사양이 다음과 같은 저역통과 필터를 생각해보자.

저지대역 감쇠를 얻을 수 있도록, 전이대역 내의 두 샘플을 사용하여 이 문제를 풀어보자.

ex7_16.cem

/*   Chapter 7: Example 7.16   Freq. Samp. Tech.: Lowpass, Optimum method T1 & T2
   wp = 0.2pi, ws=0.3pi, Rp=0.25dB, As=50dB
   T1 = 0.5925, T2=0.1099*/
 
 M = 60; alpha = (M-1)/2; l = 0:M-1; wl = (2*pi/M)*l;
 Hrs = [ones(1,7),0.5925,0.11,zeros(1,43),0.11,0.5925,ones(1,6)];
 Hdr = [1,1,0,0]; wdl = [0,0.2,0.3,1];
 k1 = 0:floor((M-1)/2); k2 = floor((M-1)/2)+1:M-1;
 angH = [-alpha*(2*pi)/M*k1, alpha*(2*pi)/M*(M-k2)];
 H = Hrs.*exp(j*angH);
 h = real(ifft(H,M));
 [db,mag,pha,grd,w] = mfreq_z(h,1);
 [Hr,ww,a,L] = Hres_t2(h);
 
 
figuresubplot(1,1,1)
subplot(2,2,3); plot(ww/pi,Hr,wl(1:31)/pi,Hrs(1:31),"o");
title("Amplitude Response")xtitle("frequency in pi units"); ytitle("Hr(w)") 
subplot(2,2,2); stem(l,h);
title("Impulse Response"); xtitle("n"); ytitle("h(n)");
 
subplot(2,2,4);plot(w/pi,db);
title("Magnitude Response");
xtitle("frequency in pi units"); ytitle("Decibels");
 
subplot(2,2,1);plot(wdl,Hdr, wl(1:31)/pi,Hrs(1:31),"o");
title("Lowpass: M=60,T1=0.59, T2=0.109")xtitle("frequency in pi units"); ytitle("Hr(k)")