The Pitch of signal

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

[Chad Greene]

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.

Answer 1

[Star Strider]

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.

[Nachiket Patki]

Ok so from the description given by you it clear that the tone will depend upon the sampling frequency. I am right? Actually in my programming what I did is I kept omega(radians) same for the sinusoid and varied the sampling frequency so I was getting the different sounds.

[Star Strider]

What you did would definitely change the pitch, with the same sampling frequency supplied to the sound function. The difference seems to be that you created sounds with different sampling frequencies and played them with the same sampling frequency in sound, while I did the opposite, and changed the sampling frequency supplied to sound The essential result is the same.

[Nachiket Patki]

No sir actually I did the same thing as yours, here is my code

i=[1:200]
s=sin(0.3*i);

figure
stem(s)
soundsc(s);

i=[1:200]
s=sin(0.3*i);

figure
stem(s)
soundsc(s,48000)

But now will you tell me what is the original Fs and f for the signal? Is it 1 samples/sec and 0.0128 Hz? I not getting that. How to calculate it if we have given only this much input.

Thank you

[Star Strider]

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

[Nachiket Patki]

Thank you sir I get that. It was very clear demonstration. Now my concepts are clear regarding Fs and f.

one more question :D.

So if you have a signal with some sampling frequency say 'x' and you are going to play it with say another sampling frequency say 'y' using 'sound' function, then we will always be able to hear the sound corresponding to 'y' frequency and not 'x' frequency? So is it true that we will not hear the sound of the original signal or the actual sound? Am I in a right track? or still am I missing something?

Please clear my doubt.

Thank you

[Star Strider]

My pleasure.

The sound you hear will be scaled in frequency by ‘Fs_sound/Fs_created’ where ‘Fs_sound’ is the sampling frequency you supply to the sound function and ‘Fs_created’ is the frequency your created or recorded it with.

So you will hear it at a higher frequency than the original frequency if the sampling frequency you give the sound function is higher than the original signal sampling frequency, and lower if you give the sound function a lower sampling frequency that the one you created or recorded the original signal with.

For example, you will hear a sound an octave higher than the original if you play it at 2 times the frequency it was recorded or created with.

[Nachiket Patki]

Ok sir I got that,

But sir matlab by default plays it at 8192hz, so anyway its not the same that we supply or create, right? So according to you if suppose we supply Fs=1000 then its scaled version will be 8192/1000? Am I correct?

[Star Strider]

That is what it will sound like.

Do that experiment to hear the result.

[Nachiket Patki]

okk sir I will try that. But really thank you for clearing all my doubts. :) Have a nice day

[Star Strider]

My pleasure.

You, too!