Usually, when some sample is played-back at twice the rate it was recorded, the pitch of the recording sounds an octave higher.

For example: $g(t) = \sin(2 \pi f t)$ and $g(2t) = \sin(2 \pi f 2 t) \implies \sin(4 \pi ft)$. Where $f$ is a given frequency.

There are some strange "sample" cases however that when played back at twice their recording rate their pitch appears to...not have changed as expected. And some even stranger "sample" cases that when played back at twice their recording rate their pitch appears to be...lower!

How do these strange "sample" cases might "look like" (in terms of a mathematical description) and what are, broadly, the conditions that can lead to such cases?

EDIT (hint 1):

Variants of the technique of pitch-shifting / time stretching have been mentioned so far by more than one persons in relation to this question.

Pitch shifting is not the core of the question and the technique of pitch-shifting does not change the nature of the question.

As it was pointed out in one of the comments to this post and as a form of a hint: Yes, pitch shifting does alter the pitch of the "sample" but a) what sort of sample would sound almost identical to its unshifted version if it was passed through a $2 \times$ pitch shifter and b) what sort of sample would sound at a lower pitch if it was passed through a $2 \times$ pitch shifter. (the $2 \times$ here is just an example).

EDIT (hint 2):

Another two things that have appeared in responses so far are aliasing and some aspect of human hearing. In other words, "Maybe you have pitch shifted in such a way that the frequency jumps back on to itself" and also "Maybe you are taking advantage of some psychoacoustic phenomenon in which the frequency has changed but the perceived frequency does not"

Aliasing and the limits of human hearing are relevant to this question but are not absolute requirements for the "paradox" that is mentioned here to occur.

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    $\begingroup$ what you are describing is called "pitch shifting" and it can be considered the combination of two operations: "time scaling" and "sample interpolation" (or "sample rate conversion" even though we are not changing the sample rate). the latter operation is mathematically well-defined. it speeds up and raises the pitch together. the time scaling operation is less well defined. it makes a sound longer (or shorter) without changing the pitch. so, to up-pitch a sound, first you have to time-scale it to be longer (and the same pitch), then apply SRC to shorten it to the original. $\endgroup$ – robert bristow-johnson Nov 19 '16 at 23:19
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    $\begingroup$ actually, what you are describing is just the "time scaling" operation. $\endgroup$ – robert bristow-johnson Nov 19 '16 at 23:21
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    $\begingroup$ It would be more clear if you add some instances of the sample cases you pointed out, as well as the way you playback with twice recording speed. $\endgroup$ – msm Nov 19 '16 at 23:54
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    $\begingroup$ imagine that do you have a signal at 5000 Hz sampled at 44100 Hz, all OK 5000 Hz is inside the nyquist boundarie, now imagine resample this signal to 8000 Hz where go this frequency if we are out of the nyquist ? maybe are you talking about cases like this. $\endgroup$ – ederwander Nov 21 '16 at 0:07
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    $\begingroup$ @A_A Oh, sorry I didn't get your message. Honestly, I am not so good in puzzles but I don't think if you pitch-shift a white noise it sounds much different. Also, a chirp signal seems to be a similar case. $\endgroup$ – msm Nov 21 '16 at 10:33
  1. White noise: sounds exactly the same if you play it back at double the speed
  2. Stack up diminished fifth starting at a low enough fundamental (say 20 Hz). Play back at double speed: sounds exactly the same.
  3. Stack octaves with equal amplitude starting with a low enough fundamental. Play back at twice the speed: sounds exactly the same
  4. Detune the octaves slightly (say multipler of 2.05 starting at 20 Hz). Play back at double the speed: perceived pitch goes DOWN.

You get plenty more when fiddling with the harmonic structure. Code example for the last one:

%% make some disharmonic sounds
fs = 44100; % sample rate
q = 2.05; % harmonic multiplier
f0 = 20;  % fundamental
n = 10; % number of harmonics
duration = 2; % length of file at fs
nx = duration*fs;
t = (0:nx-1)'/fs; % time vector
x0 = sin(2*pi*f0*t); % fundamental
f1 = f0;
for i = 2:n  % add harmonics
  f1 = f1*q;
  x0 = x0+sin(2*pi*f1*t);

x0 = 0.5.*x0/max(abs(x0)); % normalize
% and play
sound(x0,fs); pause(duration);
  • $\begingroup$ Thank you Hilmar, that is indeed the "trick" :) I would suggest that in the presence of 2,3,4, we "drop" the white noise as a "trivial" case to this puzzle (but still a valid answer). I have only added a few more pieces of information in my response which might be of interest. $\endgroup$ – A_A Nov 27 '16 at 13:23

There are a few pieces of background information that help in answering this question:

  1. Geometric series and in particular geometric series as applying to music
    • A geometric series, $x_n$, is one where each successive term is generated by multiplying the previous term by some factor. For example: $x_{n+1} = a x_n$
    • Geometric series are everywhere in music, because harmony is all about ratios between different tones (an interval) and the difference in auditory result they produce. Some are consonant, some are dissonant, some sound happy, some sound sad and so on.
    • On the keys of a piano, every successive tone results by multiplying the previous tone by a constant factor. What factor is this? Suppose two tones with frequencies $f_1, f_2$ where $f_2=2 f_1$. That would be some tone and the same tone one octave above it. The doubling of the frequency happens in 12 steps (e.g. $C, C \sharp, D, D\sharp, E, F, F\sharp, G, G\sharp, A, A\sharp, B$). This means that $a \cdot a \cdot a \cdot a \cdot a \cdot a \cdot a \cdot a \cdot a \cdot a \cdot a \cdot a = 2$ and therefore $a = \sqrt[12]{2}$ which is approximately $1.0594\ldots$. It's the same story on the fretboard of a stringed instrument. The tones produced by plucking the string in two successive frets have a ratio of $a$.
    • So, if we start with an $A=440$ Hz, we know that $A \sharp = a A \approx 466.136$ Hz and 12 steps later $440 \cdot a^{12} =880$ Hz.
    • Why is it like this? Because it is a good compromise. Because otherwise, the piano would not have this distinct shape of two black keys, gap, three black keys, gap, two black keys but a different pattern depending on the interval, chosen, to tune it. Which is what you can do with a Sitar because its frets are movable.

Now, assuming a reference tone $A_{r}$, all successive tones are produced by $A_k = a^kA_{r}$ where $k$ is the number of semitones (the smallest difference in frequency between two successive keys on a piano or two successive frets on a guitar).

So, if we were to create a signal that contains a number of successive tones in octave intervals then, assuming that we start somewhere low enough, we would end up with something like 10,20,40,80,160,320,640,1280, 2560, 5120, 10240.

Playing this back, it sounds...sort of nice. "Speeding" this up at twice the speed, we get another signal which now contains tones: 20,40,80,160,320,640,1280, 2560, 5120, 10240, 20480.

Which is (almost) exactly the same as the one we started with. Of course, if we keep "pitch shifting", all we are achieving is pushing more and more tones past the limits of human hearing or within the region of suppression of the anti-imaging filter. In both cases, after a large number of shifts, the difference is audible because the signal does not contain as many harmonics.

These are called Sheppard Tones and they sound like this.

In fact, Sheppard did specify that if you keep "feeding" the pitch shifting with low tones, then the perception is that the tones keep rising in pitch.

This is where DSP might step in and actually use the "aliasing". All that we have to do is remove the anti-imaging filter and make sure that we choose our $A_{r}$ in such a way that when the highest harmonic possible (given an $F_s$) is pitch shifted, it "folds" around the spectrum and ends up becoming $A_{r}$ again.

In such a "degenerate" case, we could keep pitch shifting the same number of harmonics again and again and again without the need to be "feeding" the shifter with "new" low frequencies.

For this to happen, given some $F_s$, we seek that geometric series which includes both $A_r, (F_s - A_r)$ and of course $\log_2\left(\frac{Fs-A_r}{A_r}\right) \in \mathbb{Z}$ and as large as possible.

This covers waveforms that do not "change" when pitch shifted by some even factor.

Pitch shifting and having the pitch go downwards is based on the same principle but is a bit more complex.

This can be done by that detuning mentioned by Hilmar in his response. The actual example that I had in mind when setting this dsp-puzzle, was that of the Weierstrass function.

The Weierstrass function is:

$$ f(t) = \sum_{k=1}^{\infty} a^k \cos(b^k \pi t)$$

Note here that the function is nothing more than a sum of cosines with an exponential amplitude profile. So, relatively low bass but lots of detail.

Things get out of hand with this function when $ab>1+\frac{3 \pi}{2}$ and $a \in \mathbb{R}, b \in \{2l: l \in \mathbb{Z}\}$ (i.e. b is odd).

In that region, the function is continuous but not differentiable. Nowhere.

The informal way to understand why (or rather, the one that works for me (?)) is to consider $g(t)=t^2$ and try to find its derivative in two successive $t, t+\Delta t$, points. The derivative would be $\lim_{\Delta t \to 0} \frac{g(t + \Delta t) - g(t)}{\Delta t}$. Now, on $g$, that is not going to be a problem, because the more the limit "zooms in" (by successively reducing the $\Delta t$), the more the function tends to become like a line. After all, all we are doing is raising a number to a power. But, the more we are zooming in on the Weierstrass function ($f(t)$), the more detail it reveals and that derivative never settles to a value. Never!. Check this out:

Visualisation of the Weierstrass function

(Visualisation from this link)

And here is what happens, when you try to listen to that $f(t)$ at a different speed. Similarly to what happened with the Shepard tones:

What if we were to speed up $f(t)$ by a factor of $b$? Then:

$$ f(b t) = \sum_{k=1}^{\infty} a^k \cos(b b^k \pi t) \implies \sum_{k=1}^{\infty} a^k \cos(b^{k+1} \pi t)$$

And if we were to ignore the amplitude profile ($a^k$) and make the sum finite to some $K$, then:

$$ f(b t) = \sum_{k=2}^{K+1} \cos(b^k \pi t)$$

Notice here how, $k+1$ was "absorbed" to show the similarity with the unshifted version.

So, pitch shifting by $b$ seems to be leaving us more or less with the same function. Now, set $b = 2^{\frac{13}{12}}$ and pitch shift by two:

$$f_{before}(t) = \sum_{k=1}^{K} \cos(2^{\frac{13}{12} k} \pi t)$$

$$\begin{matrix} f_{after}(2 t) = & \sum_{k=1}^{K} \cos(2 \cdot 2^{\frac{13}{12} k} \pi t) \implies \\ & \sum_{k=1}^{K} \cos(2^{\frac{12}{12}} \cdot 2^{\frac{13}{12} k} \pi t) \implies \\ & \sum_{k=1}^{K} \cos(2^{\frac{13}{12}-\frac{1}{12}} \cdot 2^{\frac{13}{12} k} \pi t) \implies \\ & \sum_{k=1}^{K} \cos(2^{(\frac{13}{12}-\frac{1}{12}) + \frac{13}{12} k} \pi t) \implies \\ & \sum_{k=1}^{K} \cos(2^{(k+1)\frac{13}{12}-\frac{1}{12}} \pi t) \implies \\ & \sum_{k=2}^{K+1} \cos(2^{k\frac{13}{12}-\frac{1}{12}} \pi t) \implies \\ & \sum_{k=2}^{K+1} \cos(2^{k\frac{13}{12}} 2^{-\frac{1}{12}} \pi t) \end{matrix}$$

So, the only difference between $f_{before}, f_{after}$ is that $2^{-\frac{1}{12}}$. That is, pitch shifting by two, introduces a "lower by 1 semitone" factor.

It gets freakier than this, for $b=2^{\frac{14}{12}}$ the lowering is a full tone and so on ($b=2^{\frac{q}{12}}$ with $q \in \left[13,14,15 \ldots 23\right]$) until $q=24$ where the pitch shifting now has come "full cycle".

And of course, to link it back to what Hilmar mentions, notice that $2^{\frac{13}{12}} \approx 2.1189\ldots$, that $0.1189\ldots$ "detuning" factor lands the components in such a way that the human ear choses to "latch" on the nearest tone (between $f_{before},f_{after}$) and thus perceived pitch goes down.

This covers waveforms for which the perceived pitch is lowered upon pitch shifting.

A note on sources:

  • Long time ago, I read a book on Fractals and that is where I first saw the Weierstrass function. At the time of reading the book, I did not understand fully everything I read. But I was impressed and puzzled. That book was in a different language (than English) and in fact, it was a (bit of a sloppy) "copy-paste-translate" from various books. Very recently, I found the original book those authors had used and I couldn't believe my luck, I just grabed it. That original book is "Fractals, Chaos, Power Laws: Minutes from an infinite paradise" by Manfred Schroeder. It is absolutely E P I C.

  • Sheppard tones and the Weierstrass function are mentioned in pages 95-97 (in my edition).

  • The part of the discussion on aliasing in this answer was added after quite a few people mentioned aliasing which motivated me to think about using it.

  • The breakdown of how does the $2^{\frac{1}{12}}$ emerges is not mentioned in the book. I added it here because it is not exactly obvious. To this, I was helped by a friend (M.P.L) as initially I was trying to substitute $k+1$ when it really emerges on its own.

  • The point about the Weierstrass function being continuous but not differentiable is the most difficult to grasp. Or, that has been very difficult for me. The mathematical proof, the way you go about pinpointing this fact, is not incomprehensible but it requires a lot of effort. Towards that goal, I found Johan Thim's Masters thesis incredibly useful. It contains a large number of such functions too, it's great.

  • The initial part on geometric series and music is general knowledge from various sources and experiences, cannot exactly pinpoint their source but I have added as many links as I could.


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