Page 413 - Analog and Digital Filter Design

4 1 0 Analog and Digital Filter Design

The general expression is given by:

This expression can be used to find B3(s) by letting n = 3 in the expression. The
expression will now be broken down into manageable pieces:

(2n - 1) = 5
(s) = &(S) = S2 + 3s + 3
Therefore, (2n - l)B,z-l(s) = 5 (s2 + 3s + 3) = 5? + 15s + 15
Bn-2 (s) = B, (s) = s + 1

Therefore, s’B,_,(s) = s2 (s + 1) = s3 + s’.
Finally, B,,(s) = (s3 + s‘) + (5?+ 15s + 15) = + 6s’ + 15s + 15
Considering the original expression, a3 = 1, a2 = 6, al = 15 and a. = 15.
(2n - l)!
These coefficients can be found using the equation: ai = 2(,-i) i! (n - i)!,
for i= 0,1,2.. . n

Where has this got us? Well, we have found the transfer function, which is:

15
H(s) =
s3 +6s2 +15s+15

In terms of frequency, substituting s = jw, we have:

The transfer function is used to calculate the attenuation and phase of the Bessel
response at any frequency. The denominator must be broken down into odd and
even parts. The real parts are the even powers of w and their coefficients. Imagi-
nary parts are the odd powers of w and their coefficients. The attenuation is
the zero power coefficient divided by the magnitude of real and imaginary parts.

)dB.
In decibels, the attenuation is: -20.LOGlo a0
real2 +imaginary’

The phase is given by the expression:

4 = tan-‘(Imaginary H(jw)/Real H(jw))