Hello all.
Is it possible to connect a UE to a nrGNB in both download and upload links at the same time?
In my scenario, I have to simulate UE connecting in both links at the same gNB or in different ones simultaneously, depending on some current evaluations.
In "NR Intercell Interference Modeling" and "Connect UEs to gNB" examples, I could see how to do this separately. But couldn’t understand if and, if possible, how to do it.
Any clue?
[]’s.Hello all.
Is it possible to connect a UE to a nrGNB in both download and upload links at the same time?
In my scenario, I have to simulate UE connecting in both links at the same gNB or in different ones simultaneously, depending on some current evaluations.
In "NR Intercell Interference Modeling" and "Connect UEs to gNB" examples, I could see how to do this separately. But couldn’t understand if and, if possible, how to do it.
Any clue?
[]’s. Hello all.
Is it possible to connect a UE to a nrGNB in both download and upload links at the same time?
In my scenario, I have to simulate UE connecting in both links at the same gNB or in different ones simultaneously, depending on some current evaluations.
In "NR Intercell Interference Modeling" and "Connect UEs to gNB" examples, I could see how to do this separately. But couldn’t understand if and, if possible, how to do it.
Any clue?
[]’s. 5g, sub-6 ghz, nrgnb, ue, connection MATLAB Answers — New Questions

​

Matlab 2024b。
I have installed help doc and rebooted Matlab.
However, whether I choose to open in a small window or in the help browser, it will only open in the browser.

Like this:Matlab 2024b。
I have installed help doc and rebooted Matlab.
However, whether I choose to open in a small window or in the help browser, it will only open in the browser.

Like this: Matlab 2024b。
I have installed help doc and rebooted Matlab.
However, whether I choose to open in a small window or in the help browser, it will only open in the browser.

Like this: help doc MATLAB Answers — New Questions

​

I have written MATLAB code to connect to an Oracle Database using ODBC. I have the ODBC driver installed, and the MATLAB code runs successfully in MATLAB. However, I need to compile the code for deployment as a standalone application, and the executable throws the following error when I run it:
"ODBC Driver Error: Specified driver could not be loaded due to system error 1114: A dynamic link library (DLL) initialization routine failed. (Oracle in instantclient_19_18_64, C:oracleora19c-64instantclient_19_18_64SQORA32.dll)."
I have confirmed that the ODBC Driver (ex. C:oracleora19c-64instantclient_19_18_64SQORA32.dll) exists on the machine where the application is being executed.I have written MATLAB code to connect to an Oracle Database using ODBC. I have the ODBC driver installed, and the MATLAB code runs successfully in MATLAB. However, I need to compile the code for deployment as a standalone application, and the executable throws the following error when I run it:
"ODBC Driver Error: Specified driver could not be loaded due to system error 1114: A dynamic link library (DLL) initialization routine failed. (Oracle in instantclient_19_18_64, C:oracleora19c-64instantclient_19_18_64SQORA32.dll)."
I have confirmed that the ODBC Driver (ex. C:oracleora19c-64instantclient_19_18_64SQORA32.dll) exists on the machine where the application is being executed. I have written MATLAB code to connect to an Oracle Database using ODBC. I have the ODBC driver installed, and the MATLAB code runs successfully in MATLAB. However, I need to compile the code for deployment as a standalone application, and the executable throws the following error when I run it:
"ODBC Driver Error: Specified driver could not be loaded due to system error 1114: A dynamic link library (DLL) initialization routine failed. (Oracle in instantclient_19_18_64, C:oracleora19c-64instantclient_19_18_64SQORA32.dll)."
I have confirmed that the ODBC Driver (ex. C:oracleora19c-64instantclient_19_18_64SQORA32.dll) exists on the machine where the application is being executed. database, oracle, odbc, deployed, standalone, application, compiler, driver MATLAB Answers — New Questions

​

Dear all
I am trying to code for the displacements of statically Inderterminant beam using the finite difference method. It appears that I am not that far off from achieving this, however, in one of the points that I have discretized I am getting an incorrect result.
Here, is the graph that comes from the software and what I am trying to replicate:

Here is the graph in MATAB:

As can been seen there is an anomaly at x = 4.1m at which the vertical result is zero. The displacement should only be zero at x = 0m, 5m and 8m.
I don’t know if there is an error in my code, but I can’t seem to find anything myself.

Here is my code:

clear, clc, close all

% Structural Properties
E = 2.1E+08 % Modulus of elasticity
I = 18890/100^4 % Second moment of area
EI = E*I % Flexural Rigidity
L = 8 % Beam Length
N = 161 % Number of points

x = linspace(0,L,N) % Discretise the beam
h = L/(N-1) % Step size
q = 8 % Load intensity

% Set up matrix using fourth order finite difference

mat1 = diag(ones(1,N)*6) ;
mat2 = diag(ones(1,N-1)*-4, 1);
mat3 = diag(ones(1,N-1)*-4, -1);
mat4 = diag(ones(1,N-2)*1, 2);
mat5 = diag(ones(1,N-2)*1, -2);

Mat = [mat1 + mat2 + mat3 + mat4 + mat5]

% Add boundary conditions. Slope = 0 at x = 0 – Use first order finite difference

mat6 = zeros(1,N);
mat6(1,1) = -3;
mat6(1,2) = 4;
mat6(1,3) = -1;
Mat(2,:) = mat6

% Add boundary conditions. Moment = 0 at x = L – using second order finite difference method

mat7 = zeros(1,N);
mat7(1,N-3) = 2
mat7(1,N-2) = -5;
mat7(1,N-1) = 4;
mat7(1,N) = -1
Mat(N-1,:) = mat7

% Remove rows and columns where displacement is zero
% Displacement equals zero at x = 0, x = L and x = 5/8 of beam length

Mat(:,[1,N,(N-1)*5/8]) = [];
Mat([1,N,(N-1)*5/8],:) = []

% Set up right-hand side

RHS = ones(1,N)*q*h^4/EI;
RHS(2) = 0;
RHS(N-1) = 0

% Remove rows and columns where displacement is zero
% Displacement equals zero at x = 0, x = L and x = 5/8 of beam length

RHS([1,N,(N-1)*5/8]) = []

% Solve for displacement

v = (inv(Mat)*RHS’)*1000

% Make new vector with displacements that equal zero

V = [0;v(1:(N+1)/2);0;v(((N+1)/2)+1:end);0]’

% Plot the displacement against beam length

plot(x,-V)
grid on
xlabel(‘Distance (m)’)
ylabel(‘Displacement (mm)’)
title(‘Local Displacements’)

In terms of the code, can anybody see what might be causing this anomaly?

Many thanks in advance!Dear all
I am trying to code for the displacements of statically Inderterminant beam using the finite difference method. It appears that I am not that far off from achieving this, however, in one of the points that I have discretized I am getting an incorrect result.
Here, is the graph that comes from the software and what I am trying to replicate:

Here is the graph in MATAB:

As can been seen there is an anomaly at x = 4.1m at which the vertical result is zero. The displacement should only be zero at x = 0m, 5m and 8m.
I don’t know if there is an error in my code, but I can’t seem to find anything myself.

Here is my code:

clear, clc, close all

% Structural Properties
E = 2.1E+08 % Modulus of elasticity
I = 18890/100^4 % Second moment of area
EI = E*I % Flexural Rigidity
L = 8 % Beam Length
N = 161 % Number of points

x = linspace(0,L,N) % Discretise the beam
h = L/(N-1) % Step size
q = 8 % Load intensity

% Set up matrix using fourth order finite difference

mat1 = diag(ones(1,N)*6) ;
mat2 = diag(ones(1,N-1)*-4, 1);
mat3 = diag(ones(1,N-1)*-4, -1);
mat4 = diag(ones(1,N-2)*1, 2);
mat5 = diag(ones(1,N-2)*1, -2);

Mat = [mat1 + mat2 + mat3 + mat4 + mat5]

% Add boundary conditions. Slope = 0 at x = 0 – Use first order finite difference

mat6 = zeros(1,N);
mat6(1,1) = -3;
mat6(1,2) = 4;
mat6(1,3) = -1;
Mat(2,:) = mat6

% Add boundary conditions. Moment = 0 at x = L – using second order finite difference method

mat7 = zeros(1,N);
mat7(1,N-3) = 2
mat7(1,N-2) = -5;
mat7(1,N-1) = 4;
mat7(1,N) = -1
Mat(N-1,:) = mat7

% Remove rows and columns where displacement is zero
% Displacement equals zero at x = 0, x = L and x = 5/8 of beam length

Mat(:,[1,N,(N-1)*5/8]) = [];
Mat([1,N,(N-1)*5/8],:) = []

% Set up right-hand side

RHS = ones(1,N)*q*h^4/EI;
RHS(2) = 0;
RHS(N-1) = 0

% Remove rows and columns where displacement is zero
% Displacement equals zero at x = 0, x = L and x = 5/8 of beam length

RHS([1,N,(N-1)*5/8]) = []

% Solve for displacement

v = (inv(Mat)*RHS’)*1000

% Make new vector with displacements that equal zero

V = [0;v(1:(N+1)/2);0;v(((N+1)/2)+1:end);0]’

% Plot the displacement against beam length

plot(x,-V)
grid on
xlabel(‘Distance (m)’)
ylabel(‘Displacement (mm)’)
title(‘Local Displacements’)

In terms of the code, can anybody see what might be causing this anomaly?

Many thanks in advance! Dear all
I am trying to code for the displacements of statically Inderterminant beam using the finite difference method. It appears that I am not that far off from achieving this, however, in one of the points that I have discretized I am getting an incorrect result.
Here, is the graph that comes from the software and what I am trying to replicate:

Here is the graph in MATAB:

As can been seen there is an anomaly at x = 4.1m at which the vertical result is zero. The displacement should only be zero at x = 0m, 5m and 8m.
I don’t know if there is an error in my code, but I can’t seem to find anything myself.

Here is my code:

clear, clc, close all

% Structural Properties
E = 2.1E+08 % Modulus of elasticity
I = 18890/100^4 % Second moment of area
EI = E*I % Flexural Rigidity
L = 8 % Beam Length
N = 161 % Number of points

x = linspace(0,L,N) % Discretise the beam
h = L/(N-1) % Step size
q = 8 % Load intensity

% Set up matrix using fourth order finite difference

mat1 = diag(ones(1,N)*6) ;
mat2 = diag(ones(1,N-1)*-4, 1);
mat3 = diag(ones(1,N-1)*-4, -1);
mat4 = diag(ones(1,N-2)*1, 2);
mat5 = diag(ones(1,N-2)*1, -2);

Mat = [mat1 + mat2 + mat3 + mat4 + mat5]

% Add boundary conditions. Slope = 0 at x = 0 – Use first order finite difference

mat6 = zeros(1,N);
mat6(1,1) = -3;
mat6(1,2) = 4;
mat6(1,3) = -1;
Mat(2,:) = mat6

% Add boundary conditions. Moment = 0 at x = L – using second order finite difference method

mat7 = zeros(1,N);
mat7(1,N-3) = 2
mat7(1,N-2) = -5;
mat7(1,N-1) = 4;
mat7(1,N) = -1
Mat(N-1,:) = mat7

% Remove rows and columns where displacement is zero
% Displacement equals zero at x = 0, x = L and x = 5/8 of beam length

Mat(:,[1,N,(N-1)*5/8]) = [];
Mat([1,N,(N-1)*5/8],:) = []

% Set up right-hand side

RHS = ones(1,N)*q*h^4/EI;
RHS(2) = 0;
RHS(N-1) = 0

% Remove rows and columns where displacement is zero
% Displacement equals zero at x = 0, x = L and x = 5/8 of beam length

RHS([1,N,(N-1)*5/8]) = []

% Solve for displacement

v = (inv(Mat)*RHS’)*1000

% Make new vector with displacements that equal zero

V = [0;v(1:(N+1)/2);0;v(((N+1)/2)+1:end);0]’

% Plot the displacement against beam length

plot(x,-V)
grid on
xlabel(‘Distance (m)’)
ylabel(‘Displacement (mm)’)
title(‘Local Displacements’)

In terms of the code, can anybody see what might be causing this anomaly?

Many thanks in advance! matrices, finite diffference method, boundary value problems MATLAB Answers — New Questions

​

I am retraining YAMNet for a binary classification task, operating on spectrograms of audio signals. My training audio has two classes, positive and negative. Audio is preprocessed & features extracted using yamnetPreprocess(). When training the network, trainnet() produces the following error:
Error using deep.internal.train.Trainer/train (line 74)
Number of channels in predictions (2) must match the number of
channels in the targets (3).

Error in deep.internal.train.ParallelTrainer>iTrainWithSplitCommunicator (line 227)
remoteNetwork = train(remoteTrainer, remoteNetwork, workerMbq);

Error in deep.internal.train.ParallelTrainer/computeTraining (line 127)
spmd

Error in deep.internal.train.Trainer/train (line 59)
net = computeTraining(trainer, net, mbq);

Error in deep.internal.train.trainnet (line 54)
net = train(trainer, net, mbq);

Error in trainnet (line 42)
[net,info] = deep.internal.train.trainnet(mbq, net, loss, options, …

Error in train_DenseNet_detector_from_semi_synthetic_dataset (line 192)
[trained_network, train_info] = trainnet(trainFeatures, trainLabels’, net, "crossentropy", options);
My understanding of this error is that it indicates a mismatch between the number of classes the network expects, and the number of classes in the dataset. I do not see how this can be possible, considering the number of classes in the network is explicitly set by the number of classes in the datastore:
classNames = unique(ads.Labels);
numClasses = numel(classNames);
net = audioPretrainedNetwork("yamnet", NumClasses=numClasses);
My script is based on this MATLAB tutorial: audioPretrainedNetwork and there are no functional differences in the way I’m building datastores or preprocessing the data. The training options and the call to trainnet() are configured as follows:
options = trainingOptions(‘adam’, …
InitialLearnRate = initial_learn_rate, …
MaxEpochs = max_epochs, …
MiniBatchSize = mini_batch_size, …
Shuffle = "every-epoch", …
Plots = "training-progress", …
Metrics = "accuracy", …
Verbose = 1, …
ValidationData = {single(validationFeatures), validationLabels’}, …
ValidationFrequency = validationFrequency,…
ExecutionEnvironment="parallel-auto");

[trained_network, train_info] = trainnet(trainFeatures, trainLabels’, net, "crossentropy", options);
Relevant variable dimensions are as follows:
>> unique(ads.Labels)
ans =
2×1 categorical array
negative
positiveNoisy

>> size(trainLabels)
ans =
1 16240

>> size(trainFeatures)
ans =
96 64 1 16240

>> size(validationLabels)
ans =
1 6960

>> size(validationFeatures)
ans =
96 64 1 6960

The only real differences between my script and the MATLAB tutorial are that I’m using parallel execution in the training solver, and the datastore outputEnvironment is set to "gpu" . If I set ExecutionEnvironment = "auto" instead of "parallel-auto" and set ads.OutputEnvironment = ‘cpu’ the error stack is shorter, but the problem is the same:
Error using trainnet (line 46)
Number of channels in predictions (2) must match the number of channels in
the targets (3).

Error in train_DenseNet_detector_from_semi_synthetic_dataset (line 189)
[trained_network, train_info] = trainnet(trainFeatures, trainLabels’, net, "crossentropy", options);
Please could someone give me some advice? The root cause of this is buried in the deep learning toolbox, and it’s a little beyond me right now.
Thanks,
BenI am retraining YAMNet for a binary classification task, operating on spectrograms of audio signals. My training audio has two classes, positive and negative. Audio is preprocessed & features extracted using yamnetPreprocess(). When training the network, trainnet() produces the following error:
Error using deep.internal.train.Trainer/train (line 74)
Number of channels in predictions (2) must match the number of
channels in the targets (3).

Error in deep.internal.train.ParallelTrainer>iTrainWithSplitCommunicator (line 227)
remoteNetwork = train(remoteTrainer, remoteNetwork, workerMbq);

Error in deep.internal.train.ParallelTrainer/computeTraining (line 127)
spmd

Error in deep.internal.train.Trainer/train (line 59)
net = computeTraining(trainer, net, mbq);

Error in deep.internal.train.trainnet (line 54)
net = train(trainer, net, mbq);

Error in trainnet (line 42)
[net,info] = deep.internal.train.trainnet(mbq, net, loss, options, …

Error in train_DenseNet_detector_from_semi_synthetic_dataset (line 192)
[trained_network, train_info] = trainnet(trainFeatures, trainLabels’, net, "crossentropy", options);
My understanding of this error is that it indicates a mismatch between the number of classes the network expects, and the number of classes in the dataset. I do not see how this can be possible, considering the number of classes in the network is explicitly set by the number of classes in the datastore:
classNames = unique(ads.Labels);
numClasses = numel(classNames);
net = audioPretrainedNetwork("yamnet", NumClasses=numClasses);
My script is based on this MATLAB tutorial: audioPretrainedNetwork and there are no functional differences in the way I’m building datastores or preprocessing the data. The training options and the call to trainnet() are configured as follows:
options = trainingOptions(‘adam’, …
InitialLearnRate = initial_learn_rate, …
MaxEpochs = max_epochs, …
MiniBatchSize = mini_batch_size, …
Shuffle = "every-epoch", …
Plots = "training-progress", …
Metrics = "accuracy", …
Verbose = 1, …
ValidationData = {single(validationFeatures), validationLabels’}, …
ValidationFrequency = validationFrequency,…
ExecutionEnvironment="parallel-auto");

[trained_network, train_info] = trainnet(trainFeatures, trainLabels’, net, "crossentropy", options);
Relevant variable dimensions are as follows:
>> unique(ads.Labels)
ans =
2×1 categorical array
negative
positiveNoisy

>> size(trainLabels)
ans =
1 16240

>> size(trainFeatures)
ans =
96 64 1 16240

>> size(validationLabels)
ans =
1 6960

>> size(validationFeatures)
ans =
96 64 1 6960

The only real differences between my script and the MATLAB tutorial are that I’m using parallel execution in the training solver, and the datastore outputEnvironment is set to "gpu" . If I set ExecutionEnvironment = "auto" instead of "parallel-auto" and set ads.OutputEnvironment = ‘cpu’ the error stack is shorter, but the problem is the same:
Error using trainnet (line 46)
Number of channels in predictions (2) must match the number of channels in
the targets (3).

Error in train_DenseNet_detector_from_semi_synthetic_dataset (line 189)
[trained_network, train_info] = trainnet(trainFeatures, trainLabels’, net, "crossentropy", options);
Please could someone give me some advice? The root cause of this is buried in the deep learning toolbox, and it’s a little beyond me right now.
Thanks,
Ben I am retraining YAMNet for a binary classification task, operating on spectrograms of audio signals. My training audio has two classes, positive and negative. Audio is preprocessed & features extracted using yamnetPreprocess(). When training the network, trainnet() produces the following error:
Error using deep.internal.train.Trainer/train (line 74)
Number of channels in predictions (2) must match the number of
channels in the targets (3).

Error in deep.internal.train.ParallelTrainer>iTrainWithSplitCommunicator (line 227)
remoteNetwork = train(remoteTrainer, remoteNetwork, workerMbq);

Error in deep.internal.train.ParallelTrainer/computeTraining (line 127)
spmd

Error in deep.internal.train.Trainer/train (line 59)
net = computeTraining(trainer, net, mbq);

Error in deep.internal.train.trainnet (line 54)
net = train(trainer, net, mbq);

Error in trainnet (line 42)
[net,info] = deep.internal.train.trainnet(mbq, net, loss, options, …

Error in train_DenseNet_detector_from_semi_synthetic_dataset (line 192)
[trained_network, train_info] = trainnet(trainFeatures, trainLabels’, net, "crossentropy", options);
My understanding of this error is that it indicates a mismatch between the number of classes the network expects, and the number of classes in the dataset. I do not see how this can be possible, considering the number of classes in the network is explicitly set by the number of classes in the datastore:
classNames = unique(ads.Labels);
numClasses = numel(classNames);
net = audioPretrainedNetwork("yamnet", NumClasses=numClasses);
My script is based on this MATLAB tutorial: audioPretrainedNetwork and there are no functional differences in the way I’m building datastores or preprocessing the data. The training options and the call to trainnet() are configured as follows:
options = trainingOptions(‘adam’, …
InitialLearnRate = initial_learn_rate, …
MaxEpochs = max_epochs, …
MiniBatchSize = mini_batch_size, …
Shuffle = "every-epoch", …
Plots = "training-progress", …
Metrics = "accuracy", …
Verbose = 1, …
ValidationData = {single(validationFeatures), validationLabels’}, …
ValidationFrequency = validationFrequency,…
ExecutionEnvironment="parallel-auto");

[trained_network, train_info] = trainnet(trainFeatures, trainLabels’, net, "crossentropy", options);
Relevant variable dimensions are as follows:
>> unique(ads.Labels)
ans =
2×1 categorical array
negative
positiveNoisy

>> size(trainLabels)
ans =
1 16240

>> size(trainFeatures)
ans =
96 64 1 16240

>> size(validationLabels)
ans =
1 6960

>> size(validationFeatures)
ans =
96 64 1 6960

The only real differences between my script and the MATLAB tutorial are that I’m using parallel execution in the training solver, and the datastore outputEnvironment is set to "gpu" . If I set ExecutionEnvironment = "auto" instead of "parallel-auto" and set ads.OutputEnvironment = ‘cpu’ the error stack is shorter, but the problem is the same:
Error using trainnet (line 46)
Number of channels in predictions (2) must match the number of channels in
the targets (3).

Error in train_DenseNet_detector_from_semi_synthetic_dataset (line 189)
[trained_network, train_info] = trainnet(trainFeatures, trainLabels’, net, "crossentropy", options);
Please could someone give me some advice? The root cause of this is buried in the deep learning toolbox, and it’s a little beyond me right now.
Thanks,
Ben yammnet, audio, deep learning, trainnet, classification, signal processing, machine learning, image processing MATLAB Answers — New Questions

​

I need to identify the correct maximum in a 2D image, which may not may not be a global maxima and is a kind of gaussian or asymmetric type.
Some example are attached.
A method other than imregionalmax or findpeaks is needed
Any suggestion
ThanksI need to identify the correct maximum in a 2D image, which may not may not be a global maxima and is a kind of gaussian or asymmetric type.
Some example are attached.
A method other than imregionalmax or findpeaks is needed
Any suggestion
Thanks I need to identify the correct maximum in a 2D image, which may not may not be a global maxima and is a kind of gaussian or asymmetric type.
Some example are attached.
A method other than imregionalmax or findpeaks is needed
Any suggestion
Thanks gaussian, mamima MATLAB Answers — New Questions

​

As you can see from a screen shot below, if you make a time table (or table) that has a high number of rows and columbs and open that table in the variales window it is very difficult to read due to Matlab wrapping the columbs.
I havn’t been able to find anywhere in the settings where I can enable/disable this wapping feature from happening, where when disabled there the variables window would show a horizontal scroll when a table was opened with more columbs that could be displayed in the window.
Just imagine for a min if you had a table with 100+ columbs, and 1000+ rows, that would be near imposible to read using Matlab’s built in variables window, you would need to save it off to view it in excel. This doesn’t seem very friendly for debugging.
A second note, I found that the tables will only show a limited number of rows. In this example the variables window only goes up to the 54th row. Is there a way to view a table in the matlab variables window, and be able to scrow down through all the rows. The only way I can think to see what is in the table visually is by creating a new table with only 54 user selected rows (doesn’t seem very clean).

If there is any existing features that would adress some of these issues which I am missing, please let me know.As you can see from a screen shot below, if you make a time table (or table) that has a high number of rows and columbs and open that table in the variales window it is very difficult to read due to Matlab wrapping the columbs.
I havn’t been able to find anywhere in the settings where I can enable/disable this wapping feature from happening, where when disabled there the variables window would show a horizontal scroll when a table was opened with more columbs that could be displayed in the window.
Just imagine for a min if you had a table with 100+ columbs, and 1000+ rows, that would be near imposible to read using Matlab’s built in variables window, you would need to save it off to view it in excel. This doesn’t seem very friendly for debugging.
A second note, I found that the tables will only show a limited number of rows. In this example the variables window only goes up to the 54th row. Is there a way to view a table in the matlab variables window, and be able to scrow down through all the rows. The only way I can think to see what is in the table visually is by creating a new table with only 54 user selected rows (doesn’t seem very clean).

If there is any existing features that would adress some of these issues which I am missing, please let me know. As you can see from a screen shot below, if you make a time table (or table) that has a high number of rows and columbs and open that table in the variales window it is very difficult to read due to Matlab wrapping the columbs.
I havn’t been able to find anywhere in the settings where I can enable/disable this wapping feature from happening, where when disabled there the variables window would show a horizontal scroll when a table was opened with more columbs that could be displayed in the window.
Just imagine for a min if you had a table with 100+ columbs, and 1000+ rows, that would be near imposible to read using Matlab’s built in variables window, you would need to save it off to view it in excel. This doesn’t seem very friendly for debugging.
A second note, I found that the tables will only show a limited number of rows. In this example the variables window only goes up to the 54th row. Is there a way to view a table in the matlab variables window, and be able to scrow down through all the rows. The only way I can think to see what is in the table visually is by creating a new table with only 54 user selected rows (doesn’t seem very clean).

If there is any existing features that would adress some of these issues which I am missing, please let me know. table, timetable MATLAB Answers — New Questions

​

I am trying to find symbolic solution of specific non-linear ODE and then convert it to the matlab function via flowing code:
syms y(x) f(x)
syms x x0 y0 a b c d real
eqn = diff(y,x) == f/(a+b*y+c*y^2+d*y^3);
cond = y(x0)==y0;
S(x) = dsolve(eqn,cond);
sS(x) =simplify(S)
matlabFunction(sS,"File","myfunc.m");
where
a,b,c,d are scalar real constants
f(x) is known external real function
x0,y0 define initial condition of ODE problem
During matlab code conversion I always get a several Warnings:
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FunctionNotVerifiedToBeValid∦Function ‘f’ not verified to be a valid MATLAB function.∦
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FunctionNotVerifiedToBeValid∦Function ‘f’ not verified to be a valid MATLAB function.∦
Warning: symbolic:generate:FunctionNotVerifiedToBeValid∦Function ‘f’ not verified to be a valid MATLAB function.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘(t4 + a*y0)/a’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘t14*(t6 + t20)’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a – (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a + (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
and the following matlab code:
function sS = BURh103(x,a,b,c,d,x0,y0)
%BURh103
% sS = BURh103(X,A,B,C,D,X0,Y0)
% This function was generated by the Symbolic Math Toolbox version 24.2.
% 03-Oct-2024 15:16:31
t0 = roots([d.*-3.0,c.*-4.0,b.*-6.0,a.*-1.2e+1,a.*y0.*1.2e+1+b.*y0.^2.*6.0+c.*y0.^3.*4.0+d.*y0.^4.*3.0+integral(@(u)f(u),x0,x).*1.2e+1]);
t2 = t0(1);
t0 = roots([c.*-4.0,b.*-6.0,a.*-1.2e+1,a.*y0.*1.2e+1+b.*y0.^2.*6.0+c.*y0.^3.*4.0+integral(@(u)f(u),x0,x).*1.2e+1]);
t3 = t0(1);
t4 = integral(@(u)f(u),x0,x);
t5 = b.*t4;
t6 = b.*y0;
t7 = a.^2;
t9 = (b ~= 0.0);
t10 = (c == 0.0);
t11 = (d == 0.0);
t14 = 1.0./b;
t15 = sqrt(2.0);
t13 = t5.*2.0;
t16 = a.*t6.*2.0;
t17 = a+t6;
t18 = sqrt(t5);
t19 = t6.^2;
t20 = t15.*t18;
if ~all(cellfun(@isscalar,{b,c,d,t10,t11,t17,t9}))
error(message(‘symbolic:sym:matlabFunction:ConditionsMustBeScalar’));
end
if (d ~= 0.0)
sS = t2;
elseif (t11 & (c ~= 0.0))
sS = t3;
elseif ((t10 & t11) & (b == 0.0))
sS = (t4+a.*y0)./a;
elseif (((t9 & t10) & t11) & (t17 == 0.0))
sS = t14.*(t6+t20);
elseif (((t9 & t10) & t11) & (0.0 < t17))
sS = -t14.*(a-sqrt(t7+t13+t16+t19));
elseif (((t9 & t10) & t11) & (t17 < 0.0))
sS = -t14.*(a+sqrt(t7+t13+t16+t19));
else
sS = NaN;
end
end
I did not understand the following warning messages:
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘(t4 + a*y0)/a’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘t14*(t6 + t20)’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a – (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a + (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
What exactly these messages are trying to say?
Moreover, the matlab code is far from optimal or effective code. The auxilliary variable txx are precomputed, but the same integral(@(u)f(u),x0,x) is computed three times?!I am trying to find symbolic solution of specific non-linear ODE and then convert it to the matlab function via flowing code:
syms y(x) f(x)
syms x x0 y0 a b c d real
eqn = diff(y,x) == f/(a+b*y+c*y^2+d*y^3);
cond = y(x0)==y0;
S(x) = dsolve(eqn,cond);
sS(x) =simplify(S)
matlabFunction(sS,"File","myfunc.m");
where
a,b,c,d are scalar real constants
f(x) is known external real function
x0,y0 define initial condition of ODE problem
During matlab code conversion I always get a several Warnings:
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FunctionNotVerifiedToBeValid∦Function ‘f’ not verified to be a valid MATLAB function.∦
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FunctionNotVerifiedToBeValid∦Function ‘f’ not verified to be a valid MATLAB function.∦
Warning: symbolic:generate:FunctionNotVerifiedToBeValid∦Function ‘f’ not verified to be a valid MATLAB function.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘(t4 + a*y0)/a’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘t14*(t6 + t20)’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a – (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a + (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
and the following matlab code:
function sS = BURh103(x,a,b,c,d,x0,y0)
%BURh103
% sS = BURh103(X,A,B,C,D,X0,Y0)
% This function was generated by the Symbolic Math Toolbox version 24.2.
% 03-Oct-2024 15:16:31
t0 = roots([d.*-3.0,c.*-4.0,b.*-6.0,a.*-1.2e+1,a.*y0.*1.2e+1+b.*y0.^2.*6.0+c.*y0.^3.*4.0+d.*y0.^4.*3.0+integral(@(u)f(u),x0,x).*1.2e+1]);
t2 = t0(1);
t0 = roots([c.*-4.0,b.*-6.0,a.*-1.2e+1,a.*y0.*1.2e+1+b.*y0.^2.*6.0+c.*y0.^3.*4.0+integral(@(u)f(u),x0,x).*1.2e+1]);
t3 = t0(1);
t4 = integral(@(u)f(u),x0,x);
t5 = b.*t4;
t6 = b.*y0;
t7 = a.^2;
t9 = (b ~= 0.0);
t10 = (c == 0.0);
t11 = (d == 0.0);
t14 = 1.0./b;
t15 = sqrt(2.0);
t13 = t5.*2.0;
t16 = a.*t6.*2.0;
t17 = a+t6;
t18 = sqrt(t5);
t19 = t6.^2;
t20 = t15.*t18;
if ~all(cellfun(@isscalar,{b,c,d,t10,t11,t17,t9}))
error(message(‘symbolic:sym:matlabFunction:ConditionsMustBeScalar’));
end
if (d ~= 0.0)
sS = t2;
elseif (t11 & (c ~= 0.0))
sS = t3;
elseif ((t10 & t11) & (b == 0.0))
sS = (t4+a.*y0)./a;
elseif (((t9 & t10) & t11) & (t17 == 0.0))
sS = t14.*(t6+t20);
elseif (((t9 & t10) & t11) & (0.0 < t17))
sS = -t14.*(a-sqrt(t7+t13+t16+t19));
elseif (((t9 & t10) & t11) & (t17 < 0.0))
sS = -t14.*(a+sqrt(t7+t13+t16+t19));
else
sS = NaN;
end
end
I did not understand the following warning messages:
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘(t4 + a*y0)/a’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘t14*(t6 + t20)’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a – (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a + (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
What exactly these messages are trying to say?
Moreover, the matlab code is far from optimal or effective code. The auxilliary variable txx are precomputed, but the same integral(@(u)f(u),x0,x) is computed three times?! I am trying to find symbolic solution of specific non-linear ODE and then convert it to the matlab function via flowing code:
syms y(x) f(x)
syms x x0 y0 a b c d real
eqn = diff(y,x) == f/(a+b*y+c*y^2+d*y^3);
cond = y(x0)==y0;
S(x) = dsolve(eqn,cond);
sS(x) =simplify(S)
matlabFunction(sS,"File","myfunc.m");
where
a,b,c,d are scalar real constants
f(x) is known external real function
x0,y0 define initial condition of ODE problem
During matlab code conversion I always get a several Warnings:
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FunctionNotVerifiedToBeValid∦Function ‘f’ not verified to be a valid MATLAB function.∦
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FunctionNotVerifiedToBeValid∦Function ‘f’ not verified to be a valid MATLAB function.∦
Warning: symbolic:generate:FunctionNotVerifiedToBeValid∦Function ‘f’ not verified to be a valid MATLAB function.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘(t4 + a*y0)/a’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘t14*(t6 + t20)’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a – (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a + (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
and the following matlab code:
function sS = BURh103(x,a,b,c,d,x0,y0)
%BURh103
% sS = BURh103(X,A,B,C,D,X0,Y0)
% This function was generated by the Symbolic Math Toolbox version 24.2.
% 03-Oct-2024 15:16:31
t0 = roots([d.*-3.0,c.*-4.0,b.*-6.0,a.*-1.2e+1,a.*y0.*1.2e+1+b.*y0.^2.*6.0+c.*y0.^3.*4.0+d.*y0.^4.*3.0+integral(@(u)f(u),x0,x).*1.2e+1]);
t2 = t0(1);
t0 = roots([c.*-4.0,b.*-6.0,a.*-1.2e+1,a.*y0.*1.2e+1+b.*y0.^2.*6.0+c.*y0.^3.*4.0+integral(@(u)f(u),x0,x).*1.2e+1]);
t3 = t0(1);
t4 = integral(@(u)f(u),x0,x);
t5 = b.*t4;
t6 = b.*y0;
t7 = a.^2;
t9 = (b ~= 0.0);
t10 = (c == 0.0);
t11 = (d == 0.0);
t14 = 1.0./b;
t15 = sqrt(2.0);
t13 = t5.*2.0;
t16 = a.*t6.*2.0;
t17 = a+t6;
t18 = sqrt(t5);
t19 = t6.^2;
t20 = t15.*t18;
if ~all(cellfun(@isscalar,{b,c,d,t10,t11,t17,t9}))
error(message(‘symbolic:sym:matlabFunction:ConditionsMustBeScalar’));
end
if (d ~= 0.0)
sS = t2;
elseif (t11 & (c ~= 0.0))
sS = t3;
elseif ((t10 & t11) & (b == 0.0))
sS = (t4+a.*y0)./a;
elseif (((t9 & t10) & t11) & (t17 == 0.0))
sS = t14.*(t6+t20);
elseif (((t9 & t10) & t11) & (0.0 < t17))
sS = -t14.*(a-sqrt(t7+t13+t16+t19));
elseif (((t9 & t10) & t11) & (t17 < 0.0))
sS = -t14.*(a+sqrt(t7+t13+t16+t19));
else
sS = NaN;
end
end
I did not understand the following warning messages:
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FiniteSetsRootNotSupported∦’root’ without index not supported. Using index ‘1’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘(t4 + a*y0)/a’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘t14*(t6 + t20)’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a – (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
Warning: symbolic:generate:FiniteSetsNotSupported∦Finite sets (‘DOM_SET’) not supported. Using element ‘-t14*(a + (t7 + t13 + t16 + t19)^(1/2))’ instead.∦
What exactly these messages are trying to say?
Moreover, the matlab code is far from optimal or effective code. The auxilliary variable txx are precomputed, but the same integral(@(u)f(u),x0,x) is computed three times?! matlabfunction, warnings MATLAB Answers — New Questions

​

I have this code in my matlab but it has an error as:
Incorrect number or types of inputs or outputs for function vec.
I would really appreciate it if you could help me to solve it.
I should emphasize that I installed cvx on my matla.
Np = 4;
Nsc= 4;
mu_l_n = [1.56247628361161e-06;
1.59519914659906e-06;
1.27498226759933e-06;
1.03143933197703e-05];
tilde_P1_tot = 1000;
sigma2_dBm = -104;
sigma2 = 10^(sigma2_dBm / 10 – 3);
cvx_begin quiet
variable p1_l_n(Np) % Power allocation vector, column vector
term = (mu_l_n.^2 .* p1_l_n) / sigma2; % Define term for clarity
maximize(sum(log(1 + term))) % Objective function to maximize sum rates
subject to
sum(p1_l_n) == tilde_P1_tot / Nsc; % Total power constraint across subcarriers
p1_l_n >= 0; % Power must be non-negative
cvx_end
% Display the optimal power allocation
disp(‘Optimal power allocation:’);
disp(p1_l_n);I have this code in my matlab but it has an error as:
Incorrect number or types of inputs or outputs for function vec.
I would really appreciate it if you could help me to solve it.
I should emphasize that I installed cvx on my matla.
Np = 4;
Nsc= 4;
mu_l_n = [1.56247628361161e-06;
1.59519914659906e-06;
1.27498226759933e-06;
1.03143933197703e-05];
tilde_P1_tot = 1000;
sigma2_dBm = -104;
sigma2 = 10^(sigma2_dBm / 10 – 3);
cvx_begin quiet
variable p1_l_n(Np) % Power allocation vector, column vector
term = (mu_l_n.^2 .* p1_l_n) / sigma2; % Define term for clarity
maximize(sum(log(1 + term))) % Objective function to maximize sum rates
subject to
sum(p1_l_n) == tilde_P1_tot / Nsc; % Total power constraint across subcarriers
p1_l_n >= 0; % Power must be non-negative
cvx_end
% Display the optimal power allocation
disp(‘Optimal power allocation:’);
disp(p1_l_n); I have this code in my matlab but it has an error as:
Incorrect number or types of inputs or outputs for function vec.
I would really appreciate it if you could help me to solve it.
I should emphasize that I installed cvx on my matla.
Np = 4;
Nsc= 4;
mu_l_n = [1.56247628361161e-06;
1.59519914659906e-06;
1.27498226759933e-06;
1.03143933197703e-05];
tilde_P1_tot = 1000;
sigma2_dBm = -104;
sigma2 = 10^(sigma2_dBm / 10 – 3);
cvx_begin quiet
variable p1_l_n(Np) % Power allocation vector, column vector
term = (mu_l_n.^2 .* p1_l_n) / sigma2; % Define term for clarity
maximize(sum(log(1 + term))) % Objective function to maximize sum rates
subject to
sum(p1_l_n) == tilde_P1_tot / Nsc; % Total power constraint across subcarriers
p1_l_n >= 0; % Power must be non-negative
cvx_end
% Display the optimal power allocation
disp(‘Optimal power allocation:’);
disp(p1_l_n); power_allocation, water_filling, optimization, cvx MATLAB Answers — New Questions

​

So I have an equation that goes something like N.*P + S.*P + E.*P + W.*P = Q
N S E and W are coefficients, P is pressure changing with time. In terms of [A]*[x]=[b] its supposedly [coefficients]*[Pressure]=[Q]. How would I use a solver to translate the equation into these matrix terms?So I have an equation that goes something like N.*P + S.*P + E.*P + W.*P = Q
N S E and W are coefficients, P is pressure changing with time. In terms of [A]*[x]=[b] its supposedly [coefficients]*[Pressure]=[Q]. How would I use a solver to translate the equation into these matrix terms? So I have an equation that goes something like N.*P + S.*P + E.*P + W.*P = Q
N S E and W are coefficients, P is pressure changing with time. In terms of [A]*[x]=[b] its supposedly [coefficients]*[Pressure]=[Q]. How would I use a solver to translate the equation into these matrix terms? gmres, solver MATLAB Answers — New Questions

​