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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?

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  • $\begingroup$ 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
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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.

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    $\begingroup$ The sums in your formula for $H(s)$ should be products. $\endgroup$ – Matt L. Feb 28 '19 at 8:37
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    $\begingroup$ 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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Given this is a filter and not an oscillator the correct answer is

$$H(s) = K\frac{s+2\pi15e9}{(s+2\pi20e9)(s+2\pi20e9)}$$

unless further clarification is given that there are complex poles.

When poles and zeros are described to be at a particular frequency in Hz in relation to a filter implementation, it is actually describing a system with a real negative poles and zeros unless specifically clarified otherwise. This is a sloppy approach in my opinion to describe a filter, but this is indeed what I have seen in most cases.

Consider the simple example of a first order low pass filter with a pole at 20 GHz. In this case the pole would be located at $-2 \pi 20e9$. And the transfer function would be given as

$$H(s) = \frac{K}{s+2\pi 20e9}$$

As depicted in the plot below, the frequency transfer function of this filter is determined by setting s to only be the $j\omega$ axis and sweeping it to determine the magnitude and phase of $H(s)$ for each frequency given by $s=j\omega$.

Note specifically what happens when $s = j2\pi20e9$ which is the snapshot captured in this plot: The denominator of the transfer function is a vector given by $s- (-2\pi 20e9)$. So at this particular location, the denominator is $\sqrt{2}$ larger and has an angle of 45°, relative to the starting frequency when $s=0$. Therefore overall the result will be -3 dB and -45°, which is what we expect for the cut-off frequency of such a first order filter.

single pole example

In this simpler example, when $s=0$ (DC response):

$$H(s_1) = \frac{K}{2\pi20e9}$$

And when s is "at the pole frequency" $s = j2\pi20e9$:

$$H(s_2) = \frac{K}{j2\pi20e9+ 2\pi20e9}$$

Comparing the ratio of the two results in:

$$H(s_1)/H(s_2) = \frac{1}{\sqrt{2}\angle{45°}}$$

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