​​​​​​​https://www.mathworks.com/matlabcentral/answers/2039236-how-can-i-access-a-simulink-real-time-slrt-speedgoat-target-computer-on-a-separate-network ​​​​​​​Are there any guidelines for remotely accessing a Simulink Real-Time (SLRT) Speedgoat target computer through VPN?​​​​​​​https://www.mathworks.com/matlabcentral/answers/2039236-how-can-i-access-a-simulink-real-time-slrt-speedgoat-target-computer-on-a-separate-network ​​​​​​​Are there any guidelines for remotely accessing a Simulink Real-Time (SLRT) Speedgoat target computer through VPN? ​​​​​​​https://www.mathworks.com/matlabcentral/answers/2039236-how-can-i-access-a-simulink-real-time-slrt-speedgoat-target-computer-on-a-separate-network ​​​​​​​Are there any guidelines for remotely accessing a Simulink Real-Time (SLRT) Speedgoat target computer through VPN? slrt, speedgoat, remote, access, vpn MATLAB Answers — New Questions

​

I am using following code to reconstruct a surface with poisson using pc2surfacemesh. I have reconstructed the surface twice, once with normal looking into the inside of the model. The second time mirrored so pointing outwards. There was no difference between the two mesh. I don’t know where to look at the code to determine the reconstruction. My question is does the function correct the normal when calculating.
ptcloud = pcread(‘Auto_Filter_35.ply’);
pcshow(ptcloud)
% flip normals
ptcloud.Normal = -ptcloud.Normal;
% Subsampling of ptCloud because of runtime
gridstep = 0.007;
ptCloudDownSampled = pcdownsample(ptcloud,"gridAverage",gridstep);

%% Construct surface mesh from the point cloud data using the Poisson method, and display the surface mesh.
tic
depth = 12;
mesh = pc2surfacemesh(ptCloudDownSampled,"poisson",depth);
surfaceMeshShow(mesh)
toc
writeSurfaceMesh(mesh,"AutoMesh35_flipedNormal.ply")I am using following code to reconstruct a surface with poisson using pc2surfacemesh. I have reconstructed the surface twice, once with normal looking into the inside of the model. The second time mirrored so pointing outwards. There was no difference between the two mesh. I don’t know where to look at the code to determine the reconstruction. My question is does the function correct the normal when calculating.
ptcloud = pcread(‘Auto_Filter_35.ply’);
pcshow(ptcloud)
% flip normals
ptcloud.Normal = -ptcloud.Normal;
% Subsampling of ptCloud because of runtime
gridstep = 0.007;
ptCloudDownSampled = pcdownsample(ptcloud,"gridAverage",gridstep);

%% Construct surface mesh from the point cloud data using the Poisson method, and display the surface mesh.
tic
depth = 12;
mesh = pc2surfacemesh(ptCloudDownSampled,"poisson",depth);
surfaceMeshShow(mesh)
toc
writeSurfaceMesh(mesh,"AutoMesh35_flipedNormal.ply") I am using following code to reconstruct a surface with poisson using pc2surfacemesh. I have reconstructed the surface twice, once with normal looking into the inside of the model. The second time mirrored so pointing outwards. There was no difference between the two mesh. I don’t know where to look at the code to determine the reconstruction. My question is does the function correct the normal when calculating.
ptcloud = pcread(‘Auto_Filter_35.ply’);
pcshow(ptcloud)
% flip normals
ptcloud.Normal = -ptcloud.Normal;
% Subsampling of ptCloud because of runtime
gridstep = 0.007;
ptCloudDownSampled = pcdownsample(ptcloud,"gridAverage",gridstep);

%% Construct surface mesh from the point cloud data using the Poisson method, and display the surface mesh.
tic
depth = 12;
mesh = pc2surfacemesh(ptCloudDownSampled,"poisson",depth);
surfaceMeshShow(mesh)
toc
writeSurfaceMesh(mesh,"AutoMesh35_flipedNormal.ply") surface, poisson, reconstruction, 3d MATLAB Answers — New Questions

​

I am running Simulink with Raspberry Pi and when I had hit RUN button, I realized that the model had some errors, but couldn’t find any option to TERMINATE THE COMPILATION PROCESS

Either I’d have to wait 20 minutes for the compilation process to complete or restart MATLAB.

I had to chose the 2nd one.I am running Simulink with Raspberry Pi and when I had hit RUN button, I realized that the model had some errors, but couldn’t find any option to TERMINATE THE COMPILATION PROCESS

Either I’d have to wait 20 minutes for the compilation process to complete or restart MATLAB.

I had to chose the 2nd one. I am running Simulink with Raspberry Pi and when I had hit RUN button, I realized that the model had some errors, but couldn’t find any option to TERMINATE THE COMPILATION PROCESS

Either I’d have to wait 20 minutes for the compilation process to complete or restart MATLAB.

I had to chose the 2nd one. simulink, model, compiler, raspberry pi MATLAB Answers — New Questions

​

dear all, i’m trying to create a mask for this image based on the fuction imfindingcircles. I tried setting different thresholds but I didn’t get the rigth result. Can anybody help me? than I would like to set NaN to the mask. Thank you in advance!
clear all
close all
clc
A= imread(‘image_1567.png’);
[rows, columns, numberOfColorChannels] = size(A)
%imshow(A)
[centers,radii] = imfindcircles(A,[50 100],Sensitivity=0.9);
mask = circles2mask(centers,radii,size(A));
figure
montage({A,mask})
A(~(A == mask)) = nandear all, i’m trying to create a mask for this image based on the fuction imfindingcircles. I tried setting different thresholds but I didn’t get the rigth result. Can anybody help me? than I would like to set NaN to the mask. Thank you in advance!
clear all
close all
clc
A= imread(‘image_1567.png’);
[rows, columns, numberOfColorChannels] = size(A)
%imshow(A)
[centers,radii] = imfindcircles(A,[50 100],Sensitivity=0.9);
mask = circles2mask(centers,radii,size(A));
figure
montage({A,mask})
A(~(A == mask)) = nan dear all, i’m trying to create a mask for this image based on the fuction imfindingcircles. I tried setting different thresholds but I didn’t get the rigth result. Can anybody help me? than I would like to set NaN to the mask. Thank you in advance!
clear all
close all
clc
A= imread(‘image_1567.png’);
[rows, columns, numberOfColorChannels] = size(A)
%imshow(A)
[centers,radii] = imfindcircles(A,[50 100],Sensitivity=0.9);
mask = circles2mask(centers,radii,size(A));
figure
montage({A,mask})
A(~(A == mask)) = nan imfindcircles, mask, image processing, image segmentation MATLAB Answers — New Questions

​

Hi, I am trying to create contour plot but after trying many times I am getting the same error, can anyone tell me what I am doing wrong in the code
*The code is*

s su = round(4*d/g)+1 ;
r = E((((su-1)- round(d/g)):-1:1),(su-1));
p = E((((su-1)- round(d/g)):-1:1),(su));
st = (p+r)./(1.5*(10)^(-4));
m = g*(size(r)-1);
H = 0:g:(m);
h = H./10;

writematrix(st)
type ‘st.txt’
writematrix(h)
type ‘h.txt’

plot(h,st)
xlim([0 1.5])
xlabel("X2/d",’fontsize’,14)
ylabel("Normalised Strain along vertical path", ‘fontsize’,13)
legend("d/w = 0.5")
saveas(gcf,’plot.png’)
contourf(st)
colorbar
saveas(gcf,’contour.png’)Hi, I am trying to create contour plot but after trying many times I am getting the same error, can anyone tell me what I am doing wrong in the code
*The code is*

s su = round(4*d/g)+1 ;
r = E((((su-1)- round(d/g)):-1:1),(su-1));
p = E((((su-1)- round(d/g)):-1:1),(su));
st = (p+r)./(1.5*(10)^(-4));
m = g*(size(r)-1);
H = 0:g:(m);
h = H./10;

writematrix(st)
type ‘st.txt’
writematrix(h)
type ‘h.txt’

plot(h,st)
xlim([0 1.5])
xlabel("X2/d",’fontsize’,14)
ylabel("Normalised Strain along vertical path", ‘fontsize’,13)
legend("d/w = 0.5")
saveas(gcf,’plot.png’)
contourf(st)
colorbar
saveas(gcf,’contour.png’) Hi, I am trying to create contour plot but after trying many times I am getting the same error, can anyone tell me what I am doing wrong in the code
*The code is*

s su = round(4*d/g)+1 ;
r = E((((su-1)- round(d/g)):-1:1),(su-1));
p = E((((su-1)- round(d/g)):-1:1),(su));
st = (p+r)./(1.5*(10)^(-4));
m = g*(size(r)-1);
H = 0:g:(m);
h = H./10;

writematrix(st)
type ‘st.txt’
writematrix(h)
type ‘h.txt’

plot(h,st)
xlim([0 1.5])
xlabel("X2/d",’fontsize’,14)
ylabel("Normalised Strain along vertical path", ‘fontsize’,13)
legend("d/w = 0.5")
saveas(gcf,’plot.png’)
contourf(st)
colorbar
saveas(gcf,’contour.png’) abaqus MATLAB Answers — New Questions

​

For the given system with transfer function
G(S) = 10 / (S2 + 2S + 20)
If G(S) = Y(S) / X(S), where Y(S) is the output and X(S) is the input. Compute for the output if the input is a. unit step function and b. sinusoidal input. Then simulate using MATLAB and compare the result. Plot the response of the system. Compare the result of the two different inputs.For the given system with transfer function
G(S) = 10 / (S2 + 2S + 20)
If G(S) = Y(S) / X(S), where Y(S) is the output and X(S) is the input. Compute for the output if the input is a. unit step function and b. sinusoidal input. Then simulate using MATLAB and compare the result. Plot the response of the system. Compare the result of the two different inputs. For the given system with transfer function
G(S) = 10 / (S2 + 2S + 20)
If G(S) = Y(S) / X(S), where Y(S) is the output and X(S) is the input. Compute for the output if the input is a. unit step function and b. sinusoidal input. Then simulate using MATLAB and compare the result. Plot the response of the system. Compare the result of the two different inputs. transfer function MATLAB Answers — New Questions

​

hi, as following picture shows, when after generate code, during compile how to control just let fun2 code enter compile process, fun1 code is forbidden to enter compile.hi, as following picture shows, when after generate code, during compile how to control just let fun2 code enter compile process, fun1 code is forbidden to enter compile. hi, as following picture shows, when after generate code, during compile how to control just let fun2 code enter compile process, fun1 code is forbidden to enter compile. simulink, embedded coder, conditional compilation MATLAB Answers — New Questions

​

It takes longer to start my stand-alone application the first time it is run than consecutive runs. Also if I run the application some time later, it again takes longer the first time but becomes faster on consecutive runs. I would expect it to always start quicker after the first run.It takes longer to start my stand-alone application the first time it is run than consecutive runs. Also if I run the application some time later, it again takes longer the first time but becomes faster on consecutive runs. I would expect it to always start quicker after the first run. It takes longer to start my stand-alone application the first time it is run than consecutive runs. Also if I run the application some time later, it again takes longer the first time but becomes faster on consecutive runs. I would expect it to always start quicker after the first run. standalone, mcr, slow, slower, fast, faster, second MATLAB Answers — New Questions

​

% 定义 LSTM 网络结构
numFeatures = size(X_train, 2); % 特征数量，这里是 3
numHiddenUnits = 100; % LSTM 隐藏单元数量

layers = [ …
sequenceInputLayer(numFeatures)
lstmLayer(numHiddenUnits)
fullyConnectedLayer(numFeatures)

];

maxEpochs = 50;
miniBatchSize = 3;
learningRate = 0.01;

% 初始化 LSTM 模型
net = dlnetwork(layers); % 在 CPU 上训练

% 训练循环
numObservations = size(X_train, 1);
numMiniBatches = floor(numObservations / miniBatchSize);

averageGrad = [];
averageSqGrad = [];
iteration = 0;
Velocities = zeros(size(net));
momentum = 0.9;

for epoch = 1:maxEpochs
% 打乱训练数据
idx = randperm(numObservations);
X_train_shuffled = X_train(idx, :);
Y_train_shuffled = Y_train(idx, :);

totalLoss = 0;

for i = 1:numMiniBatches
idxBatch = (i – 1) * miniBatchSize + 1 : i * miniBatchSize;
X_batch = X_train_shuffled(idxBatch, :);
Y_batch = Y_train_shuffled(idxBatch, :);

iteration = iteration + 1;

% 前向传播计算预测
Y_pred = predict(net, X_batch’);
Y_pred = Y_pred’;

Y_pred = dlarray(Y_pred, ‘TS’);
Y_batch = dlarray(Y_batch, ‘TS’);

%loss = mean(( Y_batch – Y_pred).^2, ‘all’);
[loss_1, gradients] = dlfeval(@customLoss, Y_batch, Y_pred);

gradients = extractdata(gradients);

[net, Velocities] = sgdmupdate(net, gradients, Velocities,learningRate,momentum);

% 计算损失
% loss = customLoss(Y_batch, Y_pred);
% totalLoss = totalLoss + loss;

% 反向传播更新梯度
%dL_dY = 2 * (Y_pred – Y_batch) / miniBatchSize;
%gradients = dlgradient(loss, net.LearnableParameters);
%net = updateParameters(net, gradients, learningRate);
end

% 显示每个 epoch 的平均损失
avgLoss = totalLoss / numMiniBatches;
fprintf(‘Epoch %d, Average Loss: %.4fn’, epoch, avgLoss);
end
无法执行赋值，因为此类型的变量不支持使用点进行索引。

出错 deep.internal.recording.containerfeval>iProcessNetwork_Nout_Nin (第 382 行)
protoTable.Value = zeros(height(protoTable),0);

出错 deep.internal.recording.containerfeval>iDispatch_Nout_Nin (第 214 行)
outputs = iProcessNetwork_Nout_Nin(fun, paramFun, numOut, …

出错 deep.internal.recording.containerfeval (第 38 行)
outputs = iDispatch_Nout_Nin(allowNetInput, fun, paramFun, numOut, …

出错 deep.internal.networkContainerFixedArgsFun (第 29 行)
varargout = deep.internal.recording.containerfeval(…

出错 sgdmupdate (第 126 行)
[p, vel] = deep.internal.networkContainerFixedArgsFun(func, …% 定义 LSTM 网络结构
numFeatures = size(X_train, 2); % 特征数量，这里是 3
numHiddenUnits = 100; % LSTM 隐藏单元数量

layers = [ …
sequenceInputLayer(numFeatures)
lstmLayer(numHiddenUnits)
fullyConnectedLayer(numFeatures)

];

maxEpochs = 50;
miniBatchSize = 3;
learningRate = 0.01;

% 初始化 LSTM 模型
net = dlnetwork(layers); % 在 CPU 上训练

% 训练循环
numObservations = size(X_train, 1);
numMiniBatches = floor(numObservations / miniBatchSize);

averageGrad = [];
averageSqGrad = [];
iteration = 0;
Velocities = zeros(size(net));
momentum = 0.9;

for epoch = 1:maxEpochs
% 打乱训练数据
idx = randperm(numObservations);
X_train_shuffled = X_train(idx, :);
Y_train_shuffled = Y_train(idx, :);

totalLoss = 0;

for i = 1:numMiniBatches
idxBatch = (i – 1) * miniBatchSize + 1 : i * miniBatchSize;
X_batch = X_train_shuffled(idxBatch, :);
Y_batch = Y_train_shuffled(idxBatch, :);

iteration = iteration + 1;

% 前向传播计算预测
Y_pred = predict(net, X_batch’);
Y_pred = Y_pred’;

Y_pred = dlarray(Y_pred, ‘TS’);
Y_batch = dlarray(Y_batch, ‘TS’);

%loss = mean(( Y_batch – Y_pred).^2, ‘all’);
[loss_1, gradients] = dlfeval(@customLoss, Y_batch, Y_pred);

gradients = extractdata(gradients);

[net, Velocities] = sgdmupdate(net, gradients, Velocities,learningRate,momentum);

% 计算损失
% loss = customLoss(Y_batch, Y_pred);
% totalLoss = totalLoss + loss;

% 反向传播更新梯度
%dL_dY = 2 * (Y_pred – Y_batch) / miniBatchSize;
%gradients = dlgradient(loss, net.LearnableParameters);
%net = updateParameters(net, gradients, learningRate);
end

% 显示每个 epoch 的平均损失
avgLoss = totalLoss / numMiniBatches;
fprintf(‘Epoch %d, Average Loss: %.4fn’, epoch, avgLoss);
end
无法执行赋值，因为此类型的变量不支持使用点进行索引。

出错 deep.internal.recording.containerfeval>iProcessNetwork_Nout_Nin (第 382 行)
protoTable.Value = zeros(height(protoTable),0);

出错 deep.internal.recording.containerfeval>iDispatch_Nout_Nin (第 214 行)
outputs = iProcessNetwork_Nout_Nin(fun, paramFun, numOut, …

出错 deep.internal.recording.containerfeval (第 38 行)
outputs = iDispatch_Nout_Nin(allowNetInput, fun, paramFun, numOut, …

出错 deep.internal.networkContainerFixedArgsFun (第 29 行)
varargout = deep.internal.recording.containerfeval(…

出错 sgdmupdate (第 126 行)
[p, vel] = deep.internal.networkContainerFixedArgsFun(func, … % 定义 LSTM 网络结构
numFeatures = size(X_train, 2); % 特征数量，这里是 3
numHiddenUnits = 100; % LSTM 隐藏单元数量

layers = [ …
sequenceInputLayer(numFeatures)
lstmLayer(numHiddenUnits)
fullyConnectedLayer(numFeatures)

];

maxEpochs = 50;
miniBatchSize = 3;
learningRate = 0.01;

% 初始化 LSTM 模型
net = dlnetwork(layers); % 在 CPU 上训练

% 训练循环
numObservations = size(X_train, 1);
numMiniBatches = floor(numObservations / miniBatchSize);

averageGrad = [];
averageSqGrad = [];
iteration = 0;
Velocities = zeros(size(net));
momentum = 0.9;

for epoch = 1:maxEpochs
% 打乱训练数据
idx = randperm(numObservations);
X_train_shuffled = X_train(idx, :);
Y_train_shuffled = Y_train(idx, :);

totalLoss = 0;

for i = 1:numMiniBatches
idxBatch = (i – 1) * miniBatchSize + 1 : i * miniBatchSize;
X_batch = X_train_shuffled(idxBatch, :);
Y_batch = Y_train_shuffled(idxBatch, :);

iteration = iteration + 1;

% 前向传播计算预测
Y_pred = predict(net, X_batch’);
Y_pred = Y_pred’;

Y_pred = dlarray(Y_pred, ‘TS’);
Y_batch = dlarray(Y_batch, ‘TS’);

%loss = mean(( Y_batch – Y_pred).^2, ‘all’);
[loss_1, gradients] = dlfeval(@customLoss, Y_batch, Y_pred);

gradients = extractdata(gradients);

[net, Velocities] = sgdmupdate(net, gradients, Velocities,learningRate,momentum);

% 计算损失
% loss = customLoss(Y_batch, Y_pred);
% totalLoss = totalLoss + loss;

% 反向传播更新梯度
%dL_dY = 2 * (Y_pred – Y_batch) / miniBatchSize;
%gradients = dlgradient(loss, net.LearnableParameters);
%net = updateParameters(net, gradients, learningRate);
end

% 显示每个 epoch 的平均损失
avgLoss = totalLoss / numMiniBatches;
fprintf(‘Epoch %d, Average Loss: %.4fn’, epoch, avgLoss);
end
无法执行赋值，因为此类型的变量不支持使用点进行索引。

出错 deep.internal.recording.containerfeval>iProcessNetwork_Nout_Nin (第 382 行)
protoTable.Value = zeros(height(protoTable),0);

出错 deep.internal.recording.containerfeval>iDispatch_Nout_Nin (第 214 行)
outputs = iProcessNetwork_Nout_Nin(fun, paramFun, numOut, …

出错 deep.internal.recording.containerfeval (第 38 行)
outputs = iDispatch_Nout_Nin(allowNetInput, fun, paramFun, numOut, …

出错 deep.internal.networkContainerFixedArgsFun (第 29 行)
varargout = deep.internal.recording.containerfeval(…

出错 sgdmupdate (第 126 行)
[p, vel] = deep.internal.networkContainerFixedArgsFun(func, … sgdmupdate, deeplearning, lstm, dlarray MATLAB Answers — New Questions

​

Hello friends.
A little introduction:
I do a simulation for particles moving inside a box, as part of the simulation I measure the speed of the particles, and the idea is that in the end I get the Maxwell-Boltzmann distribution of the speed.

I wanted to do a fit to the histogram of the velocity, according to the Boltzmann equation, so that I could find the temperature of the system, but for some reason the fit does not fit exactly on the data. it’s similar, but not enough.
this is the fit that I get:

I’m uploading code with the histogram data, so you can run it yourself and see what you get.
I only change the histogram() in the figure, to plot(), so that it matches the data I uploaded.
This is the equation of Boltzmann distribution that I want to fit:

I will be happy if you can understand what is the problems..thanks!

This is the code, you can simple run it and see what is happening:
%my data:
hist_x=[10 30 50 70 90 110 130 150 170 190 210 230 250,…
270 290 310 330 350 370 390 410 430 450 470 490];
hist_y=[20 60 83 88 108 117 113 94 77 63 65 51 26,…
19 9 3 1 2 0 0 0 0 0 0 1];

%fit the velocity to Boltzmann distribution:
%parameters:
K_B=1.38e-23; %[m^2Kg/s^2K]
m=39.948*1.660e-27 ; %mass of Ar, convert from atomic mass to Kg [Kg]
boltzmann = fittype…
( @(T,v) (m/K_B*T)*v.*exp(((-m)/(2*K_B*T))*v.^2)…
, ‘coefficient’, ‘T’…
, ‘independent’, ‘v’) ;

[fitted_curve] = fit(hist_x’,hist_y’,boltzmann, ‘startPoint’, 100) ;

%plot histogram and fit:
figure;
plot(hist_x,hist_y);
hold on
plot(fitted_curve)
title(‘histogram of velocity and fitted curve – Boltzmann distribution’)
legend(‘histogram’, ‘fitted curve’);
xlabel(‘velovity [m/s]’)
ylabel(‘particles amount’)
disp(fitted_curve);Hello friends.
A little introduction:
I do a simulation for particles moving inside a box, as part of the simulation I measure the speed of the particles, and the idea is that in the end I get the Maxwell-Boltzmann distribution of the speed.

I wanted to do a fit to the histogram of the velocity, according to the Boltzmann equation, so that I could find the temperature of the system, but for some reason the fit does not fit exactly on the data. it’s similar, but not enough.
this is the fit that I get:

I’m uploading code with the histogram data, so you can run it yourself and see what you get.
I only change the histogram() in the figure, to plot(), so that it matches the data I uploaded.
This is the equation of Boltzmann distribution that I want to fit:

I will be happy if you can understand what is the problems..thanks!

This is the code, you can simple run it and see what is happening:
%my data:
hist_x=[10 30 50 70 90 110 130 150 170 190 210 230 250,…
270 290 310 330 350 370 390 410 430 450 470 490];
hist_y=[20 60 83 88 108 117 113 94 77 63 65 51 26,…
19 9 3 1 2 0 0 0 0 0 0 1];

%fit the velocity to Boltzmann distribution:
%parameters:
K_B=1.38e-23; %[m^2Kg/s^2K]
m=39.948*1.660e-27 ; %mass of Ar, convert from atomic mass to Kg [Kg]
boltzmann = fittype…
( @(T,v) (m/K_B*T)*v.*exp(((-m)/(2*K_B*T))*v.^2)…
, ‘coefficient’, ‘T’…
, ‘independent’, ‘v’) ;

[fitted_curve] = fit(hist_x’,hist_y’,boltzmann, ‘startPoint’, 100) ;

%plot histogram and fit:
figure;
plot(hist_x,hist_y);
hold on
plot(fitted_curve)
title(‘histogram of velocity and fitted curve – Boltzmann distribution’)
legend(‘histogram’, ‘fitted curve’);
xlabel(‘velovity [m/s]’)
ylabel(‘particles amount’)
disp(fitted_curve); Hello friends.
A little introduction:
I do a simulation for particles moving inside a box, as part of the simulation I measure the speed of the particles, and the idea is that in the end I get the Maxwell-Boltzmann distribution of the speed.

I wanted to do a fit to the histogram of the velocity, according to the Boltzmann equation, so that I could find the temperature of the system, but for some reason the fit does not fit exactly on the data. it’s similar, but not enough.
this is the fit that I get:

I’m uploading code with the histogram data, so you can run it yourself and see what you get.
I only change the histogram() in the figure, to plot(), so that it matches the data I uploaded.
This is the equation of Boltzmann distribution that I want to fit:

I will be happy if you can understand what is the problems..thanks!

This is the code, you can simple run it and see what is happening:
%my data:
hist_x=[10 30 50 70 90 110 130 150 170 190 210 230 250,…
270 290 310 330 350 370 390 410 430 450 470 490];
hist_y=[20 60 83 88 108 117 113 94 77 63 65 51 26,…
19 9 3 1 2 0 0 0 0 0 0 1];

%fit the velocity to Boltzmann distribution:
%parameters:
K_B=1.38e-23; %[m^2Kg/s^2K]
m=39.948*1.660e-27 ; %mass of Ar, convert from atomic mass to Kg [Kg]
boltzmann = fittype…
( @(T,v) (m/K_B*T)*v.*exp(((-m)/(2*K_B*T))*v.^2)…
, ‘coefficient’, ‘T’…
, ‘independent’, ‘v’) ;

[fitted_curve] = fit(hist_x’,hist_y’,boltzmann, ‘startPoint’, 100) ;

%plot histogram and fit:
figure;
plot(hist_x,hist_y);
hold on
plot(fitted_curve)
title(‘histogram of velocity and fitted curve – Boltzmann distribution’)
legend(‘histogram’, ‘fitted curve’);
xlabel(‘velovity [m/s]’)
ylabel(‘particles amount’)
disp(fitted_curve); curve fitting MATLAB Answers — New Questions

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