We now see the great ease in utilizing classical filter results since we merely need
to rewrite the tabulated transfer functions. The bulk of the work involves determining
the acceptable type of filter (e.g. Butterworth) and its required order. The transfer
functions can *always* be easily expressed using the *frequency normalized form*
involving sn. The engineer will find that using denormalized forms involving s appear more
complicated and will obscure the pertinent information involving the wn and z of each pole
pair. It is also important to remember that each design block must be explicited labelled
with (1) order and type, (2) dc gain Ho, (3) resonant frequency of pole (wn or fn), and (4)
damping factor of pole (z). In Chap. 8, we shall utilize such block diagrams in the
design of low-pass filters.

One other comment should be made. We shall usually realize *n*th-order filters by
cascading n/2 (for n even) or (n+1)/2 (for n odd) second-order stages. The primary
advantage of cascade design is the ease of tuning and temperature compensation.

EXAMPLE 4.3.4Determine if a Butterworth filter can be designed to meet the frequency and time domain requirements shown in Fig. 4.3.7. Design the minimum filter order and draw the block diagram of the filter if it can be realized.

SolutionWe begin by determining the required filter order from the frequency response. Since Mp = 3 dB, Ms1 = 10 dB at Os1 = 2, and Ms2 = 40 dB at Os1 = 4, we see from the Butterworth filter nomograph that n1 > 1 and n2 > 3. Thus, at least a fourth-order filter is required.Now we sketch the step response of the filter noting n = 4. Since the bandwidth w3dB = 2 pi (2 KHz), we must denormalize time by w3dB in Fig. 4.3.7. We now transfer the normalized step response of Fig. 4.3.6 onto Fig. 4.3.7. We see that we have easily met the time domain requirements. Noting the time differences

Dt1 = 5 - 2.3 = 2.7 sec (r = 0.5)we see that we may reduce the time domain specifications by Dt = min(Dt1, Dt2, Dt3) = 2.7 seconds if we so desire. Alternatively, we could consider expanding the bandwidth - however, here we require the 2 KHz bandwidth shown. The transfer function of the filter equals

Dt2 = 7.5- 3.5 = 4.0 sec (r = 0.8)

Dt3 = 10 - 7.3 = 2.7 sec (r = 1.05)H(s) = 1/(sn^2 + 0.765sn + 1)(sn^2 + 1.84sn + 1)where sn = s/2 pi (2 KHz) using Table 4.3.1. The block diagram realization of the filter is shown in Fig. 4.3.8.© C.S. Lindquist,

Active Network Design with Signal Filtering Applications, vol. 1, pp. 215-216, Steward & Sons, 1977.