# Transfer function from poles and zeros

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?

• 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! – Marcus Müller Feb 27 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.
• The sums in your formula for $H(s)$ should be products. – Matt L. Feb 28 at 8:37
• 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 $H(s) =\frac{(s−j 2 \pi *15e9)}{(s−j 2 \pi * 20e9)(s−j 2 \pi * 20e9)}$ – shampar Feb 28 at 20:04