0
$\begingroup$

Digital Signal Processing by John G. Proakis, Dimitris G. Manolakis(4th Edition) gives the prototype function for Butterworth low pass filter in an analog system which is fine till the point where the cutoff radian frequency is present in the equation. But then an epsilon and pass band radian frequency come into the equation. Digital Signal Processing by Li Tan mentions this epsilon as 'Absolute ripple specification'. So, my question is, we know about the stop band attenuation, pass band ripple and the pass band, stop band frequencies, what is this epsilon and how does the equation containing it (in the image) come to be?

Butterworth low pass filter prototype for analog system

$\endgroup$
0
$\begingroup$

In the first formulation, $\Omega_c$ is the $3$ $\textrm{dB}$ cut-off frequency, because for $\Omega=\Omega_c$ you obtain $|H(\Omega_c)|^2=\frac12$. In the other formulation, the attenuation at the pass band edge $\Omega_p$ is determined by the constant $\epsilon$. For $\Omega=\Omega_p$ you have $|H(\Omega_p)|=\frac{1}{1+\epsilon^2}$. I.e., by choosing the constant $\epsilon$ appropriately, any desired attenuation at the passband edge can be achieved.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Was there any need of introducing an epsilon when we already have 1-delta1(where delta1 is a variable giving maximum deviation from pass band magnitude response) to represent pass band ripple? We could have simply put pass band frequency in the first equation(in image above) and equated it to (1-delta1)^2! What is the necessity to view the equation in terms of pass band frequency? $\endgroup$ – Prasanjit Rath Apr 4 at 13:36
  • $\begingroup$ @PrasanjitRath: I don't see any delta in your question. $\endgroup$ – Matt L. Apr 4 at 14:05
  • $\begingroup$ Yes, it is not there in the question, sorry for that, all the confusion was due to two variables pointing out to the same thing(pass band ripple), delta1 and epsilon. So, 1-delta1 is actually 1/sqrt(1+epsilon^2);delta1 as in the answer to this question- dsp.stackexchange.com/questions/38564/… $\endgroup$ – Prasanjit Rath Apr 4 at 15:15
  • $\begingroup$ @PrasanjitRath: Yes, just note that a Butterworth filter doesn't have ripples, its magnitude response decays monotonically from the passband to the stopband. $\endgroup$ – Matt L. Apr 4 at 16:00

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.