Month: June 2024
How to confirm upgraded A/D server is primary controller
I have created a new MS Server 2022 A/D server to replace the MS Server 2012 A/D. I want to ensure the new server is in control of all the features/functions of the old one. The new server has all the roles as the old one. The netdom query shows the new server in all areas:
PS C:Windowssystem32> netdom query fsmo
Schema master SVR01.mrc.net
Domain naming master SVR01.mrc.net
PDC SVR01.mrc.net
RID pool manager SVR01.mrc.net
Infrastructure master SVRV01.mrc.net
The command completed successfully.
Replication is in working as I can create a user on the new server and see it replicate to the old server as well as AAD.
When the server restarts, I see these eventIDs
2092 Microsoft-Windows-ActiveDirectory_DomainService
User Action:
This server is the owner of the following FSMO role, but does not consider it valid. For the partition which contains the FSMO, this server has not replicated successfully with any of its partners since this server has been restarted. Replication errors are preventing validation of this role.
Operations which require contacting a FSMO operation master will fail until this condition is corrected.
FSMO Role: DC=src,DC=net
4013 Microsoft-Windows-DNS-Server-Service
The DNS server is waiting for Active Directory Domain Services (AD DS) to signal that the initial synchronization of the directory has been completed. The DNS server service cannot start until the initial synchronization is complete because critical DNS data might not yet be replicated onto this domain controller. If events in the AD DS event log indicate that there is a problem with DNS name resolution, consider adding the IP address of another DNS server for this domain to the DNS server list in the Internet Protocol properties of this computer. This event will be logged every two minutes until AD DS has signaled that the initial synchronization has successfully completed.
1202 ADWS
This computer is now hosting the specified directory instance, but Active Directory Web Services could not service it. Active Directory Web Services will retry this operation periodically.
Directory instance: NTDS
Directory instance LDAP port: 389
Directory instance SSL port: 636
How are these corrected? When done, I will only need 1 A/D server in my network and that will be the new MS Server 2022. What other things should I check to make sure the new A/D server is fully in control before I remove the roles from the old server, decommission it and elevate the functional level?
I have created a new MS Server 2022 A/D server to replace the MS Server 2012 A/D. I want to ensure the new server is in control of all the features/functions of the old one. The new server has all the roles as the old one. The netdom query shows the new server in all areas:PS C:Windowssystem32> netdom query fsmoSchema master SVR01.mrc.netDomain naming master SVR01.mrc.netPDC SVR01.mrc.netRID pool manager SVR01.mrc.netInfrastructure master SVRV01.mrc.netThe command completed successfully. Replication is in working as I can create a user on the new server and see it replicate to the old server as well as AAD. When the server restarts, I see these eventIDs2092 Microsoft-Windows-ActiveDirectory_DomainServiceUser Action:This server is the owner of the following FSMO role, but does not consider it valid. For the partition which contains the FSMO, this server has not replicated successfully with any of its partners since this server has been restarted. Replication errors are preventing validation of this role.Operations which require contacting a FSMO operation master will fail until this condition is corrected.FSMO Role: DC=src,DC=net4013 Microsoft-Windows-DNS-Server-Service The DNS server is waiting for Active Directory Domain Services (AD DS) to signal that the initial synchronization of the directory has been completed. The DNS server service cannot start until the initial synchronization is complete because critical DNS data might not yet be replicated onto this domain controller. If events in the AD DS event log indicate that there is a problem with DNS name resolution, consider adding the IP address of another DNS server for this domain to the DNS server list in the Internet Protocol properties of this computer. This event will be logged every two minutes until AD DS has signaled that the initial synchronization has successfully completed. 1202 ADWSThis computer is now hosting the specified directory instance, but Active Directory Web Services could not service it. Active Directory Web Services will retry this operation periodically.Directory instance: NTDSDirectory instance LDAP port: 389Directory instance SSL port: 636 How are these corrected? When done, I will only need 1 A/D server in my network and that will be the new MS Server 2022. What other things should I check to make sure the new A/D server is fully in control before I remove the roles from the old server, decommission it and elevate the functional level? Read More
After performing stereo calibration using a checkerboard, the same checkerboard is reconstructed in 3D. I have several questions regarding the results.
Currently, I am facing three issues related to 3D reconstruction. As shown in the first attached image, I describe the x-axis as positive to the right and the y-axis as positive upwards (since the checkerboard is planar, I will not consider the z-axis in this question). The green plots mean the ground truth, and the red plots mean reconstructed values in the first attached image.
When considering the detected checkerboard corner points in the same x×y layout as the premise, the ground truth is detected and plotted as 8×11, while the reconstructed values are detected and plotted as 10×7. Why is there a difference of one cell in the detected grid? Please refer to the attached checkerboard image for this situation.
As seen in the attached image, the long side of the ground truth checkerboard is aligned along the y-axis direction, while the long side of the reconstructed checkerboard appears to be aligned along the x-axis direction. Why does it seem like the long side’s position has rotated during reconstruction?
The first detected corner point of the reconstructed values should be located at the same origin as the first detected corner point of the ground truth, but it is located in a different position. What is the cause of this? Is this due to low accuracy in the camera parameters, or is it because the reconstructed values and the ground truth are represented in different coordinate systems? Could there be other reasons for this?
The specific causes for the second and third questions might be the same, but I appreciate your response. For reference, I am attaching the code as well.
% Loading stereo images
I1 = imread(‘/Users/uchidataisei/dev/uchida/kenkyu_data/2024-06-14/C3/GX010541_frames/frame0052.png’); % 左画像
I2 = imread(‘/Users/uchidataisei/dev/uchida/kenkyu_data/2024-06-14/C2/GX010611_frames/frame0052.png’); % 右画像
% Using pre-obtained calibration parameters
load(‘/Users/uchidataisei/dev/uchida/MATLAB_data/2024_06_14/calibrationSession_0614_C3_2_.mat’);
% Rectification
% [J1, J2, reprojectionMatrix] = rectifyStereoImages(I1, I2, stereoParams_0614_C3_2);
[J1, J2, reprojectionMatrix] = rectifyStereoImages(I1, I2, stereoParams_0614_C3_2_);
% Calculating the disparity map
disparityMap = disparitySGM(rgb2gray(J1), rgb2gray(J2));
% Reconstructing the 3D scene
points3D = reconstructScene(disparityMap, reprojectionMatrix);
% Setting the ground truth of the checkerboard, [mm]
squareSize = 114;
% Detecting the corners of the checkerboard
[imagePoints, boardSize] = detectCheckerboardPoints(J1);
% Calculating the ground truth grid of the checkerboard
[worldX, worldY] = meshgrid(0:squareSize:((boardSize(1)-1)*squareSize), 0:squareSize:((boardSize(2)-1)*squareSize));
worldPoints = [worldX(:), worldY(:), zeros(numel(worldX), 1)]; % Z=0
% Matching the number of detected corners with the number of ground truth points
% if size(worldPoints, 1) > size(imagePoints, 1)
% worldPoints = worldPoints(1:size(imagePoints, 1), :);
% elseif size(worldPoints, 1) < size(imagePoints, 1)
% error(‘The number of ground truth points is less than the number of detected corners. Please check the number of ground truth points.’);
% end
% Extracting the 3D points of the reconstructed checkerboard corners
detected3DPoints = zeros(size(imagePoints, 1), 3);
validIndices = true(size(imagePoints, 1), 1);
for i = 1:size(imagePoints, 1)
x = round(imagePoints(i, 1));
y = round(imagePoints(i, 2));
if x > 0 && y > 0 && x <= size(points3D, 2) && y <= size(points3D, 1)
detected3DPoints(i, 🙂 = points3D(y, x, :);
if any(isnan(detected3DPoints(i, :)) | isinf(detected3DPoints(i, :)))
validIndices(i) = false;
end
else
validIndices(i) = false;
end
end
% Excluding invalid points
% detected3DPoints = detected3DPoints(validIndices, :); %Reconstructed grid
% worldPoints = worldPoints(validIndices, :); %Ground truth grid
% Omitting scale transformation
detected3DPoints_mm = detected3DPoints;
% Comparison of ground truth and reconstructed results
figure;
plot3(worldPoints(:,1), worldPoints(:,2), worldPoints(:,3), ‘go’);
hold on;
plot3(detected3DPoints_mm(:,1), detected3DPoints_mm(:,2), detected3DPoints_mm(:,3), ‘rx’);
% Displaying numbers on each plot
for i = 1:size(worldPoints, 1)
text(worldPoints(i, 1), worldPoints(i, 2), worldPoints(i, 3), num2str(i), ‘Color’, ‘green’);
text(detected3DPoints_mm(i, 1), detected3DPoints_mm(i, 2), detected3DPoints_mm(i, 3), num2str(i), ‘Color’, ‘red’);
end
legend(‘True Points’, ‘Reconstructed Points’);
xlabel(‘X (mm)’);
ylabel(‘Y (mm)’);
zlabel(‘Z (mm)’);
title(‘Comparison of True and Reconstructed Points’);
grid on;
% Calculating the error
errors_xy = sqrt(sum((worldPoints(:, 1:2) – detected3DPoints_mm(:, 1:2)).^2, 2));
meanError_xy = mean(errors_xy);
disp([‘Mean Error in XY plane: ‘, num2str(meanError_xy), ‘ millimeters’]);Currently, I am facing three issues related to 3D reconstruction. As shown in the first attached image, I describe the x-axis as positive to the right and the y-axis as positive upwards (since the checkerboard is planar, I will not consider the z-axis in this question). The green plots mean the ground truth, and the red plots mean reconstructed values in the first attached image.
When considering the detected checkerboard corner points in the same x×y layout as the premise, the ground truth is detected and plotted as 8×11, while the reconstructed values are detected and plotted as 10×7. Why is there a difference of one cell in the detected grid? Please refer to the attached checkerboard image for this situation.
As seen in the attached image, the long side of the ground truth checkerboard is aligned along the y-axis direction, while the long side of the reconstructed checkerboard appears to be aligned along the x-axis direction. Why does it seem like the long side’s position has rotated during reconstruction?
The first detected corner point of the reconstructed values should be located at the same origin as the first detected corner point of the ground truth, but it is located in a different position. What is the cause of this? Is this due to low accuracy in the camera parameters, or is it because the reconstructed values and the ground truth are represented in different coordinate systems? Could there be other reasons for this?
The specific causes for the second and third questions might be the same, but I appreciate your response. For reference, I am attaching the code as well.
% Loading stereo images
I1 = imread(‘/Users/uchidataisei/dev/uchida/kenkyu_data/2024-06-14/C3/GX010541_frames/frame0052.png’); % 左画像
I2 = imread(‘/Users/uchidataisei/dev/uchida/kenkyu_data/2024-06-14/C2/GX010611_frames/frame0052.png’); % 右画像
% Using pre-obtained calibration parameters
load(‘/Users/uchidataisei/dev/uchida/MATLAB_data/2024_06_14/calibrationSession_0614_C3_2_.mat’);
% Rectification
% [J1, J2, reprojectionMatrix] = rectifyStereoImages(I1, I2, stereoParams_0614_C3_2);
[J1, J2, reprojectionMatrix] = rectifyStereoImages(I1, I2, stereoParams_0614_C3_2_);
% Calculating the disparity map
disparityMap = disparitySGM(rgb2gray(J1), rgb2gray(J2));
% Reconstructing the 3D scene
points3D = reconstructScene(disparityMap, reprojectionMatrix);
% Setting the ground truth of the checkerboard, [mm]
squareSize = 114;
% Detecting the corners of the checkerboard
[imagePoints, boardSize] = detectCheckerboardPoints(J1);
% Calculating the ground truth grid of the checkerboard
[worldX, worldY] = meshgrid(0:squareSize:((boardSize(1)-1)*squareSize), 0:squareSize:((boardSize(2)-1)*squareSize));
worldPoints = [worldX(:), worldY(:), zeros(numel(worldX), 1)]; % Z=0
% Matching the number of detected corners with the number of ground truth points
% if size(worldPoints, 1) > size(imagePoints, 1)
% worldPoints = worldPoints(1:size(imagePoints, 1), :);
% elseif size(worldPoints, 1) < size(imagePoints, 1)
% error(‘The number of ground truth points is less than the number of detected corners. Please check the number of ground truth points.’);
% end
% Extracting the 3D points of the reconstructed checkerboard corners
detected3DPoints = zeros(size(imagePoints, 1), 3);
validIndices = true(size(imagePoints, 1), 1);
for i = 1:size(imagePoints, 1)
x = round(imagePoints(i, 1));
y = round(imagePoints(i, 2));
if x > 0 && y > 0 && x <= size(points3D, 2) && y <= size(points3D, 1)
detected3DPoints(i, 🙂 = points3D(y, x, :);
if any(isnan(detected3DPoints(i, :)) | isinf(detected3DPoints(i, :)))
validIndices(i) = false;
end
else
validIndices(i) = false;
end
end
% Excluding invalid points
% detected3DPoints = detected3DPoints(validIndices, :); %Reconstructed grid
% worldPoints = worldPoints(validIndices, :); %Ground truth grid
% Omitting scale transformation
detected3DPoints_mm = detected3DPoints;
% Comparison of ground truth and reconstructed results
figure;
plot3(worldPoints(:,1), worldPoints(:,2), worldPoints(:,3), ‘go’);
hold on;
plot3(detected3DPoints_mm(:,1), detected3DPoints_mm(:,2), detected3DPoints_mm(:,3), ‘rx’);
% Displaying numbers on each plot
for i = 1:size(worldPoints, 1)
text(worldPoints(i, 1), worldPoints(i, 2), worldPoints(i, 3), num2str(i), ‘Color’, ‘green’);
text(detected3DPoints_mm(i, 1), detected3DPoints_mm(i, 2), detected3DPoints_mm(i, 3), num2str(i), ‘Color’, ‘red’);
end
legend(‘True Points’, ‘Reconstructed Points’);
xlabel(‘X (mm)’);
ylabel(‘Y (mm)’);
zlabel(‘Z (mm)’);
title(‘Comparison of True and Reconstructed Points’);
grid on;
% Calculating the error
errors_xy = sqrt(sum((worldPoints(:, 1:2) – detected3DPoints_mm(:, 1:2)).^2, 2));
meanError_xy = mean(errors_xy);
disp([‘Mean Error in XY plane: ‘, num2str(meanError_xy), ‘ millimeters’]); Currently, I am facing three issues related to 3D reconstruction. As shown in the first attached image, I describe the x-axis as positive to the right and the y-axis as positive upwards (since the checkerboard is planar, I will not consider the z-axis in this question). The green plots mean the ground truth, and the red plots mean reconstructed values in the first attached image.
When considering the detected checkerboard corner points in the same x×y layout as the premise, the ground truth is detected and plotted as 8×11, while the reconstructed values are detected and plotted as 10×7. Why is there a difference of one cell in the detected grid? Please refer to the attached checkerboard image for this situation.
As seen in the attached image, the long side of the ground truth checkerboard is aligned along the y-axis direction, while the long side of the reconstructed checkerboard appears to be aligned along the x-axis direction. Why does it seem like the long side’s position has rotated during reconstruction?
The first detected corner point of the reconstructed values should be located at the same origin as the first detected corner point of the ground truth, but it is located in a different position. What is the cause of this? Is this due to low accuracy in the camera parameters, or is it because the reconstructed values and the ground truth are represented in different coordinate systems? Could there be other reasons for this?
The specific causes for the second and third questions might be the same, but I appreciate your response. For reference, I am attaching the code as well.
% Loading stereo images
I1 = imread(‘/Users/uchidataisei/dev/uchida/kenkyu_data/2024-06-14/C3/GX010541_frames/frame0052.png’); % 左画像
I2 = imread(‘/Users/uchidataisei/dev/uchida/kenkyu_data/2024-06-14/C2/GX010611_frames/frame0052.png’); % 右画像
% Using pre-obtained calibration parameters
load(‘/Users/uchidataisei/dev/uchida/MATLAB_data/2024_06_14/calibrationSession_0614_C3_2_.mat’);
% Rectification
% [J1, J2, reprojectionMatrix] = rectifyStereoImages(I1, I2, stereoParams_0614_C3_2);
[J1, J2, reprojectionMatrix] = rectifyStereoImages(I1, I2, stereoParams_0614_C3_2_);
% Calculating the disparity map
disparityMap = disparitySGM(rgb2gray(J1), rgb2gray(J2));
% Reconstructing the 3D scene
points3D = reconstructScene(disparityMap, reprojectionMatrix);
% Setting the ground truth of the checkerboard, [mm]
squareSize = 114;
% Detecting the corners of the checkerboard
[imagePoints, boardSize] = detectCheckerboardPoints(J1);
% Calculating the ground truth grid of the checkerboard
[worldX, worldY] = meshgrid(0:squareSize:((boardSize(1)-1)*squareSize), 0:squareSize:((boardSize(2)-1)*squareSize));
worldPoints = [worldX(:), worldY(:), zeros(numel(worldX), 1)]; % Z=0
% Matching the number of detected corners with the number of ground truth points
% if size(worldPoints, 1) > size(imagePoints, 1)
% worldPoints = worldPoints(1:size(imagePoints, 1), :);
% elseif size(worldPoints, 1) < size(imagePoints, 1)
% error(‘The number of ground truth points is less than the number of detected corners. Please check the number of ground truth points.’);
% end
% Extracting the 3D points of the reconstructed checkerboard corners
detected3DPoints = zeros(size(imagePoints, 1), 3);
validIndices = true(size(imagePoints, 1), 1);
for i = 1:size(imagePoints, 1)
x = round(imagePoints(i, 1));
y = round(imagePoints(i, 2));
if x > 0 && y > 0 && x <= size(points3D, 2) && y <= size(points3D, 1)
detected3DPoints(i, 🙂 = points3D(y, x, :);
if any(isnan(detected3DPoints(i, :)) | isinf(detected3DPoints(i, :)))
validIndices(i) = false;
end
else
validIndices(i) = false;
end
end
% Excluding invalid points
% detected3DPoints = detected3DPoints(validIndices, :); %Reconstructed grid
% worldPoints = worldPoints(validIndices, :); %Ground truth grid
% Omitting scale transformation
detected3DPoints_mm = detected3DPoints;
% Comparison of ground truth and reconstructed results
figure;
plot3(worldPoints(:,1), worldPoints(:,2), worldPoints(:,3), ‘go’);
hold on;
plot3(detected3DPoints_mm(:,1), detected3DPoints_mm(:,2), detected3DPoints_mm(:,3), ‘rx’);
% Displaying numbers on each plot
for i = 1:size(worldPoints, 1)
text(worldPoints(i, 1), worldPoints(i, 2), worldPoints(i, 3), num2str(i), ‘Color’, ‘green’);
text(detected3DPoints_mm(i, 1), detected3DPoints_mm(i, 2), detected3DPoints_mm(i, 3), num2str(i), ‘Color’, ‘red’);
end
legend(‘True Points’, ‘Reconstructed Points’);
xlabel(‘X (mm)’);
ylabel(‘Y (mm)’);
zlabel(‘Z (mm)’);
title(‘Comparison of True and Reconstructed Points’);
grid on;
% Calculating the error
errors_xy = sqrt(sum((worldPoints(:, 1:2) – detected3DPoints_mm(:, 1:2)).^2, 2));
meanError_xy = mean(errors_xy);
disp([‘Mean Error in XY plane: ‘, num2str(meanError_xy), ‘ millimeters’]); #reconstructscene, stereoparameters, detectcheckerboardpoints, 3d reconstruction MATLAB Answers — New Questions
Fill confidence band hexadecimal color
Hello!
I am plotting a confidence band using fill, and get an error-message when using hecadecimal color:
I have tried with both ‘Color’ and ‘FaceColor’ before the hexadecimal color, without it helping. It works when I use a default color such as ‘b’.
Also, when trying just to plot the line (not filling), it works with the html-code.
fill([0:hmax, fliplr(0:hmax)], [upper_bounds, fliplr(lower_bounds)], ‘#7E2F8E’, ‘FaceAlpha’, 0.2, ‘EdgeColor’, ‘none’);
Thanks!Hello!
I am plotting a confidence band using fill, and get an error-message when using hecadecimal color:
I have tried with both ‘Color’ and ‘FaceColor’ before the hexadecimal color, without it helping. It works when I use a default color such as ‘b’.
Also, when trying just to plot the line (not filling), it works with the html-code.
fill([0:hmax, fliplr(0:hmax)], [upper_bounds, fliplr(lower_bounds)], ‘#7E2F8E’, ‘FaceAlpha’, 0.2, ‘EdgeColor’, ‘none’);
Thanks! Hello!
I am plotting a confidence band using fill, and get an error-message when using hecadecimal color:
I have tried with both ‘Color’ and ‘FaceColor’ before the hexadecimal color, without it helping. It works when I use a default color such as ‘b’.
Also, when trying just to plot the line (not filling), it works with the html-code.
fill([0:hmax, fliplr(0:hmax)], [upper_bounds, fliplr(lower_bounds)], ‘#7E2F8E’, ‘FaceAlpha’, 0.2, ‘EdgeColor’, ‘none’);
Thanks! #plot #fill #hexadecimal #confidenceband MATLAB Answers — New Questions
Per certification designed badges
Hi
First Microsoft opted out from awesome Credly (awesome, as learners collected “all” personal certifications in one place, no matter the vendor – easy to share the Credly profile link for various reasons)
And now you have quit creating “per certification branded badge”s, and only provide standard “Associate” & “Expert” badges with a “Learn diploma) showing the name of the certification “in text” (the new Fabric exam as example)
For us globally in roles like “Alliance Managers”, “Partner Managers”, driving and summarizing partners excellence in the area of Microsoft + pushing with marketing us and Microsoft- this is bad!
Example on how we earlier are using the per certification badges
Is it just by mistake you have taken this path? Or is it just me and my learners that have missed where they can download per exam branded badges for newer certifications now?
Regards
Gabriel
HiFirst Microsoft opted out from awesome Credly (awesome, as learners collected “all” personal certifications in one place, no matter the vendor – easy to share the Credly profile link for various reasons) And now you have quit creating “per certification branded badge”s, and only provide standard “Associate” & “Expert” badges with a “Learn diploma) showing the name of the certification “in text” (the new Fabric exam as example) For us globally in roles like “Alliance Managers”, “Partner Managers”, driving and summarizing partners excellence in the area of Microsoft + pushing with marketing us and Microsoft- this is bad! Example on how we earlier are using the per certification badges Is it just by mistake you have taken this path? Or is it just me and my learners that have missed where they can download per exam branded badges for newer certifications now? RegardsGabriel Read More
Working with modified code -2024A
I am working with a slightly modified version of ode113. Until I was using the same code I copied and pasted and modified a few versions ago. Now in 2024A I got an error message, so I copied the code of ode113 again into a new file and modified again. However, a new problem has risen- it seems that 2023 is dependant on private functions, and I am now getting "unrecognized function or variable" errors. A workaround that I found is copying all the relevant private folders into a new folder. Is that the best way to solve the issue?
many thanks
NathanI am working with a slightly modified version of ode113. Until I was using the same code I copied and pasted and modified a few versions ago. Now in 2024A I got an error message, so I copied the code of ode113 again into a new file and modified again. However, a new problem has risen- it seems that 2023 is dependant on private functions, and I am now getting "unrecognized function or variable" errors. A workaround that I found is copying all the relevant private folders into a new folder. Is that the best way to solve the issue?
many thanks
Nathan I am working with a slightly modified version of ode113. Until I was using the same code I copied and pasted and modified a few versions ago. Now in 2024A I got an error message, so I copied the code of ode113 again into a new file and modified again. However, a new problem has risen- it seems that 2023 is dependant on private functions, and I am now getting "unrecognized function or variable" errors. A workaround that I found is copying all the relevant private folders into a new folder. Is that the best way to solve the issue?
many thanks
Nathan private, private functions, modified MATLAB Answers — New Questions
Why is this matlab program not able to solve accurately?
B=[E_b*I_b*(-beta^3*cos(beta*l)-beta^3*cosh(beta*l))+m2*omega^2*(sin(beta*l)-sinh(beta*l)),E_b*I_b*(beta^3*sin(beta*l)-beta^3*sinh(beta*l))+m2*omega^2*(cos(beta*l)-cosh(beta*l));
E_b*I_b*(-beta^2*sin(beta*l)-beta^2*sinh(beta*l))-J*omega^2*(sin(beta*l)-sinh(beta*l)),E_b*I_b*(-beta^2*cos(beta*l)-beta^2*cosh(beta*l))-J*omega^2*(cos(beta*l)-cosh(beta*l)) ]B=[E_b*I_b*(-beta^3*cos(beta*l)-beta^3*cosh(beta*l))+m2*omega^2*(sin(beta*l)-sinh(beta*l)),E_b*I_b*(beta^3*sin(beta*l)-beta^3*sinh(beta*l))+m2*omega^2*(cos(beta*l)-cosh(beta*l));
E_b*I_b*(-beta^2*sin(beta*l)-beta^2*sinh(beta*l))-J*omega^2*(sin(beta*l)-sinh(beta*l)),E_b*I_b*(-beta^2*cos(beta*l)-beta^2*cosh(beta*l))-J*omega^2*(cos(beta*l)-cosh(beta*l)) ] B=[E_b*I_b*(-beta^3*cos(beta*l)-beta^3*cosh(beta*l))+m2*omega^2*(sin(beta*l)-sinh(beta*l)),E_b*I_b*(beta^3*sin(beta*l)-beta^3*sinh(beta*l))+m2*omega^2*(cos(beta*l)-cosh(beta*l));
E_b*I_b*(-beta^2*sin(beta*l)-beta^2*sinh(beta*l))-J*omega^2*(sin(beta*l)-sinh(beta*l)),E_b*I_b*(-beta^2*cos(beta*l)-beta^2*cosh(beta*l))-J*omega^2*(cos(beta*l)-cosh(beta*l)) ] solve the determinant MATLAB Answers — New Questions
Training agent in reinforcement learning: reproducibility of the code
I get two different results from running this water-tank system example for reinforcement learning made by Mathworks:
https://uk.mathworks.com/help/reinforcement-learning/ug/create-simulink-environment-and-train-agent.html
This example has fixed the random number generator seed rng(0), so I expected the result to be the same on all computer. However, I ended up with two different agents on two computers:
Computer A finished training the agent after 86 episodes (just like the published example) and gave me an identical agent to the example.
Computer B needed 182 episodes to train the agent and gave me a different agent.
Both computers run MATLAB R2023b 64-bit on MS Windows 10. The code is unchanged from the example (except for changing doTraining = false to doTraining = true).
Computer A has an 8-core i7 processor. Computer B has a 6-core i7 processor.
I’m writing a tutorial for a univeristy-level course, so reproducibility is necessary so that students can follow the example. Any tip on how to facilitate this is also much appreciated.I get two different results from running this water-tank system example for reinforcement learning made by Mathworks:
https://uk.mathworks.com/help/reinforcement-learning/ug/create-simulink-environment-and-train-agent.html
This example has fixed the random number generator seed rng(0), so I expected the result to be the same on all computer. However, I ended up with two different agents on two computers:
Computer A finished training the agent after 86 episodes (just like the published example) and gave me an identical agent to the example.
Computer B needed 182 episodes to train the agent and gave me a different agent.
Both computers run MATLAB R2023b 64-bit on MS Windows 10. The code is unchanged from the example (except for changing doTraining = false to doTraining = true).
Computer A has an 8-core i7 processor. Computer B has a 6-core i7 processor.
I’m writing a tutorial for a univeristy-level course, so reproducibility is necessary so that students can follow the example. Any tip on how to facilitate this is also much appreciated. I get two different results from running this water-tank system example for reinforcement learning made by Mathworks:
https://uk.mathworks.com/help/reinforcement-learning/ug/create-simulink-environment-and-train-agent.html
This example has fixed the random number generator seed rng(0), so I expected the result to be the same on all computer. However, I ended up with two different agents on two computers:
Computer A finished training the agent after 86 episodes (just like the published example) and gave me an identical agent to the example.
Computer B needed 182 episodes to train the agent and gave me a different agent.
Both computers run MATLAB R2023b 64-bit on MS Windows 10. The code is unchanged from the example (except for changing doTraining = false to doTraining = true).
Computer A has an 8-core i7 processor. Computer B has a 6-core i7 processor.
I’m writing a tutorial for a univeristy-level course, so reproducibility is necessary so that students can follow the example. Any tip on how to facilitate this is also much appreciated. reinforcement learning, agent, training, random number generator MATLAB Answers — New Questions
Colormap a plot based on value of (x,y)
Hello,
I have a table T of dimensions x,y where each point has a value between -100 and 100.
I want to graph this data such that T(1,1) is point 1,1 on the graph and the color of that point is determined by the value of T(1,1)
I included a picture of something similar to what I want my plot to look like, along with some code that generates an example table for graphing
Thanks in advance for any help you can provide.
clear
clc
close all
%This will generate a table with example data.
x=100
y=50
data=zeros(y,x);
data(1,1)=50;
dchartdx=25/x;
dchartdy=25/y;
for ii=1:x
data(1,ii+1)=data(1,ii)+dchartdx;
end
for ii=1:x+1
for ij=1:y
data(ij+1,ii)=data(ij,ii)+dchartdy;
end
endHello,
I have a table T of dimensions x,y where each point has a value between -100 and 100.
I want to graph this data such that T(1,1) is point 1,1 on the graph and the color of that point is determined by the value of T(1,1)
I included a picture of something similar to what I want my plot to look like, along with some code that generates an example table for graphing
Thanks in advance for any help you can provide.
clear
clc
close all
%This will generate a table with example data.
x=100
y=50
data=zeros(y,x);
data(1,1)=50;
dchartdx=25/x;
dchartdy=25/y;
for ii=1:x
data(1,ii+1)=data(1,ii)+dchartdx;
end
for ii=1:x+1
for ij=1:y
data(ij+1,ii)=data(ij,ii)+dchartdy;
end
end Hello,
I have a table T of dimensions x,y where each point has a value between -100 and 100.
I want to graph this data such that T(1,1) is point 1,1 on the graph and the color of that point is determined by the value of T(1,1)
I included a picture of something similar to what I want my plot to look like, along with some code that generates an example table for graphing
Thanks in advance for any help you can provide.
clear
clc
close all
%This will generate a table with example data.
x=100
y=50
data=zeros(y,x);
data(1,1)=50;
dchartdx=25/x;
dchartdy=25/y;
for ii=1:x
data(1,ii+1)=data(1,ii)+dchartdx;
end
for ii=1:x+1
for ij=1:y
data(ij+1,ii)=data(ij,ii)+dchartdy;
end
end colormap plot MATLAB Answers — New Questions
Forgot password and lost access to phone number for old hotmail address but have access via pop3
Hi,
As stated in the title, I have an old hotmail address of which I forgot the password, also the phone number is no longer in use. I do however still have access to it on my iPhone, which can copy my pop3 accounts when transferring to a new phone.
I already tried to use the retrieve password form and filled out all the details 3 times, but I always get rejected.
Is there a way I can get my password reset by proving I have access via the pop3 on my phone?
Cheers,
Onno
Hi, As stated in the title, I have an old hotmail address of which I forgot the password, also the phone number is no longer in use. I do however still have access to it on my iPhone, which can copy my pop3 accounts when transferring to a new phone.I already tried to use the retrieve password form and filled out all the details 3 times, but I always get rejected.Is there a way I can get my password reset by proving I have access via the pop3 on my phone?Cheers,Onno Read More
Comparison chart AZD and SSMA
Since we often receive inquiries from everyone, we are releasing a comparison chart of the features and transformations of SS MA and AD. We welcome any questions, so please feel free to ask.
Since we often receive inquiries from everyone, we are releasing a comparison chart of the features and transformations of SS MA and AD. We welcome any questions, so please feel free to ask. Read More
Azure Devops certification
Hi,
I know to get the azure certificate I need to clear azure administrative exam and azure Devops exam and both exam should be active.
My condition is I have cleared the azure 103 exam and it’s expired right now.
I have cleared the azure Devops exam and did not get the certificate since azure administrative exam was expired. My question is if I give az104 exam only will I get the both certificate I mean az104 and azure Devops? Please answer Yes or No.
If No please guide me how can I get the certificate ?
Should I clear azure Devops again?
Please answer me.
Hi, I know to get the azure certificate I need to clear azure administrative exam and azure Devops exam and both exam should be active. My condition is I have cleared the azure 103 exam and it’s expired right now. I have cleared the azure Devops exam and did not get the certificate since azure administrative exam was expired. My question is if I give az104 exam only will I get the both certificate I mean az104 and azure Devops? Please answer Yes or No. If No please guide me how can I get the certificate ? Should I clear azure Devops again? Please answer me. Read More
Latest Threat Intelligence (June 2024)
Microsoft Defender for IoT has released the June 2024 Threat Intelligence package. The package is available for download from the Microsoft Defender for IoT portal (click Updates, then Download file).
Threat Intelligence updates reflect the combined impact of proprietary research and threat intelligence carried out by Microsoft security teams. Each package contains the latest CVEs (Common Vulnerabilities and Exposures), IOCs (Indicators of Compromise), and other indicators applicable to IoT/ICS/OT networks (published during the past month) researched and implemented by Microsoft Threat Intelligence Research – CPS.
The CVE scores are aligned with the National Vulnerability Database (NVD). Starting with the August 2023 threat intelligence updates, CVSSv3 scores are shown if they are relevant; otherwise the CVSSv2 scores are shown.
Guidance
Customers are recommended to update their systems with the latest TI package in order to detect potential exposure risks and vulnerabilities in their networks and on their devices. Threat Intelligence packages are updated every month with the most up-to-date security information available, ensuring that Microsoft Defender for IoT can identify malicious actors and behaviors on devices.
Update your system with the latest TI package
The package is available for download from the Microsoft Defender for IoT portal (click Updates, then Download file), for more information, please review Update threat intelligence data | Microsoft Docs.
MD5 Hash: dcdd8a2d48f81aa4df4af4c9a14652d0
For cloud connected sensors, Microsoft Defender for IoT can automatically update new threat intelligence packages following their release, click here for more information. Read More
i am working on Image Compression Using Run Length Encoding
am getting error in scanning in zigzag
the error is as follows
Undefined function or variable ‘toZigzag’.
Error in rlc_haar (line 21)
ImageArray=toZigzag(QuantizedImage);
plese help me
%% Matlab code for Image Compression Using Run Length Encoding
clc;
clear;
close all;
%% Set Quantization Parameter
quantizedvalue=10;
%% Read Input Image
InputImage=imread(‘cameraman.tif’);
[row col p]=size(InputImage);
%% Wavelet Decomposition
[LL LH HL HH]=dwt2(InputImage,’haar’);
WaveletDecomposeImage=[LL,LH;HL,HH];
imshow(WaveletDecomposeImage,[]);
%uniform quantization
QuantizedImage= WaveletDecomposeImage/quantizedvalue;
QuantizedImage= round(QuantizedImage);
% Convert the Two dimensional Image to a one dimensional Array using ZigZag Scanning
ImageArray=toZigzag(QuantizedImage);
%% Run Length Encoding
j=1;
a=length(ImageArray);
count=0;
for n=1:a
b=ImageArray(n);
if n==a
count=count+1;
c(j)=count;
s(j)=ImageArray(n);
elseif ImageArray(n)==ImageArray(n+1)
count=count+1;
elseif ImageArray(n)==b
count=count+1;
c(j)=count;
s(j)=ImageArray(n);
j=j+1;
count=0;
end
end
%% Calculation Bit Cost
InputBitcost=row*col*8;
InputBitcost=(InputBitcost);
c1=length(c);
s1=length(s);
OutputBitcost= (c1*8)+(s1*8);
OutputBitcost=(OutputBitcost);
%% Run Length Decoding g=length(s);
j=1;
l=1;
for i=1:g
v(l)=s(j);
if c(j)~=0
w=l+c(j)-1;
for p=l:w
v(l)=s(j);
l=l+1;
end
end
j=j+1;
end
ReconstructedImageArray=v;
%% Inverse ZigZag
ReconstructedImage=invZigzag(ReconstructedImageArray)
%% Inverse Quantization
ReconstructedImage=ReconstructedImage*quantizedvalue;am getting error in scanning in zigzag
the error is as follows
Undefined function or variable ‘toZigzag’.
Error in rlc_haar (line 21)
ImageArray=toZigzag(QuantizedImage);
plese help me
%% Matlab code for Image Compression Using Run Length Encoding
clc;
clear;
close all;
%% Set Quantization Parameter
quantizedvalue=10;
%% Read Input Image
InputImage=imread(‘cameraman.tif’);
[row col p]=size(InputImage);
%% Wavelet Decomposition
[LL LH HL HH]=dwt2(InputImage,’haar’);
WaveletDecomposeImage=[LL,LH;HL,HH];
imshow(WaveletDecomposeImage,[]);
%uniform quantization
QuantizedImage= WaveletDecomposeImage/quantizedvalue;
QuantizedImage= round(QuantizedImage);
% Convert the Two dimensional Image to a one dimensional Array using ZigZag Scanning
ImageArray=toZigzag(QuantizedImage);
%% Run Length Encoding
j=1;
a=length(ImageArray);
count=0;
for n=1:a
b=ImageArray(n);
if n==a
count=count+1;
c(j)=count;
s(j)=ImageArray(n);
elseif ImageArray(n)==ImageArray(n+1)
count=count+1;
elseif ImageArray(n)==b
count=count+1;
c(j)=count;
s(j)=ImageArray(n);
j=j+1;
count=0;
end
end
%% Calculation Bit Cost
InputBitcost=row*col*8;
InputBitcost=(InputBitcost);
c1=length(c);
s1=length(s);
OutputBitcost= (c1*8)+(s1*8);
OutputBitcost=(OutputBitcost);
%% Run Length Decoding g=length(s);
j=1;
l=1;
for i=1:g
v(l)=s(j);
if c(j)~=0
w=l+c(j)-1;
for p=l:w
v(l)=s(j);
l=l+1;
end
end
j=j+1;
end
ReconstructedImageArray=v;
%% Inverse ZigZag
ReconstructedImage=invZigzag(ReconstructedImageArray)
%% Inverse Quantization
ReconstructedImage=ReconstructedImage*quantizedvalue; am getting error in scanning in zigzag
the error is as follows
Undefined function or variable ‘toZigzag’.
Error in rlc_haar (line 21)
ImageArray=toZigzag(QuantizedImage);
plese help me
%% Matlab code for Image Compression Using Run Length Encoding
clc;
clear;
close all;
%% Set Quantization Parameter
quantizedvalue=10;
%% Read Input Image
InputImage=imread(‘cameraman.tif’);
[row col p]=size(InputImage);
%% Wavelet Decomposition
[LL LH HL HH]=dwt2(InputImage,’haar’);
WaveletDecomposeImage=[LL,LH;HL,HH];
imshow(WaveletDecomposeImage,[]);
%uniform quantization
QuantizedImage= WaveletDecomposeImage/quantizedvalue;
QuantizedImage= round(QuantizedImage);
% Convert the Two dimensional Image to a one dimensional Array using ZigZag Scanning
ImageArray=toZigzag(QuantizedImage);
%% Run Length Encoding
j=1;
a=length(ImageArray);
count=0;
for n=1:a
b=ImageArray(n);
if n==a
count=count+1;
c(j)=count;
s(j)=ImageArray(n);
elseif ImageArray(n)==ImageArray(n+1)
count=count+1;
elseif ImageArray(n)==b
count=count+1;
c(j)=count;
s(j)=ImageArray(n);
j=j+1;
count=0;
end
end
%% Calculation Bit Cost
InputBitcost=row*col*8;
InputBitcost=(InputBitcost);
c1=length(c);
s1=length(s);
OutputBitcost= (c1*8)+(s1*8);
OutputBitcost=(OutputBitcost);
%% Run Length Decoding g=length(s);
j=1;
l=1;
for i=1:g
v(l)=s(j);
if c(j)~=0
w=l+c(j)-1;
for p=l:w
v(l)=s(j);
l=l+1;
end
end
j=j+1;
end
ReconstructedImageArray=v;
%% Inverse ZigZag
ReconstructedImage=invZigzag(ReconstructedImageArray)
%% Inverse Quantization
ReconstructedImage=ReconstructedImage*quantizedvalue; . MATLAB Answers — New Questions
Mex-file not being found for Data Translation hardware
Hey everyone,
i recently move from MATLAB R2023b to MATLAB R2024a and must reinstall some package again.
So i reinstall Data Acquisition Toolbox and Data Acquisition Support Package for Data Translation to pursue a project i’m working on.
Everything was working fine on the previous MATLAB version but now for any try to acquire a signal for example, i receive the following error message:
‘The required MEX file to communicate with Data Translation hardware could not be loaded.
The attempt gave the Error ID of MATLAB:mex:ErrInvalidMEXFile and the message
Invalid MEX-file ‘C:Users"Username"AppDataRoamingMathWorksMATLAB Add-OnsToolboxesData Acquisition Toolbox Support Package for Data Translation Hardwareadaptorwin64mexOldaApi.mexw64′: Das angegebene Modul wurde nicht gefunden.’
reads the indicated modul ist not found. But this file mexOldaApi.mexw64 is precisely there.
I checked the vendors available and found out that ‘dt’ is set as not to be operational although the drivers are installed in MATLAB and both MATLAB and the computer have been reset.
Thanks for your suggestions to fix this.
Best regards,
GalvaniHey everyone,
i recently move from MATLAB R2023b to MATLAB R2024a and must reinstall some package again.
So i reinstall Data Acquisition Toolbox and Data Acquisition Support Package for Data Translation to pursue a project i’m working on.
Everything was working fine on the previous MATLAB version but now for any try to acquire a signal for example, i receive the following error message:
‘The required MEX file to communicate with Data Translation hardware could not be loaded.
The attempt gave the Error ID of MATLAB:mex:ErrInvalidMEXFile and the message
Invalid MEX-file ‘C:Users"Username"AppDataRoamingMathWorksMATLAB Add-OnsToolboxesData Acquisition Toolbox Support Package for Data Translation Hardwareadaptorwin64mexOldaApi.mexw64′: Das angegebene Modul wurde nicht gefunden.’
reads the indicated modul ist not found. But this file mexOldaApi.mexw64 is precisely there.
I checked the vendors available and found out that ‘dt’ is set as not to be operational although the drivers are installed in MATLAB and both MATLAB and the computer have been reset.
Thanks for your suggestions to fix this.
Best regards,
Galvani Hey everyone,
i recently move from MATLAB R2023b to MATLAB R2024a and must reinstall some package again.
So i reinstall Data Acquisition Toolbox and Data Acquisition Support Package for Data Translation to pursue a project i’m working on.
Everything was working fine on the previous MATLAB version but now for any try to acquire a signal for example, i receive the following error message:
‘The required MEX file to communicate with Data Translation hardware could not be loaded.
The attempt gave the Error ID of MATLAB:mex:ErrInvalidMEXFile and the message
Invalid MEX-file ‘C:Users"Username"AppDataRoamingMathWorksMATLAB Add-OnsToolboxesData Acquisition Toolbox Support Package for Data Translation Hardwareadaptorwin64mexOldaApi.mexw64′: Das angegebene Modul wurde nicht gefunden.’
reads the indicated modul ist not found. But this file mexOldaApi.mexw64 is precisely there.
I checked the vendors available and found out that ‘dt’ is set as not to be operational although the drivers are installed in MATLAB and both MATLAB and the computer have been reset.
Thanks for your suggestions to fix this.
Best regards,
Galvani dt_package_mexfile MATLAB Answers — New Questions
Removing outliers from the data creates gaps. Filling these gaps with missing values or the median of surrounding values does not address the issue.Why?
I am analyzing EMG data in windows. In each window, I apply z-score normalization to identify and remove outliers. To address the gaps created by removing these outliers, I attempt to fill the empty spaces with the median of the surrounding values. Additionally, I have experimented with MATLAB built-in functions such as ‘movmedian’ for this purpose.
here is my function:
function data_clean = remove_outliers_and_fill(data)
% Calculate z-scores for each column
z_scores = zscore(data);
% Define outlier threshold
threshold =3;
% Identify outliers
outliers = abs(z_scores) > threshold;
% Copy data to preserve original shape
data_clean = data;
% Loop through each column
[num_rows, num_cols] = size(data);
for col = 1:num_cols
for row = 1:num_rows
if outliers(row, col)
range_start = max(1, row-10);
range_end = min(num_rows, row+10);
neighbors = data(range_start:range_end, col);
% Exclude the outlier from median calculation
filtered_neighbors = neighbors(neighbors ~= data(row, col));
median_value = median(filtered_neighbors);
data_clean(row, col) = median_value;
end
end
end
end
here is the plot where it creates gaps after applying the above function.I am analyzing EMG data in windows. In each window, I apply z-score normalization to identify and remove outliers. To address the gaps created by removing these outliers, I attempt to fill the empty spaces with the median of the surrounding values. Additionally, I have experimented with MATLAB built-in functions such as ‘movmedian’ for this purpose.
here is my function:
function data_clean = remove_outliers_and_fill(data)
% Calculate z-scores for each column
z_scores = zscore(data);
% Define outlier threshold
threshold =3;
% Identify outliers
outliers = abs(z_scores) > threshold;
% Copy data to preserve original shape
data_clean = data;
% Loop through each column
[num_rows, num_cols] = size(data);
for col = 1:num_cols
for row = 1:num_rows
if outliers(row, col)
range_start = max(1, row-10);
range_end = min(num_rows, row+10);
neighbors = data(range_start:range_end, col);
% Exclude the outlier from median calculation
filtered_neighbors = neighbors(neighbors ~= data(row, col));
median_value = median(filtered_neighbors);
data_clean(row, col) = median_value;
end
end
end
end
here is the plot where it creates gaps after applying the above function. I am analyzing EMG data in windows. In each window, I apply z-score normalization to identify and remove outliers. To address the gaps created by removing these outliers, I attempt to fill the empty spaces with the median of the surrounding values. Additionally, I have experimented with MATLAB built-in functions such as ‘movmedian’ for this purpose.
here is my function:
function data_clean = remove_outliers_and_fill(data)
% Calculate z-scores for each column
z_scores = zscore(data);
% Define outlier threshold
threshold =3;
% Identify outliers
outliers = abs(z_scores) > threshold;
% Copy data to preserve original shape
data_clean = data;
% Loop through each column
[num_rows, num_cols] = size(data);
for col = 1:num_cols
for row = 1:num_rows
if outliers(row, col)
range_start = max(1, row-10);
range_end = min(num_rows, row+10);
neighbors = data(range_start:range_end, col);
% Exclude the outlier from median calculation
filtered_neighbors = neighbors(neighbors ~= data(row, col));
median_value = median(filtered_neighbors);
data_clean(row, col) = median_value;
end
end
end
end
here is the plot where it creates gaps after applying the above function. outliers, matlab function, emg signal MATLAB Answers — New Questions
How can i change the font size of XTick and YTick (x axis and y axis) in histogram of a image?
I have a image as lena.jpg, from which i was trying to obtain hist graph.
x=imread(‘lena.jpg’);
imhist(x);
set(gca,’FontSize’,15);
with this code i am able to change the font size of YTick only but i want to change font size of both. how can i do that????I have a image as lena.jpg, from which i was trying to obtain hist graph.
x=imread(‘lena.jpg’);
imhist(x);
set(gca,’FontSize’,15);
with this code i am able to change the font size of YTick only but i want to change font size of both. how can i do that???? I have a image as lena.jpg, from which i was trying to obtain hist graph.
x=imread(‘lena.jpg’);
imhist(x);
set(gca,’FontSize’,15);
with this code i am able to change the font size of YTick only but i want to change font size of both. how can i do that???? image processing, matlab, histogram MATLAB Answers — New Questions
Red zoom controls
Anyone else seeing zoom controls are highlighted red? Still works fine, just looks weird.
Anyone else seeing zoom controls are highlighted red? Still works fine, just looks weird. Read More
how to output capacity degradation curve based on simscape battery (table-based) block
I am using simscape battery toolbox for battery calendar aging simulation, with the battery table-based block, I enable the calendar aging function, I think which means my cell capacity now will decrease with the time
but I cannot find any port to output or somewhere to allow me to extract the battery capacity data. So I want to know, if I run a 100 weeks calendar aging simulation, and I want to obtain the capacity degradation curve (like the figure below, here I choose a cycle aging as an example, the axle x is the cycle life or time, the axle y is the remaining capacity)
how could I extract the battery capacity information from the Battery (table-based) block, if someone can help me, I will be very grateful, thanks.
(By the way, I have reviewed the help document regarding Battery (table-based) and Simscape battery, I also check the video "simscape battery essentials part 1 to 7" on Youtube, but cannot find the solution.)I am using simscape battery toolbox for battery calendar aging simulation, with the battery table-based block, I enable the calendar aging function, I think which means my cell capacity now will decrease with the time
but I cannot find any port to output or somewhere to allow me to extract the battery capacity data. So I want to know, if I run a 100 weeks calendar aging simulation, and I want to obtain the capacity degradation curve (like the figure below, here I choose a cycle aging as an example, the axle x is the cycle life or time, the axle y is the remaining capacity)
how could I extract the battery capacity information from the Battery (table-based) block, if someone can help me, I will be very grateful, thanks.
(By the way, I have reviewed the help document regarding Battery (table-based) and Simscape battery, I also check the video "simscape battery essentials part 1 to 7" on Youtube, but cannot find the solution.) I am using simscape battery toolbox for battery calendar aging simulation, with the battery table-based block, I enable the calendar aging function, I think which means my cell capacity now will decrease with the time
but I cannot find any port to output or somewhere to allow me to extract the battery capacity data. So I want to know, if I run a 100 weeks calendar aging simulation, and I want to obtain the capacity degradation curve (like the figure below, here I choose a cycle aging as an example, the axle x is the cycle life or time, the axle y is the remaining capacity)
how could I extract the battery capacity information from the Battery (table-based) block, if someone can help me, I will be very grateful, thanks.
(By the way, I have reviewed the help document regarding Battery (table-based) and Simscape battery, I also check the video "simscape battery essentials part 1 to 7" on Youtube, but cannot find the solution.) calendar aging, simscape battery, battery table-based block MATLAB Answers — New Questions
In stereocalibration, is the relationship between the ‘R and T output as PoseCamera2’ and the actual camera position the same, or does the sign of x in T reverse?
I am currently calibrating four cameras (Camera1, Camera2, Camera3, Camera4). To do this, I have created pairs (Camera1 & Camera2, Camera2 & Camera3, Camera1 & Camera4) and performed calibration to determine the relative positions of all cameras in the coordinate system of Camera1. For Camera1 and Camera2, I added about 80 images of a checkerboard taken using the stereocalibration feature of the calibration app for calibration.
As a result, I obtained the following:
R = [0.794, -0.0318, 0.605; 0.0226, 0.999, 0.0228; -0.606, -0.00446, 0.795]
T = [-2793, 44.86, 483.2] (units in [mm]).
The visual output, which I have attached as an image, shows that rotating Camera2 by R and translating it by T to align with the coordinate system of Camera1 makes it coincide with Camera1. Therefore, it can be seen that R and T correspond with the visual output.
However, the actual relative position of Camera2 to Camera1 in the coordinate system of Camera1 should be [2793, 44.86, 482.3]. Thus, I am considering that the sign of the x component of T obtained through stereocalibration might be reversed compared to the actual T. Is my understanding incorrect?I am currently calibrating four cameras (Camera1, Camera2, Camera3, Camera4). To do this, I have created pairs (Camera1 & Camera2, Camera2 & Camera3, Camera1 & Camera4) and performed calibration to determine the relative positions of all cameras in the coordinate system of Camera1. For Camera1 and Camera2, I added about 80 images of a checkerboard taken using the stereocalibration feature of the calibration app for calibration.
As a result, I obtained the following:
R = [0.794, -0.0318, 0.605; 0.0226, 0.999, 0.0228; -0.606, -0.00446, 0.795]
T = [-2793, 44.86, 483.2] (units in [mm]).
The visual output, which I have attached as an image, shows that rotating Camera2 by R and translating it by T to align with the coordinate system of Camera1 makes it coincide with Camera1. Therefore, it can be seen that R and T correspond with the visual output.
However, the actual relative position of Camera2 to Camera1 in the coordinate system of Camera1 should be [2793, 44.86, 482.3]. Thus, I am considering that the sign of the x component of T obtained through stereocalibration might be reversed compared to the actual T. Is my understanding incorrect? I am currently calibrating four cameras (Camera1, Camera2, Camera3, Camera4). To do this, I have created pairs (Camera1 & Camera2, Camera2 & Camera3, Camera1 & Camera4) and performed calibration to determine the relative positions of all cameras in the coordinate system of Camera1. For Camera1 and Camera2, I added about 80 images of a checkerboard taken using the stereocalibration feature of the calibration app for calibration.
As a result, I obtained the following:
R = [0.794, -0.0318, 0.605; 0.0226, 0.999, 0.0228; -0.606, -0.00446, 0.795]
T = [-2793, 44.86, 483.2] (units in [mm]).
The visual output, which I have attached as an image, shows that rotating Camera2 by R and translating it by T to align with the coordinate system of Camera1 makes it coincide with Camera1. Therefore, it can be seen that R and T correspond with the visual output.
However, the actual relative position of Camera2 to Camera1 in the coordinate system of Camera1 should be [2793, 44.86, 482.3]. Thus, I am considering that the sign of the x component of T obtained through stereocalibration might be reversed compared to the actual T. Is my understanding incorrect? image processing, calibratrion, stereocalibration MATLAB Answers — New Questions
A problem in using imhist to display histogram of indexed image
First i convert the image to an indexed image, and use only 5 colors to show what is my problem.
clear
img = imread(‘peppers.png’);
[x,map] = rgb2ind(img,5);
figure;
imhist(x,map)
<</matlabcentral/answers/uploaded_files/44260/imhist.PNG>>
the colorbar doesn’t match with the histogarm bars, this is my colormap:
map =
0.2784 0.1373 0.2353
0.7608 0.1686 0.1373
0.8902 0.7255 0.6353
0.4275 0.3765 0.2235
0.8471 0.5569 0.1020
the first one isn’t white, It seems that the colorbar is shifted by one.First i convert the image to an indexed image, and use only 5 colors to show what is my problem.
clear
img = imread(‘peppers.png’);
[x,map] = rgb2ind(img,5);
figure;
imhist(x,map)
<</matlabcentral/answers/uploaded_files/44260/imhist.PNG>>
the colorbar doesn’t match with the histogarm bars, this is my colormap:
map =
0.2784 0.1373 0.2353
0.7608 0.1686 0.1373
0.8902 0.7255 0.6353
0.4275 0.3765 0.2235
0.8471 0.5569 0.1020
the first one isn’t white, It seems that the colorbar is shifted by one. First i convert the image to an indexed image, and use only 5 colors to show what is my problem.
clear
img = imread(‘peppers.png’);
[x,map] = rgb2ind(img,5);
figure;
imhist(x,map)
<</matlabcentral/answers/uploaded_files/44260/imhist.PNG>>
the colorbar doesn’t match with the histogarm bars, this is my colormap:
map =
0.2784 0.1373 0.2353
0.7608 0.1686 0.1373
0.8902 0.7255 0.6353
0.4275 0.3765 0.2235
0.8471 0.5569 0.1020
the first one isn’t white, It seems that the colorbar is shifted by one. imhist, colorbar, indexed image MATLAB Answers — New Questions