Hi
I’d like to color the area in betwen the Confidence Interval after a fitlm, like shown in the attached pic
I tried patch, area, fill, but I couln’t make any of them work.
I think the issue is that the x values are not continuous (it’s not 1:19 for example)
I have 2 continuous variables
% x = rand(1,19); y = x+[1:19];
d = load(‘d.txt’) ;
x = d(:,1) ;
y = d(:,2) ;
mdl = fitlm(x,y)
mdl.plot

thanks a lotHi
I’d like to color the area in betwen the Confidence Interval after a fitlm, like shown in the attached pic
I tried patch, area, fill, but I couln’t make any of them work.
I think the issue is that the x values are not continuous (it’s not 1:19 for example)
I have 2 continuous variables
% x = rand(1,19); y = x+[1:19];
d = load(‘d.txt’) ;
x = d(:,1) ;
y = d(:,2) ;
mdl = fitlm(x,y)
mdl.plot

thanks a lot Hi
I’d like to color the area in betwen the Confidence Interval after a fitlm, like shown in the attached pic
I tried patch, area, fill, but I couln’t make any of them work.
I think the issue is that the x values are not continuous (it’s not 1:19 for example)
I have 2 continuous variables
% x = rand(1,19); y = x+[1:19];
d = load(‘d.txt’) ;
x = d(:,1) ;
y = d(:,2) ;
mdl = fitlm(x,y)
mdl.plot

thanks a lot area color, fitlm, confidence interval 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

​

Do either of these vulnerabilites affect Matlab 2025a, 2025b or 2026a as a result of the distribution of Log4j version included, e.g. for Matlab 2025a C:Program FilesMATLABR2025ajavajarextlog4j-core-2.17.1.jar.Do either of these vulnerabilites affect Matlab 2025a, 2025b or 2026a as a result of the distribution of Log4j version included, e.g. for Matlab 2025a C:Program FilesMATLABR2025ajavajarextlog4j-core-2.17.1.jar. Do either of these vulnerabilites affect Matlab 2025a, 2025b or 2026a as a result of the distribution of Log4j version included, e.g. for Matlab 2025a C:Program FilesMATLABR2025ajavajarextlog4j-core-2.17.1.jar. log4j, cve-2026-34480, cve-2026-34477 MATLAB Answers — New Questions

​

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

​

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

​

I am trying to Install <https://sites.google.com/site/srgtstoolbox/ Surrogate Toolbox 3.0>
by Dr. Viana in 64 bit machine with Windows 10. It seems to be very easy, but somehow I am getting the error while installing.

I am getting:

—————————————————————————

What would you like to do? 2

Building with ‘Microsoft Visual C++ 2013 Professional (C)’.

MEX completed successfully.

Error using fileparts

Too many output arguments.

Error in srgtsSVMGunnFiles>srgtsCompileSVMGunn (line 101)

[pathstr, name, ext, versn] = fileparts(file_source);

Error in srgtsSVMGunnFiles (line 61)

[file_source filename] = srgtsCompileSVMGunn(srgtsRootDir, interpreter);

Error in srgtsInstall (line 75)

srgtsSVMGunnFiles(srgtsRootDir, hostname, interpreter);I am trying to Install <https://sites.google.com/site/srgtstoolbox/ Surrogate Toolbox 3.0>
by Dr. Viana in 64 bit machine with Windows 10. It seems to be very easy, but somehow I am getting the error while installing.

I am getting:

—————————————————————————

What would you like to do? 2

Building with ‘Microsoft Visual C++ 2013 Professional (C)’.

MEX completed successfully.

Error using fileparts

Too many output arguments.

Error in srgtsSVMGunnFiles>srgtsCompileSVMGunn (line 101)

[pathstr, name, ext, versn] = fileparts(file_source);

Error in srgtsSVMGunnFiles (line 61)

[file_source filename] = srgtsCompileSVMGunn(srgtsRootDir, interpreter);

Error in srgtsInstall (line 75)

srgtsSVMGunnFiles(srgtsRootDir, hostname, interpreter); I am trying to Install <https://sites.google.com/site/srgtstoolbox/ Surrogate Toolbox 3.0>
by Dr. Viana in 64 bit machine with Windows 10. It seems to be very easy, but somehow I am getting the error while installing.

I am getting:

—————————————————————————

What would you like to do? 2

Building with ‘Microsoft Visual C++ 2013 Professional (C)’.

MEX completed successfully.

Error using fileparts

Too many output arguments.

Error in srgtsSVMGunnFiles>srgtsCompileSVMGunn (line 101)

[pathstr, name, ext, versn] = fileparts(file_source);

Error in srgtsSVMGunnFiles (line 61)

[file_source filename] = srgtsCompileSVMGunn(srgtsRootDir, interpreter);

Error in srgtsInstall (line 75)

srgtsSVMGunnFiles(srgtsRootDir, hostname, interpreter); surrogate toolbox MATLAB Answers — New Questions

​

How to add Simscape Electrical in Matlab Student ?
I have Simulink, but can read or reconize electrical tools.How to add Simscape Electrical in Matlab Student ?
I have Simulink, but can read or reconize electrical tools. How to add Simscape Electrical in Matlab Student ?
I have Simulink, but can read or reconize electrical tools. simscape, electrical MATLAB Answers — New Questions

​

I wanted to confirm whether the&nbsp;Streaming Data from Hardware to Software MATLAB example supports the "AMD XILINX Zynq UltraScale+ RFSoC ZCU208" Evaluation Kit. I noticed that the lower specification "ZCU111" is supported; does this example also work with the "ZCU208"?I wanted to confirm whether the&nbsp;Streaming Data from Hardware to Software MATLAB example supports the "AMD XILINX Zynq UltraScale+ RFSoC ZCU208" Evaluation Kit. I noticed that the lower specification "ZCU111" is supported; does this example also work with the "ZCU208"? I wanted to confirm whether the&nbsp;Streaming Data from Hardware to Software MATLAB example supports the "AMD XILINX Zynq UltraScale+ RFSoC ZCU208" Evaluation Kit. I noticed that the lower specification "ZCU111" is supported; does this example also work with the "ZCU208"? amd, xilinx, xilinxzynq, zcu208 MATLAB Answers — New Questions

​

clc;clf;clear;
myLegend1 = {};

rr = [ 0.4 0.5 0.6]
for i =1:numel(rr)
x=0:0.0001:9;
alfa= rr(i);
s=1;h=1;
L1=-0.5;L2=0.4;
c=30;
y=0.5;b=1;
k0=386; ce=3.831*10^2; mu=38.6*10^9;alfat=1.78*10^-5; rho=89.54*10^2; lamda=77.6*10^9;taw=0.5;Tnot=2.93*10^2;
c0=sqrt((lamda+2*mu)/(rho)); Betanot=(3*lamda+2*mu)*alfat; a1=mu/(lamda+2*mu);a2=(mu+lamda)/(lamda+2*mu);a3=(Betanot*Tnot)/(lamda+2*mu);omega=(rho*ce)/(k0);
a4=lamda/(lamda+2*mu);a5=(k0*omega*c0^2)/(k0);a6=(rho*ce*c0^2)/(k0);
a7=(Betanot*c0^2)/(k0); a8=a6*taw; a9=a7*taw; a10=rho*ce*taw*omega*c0^4/(k0); a11=Betanot*taw*omega*c0^4/(k0);w=rho*ce/(k0);
b1=s^2+a1*h^2-(x+y+b).^2;b2=a2*s*h;b3=0;b4=a3*s;b5=alfa*b1;b6=alfa*b2;b7=alfa*b3;b8=2*alfa*b4;
B1=b1*L1^2+b2*L2^2;B2=-b3*L1-b4;
B3=-b5*L1^2-b6*L2^2;
%B4=b7*L1+b8;
B4=B3./4;
sec= ((B1-B2)./(3*B4))-((B1-B2)./(3*B4))+c;
T=-((B1-B2)./(3*B4))+(sec).*exp(-(s*x+h*y+b));
% fprintf(‘%12.10f %15.8f rn’,-((B1-B2)./(3*B4)),(c+((B1-B2)./(3*B4))));
%fir=c+sec;

% N2=zeros(1,1);

% N2=(s*L1+a4*h*L2).*-1*(c+((B1-B2)./(3*B4))).*exp(-(s*x+h*y+b)).*(1-alfa.*T)+a3.*T.*(alfa.*T-1);

% N3=zeros(1,1);

% N3=a1*(h*L1+s*L2)*(1-alfa.*T).*(-1*(c+((B1-B2)./(3*B4))).*exp(-(s*x+h*y+b)));

figure(1)
plot(x,T)
grid on,hold on
myLegend1{i}=[‘alfa= ‘,num2str(rr(i))];

i=i+1;

end
figure(1)
legend(myLegend1)
hold onclc;clf;clear;
myLegend1 = {};

rr = [ 0.4 0.5 0.6]
for i =1:numel(rr)
x=0:0.0001:9;
alfa= rr(i);
s=1;h=1;
L1=-0.5;L2=0.4;
c=30;
y=0.5;b=1;
k0=386; ce=3.831*10^2; mu=38.6*10^9;alfat=1.78*10^-5; rho=89.54*10^2; lamda=77.6*10^9;taw=0.5;Tnot=2.93*10^2;
c0=sqrt((lamda+2*mu)/(rho)); Betanot=(3*lamda+2*mu)*alfat; a1=mu/(lamda+2*mu);a2=(mu+lamda)/(lamda+2*mu);a3=(Betanot*Tnot)/(lamda+2*mu);omega=(rho*ce)/(k0);
a4=lamda/(lamda+2*mu);a5=(k0*omega*c0^2)/(k0);a6=(rho*ce*c0^2)/(k0);
a7=(Betanot*c0^2)/(k0); a8=a6*taw; a9=a7*taw; a10=rho*ce*taw*omega*c0^4/(k0); a11=Betanot*taw*omega*c0^4/(k0);w=rho*ce/(k0);
b1=s^2+a1*h^2-(x+y+b).^2;b2=a2*s*h;b3=0;b4=a3*s;b5=alfa*b1;b6=alfa*b2;b7=alfa*b3;b8=2*alfa*b4;
B1=b1*L1^2+b2*L2^2;B2=-b3*L1-b4;
B3=-b5*L1^2-b6*L2^2;
%B4=b7*L1+b8;
B4=B3./4;
sec= ((B1-B2)./(3*B4))-((B1-B2)./(3*B4))+c;
T=-((B1-B2)./(3*B4))+(sec).*exp(-(s*x+h*y+b));
% fprintf(‘%12.10f %15.8f rn’,-((B1-B2)./(3*B4)),(c+((B1-B2)./(3*B4))));
%fir=c+sec;

% N2=zeros(1,1);

% N2=(s*L1+a4*h*L2).*-1*(c+((B1-B2)./(3*B4))).*exp(-(s*x+h*y+b)).*(1-alfa.*T)+a3.*T.*(alfa.*T-1);

% N3=zeros(1,1);

% N3=a1*(h*L1+s*L2)*(1-alfa.*T).*(-1*(c+((B1-B2)./(3*B4))).*exp(-(s*x+h*y+b)));

figure(1)
plot(x,T)
grid on,hold on
myLegend1{i}=[‘alfa= ‘,num2str(rr(i))];

i=i+1;

end
figure(1)
legend(myLegend1)
hold on clc;clf;clear;
myLegend1 = {};

rr = [ 0.4 0.5 0.6]
for i =1:numel(rr)
x=0:0.0001:9;
alfa= rr(i);
s=1;h=1;
L1=-0.5;L2=0.4;
c=30;
y=0.5;b=1;
k0=386; ce=3.831*10^2; mu=38.6*10^9;alfat=1.78*10^-5; rho=89.54*10^2; lamda=77.6*10^9;taw=0.5;Tnot=2.93*10^2;
c0=sqrt((lamda+2*mu)/(rho)); Betanot=(3*lamda+2*mu)*alfat; a1=mu/(lamda+2*mu);a2=(mu+lamda)/(lamda+2*mu);a3=(Betanot*Tnot)/(lamda+2*mu);omega=(rho*ce)/(k0);
a4=lamda/(lamda+2*mu);a5=(k0*omega*c0^2)/(k0);a6=(rho*ce*c0^2)/(k0);
a7=(Betanot*c0^2)/(k0); a8=a6*taw; a9=a7*taw; a10=rho*ce*taw*omega*c0^4/(k0); a11=Betanot*taw*omega*c0^4/(k0);w=rho*ce/(k0);
b1=s^2+a1*h^2-(x+y+b).^2;b2=a2*s*h;b3=0;b4=a3*s;b5=alfa*b1;b6=alfa*b2;b7=alfa*b3;b8=2*alfa*b4;
B1=b1*L1^2+b2*L2^2;B2=-b3*L1-b4;
B3=-b5*L1^2-b6*L2^2;
%B4=b7*L1+b8;
B4=B3./4;
sec= ((B1-B2)./(3*B4))-((B1-B2)./(3*B4))+c;
T=-((B1-B2)./(3*B4))+(sec).*exp(-(s*x+h*y+b));
% fprintf(‘%12.10f %15.8f rn’,-((B1-B2)./(3*B4)),(c+((B1-B2)./(3*B4))));
%fir=c+sec;

% N2=zeros(1,1);

% N2=(s*L1+a4*h*L2).*-1*(c+((B1-B2)./(3*B4))).*exp(-(s*x+h*y+b)).*(1-alfa.*T)+a3.*T.*(alfa.*T-1);

% N3=zeros(1,1);

% N3=a1*(h*L1+s*L2)*(1-alfa.*T).*(-1*(c+((B1-B2)./(3*B4))).*exp(-(s*x+h*y+b)));

figure(1)
plot(x,T)
grid on,hold on
myLegend1{i}=[‘alfa= ‘,num2str(rr(i))];

i=i+1;

end
figure(1)
legend(myLegend1)
hold on figure MATLAB Answers — New Questions

​

I have 2 datasets from a continuous signal that will display a voltage in 1 set of data (D8) and repeat that voltage in data set D2 after a randomized delay. I need to find the voltage and time these repetitions occur, but I’m unsure of how to do that.
My code finds these repetitions, removes where the 2 datasets intercept as there’s no delay going on there, and displays the time of the repetitions. I need the original voltage, the original time from D8, and the delayed time in D2 when the repeated value occurs.
data = load(‘DatasSet2_Test.txt’); %my directory is different but it includes my name so it just calls the attached file
%Characterize columns
t1 = data(:,1)/1000; %convert time from ms to s
%convert from bits to V
resolution = 2^10-1;

%Data as assigned pairs *10^8 to transfer within the MeV to GeV range
D1 = (data(:,2)*5)/resolution*10^8; D2 = (data(:,3)*5)/resolution*10^8;
D3 = (data(:,4)*5)/resolution*10^8; D4 = (data(:,5)*5)/resolution*10^8;
D5 = (data(:,6)*5)/resolution*10^8; D6 = (data(:,7)*5)/resolution*10^8;
D7 = (data(:,8)*5)/resolution*10^8; D8 = (data(:,9)*5)/resolution*10^8; %all in V

c1 = D8(:)==D2 %finds same voltage indep of time occuring
x1values1 = t1(find(c1)); %Gives time of repeated values
r1usure1 = t1(find(intersect(D8,D2))); %checks for intersections as they don’t include delay
l1 = setdiff(x1values1, r1usure1); %Gives time of repeated values excluding intersections

This is the code I have so far, but it only checks for similar values and returns repeated time values instead of including the original time, and I’ve been struggling to find the associated voltages. I have also attached the file with the signals where the first column is time, column 3 is D2 in V, and column 9 is D8 in V.

This might not be well explained so I will answer questions as quickly as possible. Thank you in advance !I have 2 datasets from a continuous signal that will display a voltage in 1 set of data (D8) and repeat that voltage in data set D2 after a randomized delay. I need to find the voltage and time these repetitions occur, but I’m unsure of how to do that.
My code finds these repetitions, removes where the 2 datasets intercept as there’s no delay going on there, and displays the time of the repetitions. I need the original voltage, the original time from D8, and the delayed time in D2 when the repeated value occurs.
data = load(‘DatasSet2_Test.txt’); %my directory is different but it includes my name so it just calls the attached file
%Characterize columns
t1 = data(:,1)/1000; %convert time from ms to s
%convert from bits to V
resolution = 2^10-1;

%Data as assigned pairs *10^8 to transfer within the MeV to GeV range
D1 = (data(:,2)*5)/resolution*10^8; D2 = (data(:,3)*5)/resolution*10^8;
D3 = (data(:,4)*5)/resolution*10^8; D4 = (data(:,5)*5)/resolution*10^8;
D5 = (data(:,6)*5)/resolution*10^8; D6 = (data(:,7)*5)/resolution*10^8;
D7 = (data(:,8)*5)/resolution*10^8; D8 = (data(:,9)*5)/resolution*10^8; %all in V

c1 = D8(:)==D2 %finds same voltage indep of time occuring
x1values1 = t1(find(c1)); %Gives time of repeated values
r1usure1 = t1(find(intersect(D8,D2))); %checks for intersections as they don’t include delay
l1 = setdiff(x1values1, r1usure1); %Gives time of repeated values excluding intersections

This is the code I have so far, but it only checks for similar values and returns repeated time values instead of including the original time, and I’ve been struggling to find the associated voltages. I have also attached the file with the signals where the first column is time, column 3 is D2 in V, and column 9 is D8 in V.

This might not be well explained so I will answer questions as quickly as possible. Thank you in advance ! I have 2 datasets from a continuous signal that will display a voltage in 1 set of data (D8) and repeat that voltage in data set D2 after a randomized delay. I need to find the voltage and time these repetitions occur, but I’m unsure of how to do that.
My code finds these repetitions, removes where the 2 datasets intercept as there’s no delay going on there, and displays the time of the repetitions. I need the original voltage, the original time from D8, and the delayed time in D2 when the repeated value occurs.
data = load(‘DatasSet2_Test.txt’); %my directory is different but it includes my name so it just calls the attached file
%Characterize columns
t1 = data(:,1)/1000; %convert time from ms to s
%convert from bits to V
resolution = 2^10-1;

%Data as assigned pairs *10^8 to transfer within the MeV to GeV range
D1 = (data(:,2)*5)/resolution*10^8; D2 = (data(:,3)*5)/resolution*10^8;
D3 = (data(:,4)*5)/resolution*10^8; D4 = (data(:,5)*5)/resolution*10^8;
D5 = (data(:,6)*5)/resolution*10^8; D6 = (data(:,7)*5)/resolution*10^8;
D7 = (data(:,8)*5)/resolution*10^8; D8 = (data(:,9)*5)/resolution*10^8; %all in V

c1 = D8(:)==D2 %finds same voltage indep of time occuring
x1values1 = t1(find(c1)); %Gives time of repeated values
r1usure1 = t1(find(intersect(D8,D2))); %checks for intersections as they don’t include delay
l1 = setdiff(x1values1, r1usure1); %Gives time of repeated values excluding intersections

This is the code I have so far, but it only checks for similar values and returns repeated time values instead of including the original time, and I’ve been struggling to find the associated voltages. I have also attached the file with the signals where the first column is time, column 3 is D2 in V, and column 9 is D8 in V.

This might not be well explained so I will answer questions as quickly as possible. Thank you in advance ! indexing, find function, repeated values MATLAB Answers — New Questions

​