Month: March 2026
[Vehicle Dynamics Blockset] Confused about coordinate system conventions
Hello everyone!
Recently I’m working on this Simulink example: https://www.mathworks.com/help/vdynblks/ug/braking-test.html
I also checked the setting of coordinate system in VDBS: https://www.mathworks.com/help/vdynblks/ug/coordinate-systems-in-vehicle-dynamics-blockset.html
In this simulation, I assume a few settings:
(1) the ground is completely flat and horizontal
(2) the origin of inertial (world-fixed) frame is attach at the ground surface, with Z-axe pointing downward.
(3) the origin of tire frame is attach at the ground surface, with Z-axe pointing upward.
I want to slightly modify the Simulink example shared above, by changing parameters in some blocks:
(1) Block 1 —- Vehicle Body 6DOF
Based on the figure below, the vehicle-fixed frame should attach at the center of mass (CoM)
In this block, I want to change the initial position of CoM in inertial frame (aka, world-fixed frame), as shown below:
Since the inertial frame is attached on the ground, for the initial vertical position of CoM, I set it to -0.30938-VEH.HeightCG = -0.44338. Here, the value 0.30938 is the unload tire radius, and VEH.HeightCG is the vertical offset between CoM and axle plane (h in figure "vehicle layout" above). From my understanding, this position can let the tire just slightly contact the ground.
(2) Block 2 —- Combined Slip Wheel 2DOF
In this block, I assign a small value to this parameter:
Based on my understanding, this value refers to the compression of tire due to vertical load. A positive value means that the tire is compressed.
(3) Block 3 —- Ground Feedback
Since I assume inertial frame is attach at the ground surface, so the ground height for all wheels G_xx_z (xx: FL, FR, RL, RR) are set to 0 all the time.
Then I start the simulation. During the first 2 seconds, the vehicle will stablize on the ground. I logged some relavent signals to check the vehicle states:
(1) Passenger Vehicle:1.Body.InertFrm.Cg.Disp.Z —- the vertical position of CoM in inertial frame
(2) Passenger Vehicle:1.Body.InertFrm.FrntAxl.Lft.Disp.Z —- the vertical position of front-left axle in inertial frame
(3) Passenger Vehicle:1.Body.BdyFrm.FrntAxl.Lft.Disp.Z —- the vertical position of front-left axle in vehicle-fixed frame
(4) Passenger Vehicle:1.Wheels.TireFrame.z(1) —- the vertical displacement of front-left axle in tire frame? (I’m not sure, just based on the description in https://www.mathworks.com/help/vdynblks/ref/combinedslipwheel2dof.html)
(5) Passenger Vehicle:1.Wheels.TireFrame.dz(1) —- the vertical displacement of front-left axle (wheel) in vehicle-fixed frame? (I’m not sure, just based on the computation in the model, as shown below)
The stablizing process for vehicle is shown below:
After the vehicle is stablized on the ground, we can see:
(1) the vertical position of CoM in inertial frame converges to around 0.0
(2) the vertical position of front-left axle in inertial frame is close to that value in vehicle-fixed frame, while the latter one is always a constant value equal to VEH.HeightCG
(3) TireFrame.z(1) and TireFrame.dz(1) converge to some small values near 0.0
(4) The initial value of TireFrame.z(1) is the opposite of the initial value of a parameter z0 in Block "Combined Slip Wheel 2DOF" (as shown above). It might be due to a flip of Z-axe direction.
It means that the CoM of the vehicle is near the ground, which means that the vehicle chassis penetrates and sinks under the ground. It doesn’t make sense to me. I have a few questions:
(1) What are the setting of coordinate systems for different components (CoM, axles, wheels, tires)? —- where should the origin of a specific frame attach?
(2) What are the exact meaning of TireFrame.z and TireFrame.dz? The vertical position of wheel center? In which frame?
I would be very grateful if someone could provide an answer.
Regards,
Zijun HeHello everyone!
Recently I’m working on this Simulink example: https://www.mathworks.com/help/vdynblks/ug/braking-test.html
I also checked the setting of coordinate system in VDBS: https://www.mathworks.com/help/vdynblks/ug/coordinate-systems-in-vehicle-dynamics-blockset.html
In this simulation, I assume a few settings:
(1) the ground is completely flat and horizontal
(2) the origin of inertial (world-fixed) frame is attach at the ground surface, with Z-axe pointing downward.
(3) the origin of tire frame is attach at the ground surface, with Z-axe pointing upward.
I want to slightly modify the Simulink example shared above, by changing parameters in some blocks:
(1) Block 1 —- Vehicle Body 6DOF
Based on the figure below, the vehicle-fixed frame should attach at the center of mass (CoM)
In this block, I want to change the initial position of CoM in inertial frame (aka, world-fixed frame), as shown below:
Since the inertial frame is attached on the ground, for the initial vertical position of CoM, I set it to -0.30938-VEH.HeightCG = -0.44338. Here, the value 0.30938 is the unload tire radius, and VEH.HeightCG is the vertical offset between CoM and axle plane (h in figure "vehicle layout" above). From my understanding, this position can let the tire just slightly contact the ground.
(2) Block 2 —- Combined Slip Wheel 2DOF
In this block, I assign a small value to this parameter:
Based on my understanding, this value refers to the compression of tire due to vertical load. A positive value means that the tire is compressed.
(3) Block 3 —- Ground Feedback
Since I assume inertial frame is attach at the ground surface, so the ground height for all wheels G_xx_z (xx: FL, FR, RL, RR) are set to 0 all the time.
Then I start the simulation. During the first 2 seconds, the vehicle will stablize on the ground. I logged some relavent signals to check the vehicle states:
(1) Passenger Vehicle:1.Body.InertFrm.Cg.Disp.Z —- the vertical position of CoM in inertial frame
(2) Passenger Vehicle:1.Body.InertFrm.FrntAxl.Lft.Disp.Z —- the vertical position of front-left axle in inertial frame
(3) Passenger Vehicle:1.Body.BdyFrm.FrntAxl.Lft.Disp.Z —- the vertical position of front-left axle in vehicle-fixed frame
(4) Passenger Vehicle:1.Wheels.TireFrame.z(1) —- the vertical displacement of front-left axle in tire frame? (I’m not sure, just based on the description in https://www.mathworks.com/help/vdynblks/ref/combinedslipwheel2dof.html)
(5) Passenger Vehicle:1.Wheels.TireFrame.dz(1) —- the vertical displacement of front-left axle (wheel) in vehicle-fixed frame? (I’m not sure, just based on the computation in the model, as shown below)
The stablizing process for vehicle is shown below:
After the vehicle is stablized on the ground, we can see:
(1) the vertical position of CoM in inertial frame converges to around 0.0
(2) the vertical position of front-left axle in inertial frame is close to that value in vehicle-fixed frame, while the latter one is always a constant value equal to VEH.HeightCG
(3) TireFrame.z(1) and TireFrame.dz(1) converge to some small values near 0.0
(4) The initial value of TireFrame.z(1) is the opposite of the initial value of a parameter z0 in Block "Combined Slip Wheel 2DOF" (as shown above). It might be due to a flip of Z-axe direction.
It means that the CoM of the vehicle is near the ground, which means that the vehicle chassis penetrates and sinks under the ground. It doesn’t make sense to me. I have a few questions:
(1) What are the setting of coordinate systems for different components (CoM, axles, wheels, tires)? —- where should the origin of a specific frame attach?
(2) What are the exact meaning of TireFrame.z and TireFrame.dz? The vertical position of wheel center? In which frame?
I would be very grateful if someone could provide an answer.
Regards,
Zijun He Hello everyone!
Recently I’m working on this Simulink example: https://www.mathworks.com/help/vdynblks/ug/braking-test.html
I also checked the setting of coordinate system in VDBS: https://www.mathworks.com/help/vdynblks/ug/coordinate-systems-in-vehicle-dynamics-blockset.html
In this simulation, I assume a few settings:
(1) the ground is completely flat and horizontal
(2) the origin of inertial (world-fixed) frame is attach at the ground surface, with Z-axe pointing downward.
(3) the origin of tire frame is attach at the ground surface, with Z-axe pointing upward.
I want to slightly modify the Simulink example shared above, by changing parameters in some blocks:
(1) Block 1 —- Vehicle Body 6DOF
Based on the figure below, the vehicle-fixed frame should attach at the center of mass (CoM)
In this block, I want to change the initial position of CoM in inertial frame (aka, world-fixed frame), as shown below:
Since the inertial frame is attached on the ground, for the initial vertical position of CoM, I set it to -0.30938-VEH.HeightCG = -0.44338. Here, the value 0.30938 is the unload tire radius, and VEH.HeightCG is the vertical offset between CoM and axle plane (h in figure "vehicle layout" above). From my understanding, this position can let the tire just slightly contact the ground.
(2) Block 2 —- Combined Slip Wheel 2DOF
In this block, I assign a small value to this parameter:
Based on my understanding, this value refers to the compression of tire due to vertical load. A positive value means that the tire is compressed.
(3) Block 3 —- Ground Feedback
Since I assume inertial frame is attach at the ground surface, so the ground height for all wheels G_xx_z (xx: FL, FR, RL, RR) are set to 0 all the time.
Then I start the simulation. During the first 2 seconds, the vehicle will stablize on the ground. I logged some relavent signals to check the vehicle states:
(1) Passenger Vehicle:1.Body.InertFrm.Cg.Disp.Z —- the vertical position of CoM in inertial frame
(2) Passenger Vehicle:1.Body.InertFrm.FrntAxl.Lft.Disp.Z —- the vertical position of front-left axle in inertial frame
(3) Passenger Vehicle:1.Body.BdyFrm.FrntAxl.Lft.Disp.Z —- the vertical position of front-left axle in vehicle-fixed frame
(4) Passenger Vehicle:1.Wheels.TireFrame.z(1) —- the vertical displacement of front-left axle in tire frame? (I’m not sure, just based on the description in https://www.mathworks.com/help/vdynblks/ref/combinedslipwheel2dof.html)
(5) Passenger Vehicle:1.Wheels.TireFrame.dz(1) —- the vertical displacement of front-left axle (wheel) in vehicle-fixed frame? (I’m not sure, just based on the computation in the model, as shown below)
The stablizing process for vehicle is shown below:
After the vehicle is stablized on the ground, we can see:
(1) the vertical position of CoM in inertial frame converges to around 0.0
(2) the vertical position of front-left axle in inertial frame is close to that value in vehicle-fixed frame, while the latter one is always a constant value equal to VEH.HeightCG
(3) TireFrame.z(1) and TireFrame.dz(1) converge to some small values near 0.0
(4) The initial value of TireFrame.z(1) is the opposite of the initial value of a parameter z0 in Block "Combined Slip Wheel 2DOF" (as shown above). It might be due to a flip of Z-axe direction.
It means that the CoM of the vehicle is near the ground, which means that the vehicle chassis penetrates and sinks under the ground. It doesn’t make sense to me. I have a few questions:
(1) What are the setting of coordinate systems for different components (CoM, axles, wheels, tires)? —- where should the origin of a specific frame attach?
(2) What are the exact meaning of TireFrame.z and TireFrame.dz? The vertical position of wheel center? In which frame?
I would be very grateful if someone could provide an answer.
Regards,
Zijun He vehicle dynamcis, coordinate system, vehicle dynamics blockset MATLAB Answers — New Questions
R2025b Update 4 does not support element-wise division (./) for tf objects?
The following should divide 1/(s+1) by 2, and 1/(s+2) by 3:
s = tf(‘s’);
sys1 = [1/(s+1), 1/(s+2)];
sys2 = [2, 3];
result = sys1 ./ sys2;
However, I get the following error message:
Error using ./
Invalid data type. Argument must be numeric, char, or logical.
Error in
test (line 4)
result = sys1 ./ sys2;The following should divide 1/(s+1) by 2, and 1/(s+2) by 3:
s = tf(‘s’);
sys1 = [1/(s+1), 1/(s+2)];
sys2 = [2, 3];
result = sys1 ./ sys2;
However, I get the following error message:
Error using ./
Invalid data type. Argument must be numeric, char, or logical.
Error in
test (line 4)
result = sys1 ./ sys2; The following should divide 1/(s+1) by 2, and 1/(s+2) by 3:
s = tf(‘s’);
sys1 = [1/(s+1), 1/(s+2)];
sys2 = [2, 3];
result = sys1 ./ sys2;
However, I get the following error message:
Error using ./
Invalid data type. Argument must be numeric, char, or logical.
Error in
test (line 4)
result = sys1 ./ sys2; control toolbox, element-wise division, transfer-function object MATLAB Answers — New Questions
How to draw the axis in the origin of the coordinate system?
Hello,
I want to make a figure that looks like this:
The code has always worked fine, but now it does not save the figure correctly anymore. (Maybe I updated MATLAB since last year, but I don’t know; I am using R2025b and don’t know, which one I used before.)
The exported figure looks like this:
The labels for f(x) and x are at the wrong places.
The error happens with exportfigure, print and manually saving the figure and png, jpg and pdf.
ChatGPT tells me, the problem is with "origin" but that there are only manual fixes which seem very complicated and unsatisfying to me.
Is there a fix for my figure / what should I do?
Thank you very much in advance!
%% Minimal Working Example
figureFontSize = 22;
set(0,’DefaultAxesFontSize’,22)
lwidth=2;
xMin=-5;
xMax=+6;
xVal=(xMin:0.05:xMax);
% Aufgabe
yVal(:,1)=-2*(xVal-4).^2+64;
% Aufgabe
fig=figure(1);
plot(xVal,yVal(:,1),’-k’,’LineWidth’,lwidth)
xlabel(‘x’)
ylA=ylabel(‘f(x)’);
xlim([xMin xMax])
a=max(abs(min(yVal(:,1))),max(yVal(:,1)));
ylim([-a-0.01*a +a])
set(gca, ‘xtick’, [])
set(gca, ‘ytick’, [])
ax = gca;
box(‘off’)
grid(‘off’)
set(gca, ‘xticklabel’, [])
set(gca, ‘yticklabel’, [])
ax.XAxisLocation = "origin";
ax.YAxisLocation = "origin";
% print(gcf,’test.png’,’-dpng’,’-r300′)
exportgraphics(gcf,’test.png’,’Resolution’,300)Hello,
I want to make a figure that looks like this:
The code has always worked fine, but now it does not save the figure correctly anymore. (Maybe I updated MATLAB since last year, but I don’t know; I am using R2025b and don’t know, which one I used before.)
The exported figure looks like this:
The labels for f(x) and x are at the wrong places.
The error happens with exportfigure, print and manually saving the figure and png, jpg and pdf.
ChatGPT tells me, the problem is with "origin" but that there are only manual fixes which seem very complicated and unsatisfying to me.
Is there a fix for my figure / what should I do?
Thank you very much in advance!
%% Minimal Working Example
figureFontSize = 22;
set(0,’DefaultAxesFontSize’,22)
lwidth=2;
xMin=-5;
xMax=+6;
xVal=(xMin:0.05:xMax);
% Aufgabe
yVal(:,1)=-2*(xVal-4).^2+64;
% Aufgabe
fig=figure(1);
plot(xVal,yVal(:,1),’-k’,’LineWidth’,lwidth)
xlabel(‘x’)
ylA=ylabel(‘f(x)’);
xlim([xMin xMax])
a=max(abs(min(yVal(:,1))),max(yVal(:,1)));
ylim([-a-0.01*a +a])
set(gca, ‘xtick’, [])
set(gca, ‘ytick’, [])
ax = gca;
box(‘off’)
grid(‘off’)
set(gca, ‘xticklabel’, [])
set(gca, ‘yticklabel’, [])
ax.XAxisLocation = "origin";
ax.YAxisLocation = "origin";
% print(gcf,’test.png’,’-dpng’,’-r300′)
exportgraphics(gcf,’test.png’,’Resolution’,300) Hello,
I want to make a figure that looks like this:
The code has always worked fine, but now it does not save the figure correctly anymore. (Maybe I updated MATLAB since last year, but I don’t know; I am using R2025b and don’t know, which one I used before.)
The exported figure looks like this:
The labels for f(x) and x are at the wrong places.
The error happens with exportfigure, print and manually saving the figure and png, jpg and pdf.
ChatGPT tells me, the problem is with "origin" but that there are only manual fixes which seem very complicated and unsatisfying to me.
Is there a fix for my figure / what should I do?
Thank you very much in advance!
%% Minimal Working Example
figureFontSize = 22;
set(0,’DefaultAxesFontSize’,22)
lwidth=2;
xMin=-5;
xMax=+6;
xVal=(xMin:0.05:xMax);
% Aufgabe
yVal(:,1)=-2*(xVal-4).^2+64;
% Aufgabe
fig=figure(1);
plot(xVal,yVal(:,1),’-k’,’LineWidth’,lwidth)
xlabel(‘x’)
ylA=ylabel(‘f(x)’);
xlim([xMin xMax])
a=max(abs(min(yVal(:,1))),max(yVal(:,1)));
ylim([-a-0.01*a +a])
set(gca, ‘xtick’, [])
set(gca, ‘ytick’, [])
ax = gca;
box(‘off’)
grid(‘off’)
set(gca, ‘xticklabel’, [])
set(gca, ‘yticklabel’, [])
ax.XAxisLocation = "origin";
ax.YAxisLocation = "origin";
% print(gcf,’test.png’,’-dpng’,’-r300′)
exportgraphics(gcf,’test.png’,’Resolution’,300) axislocation, origin MATLAB Answers — New Questions
How to use the overpotential in the parameterised battery for the simscape battery equivalent circuit? I am trying to use the molicell 21700 cell. I can run the R0 model.
The simscape equivalent circuit battery has a bunch of parameterized cells. I am using the molicelI NR_21700_P45B lithium ion battery. The in built data set has 101 state of charge breakpoints, and 7 temperature levels. So, the R0 value has a matrix of size 101×7. However, the overpotential (R1-C1, R2-C2, R3-C3) has the size 7×3. Obviously, when I run the model, I get the following error :
size of First polarization resistance, R1(SOC,T) must be equal to length of State of charge breakpoints for resistance, SOC by length of Temperature breakpoints for resistance, T.
I have read the model documentation clearly, there is a part where a hint of how to use the values properly is given: The software does not parameterize the change in series resistance with the battery cycle life and the dynamic RC network parameters. Instead, the software sums the net resistance of the RC network resistors to the series resistance of the specific parameterized data. To populate the RC parameter data, subtract the net RC network resistance from the series resistance data.
This description to me is not very clear. The model contains the following data heads: Instantaneous resistance, R0 (SOC, T), first polarization resistance R1(SOC, T) and Tau, if I use one time-constant dynamics, which are of my interest. There is nothing mentioned explicitly about a series resistance from which the net RC resistance is to be subtracted. A demonstration of how to use this will be useful.The simscape equivalent circuit battery has a bunch of parameterized cells. I am using the molicelI NR_21700_P45B lithium ion battery. The in built data set has 101 state of charge breakpoints, and 7 temperature levels. So, the R0 value has a matrix of size 101×7. However, the overpotential (R1-C1, R2-C2, R3-C3) has the size 7×3. Obviously, when I run the model, I get the following error :
size of First polarization resistance, R1(SOC,T) must be equal to length of State of charge breakpoints for resistance, SOC by length of Temperature breakpoints for resistance, T.
I have read the model documentation clearly, there is a part where a hint of how to use the values properly is given: The software does not parameterize the change in series resistance with the battery cycle life and the dynamic RC network parameters. Instead, the software sums the net resistance of the RC network resistors to the series resistance of the specific parameterized data. To populate the RC parameter data, subtract the net RC network resistance from the series resistance data.
This description to me is not very clear. The model contains the following data heads: Instantaneous resistance, R0 (SOC, T), first polarization resistance R1(SOC, T) and Tau, if I use one time-constant dynamics, which are of my interest. There is nothing mentioned explicitly about a series resistance from which the net RC resistance is to be subtracted. A demonstration of how to use this will be useful. The simscape equivalent circuit battery has a bunch of parameterized cells. I am using the molicelI NR_21700_P45B lithium ion battery. The in built data set has 101 state of charge breakpoints, and 7 temperature levels. So, the R0 value has a matrix of size 101×7. However, the overpotential (R1-C1, R2-C2, R3-C3) has the size 7×3. Obviously, when I run the model, I get the following error :
size of First polarization resistance, R1(SOC,T) must be equal to length of State of charge breakpoints for resistance, SOC by length of Temperature breakpoints for resistance, T.
I have read the model documentation clearly, there is a part where a hint of how to use the values properly is given: The software does not parameterize the change in series resistance with the battery cycle life and the dynamic RC network parameters. Instead, the software sums the net resistance of the RC network resistors to the series resistance of the specific parameterized data. To populate the RC parameter data, subtract the net RC network resistance from the series resistance data.
This description to me is not very clear. The model contains the following data heads: Instantaneous resistance, R0 (SOC, T), first polarization resistance R1(SOC, T) and Tau, if I use one time-constant dynamics, which are of my interest. There is nothing mentioned explicitly about a series resistance from which the net RC resistance is to be subtracted. A demonstration of how to use this will be useful. simulink, simscape, matrices, battery, equivalent_circuit, lithium-ion, cell modelling, matrix manipulation MATLAB Answers — New Questions
March 2026 Update for Office 365 for IT Pros
Update #129 Available for Download
The Office 365 for IT Pros eBook team is delighted to announce the availability of the March 2026 update for Office 365 for IT Pros (2026 edition). This is monthly update #129. An update (#21.2) has already been issued for the Automating Microsoft 365 with PowerShell eBook, which is available separately and as part of the Office 365 for IT Pros eBook bundle.
Current subscribers can download the updated PDF and EPUB files through their Gumroad.com account or using the link in the receipt emailed after purchase. The link always accesses the latest book files. Further details of how to access book updates are available in our FAQ. Details of the changes in update #129 are in our change log.
SharePoint Turns 25
On March 2, 2026, Microsoft will celebrate the 25th “birthday” of SharePoint with a “digital celebration event and live AMA.” Inanimate objects like software products don’t have birthdays, so the event marks when Microsoft first released the SharePoint Server 2001 product (the “release to manufacturing”, or RTM, date). In those days, releasing software was a matter of creating the master “gold” CDs and sending them to the manufacturing facility to be duplicated and made into a customer-facing product. This involved nice packaging and perhaps even printed documentation.
I worked as the Chief Technology Officer for the Compaq Global Services organization at the time and wanted to see how SharePoint Portal Server 2001 worked in production, so we installed a few servers and used them as the basis for a primitive but effective knowledge management system for Microsoft technologies.
I first wrote about SharePoint Portal Server soon after release for Windows 2000 Magazine (the article appeared in May 2001), and noted that both SharePoint and Exchange shared common storage technology. At the time, Microsoft called it WSS, or Web Storage System, but it was really just ESE, the Exchange Storage Engine. SharePoint adopted SQL Server storage for the 2003 release. At the time there was some pressure for Exchange to use SQL too (the “Kodiak” project), but that didn’t work out.
I thought SharePoint Server lost its way a little by overcomplicating things in the 2003 and 2007 releases; the online version has removed much of the complexity and made SharePoint Online into a core part of Microsoft 365. The introduction of Teams gave SharePoint Online a big lift because Teams gave SharePoint a nicer UI (the old browser UI was not great).
I find it amusing that many SharePoint fans parrot the marketing line that SharePoint Online powers Microsoft 365. SharePoint is an incredibly important part of Microsoft 365, but people can get value out of Microsoft 365 without ever touching SharePoint, directly or indirectly. In reality, SharePoint Online is the file storage service for Microsoft 365, no more and no less, a point I’ve been making for some time.
Exchange Reaches 30
Exchange Server is approaching 30 years old and will get there on April 2, 2026. I doubt that we’ll see the same kind of Microsoft celebrations for Exchange simply because Exchange Online doesn’t have the same kind of marketing heft that exists for the SharePoint and Teams organization.
At the Ignite 2025 conference, I met Iain McDonald, who did many things in the early days of Exchange (read his LinkedIn post for details), and we had a great conversation about his work with Exchange and Windows (especially Windows 2000). Iain seems to pop up every ten years because the last time we had a conversation, it was about Exchange’s 20th anniversary.
I didn’t like Exchange 4.0 very much at the start, mostly because the 1995 “Alliance for Enterprise Computing” agreement between Microsoft and Digital Equipment Corporation killed a NT-based DEC mail server. But I got used to Exchange and ended up writing ten books about the server, including the Microsoft Press Inside Out books for Exchange 2010 and Exchange 2013.
Exchange Online is the messaging service for Microsoft 365. More people use Exchange Online than use SharePoint Online. Very few of the more than 450 million paid Microsoft 365 seats (and the substantial number of free seats) don’t use Exchange Online, even if they don’t realize it. In the past, the Microsoft marketing machine has told the world that Yammer (now Viva Engage) and Teams would eliminate email. That just hasn’t happened, even if a lot of interpersonal messages flow through Teams rather than email.
The point is that people need to communicate and store knowledge. Exchange Online and SharePoint Online act as great enablers and services within Microsoft 365. Decades of development have delivered great software, even if both services have caused me to swear profusely in the past.
On to Office 365 for IT Pros Update #130
Office 365 for IT Pros started off in 2015 with a focus on Exchange Online. In those days, that’s where most of the Office 365 action was. Organizations migrated email first because the migration facilities were available. SharePoint migration tools came later and Teams arrived in 2017. Over the last 129 updates, we’ve gradually evolved our content to create balanced coverage across Entra ID, SharePoint Online, Exchange Online, Teams, OneDrive for Business, Planner, Purview, and lots more.
On we go to update #130, due on April 1. Microsoft doesn’t stop issuing updates for Microsoft 365, and we don’t stop analyzing, questioning, and reporting on what they do.
How do I find if a point is within the volume of a rotated ellipsoid
I have an ellipsoid
[ex,ey,ez]=ellipsoid(long2,lat2,h2,bb2,aa2,cc2);
direction=[1 0 0];
s=surf(ex,ey,ez,’FaceAlpha’,0.05)
rotate(s,direction,theta(i),[long2 lat2 h2])
and say I have a point ;
point = [longX latX hX];
How do I find if point is within the volume of the rotated ellipsoid?
Thanks!I have an ellipsoid
[ex,ey,ez]=ellipsoid(long2,lat2,h2,bb2,aa2,cc2);
direction=[1 0 0];
s=surf(ex,ey,ez,’FaceAlpha’,0.05)
rotate(s,direction,theta(i),[long2 lat2 h2])
and say I have a point ;
point = [longX latX hX];
How do I find if point is within the volume of the rotated ellipsoid?
Thanks! I have an ellipsoid
[ex,ey,ez]=ellipsoid(long2,lat2,h2,bb2,aa2,cc2);
direction=[1 0 0];
s=surf(ex,ey,ez,’FaceAlpha’,0.05)
rotate(s,direction,theta(i),[long2 lat2 h2])
and say I have a point ;
point = [longX latX hX];
How do I find if point is within the volume of the rotated ellipsoid?
Thanks! ellipsoid, volume MATLAB Answers — New Questions
code improvement/optimization
I would like to know if this code can be improved or it well written? any comments will be greatly appreciated. Thanks
u0 = [1; 0; 0];
tspan = [0, 1e-3]; % small final time for explicit solver
tol = 1e-2;
h0 = 1e-6;
% Execute solver
[t, u, err, h_steps, stats] = dormand_prince_solver(@robertson_ode, tspan, u0, tol, h0);
% Part (a): Plot Solutions
figure;
subplot(2,1,1);
plot(t, u(:,1), ‘b’, t, u(:,3), ‘g’, ‘LineWidth’, 1.5);
legend(‘y1 (Reactant)’, ‘y3 (Product)’);
xlabel(‘Time t’); ylabel(‘Concentration’);
%title(‘Robertson Problem: Concentrations y1 and y3’);
grid on;
subplot(2,1,2);
semilogx(t, u(:,2), ‘r’, ‘LineWidth’, 1.5);
xlabel(‘Time t (Log Scale)’); ylabel(‘Concentration y2’);
%title(‘Robertson Problem: Intermediate Species y2’);
grid on;
% Part (b): Error and Stepsize Plots
figure;
subplot(2,1,1);
semilogy(t(2:end), err);
xlabel(‘Time t’); ylabel(‘||y_n^5 – y_n^4||_{infty} / h’);
%title(‘Part (b): Estimated Error vs Time’);
grid on;
subplot(2,1,2);
semilogy(t(2:end), h_steps);
xlabel(‘Time t’); ylabel(‘Stepsize h’);
%title(‘Part (b): Stepsize h vs Time’);
grid on;
figure;
subplot(2,1,1);
semilogy(t(2:end), err, ‘LineWidth’, 1.5);
xlabel(‘Time t’);
ylabel(‘Error’);
title(‘Estimated Local Error vs Time’);
grid on;
subplot(2,1,2);
semilogy(t(2:end), h_steps, ‘LineWidth’, 1.5);
xlabel(‘Time t’);
ylabel(‘Stepsize h’);
title(‘Stepsize vs Time’);
grid on;
function dydt = robertson_ode(~, y)
% Coefficients from Screenshot 2026-02-28 at 3.38.10 PM.png
alpha = 0.04;
beta = 1e4;
gamma = 3e7;
dydt = [ -alpha*y(1) + beta*y(2)*y(3);
alpha*y(1) – beta*y(2)*y(3) – gamma*y(2)^2;
gamma*y(2)^2 ];
end
function [t_all, u_all, error_history, h_history, stats] = …
dormand_prince_solver(f, tspan, u0, tol, h0)
% Initialization
t = tspan(1);
u = u0(:);
h = h0;
t_all = t;
u_all = u.’;
error_history = [];
h_history = [];
n_accepted = 0;
n_rejected = 0;
% —- Dormand–Prince 4(5) Coefficients —-
a = [1/5, 0, 0, 0, 0, 0;
3/40, 9/40, 0, 0, 0, 0;
44/45, -56/15, 32/9, 0, 0, 0;
19372/6561, -25360/2187, 64448/6561, -212/729, 0, 0;
9017/3168, -355/33, 46732/5247, 49/176, -5103/18656, 0;
35/384, 0, 500/1113, 125/192, -2187/6784, 11/84];
b4 = [5179/57600, 0, 7571/16695, 393/640, …
-92097/339200, 187/2100, 1/40];
b5 = [35/384, 0, 500/1113, 125/192, …
-2187/6784, 11/84, 0];
c = [0, 1/5, 3/10, 4/5, 8/9, 1, 1];
safety = 0.9;
h_min = 1e-12;
h_max = 1.0;
while t < tspan(2)
if h < h_min
error(‘Step size underflow: problem likely stiff.’);
end
if t + h > tspan(2)
h = tspan(2) – t;
end
% —- Stage calculations —-
k = zeros(length(u), 7);
k(:,1) = f(t, u);
for i = 2:7
k(:,i) = f(t + c(i)*h, …
u + h * (k(:,1:i-1) * a(i-1,1:i-1)’));
end
% —- 4th and 5th order solutions —-
y4 = u + h * (k * b4′);
y5 = u + h * (k * b5′);
error_val = norm(y5 – y4, inf);
if error_val <= tol
t = t + h;
u = y5;
t_all = [t_all; t];
u_all = [u_all; u.’];
error_history = [error_history; error_val];
h_history = [h_history; h];
n_accepted = n_accepted + 1;
else
n_rejected = n_rejected + 1;
end
if error_val == 0
h_new = 2*h;
else
h_new = h * safety * (tol / error_val)^(1/5);
end
h = min(max(h_new, h_min), h_max);
end
stats.accepted = n_accepted;
stats.rejected = n_rejected;
endI would like to know if this code can be improved or it well written? any comments will be greatly appreciated. Thanks
u0 = [1; 0; 0];
tspan = [0, 1e-3]; % small final time for explicit solver
tol = 1e-2;
h0 = 1e-6;
% Execute solver
[t, u, err, h_steps, stats] = dormand_prince_solver(@robertson_ode, tspan, u0, tol, h0);
% Part (a): Plot Solutions
figure;
subplot(2,1,1);
plot(t, u(:,1), ‘b’, t, u(:,3), ‘g’, ‘LineWidth’, 1.5);
legend(‘y1 (Reactant)’, ‘y3 (Product)’);
xlabel(‘Time t’); ylabel(‘Concentration’);
%title(‘Robertson Problem: Concentrations y1 and y3’);
grid on;
subplot(2,1,2);
semilogx(t, u(:,2), ‘r’, ‘LineWidth’, 1.5);
xlabel(‘Time t (Log Scale)’); ylabel(‘Concentration y2’);
%title(‘Robertson Problem: Intermediate Species y2’);
grid on;
% Part (b): Error and Stepsize Plots
figure;
subplot(2,1,1);
semilogy(t(2:end), err);
xlabel(‘Time t’); ylabel(‘||y_n^5 – y_n^4||_{infty} / h’);
%title(‘Part (b): Estimated Error vs Time’);
grid on;
subplot(2,1,2);
semilogy(t(2:end), h_steps);
xlabel(‘Time t’); ylabel(‘Stepsize h’);
%title(‘Part (b): Stepsize h vs Time’);
grid on;
figure;
subplot(2,1,1);
semilogy(t(2:end), err, ‘LineWidth’, 1.5);
xlabel(‘Time t’);
ylabel(‘Error’);
title(‘Estimated Local Error vs Time’);
grid on;
subplot(2,1,2);
semilogy(t(2:end), h_steps, ‘LineWidth’, 1.5);
xlabel(‘Time t’);
ylabel(‘Stepsize h’);
title(‘Stepsize vs Time’);
grid on;
function dydt = robertson_ode(~, y)
% Coefficients from Screenshot 2026-02-28 at 3.38.10 PM.png
alpha = 0.04;
beta = 1e4;
gamma = 3e7;
dydt = [ -alpha*y(1) + beta*y(2)*y(3);
alpha*y(1) – beta*y(2)*y(3) – gamma*y(2)^2;
gamma*y(2)^2 ];
end
function [t_all, u_all, error_history, h_history, stats] = …
dormand_prince_solver(f, tspan, u0, tol, h0)
% Initialization
t = tspan(1);
u = u0(:);
h = h0;
t_all = t;
u_all = u.’;
error_history = [];
h_history = [];
n_accepted = 0;
n_rejected = 0;
% —- Dormand–Prince 4(5) Coefficients —-
a = [1/5, 0, 0, 0, 0, 0;
3/40, 9/40, 0, 0, 0, 0;
44/45, -56/15, 32/9, 0, 0, 0;
19372/6561, -25360/2187, 64448/6561, -212/729, 0, 0;
9017/3168, -355/33, 46732/5247, 49/176, -5103/18656, 0;
35/384, 0, 500/1113, 125/192, -2187/6784, 11/84];
b4 = [5179/57600, 0, 7571/16695, 393/640, …
-92097/339200, 187/2100, 1/40];
b5 = [35/384, 0, 500/1113, 125/192, …
-2187/6784, 11/84, 0];
c = [0, 1/5, 3/10, 4/5, 8/9, 1, 1];
safety = 0.9;
h_min = 1e-12;
h_max = 1.0;
while t < tspan(2)
if h < h_min
error(‘Step size underflow: problem likely stiff.’);
end
if t + h > tspan(2)
h = tspan(2) – t;
end
% —- Stage calculations —-
k = zeros(length(u), 7);
k(:,1) = f(t, u);
for i = 2:7
k(:,i) = f(t + c(i)*h, …
u + h * (k(:,1:i-1) * a(i-1,1:i-1)’));
end
% —- 4th and 5th order solutions —-
y4 = u + h * (k * b4′);
y5 = u + h * (k * b5′);
error_val = norm(y5 – y4, inf);
if error_val <= tol
t = t + h;
u = y5;
t_all = [t_all; t];
u_all = [u_all; u.’];
error_history = [error_history; error_val];
h_history = [h_history; h];
n_accepted = n_accepted + 1;
else
n_rejected = n_rejected + 1;
end
if error_val == 0
h_new = 2*h;
else
h_new = h * safety * (tol / error_val)^(1/5);
end
h = min(max(h_new, h_min), h_max);
end
stats.accepted = n_accepted;
stats.rejected = n_rejected;
end I would like to know if this code can be improved or it well written? any comments will be greatly appreciated. Thanks
u0 = [1; 0; 0];
tspan = [0, 1e-3]; % small final time for explicit solver
tol = 1e-2;
h0 = 1e-6;
% Execute solver
[t, u, err, h_steps, stats] = dormand_prince_solver(@robertson_ode, tspan, u0, tol, h0);
% Part (a): Plot Solutions
figure;
subplot(2,1,1);
plot(t, u(:,1), ‘b’, t, u(:,3), ‘g’, ‘LineWidth’, 1.5);
legend(‘y1 (Reactant)’, ‘y3 (Product)’);
xlabel(‘Time t’); ylabel(‘Concentration’);
%title(‘Robertson Problem: Concentrations y1 and y3’);
grid on;
subplot(2,1,2);
semilogx(t, u(:,2), ‘r’, ‘LineWidth’, 1.5);
xlabel(‘Time t (Log Scale)’); ylabel(‘Concentration y2’);
%title(‘Robertson Problem: Intermediate Species y2’);
grid on;
% Part (b): Error and Stepsize Plots
figure;
subplot(2,1,1);
semilogy(t(2:end), err);
xlabel(‘Time t’); ylabel(‘||y_n^5 – y_n^4||_{infty} / h’);
%title(‘Part (b): Estimated Error vs Time’);
grid on;
subplot(2,1,2);
semilogy(t(2:end), h_steps);
xlabel(‘Time t’); ylabel(‘Stepsize h’);
%title(‘Part (b): Stepsize h vs Time’);
grid on;
figure;
subplot(2,1,1);
semilogy(t(2:end), err, ‘LineWidth’, 1.5);
xlabel(‘Time t’);
ylabel(‘Error’);
title(‘Estimated Local Error vs Time’);
grid on;
subplot(2,1,2);
semilogy(t(2:end), h_steps, ‘LineWidth’, 1.5);
xlabel(‘Time t’);
ylabel(‘Stepsize h’);
title(‘Stepsize vs Time’);
grid on;
function dydt = robertson_ode(~, y)
% Coefficients from Screenshot 2026-02-28 at 3.38.10 PM.png
alpha = 0.04;
beta = 1e4;
gamma = 3e7;
dydt = [ -alpha*y(1) + beta*y(2)*y(3);
alpha*y(1) – beta*y(2)*y(3) – gamma*y(2)^2;
gamma*y(2)^2 ];
end
function [t_all, u_all, error_history, h_history, stats] = …
dormand_prince_solver(f, tspan, u0, tol, h0)
% Initialization
t = tspan(1);
u = u0(:);
h = h0;
t_all = t;
u_all = u.’;
error_history = [];
h_history = [];
n_accepted = 0;
n_rejected = 0;
% —- Dormand–Prince 4(5) Coefficients —-
a = [1/5, 0, 0, 0, 0, 0;
3/40, 9/40, 0, 0, 0, 0;
44/45, -56/15, 32/9, 0, 0, 0;
19372/6561, -25360/2187, 64448/6561, -212/729, 0, 0;
9017/3168, -355/33, 46732/5247, 49/176, -5103/18656, 0;
35/384, 0, 500/1113, 125/192, -2187/6784, 11/84];
b4 = [5179/57600, 0, 7571/16695, 393/640, …
-92097/339200, 187/2100, 1/40];
b5 = [35/384, 0, 500/1113, 125/192, …
-2187/6784, 11/84, 0];
c = [0, 1/5, 3/10, 4/5, 8/9, 1, 1];
safety = 0.9;
h_min = 1e-12;
h_max = 1.0;
while t < tspan(2)
if h < h_min
error(‘Step size underflow: problem likely stiff.’);
end
if t + h > tspan(2)
h = tspan(2) – t;
end
% —- Stage calculations —-
k = zeros(length(u), 7);
k(:,1) = f(t, u);
for i = 2:7
k(:,i) = f(t + c(i)*h, …
u + h * (k(:,1:i-1) * a(i-1,1:i-1)’));
end
% —- 4th and 5th order solutions —-
y4 = u + h * (k * b4′);
y5 = u + h * (k * b5′);
error_val = norm(y5 – y4, inf);
if error_val <= tol
t = t + h;
u = y5;
t_all = [t_all; t];
u_all = [u_all; u.’];
error_history = [error_history; error_val];
h_history = [h_history; h];
n_accepted = n_accepted + 1;
else
n_rejected = n_rejected + 1;
end
if error_val == 0
h_new = 2*h;
else
h_new = h * safety * (tol / error_val)^(1/5);
end
h = min(max(h_new, h_min), h_max);
end
stats.accepted = n_accepted;
stats.rejected = n_rejected;
end code generation/improvement MATLAB Answers — New Questions
Graphs display upside down
I am using matlab online, 2021b
All graphs display upside down, including numbers, axis, and toolbox
Changing the renderer causes the graph to rerender and fixes the issue for graphs currently plotted, such as
set(gcf,’renderer’,’painters’)
set(gcf,’renderer’,’opengl’)
But does not stop new graphs from being upside down again.I am using matlab online, 2021b
All graphs display upside down, including numbers, axis, and toolbox
Changing the renderer causes the graph to rerender and fixes the issue for graphs currently plotted, such as
set(gcf,’renderer’,’painters’)
set(gcf,’renderer’,’opengl’)
But does not stop new graphs from being upside down again. I am using matlab online, 2021b
All graphs display upside down, including numbers, axis, and toolbox
Changing the renderer causes the graph to rerender and fixes the issue for graphs currently plotted, such as
set(gcf,’renderer’,’painters’)
set(gcf,’renderer’,’opengl’)
But does not stop new graphs from being upside down again. graph, graphics, gui, upside down MATLAB Answers — New Questions
How can I generate a License Report?
I am aware that I can use the license center to administer my licenses. I would like to see detailed information regarding license usage, is it possible to view this information in a comprehensive report?I am aware that I can use the license center to administer my licenses. I would like to see detailed information regarding license usage, is it possible to view this information in a comprehensive report? I am aware that I can use the license center to administer my licenses. I would like to see detailed information regarding license usage, is it possible to view this information in a comprehensive report? MATLAB Answers — New Questions
searchin for an M file in all folders of my PC.
I want to search for an M file in all the folders of my PC.
I do not want to search in every folder individually.I want to search for an M file in all the folders of my PC.
I do not want to search in every folder individually. I want to search for an M file in all the folders of my PC.
I do not want to search in every folder individually. file search MATLAB Answers — New Questions
Multiple curve fitting for parameter identification: multi-objective optimization
I have 4 plots of x (eps) and y (sigmaexp) and I would like to best-fit them in the equation given by sigmanum2. The criteria to best-fit is to minimize sum of squares i.e. least sum of squares defined by function fitness. I have done this successfully for 1 plot by using fminsearch i.e. 1 objective function. Since I have 4 plots, I have 4 objective functions (or 4 values in y i.e. [F1; F2; F3; F4]) which must be minimized simultaneously. How can I do this or which function must be used for multiobjective optimization? Please help. I have done research from my end in the relevant forums, however I am not able to do it. Here is my code.
s = table2array(ExpDataMatlab); %excel data I have imported. Attached for your reference.
eps = s(:,1); % represents x
sigmaexp = s(:,2); % represent y
epsdot1 = s(:,3);
A = 332.8668; %constant
B = 128.6184; %constant
n = 0.4234; %constant
%x = [2000, 10.2234];
%y = fitness(x,A,B,n,eps,epsdot1,sigmaexp); %I used this to check if my function defined at the
%bottom works properly or not
ObjFcn = @(x)fitness(x,A,B,n,eps,epsdot1,sigmaexp);
x0 = [12000 4];
options = optimset(‘Display’,’iter’,’PlotFcns’,@optimplotfval,’TolX’,1e-10,’TolFun’,1e-10,’Algorithm’,’interior-point’);
[sol, fval, exitflag, output] = fminsearch(ObjFcn,x0,options);
C = sol(1); P = sol(2);
sigmanum4 = (A+B*(eps.^n)).*(1+(epsdot1./C).^(1/P)); %deduced value after optimization
plot(s(1:99,1),s(1:99,2),’+m’);hold on;
plot(s(1:99,1),sigmanum4(1:99,1),’m’);
hold off;
function y = fitness(x,A,B,n,eps,epsdot1,sigmaexp)
sigmanum2 = (A+B*(eps.^n)).*(1+(epsdot1./x(1)).^(1/x(2)));
F1 = sum((sigmaexp(1:99,1)-sigmanum2(1:99,1)).^2);%values for 982 /s % represent residual sum of squares
%F2 = sum((sigmaexp(100:206,1)-sigmanum2(100:206,1)).^2);%values for 2000 /s
%F3 = sum((sigmaexp(207:296,1)-sigmanum2(207:296,1)).^2);%values for 2900 /s
%F4 = sum((sigmaexp(297:368,1)-sigmanum2(297:368,1)).^2);%values for 4000 /s
%y = [F1; F2; F3; F4]; % for multiobjective minimization
y = [F1]; %for single objective minimization
%y = (0.01042*F1+0.00962*F2+0.01111*F3+0.01389*F4); %weighted residual sum of squares (Milani 2008)
end
Thank you.I have 4 plots of x (eps) and y (sigmaexp) and I would like to best-fit them in the equation given by sigmanum2. The criteria to best-fit is to minimize sum of squares i.e. least sum of squares defined by function fitness. I have done this successfully for 1 plot by using fminsearch i.e. 1 objective function. Since I have 4 plots, I have 4 objective functions (or 4 values in y i.e. [F1; F2; F3; F4]) which must be minimized simultaneously. How can I do this or which function must be used for multiobjective optimization? Please help. I have done research from my end in the relevant forums, however I am not able to do it. Here is my code.
s = table2array(ExpDataMatlab); %excel data I have imported. Attached for your reference.
eps = s(:,1); % represents x
sigmaexp = s(:,2); % represent y
epsdot1 = s(:,3);
A = 332.8668; %constant
B = 128.6184; %constant
n = 0.4234; %constant
%x = [2000, 10.2234];
%y = fitness(x,A,B,n,eps,epsdot1,sigmaexp); %I used this to check if my function defined at the
%bottom works properly or not
ObjFcn = @(x)fitness(x,A,B,n,eps,epsdot1,sigmaexp);
x0 = [12000 4];
options = optimset(‘Display’,’iter’,’PlotFcns’,@optimplotfval,’TolX’,1e-10,’TolFun’,1e-10,’Algorithm’,’interior-point’);
[sol, fval, exitflag, output] = fminsearch(ObjFcn,x0,options);
C = sol(1); P = sol(2);
sigmanum4 = (A+B*(eps.^n)).*(1+(epsdot1./C).^(1/P)); %deduced value after optimization
plot(s(1:99,1),s(1:99,2),’+m’);hold on;
plot(s(1:99,1),sigmanum4(1:99,1),’m’);
hold off;
function y = fitness(x,A,B,n,eps,epsdot1,sigmaexp)
sigmanum2 = (A+B*(eps.^n)).*(1+(epsdot1./x(1)).^(1/x(2)));
F1 = sum((sigmaexp(1:99,1)-sigmanum2(1:99,1)).^2);%values for 982 /s % represent residual sum of squares
%F2 = sum((sigmaexp(100:206,1)-sigmanum2(100:206,1)).^2);%values for 2000 /s
%F3 = sum((sigmaexp(207:296,1)-sigmanum2(207:296,1)).^2);%values for 2900 /s
%F4 = sum((sigmaexp(297:368,1)-sigmanum2(297:368,1)).^2);%values for 4000 /s
%y = [F1; F2; F3; F4]; % for multiobjective minimization
y = [F1]; %for single objective minimization
%y = (0.01042*F1+0.00962*F2+0.01111*F3+0.01389*F4); %weighted residual sum of squares (Milani 2008)
end
Thank you. I have 4 plots of x (eps) and y (sigmaexp) and I would like to best-fit them in the equation given by sigmanum2. The criteria to best-fit is to minimize sum of squares i.e. least sum of squares defined by function fitness. I have done this successfully for 1 plot by using fminsearch i.e. 1 objective function. Since I have 4 plots, I have 4 objective functions (or 4 values in y i.e. [F1; F2; F3; F4]) which must be minimized simultaneously. How can I do this or which function must be used for multiobjective optimization? Please help. I have done research from my end in the relevant forums, however I am not able to do it. Here is my code.
s = table2array(ExpDataMatlab); %excel data I have imported. Attached for your reference.
eps = s(:,1); % represents x
sigmaexp = s(:,2); % represent y
epsdot1 = s(:,3);
A = 332.8668; %constant
B = 128.6184; %constant
n = 0.4234; %constant
%x = [2000, 10.2234];
%y = fitness(x,A,B,n,eps,epsdot1,sigmaexp); %I used this to check if my function defined at the
%bottom works properly or not
ObjFcn = @(x)fitness(x,A,B,n,eps,epsdot1,sigmaexp);
x0 = [12000 4];
options = optimset(‘Display’,’iter’,’PlotFcns’,@optimplotfval,’TolX’,1e-10,’TolFun’,1e-10,’Algorithm’,’interior-point’);
[sol, fval, exitflag, output] = fminsearch(ObjFcn,x0,options);
C = sol(1); P = sol(2);
sigmanum4 = (A+B*(eps.^n)).*(1+(epsdot1./C).^(1/P)); %deduced value after optimization
plot(s(1:99,1),s(1:99,2),’+m’);hold on;
plot(s(1:99,1),sigmanum4(1:99,1),’m’);
hold off;
function y = fitness(x,A,B,n,eps,epsdot1,sigmaexp)
sigmanum2 = (A+B*(eps.^n)).*(1+(epsdot1./x(1)).^(1/x(2)));
F1 = sum((sigmaexp(1:99,1)-sigmanum2(1:99,1)).^2);%values for 982 /s % represent residual sum of squares
%F2 = sum((sigmaexp(100:206,1)-sigmanum2(100:206,1)).^2);%values for 2000 /s
%F3 = sum((sigmaexp(207:296,1)-sigmanum2(207:296,1)).^2);%values for 2900 /s
%F4 = sum((sigmaexp(297:368,1)-sigmanum2(297:368,1)).^2);%values for 4000 /s
%y = [F1; F2; F3; F4]; % for multiobjective minimization
y = [F1]; %for single objective minimization
%y = (0.01042*F1+0.00962*F2+0.01111*F3+0.01389*F4); %weighted residual sum of squares (Milani 2008)
end
Thank you. multiple curve fitting, parameter identification, multi-objective optimization MATLAB Answers — New Questions
Maltab Version R2015a required for student download
Hello,
I am a student from Heidelberg University. For my master thesis I would need the Matlab Version R2015a or older.
I could not find this version in the student download version (Free version).
Thank you for your help.
Kind regards,
Max BickelhauptHello,
I am a student from Heidelberg University. For my master thesis I would need the Matlab Version R2015a or older.
I could not find this version in the student download version (Free version).
Thank you for your help.
Kind regards,
Max Bickelhaupt Hello,
I am a student from Heidelberg University. For my master thesis I would need the Matlab Version R2015a or older.
I could not find this version in the student download version (Free version).
Thank you for your help.
Kind regards,
Max Bickelhaupt matlab version r 2015a MATLAB Answers — New Questions
How can I store data from my simulink model in and and use it during runtime.
How can I store data from my simulink model in and and use it during runtime. I used the data store blocks and trigger subsystem to enable data store only during specific events. But at a sample time of 1e-4 sec there’s a lot of data. I want to only buffer 50 points.How can I store data from my simulink model in and and use it during runtime. I used the data store blocks and trigger subsystem to enable data store only during specific events. But at a sample time of 1e-4 sec there’s a lot of data. I want to only buffer 50 points. How can I store data from my simulink model in and and use it during runtime. I used the data store blocks and trigger subsystem to enable data store only during specific events. But at a sample time of 1e-4 sec there’s a lot of data. I want to only buffer 50 points. data store simulink MATLAB Answers — New Questions
dot product for complex vector
Hello,
In the Matlab example, you have the dot product of the following two vectors A and B and its answer is vector C.
A = [1+i 1-i -1+i -1-i];
B = [3-4i 6-2i 1+2i 4+3i];
Calculate the dot product of A and B.
C = dot(A,B)
C = 1.0000 – 5.0000i
However, when I calculate it, I have vector C = 7 – 17i
That is, I have C vector results as follows below
(1+i) * (3-4i) + (1-i) * (6-2i) + (-1+i) * (1+2i) + (-1-i) * (4+3i) =
(7-i) +( 4-8i) + (-3-i) + (-1-7i) =
7 – 17i.
Hence, could you please tell me how the Matlab got the results (or show me manually how Matlab got the dot product answer) as I have different results than Matlab calculated using dot product?
Thank you,
CharlesHello,
In the Matlab example, you have the dot product of the following two vectors A and B and its answer is vector C.
A = [1+i 1-i -1+i -1-i];
B = [3-4i 6-2i 1+2i 4+3i];
Calculate the dot product of A and B.
C = dot(A,B)
C = 1.0000 – 5.0000i
However, when I calculate it, I have vector C = 7 – 17i
That is, I have C vector results as follows below
(1+i) * (3-4i) + (1-i) * (6-2i) + (-1+i) * (1+2i) + (-1-i) * (4+3i) =
(7-i) +( 4-8i) + (-3-i) + (-1-7i) =
7 – 17i.
Hence, could you please tell me how the Matlab got the results (or show me manually how Matlab got the dot product answer) as I have different results than Matlab calculated using dot product?
Thank you,
Charles Hello,
In the Matlab example, you have the dot product of the following two vectors A and B and its answer is vector C.
A = [1+i 1-i -1+i -1-i];
B = [3-4i 6-2i 1+2i 4+3i];
Calculate the dot product of A and B.
C = dot(A,B)
C = 1.0000 – 5.0000i
However, when I calculate it, I have vector C = 7 – 17i
That is, I have C vector results as follows below
(1+i) * (3-4i) + (1-i) * (6-2i) + (-1+i) * (1+2i) + (-1-i) * (4+3i) =
(7-i) +( 4-8i) + (-3-i) + (-1-7i) =
7 – 17i.
Hence, could you please tell me how the Matlab got the results (or show me manually how Matlab got the dot product answer) as I have different results than Matlab calculated using dot product?
Thank you,
Charles dot product MATLAB Answers — New Questions
