The Pitch of signal

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Nachiket Patki 2017-3-25

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评论： Star Strider 2017-3-28

Hello sir, According to my knowledge pitch of any signal is nothing but its frequency, so if i keep the same frequency of any signal then the audio that we hear should be same right? While I was working with matlab, what I did is, I kept the frequency same but changed the sampling rate(for the same sinusoid), the result is I got the different sound for 2 different sampling rates. How can that be possible? can you please explain it to me?

Thank you

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Chad Greene 2017-3-25

Pitch is perceived frequency. The two terms are often interchangeable, but the auditory system is complex, and in some situations a person can perceive two different frequencies to be the same.

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Answer 1

Star Strider 2017-3-25

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在 MATLAB Online 中打开

I am not certain what you did from your description, so I am guessing here.

The sound function (or whatever you used to play your signal) knows only what you tell it. If you sampled your signal at 10000 Hz and then told sound to play it at 10000 Hz, it will seem exactly as you recorded (or created) it. If you told sound to play it with a 12500 Hz sampling frequency, it will sound higher, and at 7500 Hz, it will sound lower.

Example:

Fs = 10000;
t = [0 : 1/Fs : 2]';
sig = sin(2*pi*t*1000);                         % Create 1000 Hz Tone
sound(5*sig, Fs)                                % Sounds Normal
pause(2)
sound(5*sig, 0.75*Fs)                           % Sounds Lower
pause(2.5)
sound(5*sig, 1.25*Fs)                           % Sounds Higher

The different pitches you hear relate to how fast sound plays your signal and sends it to your audio system. The duration will also change, so here the original sound will play for 2 seconds, the ‘lower’ sound for 2/0.75=2.7 seconds, and the ‘higher’ sound for 2/1.25=1.6 seconds.

In digital signal processing, sampling theory is not trivial. For a full understanding, please consult a detailed textbook on digital signal processing for a thorough discussion. It is much too detailed and extensive to discuss here.

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Star Strider 2017-3-26

在 MATLAB Online 中打开

The signals ‘s’ are the same for both, with the signal frequency being ‘0.3’ and a duration of 200. The sampling interval, ‘Ts’ is 1 (because the default increment in the colon operator vectors is 1), so the sampling frequency, ‘Fs’ will also be one, since ‘Fs=1/Ts’.

your original signal has a sampling frequency of 1 (assumed to be 1 second) so the sampling frequency , so the default sampling frequency in the first soundsc call is the frequency you hear: 8192 Hz. The second soundsc call with a sampling frequency of 48000 is inaudible, since the upper threshold of human nearing is about 20 kHz (assuming your sound reproduction system is even able to play that sound).

I slightly changed the stem plot x-axis scale to demonstrate the effect:

i=[1:200];
s=sin(0.3*i);
figure
stem(s)
xt = get(gca, 'XTick');
set(gca, 'XTick',xt, 'XTickLabel',xt/8192, 'XTickLabelRotation',45)
title('Fs = 8192 Hz')
xlabel('Time (seconds)')
soundsc(s)
i=[1:200];
s=sin(0.3*i);
figure
stem(s)
xt = get(gca, 'XTick');
set(gca, 'XTick',xt, 'XTickLabel',xt/48000, 'XTickLabelRotation',45)
title('Fs = 48000 Hz')
xlabel('Time (seconds)')
pause(1)
soundsc(s,48000)
signal_period = (2*pi)/0.3
signal_frequency = 1/signal_period
played_frequency = [8192  48000]
played_length = 200./played_frequency