Finding max value and its index
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Edo Christianto
il 12 Giu 2014
Risposto: Malcolm Hawksford
il 19 Dic 2017
So, here is the problem.
I have a wav file (attached), one tuts piano sound, and I analyzed it. Using this.
[wave,fs] = audioread('file01.wav');
n=length(wave)-1;
figure
t=0:1/fs:n/fs;
subplot(3,1,1), plot(t,wave)
subplot(3,1,1), xlabel('Time (Second)')
subplot(3,1,1), ylabel('Amplitude')
f=0:fs/n:fs;
wavefft=abs(fft(wave));
subplot(3,1,2), plot(f,wavefft)
subplot(3,1,2), xlabel('Frekuency (Hz)')
subplot(3,1,2), ylabel('Magnitude')
[max_value, index] = max(wavefft(:));
subplot(3,1,3), plot(f,wavefft)
subplot(3,1,3), axis([0 index 0 max_value])
The output is like this.

When I search max value using
[max_value, index] = max(wavefft(:));
I got this. Surely the peak would be on the right side, right?

When I inspect it, the max_value is in index 1111 Hz.
But when I zoom in manually, it's on index 740 Hz.

Did I do something wrong?
1 Commento
Star Strider
il 12 Giu 2014
Seems you did everything correctly. The data in ‘subplot(3,1,3)’ is correct.
Explore the documentation on fft to understand why. (See the documentation on fftshift to understand the reason subplot(3,1,2) is misleading you.)
Risposta accettata
Image Analyst
il 12 Giu 2014
Index 1111 is not at 1111 Hz necessarily. From the plot above, it looks like it's at about 750 Hz. But you msight want to use fftshift to shift the ends to the middle.
4 Commenti
Image Analyst
il 12 Giu 2014
If you shift the spectrum to put the 0 frequency at the center, you need to subtract ~2.25 * 10^4 (actually x(floor(length(x)/2)) I believe) from your x axis so that 0 shows up at the middle, with the middle index of the array you're plotting.
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Malcolm Hawksford
il 19 Dic 2017
Lets assume you have a stereo .wav file and simply want to find the peak amplitude and sample index (time domain) in each channel. Assume the file is called Cohen.wav then this works for me: [p q]=max(abs(audioread('Cohen.wav'))); p gives the peak values and q the corresponding indices.
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