在MATLAB中从频率响应中实现滤波器

Question

我一直在尝试在MATLAB上实现这个过滤器，我只知道它的频率响应：HW =（（3 * c * 1i *（J0））./（2 * rfg * W））+（（ （3 * 2 * rfg *（W. * J1）） - （3 * 3 * c * 1i * J1））./（2 * a lambda rfg * W. ^ 2））其中J0和J1分别是0和1阶的球形贝塞尔函数。

如何使用频率响应在时域中实现它？

 /Org/freedesktop/ModeManager1/Modem/12 [Generic] MBIM [8087:0911]

感谢。

Answer 1

这是你在找什么？

a = 0.1;                 %radius of a spherical zone in m

%Data
c = 344;                 %speed of sound
rf = 3;                  %location of primary sound source
phi_f = degtorad(0);     %   "
rg = 1;                  %location of secondary sound source      
phi_g = degtorad(45);    %   "
rfg = 1/((1/rg)-(1/rf));
lambda = - (2/c) * sin( (phi_f - phi_g)/2 );


Fs = 20000;           % Sampling frequency (largest frequency in the frequency domain)             
dt = 1/Fs;            % Sampling time step (period)       
N = 200000;           % Number of points in the signal
t = (0:N-1)*dt;       % Time vector

df = 1/(N*dt);        % Frequency step
f = 0:df:(Fs/2);      % Frequency vector in Hz

W = 2*pi*f;           %Angular frequency vector (rad/sec)

%Bessel Functions
J0 = sphbes(0, a*lambda*W);  
J1 = sphbes(1, a*lambda*W);

% Filter
HW = ( (3*c*1i*(J0))./(2*rfg*W) ) + (( (3*2*rfg*(W.*J1))-(3*3*c*1i*J1) ) ./ (2*a*lambda*rfg*W.^2) );
HW(1) = HW(2); % the value is NaN at f = 0

% visualising
figure(1);
cla(gca);
hold on;
plot(f, real(HW));
plot(f, imag(HW));
plot(f, abs(HW));
hold off;
xlabel('Frequency (Hz)');
ylabel('Amplitude');
legend({'Real', 'Imaginary', 'Absolute'});
box on;

% Convert from single sided to two sided
P1 = HW;
P1_flipped = fliplr(P1);
P2 = [P1(1:end-1) P1_flipped(1:end-1)];
P2 = P2/2;

% do the inverse fft
time_domain = ifft(P2);

% visualising
figure(2);
cla(gca);
hold on;
plot(t*1000, real(time_domain));
plot(t*1000, imag(time_domain));
plot(t*1000, abs(time_domain));
hold off;
xlabel('time (msec)');
ylabel('Amplitude');
legend({'Real', 'Imaginary', 'Absolute'});
box on;
xlim([0 10]);
ylim([-18*10^(-5) 18*10^(-5)]);