1
$\begingroup$

https://mathoverflow.net/q/165038/14414

Motivation : Here is a motivation as to why this problem is so important.

Let $f(t)$ be an audio signal. We can safely asume it to be bandlimited to 0-20kHz as we cannot hear anything above that. Capture this signal in digital computer with appropriate sampling frequency and denote it as $f[n]$.

Now take Discrete Hilbert transform of $f[n]$ to get $f_h[n]$, (using the code $f_h$ = imag(hilbert(f)); in Matlab).

Compute the signal $f_{\theta}[n] = f[n]\cos\theta + f_h[n]\sin\theta$ for any value of $\theta$, then listen to the signal with different values for $\theta$.

They all sound exactly identical.

Similarly our $MI_{\omega_0,\omega_1}(t)$ is same for all $f_{\theta} = f\cos\theta + f_h\sin\theta$, for any value of $\theta$.

Question :

just try it. $<f,f_h> = 0$, they why do they produce same effect in the listner? Is it some quantum mechanical effect gone wrong?

Added :

Also see this metric space : metric space

I've recently filed a patent using this metric with a slight change, instead of arccos i used sqrt(2(1-cos(theta))), which makes it a Hilbertian metric. I had then embedded this metric space into an Hilbert space isometrically, to model using vectors.

MATLAB code :

[f,fs] = wavread('audio_file.wav');

fh = imag(hilbert(f));

theta = pi/4;

f_tht = fcos(theta) + fhsin(theta);

wavplay(f,fs);

wavplay(f_tht,fs);

$\endgroup$
  • 1
    $\begingroup$ What does it mean "they sound exactly identical"? What is the sound being played on (how well is it reproduced by the equipment) and who is listening (makes the judgment)? $\endgroup$ – bright magus Jun 5 '14 at 10:59
  • $\begingroup$ @brightmagus : It could be any sound. I really mean any sound. Use the same speaker system for playing, for both the signal and hilbert transormed signal. Any speaker system/audio system that we use in daily life is ok. You may use some sophisticated Audio systems like Bose, but they still sound the same. Ofcourse I don't expect the listner to be having any hearing problems. $\endgroup$ – Rajesh Dachiraju Jun 5 '14 at 11:42
  • $\begingroup$ (What's really sophisticated about Bose is the price.) And saying that any home system is OK is equivalent to saying that in order to compare 5MP and 50MP pictures you can use any home monitor. True, they will probably look the same on any home monitor, but they will not look the same if you make really large prints out of them. So all I wanted to point out is that in general the system and the listener make the difference. I don't know about this particular case though. $\endgroup$ – bright magus Jun 5 '14 at 11:54
  • 1
    $\begingroup$ The simplest example: $\sin(x)$ and $\cos(x)$ are orthogonal (since they are Hilbert transforms of one another), but they sound the same to a human ear! That's because our brains aren't precise enough to notice the phase shift; we just detect the frequency. $\endgroup$ – DumpsterDoofus Jun 5 '14 at 13:20
  • 1
    $\begingroup$ @DumpsterDoofus: "That's because our brains aren't precise enough to notice the phase shift; we just detect the frequency". Although I don't know what particular phase shifts you are talking about, but in general human ear is able to hear phase shifts: en.wikipedia.org/wiki/Loudspeaker_time_alignment . $\endgroup$ – bright magus Jun 5 '14 at 13:51
4
$\begingroup$

To your Question:

  1. There is no quantum mechanics involved. This is essentially a signal processing question, which is rooted in calculus.

  2. Why does it sound the same?

The ear works essentially as a power spectrum analyzer, i.e. what you hear of a signal $f(t)$ is mainly determined by the powerspectrum $|{F(\omega)}|^2$, where ${F(\omega)}$ is the Fourier-transform of $f(t)$.

In your case: $F_{\theta}(\omega) = (\cos\theta) F(\omega) + (\sin\theta) F_h(\omega)$.

so that: $|F_{\theta}(\omega)|^2 = (\cos\theta)^2 |F(\omega)|^2 + (\sin\theta)^2 |F_h(\omega)|^2 + K$.

Where $K \propto F(\omega)^* F_h(\omega) + F(\omega) F_h(\omega)^*$, and $*$ denotes the complex conjugtate.

Using the relation between $F$ and $F_h$ given by the Hilbert-Transform we find that $K=0$ and $|F(\omega)|^2 = |F_h(\omega)|^2 $.

We conclude that

$|F_{\theta}(\omega)|^2 = ((\cos\theta)^2 + (\sin\theta)^2) |F(\omega)|^2 = |F(\omega)|^2 $.

(Where in the last step there is a trigonometric identity)

Summary: for all $\theta$ we find that $|F_{\theta}(\omega)|^2 = |F(\omega)|^2 $, so the ear (as a powerspectrum analyzer) hears the same.

$\endgroup$
  • $\begingroup$ @denzib : I am not to be satisfied with that statement."so the ear (as a powerspectrum analyzer) hears the same." As a genius, I see an opportunity in it. $<f,f_h> = 0$, they why do they sound same? Thats how I see opportunity to revive the way the entire mathematics being used in physics and signal processing. $\endgroup$ – Rajesh Dachiraju Jun 5 '14 at 13:38
  • $\begingroup$ please see the question I linked to, in the top of the OP. $\endgroup$ – Rajesh Dachiraju Jun 5 '14 at 13:40
  • $\begingroup$ "so the ear (as a powerspectrum analyzer) hears the same" Who told you this? Did God tell you? I want to know God's thoughts. $\endgroup$ – Rajesh Dachiraju Jun 5 '14 at 13:56
  • 1
    $\begingroup$ @RajeshD that is the way the ear works, as you can read in any anatomy text. Also, I don't think your tone is appropriate. $\endgroup$ – Davidmh Jun 5 '14 at 15:05
  • 1
    $\begingroup$ @Davidmh I was just trying to be a bit funny. No bad intentions. Apologies if that caused any trouble. $\endgroup$ – Rajesh Dachiraju Jun 5 '14 at 15:07
2
$\begingroup$

You can express your signal as the series expansion:

$$f(t) = \sum_k a_k \cos(kt) + b_k \sin(kt)$$

The Hilbert transform is a linear operator, so:

$$f_h = \sum_k a_k H[\cos(kt)] + b_k + H[\sin(kt)] = \sum -a_k \sin(kt) + b_k \cos(kt)$$

So, $f_h$ has changed the phases f the different frequencies, but leaving the mangitudes unchanged. You hear the same thing because your ear is doing a Fourier Transform of the input (each frequency is detected by a different part of the cochlea), that remains invariant (modulo some phase) under Hilbert Transform.

What your transformation is doing is shifting each frequency by a a certain phase, but not affecting the magnitudes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.