solving state equation by ODE 45 by user defined time step and also checking response at each timestep

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Kamal Bera 2016-1-27

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评论： Star Strider 2019-10-8

Hi I want to calculate response x (fixed time step=0.01 sec) by solving state equation using ODE45: xdot=A.x+B.u --> A,B,x0(initial value of x) & u are known to me, but I will decide (or modify) the input u at (i+1)-th time based on the response I am getting at i-th time step. If I use ODE45 directly then I will get the final responses at each time and I will not be able to modify u accordingly. May be some loop is required to do that but HOW!!! Please help me.Thanks in advance.

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Star Strider 2016-1-27

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The MATLAB ODE solvers return the current value of your independent variable (here, time) and the current values of the integrated function. If you have a time-dependent input, you can use the time value to change the input. If you are making small adjustments to ‘u’, this should not be a problem.

Note that the ODE solvers don’t deal well with significant discontinuities, so consider stopping the integration and re-starting it with new values for ‘u’ (using the previous final values of ‘x’ as the new initial conditions) if the discontinuities are significant.

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Star Strider 2016-1-29

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I appreciate your having Accepted my Answer. However I cannot write the exact code for you, in part because if you are doing control theory, you need to learn how to program control systems.

The ‘T’ value is a constant you define that defines the ‘tspan’ vector, and therefore the end of the next time increment. It ensures that all the time increments are the same length and duration. I should have included it in my code outline, so including it now:

tspan = linspace(0, t_end, 50);                 % Initial ‘tspan’ Vector
T = 50;                                         % Constant Chosen To Define The End Time Of Each ‘tspan’ Value
y0 = . . .;
for i=1:50  % for 50-element fixed-step time vector
    [t(:,i),y(:,:,i)]=ode45(@FN,tspan,y0);
    y0=y(end,:,i);
    tspan = linspace(t(end,i), t(end,i)+T, 50);
end

If you have a linear problem, you don’t have to use the ODE solvers. You can use the matrix exponential expm function, but you would have to use the solved version of your control equation. You control theory textbook should explain that in detail (in involves matrix Laplace transforms).

‘Now my actual problem is ydot=A*y+B*u; where u is time dependent.’ You solve this using the ‘y’ vector your ‘FN’ ODE function takes as input, doing the necessary A*y matrix multiplication (creating a column vector), then adding the column vector created by multiplying B*u. This creates the column vector ‘FN’ returns to ode45 (the usual solve for these problems).

For a time-dependent ‘u(t)’, you have to define it. I do not know what form it takes, so I can only say that if it is a continuous function, you can write it in your ‘FN’ function as an anonymous function. If it is a discontinuous step function, it is best to stop the integration, increment ‘u(t)’, and continue. To include an external input to ‘FN’, you would create it as for example:

FN = @(t,y,u) A*y + B*u;

Prototype code for your problem (that I tested to verify that it works):

A = [0.1 -0.2; 0.3 0.4];
B = [0.4; 0.5];
u = 3;
y0 = [0.1; 0.2];
FN = @(t,y,u) A*y + B*u;
tspan = linspace(0, 10, 20);
T = 10;
for i=1:10  % for 50-element fixed-step time vector
    u = i/100;
    [t(:,i),y(:,:,i)]=ode45(@(t,y) FN(t,y,u),tspan,y0);
    y0=y(end,:,i);
    tspan = linspace(t(end,i), t(end,i)+T, 20);
end
tv = reshape(t, [], 1);
yp = permute(y, [1 3 2]);
yv = reshape(yp, [], size(y,2));
figure(1)
plot(tv, yv)
grid

You will have to make the necessary changes in this code to adapt it to your control system.

Kamal Bera 2016-1-31

Please give a last try with the code ContRespNew.m that I have already sent to you on 30 Jan 2016 at 18:56. Change ' k' (that will eventually change the ' u' also) whatever you want.You will get same response each time. I think this is not correct. If correct, can you explain me!!!

Star Strider 2016-1-31

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If that’s the file with the ode45 call, I already looked at it, ran it, and commented on it. It appears to be correct to me.

The other problem was your (incorrect) use of the expm function that I also commented on in some detail.

It is quite possible to quickly converge on a similar-appearing solution. The only way you can determine if it is working correctly is to closely examine what it is doing over time, preferably by plotting the output of the plant and the control inputs over time.

You would have to save the control signals, perhaps as ‘uv’, by inserting this statement at the end of your loop (after defining the new value of ‘u’ in each iteration):

uv(:,i) = u;

Then in your plot, plot ‘uv’ against ‘t’, as well as the signals you are already plotting. I am certain you will see them change appropriately.

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