IIR filter design - Continuous and Discrete Time Signals and Systems - page 736

Fig. 16.9. Amplitude gain
response of the DT bandpass
filter designed in Example 16.6.

0

−20

−40

−60

W

− p −0.75 p −0.5 p −0.25 p 0 0.25 p 0.5 p 0.75 pp

>> [numz,denumz]=bilinear(nums,denums,0.5) % DT Filter

The resulting filter is given by

8 . 317 z 8 − 6 . 94 z 7 + 4 . 236 z 6 − 5 . 952 z 5 + 13 . 52 z 4 − 5.952 z 3 + 4 . 236 z 2 − 6 . 94 z + 8 . 317

H ( z ) = 0.001 .

z 8 − 1 . 389 z 7 + 3 . 714 z 6 − 3 . 356 z 5 + 4 . 685 z 4 − 2 . 693 z 3 + 2 . 397 z 2 − 0 . 7107 z + 0 . 4106

Figure 16.9 shows the amplitude gain response of the designed filter, which
illustrates that the pass-band and stop-band specifications are both satisfied.

Example 16.7
Example 15.7 designed a bandstop FIR filter with the following specifications:

(i) pass-band edge frequencies, Ω p1 = 0 . 25 π and Ω p2 = 0 . 625 π radians/s;

(ii) stop-band edge frequencies, Ω s1 = 0 . 375 π and Ω s2 = 0 . 5 π radians/s;

(iii) stop-band attenuations, δ s1 >50dband δ s2 > 50 dB.

Design an IIR filter with the same specifications.

Solution
Choosing k = 1 (sampling interval T = 2), step 1 transforms the pass-band and
stop-band corner frequencies into the CT frequency domain:
pass-band corner frequency I ω p1 = tan(0 . 5 Ω p1 ) = tan(0 . 125 π ) = 0 . 4142 radians/s;

pass-band corner frequency II ω p2 = tan(0 . 5 Ω p2 ) = tan(0 . 3125 π ) = 1 . 4966 radians/s;

stop-band corner frequency I ω s1 = tan(0 . 375 Ω s1 ) = tan(0 . 1875 π ) = 0 . 6682 radians/s;

stop-band corner frequency ω s2 = tan(0 . 5 Ω s2 ) = tan(0 . 25 π ) = 1 radian/s .

Step 2 designs an analog filter for the aforementioned specifications. In the fol-
lowing, we use M ATLAB to derive the analog elliptic filter for the transformed
specifications and an assumed pass-band ripple of 0.03 dB:

>> wp = [0.4142 1.4966]; ws = [0.6682 1];
>> Rp = 0.03; Rs = 50;
>> [N,wn] = ellipord(wp,ws,Rp,Rs,'s');
>> [nums,denums] = ellip(N,Rp,Rs,wn,'stop','s');

The resulting elliptic filter is of the eighth order and has the following transfer
function:

0 . 9966 s 8 + 2 . 8 s 6 + 2 . 854 s 4 + 1 . 25 s 2 + 0 . 1987
s 8 + 2 . 137 s 7 + 5 . 15 s 6 + 5 . 926 s 5 + 6 . 747 s 4 + 3 . 96 s 3 + 2 . 3 s 2 + 0 . 6377 s + 0 . 1994 .

H ( s ) =

Step 3 derives the z-transfer function of the digital filter using the bilinear
function.