EXAMPLE 5, VOL. 2

EXAMPLE 5

Designing digital filter block diagram

EXAMPLE 11.2.1 It was determined in Example 5.10.2 that a 4th-order elliptic digital filter was needed to obtain a 0.28 dB frequency = 1 KHz and a 40 dB stopband frequency > 2 KHz when sampling at 10 KHz. Design the filter in block diagram form and plot its frequency response.
Solution From Chap. 4.2, the elliptic analog filter has a transfer function of
H(pn) = k[pn^2 + 2.0173^2] / [(pn + 0.5213)^2 + 0.4828^2] [(pn + 0.1746)^2 + 1.0509^2]
= 0.1363[pn^2 + 4.0695] / [pn^2 + 1.0426 pn + 0.5048] [pn^2 + 0.3492 pn + 1.1349]

where k = 0.1363 is chosen so that H(0) = -0.8 = 0.9683. Its block diagram is shown in Fig. 11.2.1a. The 0.28 dB corner frequency vp of the analog filter is computed from the frequencies wp and ws of the digital filter as
vpT/2 = tan(wpT/2) = tan(180 fp/fs) = tan(180/10) = tan(18 deg) = 0.3249 = 1/3.078
vpT = (2/T)tan(180 fp/fs) = 2 (10 KHz) tan(18 deg) = 2 pi (1.034 KHz)
using Eq. 5.11.1b. Therefore the digital filter corner frequency of 1 KHz (or 35 deg) transforms to an analog filter corner frequency of 1.034 KHz (or 37.2 deg). They differ due to the frequency warping effects of the bilinear transform (see Fig. 5.10.3). Substituting the 0.28 dB normalized corner frequency vpT from Eq. 11.2.4 into the transformation equation of Eq. 11.2.2 shows that
pn = (2/vpT) (z-1)/(z+1) = 3.078 (z-1)/(z+1)
z = (1+pT/2)/(1-pT/2) = (1+0.3249pn)/(1-0.3249pn)

Transforming H(pn) of Eq. 11.2.3 into the digital filter gain H(z) using Eq. 11.2.2 gives
H(Z) = 0.01201 (z+1)^2 (z^2 - 0.7981z + 1)/ (z^2 - 1.360z + 0.5133)(z^2 - 1.427z + 0.8160)

The block diagram of the digital filter is shown in Fig. 11.2.1b. Each analog block is transformed according to the transform relation given by Eq. 11.2.5a to obtain each digital block. Alternatively, the analog poles and zeros of each block can be converted into the digital poles and zeros using the pn-z relation given by Eq. 11.2.5b. The dc block gains are 1 and 0.9683, respectively. The frequency responses of H(pn) and H(z) are shown in Fig. 11.2.1c. The only notable difference is that the digital filter has zero gain at the Nyquist frequency due to the bilinear transform, and there is transmission zero shifting due to frequency warping.