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Design a Butterworth lowpass filter with specification as follows     w p =0.2pi,    R p =1dB     w s =0.3pi,    A s =15dB % Butterworth Lowpass Filter Design: % % Digital Filter Specifications: wp = 0.2*pi; %digital Passband freq in Hz ws = 0.3*pi; %digital Stopband freq in Hz Rp = 1; %Passband ripple in dB As = 15; %Stopband attenuation in dB ep = sqrt(10^(Rp/10)-1); Ripple = sqrt(1/(1+ep*ep)); % Passband Ripple Attn = 1/(10^(As/20)); % Stopband Attenuation % Digital Butterworth Filter Design: [N wn]=buttord(wp/pi,ws/pi,Rp,As); [b,a]=butter(N,wn); %plotting figure(1); subplot(1,1,1) [db,mag,pha,grd,w] = freqz_m(b,a); subplot(2,2,1); plot(w/pi,mag); title('Magnitude Response') xlabel('frequency in pi units'); ylabel('|H|'); axis([0,1,0,1.1]) set(gca,'XTickMode','manual','XTick',[0,0.2,0.3,1]); set(gca,'YTickmode','manual','YTick',[0,Attn,Ripple,1]); grid subplot(2,2,3); plot(w/pi,db); ti

Thiết kế bộ lọc thông cao theo phương pháp tần số lấy mẫu với các thông số sau :             w s = 0.6 pi,     A s = 50 dB             w p = 0.8 pi,     R p = 1 dB  M = 33, T1 = 0.1095; T2 = 0.598. % 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] = freqz_m(h,1); [Hr,ww,a,L] = Hr_Type1(h); subplot(1,1,1) subplot(2,2,1);plot(wl(1:17)/pi,Hrs(1:17),'o',wdl,Hdr); axis([0,1,-0.1,1.1]); title('Highpass: M=33,T1=0.1095,T2=0.598') xlabel('frequency in pi units'); ylabel('Hr(k)') set(gca,'XTickMode','manual','XTick',[

Linear-phase FIR filter: đáp ứng xung đối xứng với M lẻ. function [Hr,w,a,L] = Hr_Type1(h); % Computes Amplitude response Hr(w) of a Type-1 LP FIR filter % ----------------------------------------------------------- % [Hr,w,a,L] = Hr_Type1(h) % Hr = Amplitude Response % w = 500 frequencies between [0 pi] over which Hr is computed % a = Type-1 LP filter coefficients % L = Order of Hr % h = Type-1 LP filter impulse response % M = length(h); L = (M-1)/2; a = [h(L+1) 2*h(L:-1:1)]; % 1x(L+1) row vector n = [0:1:L]; % (L+1)x1 column vector w = [0:1:500]'*pi/500; Hr = cos(w*n)*a';   Linear-phase FIR filter: Đáp ứng xung đối xứng với M chẵn. function [Hr,w,b,L] = Hr_Type2(h); % Computes Amplitude response of Type-2 LP FIR filter % --------------------------------------------------- % [Hr,w,b,L] = Hr_Type2(h) % Hr = Amplitude Response % w = frequencies between [0 pi] over which Hr is computed % b = Type-2 LP filter coefficients % L = Order of Hr % h

Sử dụng cửa sổ Black man, thiết kế bộ lọc thông chặn với các thông số cho dưới đây :           w 1s    = 0.2 pi,         A s  = 60 dB           w 1p    = 0.35 pi,       R p  =  1 dB           w 2p   = 0.65 pi,      R p  =  1 dB           w 2s    = 0.8 pi,         A s  = 60 dB ws1 = 0.2*pi; wp1 = 0.35*pi; wp2 = 0.65*pi; ws2 = 0.8*pi; As = 60; tr_width = min((wp1-ws1),(ws2-wp2)) M = ceil(11*pi/tr_width) + 1 %;M=68 n=[0:1:M-1]; wc1 = (ws1+wp1)/2; wc2 = (wp2+ws2)/2; hd = ideal_lp(wc2,M) - ideal_lp(wc1,M); w_bla = (blackman(M))'; h = hd .* w_bla; [db,mag,pha,grd,w] = freqz_m(h,[1]); delta_w = 2*pi/1000; Rp = -min(db(wp1/delta_w+1:1:wp2/delta_w)) % Actua; Passband Ripple As = -round(max(db(ws2/delta_w+1:1:501))) % Min Stopband Attenuation % plots subplot(1,1,1); subplot(2,2,1); stem(n,hd); title('Ideal Impulse Response') axis([0 M-1 -0.4 0.5]); xlabel('n'); ylabel('hd(n)') subplot(2,2,2); stem(n,w_bla);title('Blackman Window') a

Thiết kế bộ lọc số FIR sử dụng cửa sổ hamming với các thông số cho dưới đây :    w p  = 0.2 pi,    R p  = 0.25 dB     w s  = 0.3 pi,    A s  = 50 dB. wp = 0.2*pi; ws = 0.3*pi; tr_width = ws - wp M = ceil(6.6*pi/tr_width) + 1 n=[0:1:M-1]; wc = (ws+wp)/2 hd = ideal_lp(wc,M); w_ham = (hamming(M))'; h = hd .* w_ham; [db,mag,pha,grd,w] = freqz_m(h,[1]); delta_w = 2*pi/1000; Rp = -(min(db(1:1:wp/delta_w+1))) As = -round(max(db(ws/delta_w+1:1:501))) % vẽ subplot(1,1,1) subplot(2,2,1); stem(n,hd); title('Đáp ứng xung lý tưởng') axis([0 M-1 -0.1 0.3]); xlabel('n'); ylabel('hd(n)') subplot(2,2,2); stem(n,w_ham);title('Cửa sổ Hamming') axis([0 M-1 0 1.1]); xlabel('n'); ylabel('w(n)') subplot(2,2,3); stem(n,h);title('Đáp ứng xung thực tế') axis([0 M-1 -0.1 0.3]); xlabel('n'); ylabel('h(n)') subplot(2,2,4); plot(w/pi,db);title('Đáp ứng biên độ theo dB');grid axis([0 1 -100 10]); xlabel(&

Dưới đây là 2 ví dụ về bộc lọc FIR (MATLAB Code) function hd = ideal_lp(wc,M); % Ideal LowPass filter computation % -------------------------------- % [hd] = ideal_lp(wc,M) % hd = ideal impulse response between 0 to M-1 % wc = cutoff frequency in radians % M = length of the ideal filter % alpha = (M-1)/2; n = [0:1:(M-1)]; m = n - alpha + eps; hd = sin(wc*m) ./ (pi*m); function [db,mag,pha,grd,w] = freqz_m(b,a); % Modified version of freqz subroutine % ------------------------------------ % [db,mag,pha,grd,w] = freqz_m(b,a); % db    = Relative magnitude in dB computed over 0 to pi radians % mag = absolute magnitude computed over 0 to pi radians % pha  = Phase response in radians over 0 to pi radians % grd   = Group delay over 0 to pi radians % w     = 501 frequency samples between 0 to pi radians % b     = numerator polynomial of H(z) (for FIR: b=h) % a     = denominator polynomial of H(z) (for FIR: a=[1]) % [H,w] = freqz(b,a,1000,'whole'); H