Question

What is the general formula that represent the phase of the poles of transfer function of normalized low pass Butterworth filter of order N?

(a) \(\frac{π}{N} k+\frac{π}{2N}\) k=0,1,2…N-1

(b) \(\frac{π}{N} k+\frac{π}{2N}+\frac{π}{2}\) k=0,1,2…2N-1

(c) \(\frac{π}{N} k+\frac{π}{2N}+\frac{π}{2}\) k=0,1,2…N-1

(d) \(\frac{π}{N} k+\frac{π}{2N}\) k=0,1,2…2N-1

I got this question by my college director while I was bunking the class.

Origin of the question is Butterworth Filters in division Digital Filters Design of Digital Signal Processing

Answer 1

Right option is (d) \(\frac{π}{N} k+\frac{π}{2N}\) k=0,1,2…2N-1

The explanation: The transfer function of normalized low pass Butterworth filter is given as

HN(s).HN(-s)=\(\frac{1}{1+(\frac{s}{j})^{2N}}\)

The poles of the above equation is obtained by equating the denominator to zero.

=> \(1+(\frac{s}{j})^{2N}\)=0

=> s=(-1)^1/2N.j

=> sk=\(e^{jπ(\frac{2k+1}{2N})} e^{jπ/2}\), k=0,1,2…2N-1

The poles are therefore on a circle with radius unity and are placed at angles,

θk=\(\frac{π}{N} k+\frac{π}{2N}\) k=0,1,2…2N-1