[dx1dt;dx2dt];
end
t=0:0.01:10;
in=pi/2;
indxdt=0;
[t,x]= ode45(@(t,x) slmc(t,x),t,[in indxdt]);
plot(t,x(:,1))
采纳的回答
In the code below, x(3) = integral_{tau=0}^{tau=t} -0.01*x(1) dtau
t=0:0.01:10;
in=pi/2;
indxdt=0;
[t,x]= ode45(@(t,x) slmc(t,x),t,[in indxdt 0]);
plot(t,x(:,3))
function dxdt=slmc(t,x)
dx1dt=x(2);
e=-x(1);
de=-x(2);
ei = x(3);
u=75*e+5*ei+25*de;
dx2dt=-9.81*sin(x(1,:))+u;
dxdt=[dx1dt;dx2dt;-0.01*x(1)];
end
11 个评论
Paul
2022-9-7
Should dxdt(3) be +0.01*x(1) ?
Torsten
2022-9-7
I think -0.01*x(1) should be integrated to get ei due to the OP's setting
ei=0.01*cumtrapz(-x(1));
, shouldn't it ?
Paul
2022-9-7
Correct. My mistake.
SM
2022-9-8
SM
2022-9-8
Torsten
2022-9-8
Because out of interest, I chose x(3) to be plotted.
Be uninhibited - it's your code. Just change x(:,3) to x(:,1).
I want to integrate e=-x(1) with a time step of 0.01. Could you clarify why another variable x(3) needs to be introduced for this?
You want to integrate e = -x(1). Thus you want to get
E(t) = integral_{tau=0}^{tau=t} e(tau) dtau
If you differentiate E(t) with respect to t, you get (by the fundamental theorem of calculus)
dE/dt = e(t) = -x(1).
So you have to solve a third differential equation
dx(3)/dt = -x(1)
where E is set to x(3).
Since in your code, you chose ei = 0.01*cumtrapz(-x(1)), I integrated -0.01*x(1) instead of -x(1).
But I'm sure that after my explanation, you will be able to change this easily on your own (although it's not a mistake).
SM
2022-9-9
I don't have the time to look for what pid, ss and feedback do.
But I can list the ODE system you are trying to solve. Maybe you can find the mistake therein.
You solve
dx1/dt = -10*x1 - 20*x2 - 50*x1 - 30*(integral_{tau=0}^{tau=t} x1(tau) dtau) - 16*x2
dx2/dt = x1
with initial conditions
x1(0) = pi/2
x2(0) = 0
My guess is that it is incorrect.
Paul
2022-9-9
The initial() command is plotting the IC response of the plant, not the closed loop system.
The closed loop system formed by G and c1 is not the same as that implemented in test_pid2. In the latter, the control input is
u=50*e+30*ei+16*de = -(50*x1 + 30*intx1 + 16*x2) % that 16*x2 was probably meant to be 16*x1dot
whereas in the former the control is
u = -(50*y + 30*inty + 16*ydot) = -(50*x2 + 30*intx2 + 16*x2dot)
SM
2022-9-10
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