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and

z = e jΩ

The gain of this filter, or equivalently the magnitude of the Transfer Function, is

−
( −1 r) 1 2 r cos( Ω +2 ) r 2
He( jΩ ) = 2 3 0 4 (6.118)
+
( +1 Ar Br + Ar + r )
where

A =−4cos( )cos(Ω Ω ) (6.119)
0
B = 4cos ( )Ω + 4cos (Ω ) − 2 (6.120)
2
2
0
The lower and upper cutoff frequencies are defined, as previously noted, by
the condition:

jΩ 1
He( (, )12 ) = (6.121)
2

Substituting condition (6.121) in the gain expression (6.118) leads to the con-
clusion that the cutoff frequencies are obtained from the solutions of the fol-
lowing quadratic equation:

 (1+ r 2 )cos(Ω 
)
cos ( )Ω −  0  cos( )Ω
2
 r 
(6.122)
+ (1− r ) 2 [4r cos(2Ω ) (1− − r ) ] cos (Ω+ 2 ) = 0
2
4r 2 0 0
Adding and subtracting the roots of this equation, we deduce after some
straightforward algebra, the following determining equations for Ω and r:
0
1. r is the root in the interval [0, 1] of the following eighth-degree
polynomial:

8
5
3
4
r + ( a br− ) 6 − 8 ar + ( 14 a 2− b 2− ) r − 8 ar + ( a br− ) 2 + 1 0= (6.123)
where
a = (cos(Ω ) + cos(Ω )) 2 (6.124)
1 2

© 2001 by CRC Press LLC