How can i fix the indices on the right to fit the indices on the left?

This is the function where the error is indicated:

function AudioSections(app, signal, numSections)
signalLength = length(signal);
overlapFactor = 0.5;
%sectionLength = floor(signalLength / (numSections - overlapFactor));
sectionLength = floor(signalLength / numSections);
overlapLength = floor(sectionLength * overlapFactor);
%This part does the dividing into sections
for i = 1:numSections
    startIndex = max((i-1) * sectionLength - floor((i-2) * overlapLength), 1);
    endIndex = min(i * sectionLength + overlapFactor, signalLength);
    sect = signal(startIndex:endIndex);
    sectL = length(sect);
    sectT = transpose(sect);
    Blackman_window = blackman(sectL);
    %Sect = transpose(sectT.* Blackman_window);
    Sect = sectT.* Blackman_window;
    app.Sections(i) = Sect;                                          %Error is here
    filename = strcat(app.File_Path, 'Section_', num2str(i), '.wav');
    audiowrite(filename, app.Sections(i), app.Frequency);
end

This is how my variables are declared:

Sections = [];
Frequency = 44100;
Channels = 2;
Num_Bits = 16;
Data_Type_Audio = 'double';

This is the error:

"Error using Prac2/AudioSections

Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.

Error in Prac2/SaveWordButtonPushed (line 297)

AudioSections(app, app.RecordedAudioData, 5);"

Answer 1

[Walter Roberson]

Channels = 2;

The recorded data will have 2 channels.

signalLength = length(signal);

length() should typically only be used for vectors. Audiodata is always one column vector per channel, so you should be asking about the height of the signal array, which you can do with size(signal,1) or with height(signal)

    sect = signal(startIndex:endIndex);

You are indexing the signal as if it is only a vector. The result is going to be a column vector.

    sectT = transpose(sect);

You transpose it into a row vector.

    Blackman_window = blackman(sectL);

blackman() returns a column vector.

   Sect = sectT.* Blackman_window;

You do element-by-element multiplication between a row vector and a column vector. The result is going to be a 2D array.

   app.Sections(i) = Sect;                                          %Error is here

You try to store the entire 2D numeric array into a single location. That can only work if app.Sections were a custom class that redefined indexing for storage purposes. You should be storing into a cell array

[Zander]

Thank you for your response. I appreciate it. I do however get a ne error in my for loops when trying to write the data into a text file. For example:

for i = 1:6
    FFTsect = calculateFFT(app, app.Sections{i});    %Error is here
    app.FFTSections{i} = FFTsect;
    textFilename = strcat(app.File_Path, 'Test_Word_', num2str(app.TestWordNum), '.txt');
    % Open the text file for writing
    fileID = fopen(textFilename, 'w');
    if fileID == -1
        msgbox('Failed to create the text file.', 'Error', 'error');
        return;
    end

the error:

"Too many output arguments."

[Walter Roberson]

Your calculateFFT function is not defined as returning a value.

Note that if calculateFFT has been defined as a callback function, then it is very likely it was created to not return any output. In MATLAB, the great majority of callback functions do not return values.

(There are a few places where callbacks are expected to return values -- for example some of the routines that allow users to resize objects have callbacks to permit the program to place limitations on what new sizes are permitted.)

[Zander]

Correct me if i am wrong, but this function does return Fw as a value or does it not?

function calculateFFT(app, Data)
fftData = fft(Data); % Calculate the FFT of the audio data
N = length(fftData);
middleIndex = floor(N/2);  % Calculate the index of the middle of the FFT data
app.f = (0:middleIndex-1) * (44100 / N); %f = (0:N-1) * (44100 / N); %f = (0:-1)*48000/length(fftData);  %Calculate the frequency values for the FFT data
app.time = (0:middleIndex-1) / 44100; %time = (0:N-1)/44100; % Calculate the time values for the audio data
firstHalf = fftData(1:middleIndex);   % Create a new vector to store the first half of the FFT data
app.Fw = abs(firstHalf);
end

Thank you very much for your patience and help, i know that i am not providing everything which causes alot of constant communication. Please do not feel forced to answer. I will completely understand if you dont.

[Walter Roberson]

No, your function calculateFFT is not constructed to return any value. Instead it is setting properties of app such as setting app.Fw . Your could would have to have something that looks like an assignment between the function keyword and the function name in order to return anything. For example,

function fftdata = calculateFFT(app, Data)

but perhaps you should instead be changing from

    FFTsect = calculateFFT(app, app.Sections{i});    %Error is here

to

    calculateFFT(app, app.Sections{i});
    FFTsect = app.Fw;

[Zander]

Thank you very much my errors are now resolved.