Signals and Systems - Electrical Engineering

11.4 IIR Filter Design 669

hpf = (3.01 + alpha(1))∗ones(1,M);

% epsilon and half-power frequency

epsi = sqrt(10ˆ(0.1∗alphamax)-1);

whp = 2∗atan(tan(0.5∗wp)∗cosh(acosh(sqrt(10ˆ(0.1∗3.01)-1)/epsi)/N));

whp = whp/pi

% plotting

subplot(221); zplane(b,a)

subplot(222)

plot(w,abs(H)); grid; axis([0 max(w) 0 1.1∗max(abs(H))])

subplot(223)

plot(w,unwrap(angle(H)));grid;

subplot(224)

plot(w,alpha);

hold on; plot(w,spec0,‘r’); hold on; plot(w,spec1,‘r’)

hold on; plot(w,hpf,‘k’); hold on

plot(w,spec2,‘r’); grid; axis([0 max(w) 1.1∗min(alpha) 1.1∗(alpha(1) + 3)]);

hold off

figure(2)

end

The transfer function of the first filter is

H 1 (z)=

0.224+0.449z−^1 +0.224z−^2

1 −0.264z−^1 +0.394z−^2

and its half-power frequency isωhp=0.493πrad. The second-order filter has a transfer function of

H 2 (z)=

0.094+0.283z−^1 +0.283z−^2 +0.094z−^3

1 −0.691z−^1 +0.774z−^2 −0.327z−^3

and a half-power frequency ofωhp=0.4902π. The poles and the zeros as well as the magnitude

and the phase responses of the two filters are shown in Figure 11.16. Notice the difference in the

gain (or losses) in the passband of the two filters. In order for the dc gain to be unity, the gain in

the even-order filter reaches values above 1, while the odd-order filter does the opposite.

The cut-off frequency given as output bycheb1ordand thatcheby1uses isωp. Since the half power is

not given bycheb1ord, the half-power frequency of the filter is calculated using the minimal order

N, the ripple factorε, and the passband frequencyωpin Equation (11.37). See the script. n

nExample 11.10

Consider the following specifications of a filter that will be used to filter an acoustic signal:

dc gain= 10

Half-power frequency:fhp=4 KHz

Band-stop frequency:fst=5 Khz,αmin=60 dB

Sampling frequency:fs=20 KHz