I have a .NET Project on my Windows machine that uses MATLAB Compiler SDK .NET Data API assemblies, and I want to try building and running the project through Windows Subsystem for Linux (WSL) on the same machine. Is this possible? How can I do this?I have a .NET Project on my Windows machine that uses MATLAB Compiler SDK .NET Data API assemblies, and I want to try building and running the project through Windows Subsystem for Linux (WSL) on the same machine. Is this possible? How can I do this? I have a .NET Project on my Windows machine that uses MATLAB Compiler SDK .NET Data API assemblies, and I want to try building and running the project through Windows Subsystem for Linux (WSL) on the same machine. Is this possible? How can I do this? .net, data, api, wsl, windows, linux, subsystem, compiler, assemblies MATLAB Answers — New Questions

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I need to finish the control system analysis course for a class and the first coding task isnt being accepted by matlab i font understand why i need this to workI need to finish the control system analysis course for a class and the first coding task isnt being accepted by matlab i font understand why i need this to work I need to finish the control system analysis course for a class and the first coding task isnt being accepted by matlab i font understand why i need this to work self-paced course MATLAB Answers — New Questions

​

Dear all,
for my master’s thesis, I have to compute the standard errors of the estimated model parameters using a bootstrap approach. Trying to run the file "booti2.m", I obtain the following error:
>> booti2
Index exceeds the number of array elements. Index must not exceed 19.

Error in llfn (line 61)
sigvtr = bigthet(20);
^^^^^^^^^^^
Error in
fminunc (line 233)
f = feval(funfcn{3},x,varargin{:});
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in
booti2 (line 87)
[thetstar,fstar,exitflag] = fminunc(@llfn,bigtheto,options);
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Caused by:
Failure in initial objective function evaluation. FMINUNC cannot continue.

>>
Please note, that before I could run the original "booti.m" file with data for France successfully, while now I can’t run neither Slovenian nor French file. Thank you so much! Meancimpr.DAT file is in the folder Searching for Starting Values – Full SampleDear all,
for my master’s thesis, I have to compute the standard errors of the estimated model parameters using a bootstrap approach. Trying to run the file "booti2.m", I obtain the following error:
>> booti2
Index exceeds the number of array elements. Index must not exceed 19.

Error in llfn (line 61)
sigvtr = bigthet(20);
^^^^^^^^^^^
Error in
fminunc (line 233)
f = feval(funfcn{3},x,varargin{:});
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in
booti2 (line 87)
[thetstar,fstar,exitflag] = fminunc(@llfn,bigtheto,options);
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Caused by:
Failure in initial objective function evaluation. FMINUNC cannot continue.

>>
Please note, that before I could run the original "booti.m" file with data for France successfully, while now I can’t run neither Slovenian nor French file. Thank you so much! Meancimpr.DAT file is in the folder Searching for Starting Values – Full Sample Dear all,
for my master’s thesis, I have to compute the standard errors of the estimated model parameters using a bootstrap approach. Trying to run the file "booti2.m", I obtain the following error:
>> booti2
Index exceeds the number of array elements. Index must not exceed 19.

Error in llfn (line 61)
sigvtr = bigthet(20);
^^^^^^^^^^^
Error in
fminunc (line 233)
f = feval(funfcn{3},x,varargin{:});
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in
booti2 (line 87)
[thetstar,fstar,exitflag] = fminunc(@llfn,bigtheto,options);
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Caused by:
Failure in initial objective function evaluation. FMINUNC cannot continue.

>>
Please note, that before I could run the original "booti.m" file with data for France successfully, while now I can’t run neither Slovenian nor French file. Thank you so much! Meancimpr.DAT file is in the folder Searching for Starting Values – Full Sample standard errors MATLAB Answers — New Questions

​

Hello to everybody,
I want to calculate spray angle of this image and i use this code (that I found here) but this code evaluates an angle of 0 degrees. Can someone help me???? This is the final image
function test
clc;
close all;
workspace; % Make sure the workspace panel with all the variables is showing.
format longg;
format compact;
fontSize = 18;

% Check that user has the Image Processing Toolbox installed.
hasIPT = license(‘test’, ‘image_toolbox’);
if ~hasIPT
% User does not have the toolbox installed.
message = sprintf(‘Sorry, but you do not seem to have the Image Processing Toolbox.nDo you want to try to continue anyway?’);
reply = questdlg(message, ‘Toolbox missing’, ‘Yes’, ‘No’, ‘Yes’);
if strcmpi(reply, ‘No’)
% User said No, so exit.
return;
end
end

%===============================================================================
% Read in a standard MATLAB color demo image.
folder = pwd;
baseFileName = ‘prova.jpg’;
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
if ~exist(fullFileName, ‘file’)
% Didn’t find it there. Check the search path for it.
fullFileName = baseFileName; % No path this time.
if ~exist(fullFileName, ‘file’)
% Still didn’t find it. Alert user.
errorMessage = sprintf(‘Error: %s does not exist.’, fullFileName);
uiwait(warndlg(errorMessage));
return;
end
end
grayImage = imread(fullFileName);
% Get the dimensions of the image. numberOfColorBands should be = 3.
[rows, columns, numberOfColorBands] = size(grayImage);
if numberOfColorBands > 1
% If it’s really color, then convert to gray scale.
grayImage = grayImage(:,:,2);
end

% Display the original image.
subplot(2, 3, 1);
imshow(grayImage);
axis on;
title(‘Original Image’, ‘FontSize’, fontSize);
% Enlarge figure to full screen.
set(gcf, ‘Units’, ‘Normalized’, ‘Outerposition’, [0, 0, 1, 1]);

% Let’s compute and display the histogram.
[pixelCount, grayLevels] = imhist(grayImage);
% Suppress bins at 0 and 255 because they’re so tall we can’t see the rest of it.
pixelCount(1) = 0;
pixelCount(end) = 0;
subplot(2, 3, 2);
bar(grayLevels, pixelCount);
grid on;
title(‘Histogram of original image’, ‘FontSize’, fontSize);
xlim([0 grayLevels(end)]); % Scale x axis manually.

% [lowThreshold, highThreshold, lastThresholdedBand] = threshold(133, 255, grayImage);

% Threshold the image
thresholdValue = 50;
binaryImage = grayImage < thresholdValue;
% Fill holes
% binaryImage = imfill(binaryImage, ‘holes’);
% Display the image.
subplot(2, 3, 3);
imshow(binaryImage);
axis on;
title(‘Binary Image’, ‘FontSize’, fontSize);

%—————————————————————————
% Extract the largest area using our custom function ExtractNLargestBlobs().
numberToExtract = 1;
binaryImage = ExtractNLargestBlobs(binaryImage, numberToExtract);
%—————————————————————————
% % Fill any holes that might be present
binaryImage = imfill(binaryImage, ‘holes’);
% Do an opening to smooth out the edges.
se = strel(‘disk’, 3, 0);
binaryImage = imopen(binaryImage, se);

% Display the image.
subplot(2, 3, 4);
imshow(binaryImage);
axis on;
title(‘Biggest Blob’, ‘FontSize’, fontSize);
drawnow;

% Go down the image line by line finding the width
widths = zeros(1, rows);
leftEdge = zeros(1, rows);
rightEdge = zeros(1, rows);
for row = 1 : rows
thisRow = binaryImage(row, :);
leftIndex = find(thisRow, 1, ‘first’);
if ~isempty(leftIndex)
% There’s something there
leftEdge(row) = leftIndex;
rightEdge(row) = find(thisRow, 1, ‘last’);
widths(row) = rightEdge(row) – leftEdge(row);
end
end

% plot the widths
subplot(2, 3, 5);
plot(1:rows, widths, ‘b-‘, ‘LineWidth’, 2);
grid on;
title(‘Widths as a function of line number’, ‘FontSize’, fontSize);
axis on;

% Now find min width at top and max width.
% This could be an algorithm all by itself, but since this is just a simple demo,
% I’m going to hard code in values that I can see from the plot of the widths.
% I’m going to fit between rows 10 and 60
x = 10:60;
% Get the left angle.
y = leftEdge(x);
leftCoefficients = polyfit(x, y, 1);
leftAngle = atand(leftCoefficients(1))
subplot(2, 3, 6);
plot(x, y, ‘b-‘, ‘LineWidth’, 2);
hold on;
y = rightEdge(x);
rightCoefficients = polyfit(x, y, 1);
rightAngle = atand(rightCoefficients(1))
plot(x, y, ‘r-‘, ‘LineWidth’, 2);
grid on;
coneAngle = abs(leftAngle) + abs(rightAngle);
% Plot the fitted lines
yLeftFit = polyval(leftCoefficients, x);
plot(x, yLeftFit, ‘b-‘, ‘LineWidth’, 2);
yRightFit = polyval(rightCoefficients, x);
plot(x, yRightFit, ‘r-‘, ‘LineWidth’, 2);
legend(‘left’, ‘right’, ‘Location’, ‘east’);

message = sprintf(‘Done with demo.nThe cone angle is %.1f degrees.’, coneAngle);
uiwait(helpdlg(message));

%==============================================================================================
% Function to return the specified number of largest or smallest blobs in a binary image.
% If numberToExtract > 0 it returns the numberToExtract largest blobs.
% If numberToExtract < 0 it returns the numberToExtract smallest blobs.
% Example: return a binary image with only the largest blob:
% binaryImage = ExtractNLargestBlobs(binaryImage, 1);
% Example: return a binary image with the 3 smallest blobs:
% binaryImage = ExtractNLargestBlobs(binaryImage, -3);
function binaryImage = ExtractNLargestBlobs(binaryImage, numberToExtract)
try
% Get all the blob properties. Can only pass in originalImage in version R2008a and later.
[labeledImage, numberOfBlobs] = bwlabel(binaryImage);
blobMeasurements = regionprops(labeledImage, ‘area’);
% Get all the areas
allAreas = [blobMeasurements.Area];
if numberToExtract > length(allAreas);
% Limit the number they can get to the number that are there/available.
numberToExtract = length(allAreas);
end
if numberToExtract > 0
% For positive numbers, sort in order of largest to smallest.
% Sort them.
[sortedAreas, sortIndexes] = sort(allAreas, ‘descend’);
elseif numberToExtract < 0
% For negative numbers, sort in order of smallest to largest.
% Sort them.
[sortedAreas, sortIndexes] = sort(allAreas, ‘ascend’);
% Need to negate numberToExtract so we can use it in sortIndexes later.
numberToExtract = -numberToExtract;
else
% numberToExtract = 0. Shouldn’t happen. Return no blobs.
binaryImage = false(size(binaryImage));
return;
end
% Extract the "numberToExtract" largest blob(a)s using ismember().
biggestBlob = ismember(labeledImage, sortIndexes(1:numberToExtract));
% Convert from integer labeled image into binary (logical) image.
binaryImage = biggestBlob > 0;
catch ME
errorMessage = sprintf(‘Error in function ExtractNLargestBlobs().nnError Message:n%s’, ME.message);
fprintf(1, ‘%sn’, errorMessage);
uiwait(warndlg(errorMessage));
endHello to everybody,
I want to calculate spray angle of this image and i use this code (that I found here) but this code evaluates an angle of 0 degrees. Can someone help me???? This is the final image
function test
clc;
close all;
workspace; % Make sure the workspace panel with all the variables is showing.
format longg;
format compact;
fontSize = 18;

% Check that user has the Image Processing Toolbox installed.
hasIPT = license(‘test’, ‘image_toolbox’);
if ~hasIPT
% User does not have the toolbox installed.
message = sprintf(‘Sorry, but you do not seem to have the Image Processing Toolbox.nDo you want to try to continue anyway?’);
reply = questdlg(message, ‘Toolbox missing’, ‘Yes’, ‘No’, ‘Yes’);
if strcmpi(reply, ‘No’)
% User said No, so exit.
return;
end
end

%===============================================================================
% Read in a standard MATLAB color demo image.
folder = pwd;
baseFileName = ‘prova.jpg’;
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
if ~exist(fullFileName, ‘file’)
% Didn’t find it there. Check the search path for it.
fullFileName = baseFileName; % No path this time.
if ~exist(fullFileName, ‘file’)
% Still didn’t find it. Alert user.
errorMessage = sprintf(‘Error: %s does not exist.’, fullFileName);
uiwait(warndlg(errorMessage));
return;
end
end
grayImage = imread(fullFileName);
% Get the dimensions of the image. numberOfColorBands should be = 3.
[rows, columns, numberOfColorBands] = size(grayImage);
if numberOfColorBands > 1
% If it’s really color, then convert to gray scale.
grayImage = grayImage(:,:,2);
end

% Display the original image.
subplot(2, 3, 1);
imshow(grayImage);
axis on;
title(‘Original Image’, ‘FontSize’, fontSize);
% Enlarge figure to full screen.
set(gcf, ‘Units’, ‘Normalized’, ‘Outerposition’, [0, 0, 1, 1]);

% Let’s compute and display the histogram.
[pixelCount, grayLevels] = imhist(grayImage);
% Suppress bins at 0 and 255 because they’re so tall we can’t see the rest of it.
pixelCount(1) = 0;
pixelCount(end) = 0;
subplot(2, 3, 2);
bar(grayLevels, pixelCount);
grid on;
title(‘Histogram of original image’, ‘FontSize’, fontSize);
xlim([0 grayLevels(end)]); % Scale x axis manually.

% [lowThreshold, highThreshold, lastThresholdedBand] = threshold(133, 255, grayImage);

% Threshold the image
thresholdValue = 50;
binaryImage = grayImage < thresholdValue;
% Fill holes
% binaryImage = imfill(binaryImage, ‘holes’);
% Display the image.
subplot(2, 3, 3);
imshow(binaryImage);
axis on;
title(‘Binary Image’, ‘FontSize’, fontSize);

%—————————————————————————
% Extract the largest area using our custom function ExtractNLargestBlobs().
numberToExtract = 1;
binaryImage = ExtractNLargestBlobs(binaryImage, numberToExtract);
%—————————————————————————
% % Fill any holes that might be present
binaryImage = imfill(binaryImage, ‘holes’);
% Do an opening to smooth out the edges.
se = strel(‘disk’, 3, 0);
binaryImage = imopen(binaryImage, se);

% Display the image.
subplot(2, 3, 4);
imshow(binaryImage);
axis on;
title(‘Biggest Blob’, ‘FontSize’, fontSize);
drawnow;

% Go down the image line by line finding the width
widths = zeros(1, rows);
leftEdge = zeros(1, rows);
rightEdge = zeros(1, rows);
for row = 1 : rows
thisRow = binaryImage(row, :);
leftIndex = find(thisRow, 1, ‘first’);
if ~isempty(leftIndex)
% There’s something there
leftEdge(row) = leftIndex;
rightEdge(row) = find(thisRow, 1, ‘last’);
widths(row) = rightEdge(row) – leftEdge(row);
end
end

% plot the widths
subplot(2, 3, 5);
plot(1:rows, widths, ‘b-‘, ‘LineWidth’, 2);
grid on;
title(‘Widths as a function of line number’, ‘FontSize’, fontSize);
axis on;

% Now find min width at top and max width.
% This could be an algorithm all by itself, but since this is just a simple demo,
% I’m going to hard code in values that I can see from the plot of the widths.
% I’m going to fit between rows 10 and 60
x = 10:60;
% Get the left angle.
y = leftEdge(x);
leftCoefficients = polyfit(x, y, 1);
leftAngle = atand(leftCoefficients(1))
subplot(2, 3, 6);
plot(x, y, ‘b-‘, ‘LineWidth’, 2);
hold on;
y = rightEdge(x);
rightCoefficients = polyfit(x, y, 1);
rightAngle = atand(rightCoefficients(1))
plot(x, y, ‘r-‘, ‘LineWidth’, 2);
grid on;
coneAngle = abs(leftAngle) + abs(rightAngle);
% Plot the fitted lines
yLeftFit = polyval(leftCoefficients, x);
plot(x, yLeftFit, ‘b-‘, ‘LineWidth’, 2);
yRightFit = polyval(rightCoefficients, x);
plot(x, yRightFit, ‘r-‘, ‘LineWidth’, 2);
legend(‘left’, ‘right’, ‘Location’, ‘east’);

message = sprintf(‘Done with demo.nThe cone angle is %.1f degrees.’, coneAngle);
uiwait(helpdlg(message));

%==============================================================================================
% Function to return the specified number of largest or smallest blobs in a binary image.
% If numberToExtract > 0 it returns the numberToExtract largest blobs.
% If numberToExtract < 0 it returns the numberToExtract smallest blobs.
% Example: return a binary image with only the largest blob:
% binaryImage = ExtractNLargestBlobs(binaryImage, 1);
% Example: return a binary image with the 3 smallest blobs:
% binaryImage = ExtractNLargestBlobs(binaryImage, -3);
function binaryImage = ExtractNLargestBlobs(binaryImage, numberToExtract)
try
% Get all the blob properties. Can only pass in originalImage in version R2008a and later.
[labeledImage, numberOfBlobs] = bwlabel(binaryImage);
blobMeasurements = regionprops(labeledImage, ‘area’);
% Get all the areas
allAreas = [blobMeasurements.Area];
if numberToExtract > length(allAreas);
% Limit the number they can get to the number that are there/available.
numberToExtract = length(allAreas);
end
if numberToExtract > 0
% For positive numbers, sort in order of largest to smallest.
% Sort them.
[sortedAreas, sortIndexes] = sort(allAreas, ‘descend’);
elseif numberToExtract < 0
% For negative numbers, sort in order of smallest to largest.
% Sort them.
[sortedAreas, sortIndexes] = sort(allAreas, ‘ascend’);
% Need to negate numberToExtract so we can use it in sortIndexes later.
numberToExtract = -numberToExtract;
else
% numberToExtract = 0. Shouldn’t happen. Return no blobs.
binaryImage = false(size(binaryImage));
return;
end
% Extract the "numberToExtract" largest blob(a)s using ismember().
biggestBlob = ismember(labeledImage, sortIndexes(1:numberToExtract));
% Convert from integer labeled image into binary (logical) image.
binaryImage = biggestBlob > 0;
catch ME
errorMessage = sprintf(‘Error in function ExtractNLargestBlobs().nnError Message:n%s’, ME.message);
fprintf(1, ‘%sn’, errorMessage);
uiwait(warndlg(errorMessage));
end Hello to everybody,
I want to calculate spray angle of this image and i use this code (that I found here) but this code evaluates an angle of 0 degrees. Can someone help me???? This is the final image
function test
clc;
close all;
workspace; % Make sure the workspace panel with all the variables is showing.
format longg;
format compact;
fontSize = 18;

% Check that user has the Image Processing Toolbox installed.
hasIPT = license(‘test’, ‘image_toolbox’);
if ~hasIPT
% User does not have the toolbox installed.
message = sprintf(‘Sorry, but you do not seem to have the Image Processing Toolbox.nDo you want to try to continue anyway?’);
reply = questdlg(message, ‘Toolbox missing’, ‘Yes’, ‘No’, ‘Yes’);
if strcmpi(reply, ‘No’)
% User said No, so exit.
return;
end
end

%===============================================================================
% Read in a standard MATLAB color demo image.
folder = pwd;
baseFileName = ‘prova.jpg’;
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
if ~exist(fullFileName, ‘file’)
% Didn’t find it there. Check the search path for it.
fullFileName = baseFileName; % No path this time.
if ~exist(fullFileName, ‘file’)
% Still didn’t find it. Alert user.
errorMessage = sprintf(‘Error: %s does not exist.’, fullFileName);
uiwait(warndlg(errorMessage));
return;
end
end
grayImage = imread(fullFileName);
% Get the dimensions of the image. numberOfColorBands should be = 3.
[rows, columns, numberOfColorBands] = size(grayImage);
if numberOfColorBands > 1
% If it’s really color, then convert to gray scale.
grayImage = grayImage(:,:,2);
end

% Display the original image.
subplot(2, 3, 1);
imshow(grayImage);
axis on;
title(‘Original Image’, ‘FontSize’, fontSize);
% Enlarge figure to full screen.
set(gcf, ‘Units’, ‘Normalized’, ‘Outerposition’, [0, 0, 1, 1]);

% Let’s compute and display the histogram.
[pixelCount, grayLevels] = imhist(grayImage);
% Suppress bins at 0 and 255 because they’re so tall we can’t see the rest of it.
pixelCount(1) = 0;
pixelCount(end) = 0;
subplot(2, 3, 2);
bar(grayLevels, pixelCount);
grid on;
title(‘Histogram of original image’, ‘FontSize’, fontSize);
xlim([0 grayLevels(end)]); % Scale x axis manually.

% [lowThreshold, highThreshold, lastThresholdedBand] = threshold(133, 255, grayImage);

% Threshold the image
thresholdValue = 50;
binaryImage = grayImage < thresholdValue;
% Fill holes
% binaryImage = imfill(binaryImage, ‘holes’);
% Display the image.
subplot(2, 3, 3);
imshow(binaryImage);
axis on;
title(‘Binary Image’, ‘FontSize’, fontSize);

%—————————————————————————
% Extract the largest area using our custom function ExtractNLargestBlobs().
numberToExtract = 1;
binaryImage = ExtractNLargestBlobs(binaryImage, numberToExtract);
%—————————————————————————
% % Fill any holes that might be present
binaryImage = imfill(binaryImage, ‘holes’);
% Do an opening to smooth out the edges.
se = strel(‘disk’, 3, 0);
binaryImage = imopen(binaryImage, se);

% Display the image.
subplot(2, 3, 4);
imshow(binaryImage);
axis on;
title(‘Biggest Blob’, ‘FontSize’, fontSize);
drawnow;

% Go down the image line by line finding the width
widths = zeros(1, rows);
leftEdge = zeros(1, rows);
rightEdge = zeros(1, rows);
for row = 1 : rows
thisRow = binaryImage(row, :);
leftIndex = find(thisRow, 1, ‘first’);
if ~isempty(leftIndex)
% There’s something there
leftEdge(row) = leftIndex;
rightEdge(row) = find(thisRow, 1, ‘last’);
widths(row) = rightEdge(row) – leftEdge(row);
end
end

% plot the widths
subplot(2, 3, 5);
plot(1:rows, widths, ‘b-‘, ‘LineWidth’, 2);
grid on;
title(‘Widths as a function of line number’, ‘FontSize’, fontSize);
axis on;

% Now find min width at top and max width.
% This could be an algorithm all by itself, but since this is just a simple demo,
% I’m going to hard code in values that I can see from the plot of the widths.
% I’m going to fit between rows 10 and 60
x = 10:60;
% Get the left angle.
y = leftEdge(x);
leftCoefficients = polyfit(x, y, 1);
leftAngle = atand(leftCoefficients(1))
subplot(2, 3, 6);
plot(x, y, ‘b-‘, ‘LineWidth’, 2);
hold on;
y = rightEdge(x);
rightCoefficients = polyfit(x, y, 1);
rightAngle = atand(rightCoefficients(1))
plot(x, y, ‘r-‘, ‘LineWidth’, 2);
grid on;
coneAngle = abs(leftAngle) + abs(rightAngle);
% Plot the fitted lines
yLeftFit = polyval(leftCoefficients, x);
plot(x, yLeftFit, ‘b-‘, ‘LineWidth’, 2);
yRightFit = polyval(rightCoefficients, x);
plot(x, yRightFit, ‘r-‘, ‘LineWidth’, 2);
legend(‘left’, ‘right’, ‘Location’, ‘east’);

message = sprintf(‘Done with demo.nThe cone angle is %.1f degrees.’, coneAngle);
uiwait(helpdlg(message));

%==============================================================================================
% Function to return the specified number of largest or smallest blobs in a binary image.
% If numberToExtract > 0 it returns the numberToExtract largest blobs.
% If numberToExtract < 0 it returns the numberToExtract smallest blobs.
% Example: return a binary image with only the largest blob:
% binaryImage = ExtractNLargestBlobs(binaryImage, 1);
% Example: return a binary image with the 3 smallest blobs:
% binaryImage = ExtractNLargestBlobs(binaryImage, -3);
function binaryImage = ExtractNLargestBlobs(binaryImage, numberToExtract)
try
% Get all the blob properties. Can only pass in originalImage in version R2008a and later.
[labeledImage, numberOfBlobs] = bwlabel(binaryImage);
blobMeasurements = regionprops(labeledImage, ‘area’);
% Get all the areas
allAreas = [blobMeasurements.Area];
if numberToExtract > length(allAreas);
% Limit the number they can get to the number that are there/available.
numberToExtract = length(allAreas);
end
if numberToExtract > 0
% For positive numbers, sort in order of largest to smallest.
% Sort them.
[sortedAreas, sortIndexes] = sort(allAreas, ‘descend’);
elseif numberToExtract < 0
% For negative numbers, sort in order of smallest to largest.
% Sort them.
[sortedAreas, sortIndexes] = sort(allAreas, ‘ascend’);
% Need to negate numberToExtract so we can use it in sortIndexes later.
numberToExtract = -numberToExtract;
else
% numberToExtract = 0. Shouldn’t happen. Return no blobs.
binaryImage = false(size(binaryImage));
return;
end
% Extract the "numberToExtract" largest blob(a)s using ismember().
biggestBlob = ismember(labeledImage, sortIndexes(1:numberToExtract));
% Convert from integer labeled image into binary (logical) image.
binaryImage = biggestBlob > 0;
catch ME
errorMessage = sprintf(‘Error in function ExtractNLargestBlobs().nnError Message:n%s’, ME.message);
fprintf(1, ‘%sn’, errorMessage);
uiwait(warndlg(errorMessage));
end angle, spray, cone angle, angle spray, spray angle, spray cone MATLAB Answers — New Questions

​

Hello, I am unable to launch MATLAB R2026a due to Licensing Error 5005.
Steps I have already taken:
Attempted manual activation using activate_matlab.exe as Administrator.
Performed a clean reinstall, manually deleting the following folders:
C:Program FilesMATLABR2026a
C:UsersdanieAppDataRoamingMathWorks
I noticed a ghost entry for a previous version (R2022a) in Windows Settings that points to a non-existent path on the Desktop, which I cannot remove.
Checked the licenses folder; it was empty or contained a file that didn’t resolve the issue.
I have attached a screenshot of the error message. How can I force MATLAB to recognize my valid license?Hello, I am unable to launch MATLAB R2026a due to Licensing Error 5005.
Steps I have already taken:
Attempted manual activation using activate_matlab.exe as Administrator.
Performed a clean reinstall, manually deleting the following folders:
C:Program FilesMATLABR2026a
C:UsersdanieAppDataRoamingMathWorks
I noticed a ghost entry for a previous version (R2022a) in Windows Settings that points to a non-existent path on the Desktop, which I cannot remove.
Checked the licenses folder; it was empty or contained a file that didn’t resolve the issue.
I have attached a screenshot of the error message. How can I force MATLAB to recognize my valid license? Hello, I am unable to launch MATLAB R2026a due to Licensing Error 5005.
Steps I have already taken:
Attempted manual activation using activate_matlab.exe as Administrator.
Performed a clean reinstall, manually deleting the following folders:
C:Program FilesMATLABR2026a
C:UsersdanieAppDataRoamingMathWorks
I noticed a ghost entry for a previous version (R2022a) in Windows Settings that points to a non-existent path on the Desktop, which I cannot remove.
Checked the licenses folder; it was empty or contained a file that didn’t resolve the issue.
I have attached a screenshot of the error message. How can I force MATLAB to recognize my valid license? installation, licensing, error 5005, activation MATLAB Answers — New Questions

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I have a Simscape block that takes linPosition (physical signal) as an input and computes rotPosition (another physical signal). If I put it in my model, where linPosition is 0 for 1ms and then starts increasing smoothly, it works (rotPosition and theta are constant for 1ms and then follow the growth of linPosition). The problem arises when I try to add an additional computation: omega == der(theta) (I need it to make more calculation and provide another output). With that additional equation, I get this warning:
Equations (including nonlinear equations) of one or more components may be dependent or inconsistent. This can cause problems in transient initialization.
referred to the first and second equations. Then the simulation stops at 1ms:
An error occurred during simulation and the simulation was stopped
Caused by:
[‘testCircuit/Solver Configuration’]: Transient initialization at time 0.001, solving for consistent states and modes, failed to converge.
Nonlinear solver: Linear Algebra error. Failed to solve using iteration matrix.
all components and nodal across variables involved
Cannot solve for one or more variables, including dynamic variable derivatives:
Time derivative of ‘CRIB.CRIB_velocityBased.breaking_part.offsetSliderCrank.theta’ (rot angle with respect to horizontal)
‘CRIB.CRIB_velocityBased.breaking_part.offsetSliderCrank.omega’ (omega)
In my mind, theta and gamma in the equations are computed and then omega follows the value of theta, but I guess that this is not how it works.

component offsetSliderCrank

inputs
linPosition = {0, ‘mm’};
end

outputs
rotPosition = {0, ‘deg’};
end

parameters
lRod = { 45, ‘mm’ }; %
rJunction = { 10, ‘mm’ }; %
xC = { 2, ‘mm’ }; %
yC = { 40, ‘mm’ }; %

end

variables(Access=Protected)
theta = {value={0,’deg’},imin={-90,’deg’},imax={89,’deg’}}; % rot angle with respect to horizontal
gamma = {value={0,’deg’},imin={0,’deg’},imax={180,’deg’}}; % angle between lRod and horizontal
omega = {0,’deg/s’};

end

equations
lRod*cos(gamma)+xC == rJunction*cos(theta);
yC-linPosition – rJunction*sin(theta) == lRod*sin(gamma);
let
theta0 = asin((yC^2+rJunction^2-lRod^2+xC^2)/(2*rJunction*sqrt(xC^2+yC^2))) – atan(xC/yC) ;
in
rotPosition == – (theta- theta0);
end
omega == der(theta);
end

endI have a Simscape block that takes linPosition (physical signal) as an input and computes rotPosition (another physical signal). If I put it in my model, where linPosition is 0 for 1ms and then starts increasing smoothly, it works (rotPosition and theta are constant for 1ms and then follow the growth of linPosition). The problem arises when I try to add an additional computation: omega == der(theta) (I need it to make more calculation and provide another output). With that additional equation, I get this warning:
Equations (including nonlinear equations) of one or more components may be dependent or inconsistent. This can cause problems in transient initialization.
referred to the first and second equations. Then the simulation stops at 1ms:
An error occurred during simulation and the simulation was stopped
Caused by:
[‘testCircuit/Solver Configuration’]: Transient initialization at time 0.001, solving for consistent states and modes, failed to converge.
Nonlinear solver: Linear Algebra error. Failed to solve using iteration matrix.
all components and nodal across variables involved
Cannot solve for one or more variables, including dynamic variable derivatives:
Time derivative of ‘CRIB.CRIB_velocityBased.breaking_part.offsetSliderCrank.theta’ (rot angle with respect to horizontal)
‘CRIB.CRIB_velocityBased.breaking_part.offsetSliderCrank.omega’ (omega)
In my mind, theta and gamma in the equations are computed and then omega follows the value of theta, but I guess that this is not how it works.

component offsetSliderCrank

inputs
linPosition = {0, ‘mm’};
end

outputs
rotPosition = {0, ‘deg’};
end

parameters
lRod = { 45, ‘mm’ }; %
rJunction = { 10, ‘mm’ }; %
xC = { 2, ‘mm’ }; %
yC = { 40, ‘mm’ }; %

end

variables(Access=Protected)
theta = {value={0,’deg’},imin={-90,’deg’},imax={89,’deg’}}; % rot angle with respect to horizontal
gamma = {value={0,’deg’},imin={0,’deg’},imax={180,’deg’}}; % angle between lRod and horizontal
omega = {0,’deg/s’};

end

equations
lRod*cos(gamma)+xC == rJunction*cos(theta);
yC-linPosition – rJunction*sin(theta) == lRod*sin(gamma);
let
theta0 = asin((yC^2+rJunction^2-lRod^2+xC^2)/(2*rJunction*sqrt(xC^2+yC^2))) – atan(xC/yC) ;
in
rotPosition == – (theta- theta0);
end
omega == der(theta);
end

end I have a Simscape block that takes linPosition (physical signal) as an input and computes rotPosition (another physical signal). If I put it in my model, where linPosition is 0 for 1ms and then starts increasing smoothly, it works (rotPosition and theta are constant for 1ms and then follow the growth of linPosition). The problem arises when I try to add an additional computation: omega == der(theta) (I need it to make more calculation and provide another output). With that additional equation, I get this warning:
Equations (including nonlinear equations) of one or more components may be dependent or inconsistent. This can cause problems in transient initialization.
referred to the first and second equations. Then the simulation stops at 1ms:
An error occurred during simulation and the simulation was stopped
Caused by:
[‘testCircuit/Solver Configuration’]: Transient initialization at time 0.001, solving for consistent states and modes, failed to converge.
Nonlinear solver: Linear Algebra error. Failed to solve using iteration matrix.
all components and nodal across variables involved
Cannot solve for one or more variables, including dynamic variable derivatives:
Time derivative of ‘CRIB.CRIB_velocityBased.breaking_part.offsetSliderCrank.theta’ (rot angle with respect to horizontal)
‘CRIB.CRIB_velocityBased.breaking_part.offsetSliderCrank.omega’ (omega)
In my mind, theta and gamma in the equations are computed and then omega follows the value of theta, but I guess that this is not how it works.

component offsetSliderCrank

inputs
linPosition = {0, ‘mm’};
end

outputs
rotPosition = {0, ‘deg’};
end

parameters
lRod = { 45, ‘mm’ }; %
rJunction = { 10, ‘mm’ }; %
xC = { 2, ‘mm’ }; %
yC = { 40, ‘mm’ }; %

end

variables(Access=Protected)
theta = {value={0,’deg’},imin={-90,’deg’},imax={89,’deg’}}; % rot angle with respect to horizontal
gamma = {value={0,’deg’},imin={0,’deg’},imax={180,’deg’}}; % angle between lRod and horizontal
omega = {0,’deg/s’};

end

equations
lRod*cos(gamma)+xC == rJunction*cos(theta);
yC-linPosition – rJunction*sin(theta) == lRod*sin(gamma);
let
theta0 = asin((yC^2+rJunction^2-lRod^2+xC^2)/(2*rJunction*sqrt(xC^2+yC^2))) – atan(xC/yC) ;
in
rotPosition == – (theta- theta0);
end
omega == der(theta);
end

end simscape, derivative MATLAB Answers — New Questions

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I have a simscape model, in which a set of spheres are trying to grab a cylinder. I used the spatial contact force block between the cylinder and spheres. I set up two ways that the spheres move up: 1, Input a motion actuation to the prismatic model; 2, Input a force actuation to the prismatic model.
For setup 1: The tube does NOT move. The spheres slip along the cylinder.
For setup 2: The tube MOVE UP with the spheres.
All the other settings are the same except the input actuation. So I am wondering what is the reason for this? Can anyone help to answer? Thanks!I have a simscape model, in which a set of spheres are trying to grab a cylinder. I used the spatial contact force block between the cylinder and spheres. I set up two ways that the spheres move up: 1, Input a motion actuation to the prismatic model; 2, Input a force actuation to the prismatic model.
For setup 1: The tube does NOT move. The spheres slip along the cylinder.
For setup 2: The tube MOVE UP with the spheres.
All the other settings are the same except the input actuation. So I am wondering what is the reason for this? Can anyone help to answer? Thanks! I have a simscape model, in which a set of spheres are trying to grab a cylinder. I used the spatial contact force block between the cylinder and spheres. I set up two ways that the spheres move up: 1, Input a motion actuation to the prismatic model; 2, Input a force actuation to the prismatic model.
For setup 1: The tube does NOT move. The spheres slip along the cylinder.
For setup 2: The tube MOVE UP with the spheres.
All the other settings are the same except the input actuation. So I am wondering what is the reason for this? Can anyone help to answer? Thanks! simscape, spatial contact force, prismatic joint MATLAB Answers — New Questions

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I have a non-linear equation with 3 variables I wish to be constants and only 1 variable to be fitted to x,y data.
Using fittype, I have entered the equation and tried to specify b, c & d as constants. However, the options ‘problem’ only seems to allow single variables to be specified as constants
ft = fittype(@(a, b, c, d,x) a*c*x./(1+c*x) + b*d*x./(1+d*x),’problem’,’b,c,d’);
Error using fittype>iAssertValidVariableNames
The name ‘[b,c,d]’ is not a valid MATLAB variable name.

In the next part I use "fit" to fit the data, and specify the value of the 3 constants and starting point of the fit:
f = fit(p(1:index),q(1:index),ft,’problem’,[Qm2 b1 b2],’StartPoint’,Qm1);
Is there another way to specify constants in the fittype?I have a non-linear equation with 3 variables I wish to be constants and only 1 variable to be fitted to x,y data.
Using fittype, I have entered the equation and tried to specify b, c & d as constants. However, the options ‘problem’ only seems to allow single variables to be specified as constants
ft = fittype(@(a, b, c, d,x) a*c*x./(1+c*x) + b*d*x./(1+d*x),’problem’,’b,c,d’);
Error using fittype>iAssertValidVariableNames
The name ‘[b,c,d]’ is not a valid MATLAB variable name.

In the next part I use "fit" to fit the data, and specify the value of the 3 constants and starting point of the fit:
f = fit(p(1:index),q(1:index),ft,’problem’,[Qm2 b1 b2],’StartPoint’,Qm1);
Is there another way to specify constants in the fittype? I have a non-linear equation with 3 variables I wish to be constants and only 1 variable to be fitted to x,y data.
Using fittype, I have entered the equation and tried to specify b, c & d as constants. However, the options ‘problem’ only seems to allow single variables to be specified as constants
ft = fittype(@(a, b, c, d,x) a*c*x./(1+c*x) + b*d*x./(1+d*x),’problem’,’b,c,d’);
Error using fittype>iAssertValidVariableNames
The name ‘[b,c,d]’ is not a valid MATLAB variable name.

In the next part I use "fit" to fit the data, and specify the value of the 3 constants and starting point of the fit:
f = fit(p(1:index),q(1:index),ft,’problem’,[Qm2 b1 b2],’StartPoint’,Qm1);
Is there another way to specify constants in the fittype? curve fitting MATLAB Answers — New Questions

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I am trying to memoize a function that takes an anonymous handle with an externally-scoped parameter as input. The problem is that no two anonymous functions are ever the same. Therefore, every time the memoization handle mf is invoked, it executes in its entirety and freshly cache’s the results.
mf=memoize(@subfunc);

A=3;
for i=1:4
mf(@(z)A*z)
end
I know that the Lord punisheth those who use eval(), but I can see no other alternative but to pass the anonymous function in char/string form and use evalin. This works fine when there are no externally-scoped variables. Caching indeed only occurs on the first call:
for i=1:4
mf(‘@(z)3*z’)
end
However, when the anonymous function does contain an externally-scoped variable A, the workaround fails.
A=3;
mf(‘@(z)A*z’)
So two questions,
Why does the workaround fail to find externally scoped variables?
Is there a more robust workaround?

function out = subfunc(argFun)

disp ‘Caching’

if isa(argFun,’function_handle’) %argFun is char or string
fun=argFun;
else
fun=evalin(‘caller’,argFun);
end

out=fun(5);

endI am trying to memoize a function that takes an anonymous handle with an externally-scoped parameter as input. The problem is that no two anonymous functions are ever the same. Therefore, every time the memoization handle mf is invoked, it executes in its entirety and freshly cache’s the results.
mf=memoize(@subfunc);

A=3;
for i=1:4
mf(@(z)A*z)
end
I know that the Lord punisheth those who use eval(), but I can see no other alternative but to pass the anonymous function in char/string form and use evalin. This works fine when there are no externally-scoped variables. Caching indeed only occurs on the first call:
for i=1:4
mf(‘@(z)3*z’)
end
However, when the anonymous function does contain an externally-scoped variable A, the workaround fails.
A=3;
mf(‘@(z)A*z’)
So two questions,
Why does the workaround fail to find externally scoped variables?
Is there a more robust workaround?

function out = subfunc(argFun)

disp ‘Caching’

if isa(argFun,’function_handle’) %argFun is char or string
fun=argFun;
else
fun=evalin(‘caller’,argFun);
end

out=fun(5);

end I am trying to memoize a function that takes an anonymous handle with an externally-scoped parameter as input. The problem is that no two anonymous functions are ever the same. Therefore, every time the memoization handle mf is invoked, it executes in its entirety and freshly cache’s the results.
mf=memoize(@subfunc);

A=3;
for i=1:4
mf(@(z)A*z)
end
I know that the Lord punisheth those who use eval(), but I can see no other alternative but to pass the anonymous function in char/string form and use evalin. This works fine when there are no externally-scoped variables. Caching indeed only occurs on the first call:
for i=1:4
mf(‘@(z)3*z’)
end
However, when the anonymous function does contain an externally-scoped variable A, the workaround fails.
A=3;
mf(‘@(z)A*z’)
So two questions,
Why does the workaround fail to find externally scoped variables?
Is there a more robust workaround?

function out = subfunc(argFun)

disp ‘Caching’

if isa(argFun,’function_handle’) %argFun is char or string
fun=argFun;
else
fun=evalin(‘caller’,argFun);
end

out=fun(5);

end memoize, eval, anonymous functions MATLAB Answers — New Questions

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