ans =
結果:
Iam doing the project to find machining time for the cnc by creating a MATLAB Program ,I have G and M code in text file and the program should accept below points
- Read each line from the file.
- Extract the distance from G-code commands and the feedrate for each line.
- Calculate the time for each movement using the formula: time = distance / feedrate.
- Sum the times for all lines to get the total machining time.
Conditions:
G01 commands represent linear movements, so we calculate the distance directly.
G02 and G03 commands denote circular interpolation (clockwise and counterclockwise arcs, respectively). For these, compute the distance traveled along the arc.
For example, this below line is circular interpolation from G and M code Text file.
N1754 G03 X72.704 I10.704 J28.773 F2198.429
Could any one help me what formula should I use to get tool path for circular interpolation and linear interpolation to extract the distance.
Hello,
I am using BEAR TOOLBOX to obtain impulse response function of outcome variable to the 25 basis point monetary policy shock. The problem is there is no option in App of BEAR toolbox. how can i do it . please suggest
Hello everybody. I'm using Newton's method to solve a liner equation whose solution should be in [0 1]. Unfortunately, the coe I'm using gives NaN as a result for a specific combination of parameters and I would like to understand if I can improve the code I wrote for my Newton's method. In the specific case I'm considering, I reach the maximum iterations even if the tolerance is very low.
function [x,n,ier] = newton(f,fd,x0,nmax,tol)
% Newton's method for non-linear equations
ier = 0;
for n = 1:nmax
x = x0-f(x0)/fd(x0);
if abs(x-x0) <= tol
ier = 1;
break
end
x0 = x;
end
% % % % Script for solving NaN
mNAN= 16.1;
lNAN= 10^-4;
f= @(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN);
Fd= @(x) mNAN*(1+lNAN*exp(mNAN*x));
tolNaN=10^-1;
nmax=10^8;
AB0 = 0.5;
[amNAN,nNAN,ierNAN]=newton(f,Fd,AB0,nmax,tolNaN);
amNAN
Llimit=f(0)
Ulimit=f(1)
fplot(@(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN),[0 1.1])
This project discusses predator-prey system, particularly the Lotka-Volterra equations,which model the interaction between two sprecies: prey and predators. Let's solve the Lotka-Volterra equations numerically and visualize the results.% Define parameters
% Define parameters
alpha = 1.0; % Prey birth rate
beta = 0.1; % Predator success rate
gamma = 1.5; % Predator death rate
delta = 0.075; % Predator reproduction rate
% Define the symbolic variables
syms R W
% Define the equations
eq1 = alpha * R - beta * R * W == 0;
eq2 = delta * R * W - gamma * W == 0;
% Solve the equations
equilibriumPoints = solve([eq1, eq2], [R, W]);
% Extract the equilibrium point values
Req = double(equilibriumPoints.R);
Weq = double(equilibriumPoints.W);
% Display the equilibrium points
equilibriumPointsValues = [Req, Weq]
% Solve the differential equations using ode45
lotkaVolterra = @(t,Y)[alpha*Y(1)-beta*Y(1)*Y(2);
delta*Y(1)*Y(2)-gamma*Y(2)];
% Initial conditions
R0 = 40;
W0 = 9;
Y0 = [R0, W0];
tspan = [0, 100];
% Solve the differential equations
[t, Y] = ode45(lotkaVolterra, tspan, Y0);
% Extract the populations
R = Y(:, 1);
W = Y(:, 2);
% Plot the results
figure;
subplot(2,1,1);
plot(t, R, 'r', 'LineWidth', 1.5);
hold on;
plot(t, W, 'b', 'LineWidth', 1.5);
xlabel('Time (months)');
ylabel('Population');
legend('R', 'W');
grid on;
subplot(2,1,2);
plot(R, W, 'k', 'LineWidth', 1.5);
xlabel('R');
ylabel('W');
grid on;
hold on;
plot(Req, Weq, 'ro', 'MarkerSize', 8, 'MarkerFaceColor', 'r');
legend('Phase Trajectory', 'Equilibrium Point');
Now, we need to handle a modified version of the Lotka-Volterra equations. These modified equations incorporate logistic growth fot the prey population.
These equations are:
% Define parameters
alpha = 1.0;
K = 100; % Carrying Capacity of the prey population
beta = 0.1;
gamma = 1.5;
delta = 0.075;
% Define the symbolic variables
syms R W
% Define the equations
eq1 = alpha*R*(1 - R/K) - beta*R*W == 0;
eq2 = delta*R*W - gamma*W == 0;
% Solve the equations
equilibriumPoints = solve([eq1, eq2], [R, W]);
% Extract the equilibrium point values
Req = double(equilibriumPoints.R);
Weq = double(equilibriumPoints.W);
% Display the equilibrium points
equilibriumPointsValues = [Req, Weq]
% Solve the differential equations using ode45
modified_lv = @(t,Y)[alpha*Y(1)*(1-Y(1)/K)-beta*Y(1)*Y(2);
delta*Y(1)*Y(2)-gamma*Y(2)];
% Initial conditions
R0 = 40;
W0 = 9;
Y0 = [R0, W0];
tspan = [0, 100];
% Solve the differential equations
[t, Y] = ode45(modified_lv, tspan, Y0);
% Extract the populations
R = Y(:, 1);
W = Y(:, 2);
% Plot the results
figure;
subplot(2,1,1);
plot(t, R, 'r', 'LineWidth', 1.5);
hold on;
plot(t, W, 'b', 'LineWidth', 1.5);
xlabel('Time (months)');
ylabel('Population');
legend('R', 'W');
grid on;
subplot(2,1,2);
plot(R, W, 'k', 'LineWidth', 1.5);
xlabel('R');
ylabel('W');
grid on;
hold on;
plot(Req, Weq, 'ro', 'MarkerSize', 8, 'MarkerFaceColor', 'r');
legend('Phase Trajectory', 'Equilibrium Point');
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.)
Hi,
I have a plot as attached herewith in which the value of the point as shown by arrow mark is to be determined and compared to a reference value. It is plotted at a time step of 999 (t ranges from 1 to 1000).
global data;
cp=0;
for i=999:max(length(data.variable.t))
for j=60:max(length(data.variable.x))-1
if data.variable.curvepressure(i,j) <= -10.2661
disp(data.variable.curvepressure(i,j))
cp=1;
break
end
end
end
The above code is not working and need your advice please.
hello i found the following tools helpful to write matlab programs. copilot.microsoft.com chatgpt.com/gpts gemini.google.com and ai.meta.com. thanks a lot and best wishes.
Check out the LLMs with MATLAB project on File Exchange to access Large Language Models from MATLAB.
Along with the latest support for GPT-4o mini, you can use LLMs with MATLAB to generate images, categorize data, and provide semantic analyis.
function ans = your_fcn_name(n)
n;
j=sum(1:n);
a=zeros(1,j);
for i=1:n
a(1,((sum(1:(i-1))+1)):(sum(1:(i-1))+i))=i.*ones(1,i);
end
disp
I am trying to earn my Intro to MATLAB badge in Cody, but I cannot click the Roll the Dice! problem. It simply is not letting me click it, therefore I cannot earn my badge. Does anyone know who I should contact or what to do?
I define the class in matlab as:
classdef Myclass
properties
Content
end
methods
function obj = Myclass(content)
obj.Content = content;
end
function disp(obj)
A = symmatrix('A(1/3,[0,0,1])');
disp(A);
end
end
end
When we run this class in live editor return 'A(1/3,[0,0,1])' rather than latex form.
Myclass(1)
% return 'A(1/3,[0,0,1])'
A = symmatrix('A(1/3,[0,0,1])');
% return latrx form A(1/3,[0,0,1])
Something that had bothered me ever since I became an FEA analyst (2012) was the apparent inability of the "camera" in Matlab's 3D plot to function like the "cameras" in CAD/CAE packages.
For instance, load the ForearmLink.stl model that ships with the PDE Toolbox in Matlab and ParaView and try rotating the model.
clear
close all
gm = importGeometry( "ForearmLink.stl" );
pdegplot(gm)
Things to observe:
- Note that I cant seem to rotate continuously around the x-axis. It appears to only support rotations from [0, 360] as opposed to [-inf, inf]. So, for example, if I'm looking in the Y+ direction and rotate around X so that I'm looking at the Z- direction, and then want to look in the Y- direction, I can't simply keep rotating around the X axis... instead have to rotate 180 degrees around the Z axis and then around the X axis. I'm not aware of any data visualization applications (e.g., ParaView, VisIt, EnSight) or CAD/CAE tools with such an interaction.
- Note that at the 50 second mark, I set a view in ParaView: looking in the [X-, Y-, Z-] direction with Y+ up. Try as I might in Matlab, I'm unable to achieve that same view perspective.
Today I discovered that if one turns on the Camera Toolbar from the View menubar, then clicks the Orbit Camera icon, then the No Principal Axis icon:
That then it acts in the manner I've long desired. Oh, and also, for the interested, it is programmatically available: https://www.mathworks.com/help/matlab/ref/cameratoolbar.html
I might humbly propose this mode either be made more discoverable, similar to the little interaction widgets that pop up in figures:
Or maybe use the middle-mouse button to temporarily use this mode (a mouse setting in, e.g., Abaqus/CAE).
I've noticed is that the highly rated fonts for coding (e.g. Fira Code, Inconsolata, etc.) seem to overlook one issue that is key for coding in Matlab. While these fonts make 0 and O, as well as the 1 and l easily distinguishable, the brackets are not. Quite often the curly bracket looks similar to the curved bracket, which can lead to mistakes when coding or reviewing code.
So I was thinking: Could Mathworks put together a team to review good programming fonts, and come up with their own custom font designed specifically and optimized for Matlab syntax?
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);
Hi,
I have a condition where at the left boundary and the right boundary I need the flux to be non-zero.
Thus, the equations
pL + qL*fL = 0 and pr + qr*fr = 0 are such that fL and fr being non zero is making my equation hard to determine the values for pl, ql as well as pr, qr.
Plz tell me how should I approach for the same.
with regards,
rc
