Can the spmd parallel structure in MATLAB be used to simulate multiple independent computing nodes that can communicate with each other? That is to simulate each process as a computer.Can the spmd parallel structure in MATLAB be used to simulate multiple independent computing nodes that can communicate with each other? That is to simulate each process as a computer. Can the spmd parallel structure in MATLAB be used to simulate multiple independent computing nodes that can communicate with each other? That is to simulate each process as a computer. spmd, parallel computing MATLAB Answers — New Questions

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Hi,friend,

price2ret function is able to calculate the growth rate of two adjacent values, if it’s monthly data, it can be viewed as month over month change, can I use this function to calculate year over year change on the same monthly data series? thanksHi,friend,

price2ret function is able to calculate the growth rate of two adjacent values, if it’s monthly data, it can be viewed as month over month change, can I use this function to calculate year over year change on the same monthly data series? thanks Hi,friend,

price2ret function is able to calculate the growth rate of two adjacent values, if it’s monthly data, it can be viewed as month over month change, can I use this function to calculate year over year change on the same monthly data series? thanks mom yoy MATLAB Answers — New Questions

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I want to have a title for each column and row of subplots. I’ve tried using sgtitle but it is not giving me the effect I want. Is there another way to achieve this? I’ve included a simplifed example below of what I want to achieve.I want to have a title for each column and row of subplots. I’ve tried using sgtitle but it is not giving me the effect I want. Is there another way to achieve this? I’ve included a simplifed example below of what I want to achieve. I want to have a title for each column and row of subplots. I’ve tried using sgtitle but it is not giving me the effect I want. Is there another way to achieve this? I’ve included a simplifed example below of what I want to achieve. plot, subplot, sgtitle, format, graph, graphing, plotting MATLAB Answers — New Questions

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Hi, does anyone knows the reason why a received file (video file) have the full size however cannot playback in the windows media player?It stated..Unknown file error.
Does anyone knows why?Is it due to manipulation of my coding?Hi, does anyone knows the reason why a received file (video file) have the full size however cannot playback in the windows media player?It stated..Unknown file error.
Does anyone knows why?Is it due to manipulation of my coding? Hi, does anyone knows the reason why a received file (video file) have the full size however cannot playback in the windows media player?It stated..Unknown file error.
Does anyone knows why?Is it due to manipulation of my coding? for, data corruption, matlab, preallocation MATLAB Answers — New Questions

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While generating code Error comes as "Error:Error: File: C:Program FilesMATLABR2023brtwctlcpublic_apidiagnostics_api.tlc Line: 39 Column: 10 %exit directive: Simulink Coder Fatal: Custom Data Error: Empty DataObject record"
Please support me to resolve this issue.
Thank YouWhile generating code Error comes as "Error:Error: File: C:Program FilesMATLABR2023brtwctlcpublic_apidiagnostics_api.tlc Line: 39 Column: 10 %exit directive: Simulink Coder Fatal: Custom Data Error: Empty DataObject record"
Please support me to resolve this issue.
Thank You While generating code Error comes as "Error:Error: File: C:Program FilesMATLABR2023brtwctlcpublic_apidiagnostics_api.tlc Line: 39 Column: 10 %exit directive: Simulink Coder Fatal: Custom Data Error: Empty DataObject record"
Please support me to resolve this issue.
Thank You simulink, stateflow, embedded coder MATLAB Answers — New Questions

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Hi,
Can someone give a picture example / link for bounding box for image processing / computer vision?
Regards,Hi,
Can someone give a picture example / link for bounding box for image processing / computer vision?
Regards, Hi,
Can someone give a picture example / link for bounding box for image processing / computer vision?
Regards, image processing, computer vision MATLAB Answers — New Questions

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How can i plot my data against time on x axis which is 24 hours duration split every two hours.
Sample has been attached here.How can i plot my data against time on x axis which is 24 hours duration split every two hours.
Sample has been attached here. How can i plot my data against time on x axis which is 24 hours duration split every two hours.
Sample has been attached here. plot MATLAB Answers — New Questions

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% Parameters
L = 1; % Length of the string
c = 1; % Wave speed
num_segments = 200; % Number of segments
dx = L / num_segments; % Discretization step
dt = 0.005; % Time step
num_steps = 10 / dt; % Number of time steps
% Initial wave parameters
left_wave_length = 0.15 * L;
right_wave_length = 0.25 * L;
% Initialize the wave function
x = 0:dx:L;
u = sin(pi * x / left_wave_length) + sin(2 * pi * x / right_wave_length);
% Time integration
for step = 1:num_steps
% Compute second spatial derivative using finite differences
u_xx = (u(1:end-2) – 2*u(2:end-1) + u(3:end)) / dx^2;

% Update the wave equation using finite differences
u_new = 2*u(2:end-1) – u(2:end-1) + c^2 * dt^2 * u_xx;

% Update boundary conditions
u_new1 = [0, u_new, 0];

% Update the wave function
u = u_new1;
end
% Plot at t = 0
figure;
plot(x, u);
title(‘Wave Equation at t = 0’);
xlabel(‘Position [L]’);
ylabel(‘Amplitude’);
xlim([0, L]);
% Plot at t = 10 [t]
figure;
plot(x, u);
title(‘Wave Equation at t = 10 [t]’);
xlabel(‘Position [L]’);
ylabel(‘Amplitude’);
xlim([0, L]);% Parameters
L = 1; % Length of the string
c = 1; % Wave speed
num_segments = 200; % Number of segments
dx = L / num_segments; % Discretization step
dt = 0.005; % Time step
num_steps = 10 / dt; % Number of time steps
% Initial wave parameters
left_wave_length = 0.15 * L;
right_wave_length = 0.25 * L;
% Initialize the wave function
x = 0:dx:L;
u = sin(pi * x / left_wave_length) + sin(2 * pi * x / right_wave_length);
% Time integration
for step = 1:num_steps
% Compute second spatial derivative using finite differences
u_xx = (u(1:end-2) – 2*u(2:end-1) + u(3:end)) / dx^2;

% Update the wave equation using finite differences
u_new = 2*u(2:end-1) – u(2:end-1) + c^2 * dt^2 * u_xx;

% Update boundary conditions
u_new1 = [0, u_new, 0];

% Update the wave function
u = u_new1;
end
% Plot at t = 0
figure;
plot(x, u);
title(‘Wave Equation at t = 0’);
xlabel(‘Position [L]’);
ylabel(‘Amplitude’);
xlim([0, L]);
% Plot at t = 10 [t]
figure;
plot(x, u);
title(‘Wave Equation at t = 10 [t]’);
xlabel(‘Position [L]’);
ylabel(‘Amplitude’);
xlim([0, L]); % Parameters
L = 1; % Length of the string
c = 1; % Wave speed
num_segments = 200; % Number of segments
dx = L / num_segments; % Discretization step
dt = 0.005; % Time step
num_steps = 10 / dt; % Number of time steps
% Initial wave parameters
left_wave_length = 0.15 * L;
right_wave_length = 0.25 * L;
% Initialize the wave function
x = 0:dx:L;
u = sin(pi * x / left_wave_length) + sin(2 * pi * x / right_wave_length);
% Time integration
for step = 1:num_steps
% Compute second spatial derivative using finite differences
u_xx = (u(1:end-2) – 2*u(2:end-1) + u(3:end)) / dx^2;

% Update the wave equation using finite differences
u_new = 2*u(2:end-1) – u(2:end-1) + c^2 * dt^2 * u_xx;

% Update boundary conditions
u_new1 = [0, u_new, 0];

% Update the wave function
u = u_new1;
end
% Plot at t = 0
figure;
plot(x, u);
title(‘Wave Equation at t = 0’);
xlabel(‘Position [L]’);
ylabel(‘Amplitude’);
xlim([0, L]);
% Plot at t = 10 [t]
figure;
plot(x, u);
title(‘Wave Equation at t = 10 [t]’);
xlabel(‘Position [L]’);
ylabel(‘Amplitude’);
xlim([0, L]); matlab MATLAB Answers — New Questions

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The terminals are LC only, L=0.1H and C=0.25F with an initial voltage of 1V. The variable capacitor has an initial capacitance of 0.25F and is increased by 0.25F each time the voltage takes an extreme value. The judgment of the extreme value is made by the derivative block, which outputs an impulse signal when hitcrossing becomes 0. Q=CV, and since Q is conserved, C is increased so that V is attenuated.The circuit I am currently working on is as follows, but it does not work well.

The ideal graph is as follows.The terminals are LC only, L=0.1H and C=0.25F with an initial voltage of 1V. The variable capacitor has an initial capacitance of 0.25F and is increased by 0.25F each time the voltage takes an extreme value. The judgment of the extreme value is made by the derivative block, which outputs an impulse signal when hitcrossing becomes 0. Q=CV, and since Q is conserved, C is increased so that V is attenuated.The circuit I am currently working on is as follows, but it does not work well.

The ideal graph is as follows. The terminals are LC only, L=0.1H and C=0.25F with an initial voltage of 1V. The variable capacitor has an initial capacitance of 0.25F and is increased by 0.25F each time the voltage takes an extreme value. The judgment of the extreme value is made by the derivative block, which outputs an impulse signal when hitcrossing becomes 0. Q=CV, and since Q is conserved, C is increased so that V is attenuated.The circuit I am currently working on is as follows, but it does not work well.

The ideal graph is as follows. simulink, variable capacitor, lc MATLAB Answers — New Questions

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%How to effectively solve the problem of incompatibility of AppDesigner using symbol toolkit syms after packaging and publishing into exe and other executable files? And how can I change it in my code
cla(app.UIAxes)
P=app.PEditField.Value
P=P.*pi/180;
c=app.CKPaEditField.Value;
K=app.KEditField.Value;%侧压力系数
P_ya=app.P_yaEditField.Value;
F=app.FEditField.Value;
y=app.yKNm3EditField.Value;
R = app.RmEditField.Value;
L=app.LmEditField.Value;
L_s=app.L_smEditField.Value;
syms pha H
term1 = 9 – 12 * (1 – P_ya .* F ./ (y .* H));
term2 = R .* sqrt(term1) – 3 .* R;
denominator = 2 .* H;
% 计算 pha
pha = atan(term2 ./ denominator);
% 显示结果
% disp(pha);
pha_rad=rad2deg(pha);
H_vals = linspace(50, 250, 50);
pha_vals = double(subs(pha_rad, H, H_vals));
% Plot H versus pha
% figure(1);
plot(app.UIAxes,H_vals, pha_vals, ‘b-‘, ‘LineWidth’, 2);
xlabel(app.UIAxes,’H’);
ylabel(app.UIAxes,’pha’);
title(app.UIAxes,’Plot of pha vs H’);
% grid on;
% disp(pha_rad);
%修正破裂角计算
if K <= 1
pha_rad = 45 – P / 2;
elseif K <= 2
pha_rad = 45;
else
pha_rad = 45 + P / 2;
end
% 限制 pha_rad 的值在 15° 和 65° 之间
pha_rad = max(15, min(pha_rad, 65));
pha_rad = pha_rad * (pi / 180);
switch app.DropDown.Value
case ‘竖直滑面法’
if app.Button_2.Value==1
Vc2=4.*pi.*R.^3+pi.*R.^2*(L-2.*R);%体积
Ac2=pi.*R.^2+(L-2.*R).*2.*R;
Lc2=2*pi.*R+(L-2*R).*2;%周长
sigma2=Lc2.*H.*K.*y.*H./2;
Fbc2=Vc2.*10;%
Fbr2=Ac2.*10.*(H-L_s);
F2=Ac2.*P_ya;
Wr2=y.*H.*Ac2;%
t2=Lc2.*c.*H+sigma2.*tan(P)
eq=Wr2+t2==F.*(F2+Fbc2+Fbr2)
H_solution = double(solve(eq, H));
positiveValues = H_solution(H_solution > 0);
app.HEditField.Value=positiveValues;
app.Image.ImageSource = ‘竖直滑面法示意图.png’; % 直接用路径
app.Image_2.ImageSource = ‘竖直滑面法示意图2.png’; % 直接用路径
elseif app.Button.Value==1
Vc1=4.*pi.*R.^3;%体积
Ac1=pi.*R.^2;%表面积
Lc1=2*pi.*R;%周长
sigma1=Lc1.*H.*K.*y.*H./2;%水平应力和∑σh
Fbc1=Vc1.*10;%硐室浮力
Fbr1=Ac1.*10.*(H-L_s);%岩体浮力
F1=Ac1.*P_ya;%上抬力
Wr1=y.*H.*Ac1;%重量
t1=Lc1.*c.*H+sigma1.*tan(P)
eq=Wr1+t1==F.*(F1+Fbc1+Fbr1)
H_solution = double(solve(eq, H));
positiveValues = H_solution(H_solution > 0);
app.HEditField.Value=positiveValues;
app.Image.ImageSource = ‘竖直滑面法示意图.png’; % 直接用路径
app.Image_2.ImageSource = ‘竖直滑面法示意图2.png’; % 直接用路径
end%How to effectively solve the problem of incompatibility of AppDesigner using symbol toolkit syms after packaging and publishing into exe and other executable files? And how can I change it in my code
cla(app.UIAxes)
P=app.PEditField.Value
P=P.*pi/180;
c=app.CKPaEditField.Value;
K=app.KEditField.Value;%侧压力系数
P_ya=app.P_yaEditField.Value;
F=app.FEditField.Value;
y=app.yKNm3EditField.Value;
R = app.RmEditField.Value;
L=app.LmEditField.Value;
L_s=app.L_smEditField.Value;
syms pha H
term1 = 9 – 12 * (1 – P_ya .* F ./ (y .* H));
term2 = R .* sqrt(term1) – 3 .* R;
denominator = 2 .* H;
% 计算 pha
pha = atan(term2 ./ denominator);
% 显示结果
% disp(pha);
pha_rad=rad2deg(pha);
H_vals = linspace(50, 250, 50);
pha_vals = double(subs(pha_rad, H, H_vals));
% Plot H versus pha
% figure(1);
plot(app.UIAxes,H_vals, pha_vals, ‘b-‘, ‘LineWidth’, 2);
xlabel(app.UIAxes,’H’);
ylabel(app.UIAxes,’pha’);
title(app.UIAxes,’Plot of pha vs H’);
% grid on;
% disp(pha_rad);
%修正破裂角计算
if K <= 1
pha_rad = 45 – P / 2;
elseif K <= 2
pha_rad = 45;
else
pha_rad = 45 + P / 2;
end
% 限制 pha_rad 的值在 15° 和 65° 之间
pha_rad = max(15, min(pha_rad, 65));
pha_rad = pha_rad * (pi / 180);
switch app.DropDown.Value
case ‘竖直滑面法’
if app.Button_2.Value==1
Vc2=4.*pi.*R.^3+pi.*R.^2*(L-2.*R);%体积
Ac2=pi.*R.^2+(L-2.*R).*2.*R;
Lc2=2*pi.*R+(L-2*R).*2;%周长
sigma2=Lc2.*H.*K.*y.*H./2;
Fbc2=Vc2.*10;%
Fbr2=Ac2.*10.*(H-L_s);
F2=Ac2.*P_ya;
Wr2=y.*H.*Ac2;%
t2=Lc2.*c.*H+sigma2.*tan(P)
eq=Wr2+t2==F.*(F2+Fbc2+Fbr2)
H_solution = double(solve(eq, H));
positiveValues = H_solution(H_solution > 0);
app.HEditField.Value=positiveValues;
app.Image.ImageSource = ‘竖直滑面法示意图.png’; % 直接用路径
app.Image_2.ImageSource = ‘竖直滑面法示意图2.png’; % 直接用路径
elseif app.Button.Value==1
Vc1=4.*pi.*R.^3;%体积
Ac1=pi.*R.^2;%表面积
Lc1=2*pi.*R;%周长
sigma1=Lc1.*H.*K.*y.*H./2;%水平应力和∑σh
Fbc1=Vc1.*10;%硐室浮力
Fbr1=Ac1.*10.*(H-L_s);%岩体浮力
F1=Ac1.*P_ya;%上抬力
Wr1=y.*H.*Ac1;%重量
t1=Lc1.*c.*H+sigma1.*tan(P)
eq=Wr1+t1==F.*(F1+Fbc1+Fbr1)
H_solution = double(solve(eq, H));
positiveValues = H_solution(H_solution > 0);
app.HEditField.Value=positiveValues;
app.Image.ImageSource = ‘竖直滑面法示意图.png’; % 直接用路径
app.Image_2.ImageSource = ‘竖直滑面法示意图2.png’; % 直接用路径
end %How to effectively solve the problem of incompatibility of AppDesigner using symbol toolkit syms after packaging and publishing into exe and other executable files? And how can I change it in my code
cla(app.UIAxes)
P=app.PEditField.Value
P=P.*pi/180;
c=app.CKPaEditField.Value;
K=app.KEditField.Value;%侧压力系数
P_ya=app.P_yaEditField.Value;
F=app.FEditField.Value;
y=app.yKNm3EditField.Value;
R = app.RmEditField.Value;
L=app.LmEditField.Value;
L_s=app.L_smEditField.Value;
syms pha H
term1 = 9 – 12 * (1 – P_ya .* F ./ (y .* H));
term2 = R .* sqrt(term1) – 3 .* R;
denominator = 2 .* H;
% 计算 pha
pha = atan(term2 ./ denominator);
% 显示结果
% disp(pha);
pha_rad=rad2deg(pha);
H_vals = linspace(50, 250, 50);
pha_vals = double(subs(pha_rad, H, H_vals));
% Plot H versus pha
% figure(1);
plot(app.UIAxes,H_vals, pha_vals, ‘b-‘, ‘LineWidth’, 2);
xlabel(app.UIAxes,’H’);
ylabel(app.UIAxes,’pha’);
title(app.UIAxes,’Plot of pha vs H’);
% grid on;
% disp(pha_rad);
%修正破裂角计算
if K <= 1
pha_rad = 45 – P / 2;
elseif K <= 2
pha_rad = 45;
else
pha_rad = 45 + P / 2;
end
% 限制 pha_rad 的值在 15° 和 65° 之间
pha_rad = max(15, min(pha_rad, 65));
pha_rad = pha_rad * (pi / 180);
switch app.DropDown.Value
case ‘竖直滑面法’
if app.Button_2.Value==1
Vc2=4.*pi.*R.^3+pi.*R.^2*(L-2.*R);%体积
Ac2=pi.*R.^2+(L-2.*R).*2.*R;
Lc2=2*pi.*R+(L-2*R).*2;%周长
sigma2=Lc2.*H.*K.*y.*H./2;
Fbc2=Vc2.*10;%
Fbr2=Ac2.*10.*(H-L_s);
F2=Ac2.*P_ya;
Wr2=y.*H.*Ac2;%
t2=Lc2.*c.*H+sigma2.*tan(P)
eq=Wr2+t2==F.*(F2+Fbc2+Fbr2)
H_solution = double(solve(eq, H));
positiveValues = H_solution(H_solution > 0);
app.HEditField.Value=positiveValues;
app.Image.ImageSource = ‘竖直滑面法示意图.png’; % 直接用路径
app.Image_2.ImageSource = ‘竖直滑面法示意图2.png’; % 直接用路径
elseif app.Button.Value==1
Vc1=4.*pi.*R.^3;%体积
Ac1=pi.*R.^2;%表面积
Lc1=2*pi.*R;%周长
sigma1=Lc1.*H.*K.*y.*H./2;%水平应力和∑σh
Fbc1=Vc1.*10;%硐室浮力
Fbr1=Ac1.*10.*(H-L_s);%岩体浮力
F1=Ac1.*P_ya;%上抬力
Wr1=y.*H.*Ac1;%重量
t1=Lc1.*c.*H+sigma1.*tan(P)
eq=Wr1+t1==F.*(F1+Fbc1+Fbr1)
H_solution = double(solve(eq, H));
positiveValues = H_solution(H_solution > 0);
app.HEditField.Value=positiveValues;
app.Image.ImageSource = ‘竖直滑面法示意图.png’; % 直接用路径
app.Image_2.ImageSource = ‘竖直滑面法示意图2.png’; % 直接用路径
end matlab gui MATLAB Answers — New Questions

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