5
$\begingroup$

I have two samples which, when exposed to ultrasound, emit their unique frequency responses. As can be seen in the attached figure, where the exciting frequency is 2.25 MHz, sample 1 emits a strong subharmonic and also ultra-harmonics which is not the case for sample 2. enter image description here

My 1st question is, if I want to classify samples 1 and 2 using machine learning, how can I make use of the subharmonic and ultra-harmonics? The goal is that, when I want to know whether a signal is from sample 1 or sample 2, the ML algorithm would be able to tell that based on the presence or absence of the sub and ultra harmonics. For sample 2, the harmonics present are lower in power but I don't want to rely on amplitude.

My 2nd question is, is there any way to separate sample 1 and 2 signals from a signal coming from their mixture using machine learning?

Thanks in advance.

$\endgroup$
0

1 Answer 1

0
$\begingroup$

One simple way is to create a feature set based on the energy on the different frequency bins.

If the case above is representative, even "small guns" (Linear classifiers) will solve this.

Regarding the mixing, it depends if the mix is a sum. If it is, then it will be able to classify it if there is a difference in energy ratio between the different bins.

$\endgroup$
2
  • $\begingroup$ thanks for your answer. I know this question would sound dumb, but I was trying to figure out how to define the frequency bins in matlab. My data length is N=8751 and sampling rate is 250e6 MSa/s. Can you please help in this regard? $\endgroup$
    – nasrin
    Commented Jan 22, 2022 at 6:13
  • $\begingroup$ @nasrin, Could you please mark this answer and open a new one about binning? I would be happy to answer it. Link it here. $\endgroup$
    – Royi
    Commented Jan 22, 2022 at 13:08

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.