0
$\begingroup$

Blue part is observed, black part is not!

Edit:" please see the attached figure. The blue part is all I have. Also, you can neglect the noise term. Assume the signal is deterministic. Typical values of the unknown parameters $\alpha$ and $\omega_0$ are in the the range [0.25,5]. Note that all the parameters are unknown $A \omega_0, \alpha, \phi_0$. The setup hints at curve fitting which I did but don't prefer."

I have a small part from a chirp signal

$$s(t)= A \cos\left(\omega_0 t+ \frac\alpha2 t^2 + \phi_0\right)+n(t)$$

with very low start frequency $\omega_0$ and chirp rate $\alpha$.

The signal is sampled at high sampling frequency $f_s$. But the available time domain signal is short, i.e. barely a complete cycle. The amplitude of the signal and the phase shift are also unknown. Are there ways to figure out the chirp parameters other than curve fitting or spectrogram?

$\endgroup$
  • $\begingroup$ Noise? typically makes deterministic problems interesting $\endgroup$ – Stanley Pawlukiewicz Jan 12 at 0:01
  • 1
    $\begingroup$ i gotta paper dealing with this from 18 years ago. it is based on windowing that chirp with a gaussian window. you window your chirp, do an FFT, log that FFT, do a discrete derivative (or difference) and do a linear fit in the log domain. and somewhere i have old MATLAB code. $\endgroup$ – robert bristow-johnson Jan 12 at 5:52
  • $\begingroup$ just to be sure I get this correctly: Which of the parameters $(A, \omega_0, \phi_0)$ are known, and which are unknown, Ahmad? $\endgroup$ – Marcus Müller Jan 12 at 12:45
  • $\begingroup$ Thank you for your answer, To answer your question, nothing is known. All the parameters including $ A$, $\alpha$, $w_0$ and $ \phi_0 $ are unknown. $\endgroup$ – Ahmad Jan 12 at 23:00
  • $\begingroup$ Robert, I have gave the paper a look. It doesn't deal with low frequencies. Or am I missing something? $\endgroup$ – Ahmad Jan 12 at 23:10
0
$\begingroup$

Amplitude $A$ is trivial. Square and low-pass filter/average; then take the square root.

Without noise (or with good SNR), you could simply calculate a derivative:

\begin{align}\frac{\partial\, s\}{\partial t} = -A\sin\left(\omega_0t + \frac\alpha2t^2 +\phi_0\right)\cdot\left(\omega_o+ \alpha t\right) \end{align}

and low-pass filter the result

$$\text{LPF}\left\{\frac{\partial\, s\}{\partial t}\right\} = \omega_o + \alpha t$$

From the resulting linear function, $\alpha$ can directly be extracted as the slope, and knowing the slope, $\omega_0$ is just the constant summand.

With $A$, $\omega_0$ and $\alpha$ known, you can simply compare at any fixed $t$ to find $\phi_0$.

$\endgroup$
  • $\begingroup$ The method you proposed and called trivial for finding A is like that used to demodulate FM received signals, there we use a high pass filter instead. I tried to do it as you said but It worked only when the observed time domain signal had several cycles in it, i.e. it doesn't work for my signal that is of barely one cycle i.e. most of the time the range of the argument of the cosine doesn't complete 2 pi. Same seems to go with the derivative. The problem is, the frequency is very low. Is that relevant? does the uncertainty principle come to play here?($\Delta t \Delta f >= constant$)? $\endgroup$ – Ahmad Jan 12 at 23:08
  • $\begingroup$ no, the uncertainty principle doesn't come into play here, as far as I can see, but you're right, with less than a cycle all this gets inaccurate $\endgroup$ – Marcus Müller Jan 13 at 10:03
  • $\begingroup$ Your answer was helpful to me. Thank you! $\endgroup$ – Ahmad Jan 14 at 13:55

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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