Why the Kf_LMax value is increased beyond its limits. Is their is any logical error Kindly help me out .
% Define parameters
Kf_Max = 100; % maximum forward rate
RLC = [0.1, 0.5, 10, 5, 10, 1]; % RLC values
TauKf_ON = -0.01; % TauKf_ON
TauKf_OFF = -0.01; % TauKf_OFF
PhaseTimes = [0, 500, 1000, 2000, 3000, 4000, 5000]; % phase times (each row defines a phase)
% Generate time vector
t = 0:0.01:5000;
% Call the function to compute receptor concentrations and Kf
[ Kf_LMax] = Kf_Cal(Kf_Max, RLC, TauKf_ON, TauKf_OFF, PhaseTimes, t);
% Plotting
figure;
% Plot Kf_LMax
plot(t, Kf_LMax, ‘b’, ‘LineWidth’, 1.5);
title(‘Kf_LMax over Time’);
xlabel(‘Time’);
ylabel(‘Kf_LMax’);
grid on;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [ Kf_LMax] = Kf_Cal(Kf_Max, RLC, TauKf_ON, TauKf_OFF, PhaseTimes, t)
% Initialize output variables
Kf_LMax = zeros(size(t));

% Calculate Kf_L for each time step
for i = 1:length(t)
Kf_LMax(i) = calculate_kf(t(i), PhaseTimes, Kf_Max, RLC, TauKf_ON, TauKf_OFF);
end
% Nested function for calculating Kf_L based on time
function Kf_L = calculate_kf(t_current, PhaseTimes, Kf_Max, RLC, TauKf_ON, TauKf_OFF)
% Calculate maximum Kf_L based on RLC values using Element wise
% division
Kf_LMax_values = Kf_Max * (RLC ./ (1 + RLC));

% Default Kf_L to the maximum value of the first phase
Kf_L = Kf_LMax_values(1);
% Number of phases
num_phases = numel(RLC);
% Iterate through each phase to determine the current phase based on time
for j = 1:num_phases
T_start = PhaseTimes(j); % Start time of the current phase
if j < num_phases
T_end = PhaseTimes(j + 1); % End time of the current phase
else
T_end = inf; % Handle last phase separately
end
% Check if the current time t_current is within the current phase
if t_current >= T_start && t_current < T_end
if j == 1
% For the first phase, use the maximum value directly
Kf_L = Kf_LMax_values(j);
else
% Time at the end of the previous phase
% prev_end = PhaseTimes(j – 1);
if j < num_phases
% Check RLC conditions and compute Kf_L
if RLC(j – 1) < RLC(j) && RLC(j) > RLC(j + 1)
% Peak condition
Kf_endA = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endA – Kf_LMax_values(j – 1)) *exp(TauKf_OFF * ( t_current – T_start));
elseif RLC(j – 1) < RLC(j) && RLC(j) < RLC(j + 1)
% Increasing RLC condition
Kf_endB = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endB – Kf_LMax_values(j – 1)) *exp(TauKf_OFF * (t_current – T_start));
elseif RLC(j – 1) > RLC(j) && RLC(j) < RLC(j + 1)
% Decreasing RLC condition
Kf_endC = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endC – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start ));
elseif RLC(j – 1) > RLC(j) && RLC(j) >= RLC(j + 1)
% Flat or decreasing RLC condition
Kf_endD = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endD – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start ));
end
else
% % % Handling for the last phase
if RLC(j – 1) < RLC(j)
Kf_end1 = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start) Kf_L = Kf_LMax_values(j) + (Kf_end1 – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start));
elseif RLC(j – 1) > RLC(j)
Kf_end2 = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_end2 – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start));
end
end
end
return; % Exit the function after finding the correct phase
end
end
end
endWhy the Kf_LMax value is increased beyond its limits. Is their is any logical error Kindly help me out .
% Define parameters
Kf_Max = 100; % maximum forward rate
RLC = [0.1, 0.5, 10, 5, 10, 1]; % RLC values
TauKf_ON = -0.01; % TauKf_ON
TauKf_OFF = -0.01; % TauKf_OFF
PhaseTimes = [0, 500, 1000, 2000, 3000, 4000, 5000]; % phase times (each row defines a phase)
% Generate time vector
t = 0:0.01:5000;
% Call the function to compute receptor concentrations and Kf
[ Kf_LMax] = Kf_Cal(Kf_Max, RLC, TauKf_ON, TauKf_OFF, PhaseTimes, t);
% Plotting
figure;
% Plot Kf_LMax
plot(t, Kf_LMax, ‘b’, ‘LineWidth’, 1.5);
title(‘Kf_LMax over Time’);
xlabel(‘Time’);
ylabel(‘Kf_LMax’);
grid on;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [ Kf_LMax] = Kf_Cal(Kf_Max, RLC, TauKf_ON, TauKf_OFF, PhaseTimes, t)
% Initialize output variables
Kf_LMax = zeros(size(t));

% Calculate Kf_L for each time step
for i = 1:length(t)
Kf_LMax(i) = calculate_kf(t(i), PhaseTimes, Kf_Max, RLC, TauKf_ON, TauKf_OFF);
end
% Nested function for calculating Kf_L based on time
function Kf_L = calculate_kf(t_current, PhaseTimes, Kf_Max, RLC, TauKf_ON, TauKf_OFF)
% Calculate maximum Kf_L based on RLC values using Element wise
% division
Kf_LMax_values = Kf_Max * (RLC ./ (1 + RLC));

% Default Kf_L to the maximum value of the first phase
Kf_L = Kf_LMax_values(1);
% Number of phases
num_phases = numel(RLC);
% Iterate through each phase to determine the current phase based on time
for j = 1:num_phases
T_start = PhaseTimes(j); % Start time of the current phase
if j < num_phases
T_end = PhaseTimes(j + 1); % End time of the current phase
else
T_end = inf; % Handle last phase separately
end
% Check if the current time t_current is within the current phase
if t_current >= T_start && t_current < T_end
if j == 1
% For the first phase, use the maximum value directly
Kf_L = Kf_LMax_values(j);
else
% Time at the end of the previous phase
% prev_end = PhaseTimes(j – 1);
if j < num_phases
% Check RLC conditions and compute Kf_L
if RLC(j – 1) < RLC(j) && RLC(j) > RLC(j + 1)
% Peak condition
Kf_endA = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endA – Kf_LMax_values(j – 1)) *exp(TauKf_OFF * ( t_current – T_start));
elseif RLC(j – 1) < RLC(j) && RLC(j) < RLC(j + 1)
% Increasing RLC condition
Kf_endB = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endB – Kf_LMax_values(j – 1)) *exp(TauKf_OFF * (t_current – T_start));
elseif RLC(j – 1) > RLC(j) && RLC(j) < RLC(j + 1)
% Decreasing RLC condition
Kf_endC = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endC – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start ));
elseif RLC(j – 1) > RLC(j) && RLC(j) >= RLC(j + 1)
% Flat or decreasing RLC condition
Kf_endD = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endD – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start ));
end
else
% % % Handling for the last phase
if RLC(j – 1) < RLC(j)
Kf_end1 = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start) Kf_L = Kf_LMax_values(j) + (Kf_end1 – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start));
elseif RLC(j – 1) > RLC(j)
Kf_end2 = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_end2 – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start));
end
end
end
return; % Exit the function after finding the correct phase
end
end
end
end Why the Kf_LMax value is increased beyond its limits. Is their is any logical error Kindly help me out .
% Define parameters
Kf_Max = 100; % maximum forward rate
RLC = [0.1, 0.5, 10, 5, 10, 1]; % RLC values
TauKf_ON = -0.01; % TauKf_ON
TauKf_OFF = -0.01; % TauKf_OFF
PhaseTimes = [0, 500, 1000, 2000, 3000, 4000, 5000]; % phase times (each row defines a phase)
% Generate time vector
t = 0:0.01:5000;
% Call the function to compute receptor concentrations and Kf
[ Kf_LMax] = Kf_Cal(Kf_Max, RLC, TauKf_ON, TauKf_OFF, PhaseTimes, t);
% Plotting
figure;
% Plot Kf_LMax
plot(t, Kf_LMax, ‘b’, ‘LineWidth’, 1.5);
title(‘Kf_LMax over Time’);
xlabel(‘Time’);
ylabel(‘Kf_LMax’);
grid on;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [ Kf_LMax] = Kf_Cal(Kf_Max, RLC, TauKf_ON, TauKf_OFF, PhaseTimes, t)
% Initialize output variables
Kf_LMax = zeros(size(t));

% Calculate Kf_L for each time step
for i = 1:length(t)
Kf_LMax(i) = calculate_kf(t(i), PhaseTimes, Kf_Max, RLC, TauKf_ON, TauKf_OFF);
end
% Nested function for calculating Kf_L based on time
function Kf_L = calculate_kf(t_current, PhaseTimes, Kf_Max, RLC, TauKf_ON, TauKf_OFF)
% Calculate maximum Kf_L based on RLC values using Element wise
% division
Kf_LMax_values = Kf_Max * (RLC ./ (1 + RLC));

% Default Kf_L to the maximum value of the first phase
Kf_L = Kf_LMax_values(1);
% Number of phases
num_phases = numel(RLC);
% Iterate through each phase to determine the current phase based on time
for j = 1:num_phases
T_start = PhaseTimes(j); % Start time of the current phase
if j < num_phases
T_end = PhaseTimes(j + 1); % End time of the current phase
else
T_end = inf; % Handle last phase separately
end
% Check if the current time t_current is within the current phase
if t_current >= T_start && t_current < T_end
if j == 1
% For the first phase, use the maximum value directly
Kf_L = Kf_LMax_values(j);
else
% Time at the end of the previous phase
% prev_end = PhaseTimes(j – 1);
if j < num_phases
% Check RLC conditions and compute Kf_L
if RLC(j – 1) < RLC(j) && RLC(j) > RLC(j + 1)
% Peak condition
Kf_endA = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endA – Kf_LMax_values(j – 1)) *exp(TauKf_OFF * ( t_current – T_start));
elseif RLC(j – 1) < RLC(j) && RLC(j) < RLC(j + 1)
% Increasing RLC condition
Kf_endB = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endB – Kf_LMax_values(j – 1)) *exp(TauKf_OFF * (t_current – T_start));
elseif RLC(j – 1) > RLC(j) && RLC(j) < RLC(j + 1)
% Decreasing RLC condition
Kf_endC = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endC – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start ));
elseif RLC(j – 1) > RLC(j) && RLC(j) >= RLC(j + 1)
% Flat or decreasing RLC condition
Kf_endD = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_endD – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start ));
end
else
% % % Handling for the last phase
if RLC(j – 1) < RLC(j)
Kf_end1 = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start) Kf_L = Kf_LMax_values(j) + (Kf_end1 – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start));
elseif RLC(j – 1) > RLC(j)
Kf_end2 = Kf_LMax_values(j) – (Kf_LMax_values(j) – Kf_LMax_values(j – 1)) * exp(TauKf_ON * (t_current – T_start));
Kf_L = Kf_LMax_values(j) + (Kf_end2 – Kf_LMax_values(j – 1)) * exp(TauKf_OFF * (t_current – T_start));
end
end
end
return; % Exit the function after finding the correct phase
end
end
end
end graph MATLAB Answers — New Questions

​

Hello,

I use the rs decoder of hdl coder. the simulation in matlab simulink is correct…
but when I load this code in fpga board and capture data by using the ILA, Its not same…
I think that when generated code is converted to verilog, difference is occured??
possible waring list..
optimization option
timing closure
tool version

1. optimization option : I set correct clock, fpga board, distributed pipeline … and so on.
-> I dont know that my setting value is correct????
2. timing satisfy, fpga clock : 16Mhz
3. matlab ver : 23a, vivado ver : 20.2

please give me check point….

thank you your corporation
Je Heo.Hello,

I use the rs decoder of hdl coder. the simulation in matlab simulink is correct…
but when I load this code in fpga board and capture data by using the ILA, Its not same…
I think that when generated code is converted to verilog, difference is occured??
possible waring list..
optimization option
timing closure
tool version

1. optimization option : I set correct clock, fpga board, distributed pipeline … and so on.
-> I dont know that my setting value is correct????
2. timing satisfy, fpga clock : 16Mhz
3. matlab ver : 23a, vivado ver : 20.2

please give me check point….

thank you your corporation
Je Heo. Hello,

I use the rs decoder of hdl coder. the simulation in matlab simulink is correct…
but when I load this code in fpga board and capture data by using the ILA, Its not same…
I think that when generated code is converted to verilog, difference is occured??
possible waring list..
optimization option
timing closure
tool version

1. optimization option : I set correct clock, fpga board, distributed pipeline … and so on.
-> I dont know that my setting value is correct????
2. timing satisfy, fpga clock : 16Mhz
3. matlab ver : 23a, vivado ver : 20.2

please give me check point….

thank you your corporation
Je Heo. hdl coder, rs decoder, fpga, generated hdl code, ila MATLAB Answers — New Questions

​

Hello everyone
I am trying to predict traffic flow of future steps by previous collected data so I Use LSTM for it
but my validation loss and rmse increase and training loss and rmse decrease .because I am net to LSTM I don’t know which parameters I should check for improving model and predictions.
the picture of training progress is :

also I use different lags time for my predictions and here in my codes I have 4 step lag time
XTrain_ZaMir = (XTrain_ZaMir – mu_ZaMir)/sig_ZaMir;
YTrain_ZaMir = (YTrain_ZaMir – mu_ZaMir)/sig_ZaMir;
XTrain_ZaMir = XTrain_ZaMir(:,1:end-4);
YTrain_ZaMir = YTrain_ZaMir(:,5:end);
Test_ZaMir = [flowTe_ZaMir flowTeOther_ZaMir]’;
nt = floor(0.7*length(Test_ZaMir));
YTest_ZaMir = Test_ZaMir(1,1:end);
XTest_ZaMir = Test_ZaMir(1,1:end); %One input

% XTest_ZaMir = Test_ZaMir(:,1:end); % More than One input
XTest_ZaMir = (XTest_ZaMir – mu_ZaMir)/sig_ZaMir;
YTest_ZaMir = (YTest_ZaMir – mu_ZaMir)/sig_ZaMir;
XVal_ZaMir = XTest_ZaMir(:,1:nt-4);
YVal_ZaMir = YTest_ZaMir(:,5:nt);
XTest_ZaMir = XTest_ZaMir(:,nt+4:end-1);
YTest_ZaMir = YTest_ZaMir(:,nt+5:end);

%% Layers and Options
numResponses = 1 ;
featureDimension = 1;
numHiddenUnits =200 ;

layers = [ …
sequenceInputLayer(featureDimension)
lstmLayer(numHiddenUnits)
% dropoutLayer(0.002)
fullyConnectedLayer(numResponses)
regressionLayer
];

maxepochs = 250;
minibatchsize =128;

options = trainingOptions(‘adam’, … %%adam
‘MaxEpochs’,maxepochs, …
‘GradientThreshold’,1, …
‘InitialLearnRate’,0.005, …
‘ValidationData’,{XVal_ZaMir,YVal_ZaMir},…
‘ValidationFrequency’,20,…
‘Shuffle’,’every-epoch’,…
‘MiniBatchSize’,minibatchsize,…
‘LearnRateSchedule’,’piecewise’, …
‘LearnRateDropPeriod’,150, …
‘LearnRateDropFactor’,0.005, …
‘Verbose’,1, …
‘Plots’,’training-progress’);

%% Train the Network
[net,info] = trainNetwork(XTrain_ZaMir,YTrain_ZaMir,layers,options);
[net,YPred_ZaMir]= predictAndUpdateState(net,XTest_ZaMir);

numTimeStepsTest= (0.5*floor(length(XTest_ZaMir)));
for i = 2:numTimeStepsTest
[net,YPred_ZaMir(:,i)] = predictAndUpdateState(net,XTest_ZaMir(:,i-1),’ExecutionEnvironment’,’cpu’);
% net = resetState(net);
end
YTest_ZaMir = sig_ZaMir*YTest_ZaMir + mu_ZaMir;
YPred_ZaMir = sig_ZaMir*YPred_ZaMir + mu_ZaMir;Hello everyone
I am trying to predict traffic flow of future steps by previous collected data so I Use LSTM for it
but my validation loss and rmse increase and training loss and rmse decrease .because I am net to LSTM I don’t know which parameters I should check for improving model and predictions.
the picture of training progress is :

also I use different lags time for my predictions and here in my codes I have 4 step lag time
XTrain_ZaMir = (XTrain_ZaMir – mu_ZaMir)/sig_ZaMir;
YTrain_ZaMir = (YTrain_ZaMir – mu_ZaMir)/sig_ZaMir;
XTrain_ZaMir = XTrain_ZaMir(:,1:end-4);
YTrain_ZaMir = YTrain_ZaMir(:,5:end);
Test_ZaMir = [flowTe_ZaMir flowTeOther_ZaMir]’;
nt = floor(0.7*length(Test_ZaMir));
YTest_ZaMir = Test_ZaMir(1,1:end);
XTest_ZaMir = Test_ZaMir(1,1:end); %One input

% XTest_ZaMir = Test_ZaMir(:,1:end); % More than One input
XTest_ZaMir = (XTest_ZaMir – mu_ZaMir)/sig_ZaMir;
YTest_ZaMir = (YTest_ZaMir – mu_ZaMir)/sig_ZaMir;
XVal_ZaMir = XTest_ZaMir(:,1:nt-4);
YVal_ZaMir = YTest_ZaMir(:,5:nt);
XTest_ZaMir = XTest_ZaMir(:,nt+4:end-1);
YTest_ZaMir = YTest_ZaMir(:,nt+5:end);

%% Layers and Options
numResponses = 1 ;
featureDimension = 1;
numHiddenUnits =200 ;

layers = [ …
sequenceInputLayer(featureDimension)
lstmLayer(numHiddenUnits)
% dropoutLayer(0.002)
fullyConnectedLayer(numResponses)
regressionLayer
];

maxepochs = 250;
minibatchsize =128;

options = trainingOptions(‘adam’, … %%adam
‘MaxEpochs’,maxepochs, …
‘GradientThreshold’,1, …
‘InitialLearnRate’,0.005, …
‘ValidationData’,{XVal_ZaMir,YVal_ZaMir},…
‘ValidationFrequency’,20,…
‘Shuffle’,’every-epoch’,…
‘MiniBatchSize’,minibatchsize,…
‘LearnRateSchedule’,’piecewise’, …
‘LearnRateDropPeriod’,150, …
‘LearnRateDropFactor’,0.005, …
‘Verbose’,1, …
‘Plots’,’training-progress’);

%% Train the Network
[net,info] = trainNetwork(XTrain_ZaMir,YTrain_ZaMir,layers,options);
[net,YPred_ZaMir]= predictAndUpdateState(net,XTest_ZaMir);

numTimeStepsTest= (0.5*floor(length(XTest_ZaMir)));
for i = 2:numTimeStepsTest
[net,YPred_ZaMir(:,i)] = predictAndUpdateState(net,XTest_ZaMir(:,i-1),’ExecutionEnvironment’,’cpu’);
% net = resetState(net);
end
YTest_ZaMir = sig_ZaMir*YTest_ZaMir + mu_ZaMir;
YPred_ZaMir = sig_ZaMir*YPred_ZaMir + mu_ZaMir; Hello everyone
I am trying to predict traffic flow of future steps by previous collected data so I Use LSTM for it
but my validation loss and rmse increase and training loss and rmse decrease .because I am net to LSTM I don’t know which parameters I should check for improving model and predictions.
the picture of training progress is :

also I use different lags time for my predictions and here in my codes I have 4 step lag time
XTrain_ZaMir = (XTrain_ZaMir – mu_ZaMir)/sig_ZaMir;
YTrain_ZaMir = (YTrain_ZaMir – mu_ZaMir)/sig_ZaMir;
XTrain_ZaMir = XTrain_ZaMir(:,1:end-4);
YTrain_ZaMir = YTrain_ZaMir(:,5:end);
Test_ZaMir = [flowTe_ZaMir flowTeOther_ZaMir]’;
nt = floor(0.7*length(Test_ZaMir));
YTest_ZaMir = Test_ZaMir(1,1:end);
XTest_ZaMir = Test_ZaMir(1,1:end); %One input

% XTest_ZaMir = Test_ZaMir(:,1:end); % More than One input
XTest_ZaMir = (XTest_ZaMir – mu_ZaMir)/sig_ZaMir;
YTest_ZaMir = (YTest_ZaMir – mu_ZaMir)/sig_ZaMir;
XVal_ZaMir = XTest_ZaMir(:,1:nt-4);
YVal_ZaMir = YTest_ZaMir(:,5:nt);
XTest_ZaMir = XTest_ZaMir(:,nt+4:end-1);
YTest_ZaMir = YTest_ZaMir(:,nt+5:end);

%% Layers and Options
numResponses = 1 ;
featureDimension = 1;
numHiddenUnits =200 ;

layers = [ …
sequenceInputLayer(featureDimension)
lstmLayer(numHiddenUnits)
% dropoutLayer(0.002)
fullyConnectedLayer(numResponses)
regressionLayer
];

maxepochs = 250;
minibatchsize =128;

options = trainingOptions(‘adam’, … %%adam
‘MaxEpochs’,maxepochs, …
‘GradientThreshold’,1, …
‘InitialLearnRate’,0.005, …
‘ValidationData’,{XVal_ZaMir,YVal_ZaMir},…
‘ValidationFrequency’,20,…
‘Shuffle’,’every-epoch’,…
‘MiniBatchSize’,minibatchsize,…
‘LearnRateSchedule’,’piecewise’, …
‘LearnRateDropPeriod’,150, …
‘LearnRateDropFactor’,0.005, …
‘Verbose’,1, …
‘Plots’,’training-progress’);

%% Train the Network
[net,info] = trainNetwork(XTrain_ZaMir,YTrain_ZaMir,layers,options);
[net,YPred_ZaMir]= predictAndUpdateState(net,XTest_ZaMir);

numTimeStepsTest= (0.5*floor(length(XTest_ZaMir)));
for i = 2:numTimeStepsTest
[net,YPred_ZaMir(:,i)] = predictAndUpdateState(net,XTest_ZaMir(:,i-1),’ExecutionEnvironment’,’cpu’);
% net = resetState(net);
end
YTest_ZaMir = sig_ZaMir*YTest_ZaMir + mu_ZaMir;
YPred_ZaMir = sig_ZaMir*YPred_ZaMir + mu_ZaMir; predictions, lstm, validation, train MATLAB Answers — New Questions

​

When trying to start up MatLab, I get
> ./bin/matlab
MATLAB is selecting SOFTWARE rendering.
/home/pblase/.MathWorks/ServiceHost/clr-df9a0cbb6bd34e079ef626671d1a7b7c/_tmp_MSHI_5363-9225-767d-e56f/mci/_tempinstaller_glnxa64/bin/glnxa64/InstallMathWorksServiceHost: symbol lookup error: /usr/lib64/libcairo.so.2: undefined symbol: xcb_shm_id
/home/pblase/.MathWorks/ServiceHost/clr-df9a0cbb6bd34e079ef626671d1a7b7c/_tmp_MSHI_5363-9225-767d-e56f/mci/_tempinstaller_glnxa64/bin/glnxa64/InstallMathWorksServiceHost: symbol lookup error: /usr/lib64/libcairo.so.2: undefined symbol: xcb_shm_id
Unexpected exception: ‘N7mwboost10wrapexceptINS_16exception_detail39current_exception_std_exception_wrapperISt13runtime_errorEEEE: Error loading /home/pblase/matlab/bin/glnxa64/matlab_startup_plugins/matlab_graphics_ui/mwuixloader.so. /usr/lib64/libXt.so.6: undefined symbol: SmcModifyCallbacks: Success: Success’ in createMVMAndCallParser phase ‘Creating local MVM’When trying to start up MatLab, I get
> ./bin/matlab
MATLAB is selecting SOFTWARE rendering.
/home/pblase/.MathWorks/ServiceHost/clr-df9a0cbb6bd34e079ef626671d1a7b7c/_tmp_MSHI_5363-9225-767d-e56f/mci/_tempinstaller_glnxa64/bin/glnxa64/InstallMathWorksServiceHost: symbol lookup error: /usr/lib64/libcairo.so.2: undefined symbol: xcb_shm_id
/home/pblase/.MathWorks/ServiceHost/clr-df9a0cbb6bd34e079ef626671d1a7b7c/_tmp_MSHI_5363-9225-767d-e56f/mci/_tempinstaller_glnxa64/bin/glnxa64/InstallMathWorksServiceHost: symbol lookup error: /usr/lib64/libcairo.so.2: undefined symbol: xcb_shm_id
Unexpected exception: ‘N7mwboost10wrapexceptINS_16exception_detail39current_exception_std_exception_wrapperISt13runtime_errorEEEE: Error loading /home/pblase/matlab/bin/glnxa64/matlab_startup_plugins/matlab_graphics_ui/mwuixloader.so. /usr/lib64/libXt.so.6: undefined symbol: SmcModifyCallbacks: Success: Success’ in createMVMAndCallParser phase ‘Creating local MVM’ When trying to start up MatLab, I get
> ./bin/matlab
MATLAB is selecting SOFTWARE rendering.
/home/pblase/.MathWorks/ServiceHost/clr-df9a0cbb6bd34e079ef626671d1a7b7c/_tmp_MSHI_5363-9225-767d-e56f/mci/_tempinstaller_glnxa64/bin/glnxa64/InstallMathWorksServiceHost: symbol lookup error: /usr/lib64/libcairo.so.2: undefined symbol: xcb_shm_id
/home/pblase/.MathWorks/ServiceHost/clr-df9a0cbb6bd34e079ef626671d1a7b7c/_tmp_MSHI_5363-9225-767d-e56f/mci/_tempinstaller_glnxa64/bin/glnxa64/InstallMathWorksServiceHost: symbol lookup error: /usr/lib64/libcairo.so.2: undefined symbol: xcb_shm_id
Unexpected exception: ‘N7mwboost10wrapexceptINS_16exception_detail39current_exception_std_exception_wrapperISt13runtime_errorEEEE: Error loading /home/pblase/matlab/bin/glnxa64/matlab_startup_plugins/matlab_graphics_ui/mwuixloader.so. /usr/lib64/libXt.so.6: undefined symbol: SmcModifyCallbacks: Success: Success’ in createMVMAndCallParser phase ‘Creating local MVM’ libcairo MATLAB Answers — New Questions

​

I have to solve a dynamic programming problem using a linear programming approach. For details, please see this paper. The LP that I want to solve is:
min c’*v
s.t.
A*v>=u,
where c is n*1, v is n*1, A is n^2*n, u is n^2*1.
The min is with respect to v, the value function of the original DP problem. I have a moderate number of variables, n=300 and m=n^2*n=90000 linear inequalities as constraints. No bound constraints on v.
I use the Matlab function linprog which in turn is based on the solver HIGHS (since R2024a). The code is slow for my purposes (i.e. a brute-force value iteration is much faster). Moreover, linprog gives correct results only if I set the option ‘Algorithm’,’dual-simplex-highs’. With other algorithms, it gets stuck.
After profiling the code, it turns out that the bottleneck is line 377 of linprog:
[x, fval, exitflag, output, lambda] = run(algorithm, problem);
I was wondering if there is a way to speed up the code. Any help or suggestion is greatly appreciated! I put below a MWE to illustrate the problem.

clear,clc,close all
%% Set parameters
crra = 2;
alpha = 0.36;
beta = 0.95;
delta = 0.1;
%% Grid for capital
k_ss = ((1-beta*(1-delta))/(alpha*beta))^(1/(alpha-1));
n_k = 300;
k_grid = linspace(0.1*k_ss,1.5*k_ss,n_k)’;
%% Build current return matrix, U(k’,k)
cons = k_grid’.^alpha+(1-delta)*k_grid’-k_grid;
U_mat = f_util(cons,crra);
U_mat(cons<=0) = -inf;
%% Using LINEAR PROGRAMMING
% min c’*v
% s.t.
% A*v>=u, where c is n*1, v is n*1, A is n^2*n, u is n^2*1
n = length(k_grid);
c_vec = ones(n,1);
u_vec = U_mat(:); %% U(k’,k), stack columnwise
%% Build A matrix using cell-based method
tic
A = cell(n,1);
bigI = (-beta)*speye(n);
for i=1:n
temp = bigI;
temp(:,i) = temp(:,i)+1;
A{i} = temp;
end
A = vertcat(A{:});
disp(‘Time to build A matrix with cell method:’)
toc
%% Call linprog
% ‘dual-simplex-highs’ (default and by far the best)
options = optimoptions(‘linprog’,’Algorithm’,’dual-simplex-highs’);
tic
[V_lin,fval,exitflag,output] = linprog(c_vec,-A,-u_vec,[],[],[],[],options);
disp(‘Time linear programming:’)
toc
if exitflag<=0
warning(‘linprog did not find a solution’)
fprintf(‘exitflag = %d n’,exitflag)
end
%% Now that we solved for V, compute policy function
RHS_mat = U_mat+beta*V_lin; % (k’,k)
[V1,pol_k_ind] = max(RHS_mat,[],1);
pol_k = k_grid(pol_k_ind);
% Plots
figure
plot(k_grid,V1)
figure
plot(k_grid,k_grid,’–‘,k_grid,pol_k)

function util = f_util(c,crra)
util = c.^(1-crra)/(1-crra);
end

PROFILEI have to solve a dynamic programming problem using a linear programming approach. For details, please see this paper. The LP that I want to solve is:
min c’*v
s.t.
A*v>=u,
where c is n*1, v is n*1, A is n^2*n, u is n^2*1.
The min is with respect to v, the value function of the original DP problem. I have a moderate number of variables, n=300 and m=n^2*n=90000 linear inequalities as constraints. No bound constraints on v.
I use the Matlab function linprog which in turn is based on the solver HIGHS (since R2024a). The code is slow for my purposes (i.e. a brute-force value iteration is much faster). Moreover, linprog gives correct results only if I set the option ‘Algorithm’,’dual-simplex-highs’. With other algorithms, it gets stuck.
After profiling the code, it turns out that the bottleneck is line 377 of linprog:
[x, fval, exitflag, output, lambda] = run(algorithm, problem);
I was wondering if there is a way to speed up the code. Any help or suggestion is greatly appreciated! I put below a MWE to illustrate the problem.

clear,clc,close all
%% Set parameters
crra = 2;
alpha = 0.36;
beta = 0.95;
delta = 0.1;
%% Grid for capital
k_ss = ((1-beta*(1-delta))/(alpha*beta))^(1/(alpha-1));
n_k = 300;
k_grid = linspace(0.1*k_ss,1.5*k_ss,n_k)’;
%% Build current return matrix, U(k’,k)
cons = k_grid’.^alpha+(1-delta)*k_grid’-k_grid;
U_mat = f_util(cons,crra);
U_mat(cons<=0) = -inf;
%% Using LINEAR PROGRAMMING
% min c’*v
% s.t.
% A*v>=u, where c is n*1, v is n*1, A is n^2*n, u is n^2*1
n = length(k_grid);
c_vec = ones(n,1);
u_vec = U_mat(:); %% U(k’,k), stack columnwise
%% Build A matrix using cell-based method
tic
A = cell(n,1);
bigI = (-beta)*speye(n);
for i=1:n
temp = bigI;
temp(:,i) = temp(:,i)+1;
A{i} = temp;
end
A = vertcat(A{:});
disp(‘Time to build A matrix with cell method:’)
toc
%% Call linprog
% ‘dual-simplex-highs’ (default and by far the best)
options = optimoptions(‘linprog’,’Algorithm’,’dual-simplex-highs’);
tic
[V_lin,fval,exitflag,output] = linprog(c_vec,-A,-u_vec,[],[],[],[],options);
disp(‘Time linear programming:’)
toc
if exitflag<=0
warning(‘linprog did not find a solution’)
fprintf(‘exitflag = %d n’,exitflag)
end
%% Now that we solved for V, compute policy function
RHS_mat = U_mat+beta*V_lin; % (k’,k)
[V1,pol_k_ind] = max(RHS_mat,[],1);
pol_k = k_grid(pol_k_ind);
% Plots
figure
plot(k_grid,V1)
figure
plot(k_grid,k_grid,’–‘,k_grid,pol_k)

function util = f_util(c,crra)
util = c.^(1-crra)/(1-crra);
end

PROFILE I have to solve a dynamic programming problem using a linear programming approach. For details, please see this paper. The LP that I want to solve is:
min c’*v
s.t.
A*v>=u,
where c is n*1, v is n*1, A is n^2*n, u is n^2*1.
The min is with respect to v, the value function of the original DP problem. I have a moderate number of variables, n=300 and m=n^2*n=90000 linear inequalities as constraints. No bound constraints on v.
I use the Matlab function linprog which in turn is based on the solver HIGHS (since R2024a). The code is slow for my purposes (i.e. a brute-force value iteration is much faster). Moreover, linprog gives correct results only if I set the option ‘Algorithm’,’dual-simplex-highs’. With other algorithms, it gets stuck.
After profiling the code, it turns out that the bottleneck is line 377 of linprog:
[x, fval, exitflag, output, lambda] = run(algorithm, problem);
I was wondering if there is a way to speed up the code. Any help or suggestion is greatly appreciated! I put below a MWE to illustrate the problem.

clear,clc,close all
%% Set parameters
crra = 2;
alpha = 0.36;
beta = 0.95;
delta = 0.1;
%% Grid for capital
k_ss = ((1-beta*(1-delta))/(alpha*beta))^(1/(alpha-1));
n_k = 300;
k_grid = linspace(0.1*k_ss,1.5*k_ss,n_k)’;
%% Build current return matrix, U(k’,k)
cons = k_grid’.^alpha+(1-delta)*k_grid’-k_grid;
U_mat = f_util(cons,crra);
U_mat(cons<=0) = -inf;
%% Using LINEAR PROGRAMMING
% min c’*v
% s.t.
% A*v>=u, where c is n*1, v is n*1, A is n^2*n, u is n^2*1
n = length(k_grid);
c_vec = ones(n,1);
u_vec = U_mat(:); %% U(k’,k), stack columnwise
%% Build A matrix using cell-based method
tic
A = cell(n,1);
bigI = (-beta)*speye(n);
for i=1:n
temp = bigI;
temp(:,i) = temp(:,i)+1;
A{i} = temp;
end
A = vertcat(A{:});
disp(‘Time to build A matrix with cell method:’)
toc
%% Call linprog
% ‘dual-simplex-highs’ (default and by far the best)
options = optimoptions(‘linprog’,’Algorithm’,’dual-simplex-highs’);
tic
[V_lin,fval,exitflag,output] = linprog(c_vec,-A,-u_vec,[],[],[],[],options);
disp(‘Time linear programming:’)
toc
if exitflag<=0
warning(‘linprog did not find a solution’)
fprintf(‘exitflag = %d n’,exitflag)
end
%% Now that we solved for V, compute policy function
RHS_mat = U_mat+beta*V_lin; % (k’,k)
[V1,pol_k_ind] = max(RHS_mat,[],1);
pol_k = k_grid(pol_k_ind);
% Plots
figure
plot(k_grid,V1)
figure
plot(k_grid,k_grid,’–‘,k_grid,pol_k)

function util = f_util(c,crra)
util = c.^(1-crra)/(1-crra);
end

PROFILE linprog, performance MATLAB Answers — New Questions

​

syms V1 V2 Vp Vn Vxp Vxn Vpx Vnx
syms Rp Rx Rn positive
Ison=((Vp/((Rp*68220)/(Rp+68220)))-(Vxp/((Rx*68220)/(Rx+68220))));
equ1=Vp==(V1+V2)-Ison*Rn;
Isop=((Vn/((Rn*68220)/(Rn+68220)))-(Vxn/((Rx*68220)/(Rx+68220))));
equ2=Vn==(V1+V2)-Isop*Rp;
Isox=((Vnx/((Rn*68220)/(Rn+68220)))-(Vpx/((Rp*68220)/(Rp+68220))));
equ3= Vpx==V1+Rx*Isox;
[Rn, Rp, Rx] = solve([equ1, equ2,equ3],[Rn , Rp, Rx]);

simplify(Rn)
simplify(Rp)
simplify(Rx)
Rn=vpa(subs(Rn,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))
Rp=vpa(subs(Rp,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))
Rx=vpa(subs(Rx,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))
So I am trying to find values for Rn,Rx and Rp. I have three equations and three unknowns. When I use solve to find symbolic equation, I get equations for each variable. But when I sub real values in, I get an answer that is wrong or doesnt make sense. Adding assume doesnt seem to change anything and the return conditions is the same as the equations. The real values of Rn, Rp and Rx should be 2e6,1.5e6 and 700k. If you were to put these values in the original equations, you can verify that these values are correct. How do I get a system of equations that will give me the desired answer?syms V1 V2 Vp Vn Vxp Vxn Vpx Vnx
syms Rp Rx Rn positive
Ison=((Vp/((Rp*68220)/(Rp+68220)))-(Vxp/((Rx*68220)/(Rx+68220))));
equ1=Vp==(V1+V2)-Ison*Rn;
Isop=((Vn/((Rn*68220)/(Rn+68220)))-(Vxn/((Rx*68220)/(Rx+68220))));
equ2=Vn==(V1+V2)-Isop*Rp;
Isox=((Vnx/((Rn*68220)/(Rn+68220)))-(Vpx/((Rp*68220)/(Rp+68220))));
equ3= Vpx==V1+Rx*Isox;
[Rn, Rp, Rx] = solve([equ1, equ2,equ3],[Rn , Rp, Rx]);

simplify(Rn)
simplify(Rp)
simplify(Rx)
Rn=vpa(subs(Rn,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))
Rp=vpa(subs(Rp,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))
Rx=vpa(subs(Rx,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))
So I am trying to find values for Rn,Rx and Rp. I have three equations and three unknowns. When I use solve to find symbolic equation, I get equations for each variable. But when I sub real values in, I get an answer that is wrong or doesnt make sense. Adding assume doesnt seem to change anything and the return conditions is the same as the equations. The real values of Rn, Rp and Rx should be 2e6,1.5e6 and 700k. If you were to put these values in the original equations, you can verify that these values are correct. How do I get a system of equations that will give me the desired answer? syms V1 V2 Vp Vn Vxp Vxn Vpx Vnx
syms Rp Rx Rn positive
Ison=((Vp/((Rp*68220)/(Rp+68220)))-(Vxp/((Rx*68220)/(Rx+68220))));
equ1=Vp==(V1+V2)-Ison*Rn;
Isop=((Vn/((Rn*68220)/(Rn+68220)))-(Vxn/((Rx*68220)/(Rx+68220))));
equ2=Vn==(V1+V2)-Isop*Rp;
Isox=((Vnx/((Rn*68220)/(Rn+68220)))-(Vpx/((Rp*68220)/(Rp+68220))));
equ3= Vpx==V1+Rx*Isox;
[Rn, Rp, Rx] = solve([equ1, equ2,equ3],[Rn , Rp, Rx]);

simplify(Rn)
simplify(Rp)
simplify(Rx)
Rn=vpa(subs(Rn,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))
Rp=vpa(subs(Rp,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))
Rx=vpa(subs(Rx,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))
So I am trying to find values for Rn,Rx and Rp. I have three equations and three unknowns. When I use solve to find symbolic equation, I get equations for each variable. But when I sub real values in, I get an answer that is wrong or doesnt make sense. Adding assume doesnt seem to change anything and the return conditions is the same as the equations. The real values of Rn, Rp and Rx should be 2e6,1.5e6 and 700k. If you were to put these values in the original equations, you can verify that these values are correct. How do I get a system of equations that will give me the desired answer? solve, symbolic MATLAB Answers — New Questions

​

I’ve been trying to read data from ESP32S-Dev module to MATLAB sending over the data in bytes to make the transmission quicker. As seen below, I convert float data into bytes and then write the data. MATLAB sees the data and then creates a 1×8 array for each float value instead of just 1 value for each float. I’ve tried other methods, shown in 2nd part of MATLAB code, and it creates 1 value for each float but their wildly off (7-34 orders of magnitude)
I’ve tried different baud rates and even asked for help from the almighty ChatGPT but have not been able to crack the code. Any thoughts.
Arduino IDE code:
#include <SPI.h>

const int CS_SiPM_pot = 34;

void setup() {
// put your setup code here, to run once:
Serial.begin(115200); //Starting
delay(1000);
SPI.begin();
pinMode(CS_SiPM_pot, OUTPUT);
}

void loop() {
for (int d_pot2 = 55; d_pot2 <= 55; d_pot2 = d_pot2 + 1) //57
{
digitalWrite(CS_SiPM_pot, LOW); //LOW
delay(10); //10
SPI.transfer(d_pot2);
digitalWrite(CS_SiPM_pot, HIGH);
// put your main code here, to run repeatedly:
float newTmp[4] = { 01.00, 02.00, 03.00, 04.00 };
for (int i = 0; i < 4; i++) {
byte *byteData = (byte *)&newTmp[i];
Serial.write(byteData, sizeof(float)); // Send each float as 4 bytes
delay(10);
}
}
}

MATLAB code:
esp = serialport(‘COM5’,152000);
%%
flush(esp)
% Read 16 bytes of data (4 floats * 4 bytes per float)
numBytes = 16;
data = read(esp, numBytes, ‘uint8’);

% Convert each 4-byte sequence to a float
float1 = typecast(data(1:4)), ‘single’);
float2 = typecast(data(5:8), ‘single’);
float3 = typecast(data(9:12), ‘single’);
float4 = typecast(data(13:16), ‘single’);

disp([float1, float2, float3, float4]);

% Convert the bytes back to float values
floatValues = typecast(uint8(data), ‘single’);

% Display the result
disp(‘Float values:’);
disp(floatValues);

disp(‘Raw bytes received:’);
disp(data); % Print the raw bytes
Results from MATLAB:
Float1: 0 0 0 0 0 0 0 0
Float2: 0 3.7480 0 0 0 0 0 0
Float3: 0 2.1250 0 3.7480 0 0 0 0
Float4: 0 0 0 0 0 3.7480 0 0
data: 0 0 0 0 255 0 0 0 3 255 0 0 0 0 255 0I’ve been trying to read data from ESP32S-Dev module to MATLAB sending over the data in bytes to make the transmission quicker. As seen below, I convert float data into bytes and then write the data. MATLAB sees the data and then creates a 1×8 array for each float value instead of just 1 value for each float. I’ve tried other methods, shown in 2nd part of MATLAB code, and it creates 1 value for each float but their wildly off (7-34 orders of magnitude)
I’ve tried different baud rates and even asked for help from the almighty ChatGPT but have not been able to crack the code. Any thoughts.
Arduino IDE code:
#include <SPI.h>

const int CS_SiPM_pot = 34;

void setup() {
// put your setup code here, to run once:
Serial.begin(115200); //Starting
delay(1000);
SPI.begin();
pinMode(CS_SiPM_pot, OUTPUT);
}

void loop() {
for (int d_pot2 = 55; d_pot2 <= 55; d_pot2 = d_pot2 + 1) //57
{
digitalWrite(CS_SiPM_pot, LOW); //LOW
delay(10); //10
SPI.transfer(d_pot2);
digitalWrite(CS_SiPM_pot, HIGH);
// put your main code here, to run repeatedly:
float newTmp[4] = { 01.00, 02.00, 03.00, 04.00 };
for (int i = 0; i < 4; i++) {
byte *byteData = (byte *)&newTmp[i];
Serial.write(byteData, sizeof(float)); // Send each float as 4 bytes
delay(10);
}
}
}

MATLAB code:
esp = serialport(‘COM5’,152000);
%%
flush(esp)
% Read 16 bytes of data (4 floats * 4 bytes per float)
numBytes = 16;
data = read(esp, numBytes, ‘uint8’);

% Convert each 4-byte sequence to a float
float1 = typecast(data(1:4)), ‘single’);
float2 = typecast(data(5:8), ‘single’);
float3 = typecast(data(9:12), ‘single’);
float4 = typecast(data(13:16), ‘single’);

disp([float1, float2, float3, float4]);

% Convert the bytes back to float values
floatValues = typecast(uint8(data), ‘single’);

% Display the result
disp(‘Float values:’);
disp(floatValues);

disp(‘Raw bytes received:’);
disp(data); % Print the raw bytes
Results from MATLAB:
Float1: 0 0 0 0 0 0 0 0
Float2: 0 3.7480 0 0 0 0 0 0
Float3: 0 2.1250 0 3.7480 0 0 0 0
Float4: 0 0 0 0 0 3.7480 0 0
data: 0 0 0 0 255 0 0 0 3 255 0 0 0 0 255 0 I’ve been trying to read data from ESP32S-Dev module to MATLAB sending over the data in bytes to make the transmission quicker. As seen below, I convert float data into bytes and then write the data. MATLAB sees the data and then creates a 1×8 array for each float value instead of just 1 value for each float. I’ve tried other methods, shown in 2nd part of MATLAB code, and it creates 1 value for each float but their wildly off (7-34 orders of magnitude)
I’ve tried different baud rates and even asked for help from the almighty ChatGPT but have not been able to crack the code. Any thoughts.
Arduino IDE code:
#include <SPI.h>

const int CS_SiPM_pot = 34;

void setup() {
// put your setup code here, to run once:
Serial.begin(115200); //Starting
delay(1000);
SPI.begin();
pinMode(CS_SiPM_pot, OUTPUT);
}

void loop() {
for (int d_pot2 = 55; d_pot2 <= 55; d_pot2 = d_pot2 + 1) //57
{
digitalWrite(CS_SiPM_pot, LOW); //LOW
delay(10); //10
SPI.transfer(d_pot2);
digitalWrite(CS_SiPM_pot, HIGH);
// put your main code here, to run repeatedly:
float newTmp[4] = { 01.00, 02.00, 03.00, 04.00 };
for (int i = 0; i < 4; i++) {
byte *byteData = (byte *)&newTmp[i];
Serial.write(byteData, sizeof(float)); // Send each float as 4 bytes
delay(10);
}
}
}

MATLAB code:
esp = serialport(‘COM5’,152000);
%%
flush(esp)
% Read 16 bytes of data (4 floats * 4 bytes per float)
numBytes = 16;
data = read(esp, numBytes, ‘uint8’);

% Convert each 4-byte sequence to a float
float1 = typecast(data(1:4)), ‘single’);
float2 = typecast(data(5:8), ‘single’);
float3 = typecast(data(9:12), ‘single’);
float4 = typecast(data(13:16), ‘single’);

disp([float1, float2, float3, float4]);

% Convert the bytes back to float values
floatValues = typecast(uint8(data), ‘single’);

% Display the result
disp(‘Float values:’);
disp(floatValues);

disp(‘Raw bytes received:’);
disp(data); % Print the raw bytes
Results from MATLAB:
Float1: 0 0 0 0 0 0 0 0
Float2: 0 3.7480 0 0 0 0 0 0
Float3: 0 2.1250 0 3.7480 0 0 0 0
Float4: 0 0 0 0 0 3.7480 0 0
data: 0 0 0 0 255 0 0 0 3 255 0 0 0 0 255 0 arduino, esp32, matlab, serialport, float-to-bytes MATLAB Answers — New Questions

​

The Summery is essentially it. I created a script four months ago and accidentially used the wrong syntax (&&) and it has corrupted my Matlab Online. I can run scripts with only comments, but as soon as I add anything else I get the errors back.

Help? I am so confused

Script I am running:
%% Problem 1

k1 = 10;
k2 = 5;
k3 = 8;
k4 = 10;
k5 = 20;

L1 = [1 0 0 0;
0 1 0 0]
Error Message:

Operands to the logical AND (&&) and OR (||) operators must be convertible to logical scalar values. Use the ANY or ALL functions to
reduce operands to logical scalar values.

Error in internal.matlab.variableeditor.Actions.struct.EditActionBase/UpdateActionState (line 33)
editable = height(sRows) == 1 && (sRows(2)-sRows(1)+1 == 1);

Error in internal.matlab.variableeditor.VEActionDataService/updateActionStates (line 138)
action.UpdateActionState();

Error in internal.matlab.variableeditor.VEActionDataService/enableListenersAndUpdateActionStates (line 84)
this.updateActionStates();

Error in internal.matlab.variableeditor.VEActionDataService/initActionStates (line 54)
this.enableListenersAndUpdateActionStates();

Error in internal.matlab.datatoolsservices.actiondataservice.ActionManager/initActions (line 55)
this.ActionDataService.initActionStates();

Error in internal.matlab.variableeditor.peer.RemoteManager/callInitActionOnIdle (line 376)
ActionManager.initActions(startPath, classType);

Error in internal.matlab.variableeditor.peer.RemoteManager>@()this.callInitActionOnIdle(actionNamespace,startPath,classType,ActionManager) (line 121)
builtin(‘_dtcallback’, @() this.callInitActionOnIdle(actionNamespace, startPath, classType, ActionManager), …The Summery is essentially it. I created a script four months ago and accidentially used the wrong syntax (&&) and it has corrupted my Matlab Online. I can run scripts with only comments, but as soon as I add anything else I get the errors back.

Help? I am so confused

Script I am running:
%% Problem 1

k1 = 10;
k2 = 5;
k3 = 8;
k4 = 10;
k5 = 20;

L1 = [1 0 0 0;
0 1 0 0]
Error Message:

Operands to the logical AND (&&) and OR (||) operators must be convertible to logical scalar values. Use the ANY or ALL functions to
reduce operands to logical scalar values.

Error in internal.matlab.variableeditor.Actions.struct.EditActionBase/UpdateActionState (line 33)
editable = height(sRows) == 1 && (sRows(2)-sRows(1)+1 == 1);

Error in internal.matlab.variableeditor.VEActionDataService/updateActionStates (line 138)
action.UpdateActionState();

Error in internal.matlab.variableeditor.VEActionDataService/enableListenersAndUpdateActionStates (line 84)
this.updateActionStates();

Error in internal.matlab.variableeditor.VEActionDataService/initActionStates (line 54)
this.enableListenersAndUpdateActionStates();

Error in internal.matlab.datatoolsservices.actiondataservice.ActionManager/initActions (line 55)
this.ActionDataService.initActionStates();

Error in internal.matlab.variableeditor.peer.RemoteManager/callInitActionOnIdle (line 376)
ActionManager.initActions(startPath, classType);

Error in internal.matlab.variableeditor.peer.RemoteManager>@()this.callInitActionOnIdle(actionNamespace,startPath,classType,ActionManager) (line 121)
builtin(‘_dtcallback’, @() this.callInitActionOnIdle(actionNamespace, startPath, classType, ActionManager), … The Summery is essentially it. I created a script four months ago and accidentially used the wrong syntax (&&) and it has corrupted my Matlab Online. I can run scripts with only comments, but as soon as I add anything else I get the errors back.

Help? I am so confused

Script I am running:
%% Problem 1

k1 = 10;
k2 = 5;
k3 = 8;
k4 = 10;
k5 = 20;

L1 = [1 0 0 0;
0 1 0 0]
Error Message:

Operands to the logical AND (&&) and OR (||) operators must be convertible to logical scalar values. Use the ANY or ALL functions to
reduce operands to logical scalar values.

Error in internal.matlab.variableeditor.Actions.struct.EditActionBase/UpdateActionState (line 33)
editable = height(sRows) == 1 && (sRows(2)-sRows(1)+1 == 1);

Error in internal.matlab.variableeditor.VEActionDataService/updateActionStates (line 138)
action.UpdateActionState();

Error in internal.matlab.variableeditor.VEActionDataService/enableListenersAndUpdateActionStates (line 84)
this.updateActionStates();

Error in internal.matlab.variableeditor.VEActionDataService/initActionStates (line 54)
this.enableListenersAndUpdateActionStates();

Error in internal.matlab.datatoolsservices.actiondataservice.ActionManager/initActions (line 55)
this.ActionDataService.initActionStates();

Error in internal.matlab.variableeditor.peer.RemoteManager/callInitActionOnIdle (line 376)
ActionManager.initActions(startPath, classType);

Error in internal.matlab.variableeditor.peer.RemoteManager>@()this.callInitActionOnIdle(actionNamespace,startPath,classType,ActionManager) (line 121)
builtin(‘_dtcallback’, @() this.callInitActionOnIdle(actionNamespace, startPath, classType, ActionManager), … help, persistent error, have i broken matlab? MATLAB Answers — New Questions

​

I have a 1400*9 table (1400 rows and 9 columns) in MATLAB workspace. Each column has a different name. I want to select rows containing specific value. For example, how can I create a new table containig all rows with value of 0.456 in the column named "Biology"? I tried to use following code, but it gave error.
newData=[Data(find(Data.Biology == 0.456, :))]I have a 1400*9 table (1400 rows and 9 columns) in MATLAB workspace. Each column has a different name. I want to select rows containing specific value. For example, how can I create a new table containig all rows with value of 0.456 in the column named "Biology"? I tried to use following code, but it gave error.
newData=[Data(find(Data.Biology == 0.456, :))] I have a 1400*9 table (1400 rows and 9 columns) in MATLAB workspace. Each column has a different name. I want to select rows containing specific value. For example, how can I create a new table containig all rows with value of 0.456 in the column named "Biology"? I tried to use following code, but it gave error.
newData=[Data(find(Data.Biology == 0.456, :))] table, find MATLAB Answers — New Questions

​

On matlab, I am going to import an Excel file and work on it live in real time.

If I change the value in Excel, I want to reflect it in real time in MATLAB.

However, as far as I know, if you modify it without saving the Excel, it cannot be applied in MATLAB right away.

Is there a way to solve this?On matlab, I am going to import an Excel file and work on it live in real time.

If I change the value in Excel, I want to reflect it in real time in MATLAB.

However, as far as I know, if you modify it without saving the Excel, it cannot be applied in MATLAB right away.

Is there a way to solve this? On matlab, I am going to import an Excel file and work on it live in real time.

If I change the value in Excel, I want to reflect it in real time in MATLAB.

However, as far as I know, if you modify it without saving the Excel, it cannot be applied in MATLAB right away.

Is there a way to solve this? excel, live MATLAB Answers — New Questions

​