# Butterworth low pass filter zeros location after bilinear transformation explanation

I am studying in a text book the transformation of a continuous time Butterworth low pass filter into a discrete time filter by means of bilinear transformation: $$s = \frac{2}{T_d}*\frac{1-z^{-1}}{1+z^{-1}}$$ which imposes the following relationship between discrete and continuous frequency variables: $$w = 2*\arctan(\frac{\Omega*T_d}{2})$$ As can be clearly seen from this formulas a value of $\Omega$ = inf causes w = pi and a value of $\Omega$ = -inf causes w = -pi. However, both values of w correspond to z = -1 in the Z plane

In the example a 6th order continuous time Butterworth low pass filter is designed. Its magnitude squared function in continuous time is defined by: $$|H_c(j\Omega)|^2 = \frac{1}{1 + (\frac{j\Omega}{j\Omega_c})^{2N} }$$ From this formula it can be seen that as $\Omega$ goes to infinity the magnitude function goes to zero. So I can understand that in a 6th order filter there are 6 zeros in infinity (in S plane), which in discrete time means there are 6 zeros in w = pi and hence 6 zeros in z = -1. But:

1. What about the behavior of the system in s = -inf?
2. Isn't the magnitude squared function symmetric ?

Wouldn't it also cause the magnitude function to go to zero and should generate zeros on w = -pi and z = -1

An $N^{th}$-order analog prototype system results after bilinear transform in an $N^{th}$-order discrete-time system. So the number of zeros and poles remains the same. All zeros at $|s|\rightarrow\infty$ map to $z=-1$. In your case there will be $6$ zeros at $z=-1$ because there are $6$ zeros at $|s|\rightarrow\infty$.
Of course, the magnitude (squared) response is symmetric, but the zeros at $|s|\rightarrow\infty$ cause the zeros in the magnitude at $\Omega\rightarrow\infty$ as well as at $\Omega\rightarrow -\infty$ (you can imagine those points meet at infinity). This becomes more obvious in the $z$-domain where moving towards positive frequencies means moving counterclockwise from $z=1$ (i.e., $\omega=0$), whereas moving towards negative frequencies means moving clockwise from $z=1$. In both cases you'll end up at $z=-1$.
• @VMMF: Not sure which modification of your question you're talking about because the only thing I changed was a Latex typo in your very first equation. Concerning the limit $s\rightarrow\infty$, there's only one infinity for complex numbers. So no matter if you let the real part or the imaginary part tend to plus or minus infinity, they all meet at the same point: infinity. Have a look at the Riemann sphere. – Matt L. Jan 25 '18 at 7:19
• @VMMF: It's common to say $s\rightarrow\infty$, because, as I've mentioned before, there is only one infinity for complex numbers. You cannot say that half of the zeros are at plus infinity and the other half at minus infinity, simply because they all meet at one point, and that's (complex) infinity. (And what would you do if the number of zeros is odd ...?) – Matt L. Jan 26 '18 at 8:26