# 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! Feb 27, 2019 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. Feb 28, 2019 at 8:37
• 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. Feb 28, 2019 at 15:37
• 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 $H(s) =\frac{(s−j 2 \pi *15e9)}{(s−j 2 \pi * 20e9)(s−j 2 \pi * 20e9)}$ Feb 28, 2019 at 20:04
• @shampar The last option is the correct one. Mar 1, 2019 at 16:35
• @MattL. Thank you for the correction. I've updated the equation. Mar 1, 2019 at 16:37

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:

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