Dear friends,
I currently ran into a math-issue when trying to get the analytical roots of a polynomial of order 4.
What I want to solve is:

The solution shall run on a microcontroller with dynamically changing parameters [a,d,e]. All parameters are real. I am searching for the real results, imaginary solutions are to be prevented. When applying Matlab symbolic, The parameters [a,d,e] are simplified and represent longer sub-terms. In accordance to the article Wiki: quartic function, I will receive quite a long solution, that is futile to reproduce here. I used the following short code to generate it:
syms x a d e
y= a*x^4+d*x+e
sol=solve(y==0,x,’MaxDegree’,4,’ReturnConditions’,true)
To check the solution, I tested it on an easier sub-problem: When d=0 is chosen follows:

When chosing a>0, e<0 this clearly yields some real results, besides the imaginary ones. Now the problem: the above given long solution cannot reproduce this result. Even if I force assumptions that should solve it, follows something that does not look equal:
syms x a d e
assume(a>=0)
assume(d>=0)
assume(e<=0)
y= a*x^4+d*x+e
sol=solve(y==0,x,’MaxDegree’,4,’ReturnConditions’,true)

testresult=simplify(subs(sol.x, d,0))

latex(testresult)

However, if I change the equation before solving the quartic function, the solution of it looks as on paper:
y= a*x^4+e
assume(a>=0)
assume(d>=0)
assume(e<=0)

sol=solve(y==0,x,’MaxDegree’,4,’ReturnConditions’,true)

testresult2=sol.x
latex(testresult2)

Numerical testing via inserting numbers in the two solutions according to the assumptions a>0, e<0 confirms the difference of the results. Chosing a=1 and e=-2 yields:
1) the general solution of the quartic polynomial (inserting d=0 after solving the quartic polynomial):
vpares1=vpa(subs(testresult,{a,e},{1,-2}))

vpares1 =
0
0
0
0

2) the solution of the simplified quartic polynomial (inserting d=0 before solving):
vpares2=vpa(subs(testresult2,{a,e},{1,-2}))

vpares2 =
1.1892071150027210667174999705605
-1.1892071150027210667174999705605
-1.1892071150027210667174999705605i
1.1892071150027210667174999705605i
Visibly, the trivial solution is not permissible given the solution of case 2). Interestingly, if the requirement is switched from {a>0,e<0} to {a<0, e>0} the solutions match again.
Similar difficulties have shown, when I hand-coded the general solution from the mentioned Wikipedia article. As I need the solution for , I now have trust-issues with the long solution. Therefore, I ask for your help or comment, why the "general solution" does not end in the same result as the simplified problem.
Thank you in advance for your help!Dear friends,
I currently ran into a math-issue when trying to get the analytical roots of a polynomial of order 4.
What I want to solve is:

The solution shall run on a microcontroller with dynamically changing parameters [a,d,e]. All parameters are real. I am searching for the real results, imaginary solutions are to be prevented. When applying Matlab symbolic, The parameters [a,d,e] are simplified and represent longer sub-terms. In accordance to the article Wiki: quartic function, I will receive quite a long solution, that is futile to reproduce here. I used the following short code to generate it:
syms x a d e
y= a*x^4+d*x+e
sol=solve(y==0,x,’MaxDegree’,4,’ReturnConditions’,true)
To check the solution, I tested it on an easier sub-problem: When d=0 is chosen follows:

When chosing a>0, e<0 this clearly yields some real results, besides the imaginary ones. Now the problem: the above given long solution cannot reproduce this result. Even if I force assumptions that should solve it, follows something that does not look equal:
syms x a d e
assume(a>=0)
assume(d>=0)
assume(e<=0)
y= a*x^4+d*x+e
sol=solve(y==0,x,’MaxDegree’,4,’ReturnConditions’,true)

testresult=simplify(subs(sol.x, d,0))

latex(testresult)

However, if I change the equation before solving the quartic function, the solution of it looks as on paper:
y= a*x^4+e
assume(a>=0)
assume(d>=0)
assume(e<=0)

sol=solve(y==0,x,’MaxDegree’,4,’ReturnConditions’,true)

testresult2=sol.x
latex(testresult2)

Numerical testing via inserting numbers in the two solutions according to the assumptions a>0, e<0 confirms the difference of the results. Chosing a=1 and e=-2 yields:
1) the general solution of the quartic polynomial (inserting d=0 after solving the quartic polynomial):
vpares1=vpa(subs(testresult,{a,e},{1,-2}))

vpares1 =
0
0
0
0

2) the solution of the simplified quartic polynomial (inserting d=0 before solving):
vpares2=vpa(subs(testresult2,{a,e},{1,-2}))

vpares2 =
1.1892071150027210667174999705605
-1.1892071150027210667174999705605
-1.1892071150027210667174999705605i
1.1892071150027210667174999705605i
Visibly, the trivial solution is not permissible given the solution of case 2). Interestingly, if the requirement is switched from {a>0,e<0} to {a<0, e>0} the solutions match again.
Similar difficulties have shown, when I hand-coded the general solution from the mentioned Wikipedia article. As I need the solution for , I now have trust-issues with the long solution. Therefore, I ask for your help or comment, why the "general solution" does not end in the same result as the simplified problem.
Thank you in advance for your help! Dear friends,
I currently ran into a math-issue when trying to get the analytical roots of a polynomial of order 4.
What I want to solve is:

The solution shall run on a microcontroller with dynamically changing parameters [a,d,e]. All parameters are real. I am searching for the real results, imaginary solutions are to be prevented. When applying Matlab symbolic, The parameters [a,d,e] are simplified and represent longer sub-terms. In accordance to the article Wiki: quartic function, I will receive quite a long solution, that is futile to reproduce here. I used the following short code to generate it:
syms x a d e
y= a*x^4+d*x+e
sol=solve(y==0,x,’MaxDegree’,4,’ReturnConditions’,true)
To check the solution, I tested it on an easier sub-problem: When d=0 is chosen follows:

When chosing a>0, e<0 this clearly yields some real results, besides the imaginary ones. Now the problem: the above given long solution cannot reproduce this result. Even if I force assumptions that should solve it, follows something that does not look equal:
syms x a d e
assume(a>=0)
assume(d>=0)
assume(e<=0)
y= a*x^4+d*x+e
sol=solve(y==0,x,’MaxDegree’,4,’ReturnConditions’,true)

testresult=simplify(subs(sol.x, d,0))

latex(testresult)

However, if I change the equation before solving the quartic function, the solution of it looks as on paper:
y= a*x^4+e
assume(a>=0)
assume(d>=0)
assume(e<=0)

sol=solve(y==0,x,’MaxDegree’,4,’ReturnConditions’,true)

testresult2=sol.x
latex(testresult2)

Numerical testing via inserting numbers in the two solutions according to the assumptions a>0, e<0 confirms the difference of the results. Chosing a=1 and e=-2 yields:
1) the general solution of the quartic polynomial (inserting d=0 after solving the quartic polynomial):
vpares1=vpa(subs(testresult,{a,e},{1,-2}))

vpares1 =
0
0
0
0

2) the solution of the simplified quartic polynomial (inserting d=0 before solving):
vpares2=vpa(subs(testresult2,{a,e},{1,-2}))

vpares2 =
1.1892071150027210667174999705605
-1.1892071150027210667174999705605
-1.1892071150027210667174999705605i
1.1892071150027210667174999705605i
Visibly, the trivial solution is not permissible given the solution of case 2). Interestingly, if the requirement is switched from {a>0,e<0} to {a<0, e>0} the solutions match again.
Similar difficulties have shown, when I hand-coded the general solution from the mentioned Wikipedia article. As I need the solution for , I now have trust-issues with the long solution. Therefore, I ask for your help or comment, why the "general solution" does not end in the same result as the simplified problem.
Thank you in advance for your help! mathematics, symbolic, 4th order polynomial, polynomial MATLAB Answers — New Questions

​

I have an app built in MATLAB App Designer. the underlying code utlizes events and listeners. my problem is this: Whenever an error occurs in the code, the app needs to know about this. when the error occurs in the code itself, there are built in "catch" commands that notify the GUI.
However, when an error occurs in one of the listener callbacks, there is some built in mechanism that replaces the error with a warning. Instead of getting an error message we get a warning: "error occurred during execution of listener callback" and a description of the error message.
Is there a way to cancel this mechanism? to let the errors in listener callbacks become actual error messages? or maybe a way to notify the app in case of a warning? something like try catch that includes warnings?
many thanks
NathanI have an app built in MATLAB App Designer. the underlying code utlizes events and listeners. my problem is this: Whenever an error occurs in the code, the app needs to know about this. when the error occurs in the code itself, there are built in "catch" commands that notify the GUI.
However, when an error occurs in one of the listener callbacks, there is some built in mechanism that replaces the error with a warning. Instead of getting an error message we get a warning: "error occurred during execution of listener callback" and a description of the error message.
Is there a way to cancel this mechanism? to let the errors in listener callbacks become actual error messages? or maybe a way to notify the app in case of a warning? something like try catch that includes warnings?
many thanks
Nathan I have an app built in MATLAB App Designer. the underlying code utlizes events and listeners. my problem is this: Whenever an error occurs in the code, the app needs to know about this. when the error occurs in the code itself, there are built in "catch" commands that notify the GUI.
However, when an error occurs in one of the listener callbacks, there is some built in mechanism that replaces the error with a warning. Instead of getting an error message we get a warning: "error occurred during execution of listener callback" and a description of the error message.
Is there a way to cancel this mechanism? to let the errors in listener callbacks become actual error messages? or maybe a way to notify the app in case of a warning? something like try catch that includes warnings?
many thanks
Nathan app designer, listener, event, callback, error MATLAB Answers — New Questions

​

Since javacomponents are deprecated as of MATLAB 2025a, has anyone found a way to generate treetables similar to what was possible with jide ?
We have been relying on those for a lot of GUI and see nothing in UITABLE/UITREE that would be similar. Necessary features are
tree opening
resizable columns
cell renderer allowing push buttons and pull down menus
column sorting is a bonusSince javacomponents are deprecated as of MATLAB 2025a, has anyone found a way to generate treetables similar to what was possible with jide ?
We have been relying on those for a lot of GUI and see nothing in UITABLE/UITREE that would be similar. Necessary features are
tree opening
resizable columns
cell renderer allowing push buttons and pull down menus
column sorting is a bonus Since javacomponents are deprecated as of MATLAB 2025a, has anyone found a way to generate treetables similar to what was possible with jide ?
We have been relying on those for a lot of GUI and see nothing in UITABLE/UITREE that would be similar. Necessary features are
tree opening
resizable columns
cell renderer allowing push buttons and pull down menus
column sorting is a bonus uitable, uitree, treetable MATLAB Answers — New Questions

​

% I’d like to add the option to save backup files – earlier versions every time I save a file in the Matlab editor
editorObj = matlab.desktop.editor.getActive;
editorObj.addSaveCallback(@mySaveCallback);

function mySaveCallback(~, ~)
disp(‘File has been saved!’);
% Add your custom code here
saveBackup(originalFilePath);
end

function saveBackup(originalFilePath)
% Check if the file exists
if ~isfile(originalFilePath)
error(‘The specified file does not exist.’);
end
% Extract file parts
[filePath, fileName, fileExt] = fileparts(originalFilePath);
% Create a timestamp
timestamp = datestr(now, ‘yyyymmdd_HHMMSS’);
% Define the backup file name
backupFileName = sprintf(‘%s_backup_%s%s’, fileName, timestamp, fileExt);
backupFilePath = fullfile(filePath, backupFileName);
% Copy the file to create a backup
copyfile(originalFilePath, backupFilePath);
fprintf(‘Backup saved as: %sn’, backupFilePath);
end% I’d like to add the option to save backup files – earlier versions every time I save a file in the Matlab editor
editorObj = matlab.desktop.editor.getActive;
editorObj.addSaveCallback(@mySaveCallback);

function mySaveCallback(~, ~)
disp(‘File has been saved!’);
% Add your custom code here
saveBackup(originalFilePath);
end

function saveBackup(originalFilePath)
% Check if the file exists
if ~isfile(originalFilePath)
error(‘The specified file does not exist.’);
end
% Extract file parts
[filePath, fileName, fileExt] = fileparts(originalFilePath);
% Create a timestamp
timestamp = datestr(now, ‘yyyymmdd_HHMMSS’);
% Define the backup file name
backupFileName = sprintf(‘%s_backup_%s%s’, fileName, timestamp, fileExt);
backupFilePath = fullfile(filePath, backupFileName);
% Copy the file to create a backup
copyfile(originalFilePath, backupFilePath);
fprintf(‘Backup saved as: %sn’, backupFilePath);
end % I’d like to add the option to save backup files – earlier versions every time I save a file in the Matlab editor
editorObj = matlab.desktop.editor.getActive;
editorObj.addSaveCallback(@mySaveCallback);

function mySaveCallback(~, ~)
disp(‘File has been saved!’);
% Add your custom code here
saveBackup(originalFilePath);
end

function saveBackup(originalFilePath)
% Check if the file exists
if ~isfile(originalFilePath)
error(‘The specified file does not exist.’);
end
% Extract file parts
[filePath, fileName, fileExt] = fileparts(originalFilePath);
% Create a timestamp
timestamp = datestr(now, ‘yyyymmdd_HHMMSS’);
% Define the backup file name
backupFileName = sprintf(‘%s_backup_%s%s’, fileName, timestamp, fileExt);
backupFilePath = fullfile(filePath, backupFileName);
% Copy the file to create a backup
copyfile(originalFilePath, backupFilePath);
fprintf(‘Backup saved as: %sn’, backupFilePath);
end matlab, editor, backup MATLAB Answers — New Questions

​

The CSV file contains the following data:
Sensor names: sensorNames = data.Sensor(1:6,1)
Sensor coordinates: sensorCoordinates = data{1:6, 2:4}
Source number: (:,5)
Source coordinates: sourceCoordinates = data{:, 6:end}
I want to create a 3D plot in MATLAB that includes the following:
Plot the sensor names and their coordinates.
Plot the source coordinates. The source coordinates are categorized into two groups:
Medial: sourceCoordinates = data{7:end, 6:end}
Lateral: sourceCoordinates = data{1:6, 6:end}
3. Set the axis directions depending on the knee side (see figure. 1 below), using S2 as the reference sensor (0,0,0):

Figure.1
Right knee: X-axis positive leftward, Y-axis positive downward, Z-axis positive perpendicular to the depth of the plane.
Left knee: X-axis positive rightward, Y-axis positive downward, Z-axis positive perpendicular to the depth of the plane.
4. Generate three types of plots (similar to the figure.2):
A 3D plot with all axes.
A 2D plot of X vs. Y.
A 2D plot of X vs. Z.
5. Add a line (like the green line in the figure.2 below) that connects the medial and lateral sources. This line should represent the knee joint range in all axis directions.

Figure.2
Knee joint range determination table.

I would like to write a MATLAB script that reads the CSV file, extracts the relevant data, and generates the 3D and 2D plots including the above details.
I wrote the following code:
%% Read and Plot 3D/2D Data from CSV File

% Load the CSV file
filename = ‘3DPlot_KLII_Mori.csv’;
data = readtable(filename);

%% Extract Sensor and Source Data
sensorNames = data.Sensor(1:6, 1); % Sensor names
sensorCoordinates = data{1:6, 2:4}; % Sensor coordinates (X, Y, Z)
sourceNumbers = data{:, 5}; % Source numbers
sourceCoordinates = data{:, 6:end}; % All source coordinates

% Categorize source coordinates
lateralSources = data{1:6, 6:end}; % Lateral sources
medialSources = data{7:end, 6:end}; % Medial sources

%% User input for knee side
kneeSide = "Right"; % Change to "Left" if left knee

% Adjust axis directions based on knee side
if strcmpi(kneeSide, "Right")
% Right knee: X positive leftward, Y downward, Z downward
sensorCoordinates(:,1) = -sensorCoordinates(:,1);
lateralSources(:,1) = -lateralSources(:,1);
medialSources(:,1) = -medialSources(:,1);
elseif strcmpi(kneeSide, "Left")
% Left knee: X positive rightward, Y downward, Z downward
% No inversion needed
else
error(‘Knee side must be either "Right" or "Left".’);
end

%% Plotting – 3D Plot
figure;
hold on;

% Plot sensor coordinates
scatter3(sensorCoordinates(:,1), sensorCoordinates(:,2), sensorCoordinates(:,3), …
80, ‘filled’, ‘MarkerFaceColor’, ‘b’);

% Add sensor name labels
for i = 1:length(sensorNames)
text(sensorCoordinates(i,1), sensorCoordinates(i,2), sensorCoordinates(i,3), …
[‘ ‘ char(sensorNames{i})], ‘FontSize’, 10, ‘Color’, ‘b’);
end

% Plot lateral sources
scatter3(lateralSources(:,1), lateralSources(:,2), lateralSources(:,3), …
60, ‘o’, ‘MarkerEdgeColor’, ‘r’, ‘DisplayName’, ‘Lateral Sources’);

% Plot medial sources
scatter3(medialSources(:,1), medialSources(:,2), medialSources(:,3), …
60, ‘s’, ‘MarkerEdgeColor’, ‘g’, ‘DisplayName’, ‘Medial Sources’);

%% Formatting for 3D plot
grid on;
xlabel(‘X-axis’);
ylabel(‘Y-axis’);
zlabel(‘Z-axis’);
title([‘3D Plot of Sensors and Sources (‘ kneeSide ‘ Knee)’]);
legend;
view(3);
hold off;

%% Plotting – 2D Plot (X vs Y)
figure;
hold on;
scatter(sensorCoordinates(:,1), sensorCoordinates(:,2), 80, ‘filled’, ‘b’);
for i = 1:length(sensorNames)
text(sensorCoordinates(i,1), sensorCoordinates(i,2), [‘ ‘ char(sensorNames{i})], ‘FontSize’, 10, ‘Color’, ‘b’);
end
scatter(lateralSources(:,1), lateralSources(:,2), 60, ‘ro’, ‘DisplayName’, ‘Lateral Sources’);
scatter(medialSources(:,1), medialSources(:,2), 60, ‘gs’, ‘DisplayName’, ‘Medial Sources’);

xlabel(‘X-axis’);
ylabel(‘Y-axis’);
title([‘2D Plot (X vs Y) – ‘ kneeSide ‘ Knee’]);
legend;
grid on;
hold off;

%% Plotting – 2D Plot (X vs Z)
figure;
hold on;
scatter(sensorCoordinates(:,1), sensorCoordinates(:,3), 80, ‘filled’, ‘b’);
for i = 1:length(sensorNames)
text(sensorCoordinates(i,1), sensorCoordinates(i,3), [‘ ‘ char(sensorNames{i})], ‘FontSize’, 10, ‘Color’, ‘b’);
end
scatter(lateralSources(:,1), lateralSources(:,3), 60, ‘ro’, ‘DisplayName’, ‘Lateral Sources’);
scatter(medialSources(:,1), medialSources(:,3), 60, ‘gs’, ‘DisplayName’, ‘Medial Sources’);

xlabel(‘X-axis’);
ylabel(‘Z-axis’);
title([‘2D Plot (X vs Z) – ‘ kneeSide ‘ Knee’]);
legend;
grid on;
hold off;The CSV file contains the following data:
Sensor names: sensorNames = data.Sensor(1:6,1)
Sensor coordinates: sensorCoordinates = data{1:6, 2:4}
Source number: (:,5)
Source coordinates: sourceCoordinates = data{:, 6:end}
I want to create a 3D plot in MATLAB that includes the following:
Plot the sensor names and their coordinates.
Plot the source coordinates. The source coordinates are categorized into two groups:
Medial: sourceCoordinates = data{7:end, 6:end}
Lateral: sourceCoordinates = data{1:6, 6:end}
3. Set the axis directions depending on the knee side (see figure. 1 below), using S2 as the reference sensor (0,0,0):

Figure.1
Right knee: X-axis positive leftward, Y-axis positive downward, Z-axis positive perpendicular to the depth of the plane.
Left knee: X-axis positive rightward, Y-axis positive downward, Z-axis positive perpendicular to the depth of the plane.
4. Generate three types of plots (similar to the figure.2):
A 3D plot with all axes.
A 2D plot of X vs. Y.
A 2D plot of X vs. Z.
5. Add a line (like the green line in the figure.2 below) that connects the medial and lateral sources. This line should represent the knee joint range in all axis directions.

Figure.2
Knee joint range determination table.

I would like to write a MATLAB script that reads the CSV file, extracts the relevant data, and generates the 3D and 2D plots including the above details.
I wrote the following code:
%% Read and Plot 3D/2D Data from CSV File

% Load the CSV file
filename = ‘3DPlot_KLII_Mori.csv’;
data = readtable(filename);

%% Extract Sensor and Source Data
sensorNames = data.Sensor(1:6, 1); % Sensor names
sensorCoordinates = data{1:6, 2:4}; % Sensor coordinates (X, Y, Z)
sourceNumbers = data{:, 5}; % Source numbers
sourceCoordinates = data{:, 6:end}; % All source coordinates

% Categorize source coordinates
lateralSources = data{1:6, 6:end}; % Lateral sources
medialSources = data{7:end, 6:end}; % Medial sources

%% User input for knee side
kneeSide = "Right"; % Change to "Left" if left knee

% Adjust axis directions based on knee side
if strcmpi(kneeSide, "Right")
% Right knee: X positive leftward, Y downward, Z downward
sensorCoordinates(:,1) = -sensorCoordinates(:,1);
lateralSources(:,1) = -lateralSources(:,1);
medialSources(:,1) = -medialSources(:,1);
elseif strcmpi(kneeSide, "Left")
% Left knee: X positive rightward, Y downward, Z downward
% No inversion needed
else
error(‘Knee side must be either "Right" or "Left".’);
end

%% Plotting – 3D Plot
figure;
hold on;

% Plot sensor coordinates
scatter3(sensorCoordinates(:,1), sensorCoordinates(:,2), sensorCoordinates(:,3), …
80, ‘filled’, ‘MarkerFaceColor’, ‘b’);

% Add sensor name labels
for i = 1:length(sensorNames)
text(sensorCoordinates(i,1), sensorCoordinates(i,2), sensorCoordinates(i,3), …
[‘ ‘ char(sensorNames{i})], ‘FontSize’, 10, ‘Color’, ‘b’);
end

% Plot lateral sources
scatter3(lateralSources(:,1), lateralSources(:,2), lateralSources(:,3), …
60, ‘o’, ‘MarkerEdgeColor’, ‘r’, ‘DisplayName’, ‘Lateral Sources’);

% Plot medial sources
scatter3(medialSources(:,1), medialSources(:,2), medialSources(:,3), …
60, ‘s’, ‘MarkerEdgeColor’, ‘g’, ‘DisplayName’, ‘Medial Sources’);

%% Formatting for 3D plot
grid on;
xlabel(‘X-axis’);
ylabel(‘Y-axis’);
zlabel(‘Z-axis’);
title([‘3D Plot of Sensors and Sources (‘ kneeSide ‘ Knee)’]);
legend;
view(3);
hold off;

%% Plotting – 2D Plot (X vs Y)
figure;
hold on;
scatter(sensorCoordinates(:,1), sensorCoordinates(:,2), 80, ‘filled’, ‘b’);
for i = 1:length(sensorNames)
text(sensorCoordinates(i,1), sensorCoordinates(i,2), [‘ ‘ char(sensorNames{i})], ‘FontSize’, 10, ‘Color’, ‘b’);
end
scatter(lateralSources(:,1), lateralSources(:,2), 60, ‘ro’, ‘DisplayName’, ‘Lateral Sources’);
scatter(medialSources(:,1), medialSources(:,2), 60, ‘gs’, ‘DisplayName’, ‘Medial Sources’);

xlabel(‘X-axis’);
ylabel(‘Y-axis’);
title([‘2D Plot (X vs Y) – ‘ kneeSide ‘ Knee’]);
legend;
grid on;
hold off;

%% Plotting – 2D Plot (X vs Z)
figure;
hold on;
scatter(sensorCoordinates(:,1), sensorCoordinates(:,3), 80, ‘filled’, ‘b’);
for i = 1:length(sensorNames)
text(sensorCoordinates(i,1), sensorCoordinates(i,3), [‘ ‘ char(sensorNames{i})], ‘FontSize’, 10, ‘Color’, ‘b’);
end
scatter(lateralSources(:,1), lateralSources(:,3), 60, ‘ro’, ‘DisplayName’, ‘Lateral Sources’);
scatter(medialSources(:,1), medialSources(:,3), 60, ‘gs’, ‘DisplayName’, ‘Medial Sources’);

xlabel(‘X-axis’);
ylabel(‘Z-axis’);
title([‘2D Plot (X vs Z) – ‘ kneeSide ‘ Knee’]);
legend;
grid on;
hold off; The CSV file contains the following data:
Sensor names: sensorNames = data.Sensor(1:6,1)
Sensor coordinates: sensorCoordinates = data{1:6, 2:4}
Source number: (:,5)
Source coordinates: sourceCoordinates = data{:, 6:end}
I want to create a 3D plot in MATLAB that includes the following:
Plot the sensor names and their coordinates.
Plot the source coordinates. The source coordinates are categorized into two groups:
Medial: sourceCoordinates = data{7:end, 6:end}
Lateral: sourceCoordinates = data{1:6, 6:end}
3. Set the axis directions depending on the knee side (see figure. 1 below), using S2 as the reference sensor (0,0,0):

Figure.1
Right knee: X-axis positive leftward, Y-axis positive downward, Z-axis positive perpendicular to the depth of the plane.
Left knee: X-axis positive rightward, Y-axis positive downward, Z-axis positive perpendicular to the depth of the plane.
4. Generate three types of plots (similar to the figure.2):
A 3D plot with all axes.
A 2D plot of X vs. Y.
A 2D plot of X vs. Z.
5. Add a line (like the green line in the figure.2 below) that connects the medial and lateral sources. This line should represent the knee joint range in all axis directions.

Figure.2
Knee joint range determination table.

I would like to write a MATLAB script that reads the CSV file, extracts the relevant data, and generates the 3D and 2D plots including the above details.
I wrote the following code:
%% Read and Plot 3D/2D Data from CSV File

% Load the CSV file
filename = ‘3DPlot_KLII_Mori.csv’;
data = readtable(filename);

%% Extract Sensor and Source Data
sensorNames = data.Sensor(1:6, 1); % Sensor names
sensorCoordinates = data{1:6, 2:4}; % Sensor coordinates (X, Y, Z)
sourceNumbers = data{:, 5}; % Source numbers
sourceCoordinates = data{:, 6:end}; % All source coordinates

% Categorize source coordinates
lateralSources = data{1:6, 6:end}; % Lateral sources
medialSources = data{7:end, 6:end}; % Medial sources

%% User input for knee side
kneeSide = "Right"; % Change to "Left" if left knee

% Adjust axis directions based on knee side
if strcmpi(kneeSide, "Right")
% Right knee: X positive leftward, Y downward, Z downward
sensorCoordinates(:,1) = -sensorCoordinates(:,1);
lateralSources(:,1) = -lateralSources(:,1);
medialSources(:,1) = -medialSources(:,1);
elseif strcmpi(kneeSide, "Left")
% Left knee: X positive rightward, Y downward, Z downward
% No inversion needed
else
error(‘Knee side must be either "Right" or "Left".’);
end

%% Plotting – 3D Plot
figure;
hold on;

% Plot sensor coordinates
scatter3(sensorCoordinates(:,1), sensorCoordinates(:,2), sensorCoordinates(:,3), …
80, ‘filled’, ‘MarkerFaceColor’, ‘b’);

% Add sensor name labels
for i = 1:length(sensorNames)
text(sensorCoordinates(i,1), sensorCoordinates(i,2), sensorCoordinates(i,3), …
[‘ ‘ char(sensorNames{i})], ‘FontSize’, 10, ‘Color’, ‘b’);
end

% Plot lateral sources
scatter3(lateralSources(:,1), lateralSources(:,2), lateralSources(:,3), …
60, ‘o’, ‘MarkerEdgeColor’, ‘r’, ‘DisplayName’, ‘Lateral Sources’);

% Plot medial sources
scatter3(medialSources(:,1), medialSources(:,2), medialSources(:,3), …
60, ‘s’, ‘MarkerEdgeColor’, ‘g’, ‘DisplayName’, ‘Medial Sources’);

%% Formatting for 3D plot
grid on;
xlabel(‘X-axis’);
ylabel(‘Y-axis’);
zlabel(‘Z-axis’);
title([‘3D Plot of Sensors and Sources (‘ kneeSide ‘ Knee)’]);
legend;
view(3);
hold off;

%% Plotting – 2D Plot (X vs Y)
figure;
hold on;
scatter(sensorCoordinates(:,1), sensorCoordinates(:,2), 80, ‘filled’, ‘b’);
for i = 1:length(sensorNames)
text(sensorCoordinates(i,1), sensorCoordinates(i,2), [‘ ‘ char(sensorNames{i})], ‘FontSize’, 10, ‘Color’, ‘b’);
end
scatter(lateralSources(:,1), lateralSources(:,2), 60, ‘ro’, ‘DisplayName’, ‘Lateral Sources’);
scatter(medialSources(:,1), medialSources(:,2), 60, ‘gs’, ‘DisplayName’, ‘Medial Sources’);

xlabel(‘X-axis’);
ylabel(‘Y-axis’);
title([‘2D Plot (X vs Y) – ‘ kneeSide ‘ Knee’]);
legend;
grid on;
hold off;

%% Plotting – 2D Plot (X vs Z)
figure;
hold on;
scatter(sensorCoordinates(:,1), sensorCoordinates(:,3), 80, ‘filled’, ‘b’);
for i = 1:length(sensorNames)
text(sensorCoordinates(i,1), sensorCoordinates(i,3), [‘ ‘ char(sensorNames{i})], ‘FontSize’, 10, ‘Color’, ‘b’);
end
scatter(lateralSources(:,1), lateralSources(:,3), 60, ‘ro’, ‘DisplayName’, ‘Lateral Sources’);
scatter(medialSources(:,1), medialSources(:,3), 60, ‘gs’, ‘DisplayName’, ‘Medial Sources’);

xlabel(‘X-axis’);
ylabel(‘Z-axis’);
title([‘2D Plot (X vs Z) – ‘ kneeSide ‘ Knee’]);
legend;
grid on;
hold off; matlab, 3d-plot, customized axes MATLAB Answers — New Questions

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The Documentation for a Dropout layer states that:
"At prediction time the output of a dropout layer is equal to its input."
I assume this means that during prediction, there is no dropout.
Is there a method in MATLAB to enable Dropout during prediction time?The Documentation for a Dropout layer states that:
"At prediction time the output of a dropout layer is equal to its input."
I assume this means that during prediction, there is no dropout.
Is there a method in MATLAB to enable Dropout during prediction time? The Documentation for a Dropout layer states that:
"At prediction time the output of a dropout layer is equal to its input."
I assume this means that during prediction, there is no dropout.
Is there a method in MATLAB to enable Dropout during prediction time? deep-learing, neural network MATLAB Answers — New Questions

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I have many models which uses C-Callers and they work properly in accelerator mode with matlab 2024b and previous.
With Matlab 2025a almost all these models crash when I run them. "Validate custom code" runs fine, but when I run the model it crash. In normal as it should be it doesn’t crash.
The issue is located into the C-caller of course, but I never had problems until R2025a.
Do you have similar issues? Any advices?
Many thanks in advanced.
DI have many models which uses C-Callers and they work properly in accelerator mode with matlab 2024b and previous.
With Matlab 2025a almost all these models crash when I run them. "Validate custom code" runs fine, but when I run the model it crash. In normal as it should be it doesn’t crash.
The issue is located into the C-caller of course, but I never had problems until R2025a.
Do you have similar issues? Any advices?
Many thanks in advanced.
D I have many models which uses C-Callers and they work properly in accelerator mode with matlab 2024b and previous.
With Matlab 2025a almost all these models crash when I run them. "Validate custom code" runs fine, but when I run the model it crash. In normal as it should be it doesn’t crash.
The issue is located into the C-caller of course, but I never had problems until R2025a.
Do you have similar issues? Any advices?
Many thanks in advanced.
D c-caller, simscape MATLAB Answers — New Questions

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verver ver answers version MATLAB Answers — New Questions

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<</matlabcentral/answers/uploaded_files/108002/matlab.png>>

I mean the font size of x:5 and y:-12.38
Thanks in advance<</matlabcentral/answers/uploaded_files/108002/matlab.png>>

I mean the font size of x:5 and y:-12.38
Thanks in advance <</matlabcentral/answers/uploaded_files/108002/matlab.png>>

I mean the font size of x:5 and y:-12.38
Thanks in advance datatip, fontsize MATLAB Answers — New Questions

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