phase angle v/s frequency ratio of a damped Sdof system under harmonic force

clear all;clc
q=1;
for zeta=[0.05,0.15,.375,1];
p=1;
for r=0:0.001:5
phi(p)=atan((2*zeta*r)/(1-(r^2)));  
R(p)=r;
p=p+1;
end
plot(R,phi,'color',rand(1,3),'linewidth',2)
 pause(1) %just to see it being draw
hold on;
q=q+1;
end
grid on

% This is my code for plotting phase angle v/s frequency ratio of a damped Sdof system under harmonic force
% But the graph obtaining is wrong
% Error starts from r=1
% The graph i expect is attached with this mail.
% Pls help me in correcting the error...

% This is the graph am Expecting......

[KSSV]

YOu need not to use a loop and make code such complex: Check this code:

% zeta = [0.05,0.15,.375,1] ;
zeta = linspace(0.01,1,10) ;
r=0:0.001:5 ;
phi = zeros(length(zeta),length(r)) ;
for i = 1:length(zeta)    
    phi(i,:)=atan((2*zeta(i)*r)./(1-(r.^2)));        
end
plot(r,phi)
grid on

I am suspecting the formula....show us your reference.

[Bilal Ahmad]

from ( 0 to 1 ) graph is correct but from ( 1 to 5) graph scroll down and start from -1.5 which is not correct

[Giuseppe Carbone]

Hi. I have a question: is it possibly to plot a legend for each value of the angle?

[David Goodmanson]

If in KSSV's code you replace the atan line with

phi(i,:)=atan2((2*zeta(i)*r),(1-(r.^2)));

then the plot looks good. Compared to atan, atan2 keeps track of which quadrant to be in.

Answer 1

[Shan Min Hii]

clear all;close all;clc

for zeta=[0:0.2:1]

r=[0:0.1:3];

theta= atand((2*zeta*r)./(1-r.^2));

for j=1:31

if theta(j)<0

theta(j)=theta(j)+180;

end

end

plot(r,theta)

hold on

end

grid on

legend('\zeta=0','\zeta=0.2','\zeta=0.4','\zeta=0.6','\zeta=0.8','\zeta=1',"Location","southeast")

xlabel('w/w0')

ylabel('Phase Angle (\theta)')

title('Phase Angle')

This might help you.

[Charbell]

good

[william thomas]

what about amplitude or (whirl vector) vs. frequency ratio. should look like this..