I stumbled upon Hilbert transform when researching single sideband modulation. Apparently when the demodulator frequency is bit off by $\Delta f$, the signal after demodulation and low-pass filtering is not $m(t)$ where $m(t)$ is the original signal, but rather:
$$m(t)\cos(2 \pi \,\Delta f \,t) + \hat{m}(t)\sin(2 \pi \,\Delta f \,t)$$
...where $\hat{m}(t)$ is the Hilbert transform of the original signal. So the inaccurately demodulated signal oscillates between the original signal and its Hilbert transform.
I tried the equation for real music. I noticed that if $\Delta f$ is 100 Hz, frequencies are markedly distorted, but music at $\Delta f$ of 10 Hz sounds acceptable (but for someone with good musical ear it might be possible that 10 Hz shift wouldn't be acceptable).
So, I became interested of whether a human could hear raw Hilbert-transformed signal -- then $\Delta f$ would be 0 Hz but there would be constant phase shift in the SSB demodulation equation. It can occur in single-sideband modulation for example if the demodulator has exactly the same frequency as the modulator, but its phase is 90° off.
I can't notice any difference in the original signal (music) and Hilbert-transformed signal that I created.
I researched a bit whether there's any theoretical justification on this, and found this answer.
A Hilbert transform introduces a fixed 90° phase shift for all frequencies. Practical Hilbert transformers can only do this over a range of frequencies. Clearly a 90° phase shift represents a longer delay at low frequencies (one fourth of the period) than at higher frequencies where the period is shorter
[snip]
So a Hilbert transform filter passes high frequencies quickly while introducing progressively longer time delays for lower frequencies. In the limit where the frequency approaches zero, the time delay approaches infinity.
[snip]
However, a typical audio signal spanning a number of octaves will be distorted due to high frequencies arriving sooner than low frequencies.
Can this really be the case? I didn't notice any error in the timing of different frequencies.
I think the Quora answer might be erroneous, because Hilbert transform adds delay to the individual sine/cosine wave components of the original signal, with each component delayed by 90°. So for example a sine wave that has always been on, and will always be on, is delayed by 90° (so low frequencies are delayed more than high frequencies). But there should be no way of hearing this, right?
However, what you generally are interested is for example a sine wave that is instantly turned on, for example at the moment $t=0$. Then you no longer have $\sin(2 \pi f t)$, but $u(t) \sin(2 \pi f t)$, where $u(t)$ is the (Heaviside) unit step function that adds lots of high frequencies to the signal, so the signal is no longer a simple sine wave, and the high-frequency envelope, being high in frequencies, shouldn't be delayed as much as the low-frequency sine wave (whose delay / phase we can't hear).
Is my reasoning correct? Or could it be possible that Hilbert transform can be heard in audio by different delay for different frequencies?