[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')

 Akzeptierte Antwort

Star Strider
Star Strider am 28 Dez. 2021

1 Stimme

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 Kommentare

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

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Weitere Antworten (1)

Voss
Voss am 28 Dez. 2021

0 Stimmen

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 Kommentare

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

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