メインコンテンツ
結果:
Hello, everyone! I’m Mark Hayworth, but you might know me better in the community as Image Analyst. I've been using MATLAB since 2006 (18 years). My background spans a rich career as a former senior scientist and inventor at The Procter & Gamble Company (HQ in Cincinnati). I hold both master’s & Ph.D. degrees in optical sciences from the College of Optical Sciences at the University of Arizona, specializing in imaging, image processing, and image analysis. I have 40+ years of military, academic, and industrial experience with image analysis programming and algorithm development. I have experience designing custom light booths and other imaging systems. I also work with color and monochrome imaging, video analysis, thermal, ultraviolet, hyperspectral, CT, MRI, radiography, profilometry, microscopy, NIR, and Raman spectroscopy, etc. on a huge variety of subjects.
I'm thrilled to participate in MATLAB Central's Ask Me Anything (AMA) session, a fantastic platform for knowledge sharing and community engagement. Following Adam Danz’s insightful AMA on staff contributors in the Answers forum, I’d like to discuss topics in the area of image analysis and processing. I invite you to ask me anything related to this field, whether you're seeking recommendations on tools, looking for tips and tricks, my background, or career development advice. Additionally, I'm more than willing to share insights from my experiences in the MATLAB Answers community, File Exchange, and my role as a member of the Community Advisory Board. If you have questions related to your specific images or your custom MATLAB code though, I'll invite you to ask those in the Answers forum. It's a more appropriate forum for those kinds of questions, plus you can get the benefit of other experts offering their solutions in addition to me.
For the coming weeks, I'll be here to engage with your questions and help shed light on any topics you're curious about.
Hello, everyone!
Over the past few weeks, our community has been buzzing with activity, showcasing the incredible depth of knowledge, creativity, and innovation that makes this forum such a vibrant place. Today, we're excited to highlight some of the noteworthy contributions that have sparked discussions, offered insights, and shared knowledge across various topics. Let's dive in!
Interesting Questions
Fatima Majeed brings us a thought-provoking mathematical challenge, delving into inequalities and the realms beyond (e^e). If you're up for a mathematical journey, this question is a must-see!
lil brain tackles a practical problem many of us have faced: efficiently segmenting a CSV file based on specific criteria. This post is not only a query but a learning opportunity for anyone dealing with similar data manipulation challenges.
Popular Discussions
Discover a simple yet effective trick for digit manipulation from goc3. This tip is especially handy for those frequenting Cody challenges or anyone interested in enhancing their number handling skills in MATLAB.
Chen Lin shares an exciting update about the 'Run Code' feature in the Discussions area, highlighting how our community can now directly execute and share code snippets within discussions. This feature marks a significant enhancement in how we interact and solve problems together.
From the Blogs
A Deep Dive into EEG Analysis for Predicting Neurological Outcomes By Tanya Kuruvilla
Connell D`Souza, alongside Team Swarthbeat, explores the cutting-edge application of EEG analysis in predicting neurological outcomes post-cardiac arrest. This blog post offers an in-depth look into the challenges and methodologies of modern medical data analysis.
Mihir Acharya discusses the pivotal role of MATLAB and Simulink in the future of robotics simulation. Through an engaging conversation with industry analyst George Chowdhury, this post sheds light on overcoming simulation challenges and the exciting possibilities that lie ahead.
We encourage everyone to explore these contributions further and engage with the authors and the community. Your participation is what fuels this community's continual growth and innovation.
Here's to many more discussions, discoveries, and breakthroughs together!
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.
We're thrilled to share an exciting update with our community: the 'Run Code' feature is now available in the Discussions area!
Simply insert your code into the editor and press the green triangle button to run it. Your code will execute using the latest MATLAB R24a version, and it supports most common toolboxes. Moreover, this innovative feature allows for the running of attached files, further enhancing its utility and flexibility.
The ‘run code’ feature was first introduced in MATLAB Answers. Encouraged by the positive feedback and at the request of our community members, we are now expanding the availability of this feature to more areas within our community.
As always, your feedback is crucial to us, so please don't hesitate to share your thoughts and experiences by leaving a comment.
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, []);
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:
Many MATLAB enthusiasts come Cody to sharpen their skills, face new challenges, and engage in friendly competition. We firmly believe that learning from peers is one of the most effective ways to grow.
With this in mind, the Cody team is thrilled to unveil a new feature aimed at enriching your learning journey: the Cody Discussion Channel. This space is designed for sharing expertise, acquiring new skills, and fostering connections within our community.
On the Cody homepage, you'll now notice a Discussions section, prominently displaying the four most recent posts. For those eager to contribute, we encourage you to familiarize yourself with our posting guidelines before creating a new post. This will help maintain a constructive and valuable exchange of ideas for everyone involved.
Together, let's create an environment where every member feels empowered to share, learn, and connect.
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
Hello MATLAB Community!
We've had an exciting few weeks filled with insightful discussions, innovative tools, and engaging blog posts from our vibrant community. Here's a highlight of some noteworthy contributions that have sparked interest and inspired us all. Let's dive in!
Interesting Questions
Cindyawati explores the intriguing concept of interrupting continuous data in differential equations to study the effects of drug interventions in disease models. A thought-provoking question that bridges mathematics and medical research.
Pedro delves into the application of Linear Quadratic Regulator (LQR) for error dynamics and setpoint tracking, offering insights into control systems and their real-world implications.
Popular Discussions
Chen Lin shares an engaging interview with Zhaoxu Liu, shedding light on the creative processes behind some of the most innovative MATLAB contest entries of 2023. A must-read for anyone looking for inspiration!
Zhaoxu Liu, also known as slanderer, updates the community with the latest version of the MATLAB Plot Cheat Sheet. This resource is invaluable for anyone looking to enhance their data visualization skills.
From File Exchange
Giorgio introduces a toolbox for frequency estimation, making it simpler for users to import signals directly from the MATLAB workspace. A significant contribution for signal processing enthusiasts.
From the Blogs
Cleve Moler revisits a classic program for predicting future trends based on census data, offering a fascinating glimpse into the evolution of computational forecasting.
Boost Your App Design Efficiency – Effortless Component Swapping & Labeling in App Designer by Adam Danz
With contributions from Dinesh Kavalakuntla, Adam presents an insightful guide on improving app design workflows in MATLAB App Designer, focusing on component swapping and labeling.
We're incredibly proud of the diverse and innovative contributions our community members make every day. Each post, discussion, and tool not only enriches our knowledge but also inspires others to explore and create. Let's continue to support and learn from each other as we advance in our MATLAB journey.
Happy Coding!
quick / easy
21%
themed / in a group
20%
challenge (e.g., banned functions)
13%
puzzle / game
16%
educational
28%
other (comment below)
3%
117 票
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)|| ) )
A colleague said that you can search the Help Center using the phrase 'Introduced in' followed by a release version. Such as, 'Introduced in R2022a'. Doing this yeilds search results specific for that release.
Seems pretty handy so I thought I'd share.
Are you local to Boston?
Shape the Future of MATLAB: Join MathWorks' UX Night In-Person!
When: June 25th, 6 to 8 PM
Where: MathWorks Campus in Natick, MA
🌟 Calling All MATLAB Users! Here's your unique chance to influence the next wave of innovations in MATLAB and engineering software. MathWorks invites you to participate in our special after-hours usability studies. Dive deep into the latest MATLAB features, share your valuable feedback, and help us refine our solutions to better meet your needs.
🚀 This Opportunity Is Not to Be Missed:
- Exclusive Hands-On Experience: Be among the first to explore new MATLAB features and capabilities.
- Voice Your Expertise: Share your insights and suggestions directly with MathWorks developers.
- Learn, Discover, and Grow: Expand your MATLAB knowledge and skills through firsthand experience with unreleased features.
- Network Over Dinner: Enjoy a complimentary dinner with fellow MATLAB enthusiasts and the MathWorks team. It's a perfect opportunity to connect, share experiences, and network after work.
- Earn Rewards: Participants will not only contribute to the advancement of MATLAB but will also be compensated for their time. Plus, enjoy special MathWorks swag as a token of our appreciation!
👉 Reserve Your Spot Now: Space is limited for these after-hours sessions. If you're passionate about MATLAB and eager to contribute to its development, we'd love to hear from you.
Bringing the beauty of MathWorks Natick's tulips to life through code!
Remix challenge: create and share with us your new breeds of MATLAB tulips!
RGB triplet [0,1]
9%
RGB triplet [0,255]
12%
Hexadecimal Color Code
13%
Indexed color
16%
Truecolor array
37%
Equally unfamiliar with all-above
13%
2784 票
A high school student called for help with this physics problem:
- Car A moves with constant velocity v.
- Car B starts to move when Car A passes through the point P.
- Car B undergoes...
- uniform acc. motion from P to Q.
- uniform velocity motion from Q to R.
- uniform acc. motion from R to S.
- Car A and B pass through the point R simultaneously.
- Car A and B arrive at the point S simultaneously.
Q1. When car A passes the point Q, which is moving faster?
Q2. Solve the time duration for car B to move from P to Q using L and v.
Q3. Magnitude of acc. of car B from P to Q, and from R to S: which is bigger?
Well, it can be solved with a series of tedious equations. But... how about this?
Code below:
%% get images and prepare stuffs
figure(WindowStyle="docked"),
ax1 = subplot(2,1,1);
hold on, box on
ax1.XTick = [];
ax1.YTick = [];
A = plot(0, 1, 'ro', MarkerSize=10, MarkerFaceColor='r');
B = plot(0, 0, 'bo', MarkerSize=10, MarkerFaceColor='b');
[carA, ~, alphaA] = imread('https://cdn.pixabay.com/photo/2013/07/12/11/58/car-145008_960_720.png');
[carB, ~, alphaB] = imread('https://cdn.pixabay.com/photo/2014/04/03/10/54/car-311712_960_720.png');
carA = imrotate(imresize(carA, 0.1), -90);
carB = imrotate(imresize(carB, 0.1), 180);
alphaA = imrotate(imresize(alphaA, 0.1), -90);
alphaB = imrotate(imresize(alphaB, 0.1), 180);
carA = imagesc(carA, AlphaData=alphaA, XData=[-0.1, 0.1], YData=[0.9, 1.1]);
carB = imagesc(carB, AlphaData=alphaB, XData=[-0.1, 0.1], YData=[-0.1, 0.1]);
txtA = text(0, 0.85, 'A', FontSize=12);
txtB = text(0, 0.17, 'B', FontSize=12);
yline(1, 'r--')
yline(0, 'b--')
xline(1, 'k--')
xline(2, 'k--')
text(1, -0.2, 'Q', FontSize=20, HorizontalAlignment='center')
text(2, -0.2, 'R', FontSize=20, HorizontalAlignment='center')
% legend('A', 'B') % this make the animation slow. why?
xlim([0, 3])
ylim([-.3, 1.3])
%% axes2: plots velocity graph
ax2 = subplot(2,1,2);
box on, hold on
xlabel('t'), ylabel('v')
vA = plot(0, 1, 'r.-');
vB = plot(0, 0, 'b.-');
xline(1, 'k--')
xline(2, 'k--')
xlim([0, 3])
ylim([-.3, 1.8])
p1 = patch([0, 0, 0, 0], [0, 1, 1, 0], [248, 209, 188]/255, ...
EdgeColor = 'none', ...
FaceAlpha = 0.3);
%% solution
v = 1; % car A moves with constant speed.
L = 1; % distances of P-Q, Q-R, R-S
% acc. of car B for three intervals
a(1) = 9*v^2/8/L;
a(2) = 0;
a(3) = -1;
t_BatQ = sqrt(2*L/a(1)); % time when car B arrives at Q
v_B2 = a(1) * t_BatQ; % speed of car B between Q-R
%% patches for velocity graph
p2 = patch([t_BatQ, t_BatQ, t_BatQ, t_BatQ], [1, 1, v_B2, v_B2], ...
[248, 209, 188]/255, ...
EdgeColor = 'none', ...
FaceAlpha = 0.3);
p3 = patch([2, 2, 2, 2], [1, v_B2, v_B2, 1], [194, 234, 179]/255, ...
EdgeColor = 'none', ...
FaceAlpha = 0.3);
%% animation
tt = linspace(0, 3, 2000);
for t = tt
A.XData = v * t;
vA.XData = [vA.XData, t];
vA.YData = [vA.YData, 1];
if t < t_BatQ
B.XData = 1/2 * a(1) * t^2;
vB.XData = [vB.XData, t];
vB.YData = [vB.YData, a(1) * t];
p1.XData = [0, t, t, 0];
p1.YData = [0, vB.YData(end), 1, 1];
elseif t >= t_BatQ && t < 2
B.XData = L + (t - t_BatQ) * v_B2;
vB.XData = [vB.XData, t];
vB.YData = [vB.YData, v_B2];
p2.XData = [t_BatQ, t, t, t_BatQ];
p2.YData = [1, 1, vB.YData(end), vB.YData(end)];
else
B.XData = 2*L + v_B2 * (t - 2) + 1/2 * a(3) * (t-2)^2;
vB.XData = [vB.XData, t];
vB.YData = [vB.YData, v_B2 + a(3) * (t - 2)];
p3.XData = [2, t, t, 2];
p3.YData = [1, 1, vB.YData(end), v_B2];
end
txtA.Position(1) = A.XData(end);
txtB.Position(1) = B.XData(end);
carA.XData = A.XData(end) + [-.1, .1];
carB.XData = B.XData(end) + [-.1, .1];
drawnow
end
Are you a Simulink user eager to learn how to create apps with App Designer? Or an App Designer enthusiast looking to dive into Simulink?
Don't miss today's article on the Graphics and App Building Blog by @Robert Philbrick! Discover how to build Simulink Apps with App Designer, streamlining control of your simulations!
また、以下のリストから Web サイトを選択することもできます。
南北アメリカ
- América Latina (Español)
- Canada (English)
- United States (English)
ヨーロッパ
- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)
- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- United Kingdom(English)
アジア太平洋地域
- Australia (English)
- India (English)
- New Zealand (English)
- 中国
- 日本Japanese (日本語)
- 한국Korean (한국어)