Is it possible to change the color of a uitabgroup’s tabs as well as the font and font color? If not, is it possible to adjust the border color of the uitabgroup?
It is possible to change the background color of each tab’s pane, but I do not see a way to adjust the color or font/font color of the bar or the tab name elements on it. I am on R2023b.Is it possible to change the color of a uitabgroup’s tabs as well as the font and font color? If not, is it possible to adjust the border color of the uitabgroup?
It is possible to change the background color of each tab’s pane, but I do not see a way to adjust the color or font/font color of the bar or the tab name elements on it. I am on R2023b. Is it possible to change the color of a uitabgroup’s tabs as well as the font and font color? If not, is it possible to adjust the border color of the uitabgroup?
It is possible to change the background color of each tab’s pane, but I do not see a way to adjust the color or font/font color of the bar or the tab name elements on it. I am on R2023b. uitabgroup MATLAB Answers — New Questions

​

I have five app.Tabs.
First Tab is named app.Cooling.
Content of this app.Cooling:
1) picture of cooling system
2) app. fields with temperature value
3) plots of 4 diagrams (app.UIAxes) and legends
This app.Cooling I want to save as image (.png).
I try exportgraphics, saveas etc., but it doesn’t work.
Schema of my example:

How I can save this whole tab as image?I have five app.Tabs.
First Tab is named app.Cooling.
Content of this app.Cooling:
1) picture of cooling system
2) app. fields with temperature value
3) plots of 4 diagrams (app.UIAxes) and legends
This app.Cooling I want to save as image (.png).
I try exportgraphics, saveas etc., but it doesn’t work.
Schema of my example:

How I can save this whole tab as image? I have five app.Tabs.
First Tab is named app.Cooling.
Content of this app.Cooling:
1) picture of cooling system
2) app. fields with temperature value
3) plots of 4 diagrams (app.UIAxes) and legends
This app.Cooling I want to save as image (.png).
I try exportgraphics, saveas etc., but it doesn’t work.
Schema of my example:

How I can save this whole tab as image? image, app designer, save, tab MATLAB Answers — New Questions

​

I am trying to select these ‘flat’ peaks from a set of data and as I need the range of index values I am using islocalmax(data,’FlatSelection’,’all’) and for most of my data it does a good job as shown in the picture above but others end up looking like this.

Was wondering if there is a way to fix this? I have tried altering things like minprominence and such.I am trying to select these ‘flat’ peaks from a set of data and as I need the range of index values I am using islocalmax(data,’FlatSelection’,’all’) and for most of my data it does a good job as shown in the picture above but others end up looking like this.

Was wondering if there is a way to fix this? I have tried altering things like minprominence and such. I am trying to select these ‘flat’ peaks from a set of data and as I need the range of index values I am using islocalmax(data,’FlatSelection’,’all’) and for most of my data it does a good job as shown in the picture above but others end up looking like this.

Was wondering if there is a way to fix this? I have tried altering things like minprominence and such. matlab, graph MATLAB Answers — New Questions

​

I have an app that I developed in App Designer which performs optimization with the aid of the Matlab function lsqnonlin. A snippet:

methods (access = private)
function x=doOptimize(app,x0,lb,ub,options)
f=@(xx)Fun(app,xx);
x=lsqnonlin(f,x0,lb,ub,options);
end
function ff=Fun(app,x)
ff=someFunction(app,x);
end
end

The “someFunction” is a very complex function which utilizes many app properties and methods. The app works fine. However, I would like to take advantage of parallel processing by setting options.UseParallel = true. Doing this causes all kinds of warnings and errors, evidently in accordance with previous posts that stated that the save and load performed by the parallel workers are incompatible with app objects. My question is this: Is there any work-around for this problem? Of course, if app could be excluded altogether from the function Fun, then access=Static can be used. But in this case app is completely intertwined in the calculations. On the other hand, for the parallel processing, why should it be necessary to save the app variable? Since app-related variables only appear in someFunction, would it be possible to remove all references to app in the snippet, define access as Static, and pass app to someFunction as a global variable? Could this trick the parallel workers into thinking that there are no app objects around? (I’m willing to try anything.)I have an app that I developed in App Designer which performs optimization with the aid of the Matlab function lsqnonlin. A snippet:

methods (access = private)
function x=doOptimize(app,x0,lb,ub,options)
f=@(xx)Fun(app,xx);
x=lsqnonlin(f,x0,lb,ub,options);
end
function ff=Fun(app,x)
ff=someFunction(app,x);
end
end

The “someFunction” is a very complex function which utilizes many app properties and methods. The app works fine. However, I would like to take advantage of parallel processing by setting options.UseParallel = true. Doing this causes all kinds of warnings and errors, evidently in accordance with previous posts that stated that the save and load performed by the parallel workers are incompatible with app objects. My question is this: Is there any work-around for this problem? Of course, if app could be excluded altogether from the function Fun, then access=Static can be used. But in this case app is completely intertwined in the calculations. On the other hand, for the parallel processing, why should it be necessary to save the app variable? Since app-related variables only appear in someFunction, would it be possible to remove all references to app in the snippet, define access as Static, and pass app to someFunction as a global variable? Could this trick the parallel workers into thinking that there are no app objects around? (I’m willing to try anything.) I have an app that I developed in App Designer which performs optimization with the aid of the Matlab function lsqnonlin. A snippet:

methods (access = private)
function x=doOptimize(app,x0,lb,ub,options)
f=@(xx)Fun(app,xx);
x=lsqnonlin(f,x0,lb,ub,options);
end
function ff=Fun(app,x)
ff=someFunction(app,x);
end
end

The “someFunction” is a very complex function which utilizes many app properties and methods. The app works fine. However, I would like to take advantage of parallel processing by setting options.UseParallel = true. Doing this causes all kinds of warnings and errors, evidently in accordance with previous posts that stated that the save and load performed by the parallel workers are incompatible with app objects. My question is this: Is there any work-around for this problem? Of course, if app could be excluded altogether from the function Fun, then access=Static can be used. But in this case app is completely intertwined in the calculations. On the other hand, for the parallel processing, why should it be necessary to save the app variable? Since app-related variables only appear in someFunction, would it be possible to remove all references to app in the snippet, define access as Static, and pass app to someFunction as a global variable? Could this trick the parallel workers into thinking that there are no app objects around? (I’m willing to try anything.) app designer, parallel processing, optimization MATLAB Answers — New Questions

​

WebViews exported from R2013B work fine in Chrome for me. WebViews exported from R2014 display a blank screen. Using the debugging console, I see a javascript error:

Uncaught Error: declare: base class is not a callable constructor. webview.js:558

This is not the well-known local file reference problem (obviously, or the webview.js file would not have been loaded). The same exported WebView displays properly in Firefox and IE11. I need this to work in Chrome, as I have a project where the WebView is supposed to be displayed from inside an application using Chromium Embedded Framework. (I’m seeing the same behavior in the normal Desktop Chrome app and in CEF3). Any ideas what is wrong?

Thanks!WebViews exported from R2013B work fine in Chrome for me. WebViews exported from R2014 display a blank screen. Using the debugging console, I see a javascript error:

Uncaught Error: declare: base class is not a callable constructor. webview.js:558

This is not the well-known local file reference problem (obviously, or the webview.js file would not have been loaded). The same exported WebView displays properly in Firefox and IE11. I need this to work in Chrome, as I have a project where the WebView is supposed to be displayed from inside an application using Chromium Embedded Framework. (I’m seeing the same behavior in the normal Desktop Chrome app and in CEF3). Any ideas what is wrong?

Thanks! WebViews exported from R2013B work fine in Chrome for me. WebViews exported from R2014 display a blank screen. Using the debugging console, I see a javascript error:

Uncaught Error: declare: base class is not a callable constructor. webview.js:558

This is not the well-known local file reference problem (obviously, or the webview.js file would not have been loaded). The same exported WebView displays properly in Firefox and IE11. I need this to work in Chrome, as I have a project where the WebView is supposed to be displayed from inside an application using Chromium Embedded Framework. (I’m seeing the same behavior in the normal Desktop Chrome app and in CEF3). Any ideas what is wrong?

Thanks! simulink, webview, chrome MATLAB Answers — New Questions

​

Some icons in the Simulink Object Browser are not the correct icons. When the incorrect icon is dragged into a new model window it changes into the correct icon. Specifically the buffer icon in DSP System Toolbox>System Management>Buffers. The icon in the library window shows an icon containing the text ‘-40 db’ but when it is dragged into a new model window it changes into a buffer icon.Some icons in the Simulink Object Browser are not the correct icons. When the incorrect icon is dragged into a new model window it changes into the correct icon. Specifically the buffer icon in DSP System Toolbox>System Management>Buffers. The icon in the library window shows an icon containing the text ‘-40 db’ but when it is dragged into a new model window it changes into a buffer icon. Some icons in the Simulink Object Browser are not the correct icons. When the incorrect icon is dragged into a new model window it changes into the correct icon. Specifically the buffer icon in DSP System Toolbox>System Management>Buffers. The icon in the library window shows an icon containing the text ‘-40 db’ but when it is dragged into a new model window it changes into a buffer icon. simulink MATLAB Answers — New Questions

​

Hello.
I have 3 arrays of scalar number. I want to have some simple calculation on all numbers of these 3 arrays one by one (Permutation of all arrays numbers ), then multiplying every result (y) with rand(4,20) and finally saving every result as a excel data. I wrote the below codes but I have some problem: "I see in the output only the last multipication result of y" and therefore just last w.

clc
clear all

i=0;
j=1;
n=1;

A = [10 20 30 40 50 10];
B = [1 1.2 1.4 1.6 1.8 1];
C = [5 25 45 65 85 5];
z= rand(4,20);

for a = 1:125
i=i+1;
if i==6
j=j+1;
i=1;
end

if j==6
n=n+1;
j=1;
end

if(n==6)
n=1;
break ;
end

A_2 = A(i);
B_2 = B(j);
C_2 = C(n);

y = A_2 * B_2 * C_2 – A_2 * C_2;
w = y * z;
endHello.
I have 3 arrays of scalar number. I want to have some simple calculation on all numbers of these 3 arrays one by one (Permutation of all arrays numbers ), then multiplying every result (y) with rand(4,20) and finally saving every result as a excel data. I wrote the below codes but I have some problem: "I see in the output only the last multipication result of y" and therefore just last w.

clc
clear all

i=0;
j=1;
n=1;

A = [10 20 30 40 50 10];
B = [1 1.2 1.4 1.6 1.8 1];
C = [5 25 45 65 85 5];
z= rand(4,20);

for a = 1:125
i=i+1;
if i==6
j=j+1;
i=1;
end

if j==6
n=n+1;
j=1;
end

if(n==6)
n=1;
break ;
end

A_2 = A(i);
B_2 = B(j);
C_2 = C(n);

y = A_2 * B_2 * C_2 – A_2 * C_2;
w = y * z;
end Hello.
I have 3 arrays of scalar number. I want to have some simple calculation on all numbers of these 3 arrays one by one (Permutation of all arrays numbers ), then multiplying every result (y) with rand(4,20) and finally saving every result as a excel data. I wrote the below codes but I have some problem: "I see in the output only the last multipication result of y" and therefore just last w.

clc
clear all

i=0;
j=1;
n=1;

A = [10 20 30 40 50 10];
B = [1 1.2 1.4 1.6 1.8 1];
C = [5 25 45 65 85 5];
z= rand(4,20);

for a = 1:125
i=i+1;
if i==6
j=j+1;
i=1;
end

if j==6
n=n+1;
j=1;
end

if(n==6)
n=1;
break ;
end

A_2 = A(i);
B_2 = B(j);
C_2 = C(n);

y = A_2 * B_2 * C_2 – A_2 * C_2;
w = y * z;
end for loop, excel MATLAB Answers — New Questions

​

I’m trying to do neural network with 2 hidden layers and one neuron in the output layer without any toolboxes and just with matrix and vectors multiplications. To do this, I created fictional simple data as below to help me in this task:
%Data
x = 1:1000;
y1 = sind(x);
y2 = sind(x+30);
y3 = cosd(x);
y4 = cosd(x+30);
y5 = cosd(x+45);
% y6 will be the desired output data taht I would like my neural network
% try to predict
y6 = (y1 + y2 + y3 + y4 + y5);
Then, I coded as I thought to be be the right way, but my neural network can´t reach a good result, as below:

My doubt is if the result isn´t good because my implementation isn´t right or because I need to add more mechanisms im my neural network (like momentum, regularization and etc.) ?
I will post my code below sorry about the naem of some variables, but originally I wrote this code in portuguese. I will comment the code to help undestand it
%Nueral network achictecture
n_h1 = 10;
n_h2 = 11;
n_out = 1;

%Adjustable parameters
w1 = rand(5,n_h1);
b1 = ones(1,n_h1)*rand(1,1);
w2 = rand(n_h1,n_h2);
b2 = ones(1,n_h2)*rand(1,1);
w_out = rand(n_h2,n_out);
b_out = ones(1,n_out)*rand(1,1);
sig_a = 1;
learning_rate = 0.001;
limiar = 0.002;

%Helpful variables
max_epocas = 1000;
conj_entrada = [y1;y2;y3;y4;y5];
erros_epoca = [];

%Backpropagation

for epoch = 1:max_epocas

for i = 1:size(conj_entrada,2)
if i ==1
soma = 0;

else

end
enter = conj_entrada(:,i);

h1_in = [w1;b1]’*[enter;1];
h1_out = sig(h1_in,sig_a,’False’);

h2_in = [w2;b2]’*[h1_out;1];
h2_out = sig(h2_in,sig_a,’False’);

saida_in = [w_out;b_out]’*[h2_out;1];
saida_out = saida_in;

erro = y6(i) – saida_out;
soma = soma + (erro^2);

%Here starts the part of the code where the gradients are being
%calculated. Note that, here, I tried to folllow the chain rule.

%let me try to help in the understanding. Saida in portuguese is
%like output in english so when you read ,for example,
%d_erro_d_saida_out you need to know that this is the derivative of
%the error in relation with the output of the output layer. In the
%same way, entrada means input and pesos means weights

%output layer
%chain rule

d_erro_d_saida_out = -1*erro;
d_saida_d_entrada_out = 1; %linear
grad_saida = erro*d_saida_d_entrada_out;
d_entrada_d_pesos_out = h2_out;

d_erro_d_pesos_out = d_erro_d_saida_out*d_saida_d_entrada_out*d_entrada_d_pesos_out;

% Update the wights and bias
w_out = w_out -learning_rate*d_erro_d_pesos_out;
b_out = b_out -learning_rate*d_erro_d_saida_out*d_saida_d_entrada_out;

%Second hidden layer (The neighbor layer of the output layer)
%chain rule

d_erro_d_saida_h2 = -1*w_out*grad_saida;
d_saida_d_entrada_h2 = sig(h2_in,sig_a,’True’);
grad_h2 = sum(grad_saida)*d_saida_d_entrada_h2;
d_entrada_d_pesos_h2 = h1_out;

d_erro_d_pesos_h2 = d_entrada_d_pesos_h2*grad_h2′;

% Update the wights and bias
w2 = w2 -1*learning_rate*d_erro_d_pesos_h2;
b2 = b2 -1*learning_rate*sum(d_erro_d_saida_h2.*d_saida_d_entrada_h2,1);

%First hidden layer (The neighbor layer of the seccond hidden layer)
%chain rule

d_erro_d_saida_h1 = -1*w2*grad_h2;
d_saida_d_entrada_h1 = sig(h1_in,sig_a,’True’);
grad_h1 = sum(grad_h2)*d_saida_d_entrada_h1; %então daqui, tem que sair um 3×1
d_entrada_d_pesos_h1 = enter;

d_erro_d_pesos_h1 = d_entrada_d_pesos_h1*grad_h1′; %a segunda variável tem que resultar em um 1×3

% Update the wights and bias
w1 = w1 -1*learning_rate*d_erro_d_pesos_h1;
b1 = b1 -1*learning_rate*sum(d_erro_d_saida_h1.*d_saida_d_entrada_h1,1);
end
erro_atual = (soma/(2*size(x,2)));
erros_epoca = [erros_epoca;erro_atual];
if erros_epoca(epoch) <limiar
break
else
end
end

%testing the output of neural network
vetor_teste = 1:1000;
resposta_teste = zeros(1,size(vetor_teste,2));
for i = 1:size(vetor_teste,2)
enter_teste = conj_entrada(:,i);

h1_in_teste = [w1;b1]’*[enter_teste;1];
h1_out_teste = sig(h1_in_teste,sig_a,’False’);

h2_in_teste = [w2;b2]’*[h1_out_teste;1];
h2_out_teste = sig(h2_in_teste,sig_a,’False’);

saida_in_teste = [w_out;b_out]’*[h2_out_teste;1];
saida_out_teste = saida_in_teste; % a função de saída é linear;

resposta_teste(i) = saida_out_teste;
end
plot(1:size(erros_epoca,1),erros_epoca);
% plot(x,y3,’b’,vetor_teste,resposta_teste,’r’);
The code of my sigmoid activation function is below:
function [vetor_saida] = sig(vetor_entrada, const1, derivative)
if strcmp(derivative, ‘False’) == 1
vetor_saida = 1 ./ (1 + exp(-const1 * vetor_entrada));
else
sig_value = sig(vetor_entrada, const1, ‘False’);
vetor_saida = const1 * sig_value .* (1 – sig_value);
end
endI’m trying to do neural network with 2 hidden layers and one neuron in the output layer without any toolboxes and just with matrix and vectors multiplications. To do this, I created fictional simple data as below to help me in this task:
%Data
x = 1:1000;
y1 = sind(x);
y2 = sind(x+30);
y3 = cosd(x);
y4 = cosd(x+30);
y5 = cosd(x+45);
% y6 will be the desired output data taht I would like my neural network
% try to predict
y6 = (y1 + y2 + y3 + y4 + y5);
Then, I coded as I thought to be be the right way, but my neural network can´t reach a good result, as below:

My doubt is if the result isn´t good because my implementation isn´t right or because I need to add more mechanisms im my neural network (like momentum, regularization and etc.) ?
I will post my code below sorry about the naem of some variables, but originally I wrote this code in portuguese. I will comment the code to help undestand it
%Nueral network achictecture
n_h1 = 10;
n_h2 = 11;
n_out = 1;

%Adjustable parameters
w1 = rand(5,n_h1);
b1 = ones(1,n_h1)*rand(1,1);
w2 = rand(n_h1,n_h2);
b2 = ones(1,n_h2)*rand(1,1);
w_out = rand(n_h2,n_out);
b_out = ones(1,n_out)*rand(1,1);
sig_a = 1;
learning_rate = 0.001;
limiar = 0.002;

%Helpful variables
max_epocas = 1000;
conj_entrada = [y1;y2;y3;y4;y5];
erros_epoca = [];

%Backpropagation

for epoch = 1:max_epocas

for i = 1:size(conj_entrada,2)
if i ==1
soma = 0;

else

end
enter = conj_entrada(:,i);

h1_in = [w1;b1]’*[enter;1];
h1_out = sig(h1_in,sig_a,’False’);

h2_in = [w2;b2]’*[h1_out;1];
h2_out = sig(h2_in,sig_a,’False’);

saida_in = [w_out;b_out]’*[h2_out;1];
saida_out = saida_in;

erro = y6(i) – saida_out;
soma = soma + (erro^2);

%Here starts the part of the code where the gradients are being
%calculated. Note that, here, I tried to folllow the chain rule.

%let me try to help in the understanding. Saida in portuguese is
%like output in english so when you read ,for example,
%d_erro_d_saida_out you need to know that this is the derivative of
%the error in relation with the output of the output layer. In the
%same way, entrada means input and pesos means weights

%output layer
%chain rule

d_erro_d_saida_out = -1*erro;
d_saida_d_entrada_out = 1; %linear
grad_saida = erro*d_saida_d_entrada_out;
d_entrada_d_pesos_out = h2_out;

d_erro_d_pesos_out = d_erro_d_saida_out*d_saida_d_entrada_out*d_entrada_d_pesos_out;

% Update the wights and bias
w_out = w_out -learning_rate*d_erro_d_pesos_out;
b_out = b_out -learning_rate*d_erro_d_saida_out*d_saida_d_entrada_out;

%Second hidden layer (The neighbor layer of the output layer)
%chain rule

d_erro_d_saida_h2 = -1*w_out*grad_saida;
d_saida_d_entrada_h2 = sig(h2_in,sig_a,’True’);
grad_h2 = sum(grad_saida)*d_saida_d_entrada_h2;
d_entrada_d_pesos_h2 = h1_out;

d_erro_d_pesos_h2 = d_entrada_d_pesos_h2*grad_h2′;

% Update the wights and bias
w2 = w2 -1*learning_rate*d_erro_d_pesos_h2;
b2 = b2 -1*learning_rate*sum(d_erro_d_saida_h2.*d_saida_d_entrada_h2,1);

%First hidden layer (The neighbor layer of the seccond hidden layer)
%chain rule

d_erro_d_saida_h1 = -1*w2*grad_h2;
d_saida_d_entrada_h1 = sig(h1_in,sig_a,’True’);
grad_h1 = sum(grad_h2)*d_saida_d_entrada_h1; %então daqui, tem que sair um 3×1
d_entrada_d_pesos_h1 = enter;

d_erro_d_pesos_h1 = d_entrada_d_pesos_h1*grad_h1′; %a segunda variável tem que resultar em um 1×3

% Update the wights and bias
w1 = w1 -1*learning_rate*d_erro_d_pesos_h1;
b1 = b1 -1*learning_rate*sum(d_erro_d_saida_h1.*d_saida_d_entrada_h1,1);
end
erro_atual = (soma/(2*size(x,2)));
erros_epoca = [erros_epoca;erro_atual];
if erros_epoca(epoch) <limiar
break
else
end
end

%testing the output of neural network
vetor_teste = 1:1000;
resposta_teste = zeros(1,size(vetor_teste,2));
for i = 1:size(vetor_teste,2)
enter_teste = conj_entrada(:,i);

h1_in_teste = [w1;b1]’*[enter_teste;1];
h1_out_teste = sig(h1_in_teste,sig_a,’False’);

h2_in_teste = [w2;b2]’*[h1_out_teste;1];
h2_out_teste = sig(h2_in_teste,sig_a,’False’);

saida_in_teste = [w_out;b_out]’*[h2_out_teste;1];
saida_out_teste = saida_in_teste; % a função de saída é linear;

resposta_teste(i) = saida_out_teste;
end
plot(1:size(erros_epoca,1),erros_epoca);
% plot(x,y3,’b’,vetor_teste,resposta_teste,’r’);
The code of my sigmoid activation function is below:
function [vetor_saida] = sig(vetor_entrada, const1, derivative)
if strcmp(derivative, ‘False’) == 1
vetor_saida = 1 ./ (1 + exp(-const1 * vetor_entrada));
else
sig_value = sig(vetor_entrada, const1, ‘False’);
vetor_saida = const1 * sig_value .* (1 – sig_value);
end
end I’m trying to do neural network with 2 hidden layers and one neuron in the output layer without any toolboxes and just with matrix and vectors multiplications. To do this, I created fictional simple data as below to help me in this task:
%Data
x = 1:1000;
y1 = sind(x);
y2 = sind(x+30);
y3 = cosd(x);
y4 = cosd(x+30);
y5 = cosd(x+45);
% y6 will be the desired output data taht I would like my neural network
% try to predict
y6 = (y1 + y2 + y3 + y4 + y5);
Then, I coded as I thought to be be the right way, but my neural network can´t reach a good result, as below:

My doubt is if the result isn´t good because my implementation isn´t right or because I need to add more mechanisms im my neural network (like momentum, regularization and etc.) ?
I will post my code below sorry about the naem of some variables, but originally I wrote this code in portuguese. I will comment the code to help undestand it
%Nueral network achictecture
n_h1 = 10;
n_h2 = 11;
n_out = 1;

%Adjustable parameters
w1 = rand(5,n_h1);
b1 = ones(1,n_h1)*rand(1,1);
w2 = rand(n_h1,n_h2);
b2 = ones(1,n_h2)*rand(1,1);
w_out = rand(n_h2,n_out);
b_out = ones(1,n_out)*rand(1,1);
sig_a = 1;
learning_rate = 0.001;
limiar = 0.002;

%Helpful variables
max_epocas = 1000;
conj_entrada = [y1;y2;y3;y4;y5];
erros_epoca = [];

%Backpropagation

for epoch = 1:max_epocas

for i = 1:size(conj_entrada,2)
if i ==1
soma = 0;

else

end
enter = conj_entrada(:,i);

h1_in = [w1;b1]’*[enter;1];
h1_out = sig(h1_in,sig_a,’False’);

h2_in = [w2;b2]’*[h1_out;1];
h2_out = sig(h2_in,sig_a,’False’);

saida_in = [w_out;b_out]’*[h2_out;1];
saida_out = saida_in;

erro = y6(i) – saida_out;
soma = soma + (erro^2);

%Here starts the part of the code where the gradients are being
%calculated. Note that, here, I tried to folllow the chain rule.

%let me try to help in the understanding. Saida in portuguese is
%like output in english so when you read ,for example,
%d_erro_d_saida_out you need to know that this is the derivative of
%the error in relation with the output of the output layer. In the
%same way, entrada means input and pesos means weights

%output layer
%chain rule

d_erro_d_saida_out = -1*erro;
d_saida_d_entrada_out = 1; %linear
grad_saida = erro*d_saida_d_entrada_out;
d_entrada_d_pesos_out = h2_out;

d_erro_d_pesos_out = d_erro_d_saida_out*d_saida_d_entrada_out*d_entrada_d_pesos_out;

% Update the wights and bias
w_out = w_out -learning_rate*d_erro_d_pesos_out;
b_out = b_out -learning_rate*d_erro_d_saida_out*d_saida_d_entrada_out;

%Second hidden layer (The neighbor layer of the output layer)
%chain rule

d_erro_d_saida_h2 = -1*w_out*grad_saida;
d_saida_d_entrada_h2 = sig(h2_in,sig_a,’True’);
grad_h2 = sum(grad_saida)*d_saida_d_entrada_h2;
d_entrada_d_pesos_h2 = h1_out;

d_erro_d_pesos_h2 = d_entrada_d_pesos_h2*grad_h2′;

% Update the wights and bias
w2 = w2 -1*learning_rate*d_erro_d_pesos_h2;
b2 = b2 -1*learning_rate*sum(d_erro_d_saida_h2.*d_saida_d_entrada_h2,1);

%First hidden layer (The neighbor layer of the seccond hidden layer)
%chain rule

d_erro_d_saida_h1 = -1*w2*grad_h2;
d_saida_d_entrada_h1 = sig(h1_in,sig_a,’True’);
grad_h1 = sum(grad_h2)*d_saida_d_entrada_h1; %então daqui, tem que sair um 3×1
d_entrada_d_pesos_h1 = enter;

d_erro_d_pesos_h1 = d_entrada_d_pesos_h1*grad_h1′; %a segunda variável tem que resultar em um 1×3

% Update the wights and bias
w1 = w1 -1*learning_rate*d_erro_d_pesos_h1;
b1 = b1 -1*learning_rate*sum(d_erro_d_saida_h1.*d_saida_d_entrada_h1,1);
end
erro_atual = (soma/(2*size(x,2)));
erros_epoca = [erros_epoca;erro_atual];
if erros_epoca(epoch) <limiar
break
else
end
end

%testing the output of neural network
vetor_teste = 1:1000;
resposta_teste = zeros(1,size(vetor_teste,2));
for i = 1:size(vetor_teste,2)
enter_teste = conj_entrada(:,i);

h1_in_teste = [w1;b1]’*[enter_teste;1];
h1_out_teste = sig(h1_in_teste,sig_a,’False’);

h2_in_teste = [w2;b2]’*[h1_out_teste;1];
h2_out_teste = sig(h2_in_teste,sig_a,’False’);

saida_in_teste = [w_out;b_out]’*[h2_out_teste;1];
saida_out_teste = saida_in_teste; % a função de saída é linear;

resposta_teste(i) = saida_out_teste;
end
plot(1:size(erros_epoca,1),erros_epoca);
% plot(x,y3,’b’,vetor_teste,resposta_teste,’r’);
The code of my sigmoid activation function is below:
function [vetor_saida] = sig(vetor_entrada, const1, derivative)
if strcmp(derivative, ‘False’) == 1
vetor_saida = 1 ./ (1 + exp(-const1 * vetor_entrada));
else
sig_value = sig(vetor_entrada, const1, ‘False’);
vetor_saida = const1 * sig_value .* (1 – sig_value);
end
end neural network, backpropagation MATLAB Answers — New Questions

​

As I asked, I want to design a fractional pid controller for a model. Unfortunately, the results I got with step in matlab are quite different from those in simulink
.
Here the orange line is the output and the blue line is the offset,Now let me show you the situation of matlab.The specific parameters of the controller are consistent with the code below
cs=fopid(9.503084707494814e+03,1.889631254823942e+02,2.304958553299564e+02,0.058440555124293,1.558103446462550);%This is parameter of controller
gs=tf(118700,[1 1085 1224000]);%This is the transfer function of the controlled object
step(feedback(cs*gs,1))%Step Response of Closed-loop Transfer Function

This is the general situation. I feel confused now.As I asked, I want to design a fractional pid controller for a model. Unfortunately, the results I got with step in matlab are quite different from those in simulink
.
Here the orange line is the output and the blue line is the offset,Now let me show you the situation of matlab.The specific parameters of the controller are consistent with the code below
cs=fopid(9.503084707494814e+03,1.889631254823942e+02,2.304958553299564e+02,0.058440555124293,1.558103446462550);%This is parameter of controller
gs=tf(118700,[1 1085 1224000]);%This is the transfer function of the controlled object
step(feedback(cs*gs,1))%Step Response of Closed-loop Transfer Function

This is the general situation. I feel confused now. As I asked, I want to design a fractional pid controller for a model. Unfortunately, the results I got with step in matlab are quite different from those in simulink
.
Here the orange line is the output and the blue line is the offset,Now let me show you the situation of matlab.The specific parameters of the controller are consistent with the code below
cs=fopid(9.503084707494814e+03,1.889631254823942e+02,2.304958553299564e+02,0.058440555124293,1.558103446462550);%This is parameter of controller
gs=tf(118700,[1 1085 1224000]);%This is the transfer function of the controlled object
step(feedback(cs*gs,1))%Step Response of Closed-loop Transfer Function

This is the general situation. I feel confused now. matlab, simulink MATLAB Answers — New Questions

​

area = roc(fileName,exp)
%
% fileName: The name of the output file. It is an image
% exp: a structure representing an experiment with three fields
% – exp.name: a string specifying the experiment name
% – exp.positive_n: the number of positive samples in the experiment
% – exp.likelihoods: likelihoods of each sample where the first
% exp.positive_n number of likelihoods belongs to the positive
% samples while others belong to the negative samples.
% area: a 1×1-dimensional cell array of the area under the roc curve.

true positive,false positive values of ground truth,edge detected image was calculated…then how apply these values to roc program…??????area = roc(fileName,exp)
%
% fileName: The name of the output file. It is an image
% exp: a structure representing an experiment with three fields
% – exp.name: a string specifying the experiment name
% – exp.positive_n: the number of positive samples in the experiment
% – exp.likelihoods: likelihoods of each sample where the first
% exp.positive_n number of likelihoods belongs to the positive
% samples while others belong to the negative samples.
% area: a 1×1-dimensional cell array of the area under the roc curve.

true positive,false positive values of ground truth,edge detected image was calculated…then how apply these values to roc program…?????? area = roc(fileName,exp)
%
% fileName: The name of the output file. It is an image
% exp: a structure representing an experiment with three fields
% – exp.name: a string specifying the experiment name
% – exp.positive_n: the number of positive samples in the experiment
% – exp.likelihoods: likelihoods of each sample where the first
% exp.positive_n number of likelihoods belongs to the positive
% samples while others belong to the negative samples.
% area: a 1×1-dimensional cell array of the area under the roc curve.

true positive,false positive values of ground truth,edge detected image was calculated…then how apply these values to roc program…?????? roc MATLAB Answers — New Questions

​