CHAPTER 6 Application to Control and Communications
6.5.2 Butterworth Low-Pass Filter Design
The magnitude-squared approximation of a low-passNth-order Butterworth filter is given by
|HN(j0)|2= 1
1+ /hp2N 0=
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 is0=/hp) and normalized in mag- nitude as the dc gain is|H(j0)| =1. The frequency0=/hp=1 is the normalized half-power frequency since|HN(j1)|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(j0)|2
2 = 1
2. (6.36)
In the logarithmic scale we have
10 log10(|H(jhp)|2)= −10 log10(2)≈ −3 (dB) (6.37) This corresponds to a loss of3dB.
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=1it can be obtained by multiplying the magnitude-normalized filter by the constant K. Likewise, a filter H(S)designed with a normalized frequency, say0=/hpso that the normalized half-power frequency is1, is converted into a denormalized filter H(s)with a desiredhp by 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=0=
/hpand
H(S)H(−S)= 1 1+(−S2)N If the denominator can be factorized as
D(S)D(−S)=1+(−S2)N (6.38)
we letH(S)=1/D(S)—that is, we assign toH(S)the poles in the left-hands-plane so that the resulting filter is stable. The roots ofD(S)in Equation (6.38) are
S2Nk =ej(2k−1)π
e−jπN =ej(2k−1+N)π for integersk=1,. . ., 2N after replacing−1=ej(2k−1)π and(−1)N=e−jπN. The 2Nroots are then
Sk=ej(2k−1+N)π/(2N) k=1,. . ., 2N (6.39) Remarks
n Since|Sk| =1, the poles of the Butterworth filter are on a circle of unit radius. De Moivre’s theorem guar- antees that the poles are also symmetrically distributed around the circle, and because of the condition that complex poles should be complex conjugate pairs, the poles are symmetrically distributed with respect to the σ axis. Letting S=s/hpbe the normalized Laplace variable, then s=Shp, so that the denormalized filter H(s)has its poles in a circle of radiushp.
n No poles are on the j0axis, as can be seen by showing that the angle of the poles are not equal toπ/2or 3π/2. In fact, for1≤k≤N, the angle of the poles are bounded below and above by letting1≤k and then k≤N to get
π 2
1+ 1
N
≤ (2k−1+N)π
2N ≤ π
2
3− 1 N
and for integers N≥1the above indicates that the angle will not be equal to eitherπ/2or3π/2, or on the j0axis.
n Consecutive poles are separated byπ/N radians from each other. In fact, subtracting the angles of two consecutive poles can be shown to give±π/N.
Using the above remarks and the fact that the poles must be in conjugate pairs, since the coefficients of the filter are real-valued, it is easy to determine the location of the poles geometrically.
nExample 6.7
A second-order low-pass Butterworth filter, normalized in magnitude and in frequency, has a transfer function of
H(S)= 1 S2+√
2S+1
We would like to obtain a new filterH(s)with a dc gain of 10 and a half-power frequencyhp= 100 rad/sec.
The DC gain of H(S) is unity—in fact, when =0, S=j0 gives H(j0)=1. The half-power frequency ofH(S)is unity, indeed letting0=1, thenS=j1 and
H(j1)=
1 j2+j√
2+1
= 1 j√ 2
so that|H(j1)|2= |H(j0)|2/2=1/2, or0=1 is the half-power frequency.
6.5 Analog Filtering 395
Thus, the desired filter with a dc gain of 10 is obtained by multiplyingH(S)by 10. Furthermore, if we letS=s/100 be the normalized Laplace variable whenS=j0hp=j1, we get thats=jhp= j100, orhp=100, the desired half-power frequency. Thus, the denormalized filter in frequency H(s)is obtained by replacing S=s/100. The denormalized filter in magnitude and frequency is then
H(s)= 10
(s/100)2+√
2(s/100)+1 = 105 s2+100√
2s+104 n
Design
For the Butterworth low-pass filter, the design consists in finding the parametersN, the minimum order, andhp, the half-power frequency, of the filter from the constrains in the passband and in the stopband.
The loss function for the low-pass Butterworth is
α()= −10 log10|HN(/hp)|2=10 log10(1+ /hp2N) The loss specifications are
0≤α()≤αmax 0≤≤p
αmin ≤α() <∞ ≥s
At=p, we have that
10 log10(1+ p/hp2N
)≤αmax
so that
p
hp
2N
≤100.1αmax −1 (6.40)
and similarly for=s, we have that
100.1αmin−1≤ s
hp
2N
(6.41) We then have that from Equation (6.40) and (6.41), the half-power frequency is in the range
p
(100.1αmax−1)1/2N ≤hp≤ s
(100.1αmin−1)1/2N (6.42) and from the log of the two extremes of Equation (6.42), we have that
N≥ log10[(100.1αmin−1)/(100.1αmax−1)]
2 log10(s/p) (6.43)
Remarks
n According to Equation (6.43) when either
n The transition band is narrowed (i.e.,p→s), or
n The lossαminis increased, or
n The lossαmaxis decreased
the quality of the filter is improved at the cost of having to implement a filter with a high order N.
n The minimum order N is an integer larger or equal to the right side of Equation (6.43). Any integer larger than the minimum N also satisfies the specifications but increases the complexity of the filter.
n Although there is a range of possible values for the half-power frequency, it is typical to make the frequency response coincide with either the passband or the stopband specifications giving a value for the half-power frequency in the range. Thus, we can have either
hp= p
(100.1αmax−1)1/2N (6.44)
or
hp= s
(100.1αmin−1)1/2N (6.45)
as possible values for the half-power frequency.
n The design aspect is clearly seen in the flexibility given by the equations. We can select out of an infinite possible set of values of N and of half-power frequencies. The optimal order is the smallest value of N and the half-power frequency can be taken as one of the extreme values.
n After the factorization, or the formation of D(S)from the poles, we need to denormalize the obtained transfer function HN(S)=1/D(S) by letting S=s/hp to get HN(s)=1/D(s/hp), the filter that satisfies the specifications. If the desired DC gain is not unit, the filter needs to be denormalized in magnitude by multiplying it by an appropriate gain K.