Page 430 - Analog and Digital Filter Design

427
Design Equations

The third pole, with K= 3, is at -sin(5d10) +jcos(5dlO), or -1.0

The fourth and fifth poles are at the complex conjugate positions relative to the
first and second poles. The complex conjugate has the same real part, but the
imaginary part has the opposite sign. That is -0.309 - j0.9511 and -0.809
-j0.5878, which are the negative frequency complements to the first and second
poles. Note that the third pole is real, is on the --(r axis, and has a magnitude of
one. All odd-order Butterworth responses have a pole in this position.

If a cutoff point other than the 3dB attenuation frequency is required there is
a simple formula that can be used to scale the pole positions given:

I
WKP = (10" 'KP - 1)F

KP is the desired cutoff point attenuation.

For example, say we want a fifth-order response with a 1 dB cutoff, then:
WtiP = (100.' - $.'

= 0.2589"'
= 0.8736

The new cutoff point occurs at a lower frequency (as you would expect, since
the attenuation is lower), so the pole magnitudes have to be divided by mflP,
which increases their value and moves them away from the origin of the S-plane.
The real pole moves from -1.0 to -1.14467. Both real and imaginary parts of
the other poles are increased in magnitude by 1.14467. In fact, all poles will lie
on a circle that is 1.14467 in diameter. Now the 1 dB point is at w = 1 and the
3 dB point is at w = 1.14467.

Do not worry about scaling the normalized Butterworth response unless you
have some particular reason for using anything other than a 3 dB cutoff point.

Pole locations for the Butterworth response are given by the formula:
(2K -l>n +jcos (2K - 1)lr
-sin
2n 2n

for K = 1, 2, . . . , ri and where n is the required filter order.

Chebyshev Pole Locations
Considering the comment made earlier about the relationship between
Butterworth and Chebyshev response pole locations, close correspondences