Generally, more insight can be gained in filter order by using a nomograph (i.e., a graphical aid) rather than an equation (in the Butterworth case, Eq. 4.3.13). Nomographs derive their name from the Greek words nomos (law) and graphein (to write). They are literally graphical representations of equations and were introduced in 1891 by d'Ocagne and recently applied to filters by Kawakami. They form marvelous design aids as we shall soon see. The nomograph for Butterworth filters is shown in Fig. 4.4.3. A monograph on nomographs will not be presented here, but we shall utilize their results. The interested engineer will find it enlightening to review the theoretical basis of nomographs.
The procedure for determining filter order begins by first expressing the required frequency response in the form shown in Fig. 4.3.4a. The pertinent parameters equal:
Mp = maximum attenuation (in dB) in the passband
Ms = minimum attenuation (in dB) in the stopband
Os = normalized stopband frequency
Os = fs/fp = ws/wp
fp (or wp) = band-edge frequency of passband in Hz (or rad/sec) fs (or ws) = band-edge frequency of stopband in Hz (or rad/sec)
This data is then entered on the nomograph as shown in Fig. 4.3.4b. A line is drawn through Mp (point P1) and Ms (point P2) until it intersects the graph at P3. A horizontal line is then drawn across the graph from P3. A vertical line is drawn through the stopband frequency Os. The horizontal line and vertical line intersect at P5. The minimum required filter order is given by the first curve lying above P5. After practice, this procedure will become second-nature to the engineer.
EXAMPLE 4.3.1 Determine the order required for a Butterworth filter to meet the specification shown in Fig. 4.3.5a using the Butterworth nomograph. Here 40 dB and 60 dB of rejection is required at frequencies of twice and three times the 1.25 dB bandwidth, respectively.
Solution To use the nomograph, we first write thatMp = 1.25 dB (at f = 1 KHz)
Ms1 = 40 dB (at f = 2 KHz), Os1 = 2 KHz/1 KHz = 2
Ms2 = 60 dB (at f = 3 KHz), Os2 = 3 KHz/1 KHz = 3
Entering this data on the nomograph as shown in Fig. 4.3.4b, we find n1 > 7 and n2 > 6. Thus, an eighth-order Butterworth filter is required.
© C.S. Lindquist, Active Network Design with Signal Filtering Applications, vol. 1, pp. 211-212, Steward & Sons, 1977.