0
$\begingroup$

If we know that a filter (or a system) has two poles at 20 GHz and one zero at 15 GHz, then how do you write the transfer function $H(s)$ for such a system?

I am wondering why sometimes the poles and zeros are given in Hz and other times as complex numbers?

|improve this question
$\endgroup$
  • $\begingroup$ Trying to narrow down your question: Do you know what a pole and a zero is in terms of a polynomial (or a fraction of polyomials)? What you're essentially asking what the $s$ in a Laplace or the $\omega$ or $f$ in a Fourier transform mean, and that's actually quite an invitation to write a book! $\endgroup$ – Marcus Müller Feb 27 at 18:48
0
$\begingroup$

You must remember that $s=j\omega$ and that $\omega=2\pi f$ and the transfer function is:

$H(s)={\prod_n (s-z_n)}/{\prod_m(s-p_m)}$

Where $z$ and $p$ are the zeros and poles respectively. Then you just need to adapt to your case.

|improve this answer
$\endgroup$
  • 3
    $\begingroup$ The sums in your formula for $H(s)$ should be products. $\endgroup$ – Matt L. Feb 28 at 8:37
  • $\begingroup$ Besides what Matt L. said it can also be mentioned that a transfer function can also be multiplied by any nonzero gain. So the poles and zeros alone are not enough information to define an unique transfer function. $\endgroup$ – fibonatic Feb 28 at 15:37
  • $\begingroup$ Yes, you are right. I am aware of this general formula, but my question was about the details of fitting those poles and zeros given in GHz into the H(s) equation. Should I write it as $H(s) = \frac{(s−15e9)}{(s−20e9)(s−20e9)}$ or should I write it as $H(s) = \frac{(s−j*15e9)}{(s−j*20e9)(s−j*20e9)}$ or $H(s) =\frac{(s−j 2 \pi *15e9)}{(s−j 2 \pi * 20e9)(s−j 2 \pi * 20e9)}$ $\endgroup$ – shampar Feb 28 at 20:04
  • $\begingroup$ @shampar The last option is the correct one. $\endgroup$ – Filipe Pinto Mar 1 at 16:35
  • $\begingroup$ @MattL. Thank you for the correction. I've updated the equation. $\endgroup$ – Filipe Pinto Mar 1 at 16:37

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.