[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')

 Risposta accettata

Star Strider
Star Strider il 28 Dic 2021
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 Commenti

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

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Più risposte (1)

Voss
Voss il 28 Dic 2021
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 Commenti

burak Kalayoglu
burak Kalayoglu il 28 Dic 2021
Modificato: burak Kalayoglu il 28 Dic 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
What does this do?
f(1698)
PSD(1698)
burak Kalayoglu
burak Kalayoglu il 28 Dic 2021
Because I want to see what corresponds to the same psd value in different signals

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