not enough input arguments
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hi, I'm trying to write a code about cstr simulation and my loop cant continue due to this error, I'm sure it has what it needs for just first iteration but it does not calculate after the k1 part, can you help me please.
clear; clc;
Ts = 0.1;
%t = [1:0.1:1000];
Da1 = 3;
Da2 = 0.5;
Da3 = 1;
x1(1) = 0.5;
x2(1) = 0.5;
x3(1) = 1.5;
u = 0.45;
d2 = 1;
for n=1:1000
k1x1(n) = Ts*f1_func(x1(n),x2(n),Da1,Da2)
k1x2(n) = Ts*f2_func(x1(n),x2(n),Da1,Da2,Da3,d2,u)
k1x3(n) = Ts*f3_func(x2(n),x3(n),Da3,d2)
k2x1(n) = Ts*f1_func(x1(n)+k1x1(n)/2,x2(n)+k1x2(n)/2,u);
k2x2(n) = Ts*f2_func(x1(n)+k1x1(n)/2,x2(n)+k1x2(n)/2,u);
k2x3(n) = Ts*f3_func(x1(n)+k1x1(n)/2,x2(n)+k1x2(n)/2,x3(n)+k1x3(n)/2);
k3x1(n) = Ts*f1_func(x1(n)+k2x1(n)/2,x2(n)+k2x2(n)/2,u);
k3x2(n) = Ts*f2_func(x1(n)+k2x1(n)/2,x2(n)+k2x2(n)/2,u);
%k3x3
k4x1(n) = Ts*f1_func(x1(n)+k3x1(n),x2(n)+k3x2(n),u);
k4x2(n) = Ts*f2_func(x1(n)+k3x1(n),x2(n)+k3x2(n),u);
%k4x3
x1(n+1) = x1(n)+1/6*(k1x1(n)+2*k2x1(n)+2*k3x1(n)+k4x1(n));
x2(n+1) = x2(n)+1/6*(k1x2(n)+2*k2x2(n)+2*k3x2(n)+k4x2(n));
%x3
end
采纳的回答
Walter Roberson
2024-1-7
You declare f1_func as expecting f1_func(x1,x2,Da1,Da2) but starting from
k2x1(n) = Ts*f1_func(x1(n)+k1x1(n)/2,x2(n)+k1x2(n)/2,u);
you only pass in three parameters.
Also
k1x2(n) = Ts*f2_func(x1(n),x2(n),Da1,Da2,Da3,d2,u)
you are passing in 7 parameters there, but
k2x2(n) = Ts*f2_func(x1(n)+k1x1(n)/2,x2(n)+k1x2(n)/2,u);
you are passing in 3 parameters there.
9 个评论
Halil Ibrahim
2024-1-7
Sam Chak
2024-1-7
If the dynamics (ordinary differential equations) of the Continuous Stirred-Tank Reactor (CSTR) are given, would you like to run the simulation using the built-in ode45() solver? This way, you can focus on ensuring that the CSTR equations and parameters are designed or described correctly.
The code snippet is as follows:
%% Simulation settings and call ode45 solver
tspan = 0:0.1:1000; % simulation time from 0 to 1000
x10 = 0.5;
x20 = 0.5;
x30 = 1.5;
x0 = [x10; x20; x30]; % initial values of states x1, x2, x3
[t, x] = ode45(@cstr, tspan, x0);
%% Plot results
plot(t, x), grid, xlabel Time
%% Continuous Stirred-Tank Reactor
function dxdt = cstr(t, x)
dxdt = zeros(3, 1);
% Definitions of the states
cA = x(1); % concentration of A
cB = x(2); % concentration of B
cC = x(3); % concentration of C
% Constants
fAi = ...; % molar flow rate inlet of species A
fAo = ...; % molar flow rate outlet of species A
vA = ...; % stoichiometric coefficient
rA = ...; % reaction rate
% Parameters that depend on the constants or states (if any)
p1 = ...;
p2 = ...;
p3 = ...;
% Differential equations
dxdt(1) = ...;
dxdt(2) = ...;
dxdt(3) = ...;
end
Walter Roberson
2024-1-7
when i input (x1(n)+k1x1(n)/2,x2(n)+k1x2(n)/2,u), is it directly replacing (x1) with (x1(n)+k1x1(n)/2), x2 with (x2(n)+k1x2(n)/2) and so on ?
Yes. The replacements are positional. Whatever is passed first will be associated with the first variable in the function() definition; whatever is passed second will be associated with the second variable in the function() definition, and so on
Halil Ibrahim
2024-1-7
编辑：Halil Ibrahim
2024-1-7
I'm unfamiliar with your version of the Runge–Kutta 4th-order (RK4) Solver. The RK4 method is typically covered in the Numerical Methods course. In my Numerical Methods course, I solved Mass-Spring-Damper, Pendulum motion, and RLC electrical circuits.
Considering it's merely a homework assignment, I believe this might be too challenging for both of us. Your professor has instructed you to implement a custom RK4 algorithm to solve the CSTR dynamics, (possibly taught at the undergraduate level in Chemical Engineering). Additionally, you are expected to use an Adaptive PID controller, a topic usually encountered at the postgraduate level.
Moreover, you may find it challenging to assess the accuracy of your simulation results obtained with the custom RK4 algorithm unless you perform a comparison and validation. If you opt for the built-in ode45() solver, the code may look something like the following:
%% Simulation settings and call ode45 solver
tspan = 0:0.01:20; % simulation time from 0 to 20
x10 = 0.5;
x20 = 0.5;
x30 = 1.5;
x0 = [x10; x20; x30]; % initial values of states x1, x2, x3
[t, x] = ode45(@cstr, tspan, x0);
%% Plot results
plot(t, x), grid, xlabel Time
legend('x_{1}', 'x_{2}', 'x_{3}')
%% Continuous Stirred-Tank Reactor
function dxdt = cstr(t, x)
dxdt = zeros(3, 1);
% Parameters
d2 = 1; % this value is hidden somewhere
u2 = 0; % maybe the Adaptive PID controller
% Differential equations
dxdt(1) = 1 - x(1) - 3*x(1) + 0.5*x(2)*x(2);
dxdt(2) = - x(2) + 3*x(1) - 0.5*x(2)*x(2) - 1*d2*x(2)*x(2) + u2;
dxdt(3) = - x(3) + 1*d2*x(2)*x(2);
end
Halil Ibrahim
2024-1-7
I'm glad that custom RK4 for Adaptive PID Controller works out for you. However, when I ran the code, it threw the "Index exceeds" error message.
Could you please update the code? This will make your post more meaningful, benefiting users who wish to work with the coding-based Adaptive PID Controller.
clear; clc; close all;
% Initialization of Adaptive PID Control gains
Kp0 = 0.75;
Kp(3) = Kp0;
Ki0 = 0.1;
Ki(3) = Ki0;
Kd0 = 0.25;
Kd(3) = Kd0;
% Control signal saturation parameters
umax = 1;
umin = 0;
u(2) = 0.1;
% CSTR system parameters
Ts = 0.1;
Da1 = 3;
Da2 = 0.5;
Da3 = 1;
x1(1) = 0.5;
x2(1) = 0.5;
x3(1) = 1.5;
d2 = 1;
for n = 3:100
% Compute error
e(n-2) = x1(n-2) - x3(n-2);
% Error-based Adaptive PID Control Signal
u(n) = u(n-1) + Kp(n)*(e(n) - e(n-1)) + Ki(n)*e(n) + Kd(n)*(e(n) - 2*e(n-1) + e(n-2));
% Saturate the Control Signal
if u(n) > umax
u(n) = umax;
elseif u(n) < umin
u(n) = umin;
end
% Apply Adaptive PID Control Signal to CSTR via custom RK4 algorithm
k1x1 = Ts*f1_func(x1(n), x2(n));
k1x2 = Ts*f2_func(x1(n), x2(n), d2, u(n));
k1x3 = Ts*f3_func(x2(n), x3(n), d2);
k2x1 = Ts*f1_func(x1(n) + k1x1/2, x2(n) + k1x2/2);
k2x2 = Ts*f2_func(x1(n) + k1x1/2, x2(n) + k1x2/2, d2, u(n));
k2x3 = Ts*f3_func(x2(n) + k1x2/2, x3(n) + k1x3/2, d2);
k3x1 = Ts*f1_func(x1(n) + k2x1/2, x2(n) + k2x2/2);
k3x2 = Ts*f2_func(x1(n) + k2x1/2, x2(n) + k2x2/2, d2, u(n));
k3x3 = Ts*f3_func(x2(n) + k2x2/2, x3(n) + k2x3/2, d2);
k4x1 = Ts*f1_func(x1(n) + k3x1, x2(n) + k3x2);
k4x2 = Ts*f2_func(x1(n) + k3x1, x2(n) + k3x2, d2, u(n));
k4x3 = Ts*f3_func(x2(n) + k3x2, x3(n) + k3x3, d2);
x1(n + 1) = x1(n) + 1/6 * (k1x1 + 2*k2x1 + 2*k3x1 + k4x1);
x2(n + 1) = x2(n) + 1/6 * (k1x2 + 2*k2x2 + 2*k3x2 + k4x2);
x3(n + 1) = x3(n) + 1/6 * (k1x3 + 2*k2x3 + 2*k3x3 + k4x3);
% Update PID gains by adapting to the Error signal
Kp(n+1) = Kp0 + e(n);
Ki(n+1) = Ki0 + e(n);
Kd(n+1) = Kd0 + e(n);
% Plotting
figure(1);
subplot(3,1,1);
plot(n, x1(n), '-o');
title('x1');
hold on;
subplot(3,1,2);
plot(n, x2(n), '-o');
title('x2');
hold on;
subplot(3,1,3);
plot(n, x3(n), '-o');
title('x3');
hold on;
end
%% Local functions
% 1st CSTR ODE
function dx1 = f1_func(x1, x2)
t = 1;
dx1 = 1 - x1(t) - 3*x1(t) + 0.5*x2(t)*x2(t);
end
% 2nd CSTR ODE
function dx2 = f2_func(x1, x2, d2, u2)
t = 1;
dx2 = -x2(t) + 3*x1(t) - 0.5*x2(t)*x2(t) - 1*d2(t)*x2(t)*x2(t) + u2;
end
% 3rd CSTR ODE
function dx3 = f3_func(x2, x3, d2)
t = 1;
dx3 = -x3(t) + 1*d2(t)*x2(t)*x2(t);
end
Halil Ibrahim
2024-1-8
Sam Chak
2024-1-8
I understand that the simulation code for the Adaptive PID controller is not completely fixed. However, I am not familiar with the RK4 version you coded. Now that the issue of passing extra parameters mentioned in your initial question has been resolved, it might be helpful to create a new question specifically addressing the "Index exceeds..." issue.
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