Page 260 - Analog and Digital Filter Design

Phase-Shift Networks (All-Pass Filters) 257

cos[(k - 1)7c/2n]
U/< = .uk-l k = 1,2,. . . n
sin(kxi2n)
no = a,, = 1
ai = Q,z-l
0-2 = 4-2
The coefficients obtained using this formula are given in Table 9.1 for up to
twelfth-order filters. The number of coefficients given is no more than half of
the filter-order; this is because the coefficients are symmetric. Take the example
of a third-order filter (12 = 3); only one coefficient is given: al = 2.000. However,
no = 1.000 and ai= n,, = 1.000. Also a2 = which equals nl = 2.000.

2 1.414214
3 2.000000
4 2.613126 3.4 142 14
5 3.236068 5.236068
6 3.863703 7.464102 9.141620
7 4.493959 10.097835 14.591 794
a 5.12583 1 13.137071 21.846151 25.688356
9 5.758770 16.581719 31.163437 41.986386
io 6.392453 20.431729 42.802061 64.882396 74.233429
li 7.026675 24.687075 57.020267 95.937001 123.24352
12 7.66 11297 29.347740 74.076215 136.87499 194.71869 218.46873

Table 9.1
Gutterworth Transfer Function Denominator Coefficients

STEP TWO: The transfer function is the reciprocal of the coefficient and fre-
quency variable products. So, again for the third-order filter, the denominator
is the sum of:

STEP THREE: Now the phase-shift function needs to be obtained. Having
found the denominator of the transfer function, you now need to separate it
into odd and even powers of frequency. Odd powers and their associated coef-
ficients are summed and used as a numerator, leaving the even powers in the