Page 214 - Analog and Digital Filter Design

Bandstop Filters 2

Suppose the filter you want is required to have a stopband between 45 Hz and
55 Hz. This is for illustration only but could be used to remove power line fre-
quencies (50 Hz in Europe). This specification gives B @' = 10, = 50 Hz, and
Q,, = 5. Taking one pole from the first pair: s = -0.9239 + j0.3827, (r= 0.9239,
and u) = 0.3827.

The two bandstop poles produced from this are found from the following
equations:
@2- - (r' + w' = 1 (since the poles are on a unit circle for the Butter-
worth response).

/I=------- 0 - 0.18478
00 .QBS
w
B=- = 0.07654
wu .Qss
f= B' - A' + 4 = 3.89176
s= , " ' d f+ f2-4A'B' = 1.9727414

This gives K' = 0.5d(A +/I)' + (B +g)' = 1.0291 25693.
f;,
The frequencies are .f., = - = 48.58492 andU& = Pf.r:f) = 51.456284
It'
m '
The pole's Q factor is given by p = ~ = 5.361447.
A+h
The second pair of poles can be found in a similar way. Due to symmetry,
G= 0.3927 and o = 0.9239:

.f = B' - A' + 4 = 4.028285277

'4 B
h = ~ = 0.00704670333
g
This gives /F'=0.5J(d+/i)' +(B+g)' = 1.096709865.