# Find Transfer Function and Appropriate Coefficients of the Transfer Functions from Pole Zero Plot

I got a Transfer function problem and I am confused in finding a solid solution step. Below is the problem description:

1st and 2nd order discrete-time filters with different pole-zero locations shall be considered for below:

Determine the transfer function and compute the orders M, N of the numerator and denominator, respectively. Choose appropriate coefficients of the transfer functions which approximate the given locations.

I had looked for information about get the transfer function from Pole Zero Plot. Unfortunately I only got search results of plotting Pole Zero Plot from example transfer function. Therefore, I try to estimate the pole-zero locations and plot it by Matlab. The result look similar with the given plot. Below are the result:

As of now I estimate the transfer function in below:

Transfer Function For the left plot $$H(z)=\frac{1}{z-0.8}$$

Transfer Function For the right plot $$H(z)=\frac{{(z-0)}{(z-0.8)}}{{(z+0.5+0.5i)}{(z+0.5-0.5i)}}$$

Also, below is my step to compute the orders M, N of the numerator and denominator, and appropriate coefficients for the left plot

\begin{align} H(Z) = \frac{b_0}{a_0} z \frac{1}{z-0.8} \\ = \frac{b_0}{a_0} \frac{z}{z-0.8} \\ = \frac{b_0}{a_0} \frac{1}{1-0.8z^{-1}} \end{align}

I hope I did it right for the above solution. I am confused to do the same for $$H(z) = \frac{{(z-0)}{(z-0.8)}}{{(z+0.5+0.5i)}{(z+0.5-0.5i)}}$$ since it is complex Pole Zero Plot. Therefore I would like to ask how can I compute and get the coefficient for the right plot?

Updated:

Solution steps for the right plot:

\begin{align} H(Z) =\frac{{(z-0)}{(z-0.8)}}{{(z+0.5+0.5i)}{(z+0.5-0.5i)}} \\ = \frac{z^2-0.8z}{z^2+0.5z-0.5iz+0.5z+0.25-0.25i+0.5iz+0.25i-0.25i^2} \\ = \frac{z^2-0.8z}{z^2+z+0.25-0.25i^2} \\ = \frac{z^2-0.8z}{z^2+z+0.25-(-0.25)} \\ = \frac{z^2-0.8z}{z^2+z+0.5} \\ \end{align}

$$(z-z_{\infty})(z-z_{\infty}^*)=z^2-2\textrm{Re}\{z_{\infty}\}z+|z_{\infty}|^2$$
• @HaRLoFei: Not bad, but I'm sure you know how to simplify $0.25-0.25i^2$. Dec 27, 2021 at 22:32
• Thank you. Yes and it should be $0.5$ Dec 30, 2021 at 7:17