6.5 Analog Filtering 393
- Factorize the magnitude-squared function and choose the poles on the left-hand s-plane,
guaranteeing the filter stability, to obtain the transfer functionHN(s)of the filter.
6.5.2 Butterworth Low-Pass Filter Design
The magnitude-squared approximation of a low-passNth-order Butterworth filter is given by
|HN(j′)|^2 =
1
1 +
(
/hp
) 2 N
′=
hp
(6.35)
wherehpis the half-power or−3-dB frequency. This frequency response is normalized with respect
to the half-power frequency (i.e., the normalized frequency is′=/hp) and normalized in mag-
nitude as the dc gain is|H(j 0 )|=1. The frequency′=/hp=1 is the normalized half-power
frequency since|HN(j 1 )|^2 = 1 /2. The given magnitude-squared function is thus normalized with
respect to frequency (giving a unity half-power frequency) and in magnitude (giving a unity DC gain
for the low-pass filter). The approximation improves (i.e., gets closer to the ideal filter) as the order
Nincreases.
Remarks
n The half-power frequency is called the− 3 -dB frequency because in the case of the low-pass filter with a
dc gain of 1, at the half-power frequencyhpthe magnitude-squared function is
|H(jhp)|^2 =
|H(j 0 )|^2
2
=
1
2
. (6.36)
In the logarithmic scale we have
10 log 10 (|H(jhp)|^2 )=−10 log 10 ( 2 )≈− 3 (dB) (6.37)
This corresponds to a loss of 3 dB.
n It is important to understand the significance of the frequency and magnitude normalizations typical in
filter design. Having a low-pass filter with normalized magnitude, its dc gain is 1, if one desires a filter
with a DC gain K6= 1 it can be obtained by multiplying the magnitude-normalized filter by the constant
K. Likewise, a filter H(S)designed with a normalized frequency, say′=/hpso that the normalized
half-power frequency is 1 , is converted into a denormalized filter H(s)with a desiredhpby replacing
S=s/hpin H(S).
Factorization
To obtain a filter that satisfies the specifications and that is stable we need to factorize the
magnitude-squared function. By lettingS=s/hpbe a normalized Laplace variable, thenS/j=′=
/hpand
H(S)H(−S)=
1
1 +(−S^2 )N
If the denominator can be factorized as
D(S)D(−S)= 1 +(−S^2 )N (6.38)
