Hi everybody!
I’m trying to solve a fitting curve problem concerning a double-exponential fit.
I have this data set (Data.txt) that behaves like a bi-exponential function following a resting condition. Therefore, since x<0 the signal is almost steady whereas, for x>0 the signal presents a first exponential phase (lasting about 20 sec) followed by a second exponential phase.
The problem is that the resulting fitting does not model the signal behavior in the correct way. Specifically, the first exp phase is too short and/or too low even though correct StartingPoint and Upper and Lower limits are set.
Lastly, I’ve written two algorithms with and without "Exclude" options. The second one presents a strange pattern at the onset of the second exp function.
I’ll appreciate any suggestions.

This is the code.

[x, y] = readvars(‘prova.txt’);

tRest = abs(min(x)); % Resting phase duration (sec)
cdp = 20; % estimated duration (sec) of CardioDynamicPhase (cdp)
cdpEpoch = 5; % epoch duration (sec) to compute mean value of amplitude of cdp
plot(x,y);

% Fitting StartingPoint
yRest = mean( y(1:tRest-1) ); % y value at resting condition before exp onset

A1start = mean( y(1:tRest+cdp-cdpEpoch):y(tRest+cdp) )-yRest; % mean y value of the last 5sec of the cardiodynamic phase
TD1start = 0; % starting value (sec) of the first exponential phase onset
tau1Start = 8; % starting value (sec) of the first exponential phase time constant

A2start = max(y)-A1start-yRest; % starting value of the second exponential phase amplitude
TD2start = 30; % starting value (sec) of the second exponential phase onset
tau2Start = 30; % starting value (sec) of the second exponential phase time constant

% Fitting from exponential onset
g = fittype( @(A1, TD1, tau1, A2, TD2, tau2, yRest, x) yRest+A1.*(1-exp(-(x-TD1)/tau1))+A2.*(1-exp(-(x-TD2)/tau2)), ‘problem’, ‘yRest’);
f1 = fit( x(tRest:size(x,1)), y(tRest:size(x,1)), g, ‘problem’, yRest, ‘StartPoint’, [A1start, TD1start, tau1Start, A2start, TD2start, tau2Start], ‘lower’ , [ A1start-100, 0, 5, A2start-200, 10, 10], ‘upper’, [ A1start+200, 20, 40, max(y), 60, 40], ‘MaxIter’, 10000);
pg =coeffvalues(f1);
A1 = pg(1);
TD1 = pg(2);
tau1 = pg(3);
A2 = pg(4);
TD2 = pg(5);
tau2 = pg(6);

yFit( 1:(tRest+floor(TD1)) ) = yRest;
yFit( tRest+floor(TD1)+1 : tRest+floor(TD2) ) = yRest + A1.*(1-exp(-( x(tRest+floor(TD1)+1 : tRest+floor(TD2))-TD1 )/tau1));
yFit( tRest+floor(TD2)+1 : size(y,1)) = yFit(tRest+floor(TD2))+ A2.*(1-exp(-( x(tRest+floor(TD2)+1 : size(y,1))-TD2 )/tau2));

% fitting using Exclude option
f2 = fit( x, y, g, ‘problem’, yRest, ‘StartPoint’, [A1start, TD1start, tau1Start, A2start, TD2start, tau2Start], ‘lower’ , [ 390, 0, 5, A2start-200, 10, 10], ‘upper’, [ 400, 20, 40, max(y), 60, 40], ‘Exclude’, x<0); %Same fitting algortihm that exclude negative x values
pg2=coeffvalues(f2);

A1e = pg2(1);
TD1e = pg2(2);
tau1e = pg2(3);
A2e = pg2(4);
TD2e = pg2(5);
tau2e = pg2(6);

yFite( 1:(tRest+floor(TD1e)) ) = yRest;
yFite( tRest+floor(TD1e)+1 : tRest+floor(TD2e) ) = yRest + A1e.*(1-exp(-( x(tRest+floor(TD1e)+1 : tRest+floor(TD2e))-TD1e )/tau1e));
yFite( tRest+floor(TD2e)+1 : size(y,1)) = yFit(tRest+floor(TD2e))+ A2e.*(1-exp(-( x(tRest+floor(TD2e)+1 : size(y,1))-TD2e )/tau2e));

% Plot the fitting curves
hold on
plot (x, yFit, ‘r’);
plot (x, yFite, ‘k’);Hi everybody!
I’m trying to solve a fitting curve problem concerning a double-exponential fit.
I have this data set (Data.txt) that behaves like a bi-exponential function following a resting condition. Therefore, since x<0 the signal is almost steady whereas, for x>0 the signal presents a first exponential phase (lasting about 20 sec) followed by a second exponential phase.
The problem is that the resulting fitting does not model the signal behavior in the correct way. Specifically, the first exp phase is too short and/or too low even though correct StartingPoint and Upper and Lower limits are set.
Lastly, I’ve written two algorithms with and without "Exclude" options. The second one presents a strange pattern at the onset of the second exp function.
I’ll appreciate any suggestions.

This is the code.

[x, y] = readvars(‘prova.txt’);

tRest = abs(min(x)); % Resting phase duration (sec)
cdp = 20; % estimated duration (sec) of CardioDynamicPhase (cdp)
cdpEpoch = 5; % epoch duration (sec) to compute mean value of amplitude of cdp
plot(x,y);

% Fitting StartingPoint
yRest = mean( y(1:tRest-1) ); % y value at resting condition before exp onset

A1start = mean( y(1:tRest+cdp-cdpEpoch):y(tRest+cdp) )-yRest; % mean y value of the last 5sec of the cardiodynamic phase
TD1start = 0; % starting value (sec) of the first exponential phase onset
tau1Start = 8; % starting value (sec) of the first exponential phase time constant

A2start = max(y)-A1start-yRest; % starting value of the second exponential phase amplitude
TD2start = 30; % starting value (sec) of the second exponential phase onset
tau2Start = 30; % starting value (sec) of the second exponential phase time constant

% Fitting from exponential onset
g = fittype( @(A1, TD1, tau1, A2, TD2, tau2, yRest, x) yRest+A1.*(1-exp(-(x-TD1)/tau1))+A2.*(1-exp(-(x-TD2)/tau2)), ‘problem’, ‘yRest’);
f1 = fit( x(tRest:size(x,1)), y(tRest:size(x,1)), g, ‘problem’, yRest, ‘StartPoint’, [A1start, TD1start, tau1Start, A2start, TD2start, tau2Start], ‘lower’ , [ A1start-100, 0, 5, A2start-200, 10, 10], ‘upper’, [ A1start+200, 20, 40, max(y), 60, 40], ‘MaxIter’, 10000);
pg =coeffvalues(f1);
A1 = pg(1);
TD1 = pg(2);
tau1 = pg(3);
A2 = pg(4);
TD2 = pg(5);
tau2 = pg(6);

yFit( 1:(tRest+floor(TD1)) ) = yRest;
yFit( tRest+floor(TD1)+1 : tRest+floor(TD2) ) = yRest + A1.*(1-exp(-( x(tRest+floor(TD1)+1 : tRest+floor(TD2))-TD1 )/tau1));
yFit( tRest+floor(TD2)+1 : size(y,1)) = yFit(tRest+floor(TD2))+ A2.*(1-exp(-( x(tRest+floor(TD2)+1 : size(y,1))-TD2 )/tau2));

% fitting using Exclude option
f2 = fit( x, y, g, ‘problem’, yRest, ‘StartPoint’, [A1start, TD1start, tau1Start, A2start, TD2start, tau2Start], ‘lower’ , [ 390, 0, 5, A2start-200, 10, 10], ‘upper’, [ 400, 20, 40, max(y), 60, 40], ‘Exclude’, x<0); %Same fitting algortihm that exclude negative x values
pg2=coeffvalues(f2);

A1e = pg2(1);
TD1e = pg2(2);
tau1e = pg2(3);
A2e = pg2(4);
TD2e = pg2(5);
tau2e = pg2(6);

yFite( 1:(tRest+floor(TD1e)) ) = yRest;
yFite( tRest+floor(TD1e)+1 : tRest+floor(TD2e) ) = yRest + A1e.*(1-exp(-( x(tRest+floor(TD1e)+1 : tRest+floor(TD2e))-TD1e )/tau1e));
yFite( tRest+floor(TD2e)+1 : size(y,1)) = yFit(tRest+floor(TD2e))+ A2e.*(1-exp(-( x(tRest+floor(TD2e)+1 : size(y,1))-TD2e )/tau2e));

% Plot the fitting curves
hold on
plot (x, yFit, ‘r’);
plot (x, yFite, ‘k’); Hi everybody!
I’m trying to solve a fitting curve problem concerning a double-exponential fit.
I have this data set (Data.txt) that behaves like a bi-exponential function following a resting condition. Therefore, since x<0 the signal is almost steady whereas, for x>0 the signal presents a first exponential phase (lasting about 20 sec) followed by a second exponential phase.
The problem is that the resulting fitting does not model the signal behavior in the correct way. Specifically, the first exp phase is too short and/or too low even though correct StartingPoint and Upper and Lower limits are set.
Lastly, I’ve written two algorithms with and without "Exclude" options. The second one presents a strange pattern at the onset of the second exp function.
I’ll appreciate any suggestions.

This is the code.

[x, y] = readvars(‘prova.txt’);

tRest = abs(min(x)); % Resting phase duration (sec)
cdp = 20; % estimated duration (sec) of CardioDynamicPhase (cdp)
cdpEpoch = 5; % epoch duration (sec) to compute mean value of amplitude of cdp
plot(x,y);

% Fitting StartingPoint
yRest = mean( y(1:tRest-1) ); % y value at resting condition before exp onset

A1start = mean( y(1:tRest+cdp-cdpEpoch):y(tRest+cdp) )-yRest; % mean y value of the last 5sec of the cardiodynamic phase
TD1start = 0; % starting value (sec) of the first exponential phase onset
tau1Start = 8; % starting value (sec) of the first exponential phase time constant

A2start = max(y)-A1start-yRest; % starting value of the second exponential phase amplitude
TD2start = 30; % starting value (sec) of the second exponential phase onset
tau2Start = 30; % starting value (sec) of the second exponential phase time constant

% Fitting from exponential onset
g = fittype( @(A1, TD1, tau1, A2, TD2, tau2, yRest, x) yRest+A1.*(1-exp(-(x-TD1)/tau1))+A2.*(1-exp(-(x-TD2)/tau2)), ‘problem’, ‘yRest’);
f1 = fit( x(tRest:size(x,1)), y(tRest:size(x,1)), g, ‘problem’, yRest, ‘StartPoint’, [A1start, TD1start, tau1Start, A2start, TD2start, tau2Start], ‘lower’ , [ A1start-100, 0, 5, A2start-200, 10, 10], ‘upper’, [ A1start+200, 20, 40, max(y), 60, 40], ‘MaxIter’, 10000);
pg =coeffvalues(f1);
A1 = pg(1);
TD1 = pg(2);
tau1 = pg(3);
A2 = pg(4);
TD2 = pg(5);
tau2 = pg(6);

yFit( 1:(tRest+floor(TD1)) ) = yRest;
yFit( tRest+floor(TD1)+1 : tRest+floor(TD2) ) = yRest + A1.*(1-exp(-( x(tRest+floor(TD1)+1 : tRest+floor(TD2))-TD1 )/tau1));
yFit( tRest+floor(TD2)+1 : size(y,1)) = yFit(tRest+floor(TD2))+ A2.*(1-exp(-( x(tRest+floor(TD2)+1 : size(y,1))-TD2 )/tau2));

% fitting using Exclude option
f2 = fit( x, y, g, ‘problem’, yRest, ‘StartPoint’, [A1start, TD1start, tau1Start, A2start, TD2start, tau2Start], ‘lower’ , [ 390, 0, 5, A2start-200, 10, 10], ‘upper’, [ 400, 20, 40, max(y), 60, 40], ‘Exclude’, x<0); %Same fitting algortihm that exclude negative x values
pg2=coeffvalues(f2);

A1e = pg2(1);
TD1e = pg2(2);
tau1e = pg2(3);
A2e = pg2(4);
TD2e = pg2(5);
tau2e = pg2(6);

yFite( 1:(tRest+floor(TD1e)) ) = yRest;
yFite( tRest+floor(TD1e)+1 : tRest+floor(TD2e) ) = yRest + A1e.*(1-exp(-( x(tRest+floor(TD1e)+1 : tRest+floor(TD2e))-TD1e )/tau1e));
yFite( tRest+floor(TD2e)+1 : size(y,1)) = yFit(tRest+floor(TD2e))+ A2e.*(1-exp(-( x(tRest+floor(TD2e)+1 : size(y,1))-TD2e )/tau2e));

% Plot the fitting curves
hold on
plot (x, yFit, ‘r’);
plot (x, yFite, ‘k’); signal fitting MATLAB Answers — New Questions

​

Error in port widths or dimensions. Output port 1 of ‘last/Subsystem/powergui/EquivalentModel1/Gates/From2’ is a one dimensional vector with 2 elements.

AND

Error in port widths or dimensions. Invalid dimension has been specified for input port 2 of ‘last/Subsystem/powergui/EquivalentModel1/Gates/Mux’.

I am using MATLAB 2010a. problem occurred in the switching circuit of invert.
Anticipating your answers…

<</matlabcentral/answers/uploaded_files/38534/1.png>>

<</matlabcentral/answers/uploaded_files/38535/2.png>>Error in port widths or dimensions. Output port 1 of ‘last/Subsystem/powergui/EquivalentModel1/Gates/From2’ is a one dimensional vector with 2 elements.

AND

Error in port widths or dimensions. Invalid dimension has been specified for input port 2 of ‘last/Subsystem/powergui/EquivalentModel1/Gates/Mux’.

I am using MATLAB 2010a. problem occurred in the switching circuit of invert.
Anticipating your answers…

<</matlabcentral/answers/uploaded_files/38534/1.png>>

<</matlabcentral/answers/uploaded_files/38535/2.png>> Error in port widths or dimensions. Output port 1 of ‘last/Subsystem/powergui/EquivalentModel1/Gates/From2’ is a one dimensional vector with 2 elements.

AND

Error in port widths or dimensions. Invalid dimension has been specified for input port 2 of ‘last/Subsystem/powergui/EquivalentModel1/Gates/Mux’.

I am using MATLAB 2010a. problem occurred in the switching circuit of invert.
Anticipating your answers…

<</matlabcentral/answers/uploaded_files/38534/1.png>>

<</matlabcentral/answers/uploaded_files/38535/2.png>>  MATLAB Answers — New Questions

​

Hello everyone,
I made a state space representation with disturbances using blocks in simulink, then I connected this state space to model predictive control, and when I run the system it gives an error as shown below: the red spot in the figure is where the error is,

and my state space model is as follow:

so A, B, C and D and Bd should be correct as I made them, right?Hello everyone,
I made a state space representation with disturbances using blocks in simulink, then I connected this state space to model predictive control, and when I run the system it gives an error as shown below: the red spot in the figure is where the error is,

and my state space model is as follow:

so A, B, C and D and Bd should be correct as I made them, right? Hello everyone,
I made a state space representation with disturbances using blocks in simulink, then I connected this state space to model predictive control, and when I run the system it gives an error as shown below: the red spot in the figure is where the error is,

and my state space model is as follow:

so A, B, C and D and Bd should be correct as I made them, right? matlab, simulink, model predictive controller, state space representation MATLAB Answers — New Questions

​

Hi i am looking to try calculate the latitude and longitude of a point on the ground with respect to an aircraft which has a lat lon but obviously at a specific height, a azmiuth and elevation look angle and also range to the point on ground ( all known ) .

[latout, lonout] = reckon(lat, lon, rng, az)

e.g from help section

dist = nm2deg(600) % convert nm distance to degrees

dist =
9.9933

pt1 = reckon(51.5,0,dist,315) % northwest is 315 degrees

pt1 =
57.8999 -13.3507

The reckon function seems to do near enough exactly what i want but does not take in account elevation angle ( i.e assumes both points are on the ground)

Is there any function similar that i can achieve being able to add in an elevation angle to the calculation, or change the reckon function in anyway?

Any help much appreciatedHi i am looking to try calculate the latitude and longitude of a point on the ground with respect to an aircraft which has a lat lon but obviously at a specific height, a azmiuth and elevation look angle and also range to the point on ground ( all known ) .

[latout, lonout] = reckon(lat, lon, rng, az)

e.g from help section

dist = nm2deg(600) % convert nm distance to degrees

dist =
9.9933

pt1 = reckon(51.5,0,dist,315) % northwest is 315 degrees

pt1 =
57.8999 -13.3507

The reckon function seems to do near enough exactly what i want but does not take in account elevation angle ( i.e assumes both points are on the ground)

Is there any function similar that i can achieve being able to add in an elevation angle to the calculation, or change the reckon function in anyway?

Any help much appreciated Hi i am looking to try calculate the latitude and longitude of a point on the ground with respect to an aircraft which has a lat lon but obviously at a specific height, a azmiuth and elevation look angle and also range to the point on ground ( all known ) .

[latout, lonout] = reckon(lat, lon, rng, az)

e.g from help section

dist = nm2deg(600) % convert nm distance to degrees

dist =
9.9933

pt1 = reckon(51.5,0,dist,315) % northwest is 315 degrees

pt1 =
57.8999 -13.3507

The reckon function seems to do near enough exactly what i want but does not take in account elevation angle ( i.e assumes both points are on the ground)

Is there any function similar that i can achieve being able to add in an elevation angle to the calculation, or change the reckon function in anyway?

Any help much appreciated reckon MATLAB Answers — New Questions

​

Hi, Simulink supporters team
I want to print my Entire model as pdf by MATLAB script instead of By Print Option in Simulnik Toolstrip. I have tried 2 ways:
1.Using print( ) function but it only print first page of my model
2. Using exportgraphics( ) but it does not support for Simulink model.
So, is there any way to solve this problem?Hi, Simulink supporters team
I want to print my Entire model as pdf by MATLAB script instead of By Print Option in Simulnik Toolstrip. I have tried 2 ways:
1.Using print( ) function but it only print first page of my model
2. Using exportgraphics( ) but it does not support for Simulink model.
So, is there any way to solve this problem? Hi, Simulink supporters team
I want to print my Entire model as pdf by MATLAB script instead of By Print Option in Simulnik Toolstrip. I have tried 2 ways:
1.Using print( ) function but it only print first page of my model
2. Using exportgraphics( ) but it does not support for Simulink model.
So, is there any way to solve this problem? print, model, matlab, command MATLAB Answers — New Questions

​

How can I get the ADSR sample points of a audio file…Like:

"Fake" example:

[Amp, SR] = audioread(audio.file);

[a d s r] = get_ADSR(Amp);

% Results: The sample position of the Amp vector:
a = 85 % End of Attack -> Attack start = Amp(0)
d = 300 % End of Decay
s = 2456 % End of Sustain
r = 50000 % End of ReleaseHow can I get the ADSR sample points of a audio file…Like:

"Fake" example:

[Amp, SR] = audioread(audio.file);

[a d s r] = get_ADSR(Amp);

% Results: The sample position of the Amp vector:
a = 85 % End of Attack -> Attack start = Amp(0)
d = 300 % End of Decay
s = 2456 % End of Sustain
r = 50000 % End of Release How can I get the ADSR sample points of a audio file…Like:

"Fake" example:

[Amp, SR] = audioread(audio.file);

[a d s r] = get_ADSR(Amp);

% Results: The sample position of the Amp vector:
a = 85 % End of Attack -> Attack start = Amp(0)
d = 300 % End of Decay
s = 2456 % End of Sustain
r = 50000 % End of Release envelope, adsr, audio, wave MATLAB Answers — New Questions

​

I’m now trying to setup the pixhawk 6X using UAV Toolbox Support Package for PX4 Autopilots (ver. 24.12).
I successfully finished the hardware setup.
Next, I created a simple simulink file as below.
I connected the hardware and tried to see the measurement values, but I could not find the values.
(The displayed values are zero.)
Please tell me how to solve this problem.I’m now trying to setup the pixhawk 6X using UAV Toolbox Support Package for PX4 Autopilots (ver. 24.12).
I successfully finished the hardware setup.
Next, I created a simple simulink file as below.
I connected the hardware and tried to see the measurement values, but I could not find the values.
(The displayed values are zero.)
Please tell me how to solve this problem. I’m now trying to setup the pixhawk 6X using UAV Toolbox Support Package for PX4 Autopilots (ver. 24.12).
I successfully finished the hardware setup.
Next, I created a simple simulink file as below.
I connected the hardware and tried to see the measurement values, but I could not find the values.
(The displayed values are zero.)
Please tell me how to solve this problem. pixhawk 6x, matlab, simulink, sensor, uav toolbox support package MATLAB Answers — New Questions

​

Hi
I’m studying neural networks and i’m doing a NN with 2 hidden layers and one neuron in the output layer
While I was studying and coding my NN, I faced one doubt.

In the backward step, the math behind this is clear:
to the output layer we have:
, where ⊙ is the hadamard product.
and to hidden layers, we have:

my problem is that, when I code this formulas, I need to change the second equation when I will calculate the the gradient of the first hidden layer (the hidden layer next to the input layer) to match the dimensions of the matrix as follow below:
% Backpropagation
delta_saida = erro_estim.*selecionar_funcao(saida_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h2 = (w_out’*delta_saida).*selecionar_funcao(h2_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h1 = (w2*delta_h2′)’.*selecionar_funcao(h1_in_estim,ativ_h1,sig_a,tanh_a,tanh_b,’True’);

%update weights and biases
w_out = w_out + learning_rate*delta_saida*h2_out_estim’;
b_out = b_out + learning_rate*delta_saida;

w2 = w2 + learning_rate*(delta_h2’*h1_out_estim)’;
b2 = b2 + learning_rate*sum(delta_h2);

w1 = w1 + learning_rate*delta_h1’*enter_estim;
b1 = b1 + learning_rate*sum(delta_h2);

% I wrote this code partially in portuguese so let me explain a litlle.
%’delta_saida’ is the gradient of the output layer
%delta_h2 is the gradient of the second hidden layer
%delta_h1 is the gradient of the first hidden layer
%w_out,w2 and w1 are the weights of output, second hidden layer and first hidden layer, respectively.
%b_out,b2 and b1 are the biases of output, second hidden layer and first hidden layer, respectively.
% the function selecionar_funcao() is just to calculate the derivative accordingly with the activation function of the layer

%As you can see, I need to change delta_h1 to match the matrix dimensions
It is right to change the formula like i’m doing in my code ? I’m asking it because in my mind, the way that we calculate the gradient of all hidden layers must be the same, but in my case it isn’t true. I will share part of my code here to help anyone to see if i’m doing some mistake
%weights and biases initialization
w1 = randn(num_entradas,n_h1)*sqrt(2/num_entradas);
w2 = randn(n_h1,n_h2) *sqrt(2/n_h1);
w_out = randn(n_h2,n_out) *sqrt(2/n_h2);

b1 = randn(1, n_h1) * sqrt(2/num_entradas);
b2 = randn(1, n_h2) * sqrt(2/n_h1);
b_out = randn(1,n_out) * sqrt(2/n_h2);

%backpropagation
for epoch =1:max_epocas
soma_valid = 0;
soma_estim = 0;
%embaralhar os dados
conj_estim = embaralhar(conj_estim);
% conj_valid = embaralhar(conj_valid);

%Validating
for j=1:size(conj_valid,1)
enter_valid = conj_valid(j,2:end);

h1_in_valid = [enter_valid,1]*[w1;b1];
h1_out_valid = selecionar_funcao(h1_in_valid,ativ_h1,sig_a,tanh_a,tanh_b,’False’);

h2_in_valid = [h1_out_valid,1]*[w2;b2];
h2_out_valid = selecionar_funcao(h2_in_valid,ativ_h2,sig_a,tanh_a,tanh_b,’False’);

saida_in_valid = [h2_out_valid,1]*[w_out;b_out];
saida_out_valid = selecionar_funcao(saida_in_valid,ativ_out,sig_a,tanh_a,tanh_b,’False’);

erro_valid = conj_valid(j,1) – saida_out_valid;

soma_valid = soma_valid + (erro_valid^2);
end
erro_atual_valid = (soma_valid/(2*size(conj_valid,1)));
erros_epoca_valid = [erros_epoca_valid;erro_atual_valid];

%trainning
for i =1:size(conj_estim,1)
enter_estim = conj_estim(i,2:end);

h1_in_estim = [enter_estim,1]*[w1;b1];
h1_out_estim = selecionar_funcao(h1_in_estim,ativ_h1,sig_a,tanh_a,tanh_b,’False’);

h2_in_estim = [h1_out_estim,1]*[w2;b2];
h2_out_estim = selecionar_funcao(h2_in_estim,ativ_h2,sig_a,tanh_a,tanh_b,’False’);

saida_in_estim = [h2_out_estim,1]*[w_out;b_out];
saida_out_estim = selecionar_funcao(saida_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’False’);

erro_estim = conj_estim(i,1) – saida_out_estim;
soma_estim = soma_estim + (erro_estim^2);

% Backpropagation
delta_saida = erro_estim.*selecionar_funcao(saida_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h2 = (w_out’*delta_saida).*selecionar_funcao(h2_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h1 = (w2*delta_h2′)’.*selecionar_funcao(h1_in_estim,ativ_h1,sig_a,tanh_a,tanh_b,’True’);

%atualizar pesos e biases
w_out = w_out + learning_rate*delta_saida*h2_out_estim’;
b_out = b_out + learning_rate*delta_saida;

w2 = w2 + learning_rate*(delta_h2’*h1_out_estim)’;
b2 = b2 + learning_rate*sum(delta_h2);

w1 = w1 + learning_rate*delta_h1’*enter_estim;
b1 = b1 + learning_rate*sum(delta_h2);

end
erro_atual_estim = (soma_estim/(2*size(conj_estim,1)));
erros_epoca_estim = [erros_epoca_estim;erro_atual_estim];
if erros_epoca_estim(epoch) <limiar
break
else
end
endHi
I’m studying neural networks and i’m doing a NN with 2 hidden layers and one neuron in the output layer
While I was studying and coding my NN, I faced one doubt.

In the backward step, the math behind this is clear:
to the output layer we have:
, where ⊙ is the hadamard product.
and to hidden layers, we have:

my problem is that, when I code this formulas, I need to change the second equation when I will calculate the the gradient of the first hidden layer (the hidden layer next to the input layer) to match the dimensions of the matrix as follow below:
% Backpropagation
delta_saida = erro_estim.*selecionar_funcao(saida_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h2 = (w_out’*delta_saida).*selecionar_funcao(h2_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h1 = (w2*delta_h2′)’.*selecionar_funcao(h1_in_estim,ativ_h1,sig_a,tanh_a,tanh_b,’True’);

%update weights and biases
w_out = w_out + learning_rate*delta_saida*h2_out_estim’;
b_out = b_out + learning_rate*delta_saida;

w2 = w2 + learning_rate*(delta_h2’*h1_out_estim)’;
b2 = b2 + learning_rate*sum(delta_h2);

w1 = w1 + learning_rate*delta_h1’*enter_estim;
b1 = b1 + learning_rate*sum(delta_h2);

% I wrote this code partially in portuguese so let me explain a litlle.
%’delta_saida’ is the gradient of the output layer
%delta_h2 is the gradient of the second hidden layer
%delta_h1 is the gradient of the first hidden layer
%w_out,w2 and w1 are the weights of output, second hidden layer and first hidden layer, respectively.
%b_out,b2 and b1 are the biases of output, second hidden layer and first hidden layer, respectively.
% the function selecionar_funcao() is just to calculate the derivative accordingly with the activation function of the layer

%As you can see, I need to change delta_h1 to match the matrix dimensions
It is right to change the formula like i’m doing in my code ? I’m asking it because in my mind, the way that we calculate the gradient of all hidden layers must be the same, but in my case it isn’t true. I will share part of my code here to help anyone to see if i’m doing some mistake
%weights and biases initialization
w1 = randn(num_entradas,n_h1)*sqrt(2/num_entradas);
w2 = randn(n_h1,n_h2) *sqrt(2/n_h1);
w_out = randn(n_h2,n_out) *sqrt(2/n_h2);

b1 = randn(1, n_h1) * sqrt(2/num_entradas);
b2 = randn(1, n_h2) * sqrt(2/n_h1);
b_out = randn(1,n_out) * sqrt(2/n_h2);

%backpropagation
for epoch =1:max_epocas
soma_valid = 0;
soma_estim = 0;
%embaralhar os dados
conj_estim = embaralhar(conj_estim);
% conj_valid = embaralhar(conj_valid);

%Validating
for j=1:size(conj_valid,1)
enter_valid = conj_valid(j,2:end);

h1_in_valid = [enter_valid,1]*[w1;b1];
h1_out_valid = selecionar_funcao(h1_in_valid,ativ_h1,sig_a,tanh_a,tanh_b,’False’);

h2_in_valid = [h1_out_valid,1]*[w2;b2];
h2_out_valid = selecionar_funcao(h2_in_valid,ativ_h2,sig_a,tanh_a,tanh_b,’False’);

saida_in_valid = [h2_out_valid,1]*[w_out;b_out];
saida_out_valid = selecionar_funcao(saida_in_valid,ativ_out,sig_a,tanh_a,tanh_b,’False’);

erro_valid = conj_valid(j,1) – saida_out_valid;

soma_valid = soma_valid + (erro_valid^2);
end
erro_atual_valid = (soma_valid/(2*size(conj_valid,1)));
erros_epoca_valid = [erros_epoca_valid;erro_atual_valid];

%trainning
for i =1:size(conj_estim,1)
enter_estim = conj_estim(i,2:end);

h1_in_estim = [enter_estim,1]*[w1;b1];
h1_out_estim = selecionar_funcao(h1_in_estim,ativ_h1,sig_a,tanh_a,tanh_b,’False’);

h2_in_estim = [h1_out_estim,1]*[w2;b2];
h2_out_estim = selecionar_funcao(h2_in_estim,ativ_h2,sig_a,tanh_a,tanh_b,’False’);

saida_in_estim = [h2_out_estim,1]*[w_out;b_out];
saida_out_estim = selecionar_funcao(saida_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’False’);

erro_estim = conj_estim(i,1) – saida_out_estim;
soma_estim = soma_estim + (erro_estim^2);

% Backpropagation
delta_saida = erro_estim.*selecionar_funcao(saida_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h2 = (w_out’*delta_saida).*selecionar_funcao(h2_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h1 = (w2*delta_h2′)’.*selecionar_funcao(h1_in_estim,ativ_h1,sig_a,tanh_a,tanh_b,’True’);

%atualizar pesos e biases
w_out = w_out + learning_rate*delta_saida*h2_out_estim’;
b_out = b_out + learning_rate*delta_saida;

w2 = w2 + learning_rate*(delta_h2’*h1_out_estim)’;
b2 = b2 + learning_rate*sum(delta_h2);

w1 = w1 + learning_rate*delta_h1’*enter_estim;
b1 = b1 + learning_rate*sum(delta_h2);

end
erro_atual_estim = (soma_estim/(2*size(conj_estim,1)));
erros_epoca_estim = [erros_epoca_estim;erro_atual_estim];
if erros_epoca_estim(epoch) <limiar
break
else
end
end Hi
I’m studying neural networks and i’m doing a NN with 2 hidden layers and one neuron in the output layer
While I was studying and coding my NN, I faced one doubt.

In the backward step, the math behind this is clear:
to the output layer we have:
, where ⊙ is the hadamard product.
and to hidden layers, we have:

my problem is that, when I code this formulas, I need to change the second equation when I will calculate the the gradient of the first hidden layer (the hidden layer next to the input layer) to match the dimensions of the matrix as follow below:
% Backpropagation
delta_saida = erro_estim.*selecionar_funcao(saida_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h2 = (w_out’*delta_saida).*selecionar_funcao(h2_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h1 = (w2*delta_h2′)’.*selecionar_funcao(h1_in_estim,ativ_h1,sig_a,tanh_a,tanh_b,’True’);

%update weights and biases
w_out = w_out + learning_rate*delta_saida*h2_out_estim’;
b_out = b_out + learning_rate*delta_saida;

w2 = w2 + learning_rate*(delta_h2’*h1_out_estim)’;
b2 = b2 + learning_rate*sum(delta_h2);

w1 = w1 + learning_rate*delta_h1’*enter_estim;
b1 = b1 + learning_rate*sum(delta_h2);

% I wrote this code partially in portuguese so let me explain a litlle.
%’delta_saida’ is the gradient of the output layer
%delta_h2 is the gradient of the second hidden layer
%delta_h1 is the gradient of the first hidden layer
%w_out,w2 and w1 are the weights of output, second hidden layer and first hidden layer, respectively.
%b_out,b2 and b1 are the biases of output, second hidden layer and first hidden layer, respectively.
% the function selecionar_funcao() is just to calculate the derivative accordingly with the activation function of the layer

%As you can see, I need to change delta_h1 to match the matrix dimensions
It is right to change the formula like i’m doing in my code ? I’m asking it because in my mind, the way that we calculate the gradient of all hidden layers must be the same, but in my case it isn’t true. I will share part of my code here to help anyone to see if i’m doing some mistake
%weights and biases initialization
w1 = randn(num_entradas,n_h1)*sqrt(2/num_entradas);
w2 = randn(n_h1,n_h2) *sqrt(2/n_h1);
w_out = randn(n_h2,n_out) *sqrt(2/n_h2);

b1 = randn(1, n_h1) * sqrt(2/num_entradas);
b2 = randn(1, n_h2) * sqrt(2/n_h1);
b_out = randn(1,n_out) * sqrt(2/n_h2);

%backpropagation
for epoch =1:max_epocas
soma_valid = 0;
soma_estim = 0;
%embaralhar os dados
conj_estim = embaralhar(conj_estim);
% conj_valid = embaralhar(conj_valid);

%Validating
for j=1:size(conj_valid,1)
enter_valid = conj_valid(j,2:end);

h1_in_valid = [enter_valid,1]*[w1;b1];
h1_out_valid = selecionar_funcao(h1_in_valid,ativ_h1,sig_a,tanh_a,tanh_b,’False’);

h2_in_valid = [h1_out_valid,1]*[w2;b2];
h2_out_valid = selecionar_funcao(h2_in_valid,ativ_h2,sig_a,tanh_a,tanh_b,’False’);

saida_in_valid = [h2_out_valid,1]*[w_out;b_out];
saida_out_valid = selecionar_funcao(saida_in_valid,ativ_out,sig_a,tanh_a,tanh_b,’False’);

erro_valid = conj_valid(j,1) – saida_out_valid;

soma_valid = soma_valid + (erro_valid^2);
end
erro_atual_valid = (soma_valid/(2*size(conj_valid,1)));
erros_epoca_valid = [erros_epoca_valid;erro_atual_valid];

%trainning
for i =1:size(conj_estim,1)
enter_estim = conj_estim(i,2:end);

h1_in_estim = [enter_estim,1]*[w1;b1];
h1_out_estim = selecionar_funcao(h1_in_estim,ativ_h1,sig_a,tanh_a,tanh_b,’False’);

h2_in_estim = [h1_out_estim,1]*[w2;b2];
h2_out_estim = selecionar_funcao(h2_in_estim,ativ_h2,sig_a,tanh_a,tanh_b,’False’);

saida_in_estim = [h2_out_estim,1]*[w_out;b_out];
saida_out_estim = selecionar_funcao(saida_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’False’);

erro_estim = conj_estim(i,1) – saida_out_estim;
soma_estim = soma_estim + (erro_estim^2);

% Backpropagation
delta_saida = erro_estim.*selecionar_funcao(saida_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h2 = (w_out’*delta_saida).*selecionar_funcao(h2_in_estim,ativ_out,sig_a,tanh_a,tanh_b,’True’);
delta_h1 = (w2*delta_h2′)’.*selecionar_funcao(h1_in_estim,ativ_h1,sig_a,tanh_a,tanh_b,’True’);

%atualizar pesos e biases
w_out = w_out + learning_rate*delta_saida*h2_out_estim’;
b_out = b_out + learning_rate*delta_saida;

w2 = w2 + learning_rate*(delta_h2’*h1_out_estim)’;
b2 = b2 + learning_rate*sum(delta_h2);

w1 = w1 + learning_rate*delta_h1’*enter_estim;
b1 = b1 + learning_rate*sum(delta_h2);

end
erro_atual_estim = (soma_estim/(2*size(conj_estim,1)));
erros_epoca_estim = [erros_epoca_estim;erro_atual_estim];
if erros_epoca_estim(epoch) <limiar
break
else
end
end machine learning, backpropagation MATLAB Answers — New Questions

​

I have a line in my code where data is imported, and I wantto add a directory that can import the data from SharePoint so that anyone who is in the sharepoint can use the same code without having to change the directory. I can’t seem to figur eout how to do this without adding a shortcut, becaus enot everyone would have the same shortcut. If anyone could help with how I can set up the import so that anyone can use the code that would be great! Thanks!I have a line in my code where data is imported, and I wantto add a directory that can import the data from SharePoint so that anyone who is in the sharepoint can use the same code without having to change the directory. I can’t seem to figur eout how to do this without adding a shortcut, becaus enot everyone would have the same shortcut. If anyone could help with how I can set up the import so that anyone can use the code that would be great! Thanks! I have a line in my code where data is imported, and I wantto add a directory that can import the data from SharePoint so that anyone who is in the sharepoint can use the same code without having to change the directory. I can’t seem to figur eout how to do this without adding a shortcut, becaus enot everyone would have the same shortcut. If anyone could help with how I can set up the import so that anyone can use the code that would be great! Thanks! data import, sharepoint, import MATLAB Answers — New Questions

​

I’m trying to speed up this part of code. I have a constraint: the inputs of ReturnFn must all be scalars. If it was not for this restriction, I could easily vectorize the code. So I would like to know if there is a way to make the code below faster, while still satisfying this restriction of the inputs of ReturnFn.
Any help is really appreciated!

N_d = 50;
N_a = 300;
N_z = 10;

% ParamCell contains: K_to_L,alpha,delta,pen,gamma,crra
% I need the cell array to handle variable number of inputs in ReturnFn

Fmatrix=zeros(N_d*N_a,N_a,N_z);
parfor i4i5=1:N_z
Fmatrix_z=zeros(N_d*N_a,N_a);
for i3=1:N_a % a today
for i2=1:N_a % a’ tomorrow
for i1=1:N_d % d choice
Fmatrix_z(i1+(i2-1)*N_d,i3)=ReturnFn(d_gridvals(i1),a_gridvals(i2),a_gridvals(i3),z_gridvals(i4i5,1),z_gridvals(i4i5,2),ParamCell{:});
end
end
end
Fmatrix(:,:,i4i5)=Fmatrix_z;
end

function F = f_ReturnFn(d,aprime,a,e,age,K_to_L,alpha,delta,pen,gamma,crra)
% INPUTS (always 5 inputs, plus some extra parameter inputs)
% d: Hours worked
% aprime: Next-period’s assets
% a: Current period assets
% e: Labor efficiency shock
% age: Age of individual: young or old
% TOOLKIT NOTATION
% (d,aprime,a,z), where z = [e;age]
F = -inf;
r = alpha*K_to_L^(alpha-1)-delta;
w = (1-alpha)*K_to_L^alpha;
income = (w*e*d)*(age==1)+pen*(age==2)+r*a;
c = income+a-aprime; % Budget Constraint
if c>0
% NOTE: 0<d<1 is already built into the grid
% WARNING: this will not work if crra=1
inside = (c^gamma)*((1-d)^(1-gamma));
F = inside^(1-crra)/(1-crra);
end
endI’m trying to speed up this part of code. I have a constraint: the inputs of ReturnFn must all be scalars. If it was not for this restriction, I could easily vectorize the code. So I would like to know if there is a way to make the code below faster, while still satisfying this restriction of the inputs of ReturnFn.
Any help is really appreciated!

N_d = 50;
N_a = 300;
N_z = 10;

% ParamCell contains: K_to_L,alpha,delta,pen,gamma,crra
% I need the cell array to handle variable number of inputs in ReturnFn

Fmatrix=zeros(N_d*N_a,N_a,N_z);
parfor i4i5=1:N_z
Fmatrix_z=zeros(N_d*N_a,N_a);
for i3=1:N_a % a today
for i2=1:N_a % a’ tomorrow
for i1=1:N_d % d choice
Fmatrix_z(i1+(i2-1)*N_d,i3)=ReturnFn(d_gridvals(i1),a_gridvals(i2),a_gridvals(i3),z_gridvals(i4i5,1),z_gridvals(i4i5,2),ParamCell{:});
end
end
end
Fmatrix(:,:,i4i5)=Fmatrix_z;
end

function F = f_ReturnFn(d,aprime,a,e,age,K_to_L,alpha,delta,pen,gamma,crra)
% INPUTS (always 5 inputs, plus some extra parameter inputs)
% d: Hours worked
% aprime: Next-period’s assets
% a: Current period assets
% e: Labor efficiency shock
% age: Age of individual: young or old
% TOOLKIT NOTATION
% (d,aprime,a,z), where z = [e;age]
F = -inf;
r = alpha*K_to_L^(alpha-1)-delta;
w = (1-alpha)*K_to_L^alpha;
income = (w*e*d)*(age==1)+pen*(age==2)+r*a;
c = income+a-aprime; % Budget Constraint
if c>0
% NOTE: 0<d<1 is already built into the grid
% WARNING: this will not work if crra=1
inside = (c^gamma)*((1-d)^(1-gamma));
F = inside^(1-crra)/(1-crra);
end
end I’m trying to speed up this part of code. I have a constraint: the inputs of ReturnFn must all be scalars. If it was not for this restriction, I could easily vectorize the code. So I would like to know if there is a way to make the code below faster, while still satisfying this restriction of the inputs of ReturnFn.
Any help is really appreciated!

N_d = 50;
N_a = 300;
N_z = 10;

% ParamCell contains: K_to_L,alpha,delta,pen,gamma,crra
% I need the cell array to handle variable number of inputs in ReturnFn

Fmatrix=zeros(N_d*N_a,N_a,N_z);
parfor i4i5=1:N_z
Fmatrix_z=zeros(N_d*N_a,N_a);
for i3=1:N_a % a today
for i2=1:N_a % a’ tomorrow
for i1=1:N_d % d choice
Fmatrix_z(i1+(i2-1)*N_d,i3)=ReturnFn(d_gridvals(i1),a_gridvals(i2),a_gridvals(i3),z_gridvals(i4i5,1),z_gridvals(i4i5,2),ParamCell{:});
end
end
end
Fmatrix(:,:,i4i5)=Fmatrix_z;
end

function F = f_ReturnFn(d,aprime,a,e,age,K_to_L,alpha,delta,pen,gamma,crra)
% INPUTS (always 5 inputs, plus some extra parameter inputs)
% d: Hours worked
% aprime: Next-period’s assets
% a: Current period assets
% e: Labor efficiency shock
% age: Age of individual: young or old
% TOOLKIT NOTATION
% (d,aprime,a,z), where z = [e;age]
F = -inf;
r = alpha*K_to_L^(alpha-1)-delta;
w = (1-alpha)*K_to_L^alpha;
income = (w*e*d)*(age==1)+pen*(age==2)+r*a;
c = income+a-aprime; % Budget Constraint
if c>0
% NOTE: 0<d<1 is already built into the grid
% WARNING: this will not work if crra=1
inside = (c^gamma)*((1-d)^(1-gamma));
F = inside^(1-crra)/(1-crra);
end
end nested loops, parfor MATLAB Answers — New Questions

​