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% Chapter 7: Example 7.15 % Freq. Samp. Tech.: Lowpass,Optimum method % % (a) T1 = 0.5 M = 40; alpha = (M-1)/2; l = 0:M-1; wl = (2*pi/M)*l; Hrs = [ones(1,5),0.5,zeros(1,29),0.5,ones(1,4)];

% Chapter 7 Figure mfile: % Mfile to draw figures on equiripplr amplitude response, % error function, and weighted error function. % a) equiripple amplitude response figure(1

% Chapter 3: Example 3.4: % DTFT of finite duration seq using mat-vec operation % n = -1:3; x = 1:5; % sequence x(n) k = 0:500; w = (pi/500)*k; % [0, pi] axis divided into 5

subplot(1,1,1) n = -5:5; x = (-0.9).^n; k = -200:200; w = (pi/100)*k; X = x * (exp(-j*pi/100)) .^ (n'*k); magX = abs(X); angX =angle(X); subplot(2,1,1); plot(w/pi,magX);grid axis([-2,2,0,15]) xlabel('

n = 0:100; x = cos(pi*n/2); k = -100:100; w = (pi/100)*k; % frequency between -pi and +pi X = x * (exp(-j*pi/100)).^(n'*k); % DTFT of x % y = exp(j*pi*n/4).*x; % signal multip

w = [0:1:500]*pi/500; % [0, pi] axis divided into 501 points. X = exp(j*w) ./ (exp(j*w) - 0.5*ones(1,501)); magX = abs(X); angX = angle(X); realX = real(X); imagX = imag(X); subplot(2,2,1); plot(w/pi

w = [-100:100]*pi/100; % frequency between -pi and +pi N = 5; X1 = sin((w+eps)*N/2)./sin((w+eps)/2); N = 15; X2 = sin((w+eps)*N/2)./sin((w+eps)/2); N = 25; X3 = sin((w+eps)*N/2)./sin((w+eps)/2); N = 1

subplot(1,1,1) n = 0:10; x = (0.9*exp(j*pi/3)).^n; k = -200:200; w = (pi/100)*k; X = x * (exp(-j*pi/100)) .^ (n'*k); magX = abs(X); angX =angle(X); subplot(2,1,1); plot(w/pi,magX);grid axis([-2,2,0,8]

w = [0:1:500]*pi/500; % [0, pi] axis divided into 501 points. X = exp(j*w) ./ (exp(j*w) - 0.9*ones(1,501)); magX = abs(X); angX = angle(X); subplot(2,1,1); plot(w/pi,magX); grid; axis([0,1,0,10]) xla

n = -5:10; x = rand(1,length(n)) + j*rand(1,length(n)); k = -100:100; w = (pi/100)*k; % frequency between -pi and +pi X = x * (exp(-j*pi/100)).^(n'*k); % DTFT of x % conjugation property y =