0
$\begingroup$

I have recently created a real-time STFT with 50% overlap.

enter image description here

I wanted to know if this window-based is possible for scalogram, especially continuous wavelet transform. I haven't found anyone implementing it real-time.

Also, if it is possible, does one need to compute the CWT of the whole window signal or simply a portion of it, like in a recursive way. I started learning wavelets and I look forward to any hints.

$\endgroup$

1 Answer 1

1
$\begingroup$

Whether it's "real-time" depends on sampling rate. What's true is that most implementations fall short for realistic $f_s$.

I am currently working on CWT that will be faster than any other I know of, best case by several times. Though, it may be a while until it's open-sourced.

An often overlooked point is, CWT has hop_size just like STFT. Especially if using a scalogram (modulus), there's very little to lose from hop_size=2 or 4, and sometimes we can do 100+. This is implemented easily with Fourier-domain subsampling:

ifft(a * b)[::2] == ifft((a * b).reshape(2, -1).mean(axis=0))

which reduces FFT size by hop_size. The relevant question is, what hop_size is safe? This can be measured directly, as discussed in this answer. Note, perfect inversion is possible for some CWT with complex-valued outputs and hop_size>1, but max permissible hop_size is fairly small and the inversion algorithm not straightforward. However, we don't need complex for inversion; inversion is still possible within a global phase-shift.

Another major slowdown is in low-support wavelets, which are better off with overlap-add/overlap-save.

Putting all these together elevates max supported $f_s$ significantly.

$\endgroup$
2
  • $\begingroup$ Is there a way to determine this realistic $f_s$? Is there a way to know the big-$O$ notation of CWT? $\endgroup$ Commented Jun 18, 2022 at 9:09
  • $\begingroup$ @EddyPiedad If it processes 1 second of data in 1 second, it's "real-time" with 1 sec latency. What qualifies as "real-time" varies by application, sometimes we want 1 sec in 0.001 sec. $\endgroup$ Commented Jun 18, 2022 at 21:10

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.