0
$\begingroup$

The antitransform of the function is given:

$ \hat{x}(f) = \frac{123 + i246\pi f}{246 - 24600 \pi^2 f^2 + i4920\pi f} $

I'm asked to determine which frequency can I sample using the function x(t) avoiding aliasing? (or which range of values)

Update the problem actually suggest five answers: 100, 10, 1, 0.4 or none of these. Turns out the answer is none of these.

$\endgroup$
  • $\begingroup$ draw a bode plot for given transfer function and the bandwidth comes close to 0.275 Hz hence sampling frequency is grater than 0.55 Hz, if your calculations is based on cutoff frequency (0.2 Hz) then sampling frequency becomes 0.4 Hz. $\endgroup$ – Ch.Siva Ram Kishore Dec 12 '19 at 16:47
  • $\begingroup$ “losing aliasing” is a peculiar usage. do you mean avoiding aliasing? $\endgroup$ – user28715 Dec 12 '19 at 18:08
  • $\begingroup$ @StanleyPawlukiewicz yes, sorry I'm correcting it. $\endgroup$ – Someone Dec 12 '19 at 18:12
1
$\begingroup$

The function you show has a frequency range of $-\infty$ to $\infty$ so it isn’t strictly band limited. To completely avoid aliasing, it needs infinite sampling.

Of course, you can use tables or Fourier identities to obtain the exact inverse.

Usually some acceptable criteria is specified for tolerable aliasing when specifying a sample rate.

$\endgroup$
  • $\begingroup$ How did you determine the frequency was -\infty to \infty $\endgroup$ – Someone Dec 12 '19 at 18:30
  • 1
    $\begingroup$ Probably by inspection. But the formal way to do it is to note that for any finite $f$, $\left | \hat x (f) \right | > 0$. $\endgroup$ – TimWescott Dec 12 '19 at 18:31
  • 1
    $\begingroup$ @Someone, by inspection, $\endgroup$ – user28715 Dec 12 '19 at 18:38
  • $\begingroup$ Thank you I accepted your answer. Now what if my function was actually finite, is there a way to obtain the frequencies using the fourier transform? $\endgroup$ – Someone Dec 12 '19 at 19:03
  • 1
    $\begingroup$ @Someone you would use the Nyquist criteria. $\endgroup$ – user28715 Dec 12 '19 at 19:30

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.