In this image, the leaves are on a piece of white paper. I wonder in this particular case if there are any effective methods for shadow removal or reduction using Matlab. If it’s not possible to completely remove the shadows, reducing them is also OK. Thank you!In this image, the leaves are on a piece of white paper. I wonder in this particular case if there are any effective methods for shadow removal or reduction using Matlab. If it’s not possible to completely remove the shadows, reducing them is also OK. Thank you! In this image, the leaves are on a piece of white paper. I wonder in this particular case if there are any effective methods for shadow removal or reduction using Matlab. If it’s not possible to completely remove the shadows, reducing them is also OK. Thank you! shadow-removal, shadow-reduction, image enhancement, image processing, image segmentation MATLAB Answers — New Questions

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I have build the DDS application shapesdemo.exe with DDS Blockset. I have test the communication between shapesdemo.exe with OpenDDS Shapes Demo Application successfully. However, these two DDS Application run on the same desktop computer, I want to communicate shapesdemo.exe with another OpenDDS Shapes Demo Application runing on the different computer.
I want to configure the shapesdemo.exe with the another computer’s ip address/port, such that it can communicate with another computer’s DDS Application, where is the DDS Blockset ip address/port informations saved?I have build the DDS application shapesdemo.exe with DDS Blockset. I have test the communication between shapesdemo.exe with OpenDDS Shapes Demo Application successfully. However, these two DDS Application run on the same desktop computer, I want to communicate shapesdemo.exe with another OpenDDS Shapes Demo Application runing on the different computer.
I want to configure the shapesdemo.exe with the another computer’s ip address/port, such that it can communicate with another computer’s DDS Application, where is the DDS Blockset ip address/port informations saved? I have build the DDS application shapesdemo.exe with DDS Blockset. I have test the communication between shapesdemo.exe with OpenDDS Shapes Demo Application successfully. However, these two DDS Application run on the same desktop computer, I want to communicate shapesdemo.exe with another OpenDDS Shapes Demo Application runing on the different computer.
I want to configure the shapesdemo.exe with the another computer’s ip address/port, such that it can communicate with another computer’s DDS Application, where is the DDS Blockset ip address/port informations saved? dds blockset, shapes demo, simulink MATLAB Answers — New Questions

​

The scanning tunneling microscope tip is moved to obtain the tunneling current z. The tip is scanned with raster scanning (triangular wave-like behavior) + dither circular motion behavior.
This tunneling current z is then scanned for the time interval Δtm=N(1/fm+j/fm)(N=0, 1, 2, ….. , j=0, 1, 2, 3, 4, 5), we want to extract and draw. j=0, 1, 2, 3, 4, 5 represent the 6 equal positions of the dither circumference. These 6 points are points A, B, C, D, E, and F, which are 210, 330, 90, 30, 150, and 270° of the dither circle, respectively.
How can I extract each tunnel current at each point?

Attached below are images of raster scanning and dither circular motion.
　　

%Parameter Setting
pixel_image = 256; %Input the number of pixels in the image obtained by raster scanning (input 2^n)
dr = 1/(2*sqrt(3)); %Enter dither circle radius [grid].
a_fast_grid = 10; %fast axis scanning range [grid]
a_slow_grid = 10; %Slow axis scanning range [grid]
fm=5000; %Dither circle modulation frequency [Hz]
fs= fm*240 ; %Sampling frequency [Hz]
f_fast = 10.2; %Input scanning frequency [Hz] (1 line scanning count in 1[s])
start_point_x = 0; %Input x-coordinate of scanning start point (input 1 if you want to move by 1[grid])
start_point_y = 0; %Input y-coordinate of scanning start point (input 1 if you want to move by 1[grid])

%Parameter setting for fast-axis triangular wave
amplitude_fast = a_fast_grid/2; %fast axis amplitude

%Parameter setting for slow-axis triangular wave
amplitude_slow = a_slow_grid/2; %slow axis amplitude
f_slow = (f_fast)/(2*pixel_image); %Slow axis triangular wave frequency

% Generation of time vectors
total_time=256/f_fast; %Total Scan Time
t = linspace(0, total_time, fs * total_time);

x_raster = start_point_x + amplitude_fast*(2/pi)*acos(cos(2*pi*f_fast*t));
y_raster = start_point_y + amplitude_slow*(2/pi)*acos(cos(2*pi*f_slow*t));

x_dither = dr*cos(2*pi*fm*t);
y_dither = dr*sin(2*pi*fm*t);

x = x_raster + x_dither;
y = y_raster + y_dither;

z1 = cos(2*pi*((x-y)/(sqrt(3))));
z2 = cos(2*pi*(2*y/(sqrt(3))));
z3 = cos(2*pi*((x+y)/(sqrt(3))));
z = (z1 + z2 + z3);

% Plotting Raster Scanning
figure;
plot(x_raster, y_raster, ‘b.-‘);
title(‘Raster Scanning’);
xlabel(‘x coordinate’);
ylabel(‘y coordinate’);
axis equal;
grid on;

% Dither circle plot
figure;
plot(x_dither, y_dither, ‘b.-‘);
title(‘dither circle scanning locus’);
xlabel(‘x coordinate’);
ylabel(‘y coordinate’);
axis equal;
grid on;

%Raster Scanning + dither circle scanning Plot
figure;
plot(x, y, ‘b.-‘);
title(‘Raster scanning + dither circle scanning locus’);
xlabel(‘x coordinate’);
ylabel(‘y coordinate’);
axis equal;
grid on;The scanning tunneling microscope tip is moved to obtain the tunneling current z. The tip is scanned with raster scanning (triangular wave-like behavior) + dither circular motion behavior.
This tunneling current z is then scanned for the time interval Δtm=N(1/fm+j/fm)(N=0, 1, 2, ….. , j=0, 1, 2, 3, 4, 5), we want to extract and draw. j=0, 1, 2, 3, 4, 5 represent the 6 equal positions of the dither circumference. These 6 points are points A, B, C, D, E, and F, which are 210, 330, 90, 30, 150, and 270° of the dither circle, respectively.
How can I extract each tunnel current at each point?

Attached below are images of raster scanning and dither circular motion.
　　

%Parameter Setting
pixel_image = 256; %Input the number of pixels in the image obtained by raster scanning (input 2^n)
dr = 1/(2*sqrt(3)); %Enter dither circle radius [grid].
a_fast_grid = 10; %fast axis scanning range [grid]
a_slow_grid = 10; %Slow axis scanning range [grid]
fm=5000; %Dither circle modulation frequency [Hz]
fs= fm*240 ; %Sampling frequency [Hz]
f_fast = 10.2; %Input scanning frequency [Hz] (1 line scanning count in 1[s])
start_point_x = 0; %Input x-coordinate of scanning start point (input 1 if you want to move by 1[grid])
start_point_y = 0; %Input y-coordinate of scanning start point (input 1 if you want to move by 1[grid])

%Parameter setting for fast-axis triangular wave
amplitude_fast = a_fast_grid/2; %fast axis amplitude

%Parameter setting for slow-axis triangular wave
amplitude_slow = a_slow_grid/2; %slow axis amplitude
f_slow = (f_fast)/(2*pixel_image); %Slow axis triangular wave frequency

% Generation of time vectors
total_time=256/f_fast; %Total Scan Time
t = linspace(0, total_time, fs * total_time);

x_raster = start_point_x + amplitude_fast*(2/pi)*acos(cos(2*pi*f_fast*t));
y_raster = start_point_y + amplitude_slow*(2/pi)*acos(cos(2*pi*f_slow*t));

x_dither = dr*cos(2*pi*fm*t);
y_dither = dr*sin(2*pi*fm*t);

x = x_raster + x_dither;
y = y_raster + y_dither;

z1 = cos(2*pi*((x-y)/(sqrt(3))));
z2 = cos(2*pi*(2*y/(sqrt(3))));
z3 = cos(2*pi*((x+y)/(sqrt(3))));
z = (z1 + z2 + z3);

% Plotting Raster Scanning
figure;
plot(x_raster, y_raster, ‘b.-‘);
title(‘Raster Scanning’);
xlabel(‘x coordinate’);
ylabel(‘y coordinate’);
axis equal;
grid on;

% Dither circle plot
figure;
plot(x_dither, y_dither, ‘b.-‘);
title(‘dither circle scanning locus’);
xlabel(‘x coordinate’);
ylabel(‘y coordinate’);
axis equal;
grid on;

%Raster Scanning + dither circle scanning Plot
figure;
plot(x, y, ‘b.-‘);
title(‘Raster scanning + dither circle scanning locus’);
xlabel(‘x coordinate’);
ylabel(‘y coordinate’);
axis equal;
grid on; The scanning tunneling microscope tip is moved to obtain the tunneling current z. The tip is scanned with raster scanning (triangular wave-like behavior) + dither circular motion behavior.
This tunneling current z is then scanned for the time interval Δtm=N(1/fm+j/fm)(N=0, 1, 2, ….. , j=0, 1, 2, 3, 4, 5), we want to extract and draw. j=0, 1, 2, 3, 4, 5 represent the 6 equal positions of the dither circumference. These 6 points are points A, B, C, D, E, and F, which are 210, 330, 90, 30, 150, and 270° of the dither circle, respectively.
How can I extract each tunnel current at each point?

Attached below are images of raster scanning and dither circular motion.
　　

%Parameter Setting
pixel_image = 256; %Input the number of pixels in the image obtained by raster scanning (input 2^n)
dr = 1/(2*sqrt(3)); %Enter dither circle radius [grid].
a_fast_grid = 10; %fast axis scanning range [grid]
a_slow_grid = 10; %Slow axis scanning range [grid]
fm=5000; %Dither circle modulation frequency [Hz]
fs= fm*240 ; %Sampling frequency [Hz]
f_fast = 10.2; %Input scanning frequency [Hz] (1 line scanning count in 1[s])
start_point_x = 0; %Input x-coordinate of scanning start point (input 1 if you want to move by 1[grid])
start_point_y = 0; %Input y-coordinate of scanning start point (input 1 if you want to move by 1[grid])

%Parameter setting for fast-axis triangular wave
amplitude_fast = a_fast_grid/2; %fast axis amplitude

%Parameter setting for slow-axis triangular wave
amplitude_slow = a_slow_grid/2; %slow axis amplitude
f_slow = (f_fast)/(2*pixel_image); %Slow axis triangular wave frequency

% Generation of time vectors
total_time=256/f_fast; %Total Scan Time
t = linspace(0, total_time, fs * total_time);

x_raster = start_point_x + amplitude_fast*(2/pi)*acos(cos(2*pi*f_fast*t));
y_raster = start_point_y + amplitude_slow*(2/pi)*acos(cos(2*pi*f_slow*t));

x_dither = dr*cos(2*pi*fm*t);
y_dither = dr*sin(2*pi*fm*t);

x = x_raster + x_dither;
y = y_raster + y_dither;

z1 = cos(2*pi*((x-y)/(sqrt(3))));
z2 = cos(2*pi*(2*y/(sqrt(3))));
z3 = cos(2*pi*((x+y)/(sqrt(3))));
z = (z1 + z2 + z3);

% Plotting Raster Scanning
figure;
plot(x_raster, y_raster, ‘b.-‘);
title(‘Raster Scanning’);
xlabel(‘x coordinate’);
ylabel(‘y coordinate’);
axis equal;
grid on;

% Dither circle plot
figure;
plot(x_dither, y_dither, ‘b.-‘);
title(‘dither circle scanning locus’);
xlabel(‘x coordinate’);
ylabel(‘y coordinate’);
axis equal;
grid on;

%Raster Scanning + dither circle scanning Plot
figure;
plot(x, y, ‘b.-‘);
title(‘Raster scanning + dither circle scanning locus’);
xlabel(‘x coordinate’);
ylabel(‘y coordinate’);
axis equal;
grid on; signal processing, image, matlab MATLAB Answers — New Questions

​

I’m trying to model a system of LEDs with gradually increasing and decreasing intensities. Are there any blocks or add-ons that anyone is aware of that can help me achieve this behavior? Essentially, I want something similar to the Lamp block, but rather than set values and their accompanying colors, I want something that can change based on the value from 0% to 100% intensity. Thanks in advance!I’m trying to model a system of LEDs with gradually increasing and decreasing intensities. Are there any blocks or add-ons that anyone is aware of that can help me achieve this behavior? Essentially, I want something similar to the Lamp block, but rather than set values and their accompanying colors, I want something that can change based on the value from 0% to 100% intensity. Thanks in advance! I’m trying to model a system of LEDs with gradually increasing and decreasing intensities. Are there any blocks or add-ons that anyone is aware of that can help me achieve this behavior? Essentially, I want something similar to the Lamp block, but rather than set values and their accompanying colors, I want something that can change based on the value from 0% to 100% intensity. Thanks in advance! simulink, lamp, intensity, gradual MATLAB Answers — New Questions

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@Torsten
I asked this question on URL: https://www.mathworks.com/matlabcentral/answers/2172760-how-to-find-the-rmse
Even I responded there several times but then I became busy and when I got free to respond there, my response was not recorded. I tried sevral times but in vain. That’s why I am re-posting here.
I used the "rmse code" with my previous code (just added awgn() to add noise ) but it gives me error:
clear;clc
ula = phased.ULA(‘NumElements’,10,’ElementSpacing’,0.5);
angs = [40 -20 20; 0 0 0];% angs=[azimuth; elevation]; My desired angles are 40, -20 and 20.
NumSignals = size(angs,2);
c = physconst(‘LightSpeed’);
fc = 300e6; % Operating frequency
lambda = c/fc;
pos = getElementPosition(ula)/lambda;
Nsamp = 10;%1000;
nPower = 0.01;
rs = rng(2007);
signal = sensorsig(pos,Nsamp,angs,nPower);
signal = awgn(signal,-15);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% MUSIC Algorithm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
musicspatialspect = phased.MUSICEstimator(‘SensorArray’,ula,…
‘OperatingFrequency’,fc,’ScanAngles’,-90:90,…
‘DOAOutputPort’,true,’NumSignalsSource’,’Property’,’NumSignals’,NumSignals);
[~,ang] = musicspatialspect(signal)
ymusic = musicspatialspect(signal);
helperPlotDOASpectra(musicspatialspect.ScanAngles,ymusic)% Changed by Me

function helperPlotDOASpectra(x2,y2)% Changed by Me
% Plot spectra in dB normalized to 0 dB at the peak
y2_dB = 20*log10(y2) – max(20*log10(y2));
plot(x2,y2_dB)
xlabel(‘Broadside Angle (degrees)’);
ylabel(‘Power (dB)’);
title(‘DOA Spatial Spectra’)
end
%%%%%%% RMSE
num_runs = 100;
rmse_values = zeros(1, num_runs);
for idx = 1:num_runs
error = angs – ang;
rmse = sqrt(mean(error.^2));
rmse_values(idx) = rmse;
end
figure;
plot(1:num_runs, rmse_values, ‘-o’);
xlabel(‘Run Number’);
ylabel(‘RMSE’);
title(‘RMSE vs Number of Runs’);
grid on;@Torsten
I asked this question on URL: https://www.mathworks.com/matlabcentral/answers/2172760-how-to-find-the-rmse
Even I responded there several times but then I became busy and when I got free to respond there, my response was not recorded. I tried sevral times but in vain. That’s why I am re-posting here.
I used the "rmse code" with my previous code (just added awgn() to add noise ) but it gives me error:
clear;clc
ula = phased.ULA(‘NumElements’,10,’ElementSpacing’,0.5);
angs = [40 -20 20; 0 0 0];% angs=[azimuth; elevation]; My desired angles are 40, -20 and 20.
NumSignals = size(angs,2);
c = physconst(‘LightSpeed’);
fc = 300e6; % Operating frequency
lambda = c/fc;
pos = getElementPosition(ula)/lambda;
Nsamp = 10;%1000;
nPower = 0.01;
rs = rng(2007);
signal = sensorsig(pos,Nsamp,angs,nPower);
signal = awgn(signal,-15);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% MUSIC Algorithm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
musicspatialspect = phased.MUSICEstimator(‘SensorArray’,ula,…
‘OperatingFrequency’,fc,’ScanAngles’,-90:90,…
‘DOAOutputPort’,true,’NumSignalsSource’,’Property’,’NumSignals’,NumSignals);
[~,ang] = musicspatialspect(signal)
ymusic = musicspatialspect(signal);
helperPlotDOASpectra(musicspatialspect.ScanAngles,ymusic)% Changed by Me

function helperPlotDOASpectra(x2,y2)% Changed by Me
% Plot spectra in dB normalized to 0 dB at the peak
y2_dB = 20*log10(y2) – max(20*log10(y2));
plot(x2,y2_dB)
xlabel(‘Broadside Angle (degrees)’);
ylabel(‘Power (dB)’);
title(‘DOA Spatial Spectra’)
end
%%%%%%% RMSE
num_runs = 100;
rmse_values = zeros(1, num_runs);
for idx = 1:num_runs
error = angs – ang;
rmse = sqrt(mean(error.^2));
rmse_values(idx) = rmse;
end
figure;
plot(1:num_runs, rmse_values, ‘-o’);
xlabel(‘Run Number’);
ylabel(‘RMSE’);
title(‘RMSE vs Number of Runs’);
grid on; @Torsten
I asked this question on URL: https://www.mathworks.com/matlabcentral/answers/2172760-how-to-find-the-rmse
Even I responded there several times but then I became busy and when I got free to respond there, my response was not recorded. I tried sevral times but in vain. That’s why I am re-posting here.
I used the "rmse code" with my previous code (just added awgn() to add noise ) but it gives me error:
clear;clc
ula = phased.ULA(‘NumElements’,10,’ElementSpacing’,0.5);
angs = [40 -20 20; 0 0 0];% angs=[azimuth; elevation]; My desired angles are 40, -20 and 20.
NumSignals = size(angs,2);
c = physconst(‘LightSpeed’);
fc = 300e6; % Operating frequency
lambda = c/fc;
pos = getElementPosition(ula)/lambda;
Nsamp = 10;%1000;
nPower = 0.01;
rs = rng(2007);
signal = sensorsig(pos,Nsamp,angs,nPower);
signal = awgn(signal,-15);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% MUSIC Algorithm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
musicspatialspect = phased.MUSICEstimator(‘SensorArray’,ula,…
‘OperatingFrequency’,fc,’ScanAngles’,-90:90,…
‘DOAOutputPort’,true,’NumSignalsSource’,’Property’,’NumSignals’,NumSignals);
[~,ang] = musicspatialspect(signal)
ymusic = musicspatialspect(signal);
helperPlotDOASpectra(musicspatialspect.ScanAngles,ymusic)% Changed by Me

function helperPlotDOASpectra(x2,y2)% Changed by Me
% Plot spectra in dB normalized to 0 dB at the peak
y2_dB = 20*log10(y2) – max(20*log10(y2));
plot(x2,y2_dB)
xlabel(‘Broadside Angle (degrees)’);
ylabel(‘Power (dB)’);
title(‘DOA Spatial Spectra’)
end
%%%%%%% RMSE
num_runs = 100;
rmse_values = zeros(1, num_runs);
for idx = 1:num_runs
error = angs – ang;
rmse = sqrt(mean(error.^2));
rmse_values(idx) = rmse;
end
figure;
plot(1:num_runs, rmse_values, ‘-o’);
xlabel(‘Run Number’);
ylabel(‘RMSE’);
title(‘RMSE vs Number of Runs’);
grid on; rmse, root mean square error, code error MATLAB Answers — New Questions

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Why does checking for equality with ‘isequal’ sometimes fail when using the Symbolic Toolbox? For example:

>> syms x y
>> isequal(x*y+x, x*(y+1))
ans =
logical
0Why does checking for equality with ‘isequal’ sometimes fail when using the Symbolic Toolbox? For example:

>> syms x y
>> isequal(x*y+x, x*(y+1))
ans =
logical
0 Why does checking for equality with ‘isequal’ sometimes fail when using the Symbolic Toolbox? For example:

>> syms x y
>> isequal(x*y+x, x*(y+1))
ans =
logical
0 symbolic, comparison MATLAB Answers — New Questions

​

I am looking for a MATLAB programmer who has skills in numerical methods. To be more specific, it is about

1) Solving Poisson Equation with some relaxation method

2) Implementing the multigrid method in matlab

If you are interested earning some extra money with that please write me a mail, I will send you all the details.I am looking for a MATLAB programmer who has skills in numerical methods. To be more specific, it is about

1) Solving Poisson Equation with some relaxation method

2) Implementing the multigrid method in matlab

If you are interested earning some extra money with that please write me a mail, I will send you all the details. I am looking for a MATLAB programmer who has skills in numerical methods. To be more specific, it is about

1) Solving Poisson Equation with some relaxation method

2) Implementing the multigrid method in matlab

If you are interested earning some extra money with that please write me a mail, I will send you all the details. programmer, numerical analysis MATLAB Answers — New Questions

​

Hello,
I tried to simulate a very similar PLC model like presented in here https://nl.mathworks.com/help/plccoder/ug/ladder-logic-using-timers.html . The differences are Im using step signal which is converted into boolean for the input, you can see my model in the attached file.
However, it seems like the preset time just doesn’t work. Everytime the switch is on, then the lamp is automatically "on", while it is supposed to be delayed for around 6 seconds (the preset time). I have tried everything, even mirroring every setting and values set in the file from here https://nl.mathworks.com/help/plccoder/ug/ladder-logic-using-timers.html. I have also ensured the initfcn, etc. But still doesnt work as it should be. Im using matlab R2024b. I also put the sceenshots of the windows below.
Thanks for the help.Hello,
I tried to simulate a very similar PLC model like presented in here https://nl.mathworks.com/help/plccoder/ug/ladder-logic-using-timers.html . The differences are Im using step signal which is converted into boolean for the input, you can see my model in the attached file.
However, it seems like the preset time just doesn’t work. Everytime the switch is on, then the lamp is automatically "on", while it is supposed to be delayed for around 6 seconds (the preset time). I have tried everything, even mirroring every setting and values set in the file from here https://nl.mathworks.com/help/plccoder/ug/ladder-logic-using-timers.html. I have also ensured the initfcn, etc. But still doesnt work as it should be. Im using matlab R2024b. I also put the sceenshots of the windows below.
Thanks for the help. Hello,
I tried to simulate a very similar PLC model like presented in here https://nl.mathworks.com/help/plccoder/ug/ladder-logic-using-timers.html . The differences are Im using step signal which is converted into boolean for the input, you can see my model in the attached file.
However, it seems like the preset time just doesn’t work. Everytime the switch is on, then the lamp is automatically "on", while it is supposed to be delayed for around 6 seconds (the preset time). I have tried everything, even mirroring every setting and values set in the file from here https://nl.mathworks.com/help/plccoder/ug/ladder-logic-using-timers.html. I have also ensured the initfcn, etc. But still doesnt work as it should be. Im using matlab R2024b. I also put the sceenshots of the windows below.
Thanks for the help. plc, ton, tof, simulink, studio5000, timer MATLAB Answers — New Questions

​

I need the help of people who have excellent talents in calculating using Matlab.
The equations to be implemented through Matlab coding is as follows.

Answer) β = 5.874395, A = 0.000060, b = 0.280599, φ = 5630.329851

The results to be obtained through this coding is written in "Answer) ~" above.
And, the data related to this problem(or eqution) is below.

I have tried with the following coding but kept getting the same results as the title. I would like to get helpful advice on this part.
I would appreciate it if you could give me a guide on the solution or error cause.
clear all
clc

th1 = [378 378 378 378; 0.4 0.4 0.4 0.4; 310 316 329 411]; % [V_i ; U_i ; T_i]
th2 = [378 378 378 378; 0.8 0.8 0.8 0.8; 190 208 230 298];
th3 = [398 398 398 398; 0.4 0.4 0.4 0.4; 108 123 166 200];

syms B;
syms A;
syms phi;
syms b;

%————- equation 1 ————-%
a1_th1 = 0;
a1_th2 = 0;
a1_th3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂β(equation 1)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));

a1_th1 = N*(1-(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B)*(log(th1(i+2,j)/A)-(phi/th1(i,j)+b/th1(i+1,j))) + a1_th1 ;

end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂β(equation 1)
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));

a1_th2 = N*(1-(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B)*(log(th2(i+2,j)/A)-(phi/th2(i,j)+b/th2(i+1,j))) + a1_th2 ;

end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂β(equation 1)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));

a1_th3 = N*(1-(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B)*(log(th3(i+2,j)/A)-(phi/th3(i,j)+b/th3(i+1,j))) + a1_th3 ;

end

a1 = a1_th1 + a1_th2 + a1_th3;
N = numel(th1(1,:)) + numel(th2(1,:)) + numel(th3(1,:)); % ∂Λ/∂β first term

one = N/B + a1;
% vpa(one,3);

%————- equation 2 ————-%
b1 = 0;
b2 = 0;
b3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂A(equation 2)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));
b1 = N*(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B+b1;
end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂A(equation 2)
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));
b2 = N*(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B+b2;
end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂A(equation 2)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));
b3 = N*(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B+b3;
end

N = numel(th1(1,:)) + numel(th2(1,:)) + numel(th3(1,:)); % ∂Λ/∂A first term

two = -B/A*(N + (b1 + b2 + b3));
% vpa(two,3);

%————- equation 3 ————-%
c1 = 0;
c2 = 0;
c3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂φ(equation 3)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));
c1 = N/th1(i,j)*(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B+c1;
end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂φ(equation 3)
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));
c2 = N/th2(i,j)*(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B+c2;
end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂φ(equation 3)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));
c3 = N/th3(i,j)*(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B+c3;
end

N_c1 = 0;
N_c2 = 0;
N_c3 = 0;

i = 1;
for j = 1:(numel(th1(i, :)))
N_c1 = numel(th1(i,j))/th1(i,j)+N_c1; % "Stress Lv. 1" in the first term of ∂Λ/∂φ(equation 3)
end

i = 1;
for j = 1:(numel(th2(i, :)))
N_c2 = numel(th2(i,j))/th2(i,j)+N_c2; % "Stress Lv. 2" in the first term of ∂Λ/∂φ(equation 3)
end

i = 1;
for j = 1:(numel(th3(i, :)))
N_c3 = numel(th3(i,j))/th3(i,j)+N_c3; % "Stress Lv. 3" in the first term of ∂Λ/∂φ(equation 3)
end

N_c = N_c1 + N_c2 + N_c3; % ∂Λ/∂φ first term

three = -B*(N_c – (c1 + c2 + c3));
% vpa(three,3);

%————- equation 3 ————-%
d1 = 0;
d2 = 0;
d3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂b(equation 4)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));
d1 = N/th1(i+1,j)*(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B+d1;
end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂b(equation 4)"
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));
d2 = N/th2(i+1,j)*(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B+d2;
end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂b(equation 4)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));
d3 = N/th3(i+1,j)*(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B+d3;
end

N_d1 = 0;
N_d2 = 0;
N_d3 = 0;

i = 1;
for j = 1:(numel(th1(i, :)))
N_d1 = numel(th1(i,j))/th1(i+1,j)+N_d1; % "Stress Lv. 1" in the first term of ∂Λ/∂b(equation 4)
end

i = 1;
for j = 1:(numel(th2(i, :)))
N_d2 = numel(th2(i,j))/th2(i+1,j)+N_d2; % "Stress Lv. 2" in the first term of ∂Λ/∂b(equation 4)
end

i = 1;
for j = 1:(numel(th3(i, :)))
N_d3 = numel(th3(i,j))/th3(i+1,j)+N_d3; % "Stress Lv. 3" in the first term of ∂Λ/∂b(equation 4)
end

N_d = N_d1 + N_d2 + N_d3; % ∂Λ/∂b first term

four = -B*(N_d – (d1 + d2 + d3));
% vpa(four,3);

[B, A, phi, b] = solve(one,two,three,four,’Real’,true)

% eqns = [one == 0, two == 0, three == 0, four == 0];
% vars = [B, A, phi, b];
% [solB, solA, solphi, solb] = solve(eqns, vars, ‘Real’, true)I need the help of people who have excellent talents in calculating using Matlab.
The equations to be implemented through Matlab coding is as follows.

Answer) β = 5.874395, A = 0.000060, b = 0.280599, φ = 5630.329851

The results to be obtained through this coding is written in "Answer) ~" above.
And, the data related to this problem(or eqution) is below.

I have tried with the following coding but kept getting the same results as the title. I would like to get helpful advice on this part.
I would appreciate it if you could give me a guide on the solution or error cause.
clear all
clc

th1 = [378 378 378 378; 0.4 0.4 0.4 0.4; 310 316 329 411]; % [V_i ; U_i ; T_i]
th2 = [378 378 378 378; 0.8 0.8 0.8 0.8; 190 208 230 298];
th3 = [398 398 398 398; 0.4 0.4 0.4 0.4; 108 123 166 200];

syms B;
syms A;
syms phi;
syms b;

%————- equation 1 ————-%
a1_th1 = 0;
a1_th2 = 0;
a1_th3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂β(equation 1)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));

a1_th1 = N*(1-(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B)*(log(th1(i+2,j)/A)-(phi/th1(i,j)+b/th1(i+1,j))) + a1_th1 ;

end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂β(equation 1)
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));

a1_th2 = N*(1-(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B)*(log(th2(i+2,j)/A)-(phi/th2(i,j)+b/th2(i+1,j))) + a1_th2 ;

end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂β(equation 1)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));

a1_th3 = N*(1-(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B)*(log(th3(i+2,j)/A)-(phi/th3(i,j)+b/th3(i+1,j))) + a1_th3 ;

end

a1 = a1_th1 + a1_th2 + a1_th3;
N = numel(th1(1,:)) + numel(th2(1,:)) + numel(th3(1,:)); % ∂Λ/∂β first term

one = N/B + a1;
% vpa(one,3);

%————- equation 2 ————-%
b1 = 0;
b2 = 0;
b3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂A(equation 2)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));
b1 = N*(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B+b1;
end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂A(equation 2)
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));
b2 = N*(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B+b2;
end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂A(equation 2)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));
b3 = N*(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B+b3;
end

N = numel(th1(1,:)) + numel(th2(1,:)) + numel(th3(1,:)); % ∂Λ/∂A first term

two = -B/A*(N + (b1 + b2 + b3));
% vpa(two,3);

%————- equation 3 ————-%
c1 = 0;
c2 = 0;
c3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂φ(equation 3)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));
c1 = N/th1(i,j)*(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B+c1;
end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂φ(equation 3)
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));
c2 = N/th2(i,j)*(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B+c2;
end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂φ(equation 3)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));
c3 = N/th3(i,j)*(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B+c3;
end

N_c1 = 0;
N_c2 = 0;
N_c3 = 0;

i = 1;
for j = 1:(numel(th1(i, :)))
N_c1 = numel(th1(i,j))/th1(i,j)+N_c1; % "Stress Lv. 1" in the first term of ∂Λ/∂φ(equation 3)
end

i = 1;
for j = 1:(numel(th2(i, :)))
N_c2 = numel(th2(i,j))/th2(i,j)+N_c2; % "Stress Lv. 2" in the first term of ∂Λ/∂φ(equation 3)
end

i = 1;
for j = 1:(numel(th3(i, :)))
N_c3 = numel(th3(i,j))/th3(i,j)+N_c3; % "Stress Lv. 3" in the first term of ∂Λ/∂φ(equation 3)
end

N_c = N_c1 + N_c2 + N_c3; % ∂Λ/∂φ first term

three = -B*(N_c – (c1 + c2 + c3));
% vpa(three,3);

%————- equation 3 ————-%
d1 = 0;
d2 = 0;
d3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂b(equation 4)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));
d1 = N/th1(i+1,j)*(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B+d1;
end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂b(equation 4)"
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));
d2 = N/th2(i+1,j)*(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B+d2;
end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂b(equation 4)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));
d3 = N/th3(i+1,j)*(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B+d3;
end

N_d1 = 0;
N_d2 = 0;
N_d3 = 0;

i = 1;
for j = 1:(numel(th1(i, :)))
N_d1 = numel(th1(i,j))/th1(i+1,j)+N_d1; % "Stress Lv. 1" in the first term of ∂Λ/∂b(equation 4)
end

i = 1;
for j = 1:(numel(th2(i, :)))
N_d2 = numel(th2(i,j))/th2(i+1,j)+N_d2; % "Stress Lv. 2" in the first term of ∂Λ/∂b(equation 4)
end

i = 1;
for j = 1:(numel(th3(i, :)))
N_d3 = numel(th3(i,j))/th3(i+1,j)+N_d3; % "Stress Lv. 3" in the first term of ∂Λ/∂b(equation 4)
end

N_d = N_d1 + N_d2 + N_d3; % ∂Λ/∂b first term

four = -B*(N_d – (d1 + d2 + d3));
% vpa(four,3);

[B, A, phi, b] = solve(one,two,three,four,’Real’,true)

% eqns = [one == 0, two == 0, three == 0, four == 0];
% vars = [B, A, phi, b];
% [solB, solA, solphi, solb] = solve(eqns, vars, ‘Real’, true) I need the help of people who have excellent talents in calculating using Matlab.
The equations to be implemented through Matlab coding is as follows.

Answer) β = 5.874395, A = 0.000060, b = 0.280599, φ = 5630.329851

The results to be obtained through this coding is written in "Answer) ~" above.
And, the data related to this problem(or eqution) is below.

I have tried with the following coding but kept getting the same results as the title. I would like to get helpful advice on this part.
I would appreciate it if you could give me a guide on the solution or error cause.
clear all
clc

th1 = [378 378 378 378; 0.4 0.4 0.4 0.4; 310 316 329 411]; % [V_i ; U_i ; T_i]
th2 = [378 378 378 378; 0.8 0.8 0.8 0.8; 190 208 230 298];
th3 = [398 398 398 398; 0.4 0.4 0.4 0.4; 108 123 166 200];

syms B;
syms A;
syms phi;
syms b;

%————- equation 1 ————-%
a1_th1 = 0;
a1_th2 = 0;
a1_th3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂β(equation 1)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));

a1_th1 = N*(1-(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B)*(log(th1(i+2,j)/A)-(phi/th1(i,j)+b/th1(i+1,j))) + a1_th1 ;

end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂β(equation 1)
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));

a1_th2 = N*(1-(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B)*(log(th2(i+2,j)/A)-(phi/th2(i,j)+b/th2(i+1,j))) + a1_th2 ;

end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂β(equation 1)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));

a1_th3 = N*(1-(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B)*(log(th3(i+2,j)/A)-(phi/th3(i,j)+b/th3(i+1,j))) + a1_th3 ;

end

a1 = a1_th1 + a1_th2 + a1_th3;
N = numel(th1(1,:)) + numel(th2(1,:)) + numel(th3(1,:)); % ∂Λ/∂β first term

one = N/B + a1;
% vpa(one,3);

%————- equation 2 ————-%
b1 = 0;
b2 = 0;
b3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂A(equation 2)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));
b1 = N*(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B+b1;
end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂A(equation 2)
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));
b2 = N*(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B+b2;
end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂A(equation 2)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));
b3 = N*(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B+b3;
end

N = numel(th1(1,:)) + numel(th2(1,:)) + numel(th3(1,:)); % ∂Λ/∂A first term

two = -B/A*(N + (b1 + b2 + b3));
% vpa(two,3);

%————- equation 3 ————-%
c1 = 0;
c2 = 0;
c3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂φ(equation 3)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));
c1 = N/th1(i,j)*(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B+c1;
end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂φ(equation 3)
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));
c2 = N/th2(i,j)*(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B+c2;
end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂φ(equation 3)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));
c3 = N/th3(i,j)*(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B+c3;
end

N_c1 = 0;
N_c2 = 0;
N_c3 = 0;

i = 1;
for j = 1:(numel(th1(i, :)))
N_c1 = numel(th1(i,j))/th1(i,j)+N_c1; % "Stress Lv. 1" in the first term of ∂Λ/∂φ(equation 3)
end

i = 1;
for j = 1:(numel(th2(i, :)))
N_c2 = numel(th2(i,j))/th2(i,j)+N_c2; % "Stress Lv. 2" in the first term of ∂Λ/∂φ(equation 3)
end

i = 1;
for j = 1:(numel(th3(i, :)))
N_c3 = numel(th3(i,j))/th3(i,j)+N_c3; % "Stress Lv. 3" in the first term of ∂Λ/∂φ(equation 3)
end

N_c = N_c1 + N_c2 + N_c3; % ∂Λ/∂φ first term

three = -B*(N_c – (c1 + c2 + c3));
% vpa(three,3);

%————- equation 3 ————-%
d1 = 0;
d2 = 0;
d3 = 0;

i = 1; % "Stress Lv. 1" in the second term of ∂Λ/∂b(equation 4)
for j = 1:(numel(th1(i, :)))
N = numel(th1(i,j));
d1 = N/th1(i+1,j)*(th1(i+2,j)/A*exp(-(phi/th1(i,j)+b/th1(i+1,j))))^B+d1;
end

i = 1; % "Stress Lv. 2" in the second term of ∂Λ/∂b(equation 4)"
for j = 1:(numel(th2(i, :)))
N = numel(th2(i,j));
d2 = N/th2(i+1,j)*(th2(i+2,j)/A*exp(-(phi/th2(i,j)+b/th2(i+1,j))))^B+d2;
end

i = 1; % "Stress Lv. 3" in the second term of ∂Λ/∂b(equation 4)
for j = 1:(numel(th3(i, :)))
N = numel(th3(i,j));
d3 = N/th3(i+1,j)*(th3(i+2,j)/A*exp(-(phi/th3(i,j)+b/th3(i+1,j))))^B+d3;
end

N_d1 = 0;
N_d2 = 0;
N_d3 = 0;

i = 1;
for j = 1:(numel(th1(i, :)))
N_d1 = numel(th1(i,j))/th1(i+1,j)+N_d1; % "Stress Lv. 1" in the first term of ∂Λ/∂b(equation 4)
end

i = 1;
for j = 1:(numel(th2(i, :)))
N_d2 = numel(th2(i,j))/th2(i+1,j)+N_d2; % "Stress Lv. 2" in the first term of ∂Λ/∂b(equation 4)
end

i = 1;
for j = 1:(numel(th3(i, :)))
N_d3 = numel(th3(i,j))/th3(i+1,j)+N_d3; % "Stress Lv. 3" in the first term of ∂Λ/∂b(equation 4)
end

N_d = N_d1 + N_d2 + N_d3; % ∂Λ/∂b first term

four = -B*(N_d – (d1 + d2 + d3));
% vpa(four,3);

[B, A, phi, b] = solve(one,two,three,four,’Real’,true)

% eqns = [one == 0, two == 0, three == 0, four == 0];
% vars = [B, A, phi, b];
% [solB, solA, solphi, solb] = solve(eqns, vars, ‘Real’, true) symbolic, emptysym, 0-by-1, temperature-humidity, partialdifferentiation MATLAB Answers — New Questions

​

I am creating an app that will display plots on top of an image in a UIAxes object. The issue I’m running into is that once I load the image into UIAxes using imshow, the X and Y Grid lines are covered by the image. If I increase the transparency of the image while debugging I can see that they are still there, but I haven’t been able to figure out how to move them in front of the image.
Data I plot will appear in front of the image, so I feel like there is a layering issue I can’t figure out. I created a quick test app to demonstrate.

The code that creates that updates UIAxes when the button is pressed is:
% Button pushed function: Button
function ButtonPushed(app, event)
%Load Test Image
rgbImage = imread("peppers.png");
%Flip Image so it displays properly
rgbImage = flip(rgbImage);

hold(app.UIAxes,"on");
%Display image in test axes
hImage = imshow(rgbImage, ‘XData’, [0.0, 1.0], ‘YData’, [0.0, 1.0],’Parent’, app.UIAxes);

%return axes to normal position (imshow reverses the direction
%automatically when called)
app.UIAxes.YDir = ‘normal’;

%Turn the axes back on
app.UIAxes.XGrid = ‘on’;
app.UIAxes.YGrid = ‘on’;

%Plot test data to show it appears on top of the image
xData = [0.4,0.6,0.6,0.4];
yData = [0.4,0.6,0.6,0.4];
plot(xData,yData,’parent’,app.UIAxes);
end
Thanks for your help!I am creating an app that will display plots on top of an image in a UIAxes object. The issue I’m running into is that once I load the image into UIAxes using imshow, the X and Y Grid lines are covered by the image. If I increase the transparency of the image while debugging I can see that they are still there, but I haven’t been able to figure out how to move them in front of the image.
Data I plot will appear in front of the image, so I feel like there is a layering issue I can’t figure out. I created a quick test app to demonstrate.

The code that creates that updates UIAxes when the button is pressed is:
% Button pushed function: Button
function ButtonPushed(app, event)
%Load Test Image
rgbImage = imread("peppers.png");
%Flip Image so it displays properly
rgbImage = flip(rgbImage);

hold(app.UIAxes,"on");
%Display image in test axes
hImage = imshow(rgbImage, ‘XData’, [0.0, 1.0], ‘YData’, [0.0, 1.0],’Parent’, app.UIAxes);

%return axes to normal position (imshow reverses the direction
%automatically when called)
app.UIAxes.YDir = ‘normal’;

%Turn the axes back on
app.UIAxes.XGrid = ‘on’;
app.UIAxes.YGrid = ‘on’;

%Plot test data to show it appears on top of the image
xData = [0.4,0.6,0.6,0.4];
yData = [0.4,0.6,0.6,0.4];
plot(xData,yData,’parent’,app.UIAxes);
end
Thanks for your help! I am creating an app that will display plots on top of an image in a UIAxes object. The issue I’m running into is that once I load the image into UIAxes using imshow, the X and Y Grid lines are covered by the image. If I increase the transparency of the image while debugging I can see that they are still there, but I haven’t been able to figure out how to move them in front of the image.
Data I plot will appear in front of the image, so I feel like there is a layering issue I can’t figure out. I created a quick test app to demonstrate.

The code that creates that updates UIAxes when the button is pressed is:
% Button pushed function: Button
function ButtonPushed(app, event)
%Load Test Image
rgbImage = imread("peppers.png");
%Flip Image so it displays properly
rgbImage = flip(rgbImage);

hold(app.UIAxes,"on");
%Display image in test axes
hImage = imshow(rgbImage, ‘XData’, [0.0, 1.0], ‘YData’, [0.0, 1.0],’Parent’, app.UIAxes);

%return axes to normal position (imshow reverses the direction
%automatically when called)
app.UIAxes.YDir = ‘normal’;

%Turn the axes back on
app.UIAxes.XGrid = ‘on’;
app.UIAxes.YGrid = ‘on’;

%Plot test data to show it appears on top of the image
xData = [0.4,0.6,0.6,0.4];
yData = [0.4,0.6,0.6,0.4];
plot(xData,yData,’parent’,app.UIAxes);
end
Thanks for your help! appdesigner, imshow, uiaxes, layers MATLAB Answers — New Questions

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