1
\$\begingroup\$

I have a question. I'm studying filters and I came up with an apparent contradiction.

I have learned that a lowpass biquadratic section:

$$T(s)=\frac{\omega_0^2}{s^2+\frac{\omega_0}{Q_0}s+\omega_0^2}$$

Has 2 poles and 2 zeros at infinite frequency.

Then I have studied the various approximation methods and I have read that the Inverse-Chebyshev has finite transmission zeros (visible as zeros in the frequency response, in the attenuation band). My question now is: are these the previous infinite zeros, that were brought from infinite frequency to finite frequency, or are these "new" zeros that appear. It doesn't make sense to be additional zeros since that would lead to a numerator with a greater order than the denominator. Also Itested for a 2nd order filter and I found out just one finite zero so I assume the other one has infinite frequency? I'm a bit confused.

\$\endgroup\$
5
  • \$\begingroup\$ Suppose I had a bunch of 2nd order bandpass filters chained together in series with each other. Would not each one of these have one finite zero in their numerator? When multiplied, would not this mean many more than one finite zero in the numerator? Perhaps the only thing you may be missing is that the final denominator isn't stuck at 2nd order as the numerator order climbs? \$\endgroup\$
    – jonk
    Jul 16 '21 at 4:45
  • \$\begingroup\$ Yeah I get that but let's think just lowpass first. What I kind is that for example Butterworth just has a second order polinomial on the denominator, while Inverse Chebyshev has a polynomial in the numerator and denominator. So finite zeros appear. \$\endgroup\$ Jul 16 '21 at 4:48
  • \$\begingroup\$ I think you'll need to provide at least one example to work with. Pick a value of \$\epsilon\$ and a low pass filter order of your choosing, even or odd, and develop the exact transfer function under discussion. For 2nd order, I'm not sure you'll be able to make much of a distinction beyond picking some \$Q\$ or \$\zeta\$. It's in higher orders where it distinguishes itself, isn't it? However, you are discussing inverse functions (type II) and there I admit I'm more at a loss. So we may be at cross-purposes and that's entirely my fault, then. \$\endgroup\$
    – jonk
    Jul 16 '21 at 5:02
  • \$\begingroup\$ I followed along with a type II design, just now. I can't say I recall doing one of these designs before and I'm betting they aren't used nearly so much. But a low-pass filter most certainly has a numerator with zeros. Which always means the denominator has a order higher than 2, as well. The one I just went through was 5th order in the denominator and 4th order in the numerator. One of the zeros was out at infinity, too. And four as two pairs of finite conjugates. No 3rd or 1st order terms. I think I'll need to see a case example from you, I guess, that makes your point clearer. \$\endgroup\$
    – jonk
    Jul 16 '21 at 5:25
  • 1
    \$\begingroup\$ @GrangerObliviate I would have posted an answer but I'm on the run. You're probably confused by the order of the numerator. For a Butterworth, the order is 0, but the overall transfer function is a 2nd order. That's why the zeroes are implicitly at infinity, as if the numerator would have a 2nd order polynomial with null coefficients for \$s^2\$ and \$s\$. For an inverse Chebyshev, the coefficients are finite. When you'll get to FIR filters you'll see that their transfer function has no denominator, but that doesn't mean there are no poles: there are, as many as zeroes, and all of them at DC. \$\endgroup\$ Jul 16 '21 at 5:40
1
\$\begingroup\$

"I have read that the Inverse-Chebyshev has finite transmission zeros (visible as zeros in the frequency response, in the attenuation band). My question now is: are these the previous infinite zeros, that were brought from infinite frequency to finite frequency, or are these "new" zeros that appear. It doesn't make sense to be additional zeros since that would lead to a numerator with a greater order than the denominator."

For all classical 2nd-order lowpass approximations there are 2 zeros at infinite frequencies (exception: Even order Chebyshev, invers). These zeros are NOT caused by the numerator, but by corresponding poles of the denominator.

For some of these approximations (inverse Chebyshev, Cauer/elliptical,...) we have in addition 2 real zeros which are introduced by the numerator polynom. So, the numerator degree will not exceed the degree of the denominator.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.