Transfer function from poles and zeros

Question

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?

Answer 1

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.

Answer 2

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.

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:

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