Hello,
How are you? I am using MATLAB to work on a research project. Our lab previously created a code to generate a GUI, enabling us (non-coders) to use MATLAB and carry out the research. The code was working fine, but then I tried to copy the MATLAB file from the Documents folder to a different folder. It worked fine for one case but then stopped working. I am getting this error:
X.m Line:709 Column:1 Invalid use of operator.
We have checked the code and there is no issue, especially in line 709 and its surroundings. I have tried the following steps:
Putting the MATLAB folder back in Documents -> same error.
Uninstalling MATLAB from my Mac and reinstalling the same version -> same error.
Using a previous code for the GUI that has exactly the same code in line 709 -> works fine. (I can’t use this code as the latest one has functions that I need for my work.)
Seeing that the only thing that had changed was the directory, I am assuming it has something to do with that. I must note that even after changing the directory, it worked for one sample and then stopped working. I am really stuck and would appreciate anyone’s feedback.
Thank you and kind regards,
SouvikHello,
How are you? I am using MATLAB to work on a research project. Our lab previously created a code to generate a GUI, enabling us (non-coders) to use MATLAB and carry out the research. The code was working fine, but then I tried to copy the MATLAB file from the Documents folder to a different folder. It worked fine for one case but then stopped working. I am getting this error:
X.m Line:709 Column:1 Invalid use of operator.
We have checked the code and there is no issue, especially in line 709 and its surroundings. I have tried the following steps:
Putting the MATLAB folder back in Documents -> same error.
Uninstalling MATLAB from my Mac and reinstalling the same version -> same error.
Using a previous code for the GUI that has exactly the same code in line 709 -> works fine. (I can’t use this code as the latest one has functions that I need for my work.)
Seeing that the only thing that had changed was the directory, I am assuming it has something to do with that. I must note that even after changing the directory, it worked for one sample and then stopped working. I am really stuck and would appreciate anyone’s feedback.
Thank you and kind regards,
Souvik Hello,
How are you? I am using MATLAB to work on a research project. Our lab previously created a code to generate a GUI, enabling us (non-coders) to use MATLAB and carry out the research. The code was working fine, but then I tried to copy the MATLAB file from the Documents folder to a different folder. It worked fine for one case but then stopped working. I am getting this error:
X.m Line:709 Column:1 Invalid use of operator.
We have checked the code and there is no issue, especially in line 709 and its surroundings. I have tried the following steps:
Putting the MATLAB folder back in Documents -> same error.
Uninstalling MATLAB from my Mac and reinstalling the same version -> same error.
Using a previous code for the GUI that has exactly the same code in line 709 -> works fine. (I can’t use this code as the latest one has functions that I need for my work.)
Seeing that the only thing that had changed was the directory, I am assuming it has something to do with that. I must note that even after changing the directory, it worked for one sample and then stopped working. I am really stuck and would appreciate anyone’s feedback.
Thank you and kind regards,
Souvik invalid use of operator MATLAB Answers — New Questions

​

Hello everyone,
I am trying to solve some differential equation in matlab using ode45. The equations are as shown below:

I have written the equations in state space form where ydot_1(i) is , ydot_2(i) is , ydot_3(i) is and ydot_4(i) is . In the equation of motion of x, eq(2), x terms are not in terms of “ i “ but I have written them like that (so ydot_3(i) is and ydot_4(i) is )b ecause of the loop that I use on the function below. So the “for” loop, for each time step, lets say starting at solves first the ydot terms for each mode for i = 1:n and then gets out of the loop, goes to the new time step and then solve again for each mode. However, since for each time step it will start from the first mode again, and ydot_1(i) and ydot_2(i) depend on ydot_3(i) and ydot_4(i) , then I am solving the equation of x, eq.2, for each mode. For the new time step, before starting with the first mode again I am defining the
ydot_1=y(1:n); %displacement beam
ydot_2=y(n+1:2*n); %velocity beam
ydot_3=y(2*n+1:3*n); %displacement mass
ydot_4=y(3*n+1:4*n); %velocity mass
v_m=0;
x_m=0;
x_c_prev = x_c_prev_modes{1};

so it can start with the data from the previous first mode (first mode of the previous time step). My problem is that when I solve for 1 mode only, so for n=1 and get the solution du_1= y(:,n+i).*sin(i*pi.* x_c_values_new); and then solve for n=2 but still get the solution du_1= y(:,n+i).*sin(i*pi.* x_c_values_new); for the first mode only, these first modes do not match. The solution have the same shape but differ in amplitude. The same happens with I use n=20, for example, but again plot only the first mode du_1. The solution changes from the first mode obtained when n=2 and from the first mode obtained for n=1. I have checked for days and cant see my mistake so have reached desperation. Would really appreciate your help! thank youuu!

clear all
clc
global xi xi_b r_w r_m mu_s mu_k v alpha P l a w l_0 n x_c_prev x_c_prev_modes

dt=0.01;
tspan = (0:dt:150).’;

% parameters
xi=0.14;
xi_b=0.0025;
r_w=4.5;
r_m=4.8;
mu_s=0.5;
mu_k=0.2 ;
alpha=300;
P=0.002;
l=1/75;
l_0=0.25;
a=0;
w=9;

v=0.005;
% Initialize y_c_prev and the array to store x_c values
x_c_prev = 0;
global t_values x_c_values; % Declare t_values and x_c_values as global
t_values = tspan; % Store time values
x_c_values = zeros(size(tspan)); % Initialize array to store x_c values

n=1;
y0 = zeros(4*n,1);
y0(3*n+1) = -v;

% Initialize y_c_prev
x_c_prev = 0;
% Define your initial value for x_c_prev
x_c_prev_initial = 0;

% Initialize x_c_prev_modes, assuming you have ‘n’ modes
x_c_prev_modes = cell(1, n);

% Initialize the first mode’s x_c_prev
x_c_prev_modes{1} = x_c_prev_initial;

[t,y] = ode45(‘rhs_cont4’,tspan,y0);
for i=1:1

u_1= y(:,i).*sin(i*pi.* x_c_values_new);
du_1= y(:,n+i).*sin(i*pi.* x_c_values_new);

end

figure
plot(t,du_1)
legend("beam vel mode 1")

function ydot=rhs_cont4(t,y)
global xi xi_b r_w r_m mu_s mu_k v alpha P a w l_0 n x_c_prev x_c_prev_modes

% ydot_1=y(1:n); %displacement beam
% ydot_2=y(n+1:2*n); %velocity beam
% ydot_3=y(2*n+1:3*n); %displacement mass
% ydot_4=y(3*n+1:4*n); %velocity mass

ydot_1=zeros(n,1); %displacement beam
ydot_2=zeros(n,1); %velocity beam
ydot_3=zeros(n,1); %displacement mass
ydot_4=zeros(n,1); %velocity mass%

v_m=0;
x_m=0;
x_c_prev = x_c_prev_modes{1};

% fprintf(‘x_c_prev %.7fn ‘, x_c_prev);

for i = 1:n

x_m=y(i)*sin(i*pi*x_c_prev)+x_m;
x_c=y(2*n+i)+v*t+l_0-x_m;
x_c_prev = x_c; % Update y_c_prev for the next iteration
x_c_prev_modes{i} = x_c_prev;
%Store y_c values at the corresponding time step
global t_values x_c_values; % Access global variables
[~, idx] = min(abs(t – t_values)); % Find the index of the closest time step
x_c_values(idx) = x_c; % Store y_c at the corresponding time step

v_m=y(n+i)*sin(i*pi*x_c)+v_m;

ydot_1(i) = y(n+i);
ydot_2(i) = -2*xi_b*r_w*y(n+i)-(r_w*i)^2*y(i)+2*sign(y(3*n+i)+v-v_m)*(mu_k+(mu_s-mu_k)*exp(-alpha*abs(y(3*n+i)+v-v_m)))*P*sin(i*pi*x_c);
ydot_3(i) =y(3*n+i);
ydot_4(i) =-2*xi*y(3*n+i)-y(2*n+i)-r_m*sign(y(3*n+i)+v-v_m)*(mu_k+(mu_s-mu_k)*exp(-alpha*abs(y(3*n+i)+v-v_m)))*P+a*w^2*sin(w*t);
%
% if i == 1
% fprintf(‘Time: %.4f, ydot_1: %.7fn, ydot_2: %.7fn, ydot_3: %.7fn, ydot_4: %.7fn ‘, t, ydot_1(i), ydot_2(i), ydot_3(i), ydot_4(i));
% end

end

ydot = [ydot_1; ydot_2; ydot_3; ydot_4];

endHello everyone,
I am trying to solve some differential equation in matlab using ode45. The equations are as shown below:

I have written the equations in state space form where ydot_1(i) is , ydot_2(i) is , ydot_3(i) is and ydot_4(i) is . In the equation of motion of x, eq(2), x terms are not in terms of “ i “ but I have written them like that (so ydot_3(i) is and ydot_4(i) is )b ecause of the loop that I use on the function below. So the “for” loop, for each time step, lets say starting at solves first the ydot terms for each mode for i = 1:n and then gets out of the loop, goes to the new time step and then solve again for each mode. However, since for each time step it will start from the first mode again, and ydot_1(i) and ydot_2(i) depend on ydot_3(i) and ydot_4(i) , then I am solving the equation of x, eq.2, for each mode. For the new time step, before starting with the first mode again I am defining the
ydot_1=y(1:n); %displacement beam
ydot_2=y(n+1:2*n); %velocity beam
ydot_3=y(2*n+1:3*n); %displacement mass
ydot_4=y(3*n+1:4*n); %velocity mass
v_m=0;
x_m=0;
x_c_prev = x_c_prev_modes{1};

so it can start with the data from the previous first mode (first mode of the previous time step). My problem is that when I solve for 1 mode only, so for n=1 and get the solution du_1= y(:,n+i).*sin(i*pi.* x_c_values_new); and then solve for n=2 but still get the solution du_1= y(:,n+i).*sin(i*pi.* x_c_values_new); for the first mode only, these first modes do not match. The solution have the same shape but differ in amplitude. The same happens with I use n=20, for example, but again plot only the first mode du_1. The solution changes from the first mode obtained when n=2 and from the first mode obtained for n=1. I have checked for days and cant see my mistake so have reached desperation. Would really appreciate your help! thank youuu!

clear all
clc
global xi xi_b r_w r_m mu_s mu_k v alpha P l a w l_0 n x_c_prev x_c_prev_modes

dt=0.01;
tspan = (0:dt:150).’;

% parameters
xi=0.14;
xi_b=0.0025;
r_w=4.5;
r_m=4.8;
mu_s=0.5;
mu_k=0.2 ;
alpha=300;
P=0.002;
l=1/75;
l_0=0.25;
a=0;
w=9;

v=0.005;
% Initialize y_c_prev and the array to store x_c values
x_c_prev = 0;
global t_values x_c_values; % Declare t_values and x_c_values as global
t_values = tspan; % Store time values
x_c_values = zeros(size(tspan)); % Initialize array to store x_c values

n=1;
y0 = zeros(4*n,1);
y0(3*n+1) = -v;

% Initialize y_c_prev
x_c_prev = 0;
% Define your initial value for x_c_prev
x_c_prev_initial = 0;

% Initialize x_c_prev_modes, assuming you have ‘n’ modes
x_c_prev_modes = cell(1, n);

% Initialize the first mode’s x_c_prev
x_c_prev_modes{1} = x_c_prev_initial;

[t,y] = ode45(‘rhs_cont4’,tspan,y0);
for i=1:1

u_1= y(:,i).*sin(i*pi.* x_c_values_new);
du_1= y(:,n+i).*sin(i*pi.* x_c_values_new);

end

figure
plot(t,du_1)
legend("beam vel mode 1")

function ydot=rhs_cont4(t,y)
global xi xi_b r_w r_m mu_s mu_k v alpha P a w l_0 n x_c_prev x_c_prev_modes

% ydot_1=y(1:n); %displacement beam
% ydot_2=y(n+1:2*n); %velocity beam
% ydot_3=y(2*n+1:3*n); %displacement mass
% ydot_4=y(3*n+1:4*n); %velocity mass

ydot_1=zeros(n,1); %displacement beam
ydot_2=zeros(n,1); %velocity beam
ydot_3=zeros(n,1); %displacement mass
ydot_4=zeros(n,1); %velocity mass%

v_m=0;
x_m=0;
x_c_prev = x_c_prev_modes{1};

% fprintf(‘x_c_prev %.7fn ‘, x_c_prev);

for i = 1:n

x_m=y(i)*sin(i*pi*x_c_prev)+x_m;
x_c=y(2*n+i)+v*t+l_0-x_m;
x_c_prev = x_c; % Update y_c_prev for the next iteration
x_c_prev_modes{i} = x_c_prev;
%Store y_c values at the corresponding time step
global t_values x_c_values; % Access global variables
[~, idx] = min(abs(t – t_values)); % Find the index of the closest time step
x_c_values(idx) = x_c; % Store y_c at the corresponding time step

v_m=y(n+i)*sin(i*pi*x_c)+v_m;

ydot_1(i) = y(n+i);
ydot_2(i) = -2*xi_b*r_w*y(n+i)-(r_w*i)^2*y(i)+2*sign(y(3*n+i)+v-v_m)*(mu_k+(mu_s-mu_k)*exp(-alpha*abs(y(3*n+i)+v-v_m)))*P*sin(i*pi*x_c);
ydot_3(i) =y(3*n+i);
ydot_4(i) =-2*xi*y(3*n+i)-y(2*n+i)-r_m*sign(y(3*n+i)+v-v_m)*(mu_k+(mu_s-mu_k)*exp(-alpha*abs(y(3*n+i)+v-v_m)))*P+a*w^2*sin(w*t);
%
% if i == 1
% fprintf(‘Time: %.4f, ydot_1: %.7fn, ydot_2: %.7fn, ydot_3: %.7fn, ydot_4: %.7fn ‘, t, ydot_1(i), ydot_2(i), ydot_3(i), ydot_4(i));
% end

end

ydot = [ydot_1; ydot_2; ydot_3; ydot_4];

end Hello everyone,
I am trying to solve some differential equation in matlab using ode45. The equations are as shown below:

I have written the equations in state space form where ydot_1(i) is , ydot_2(i) is , ydot_3(i) is and ydot_4(i) is . In the equation of motion of x, eq(2), x terms are not in terms of “ i “ but I have written them like that (so ydot_3(i) is and ydot_4(i) is )b ecause of the loop that I use on the function below. So the “for” loop, for each time step, lets say starting at solves first the ydot terms for each mode for i = 1:n and then gets out of the loop, goes to the new time step and then solve again for each mode. However, since for each time step it will start from the first mode again, and ydot_1(i) and ydot_2(i) depend on ydot_3(i) and ydot_4(i) , then I am solving the equation of x, eq.2, for each mode. For the new time step, before starting with the first mode again I am defining the
ydot_1=y(1:n); %displacement beam
ydot_2=y(n+1:2*n); %velocity beam
ydot_3=y(2*n+1:3*n); %displacement mass
ydot_4=y(3*n+1:4*n); %velocity mass
v_m=0;
x_m=0;
x_c_prev = x_c_prev_modes{1};

so it can start with the data from the previous first mode (first mode of the previous time step). My problem is that when I solve for 1 mode only, so for n=1 and get the solution du_1= y(:,n+i).*sin(i*pi.* x_c_values_new); and then solve for n=2 but still get the solution du_1= y(:,n+i).*sin(i*pi.* x_c_values_new); for the first mode only, these first modes do not match. The solution have the same shape but differ in amplitude. The same happens with I use n=20, for example, but again plot only the first mode du_1. The solution changes from the first mode obtained when n=2 and from the first mode obtained for n=1. I have checked for days and cant see my mistake so have reached desperation. Would really appreciate your help! thank youuu!

clear all
clc
global xi xi_b r_w r_m mu_s mu_k v alpha P l a w l_0 n x_c_prev x_c_prev_modes

dt=0.01;
tspan = (0:dt:150).’;

% parameters
xi=0.14;
xi_b=0.0025;
r_w=4.5;
r_m=4.8;
mu_s=0.5;
mu_k=0.2 ;
alpha=300;
P=0.002;
l=1/75;
l_0=0.25;
a=0;
w=9;

v=0.005;
% Initialize y_c_prev and the array to store x_c values
x_c_prev = 0;
global t_values x_c_values; % Declare t_values and x_c_values as global
t_values = tspan; % Store time values
x_c_values = zeros(size(tspan)); % Initialize array to store x_c values

n=1;
y0 = zeros(4*n,1);
y0(3*n+1) = -v;

% Initialize y_c_prev
x_c_prev = 0;
% Define your initial value for x_c_prev
x_c_prev_initial = 0;

% Initialize x_c_prev_modes, assuming you have ‘n’ modes
x_c_prev_modes = cell(1, n);

% Initialize the first mode’s x_c_prev
x_c_prev_modes{1} = x_c_prev_initial;

[t,y] = ode45(‘rhs_cont4’,tspan,y0);
for i=1:1

u_1= y(:,i).*sin(i*pi.* x_c_values_new);
du_1= y(:,n+i).*sin(i*pi.* x_c_values_new);

end

figure
plot(t,du_1)
legend("beam vel mode 1")

function ydot=rhs_cont4(t,y)
global xi xi_b r_w r_m mu_s mu_k v alpha P a w l_0 n x_c_prev x_c_prev_modes

% ydot_1=y(1:n); %displacement beam
% ydot_2=y(n+1:2*n); %velocity beam
% ydot_3=y(2*n+1:3*n); %displacement mass
% ydot_4=y(3*n+1:4*n); %velocity mass

ydot_1=zeros(n,1); %displacement beam
ydot_2=zeros(n,1); %velocity beam
ydot_3=zeros(n,1); %displacement mass
ydot_4=zeros(n,1); %velocity mass%

v_m=0;
x_m=0;
x_c_prev = x_c_prev_modes{1};

% fprintf(‘x_c_prev %.7fn ‘, x_c_prev);

for i = 1:n

x_m=y(i)*sin(i*pi*x_c_prev)+x_m;
x_c=y(2*n+i)+v*t+l_0-x_m;
x_c_prev = x_c; % Update y_c_prev for the next iteration
x_c_prev_modes{i} = x_c_prev;
%Store y_c values at the corresponding time step
global t_values x_c_values; % Access global variables
[~, idx] = min(abs(t – t_values)); % Find the index of the closest time step
x_c_values(idx) = x_c; % Store y_c at the corresponding time step

v_m=y(n+i)*sin(i*pi*x_c)+v_m;

ydot_1(i) = y(n+i);
ydot_2(i) = -2*xi_b*r_w*y(n+i)-(r_w*i)^2*y(i)+2*sign(y(3*n+i)+v-v_m)*(mu_k+(mu_s-mu_k)*exp(-alpha*abs(y(3*n+i)+v-v_m)))*P*sin(i*pi*x_c);
ydot_3(i) =y(3*n+i);
ydot_4(i) =-2*xi*y(3*n+i)-y(2*n+i)-r_m*sign(y(3*n+i)+v-v_m)*(mu_k+(mu_s-mu_k)*exp(-alpha*abs(y(3*n+i)+v-v_m)))*P+a*w^2*sin(w*t);
%
% if i == 1
% fprintf(‘Time: %.4f, ydot_1: %.7fn, ydot_2: %.7fn, ydot_3: %.7fn, ydot_4: %.7fn ‘, t, ydot_1(i), ydot_2(i), ydot_3(i), ydot_4(i));
% end

end

ydot = [ydot_1; ydot_2; ydot_3; ydot_4];

end ode45, differential equations MATLAB Answers — New Questions

​

I’m using two (2) uidropdowns with a valuechangedfcn to live update the x-y axes for a plot based on the selected variable. I can’t figure out how to simultaneously pass the information from both dropdowns to my plotting function. For example, say I have data for three variables: time, position, and velocity, and I have two drop downs each containing the same three variables for the x and y axes. I’m unable to retain a previous variable selection for the x-axis once I select a new variable for the y-axis, so it just reverts to the initialized x-variable. Is there a way to store the selection from each dropdown simultaneouslyI’m using two (2) uidropdowns with a valuechangedfcn to live update the x-y axes for a plot based on the selected variable. I can’t figure out how to simultaneously pass the information from both dropdowns to my plotting function. For example, say I have data for three variables: time, position, and velocity, and I have two drop downs each containing the same three variables for the x and y axes. I’m unable to retain a previous variable selection for the x-axis once I select a new variable for the y-axis, so it just reverts to the initialized x-variable. Is there a way to store the selection from each dropdown simultaneously I’m using two (2) uidropdowns with a valuechangedfcn to live update the x-y axes for a plot based on the selected variable. I can’t figure out how to simultaneously pass the information from both dropdowns to my plotting function. For example, say I have data for three variables: time, position, and velocity, and I have two drop downs each containing the same three variables for the x and y axes. I’m unable to retain a previous variable selection for the x-axis once I select a new variable for the y-axis, so it just reverts to the initialized x-variable. Is there a way to store the selection from each dropdown simultaneously uidropdrown, appdesigner, plot, gui MATLAB Answers — New Questions

​

[tform, peakregcorr] = imregcorr(SourceImageAdj,SourceRef, …
TargetImageAdj,TargetRef, …
‘transformType’, ‘similarity’, …
‘Window’, true);
RegisteredImage = imwarp(SourceImageAdj,SourceRef, tform, ‘linear’, …
‘OutputView’, TargetRef);

disp(peakregcorr)
figure, imshowpair(TargetImageAdj,TargetRef,RegisteredImage, TargetRef)
above is my code to correlate 2 images of different sizes. the goal is to find the correlation between the images to determine what pixel location a point of interest might lie in both images. However, imregcorr consistently misses the mark badly.
Whereas the code above might output a tform object – Dimensionality: 2, Scale: 0.4254, RotationAngle: 2.4851, Translation: [-0.3269 -0.0904]
A more correct tform object (based upon the output image would look like – Dimensionality: 2, Scale: 1, RotationAngle: -1, Translation: [0.0640 0]
The SourceRef and TargetRef seem correct, so Im not sure what is causing the issue[tform, peakregcorr] = imregcorr(SourceImageAdj,SourceRef, …
TargetImageAdj,TargetRef, …
‘transformType’, ‘similarity’, …
‘Window’, true);
RegisteredImage = imwarp(SourceImageAdj,SourceRef, tform, ‘linear’, …
‘OutputView’, TargetRef);

disp(peakregcorr)
figure, imshowpair(TargetImageAdj,TargetRef,RegisteredImage, TargetRef)
above is my code to correlate 2 images of different sizes. the goal is to find the correlation between the images to determine what pixel location a point of interest might lie in both images. However, imregcorr consistently misses the mark badly.
Whereas the code above might output a tform object – Dimensionality: 2, Scale: 0.4254, RotationAngle: 2.4851, Translation: [-0.3269 -0.0904]
A more correct tform object (based upon the output image would look like – Dimensionality: 2, Scale: 1, RotationAngle: -1, Translation: [0.0640 0]
The SourceRef and TargetRef seem correct, so Im not sure what is causing the issue [tform, peakregcorr] = imregcorr(SourceImageAdj,SourceRef, …
TargetImageAdj,TargetRef, …
‘transformType’, ‘similarity’, …
‘Window’, true);
RegisteredImage = imwarp(SourceImageAdj,SourceRef, tform, ‘linear’, …
‘OutputView’, TargetRef);

disp(peakregcorr)
figure, imshowpair(TargetImageAdj,TargetRef,RegisteredImage, TargetRef)
above is my code to correlate 2 images of different sizes. the goal is to find the correlation between the images to determine what pixel location a point of interest might lie in both images. However, imregcorr consistently misses the mark badly.
Whereas the code above might output a tform object – Dimensionality: 2, Scale: 0.4254, RotationAngle: 2.4851, Translation: [-0.3269 -0.0904]
A more correct tform object (based upon the output image would look like – Dimensionality: 2, Scale: 1, RotationAngle: -1, Translation: [0.0640 0]
The SourceRef and TargetRef seem correct, so Im not sure what is causing the issue image processing MATLAB Answers — New Questions

​

I am solving a time-dependent finite element problem with nonlinearities in the stiffness matrix which takes the following form:

where the dot symbolizes derivative with respect to time, U is the displacement vector, M is the mass matrix, C is the damping matrix, K(U) is the stiffness matrix which depends on U, and F is an external force. I would like to use ode45 to solve the dynamic problem. The general flow for solving this problem would be the following:
Calculate the stiffness matrix K using the latest displacement U (Initial condition if it is the first time step).
Formulate the state-space equations using K matrix, M matrix, C matrix, and F vector.
Integrate the state-space equations with ode45 to caclulate the new displacement U.
Recalculate the K matrix with the new displacement U. Check if the residual error meets the preselected tolerance.
If the error is met, update velocities and accelerations, and go to the next time step. If the error is not met, then go back to step 1 using this latest K matrix.
I am familiar with solving this problem without the nonlinearity in K. However, I am unsure how to update the K matrix at every time step within the ode45 framework.
Thank you for the guidance!I am solving a time-dependent finite element problem with nonlinearities in the stiffness matrix which takes the following form:

where the dot symbolizes derivative with respect to time, U is the displacement vector, M is the mass matrix, C is the damping matrix, K(U) is the stiffness matrix which depends on U, and F is an external force. I would like to use ode45 to solve the dynamic problem. The general flow for solving this problem would be the following:
Calculate the stiffness matrix K using the latest displacement U (Initial condition if it is the first time step).
Formulate the state-space equations using K matrix, M matrix, C matrix, and F vector.
Integrate the state-space equations with ode45 to caclulate the new displacement U.
Recalculate the K matrix with the new displacement U. Check if the residual error meets the preselected tolerance.
If the error is met, update velocities and accelerations, and go to the next time step. If the error is not met, then go back to step 1 using this latest K matrix.
I am familiar with solving this problem without the nonlinearity in K. However, I am unsure how to update the K matrix at every time step within the ode45 framework.
Thank you for the guidance! I am solving a time-dependent finite element problem with nonlinearities in the stiffness matrix which takes the following form:

where the dot symbolizes derivative with respect to time, U is the displacement vector, M is the mass matrix, C is the damping matrix, K(U) is the stiffness matrix which depends on U, and F is an external force. I would like to use ode45 to solve the dynamic problem. The general flow for solving this problem would be the following:
Calculate the stiffness matrix K using the latest displacement U (Initial condition if it is the first time step).
Formulate the state-space equations using K matrix, M matrix, C matrix, and F vector.
Integrate the state-space equations with ode45 to caclulate the new displacement U.
Recalculate the K matrix with the new displacement U. Check if the residual error meets the preselected tolerance.
If the error is met, update velocities and accelerations, and go to the next time step. If the error is not met, then go back to step 1 using this latest K matrix.
I am familiar with solving this problem without the nonlinearity in K. However, I am unsure how to update the K matrix at every time step within the ode45 framework.
Thank you for the guidance! nonlinear-dynamics, ode45 MATLAB Answers — New Questions

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function distance = get_distance(x,y)
[~,~,raw] = xlsread(‘Distances.xlsx’);
col_labels = raw(1,:);
row_labels = raw(:,1);
try
distance = raw{contains(row_labels,y),contains(col_labels,x)};
catch
distance = -1;
end
end
error

Assessment result: incorrectNashville, TN and Las Vegas, NV
Variable distance has an incorrect value.
Assessment result: incorrectRandom city pairs
Variable distance has an incorrect value.
get_distance(‘Chattanooga, TN’,’Meads, KY’) returned -1 which is incorrect.function distance = get_distance(x,y)
[~,~,raw] = xlsread(‘Distances.xlsx’);
col_labels = raw(1,:);
row_labels = raw(:,1);
try
distance = raw{contains(row_labels,y),contains(col_labels,x)};
catch
distance = -1;
end
end
error

Assessment result: incorrectNashville, TN and Las Vegas, NV
Variable distance has an incorrect value.
Assessment result: incorrectRandom city pairs
Variable distance has an incorrect value.
get_distance(‘Chattanooga, TN’,’Meads, KY’) returned -1 which is incorrect. function distance = get_distance(x,y)
[~,~,raw] = xlsread(‘Distances.xlsx’);
col_labels = raw(1,:);
row_labels = raw(:,1);
try
distance = raw{contains(row_labels,y),contains(col_labels,x)};
catch
distance = -1;
end
end
error

Assessment result: incorrectNashville, TN and Las Vegas, NV
Variable distance has an incorrect value.
Assessment result: incorrectRandom city pairs
Variable distance has an incorrect value.
get_distance(‘Chattanooga, TN’,’Meads, KY’) returned -1 which is incorrect. homework, city distance, no more answers please! MATLAB Answers — New Questions

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Hello, the TLDR of this post is that I am trying to learn how to implement a sensorless control for a 3 phase PMSM using the EKF Simulink block, but I can’t seem to make it work. I’m not sure if I implemented the State Transition Function or the Measurement Functions correctly.
The EKF Simulink block is set up as such

The equivalent Simulink implementation is as such

I have set the Covariance Matrix Q as NoiseMatrix(1:4) and Covariance Matrix R as NoiseMatrix(5:6) in the EKF block parameter below

StateTransitionFunction
The relevant state space equations of the 3 phase PMSM is derived as such

and I implemented the State Transition function as such
function x_next = pmsmStateTransition_3p(x,u)
% represents the dynaics of the PMSM and how the states of the …
% evolves over time
% x=[i_d; i_q; omega_e; theta_e] (state vector)
% u=[v_alpha; v_beta] (input vector)

% ——– Inputs ——– %

i_d=x(1);
i_q=x(2);
omega_e=x(3);
theta_e=x(4);

v_alpha=u(1);
v_beta=u(2);

% ——– Motor Parameters ——– %

R = 0.007184;
L = 19.5e-6;
pole_pairs = 6; %pole pairs
J = 3.37862;
B = 0;
lambda = 0.0623;

dt = 9e-5; % Ts

% ——– Equations ——– %
lambda_q = i_q*L;
lambda_d = i_d*L+lambda;
T_e = 3/2*pole_pairs*(i_q*(i_d*L + lambda) – i_d*i_q*L);
v_d = cos(theta_e)*v_alpha + sin(theta_e)*v_beta;
v_q = -sin(theta_e)*v_alpha + cos(theta_e)*v_beta;

% State Transition Equations
% Euler Discretization is used to calculate the next states
id_next = i_d + (v_d – R*i_d + omega_e*lambda_q)/L*dt;
iq_next = i_q + (v_q – R*i_q – omega_e*lambda_d)/L*dt;
omega_next = omega_e + (T_e – B*omega_e)/J*dt;
theta_next = theta_e + omega_e*dt;

% ——– Output ——– %

x_next=[id_next; iq_next; omega_next; theta_next];

end
Measurement Function
function y = pmsmMeasurement_3p(x)
% x=[i_d; i_q; omega_e; theta_e] (state vector)
% y=[i_alpha; i_beta] (output vector)

% ——— Inputs ———- %

i_d=x(1);
i_q=x(2);
theta_e=x(4);

% ——– Equations ——– %

i_alpha = i_d*cos(theta_e) – i_q*sin(theta_e);
i_beta = i_d*sin(theta_e) + i_q*cos(theta_e);

% ——— Output ———- %

y=[i_alpha; i_beta];

end
Results

I understand that the covariance matrices determines the results. After I set everything up, I would find the optimal values using the Genetic Algorithm, but even then the results from it don’t differ so much from the estimated we_hat in the bottom figure for results, so I just used a generic placeholder value of 100 for all values of the Noise Covariance Matrices (i.e. Q = [100, 100, 100, 100] and R = [100, 100]).

Have I implemented the functions properly? I would appreciate any help at all to help lessen the amount of things I need to debug. Thank you so much and I look forward to your reply.

I have attached all the relevant files in this post, so please let me know what you need and I can direct you to the relevant file.Hello, the TLDR of this post is that I am trying to learn how to implement a sensorless control for a 3 phase PMSM using the EKF Simulink block, but I can’t seem to make it work. I’m not sure if I implemented the State Transition Function or the Measurement Functions correctly.
The EKF Simulink block is set up as such

The equivalent Simulink implementation is as such

I have set the Covariance Matrix Q as NoiseMatrix(1:4) and Covariance Matrix R as NoiseMatrix(5:6) in the EKF block parameter below

StateTransitionFunction
The relevant state space equations of the 3 phase PMSM is derived as such

and I implemented the State Transition function as such
function x_next = pmsmStateTransition_3p(x,u)
% represents the dynaics of the PMSM and how the states of the …
% evolves over time
% x=[i_d; i_q; omega_e; theta_e] (state vector)
% u=[v_alpha; v_beta] (input vector)

% ——– Inputs ——– %

i_d=x(1);
i_q=x(2);
omega_e=x(3);
theta_e=x(4);

v_alpha=u(1);
v_beta=u(2);

% ——– Motor Parameters ——– %

R = 0.007184;
L = 19.5e-6;
pole_pairs = 6; %pole pairs
J = 3.37862;
B = 0;
lambda = 0.0623;

dt = 9e-5; % Ts

% ——– Equations ——– %
lambda_q = i_q*L;
lambda_d = i_d*L+lambda;
T_e = 3/2*pole_pairs*(i_q*(i_d*L + lambda) – i_d*i_q*L);
v_d = cos(theta_e)*v_alpha + sin(theta_e)*v_beta;
v_q = -sin(theta_e)*v_alpha + cos(theta_e)*v_beta;

% State Transition Equations
% Euler Discretization is used to calculate the next states
id_next = i_d + (v_d – R*i_d + omega_e*lambda_q)/L*dt;
iq_next = i_q + (v_q – R*i_q – omega_e*lambda_d)/L*dt;
omega_next = omega_e + (T_e – B*omega_e)/J*dt;
theta_next = theta_e + omega_e*dt;

% ——– Output ——– %

x_next=[id_next; iq_next; omega_next; theta_next];

end
Measurement Function
function y = pmsmMeasurement_3p(x)
% x=[i_d; i_q; omega_e; theta_e] (state vector)
% y=[i_alpha; i_beta] (output vector)

% ——— Inputs ———- %

i_d=x(1);
i_q=x(2);
theta_e=x(4);

% ——– Equations ——– %

i_alpha = i_d*cos(theta_e) – i_q*sin(theta_e);
i_beta = i_d*sin(theta_e) + i_q*cos(theta_e);

% ——— Output ———- %

y=[i_alpha; i_beta];

end
Results

I understand that the covariance matrices determines the results. After I set everything up, I would find the optimal values using the Genetic Algorithm, but even then the results from it don’t differ so much from the estimated we_hat in the bottom figure for results, so I just used a generic placeholder value of 100 for all values of the Noise Covariance Matrices (i.e. Q = [100, 100, 100, 100] and R = [100, 100]).

Have I implemented the functions properly? I would appreciate any help at all to help lessen the amount of things I need to debug. Thank you so much and I look forward to your reply.

I have attached all the relevant files in this post, so please let me know what you need and I can direct you to the relevant file. Hello, the TLDR of this post is that I am trying to learn how to implement a sensorless control for a 3 phase PMSM using the EKF Simulink block, but I can’t seem to make it work. I’m not sure if I implemented the State Transition Function or the Measurement Functions correctly.
The EKF Simulink block is set up as such

The equivalent Simulink implementation is as such

I have set the Covariance Matrix Q as NoiseMatrix(1:4) and Covariance Matrix R as NoiseMatrix(5:6) in the EKF block parameter below

StateTransitionFunction
The relevant state space equations of the 3 phase PMSM is derived as such

and I implemented the State Transition function as such
function x_next = pmsmStateTransition_3p(x,u)
% represents the dynaics of the PMSM and how the states of the …
% evolves over time
% x=[i_d; i_q; omega_e; theta_e] (state vector)
% u=[v_alpha; v_beta] (input vector)

% ——– Inputs ——– %

i_d=x(1);
i_q=x(2);
omega_e=x(3);
theta_e=x(4);

v_alpha=u(1);
v_beta=u(2);

% ——– Motor Parameters ——– %

R = 0.007184;
L = 19.5e-6;
pole_pairs = 6; %pole pairs
J = 3.37862;
B = 0;
lambda = 0.0623;

dt = 9e-5; % Ts

% ——– Equations ——– %
lambda_q = i_q*L;
lambda_d = i_d*L+lambda;
T_e = 3/2*pole_pairs*(i_q*(i_d*L + lambda) – i_d*i_q*L);
v_d = cos(theta_e)*v_alpha + sin(theta_e)*v_beta;
v_q = -sin(theta_e)*v_alpha + cos(theta_e)*v_beta;

% State Transition Equations
% Euler Discretization is used to calculate the next states
id_next = i_d + (v_d – R*i_d + omega_e*lambda_q)/L*dt;
iq_next = i_q + (v_q – R*i_q – omega_e*lambda_d)/L*dt;
omega_next = omega_e + (T_e – B*omega_e)/J*dt;
theta_next = theta_e + omega_e*dt;

% ——– Output ——– %

x_next=[id_next; iq_next; omega_next; theta_next];

end
Measurement Function
function y = pmsmMeasurement_3p(x)
% x=[i_d; i_q; omega_e; theta_e] (state vector)
% y=[i_alpha; i_beta] (output vector)

% ——— Inputs ———- %

i_d=x(1);
i_q=x(2);
theta_e=x(4);

% ——– Equations ——– %

i_alpha = i_d*cos(theta_e) – i_q*sin(theta_e);
i_beta = i_d*sin(theta_e) + i_q*cos(theta_e);

% ——— Output ———- %

y=[i_alpha; i_beta];

end
Results

I understand that the covariance matrices determines the results. After I set everything up, I would find the optimal values using the Genetic Algorithm, but even then the results from it don’t differ so much from the estimated we_hat in the bottom figure for results, so I just used a generic placeholder value of 100 for all values of the Noise Covariance Matrices (i.e. Q = [100, 100, 100, 100] and R = [100, 100]).

Have I implemented the functions properly? I would appreciate any help at all to help lessen the amount of things I need to debug. Thank you so much and I look forward to your reply.

I have attached all the relevant files in this post, so please let me know what you need and I can direct you to the relevant file. ekf, foc, pmsm, sensorless, state transition function, measurement function, simulink, r2023a MATLAB Answers — New Questions

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Hi guys, I am new to MATLAB. I need help on how to get FFT plot from an audio file that i have?Hi guys, I am new to MATLAB. I need help on how to get FFT plot from an audio file that i have? Hi guys, I am new to MATLAB. I need help on how to get FFT plot from an audio file that i have? audioread noise fast fourier MATLAB Answers — New Questions

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Hi, I have a rather simple model with a battery, an dynamical load and a source representing the grid. Now I want the battery to shut off at 20%. I am using two ideal switches for it but I just cannot seem to get it to work. The battery is discharging but it does not stop at 20%.Hi, I have a rather simple model with a battery, an dynamical load and a source representing the grid. Now I want the battery to shut off at 20%. I am using two ideal switches for it but I just cannot seem to get it to work. The battery is discharging but it does not stop at 20%. Hi, I have a rather simple model with a battery, an dynamical load and a source representing the grid. Now I want the battery to shut off at 20%. I am using two ideal switches for it but I just cannot seem to get it to work. The battery is discharging but it does not stop at 20%. battery_system_management, switch MATLAB Answers — New Questions

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In the coverage report, it lists a condition under decisisions analyzed (example: if A > x) with the 2 possible outcomes true and false, with the totals for each true and false. However it then lists that identical condition a 2nd time with different totals for the outcomes again.
A visual example of the table that shows:
Decisions Analyzed:

What is the difference between the first set of 3 (labeled as 50% covered) and the second set of 230 (labeled as 100% covered)? This issue is present for every condition in my model. It is also what occurs for other coverage types, like MCDC. Here’s an MCDC example:

Again, the top half is the identical text to the bottom half, but the bottom half shows 1 of the 3 rows red (so more coverage is complete) whereas the top half shows all 3 in red (missing more coverage). What is the difference between the halves?
Thank you.
CodyIn the coverage report, it lists a condition under decisisions analyzed (example: if A > x) with the 2 possible outcomes true and false, with the totals for each true and false. However it then lists that identical condition a 2nd time with different totals for the outcomes again.
A visual example of the table that shows:
Decisions Analyzed:

What is the difference between the first set of 3 (labeled as 50% covered) and the second set of 230 (labeled as 100% covered)? This issue is present for every condition in my model. It is also what occurs for other coverage types, like MCDC. Here’s an MCDC example:

Again, the top half is the identical text to the bottom half, but the bottom half shows 1 of the 3 rows red (so more coverage is complete) whereas the top half shows all 3 in red (missing more coverage). What is the difference between the halves?
Thank you.
Cody In the coverage report, it lists a condition under decisisions analyzed (example: if A > x) with the 2 possible outcomes true and false, with the totals for each true and false. However it then lists that identical condition a 2nd time with different totals for the outcomes again.
A visual example of the table that shows:
Decisions Analyzed:

What is the difference between the first set of 3 (labeled as 50% covered) and the second set of 230 (labeled as 100% covered)? This issue is present for every condition in my model. It is also what occurs for other coverage types, like MCDC. Here’s an MCDC example:

Again, the top half is the identical text to the bottom half, but the bottom half shows 1 of the 3 rows red (so more coverage is complete) whereas the top half shows all 3 in red (missing more coverage). What is the difference between the halves?
Thank you.
Cody test coverage, double reported test coverage, mcdc, decision, coverage report MATLAB Answers — New Questions

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