Non-Negative Solution with Integrator in Simulink
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When solving ode in Matlab, the ode solver has an option that allows setting the solution non-negative property to be true. For instance -
x = odeset;
x.NonNegative = ones(numStates,1);
Is there an equivalent setting in Simulink? Something that can be adjusted in either model parameters or the [1/s] block?
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採用された回答
Paul
2024 年 5 月 7 日
Hi Rajmohan,
Do the parameters Limit Output along with Upper and Lower Saturation Limit do what's needed? Integrator
3 件のコメント
Paul
2024 年 5 月 8 日
"The saturation modifies the output only, it does not limit the state value"
I'm quite sure that's incorrect. I ran the equivalent of this model in Simulink, with Limit Output checked and Lower Saturation Limit set to 0 in the Integrator parameters and got the same result.
[t,x]=ode45(@(t,x) -t.*sin(t),[0 30],1,odeset('MaxStep',0.01,'NonNegative',1));
figure
plot(t,x)
Sam Chak
2024 年 5 月 9 日
Hi @Rajmohan
Now that I've had the chance to test it in Simulink, I can confirm that @Paul's suggestion of setting the Lower Saturation Limit to 0 does indeed enforce the nonnegativity constraint and yield the correct solution. This option is also directly available in the Integrator Limited block. Here's a demo using the liquid level drainage system for reference.
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If we don't impose the nonnegativity constraint (either mathematically or technically), the Simulink solver will terminate at 2 seconds, and the ode45 solver will provide an incorrect solution.
ode = @(t, h) - sqrt(h);
[t, h] = ode45(ode, [0 10], 1);
plot(t, h), grid on, xlabel('t'), ylabel('h(t)')
title('Solution without nonnegativity constraint')
その他の回答 (1 件)
Sam Chak
2024 年 5 月 8 日
編集済み: Sam Chak
2024 年 5 月 8 日
Hi @Rajmohan
I'm not familiar with all types of settings, codes, and tools in MATLAB/Simulink. Therefore, I usually resolve numerical issues using my own mathematical tricks, whenever possible.
In this case, you can attempt to address the ODE by incorporating a signum function in the Simulink model. You may find the demonstration below helpful, which is obtained from the "Nonnegative ODE Solution" article. However, I have also devised my own solution without relying on the built-in 'NonNegative' option.
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ode = @(t, y) - abs(y);
% Analytic solution
tspan = [0 40];
t = linspace(tspan(1), tspan(2), 1001);
y = exp(-t);
%% Case 1: Standard solution with ode45
options1 = odeset('Refine',1);
[t1, y1] = ode45(ode, tspan, 1, options1);
%% Case 2: Solution with nonnegative constraint
options2 = odeset(options1, 'NonNegative', 1);
[t2, y2] = ode45(ode, tspan, 1, options2);
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%% Case 3: Solution with signum function
odefix = @(t, y) - abs(y)*sign(y);
[t3, y3] = ode45(odefix, tspan, 1, options1);
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%% Case 4: Solution with ReLU function
odeReLU = @(t, y) - abs((sqrt(y.^2) + y)/2); % ReLU(x) = (√(x²) + x)/2
[t4, y4] = ode45(odeReLU, tspan, 1, options1);
%% Plot results
plot(t, y, '-', t1, y1, 'o', t2, y2, '*', t3, y3, 'x', t4, y4, 'p'), grid on, ylim([-5 2])
xlabel('t'), ylabel('y(t)')
legend('Exact solution', 'No constraints', 'Nonnegativity', 'Signum Fix', 'ReLU Fix', 'Location', 'best')
1 件のコメント
Sam Chak
2024 年 5 月 8 日
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- The output is true (equal to 1) when the input signal is greater than or equal to zero, and its previous value was less than zero.
- The output is false (equal to 0) when the input signal is less than zero, or if the input signal is nonnegative, its previous value was also nonnegative.
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