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,


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.

  • $\begingroup$ 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. $\endgroup$ Commented May 13, 2021 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. $\endgroup$
    – scho
    Commented May 13, 2021 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? $\endgroup$
    – scho
    Commented May 13, 2021 at 15:54
  • $\begingroup$ Ignore my deleted reply, I get where the confusion is - am writing an answer. $\endgroup$ Commented May 13, 2021 at 15:55
  • $\begingroup$ Oh okay, thank you very much! $\endgroup$
    – scho
    Commented May 13, 2021 at 15:55

1 Answer 1


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).

  • $\begingroup$ 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? $\endgroup$
    – scho
    Commented May 13, 2021 at 16:25
  • $\begingroup$ Right, typo (fixed), also added a bullet. And you're right on scaled sigma. $\endgroup$ Commented May 13, 2021 at 16:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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