A few conceptual questions about filter, pole, and bilinear

Question

I am working on a school project on converting a 6th order butterworth high pass filter to digital filter using bilinear transformation.

Just got a couple conceptual questions need to be clarified before I continue.

1. In an analog 6th order butterworth filter, the poles are the same for highpass and lowpass since poles are found in the denominator?

2. I have found 6 poles, and say my first pole is -0.259+ 0.966i if my cut off frequency is $1 \textrm{ rad/s}$. If my cut off frequency is changed to $10000 \textrm{ rad/sec}$. The radius of the unit circle would be $10000$ and my pole would be -2590 + 9660i?

3. Is there anyway someone could verify that once applied bilinear transformation without pre-wraping the frequency (professor said do it without pre-wraping first and compare to wrapped after), in the $z$-plane the poles would be:

0.1894 + 0.7499i
0.1894 - 0.7499i
0.1405 + 0.4077i
0.1405 - 0.4077i
0.1222 + 0.1283i
0.1222 - 0.1283i 

I'm quite new on DSP, any help is appreciated.

Answer 1

The poles of low pass and high pass Butterworth filters are indeed the same if both filters have the same cut-off frequency. The difference between the two lies in the numerator. A low pass filter has a constant in the numerator, whereas a high pass filter has a constant times $s^N$, where $N$ is the filter order.

The poles of a normalized Butterworth low pass filter (i.e., one with cut-off frequency $\omega_c=1$) lie on a circle with center $s=0$ and radius $1$ in the complex $s$-plane. Of course they are all in the left-half plane:

$$p_k=-e^{j\frac{k\pi}{2N}}\tag{1}\\ k=\pm 1,\pm 3,\ldots,\pm (N-1),\quad N\text{ even}\\ k=0 ,\pm 2,\ldots,\pm (N-1),\quad N\text{ odd}$$

If you have a cut-off frequency $\omega_c\neq 1$, the variable $s$ is transformed as $s\rightarrow \frac{s}{\omega_c}$, so a factor $\frac{1}{s-p_k}$ of the transfer function becomes $\frac{1}{s/\omega_c-p_k}=\frac{1}{\omega_c}\frac{1}{s-\omega_cp_k}$. So the poles are indeed multiplied by the cut-off frequency $\omega_c$, i.e., they lie on a circle with radius $\omega_c$. Their angles of course remain unchanged.

Without pre-warping, the bilinear transformation is usually implemented as

$$s=2f_s\frac{z-1}{z+1}\tag{2}$$

where $f_s$ is the sampling frequency. Expressing $z$ in terms of $s$ gives

$$z=\frac{2f_s+s}{2f_s-s}\tag{3}$$

Equation $(3)$ can be used to determine the pole locations in the complex $z$-plane. If the cut-off frequency of the analog filter is $\omega_c$, its poles are located at $\omega_cp_k$, with $p_k$ given by $(1)$. So the poles in the complex $z$-plane are given by

$$z_{\infty,k}=\frac{2f_s+\omega_cp_k}{2f_s-\omega_cp_k}\tag{4}$$

You can use $(4)$ to check the pole locations you obtained from Matlab (and I don't doubt that they are identical up to numerical inaccuracies).