結果:
could you explain me how to calculate the gain values for different types of controllers (Conventional Sliding Mode Control, Third Order Sliding Mode Control, Variable Gain Super Twisting Algorithm.
Could you, assist me in providing a mathematical method, for example, to calculate the gains of the above-mentioned controllers?
Thank you
M. Itouchene
Hi,
I'm trying to write a code which can determine the gradient change in a given profile. However the code is unable to determine the same correctly and giving incorrect results. For the pic1 below it should give me the width of the region where it is a steep gradient determined.
However for pic2 below it shouldn't as there is no steep gradient compared to pic1
The code is as below:
clear width;
global width;
global data;
for i=1:max(length(data.variable.t))
for j=1:max(length(data.variable.x))-1
change_p = (abs(data.variable.gradpressure(i,j+1))-abs(data.variable.gradpressure(i,j)))/abs(data.variable.gradpressure(i,j));
%disp(change_p)
if change_p > 0.1
disp("steep gradient found")
width(j)=1-data.variable.x(j);
disp(width)
else
disp("no steep gradient found")
end
end
end
the data.variable.gradpressure is a 1000x100 matrix with t along the vertical and x along the horizontal.
with regards,
rc
Sub aspenaAdorption()
' Declare variables for the ACM application, document, and simulation
Dim ACMApp As Object
Dim ACMDocument As Object
Dim ACMSimulation As Object
' Create an instance of the ACM application
Set ACMApp = CreateObject("ACM Application")
' Use "ACM Application" for Aspen Custom Modeler
' Use "ADS Application" for Aspen Adsorption
' Make the ACM application visible
ACMApp.Visible = True
' Open the specified simulation document
Set ACMDocument = ACMApp.OpenDocument("C:\Users\user\Desktop\H2_Purification.acmf")
' Set the simulation object to the current simulation in the application
Set ACMSimulation = ACMApp.Simulation
' Set the simulation to run in dynamic mode
ACMSimulation.RunMode = "Dynamic"
' Run the simulation
ACMSimulation.Run (True)
' Check if the simulation was successful and display a message box
If ACMSimulation.Successful Then
MsgBox "Simulation Complete"
Else
MsgBox "Simulation Failed"
End If
' Quit the ACM application
ACMApp.Quit
End Sub
I have an old application that gives me an error when I run it. The error message states: "Could not find version 7.13 of the MCR. Attempting to load mclmcrrt713.dll. Please install the correct version of the MCR." I tried to install this version, but it is no longer available. Any help would be highly appreciated. Thanks!
In the program given below I fail to obtaine real pole as title in intger format if anyone know please guide me
num1=[1 -1];
den1=conv([1 1],conv([1 2+2j],[1 2-2j]));
G=tf(num1,den1);
P=pole(G);
Z = zero(G);
formatSpec='%s,%i,%f+%fi,%i';
a="Root Locus of ";
b='step response of';
figure(17)
rlocus(G)
p=sprintf(formatSpec,a,Z,P/1i,P(3,1));
title(p);
An option for 10th degree polynomials but no weighted linear least squares. Seriously? Jesse
What do you think about the NVIDIA's achivement of becoming the top giant of manufacturing chips, especially for AI world?
Twitch built an entire business around letting you watch over someone's shoulder while they play video games. I feel like we should be able to make at least a few videos where we get to watch over someone's shoulder while they solve Cody problems. I would pay good money for a front-row seat to watch some of my favorite solvers at work. Like, I want to know, did Alfonso Nieto-Castonon just sit down and bang out some of those answers, or did he have to think about it for a while? What was he thinking about while he solved it? What resources was he drawing on? There's nothing like watching a master craftsman at work.
I can imagine a whole category of Cody videos called "How I Solved It". I tried making one of these myself a while back, but as far as I could tell, nobody else made one.
Here's the direct link to the video: https://www.youtube.com/watch?v=hoSmO1XklAQ
I hereby challenge you to make a "How I Solved It" video and post it here. If you make one, I'll make another one.
The Ans Hack is a dubious way to shave a few points off your solution score. Instead of a standard answer like this
function y = times_two(x)
y = 2*x;
end
you would do this
function ans = times_two(x)
2*x;
end
The ans variable is automatically created when there is no left-hand side to an evaluated expression. But it makes for an ugly function. I don't think anyone actually defends it as a good practice. The question I would ask is: is it so offensive that it should be specifically disallowed by the rules? Or is it just one of many little hacks that you see in Cody, inelegant but tolerable in the context of the surrounding game?
Incidentally, I wrote about the Ans Hack long ago on the Community Blog. Dealing with user-unfriendly code is also one of the reasons we created the Head-to-Head voting feature. Some techniques are good for your score, and some are good for your code readability. You get to decide with you care about.
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.
There are a host of problems on Cody that require manipulation of the digits of a number. Examples include summing the digits of a number, separating the number into its powers, and adding very large numbers together.
If you haven't come across this trick yet, you might want to write it down (or save it electronically):
digits = num2str(4207) - '0'
That code results in the following:
digits =
4 2 0 7
Now, summing the digits of the number is easy:
sum(digits)
ans =
13
Hello and a warm welcome to everyone! We're excited to have you in the Cody Discussion Channel. To ensure the best possible experience for everyone, it's important to understand the types of content that are most suitable for this channel.
Content that belongs in the Cody Discussion Channel:
- Tips & tricks: Discuss strategies for solving Cody problems that you've found effective.
- Ideas or suggestions for improvement: Have thoughts on how to make Cody better? We'd love to hear them.
- Issues: Encountering difficulties or bugs with Cody? Let us know so we can address them.
- Requests for guidance: Stuck on a Cody problem? Ask for advice or hints, but make sure to show your efforts in attempting to solve the problem first.
- General discussions: Anything else related to Cody that doesn't fit into the above categories.
Content that does not belong in the Cody Discussion Channel:
- Comments on specific Cody problems: Examples include unclear problem descriptions or incorrect testing suites.
- Comments on specific Cody solutions: For example, you find a solution creative or helpful.
Please direct such comments to the Comments section on the problem or solution page itself.
We hope the Cody discussion channel becomes a vibrant space for sharing expertise, learning new skills, and connecting with others.
It is necessary to find a solution that satisfies the boundary conditions:
u′′′+u′+λu=f.
initial conditions:
u′′(0)=
u′(0)+
u(0),
u′(0)+u(0),u′′(1)=
u(1),
u(1), u′(1)=
u(1),
u(1),How to define u and fin matlab?
where λ>0, 
;
, 
, 
and 
- given numbers.
;, , and - given numbers.
≤ 
where C is a positive constant independent of 𝑢 and f
clear;
syms y(x) y0 lambda u
Dy = diff(y);
D2y = diff(y,2);
D2y_2 = diff(y,2);
D3y = diff(y,3);
ode = D3y + lambda*u == y(x);
ySol(x) = dsolve(ode);
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!
One of the starter prompts is about rolling two six-sided dice and plot the results. As a hobby, I create my own board games. I was able to use the dice rolling prompt to show how a simple roll and move game would work. That was a great surprise!
Drumlin Farm has welcomed MATLAMB, named in honor of MathWorks, among ten adorable new lambs this season!
📚 New Book Announcement: "Image Processing Recipes in MATLAB" 📚
I am delighted to share the release of my latest book, "Image Processing Recipes in MATLAB," co-authored by my dear friend and colleague Gustavo Benvenutti Borba.
This 'cookbook' contains 30 practical recipes for image processing, ranging from foundational techniques to recently published algorithms. It serves as a concise and readable reference for quickly and efficiently deploying image processing pipelines in MATLAB.
Gustavo and I are immensely grateful to the MathWorks Book Program for their support. We also want to thank Randi Slack and her fantastic team at CRC Press for their patience, expertise, and professionalism throughout the process.
___________
Hello MATLAB community,
I am doing some image processing with MATLAB and some issues with my coding. I just like to warn you that I am very new at coding and MATLAB so I apologise in advance for my low level and I would be very glad to have some help as I have hitted a wall, and can't find a solution to my problem.
Context: I have a video of beams, that move right to left over time. The base is fixed, only the beam moves. I converted the video to images, and my MATLAB program is going through the image file and treating every image in it. Here are two image examples:
and 
I want to measure the following things:
a. The coordinates between the 2 extremities of the beam (length of the beam, without its base), let's call them A and B.
b. The bending deformation E (L0-Lt/L0 *100), obtained by calculating the distance between A and B, called Lt.
c. the curvature of the beam (1/R), obtained by extracting the radius R of a circle fitting the curvature of the beam.
d. The angle between a vertical line passing through A, and the line AB.
What I have done so far:
My approach has been to transform my image into an rgbimage, then binaryImage, then have the complementary image, apply some modifications/corrections to the image, and then skeletonize it. And from then, I extract the coordinates of A and B, the distance between A&B (Lt), the radius of the beam R, and the angle between A&B (T).
My main issue is the skeletonisation. Because my beam is quite thick, it shortens up too much my beam, and in an inconcistent manner. So then my results are completly wrong. Here is an image of the different images and operation I have done and the result:
So as you can see, the length is shorter. I would like to have a skeleton that meets the edges of the beam to calculate the end points.
I have tried "bwskel(BW, 'MinBranchLength', 30)" and "bwmorph(BW, 'thin', inf)", and this: https://uk.mathworks.com/matlabcentral/fileexchange/11123-better-skeletonization. But the problem remains the same. I have tried regionpropos, but the major axis they return is too long, I have tried bwferet(), but the maxlength is in diagonal of the beam... I have running out of ideas.
Problem: So I guess my main problem is how can I get a skeletonisation that goes to the edges of the beam?
Here is my code:
for i = TrackingStart:TrackingEnd
FileRGB(:,:,i) = rgb2gray(imread(IMG)); % Convert to grayscale
croppedRGB = FileRGB(y3left:y3right, x3left:x3right, i);
binaryImage = imbinarize(croppedRGB, 'adaptive', 'ForegroundPolarity','dark','Sensitivity', 0.50);
out = nnz(~binaryImage);
while out <= 4300 % Change threshold if needed
for j = 1:50
sensitivity = 0.50 + j * 0.01;
binaryImage = imbinarize(croppedRGB, 'adaptive', 'ForegroundPolarity', 'dark', 'Sensitivity', sensitivity);
out = nnz(~binaryImage);
if out >= 4325
break; % Exit the loop if the condition is met
end
end
end
% Create a line Model
BW = imcomplement(binaryImage);
BW(y1left:y1right, x1left:x1right) = 1; % there is always sample at the junction area (between beam and base)
BW(y2left:y2right, x2left:x2right) = 0; % Always = 0 if no sample here
BW = bwmorph(BW, 'close', Inf);
BW = bwmorph(BW, 'bridge');
BW = bwareafilt(BW, 1);
s = regionprops(BW, 'FilledImage');
BW = s.FilledImage;
BW = bwskel(BW, 'MinBranchLength', 30);
endpoints = bwmorph(BW, 'endpoints');
[y_end, x_end] = find(endpoints == 1);
%Degree of bending deformation method
Lt = sqrt(power(x_end(1)-x_end(2),2)+power(y_end(1)-y_end(2),2));
if x_end(2) > x_end(1)
Lt = -Lt;
end
Lstore(i) = Lt;
%Curvature method
[row_dots_cir, col_dots_cir, val] = find(BW == 1);
[xc(i),yc(i),Rstore(i),a] = circfit(col_dots_cir,row_dots_cir);
%Angle method
slope_endpoints = (x_end(1) - x_end(2)) / (y_end(1) - y_end(2));
angle_radians = atan(slope_endpoints);
angle_degrees = rad2deg(angle_radians);
if x_end(2) > x_end(1)
angle_degrees = -angle_degrees;
end
Tstore(i) = angle_degrees;
i
end
I found this plot of words said by different characters on the US version of The Office sitcom. There's a sparkline for each character from pilot to finale episode.
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
