ans =
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Does your company or organization require that all your Word Documents and Excel workbooks be labeled with a Microsoft Azure Information Protection label or else they can't be saved? These are the labels that are right below the tool ribbon that apply a category label such as "Public", "Business Use", or "Highly Restricted". If so, you can either
- Create and save a "template file" with the desired label and then call copyfile to make a copy of that file and then write your results to the new copy, or
- If using Windows you can create and/or open the file using ActiveX and then apply the desired label from your MATLAB program's code.
For #1 you can do
copyfile(templateFileName, newDataFileName);
writematrix(myData, newDataFileName);
If the template has the AIP label applied to it, then the copy will also inherit the same label.
For #2, here is a demo for how to apply the code using ActiveX.
% Test to set the Microsoft Azure Information Protection label on an Excel workbook.
% Reference support article:
% https://www.mathworks.com/matlabcentral/answers/1901140-why-does-azure-information-protection-popup-pause-the-matlab-script-when-i-use-actxserver?s_tid=ta_ans_results
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format compact;
% Define your workbook file name.
excelFullFileName = fullfile(pwd, '\testAIP.xlsx');
% Make sure it exists. Open Excel as an ActiveX server if it does.
if isfile(excelFullFileName)
% If the workbook exists, launch Excel as an ActiveX server.
Excel = actxserver('Excel.Application');
Excel.visible = true; % Make the server visible.
fprintf('Excel opened successfully.\n');
fprintf('Your workbook file exists:\n"%s".\nAbout to try to open it.\n', excelFullFileName);
% Open up the existing workbook named in the variable fullFileName.
Excel.Workbooks.Open(excelFullFileName);
fprintf('Excel opened file successfully.\n');
else
% File does not exist. Alert the user.
warningMessage = sprintf('File does not exist:\n\n"%s"\n', excelFullFileName);
fprintf('%s\n', warningMessage);
errordlg(warningMessage);
return;
end
% If we get here, the workbook file exists and has been opened by Excel.
% Ask Excel for the Microsoft Azure Information Protection (AIP) label of the workbook we just opened.
label = Excel.ActiveWorkbook.SensitivityLabel.GetLabel
% See if there is a label already. If not, these will be null:
existingLabelID = label.LabelId
existingLabelName = label.LabelName
% Create a label.
label = Excel.ActiveWorkbook.SensitivityLabel.CreateLabelInfo
label.LabelId = "a518e53f-798e-43aa-978d-c3fda1f3a682";
label.LabelName = "Business Use";
% Assign the label to the workbook.
fprintf('Setting Microsoft Azure Information Protection to "Business Use", GUID of a518e53f-798e-43aa-978d-c3fda1f3a682\n');
Excel.ActiveWorkbook.SensitivityLabel.SetLabel(label, label);
% Save this workbook with the new AIP setting we just created.
Excel.ActiveWorkbook.Save;
% Shut down Excel.
Excel.ActiveWorkbook.Close;
Excel.Quit;
% Excel is now closed down. Delete the variable from the MATLAB workspace.
clear Excel;
% Now check to see if the AIP label has been set
% by opening up the file in Excel and looking at the AIP banner.
winopen(excelFullFileName)
Note that there is a line in there that gets an AIP label from the existing workbook, if there is one at all. If there is not one, you can set one. But to determine what the proper LabelId (that crazy long hexadecimal number) should be, you will probably need to open an existing document that already has the label that you want set (applied to it) and then read that label with this line:
label = Excel.ActiveWorkbook.SensitivityLabel.GetLabel
This stems purely from some play on my part. Suppose I asked you to work with the sequence formed as 2*n*F_n + 1, where F_n is the n'th Fibonacci number? Part of me would not be surprised to find there is nothing simple we could do. But, then it costs nothing to try, to see where MATLAB can take me in an explorative sense.
n = sym(0:100).';
Fn = fibonacci(n);
Sn = 2*n.*Fn + 1;
Sn(1:10) % A few elements
For kicks, I tried asking ChatGPT. Giving it nothing more than the first 20 members of thse sequence as integers, it decided this is a Perrin sequence, and gave me a recurrence relation, but one that is in fact incorrect. Good effort from the Ai, but a fail in the end.
Is there anything I can do? Try null! (Look carefully at the array generated by Toeplitz. It is at least a pretty way to generate the matrix I needed.)
X = toeplitz(Sn,[1,zeros(1,4)]);
rank(X(5:end,:))
Hmm. So there is no linear combination of those columns that yields all zeros, since the resulting matrix was full rank.
X = toeplitz(Sn,[1,zeros(1,5)]);
rank(X(6:end,:))
But if I take it one step further, we see the above matrix is now rank deficient. What does that tell me? It says there is some simple linear combination of the columns of X(6:end,:) that always yields zero. The previous test tells me there is no shorter constant coefficient recurrence releation, using fewer terms.
null(X(6:end,:))
Let me explain what those coefficients tell me. In fact, they yield a very nice recurrence relation for the sequence S_n, not unlike the original Fibonacci sequence it was based upon.
S(n+1) = 3*S(n) - S_(n-1) - 3*S(n-2) + S(n-3) + S(n-4)
where the first 5 members of that sequence are given as [1 3 5 13 25]. So a 6 term linear constant coefficient recurrence relation. If it reminds you of the generating relation for the Fibonacci sequence, that is good, because it should. (Remember I started the sequence at n==0, IF you decide to test it out.) We can test it out, like this:
SfunM = memoize(@(N) Sfun(N));
SfunM(25)
2*25*fibonacci(sym(25)) + 1
And indeed, it works as expected.
function Sn = Sfun(n)
switch n
case 0
Sn = 1;
case 1
Sn = 3;
case 2
Sn = 5;
case 3
Sn = 13;
case 4
Sn = 25;
otherwise
Sn = Sfun(n-5) + Sfun(n-4) - 3*Sfun(n-3) - Sfun(n-2) +3*Sfun(n-1);
end
end
A beauty of this, is I started from nothing but a sequence of integers, derived from an expression where I had no rational expectation of finding a formula, and out drops something pretty. I might call this explorational mathematics.
The next step of course is to go in the other direction. That is, given the derived recurrence relation, if I substitute the formula for S_n in terms of the Fibonacci numbers, can I prove it is valid in general? (Yes.) After all, without some proof, it may fail for n larger than 100. (I'm not sure how much I can cram into a single discussion, so I'll stop at this point for now. If I see interest in the ideas here, I can proceed further. For example, what was I doing with that sequence in the first place? And of course, can I prove the relation is valid? Can I do so using MATLAB?)
(I'll be honest, starting from scratch, I'm not sure it would have been obvious to find that relation, so null was hugely useful here.)
I have picked the title but don't know which direction to take it. Looking for any and all inspiration. I took the project as it sounded interesting when reading into it, but I'm a satellite novice, and my degree is in electronics.
Gabriel's horn is a shape with the paradoxical property that it has infinite surface area, but a finite volume.
Gabriel’s horn is formed by taking the graph of 
with the domain 
and rotating it in three dimensions about the 
axis.
with the domain and rotating it in three dimensions about the axis.There is a standard formula for calculating the volume of this shape, for a general function 
.Wwe will just state that the volume of the 
solid between a and b is:
.Wwe will just state that the volume of the solid between a and b is:
The surface area of the solid is given by:
One other thing we need to consider is that we are trying to find the value of these integrals between 1 and ∞. An integral with a limit of infinity is called an improper integral and we can't evaluate it simply by plugging the value infinity into the normal equation for a definite integral. Instead, we must first calculate the definite integral up to some finite limit b and then calculate the limit of the result as b tends to ∞:
Volume
We can calculate the horn's volume using the volume integral above, so
The total volume of this infinitely long trumpet isπ.
Surface Area
To determine the surface area, we first need the function’s derivative:
Now plug it into the surface area formula and we have:
This is an improper integral and it's hard to evaluate, but since in our interval 
So, we have :
Now,we evaluate this last integral
So the surface are is infinite.
% Define the function for Gabriel's Horn
gabriels_horn = @(x) 1 ./ x;
% Create a range of x values
x = linspace(1, 40, 4000); % Increase the number of points for better accuracy
y = gabriels_horn(x);
% Create the meshgrid
theta = linspace(0, 2 * pi, 6000); % Increase theta points for a smoother surface
[X, T] = meshgrid(x, theta);
Y = gabriels_horn(X) .* cos(T);
Z = gabriels_horn(X) .* sin(T);
% Plot the surface of Gabriel's Horn
figure('Position', [200, 100, 1200, 900]);
surf(X, Y, Z, 'EdgeColor', 'none', 'FaceAlpha', 0.9);
hold on;
% Plot the central axis
plot3(x, zeros(size(x)), zeros(size(x)), 'r', 'LineWidth', 2);
% Set labels
xlabel('x');
ylabel('y');
zlabel('z');
% Adjust colormap and axis properties
colormap('gray');
shading interp; % Smooth shading
% Adjust the view
view(3);
axis tight;
grid on;
% Add formulas as text annotations
dim1 = [0.4 0.7 0.3 0.2];
annotation('textbox',dim1,'String',{'$$V = \pi \int_{1}^{a} \left( \frac{1}{x} \right)^2 dx = \pi \left( 1 - \frac{1}{a} \right)$$', ...
'', ... % Add an empty line for larger gap
'$$\lim_{a \to \infty} V = \lim_{a \to \infty} \pi \left( 1 - \frac{1}{a} \right) = \pi$$'}, ...
'Interpreter','latex','FontSize',12, 'EdgeColor','none', 'FitBoxToText', 'on');
dim2 = [0.4 0.5 0.3 0.2];
annotation('textbox',dim2,'String',{'$$A = 2\pi \int_{1}^{a} \frac{1}{x} \sqrt{1 + \left( -\frac{1}{x^2} \right)^2} dx > 2\pi \int_{1}^{a} \frac{dx}{x} = 2\pi \ln(a)$$', ...
'', ... % Add an empty line for larger gap
'$$\lim_{a \to \infty} A \geq \lim_{a \to \infty} 2\pi \ln(a) = \infty$$'}, ...
'Interpreter','latex','FontSize',12, 'EdgeColor','none', 'FitBoxToText', 'on');
% Add Gabriel's Horn label
dim3 = [0.3 0.9 0.3 0.1];
annotation('textbox',dim3,'String','Gabriel''s Horn', ...
'Interpreter','latex','FontSize',14, 'EdgeColor','none', 'HorizontalAlignment', 'center');
hold off
daspect([3.5 1 1]) % daspect([x y z])
view(-27, 15)
lightangle(-50,0)
lighting('gouraud')
The properties of this figure were first studied by Italian physicist and mathematician Evangelista Torricelli in the 17th century.
Acknowledgment
I would like to express my sincere gratitude to all those who have supported and inspired me throughout this project.
First and foremost, I would like to thank the mathematician and my esteemed colleague, Stavros Tsalapatis, for inspiring me with the fascinating subject of Gabriel's Horn.
I am also deeply thankful to Mr. @Star Strider for his invaluable assistance in completing the final code.
References:
When it comes to MOS tube burnout, it is usually because it is not working in the SOA workspace, and there is also a case where the MOS tube is overcurrent.
For example, the maximum allowable current of the PMOS transistor in this circuit is 50A, and the maximum current reaches 80+ at the moment when the MOS transistor is turned on, then the current is very large.
At this time, the PMOS is over-specified, and we can see on the SOA curve that it is not working in the SOA range, which will cause the PMOS to be damaged.
So what if you choose a higher current PMOS? Of course you can, but the cost will be higher.
We can choose to adjust the peripheral resistance or capacitor to make the PMOS turn on more slowly, so that the current can be lowered.
For example, when adjusting R1, R2, and the jumper capacitance between gs, when Cgs is adjusted to 1uF, The Ids are only 40A max, which is fine in terms of current, and meets the 80% derating.
(50 amps * 0.8 = 40 amps).
Next, let’s look at the power, from the SOA curve, the opening time of the MOS tube is about 1ms, and the maximum power at this time is 280W.
The normal thermal resistance of the chip is 50°C/W, and the maximum junction temperature can be 302°F.
Assuming the ambient temperature is 77°F, then the instantaneous power that 1ms can withstand is about 357W.
The actual power of PMOS here is 280W, which does not exceed the limit, which means that it works normally in the SOA area.
Therefore, when the current impact of the MOS transistor is large at the moment of turning, the Cgs capacitance can be adjusted appropriately to make the PMOS Working in the SOA area, you can avoid the problem of MOS corruption.
Hello everyone, i hope you all are in good health. i need to ask you about the help about where i should start to get indulge in matlab. I am an electrical engineer but having experience of construction field. I am new here. Please do help me. I shall be waiting forward to hear from you. I shall be grateful to you. Need recommendations and suggestions from experienced members. Thank you.
I recently wrote up a document which addresses the solution of ordinary and partial differential equations in Matlab (with some Python examples thrown in for those who are interested). For ODEs, both initial and boundary value problems are addressed. For PDEs, it addresses parabolic and elliptic equations. The emphasis is on finite difference approaches and built-in functions are discussed when available. Theory is kept to a minimum. I also provide a discussion of strategies for checking the results, because I think many students are too quick to trust their solutions. For anyone interested, the document can be found at https://blanchard.neep.wisc.edu/SolvingDifferentialEquationsWithMatlab.pdf
Kindly link me to the Channel Modeling Group.
I read and compreheneded a paper on channel modeling "An Adaptive Geometry-Based Stochastic Model for Non-Isotropic MIMO Mobile-to-Mobile Channels" except the graphical results obtained from the MATLAB codes. I have tried to replicate the same graphs but to no avail from my codes. And I am really interested in the topic, i have even written to the authors of the paper but as usual, there is no reply from them. Kindly assist if possible.
Hi, I'm looking for sites where I can find coding & algorithms problems and their solutions. I'm doing this workshop in college and I'll need some problems to go over with the students and explain how Matlab works by solving the problems with them and then reviewing and going over different solution options. Does anyone know a website like that? I've tried looking in the Matlab Cody By Mathworks, but didn't exactly find what I'm looking for. Thanks in advance.
We are modeling the introduction of a novel pathogen into a completely susceptible population. In the cells below, I have provided you with the Matlab code for a simple stochastic SIR model, implemented using the "GillespieSSA" function
Simulating the stochastic model 100 times for 
Since γ is 0.4 per day, 
per day
per day% Define the parameters
beta = 0.36;
gamma = 0.4;
n_sims = 100;
tf = 100; % Time frame changed to 100
% Calculate R0
R0 = beta / gamma
% Initial state values
initial_state_values = [1000000; 1; 0; 0]; % S, I, R, cum_inc
% Define the propensities and state change matrix
a = @(state) [beta * state(1) * state(2) / 1000000, gamma * state(2)];
nu = [-1, 0; 1, -1; 0, 1; 0, 0];
% Define the Gillespie algorithm function
function [t_values, state_values] = gillespie_ssa(initial_state, a, nu, tf)
t = 0;
state = initial_state(:); % Ensure state is a column vector
t_values = t;
state_values = state';
while t < tf
rates = a(state);
rate_sum = sum(rates);
if rate_sum == 0
break;
end
tau = -log(rand) / rate_sum;
t = t + tau;
r = rand * rate_sum;
cum_sum_rates = cumsum(rates);
reaction_index = find(cum_sum_rates >= r, 1);
state = state + nu(:, reaction_index);
% Update cumulative incidence if infection occurred
if reaction_index == 1
state(4) = state(4) + 1; % Increment cumulative incidence
end
t_values = [t_values; t];
state_values = [state_values; state'];
end
end
% Function to simulate the stochastic model multiple times and plot results
function simulate_stoch_model(beta, gamma, n_sims, tf, initial_state_values, R0, plot_type)
% Define the propensities and state change matrix
a = @(state) [beta * state(1) * state(2) / 1000000, gamma * state(2)];
nu = [-1, 0; 1, -1; 0, 1; 0, 0];
% Set random seed for reproducibility
rng(11);
% Initialize plot
figure;
hold on;
for i = 1:n_sims
[t, output] = gillespie_ssa(initial_state_values, a, nu, tf);
% Check if the simulation had only one step and re-run if necessary
while length(t) == 1
[t, output] = gillespie_ssa(initial_state_values, a, nu, tf);
end
if strcmp(plot_type, 'cumulative_incidence')
plot(t, output(:, 4), 'LineWidth', 2, 'Color', rand(1, 3));
elseif strcmp(plot_type, 'prevalence')
plot(t, output(:, 2), 'LineWidth', 2, 'Color', rand(1, 3));
end
end
xlabel('Time (days)');
if strcmp(plot_type, 'cumulative_incidence')
ylabel('Cumulative Incidence');
ylim([0 inf]);
elseif strcmp(plot_type, 'prevalence')
ylabel('Prevalence of Infection');
ylim([0 50]);
end
title(['Stochastic model output for R0 = ', num2str(R0)]);
subtitle([num2str(n_sims), ' simulations']);
xlim([0 tf]);
grid on;
hold off;
end
% Simulate the model 100 times and plot cumulative incidence
simulate_stoch_model(beta, gamma, n_sims, tf, initial_state_values, R0, 'cumulative_incidence');
% Simulate the model 100 times and plot prevalence
simulate_stoch_model(beta, gamma, n_sims, tf, initial_state_values, R0, 'prevalence');
Base case:
Suppose you need to do a computation many times. We are going to assume that this computation cannot be vectorized. The simplest case is to use a for loop:
number_of_elements = 1e6;
test_fcn = @(x) sqrt(x) / x;
tic
for i = 1:number_of_elements
x(i) = test_fcn(i);
end
t_forward = toc;
disp(t_forward + " seconds")
Preallocation:
This can easily be sped up by preallocating the variable that houses results:
tic
x = zeros(number_of_elements, 1);
for i = 1:number_of_elements
x(i) = test_fcn(i);
end
t_forward_prealloc = toc;
disp(t_forward_prealloc + " seconds")
In this example, preallocation speeds up the loop by a factor of about three to four (running in R2024a). Comment below if you get dramatically different results.
disp(sprintf("%.1f", t_forward / t_forward_prealloc))
Run it in reverse:
Is there a way to skip the explicit preallocation and still be fast? Indeed, there is.
clear x
tic
for i = number_of_elements:-1:1
x(i) = test_fcn(i);
end
t_backward = toc;
disp(t_backward + " seconds")
By running the loop backwards, the preallocation is implicitly performed during the first iteration and the loop runs in about the same time (within statistical noise):
disp(sprintf("%.2f", t_forward_prealloc / t_backward))
Do you get similar results when running this code? Let us know your thoughts in the comments below.
Beneficial side effect:
Have you ever had to use a for loop to delete elements from a vector? If so, keeping track of index offsets can be tricky, as deleting any element shifts all those that come after. By running the for loop in reverse, you don't need to worry about index offsets while deleting elements.
Many times when ploting, we not only need to set the color of the plot, but also its
transparency, Then how we set the alphaData of colorbar at the same time ?
It seems easy to do so :
data = rand(12,12);
% Transparency range 0-1, .3-1 for better appearance here
AData = rescale(- data, .3, 1);
% Draw an imagesc with numerical control over colormap and transparency
imagesc(data, 'AlphaData',AData);
colormap(jet);
ax = gca;
ax.DataAspectRatio = [1,1,1];
ax.TickDir = 'out';
ax.Box = 'off';
% get colorbar object
CBarHdl = colorbar;
pause(1e-16)
% Modify the transparency of the colorbar
CData = CBarHdl.Face.Texture.CData;
ALim = [min(min(AData)), max(max(AData))];
CData(4,:) = uint8(255.*rescale(1:size(CData, 2), ALim(1), ALim(2)));
CBarHdl.Face.Texture.ColorType = 'TrueColorAlpha';
CBarHdl.Face.Texture.CData = CData;
But !!!!!!!!!!!!!!! We cannot preserve the changes when saving them as images ：
It seems that when saving plots, the `Texture` will be refresh, but the `Face` will not :
however, object Face only have 4 colors to change(The four corners of a quadrilateral), how
can we set more colors ??
`Face` is a quadrilateral object, and we can change the `VertexData` to draw more than one little quadrilaterals:
data = rand(12,12);
% Transparency range 0-1, .3-1 for better appearance here
AData = rescale(- data, .3, 1);
%Draw an imagesc with numerical control over colormap and transparency
imagesc(data, 'AlphaData',AData);
colormap(jet);
ax = gca;
ax.DataAspectRatio = [1,1,1];
ax.TickDir = 'out';
ax.Box = 'off';
% get colorbar object
CBarHdl = colorbar;
pause(1e-16)
% Modify the transparency of the colorbar
CData = CBarHdl.Face.Texture.CData;
ALim = [min(min(AData)), max(max(AData))];
CData(4,:) = uint8(255.*rescale(1:size(CData, 2), ALim(1), ALim(2)));
warning off
CBarHdl.Face.ColorType = 'TrueColorAlpha';
VertexData = CBarHdl.Face.VertexData;
tY = repmat((1:size(CData,2))./size(CData,2), [4,1]);
tY1 = tY(:).'; tY2 = tY - tY(1,1); tY2(3:4,:) = 0; tY2 = tY2(:).';
tM1 = [tY1.*0 + 1; tY1; tY1.*0 + 1];
tM2 = [tY1.*0; tY2; tY1.*0];
CBarHdl.Face.VertexData = repmat(VertexData, [1,size(CData,2)]).*tM1 + tM2;
CBarHdl.Face.ColorData = reshape(repmat(CData, [4,1]), 4, []);
The higher the value, the more transparent it becomes
data = rand(12,12);
AData = rescale(- data, .3, 1);
imagesc(data, 'AlphaData',AData);
colormap(jet);
ax = gca;
ax.DataAspectRatio = [1,1,1];
ax.TickDir = 'out';
ax.Box = 'off';
CBarHdl = colorbar;
pause(1e-16)
CData = CBarHdl.Face.Texture.CData;
ALim = [min(min(AData)), max(max(AData))];
CData(4,:) = uint8(255.*rescale(size(CData, 2):-1:1, ALim(1), ALim(2)));
warning off
CBarHdl.Face.ColorType = 'TrueColorAlpha';
VertexData = CBarHdl.Face.VertexData;
tY = repmat((1:size(CData,2))./size(CData,2), [4,1]);
tY1 = tY(:).'; tY2 = tY - tY(1,1); tY2(3:4,:) = 0; tY2 = tY2(:).';
tM1 = [tY1.*0 + 1; tY1; tY1.*0 + 1];
tM2 = [tY1.*0; tY2; tY1.*0];
CBarHdl.Face.VertexData = repmat(VertexData, [1,size(CData,2)]).*tM1 + tM2;
CBarHdl.Face.ColorData = reshape(repmat(CData, [4,1]), 4, []);
More transparent in the middle
data = rand(12,12) - .5;
AData = rescale(abs(data), .1, .9);
imagesc(data, 'AlphaData',AData);
colormap(jet);
ax = gca;
ax.DataAspectRatio = [1,1,1];
ax.TickDir = 'out';
ax.Box = 'off';
CBarHdl = colorbar;
pause(1e-16)
CData = CBarHdl.Face.Texture.CData;
ALim = [min(min(AData)), max(max(AData))];
CData(4,:) = uint8(255.*rescale(abs((1:size(CData, 2)) - (1 + size(CData, 2))/2), ALim(1), ALim(2)));
warning off
CBarHdl.Face.ColorType = 'TrueColorAlpha';
VertexData = CBarHdl.Face.VertexData;
tY = repmat((1:size(CData,2))./size(CData,2), [4,1]);
tY1 = tY(:).'; tY2 = tY - tY(1,1); tY2(3:4,:) = 0; tY2 = tY2(:).';
tM1 = [tY1.*0 + 1; tY1; tY1.*0 + 1];
tM2 = [tY1.*0; tY2; tY1.*0];
CBarHdl.Face.VertexData = repmat(VertexData, [1,size(CData,2)]).*tM1 + tM2;
CBarHdl.Face.ColorData = reshape(repmat(CData, [4,1]), 4, []);
The code will work if the plot have AlphaData property
data = peaks(30);
AData = rescale(data, .2, 1);
surface(data, 'FaceAlpha','flat','AlphaData',AData);
colormap(jet(100));
ax = gca;
ax.DataAspectRatio = [1,1,1];
ax.TickDir = 'out';
ax.Box = 'off';
view(3)
CBarHdl = colorbar;
pause(1e-16)
CData = CBarHdl.Face.Texture.CData;
ALim = [min(min(AData)), max(max(AData))];
CData(4,:) = uint8(255.*rescale(1:size(CData, 2), ALim(1), ALim(2)));
warning off
CBarHdl.Face.ColorType = 'TrueColorAlpha';
VertexData = CBarHdl.Face.VertexData;
tY = repmat((1:size(CData,2))./size(CData,2), [4,1]);
tY1 = tY(:).'; tY2 = tY - tY(1,1); tY2(3:4,:) = 0; tY2 = tY2(:).';
tM1 = [tY1.*0 + 1; tY1; tY1.*0 + 1];
tM2 = [tY1.*0; tY2; tY1.*0];
CBarHdl.Face.VertexData = repmat(VertexData, [1,size(CData,2)]).*tM1 + tM2;
CBarHdl.Face.ColorData = reshape(repmat(CData, [4,1]), 4, []);
While searching the internet for some books on ordinary differential equations, I came across a link that I believe is very useful for all math students and not only. If you are interested in ODEs, it's worth taking the time to study it.
A First Look at Ordinary Differential Equations by Timothy S. Judson is an excellent resource for anyone looking to understand ODEs better. Here's a brief overview of the main topics covered:
- Introduction to ODEs: Basic concepts, definitions, and initial differential equations.
- Methods of Solution:
- Separable equations
- First-order linear equations
- Exact equations
- Transcendental functions
- Applications of ODEs: Practical examples and applications in various scientific fields.
- Systems of ODEs: Analysis and solutions of systems of differential equations.
- Series and Numerical Methods: Use of series and numerical methods for solving ODEs.
This book provides a clear and comprehensive introduction to ODEs, making it suitable for students and new researchers in mathematics. If you're interested, you can explore the book in more detail here: A First Look at Ordinary Differential Equations.
The study of the dynamics of the discrete Klein - Gordon equation (DKG) with friction is given by the equation :
In the above equation, W describes the potential function:
to which every coupled unit 
adheres. In Eq. (1), the variable $
$ is the unknown displacement of the oscillator occupying the n-th position of the lattice, and 
is the discretization parameter. We denote by h the distance between the oscillators of the lattice. The chain (DKG) contains linear damping with a damping coefficient 
, while
is the coefficient of the nonlinear cubic term.
adheres. In Eq. (1), the variable $$ is the unknown displacement of the oscillator occupying the n-th position of the lattice, and is the discretization parameter. We denote by h the distance between the oscillators of the lattice. The chain (DKG) contains linear damping with a damping coefficient , whileis the coefficient of the nonlinear cubic term.For the DKG chain (1), we will consider the problem of initial-boundary values, with initial conditions
and Dirichlet boundary conditions at the boundary points 
and 
, that is,
and , that is,
Therefore, when necessary, we will use the short notation 
for the one-dimensional discrete Laplacian
for the one-dimensional discrete Laplacian
Now we want to investigate numerically the dynamics of the system (1)-(2)-(3). Our first aim is to conduct a numerical study of the property of Dynamic Stability of the system, which directly depends on the existence and linear stability of the branches of equilibrium points.
For the discussion of numerical results, it is also important to emphasize the role of the parameter 
. By changing the time variable 
, we rewrite Eq. (1) in the form
. By changing the time variable , we rewrite Eq. (1) in the form
. We consider spatially extended initial conditions of the form:
where 
is the distance of the grid and 
is the amplitude of the initial condition We also assume zero initial velocity:
the following graphs for 
and 
and % Parameters
L = 200; % Length of the system
K = 99; % Number of spatial points
j = 2; % Mode number
omega_d = 1; % Characteristic frequency
beta = 1; % Nonlinearity parameter
delta = 0.05; % Damping coefficient
% Spatial grid
h = L / (K + 1);
n = linspace(-L/2, L/2, K+2); % Spatial points
N = length(n);
omegaDScaled = h * omega_d;
deltaScaled = h * delta;
% Time parameters
dt = 1; % Time step
tmax = 3000; % Maximum time
tspan = 0:dt:tmax; % Time vector
% Values of amplitude 'a' to iterate over
a_values = [2, 1.95, 1.9, 1.85, 1.82]; % Modify this array as needed
% Differential equation solver function
function dYdt = odefun(~, Y, N, h, omegaDScaled, deltaScaled, beta)
U = Y(1:N);
Udot = Y(N+1:end);
Uddot = zeros(size(U));
% Laplacian (discrete second derivative)
for k = 2:N-1
Uddot(k) = (U(k+1) - 2 * U(k) + U(k-1)) ;
end
% System of equations
dUdt = Udot;
dUdotdt = Uddot - deltaScaled * Udot + omegaDScaled^2 * (U - beta * U.^3);
% Pack derivatives
dYdt = [dUdt; dUdotdt];
end
% Create a figure for subplots
figure;
% Initial plot
a_init = 2; % Example initial amplitude for the initial condition plot
U0_init = a_init * sin((j * pi * h * n) / L); % Initial displacement
U0_init(1) = 0; % Boundary condition at n = 0
U0_init(end) = 0; % Boundary condition at n = K+1
subplot(3, 2, 1);
plot(n, U0_init, 'r.-', 'LineWidth', 1.5, 'MarkerSize', 10); % Line and marker plot
xlabel('$x_n$', 'Interpreter', 'latex');
ylabel('$U_n$', 'Interpreter', 'latex');
title('$t=0$', 'Interpreter', 'latex');
set(gca, 'FontSize', 12, 'FontName', 'Times');
xlim([-L/2 L/2]);
ylim([-3 3]);
grid on;
% Loop through each value of 'a' and generate the plot
for i = 1:length(a_values)
a = a_values(i);
% Initial conditions
U0 = a * sin((j * pi * h * n) / L); % Initial displacement
U0(1) = 0; % Boundary condition at n = 0
U0(end) = 0; % Boundary condition at n = K+1
Udot0 = zeros(size(U0)); % Initial velocity
% Pack initial conditions
Y0 = [U0, Udot0];
% Solve ODE
opts = odeset('RelTol', 1e-5, 'AbsTol', 1e-6);
[t, Y] = ode45(@(t, Y) odefun(t, Y, N, h, omegaDScaled, deltaScaled, beta), tspan, Y0, opts);
% Extract solutions
U = Y(:, 1:N);
Udot = Y(:, N+1:end);
% Plot final displacement profile
subplot(3, 2, i+1);
plot(n, U(end,:), 'b.-', 'LineWidth', 1.5, 'MarkerSize', 10); % Line and marker plot
xlabel('$x_n$', 'Interpreter', 'latex');
ylabel('$U_n$', 'Interpreter', 'latex');
title(['$t=3000$, $a=', num2str(a), '$'], 'Interpreter', 'latex');
set(gca, 'FontSize', 12, 'FontName', 'Times');
xlim([-L/2 L/2]);
ylim([-2 2]);
grid on;
end
% Adjust layout
set(gcf, 'Position', [100, 100, 1200, 900]); % Adjust figure size as needed
Dynamics for the initial condition , 
, for 
, for different amplitude values. By reducing the amplitude values, we observe the convergence to equilibrium points of different branches from 
and the appearance of values 
for which the solution converges to a non-linear equilibrium point 
Parameters: 
, for , for different amplitude values. By reducing the amplitude values, we observe the convergence to equilibrium points of different branches from and the appearance of values for which the solution converges to a non-linear equilibrium point Parameters: 
Detection of a stability threshold 
: For 
, the initial condition , 
, converges to a non-linear equilibrium point
.
: For , the initial condition , , converges to a non-linear equilibrium point.Characteristics for 
, with corresponding norm 
where the dynamics appear in the first image of the third row, we observe convergence to a non-linear equilibrium point of branch 
This has the same norm and the same energy as the previous case but the final state has a completely different profile. This result suggests secondary bifurcations have occurred in branch 
, with corresponding norm where the dynamics appear in the first image of the third row, we observe convergence to a non-linear equilibrium point of branch This has the same norm and the same energy as the previous case but the final state has a completely different profile. This result suggests secondary bifurcations have occurred in branch By further reducing the amplitude, distinct values of 
are discerned: 1.9, 1.85, 1.81 for which the initial condition 
with norms 
respectively, converges to a non-linear equilibrium point of branch 
This equilibrium point has norm 
and energy 
. The behavior of this equilibrium is illustrated in the third row and in the first image of the third row of Figure 1, and also in the first image of the third row of Figure 2. For all the values between the aforementioned a, the initial condition 
converges to geometrically different non-linear states of branch 
as shown in the second image of the first row and the first image of the second row of Figure 2, for amplitudes 
and 
respectively.
are discerned: 1.9, 1.85, 1.81 for which the initial condition with norms respectively, converges to a non-linear equilibrium point of branch This equilibrium point has norm and energy . The behavior of this equilibrium is illustrated in the third row and in the first image of the third row of Figure 1, and also in the first image of the third row of Figure 2. For all the values between the aforementioned a, the initial condition converges to geometrically different non-linear states of branch as shown in the second image of the first row and the first image of the second row of Figure 2, for amplitudes and respectively. Refference:
Spring is here in Natick and the tulips are blooming! While tulips appear only briefly here in Massachusetts, they provide a lot of bright and diverse colors and shapes. To celebrate this cheerful flower, here's some code to create your own tulip!
Check out this episode about PIVLab: https://www.buzzsprout.com/2107763/15106425
Join the conversation with William Thielicke, the developer of PIVlab, as he shares insights into the world of particle image velocimetery (PIV) and its applications. Discover how PIV accurately measures fluid velocities, non invasively revolutionising research across the industries. Delve into the development journey of PI lab, including collaborations, key features and future advancements for aerodynamic studies, explore the advanced hardware setups camera technologies, and educational prospects offered by PIVlab, for enhanced fluid velocity measurements. If you are interested in the hardware he speaks of check out the company: Optolution.
How to leave feedback on a doc page
Leaving feedback is a two-step process. At the bottom of most pages in the MATLAB documentation is a star rating.
Start by selecting a star that best answers the question. After selecting a star rating, an edit box appears where you can offer specific feedback.
When you press "Submit" you'll see the confirmation dialog below. You cannot go back and edit your content, although you can refresh the page to go through that process again.
Tips on leaving feedback
- Be productive. The reader should clearly understand what action you'd like to see, what was unclear, what you think needs work, or what areas were really helpful.
- Positive feedback is also helpful. By nature, feedback often focuses on suggestions for changes but it also helps to know what was clear and what worked well.
- Point to specific areas of the page. This helps the reader to narrow the focus of the page to the area described by your feedback.
What happens to that feedback?
Before working at MathWorks I often left feedback on documentation pages but I never knew what happens after that. One day in 2021 I shared my speculation on the process:
> That feedback is received by MathWorks Gnomes which are never seen nor heard but visit the MathWorks documentation team at night while they are sleeping and whisper selected suggestions into their ears to manipulate their dreams. Occassionally this causes them to wake up with a Eureka moment that leads to changes in the documentation.
I'd like to let you in on the secret which is much less fanciful. Feedback left in the star rating and edit box are collected and periodically reviewed by the doc writers who look for trends on highly trafficked pages and finer grain feedback on less visited pages. Your feedback is important and often results in improvements.
Let's talk about probability theory in Matlab.
Conditions of the problem - how many more letters do I need to write to the sales department to get an answer?
To get closer to the problem, I need to buy a license under a contract. Maybe sometimes there are responsible employees sitting here who will give me an answer.
Thank you
In the MATLAB description of the algorithm for Lyapunov exponents, I believe there is ambiguity and misuse.
The lambda(i) in the reference literature signifies the Lyapunov exponent of the entire phase space data after expanding by i time steps, but in the calculation formula provided in the MATLAB help documentation, Y_(i+K) represents the data point at the i-th point in the reconstructed data Y after K steps, and this calculation formula also does not match the calculation code given by MATLAB. I believe there should be some misguidance and misunderstanding here.
According to the symbol regulations in the algorithm description and the MATLAB code, I think the correct formula might be y(i) = 1/dt * 1/N * sum_j( log( ||Y_(j+i) - Y_(j*+i)|| ) )
Drumlin Farm has welcomed MATLAMB, named in honor of MathWorks, among ten adorable new lambs this season!
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