Download Matlab 2019a

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Quang Thuc Ha
Quang Thuc Ha 2020-9-28
评论： Walter Roberson 2025-11-7，19:27
Hi mates,
I need to install Matlab 2019a for academic purpose but could not find out how to download. The 2020b version is too heavy. Could you please help me out ?
Cheers
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shashank
shashank 2025-10-29，13:09
pls download
Rik
Rik 2025-10-29，13:23
@shashank, did you see the answer below? Did you try it?

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回答（2 个）

Walter Roberson
Walter Roberson 2020-9-28
If you have authorization to download MATLAB then
https://www.mathworks.com/downloads/web_downloads/download_release?release=R2019a
However, license adminstrators can block end users from downloads. Also, my understanding is that Student licenses do not have access to download previous releases.
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nibras ounissi
nibras ounissi 2022-8-16
Hi, can you explain how you installed a previous version?
Walter Roberson
Walter Roberson 2022-8-16
@nibras ounissi
If your license is already authorized:
https://www.mathworks.com/downloads/ and on the left hand side there is a Select Release column. If necessary click on Show All there to see older versions. Click on the version on the left that you want.
Depending on how old the version is, you will either be brought to a page to download an installer, or else you will need to download a number of different files
If that Downloads page does not offer you a Select Release then you are not authorized to download old releases with the license you have associated with your MATLAB Central account.

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Fadila
Fadila 2025-11-7，12:20
编辑：Walter Roberson 2025-11-7，19:24
%% --- 1) Lecture du signal audio ---
[x, fs] = audioread('voix.wav'); % x est le signal audio, fs est la fréquence d'échantillonnage
x = x(:,1); % si le signal est stéréo, on prend une seule voie
N = length(x);
t = (0:N-1)/fs; % axe du temps
dt = 1/fs; % intervalle de temps
%% --- 2) Calcul manuel de la Transformée de Fourier continue approchée ---
f_max = 500; % fréquence max à afficher (Hz)
df = 1; % résolution en fréquence (Hz)
f = 0:df:f_max; % axe des fréquences
Xf = zeros(size(f));
for k = 1:length(f)
Xf(k) = sum(x .* exp(-1i*2*pi*f(k)*t)) * dt;
end
Xf_mag = abs(Xf);
%% --- 3) Détection des fréquences principales (peaks) ---
% on choisit un seuil pour détecter les pics significatifs
seuil = max(Xf_mag) * 0.05; % 5% du maximum
indices_peaks = find(Xf_mag > seuil);
% garder seulement les pics locaux
peaks = [];
for i = 2:length(indices_peaks)-1
if Xf_mag(indices_peaks(i)) > Xf_mag(indices_peaks(i-1)) && ...
Xf_mag(indices_peaks(i)) > Xf_mag(indices_peaks(i+1))
peaks(end+1) = indices_peaks(i);
end
end
nombre_frequences = length(peaks);
disp(['Nombre de fréquences principales: ', num2str(nombre_frequences)]);
%% --- 4) Affichage ---
figure;
subplot(2,1,1);
plot(t, x);
title('Signal audio dans le domaine temporel');
xlabel('Temps (s)');
ylabel('Amplitude');
subplot(2,1,2);
plot(f, Xf_mag);
hold on;
plot(f(peaks), Xf_mag(peaks), 'ro'); % marquer les pics
title('Transformée de Fourier continue approximée du signal audio');
xlabel('Fréquence (Hz)');
ylabel('|X(f)|');
grid on;
  1 个评论
Walter Roberson
Walter Roberson 2025-11-7，19:27
That's really some exploit you found, that running that code ends up by downloading MATLAB R2019a! Please expand on how the exploit works ??

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