The BW bandreject filter Hbr(s) is derived from a lowpass filter Hlp(s)   through the transform Hbr(s) = Hlp( B s / (s^2 + w0^2) )  resulting in a 4th order numerator and denominator for Hbr(s). Is the cutoff frequency of the lowpass filter the same as the W0 used for the transfomation? If so, why would they be different? This question is the result of an argument I am having with Grok.

You are asking about frequency transformation in the analog domain.

In general, you can take a lowpass filter design in the analog domain, and convert it to a different filter: lowpass (with different corner), highpass, bandpass, or bandstop.

In general, w0 is not the same as the bandwidth of the original analog LPF. The original bandwidth of the analog LPF needs to get stretched/scaled to achieve the desired bandwidth of stopband. What you are calling w0 corresponds to the center frequency of where the notch is placed. This is independent from the bandwidth.

References:

Proakis and Manolakis. "Digital Signal Processing, Principles, Algorithms and Applications" 4th ed. Section 10.4.1: Frequency Transformations in the Analog Domain (p.730).

WPI lecture notes: https://spinlab.wpi.edu/courses/ece503_2014/10-5fr...

When the transform Hbr(s) = Hlp( B s / (s^2 + w0^2) ) is used, where w0 is the target center frequency and B is the target bandwidth, the cutoff frequency of the prototype low-pass filter is assumed to be 1 radian/second.

B s / (s^2 + w0^2 )= B/2(1/(s+w0)+(1/(s-w0))

s+w0 shifts the frequency response of the prototype low-pass filter to the left by w0, and s-w0 shifts it to the right by w0.

If s=w0, the magnitude of Hbr is zero, as in the prototype low-pass filter for s → infinity.

If s= 0 or s → infinity, the magnitude of Hbr is one, as in the prototype low-pass filter for s=0.

The cutoff frequencies of the band-stop filter are wc1 and wc2, where wc2-wc1=B and wc1*wc2=w0^2.