1
$\begingroup$

Question

I'm trying to figure out the sampling rate for my ADC to sample essentially signal essentially of the form: $$y(t) = \sin(\max(t, \omega_{max})\times t) + n$$ where $n$ is noise.

Context

This signal is from a motor, where the rotational speed linearly ramps up to some $\omega_{max}$. To figure out the required sampling rate, the only method I know of is to use the Nyquist rate of 2 times the maximum frequency content. The above signal can be re-written as

$y(t) = rectangularPulse(t \epsilon [0, \omega_{max}]) \times \sin(t^2) + heaviside(t-\omega_{max}) \times \sin(\omega_{max}t)$

This isn't bandlimited since a rectangular pulse is a $\text{sinc}$ in the frequency domain. What I want to do is low pass filter the signal first as described here. But I'm at a loss as to figure out what I should set my cut off frequency to, as the fourier transform of the signal doesn't die off at any frequency (unless I'm mistaken).

I feel like this should be a classical problem, but I can't find anything online. If you know of how to solve this problem, or you could point me to where to look, it would be greatly appreciated!

|improve this question
$\endgroup$
  • $\begingroup$ I don't see how your first equation corresponds to a motor ramping up. I would write it as $$y(t) = \begin{cases} \sin(r(t) t), \,\text{ $t \leq t_R$} \\ \sin(\omega_\text{max} t), \, \text{ $t > t_R$} \end{cases},$$ where $r(t)$ is a linear ramp and $t_R$ is the time it takes the motor to reach $\omega_\text{max}$. What do you think? $\endgroup$ – MBaz Oct 25 '19 at 0:23
  • $\begingroup$ That works! I think the above should be written as $min(t, $\omega_{max})$. I'll correct it to yours to be safe thanks! $\endgroup$ – YYH Oct 25 '19 at 1:12
1
$\begingroup$

If your frequency increase linearly, then you have a chirp signal with linearly increasing frequency.

https://en.wikipedia.org/wiki/Chirp_spectrum

There is a closed-form solution for the Fourier transform. I'm not sure how that will help you. Why not pick a sampling frequency

$\omega_s \gt 2 \omega_{max}$

Since your highest frequency is $\omega_{max}$, no point in considering lower frequencies for your sampling rate. Is this for a control-loop algorithm? In that case, I would use a more rule-of-thumb criterion.

$\omega_s \gt 20 \omega_{max}$

This makes you less susceptible to sampling delays (among other things)

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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