[y,Fs] = audioread('C:\Users\Casper\Desktop\3.3V.wav');
L=length(y); % series length
;
f = Fs/2*linspace(0,1,L/2+1); % single-sided positive frequency
X = fft(y)/L; % normalized fft
PSD=2*abs(X(1:L/2+1)); % one-sided amplitude spectrum
figure,plot(f,PSD);
grid
xlabel('freq(Hz)')
ylabel('amplitude')

 Réponse acceptée

Star Strider
Star Strider le 28 Déc 2021

1 vote

Try this —
idx = (f <= 1500) & (f <= 1800)
[rpm, freq] = findpeaks(PSD(idx), f(idx), 'MinPeakHeight',0.75E-3)
Without the data, I cannot experiment with that, however it should return the peak value as ‘rpm’ and the frequency as ‘freq’. If it does not, experiment with 'MinPeakProminence' instead of 'MinPeakHeight' since it cannot be determined from the plot that there are not more components to the desired peak.
.

2 commentaires

burak Kalayoglu
burak Kalayoglu le 28 Déc 2021
Thank you king
Star Strider
Star Strider le 28 Déc 2021
As always, my pleasure!
.

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Plus de réponses (1)

Voss
Voss le 28 Déc 2021

0 votes

You could use Data Cursor to create a datatip, or you could try something like this:
idx = find(f(:) > 1000 & f(:) < 2500 & PSD(:) > 0.00075);

3 commentaires

burak Kalayoglu
burak Kalayoglu le 28 Déc 2021
Modifié(e) : burak Kalayoglu le 28 Déc 2021
I want to find a value1698 (signed point in graph) but ı cant , ı want to find signed point and after ı can find rpm this signal
Voss
Voss le 28 Déc 2021
What does this do?
f(1698)
PSD(1698)
burak Kalayoglu
burak Kalayoglu le 28 Déc 2021
Because I want to see what corresponds to the same psd value in different signals

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