Craig, thank you for your help. It's a good way of reuse the bode command that works only for positive frequencies.
I solved my problem of seeing the frequency response, first realising that logaritmic scale for negative frequencies makes no sense, and that the bode diagram shows the modulus and phase of the complex number that transfer function represent. So, i evaluated both for different frequencies, but plotting them in linear axes.
This is for two complex transfer functions:
f=[-1000:1:1000]; %Vector of frequencies [Hz]
w=2*pi*f; %Vector of frequencies [rad]
WC=2*pi*50; %Parameters of filter
WO=2*pi*50;
f1=(WC+1i*0)./(1i*w-1i*WO+WC); %Complex filters
f2=(WC+1i*0)./(1i*w+1i*WO+WC);
AMP_f1=abs(f1); %Evaluation of modulus
AMP_f2=abs(f2);
ANG_f1=angle(f1)*(180/pi); %Evaluation of phase (changed to degrees)
ANG_f2=angle(f2)*(180/pi);
close all
figure
subplot(2,1,1),plot(f,AMP_f1,f,AMP_f2),grid on
xlabel('Frequency [Hz]')
ylabel('Gain')
subplot(2,1,2),plot(f,ANG_f1,f,ANG_f2),grid on
xlabel('Frequency [Hz]')
ylabel('Phase [º]')
