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While using a continuous wavelet transform for my research project, I came up with some questions.

Having a set of equations for a Morlet wavelet,

$\psi(t)=\sqrt{\frac{2}{\pi}}e^{-t^2/2}e^{i6t}$
$\hat{\psi}(\omega)=2e^{-(\omega-6)^2/2}$

I have calculated the time and frequency resolution (i.e., $\Delta t$ and $\Delta \omega$, respecitvely) such that

$\sigma_\psi^2=\frac{\int_{-\infty}^{\infty}(t-t_0)^2|\psi(t)|^2dt}{\int_{-\infty}^{\infty}|\psi(t)|^2dt}=\frac{1}{2}$ and $\sigma_\hat{\psi}^2=\frac{\int_{-\infty}^{\infty}(\omega-\omega_0)^2|\hat{\psi}(\omega)|^2d\omega}{\int_{-\infty}^{\infty}|\hat{\psi}(\omega)|^2d\omega}=\frac{1}{2}$ where $t_0=0$ and $\omega_0=6=2\pi f_0$ $\Delta t=\sigma_t=s\sigma_\psi$ and $\Delta \omega=\sigma_\omega=\frac{\sigma_\hat{\psi}}{s}$ where the scale $s$ is approximated with $\frac{f_0}{f\cdot \text{sampling period}}$

I have been trying to calculate the Heisenberg box as the intervals $[t-\frac{\sigma_t}{2},t+\frac{\sigma_t}{2}]$ and $[\omega-\frac{\sigma_\omega}{2},\omega+\frac{\sigma_\omega}{2}]$, but the computation resulted with counterintuitive values, with $\Delta t$ being extremely large and $\Delta \omega$ being extremely small for almost every $t$ and $\omega$ values I plugged in.

As I read some previous posts (including here), it seems the problem may be coming from the units of the resolutions. I read that the units of $\Delta t$ and $\Delta \omega$ are

$[\Delta t] = \frac{\text{samples}}{\text{cycles}\cdot\text{radians}}$
$[\Delta \omega] = \frac{\text{cycles}\cdot\text{radians}}{\text{samples}}$

If I convert samples to seconds, the resulting intervals for the Heisenberg box seem to make sense. However, even if I change samples to seconds, the units still include cycles in them, and I do not know how I should interpret the values qualitatively.

My main questions therefore are (1) if the equations defining $\Delta t$, $\Delta \omega$, and $s$ are mathematically correct, and (2) how I should change the units of time and frequency resolution so I can interpret them in seconds and Hz.

Thank you very much.

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  • $\begingroup$ Hz = cycles / second. (1) are correct. If your question is rather why the values are counterintuitive, it'd help to see a concrete example, and what you expect values instead to be. $\endgroup$ – OverLordGoldDragon May 13 at 14:42
  • $\begingroup$ @OverLordGoldDragon Thank you for your comment. For example, having $f_0=\frac{\pi}{3}$ and sampling period of $\frac{1}{512}$, if I test on $f=40$ Hz, I get $s=13.4041$, $\sigma_t = 9.4782$, and $\sigma_\omega=0.0528$. What I am confused about is if I can interpret $\sigma_t$ as 9.4782 seconds width and $\sigma_\omega$ as 0.0528 Hz width on a time-frequency plane, or if this is not a correct interpretation. $\endgroup$ – SungJun Cho May 13 at 15:36
  • $\begingroup$ @OverLordGoldDragon Thank you. And sorry for my confusions, but wouldn't $[\sigma_\omega\cdot f_s]=\frac{\text{cycles}\cdot\text{radians}}{\text{samples}}\frac{samples}{sec}=$rad-Hz? Also, how does $\sigma_t$ end up with the seconds unit? Are the units I wrote above incorrect? $\endgroup$ – SungJun Cho May 13 at 15:54
  • $\begingroup$ Ignore my deleted reply, I get where the confusion is - am writing an answer. $\endgroup$ – OverLordGoldDragon May 13 at 15:55
  • $\begingroup$ Oh okay, thank you very much! $\endgroup$ – SungJun Cho May 13 at 15:55
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Confusion seems rooted in notation:

  • In $s = f_0 / (f \cdot \text{sampling rate})$, $f_0$ is the center frequency of mother $\psi$, and $f$ is the center frequency of the scaled $\psi$, i.e. $\psi_s$.
  • In $\sigma_{\hat\psi}=$ ..., $\omega_0$ is the (radian) center frequency of the wavelet of interest, or $\psi_s$, rather than $\omega_0=2\pi f_0$.
  • However, you can compute sigma of scaled from sigma of mother via: $\sigma_{\hat\psi_s} = \sigma_{\hat\psi} / s$ (see under 4.54).
  • $\text{sampling rate}$ is only relevant in discretized computations and must get canceled to yield physical units (Hz, sec, etc), so if you seek only physical use $s = f_0 / f_s$ (adjusted $f\rightarrow f_s$ to avoid confusion; recall that $f$ (or $\omega$) is what we integrate over, rather than plug in)
  • Do not involve $\text{sampling rate}$ in defining $\psi, \hat\psi$, scaled, or computing $\sigma$, or effectively most formulas you come across unless they explicitly include it, it's only to be used after computation, e.g. you first find $\sigma_{\hat\psi}$ without $\text{sampling rate}$, and then rescale as in $(4)$ here.
  • You commented "$\sigma_\omega$ Hz", but it's actually "rad-Hz"; to get Hz do $f=\omega / (2\pi)$.

Let me know if this makes sense (if not, again with a specific example).

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  • $\begingroup$ Thank you for the answer. Now it makes much more sense to me. I have some clarification questions for the third bullet point. First of all, I think $\psi$ should be $\hat{\psi}$ such that $\sigma_{\hat{\psi}_s}=\sigma_{\hat{\psi}}/s$? Please let me know if I am misunderstanding this point. Also, since I have $\sigma_\hat{\psi} = 1/\sqrt{2}$, for the sigma of scaled, I should calculate $\sigma_{\hat{\psi}_s} = \frac{1}{\sqrt{2}s}$. Am I right? $\endgroup$ – SungJun Cho May 13 at 16:25
  • $\begingroup$ Right, typo (fixed), also added a bullet. And you're right on scaled sigma. $\endgroup$ – OverLordGoldDragon May 13 at 16:58

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