A few conceptual questions about filter, pole, and bilinear

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

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).