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?

  • $\begingroup$ 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 '19 at 18:48

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.

  • 3
    $\begingroup$ The sums in your formula for $H(s)$ should be products. $\endgroup$ – Matt L. Feb 28 '19 at 8:37
  • 1
    $\begingroup$ 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 '19 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 '19 at 20:04
  • $\begingroup$ @shampar The last option is the correct one. $\endgroup$ – Filipe Pinto Mar 1 '19 at 16:35
  • $\begingroup$ @MattL. Thank you for the correction. I've updated the equation. $\endgroup$ – Filipe Pinto Mar 1 '19 at 16:37

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