사양이 다음과 같은 고역통과 필터를 설계하라.

• 저지대역의 끝: 
• 통과대역의 끝: 

ex7_18.cem

/*   Chapter 7: Example 7.18   Freq. Samp. Tech.: Highpass, Optimum method T1
   ws=0.6pi, wp=0.8pi, Rp=1dB, As=50dB
   M=33, T1 = 0.1095; T2 = 0.598;
*/
 
 M = 33; alpha = (M-1)/2; l = 0:M-1; wl = (2*pi/M)*l;
 T1 = 0.1095; T2 = 0.598;
 Hrs = [zeros(1,11),T1,T2,ones(1,8),T2,T1,zeros(1,10)];
 Hdr = [0,0,1,1]; wdl = [0,0.6,0.8,1];
 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_t1(h);
 
 
figuresubplot(1,1,1)subplot(2,2,3); plot(ww/pi,Hr,wl(1:17)/pi,Hrs(1:17),"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(1:17)/pi,Hrs(1:17),"o");
title("Bandpass: M=40,T1=1095,T2=0.598")xtitle("frequency in pi units"); ytitle("Hr(k)")