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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: $$ w = 2*\arctan(\frac{\Omega*T_d}{2}) $$

bilinear transformation

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

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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 $z=-1$.

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  • $\begingroup$ I'm sorry but I don't understand your answer. Isn't s = sigma + j*omega? and |s| = sqrt(sigma^2 + omega^2)? Why is s = inf equivalent to omega = inf (you modified my original question making this assumption without explaining why) and |s| = inf equivalent to both omega = inf and omega = -inf (in your answer). When you say that s = inf why are you assuming that is only the omega component the responsible for this? Couldn't you also have a s = inf if omega = 0 but sigma = inf? Please if it is not too much asking could you edit your answer to include these explanations? Thank you very much $\endgroup$
    – VMMF
    Jan 24, 2018 at 21:24
  • $\begingroup$ @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. $\endgroup$
    – Matt L.
    Jan 25, 2018 at 7:19
  • $\begingroup$ I'm so sorry you didn't modified my question. What I can actually see is that it is not S but Omega in the equation the one that when tending to infinity generates the zeros $\endgroup$
    – VMMF
    Jan 26, 2018 at 5:16
  • $\begingroup$ Would it be accurate to say that since the magnitude (squared) response is symmetric if I have N zeros, there will be N/2 at inf and N/2 at -inf? I'm studying Alan Oppenheim Discrete-Time_Signal_Processing 3rd edition and at page 509 ch 7 it says "if s = inf maps to z = -1 in bilinear transformation" I believe it should have said |s| = inf like in your answer $\endgroup$
    – VMMF
    Jan 26, 2018 at 5:31
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    $\begingroup$ @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 ...?) $\endgroup$
    – Matt L.
    Jan 26, 2018 at 8:26

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