How can i plot the magnitude of displacement ?

Hi, i want to plot the magnitude of displacement x(:,1) and i have this program but the curve displacement =f(rload) mentioned is false. Can someone help me please.

function xdot = equacte(t,x)

% Function file for mass with spring.

% Position is first variable, velocity is second variable,load is the third variable

freq=100; %frequency (Hz)

w=2*pi*freq;

m=0.0112;

k=262.8257 ;

teta =-4.40e+03;

cp= 1.8492*10^-7;

%fre=100/(2*pi);

%tandelta=(-4.5235*10^-6)*fre^2 +0.001*fre + 0.13556;

%rload=tandelta/(100*cp) 

for rload =0:20:50000

A = [0,1,0;-k/m,0,-teta/m;(-teta/rload),0,-1/(rload*cp)];

B = [0;1/m;0];

f =sin(w*t);

xdot = A*x+B*f;

end

end

d31=-320*10^-12;               

y11e= 62*10^9;                 

e33sig= 33.65*10^-9;            

rop= 7800;                       

tp= 0.1905*10^-3;                

lp= 50.8*10^-3;                  

bp= 25.4*10^-3;                  

l1= 6.35*10^-3;                

l2= 44.45*10^-3;                

k31carre= ((d31^2)*y11e)/e33sig;  

y11d = y11e /(1-k31carre);     

e33eps = e33sig*(1-k31carre); 

ap = lp*bp;       

cp = (e33eps*ap)/tp;  

%%%%%%%%%%%%%%%%%%%  beam  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
yb = 69*10^9;     

rob= 2700;         

tb=0.79375*10^-3;   

lb=152.4*10^-3;          

mtip=0.01;            

bb=25.4*10^-3;             

%%%%%%%%%%%%%%%%%calcul des termes de l'equation de mvt%%%%%%%%%%%%%%%%%%%

k= ((yb*bb*tb^3)/lb^3 )+( (8*y11d*bp*(l2^3-l1^3)*(((tb/2)+tp)^3-(tb/2)^3)) / lb^6); 

teta=( 3*d31*y11e*bp*tp*(tb+tp)*(l2^2-l1^2) )/(lb^3*cp*tp);                     

m= ((rob*bb*tb*lb)/7)+((2*rop*bp*tp*(l2^7-l1^7))/(7*lb^6))+mtip;               

[t,x]=ode45(@equacte,[0:0.0004:1],[0,0,0]);     

index=1;

for rload=0:20:50000

     if(x(index,1)>=0)

         fg=x(index,1);

     else

          fg=abs(x(index,1));

     end
  fk(index)=fg;

index= index + 1;

end

figure;

plot([0:20:50000],fk,'r')

xlabel('rload(Ohm)');

ylabel('displacement(m)');

Answer 1

[Walter Roberson]

Your code has

for rload=0:20:50000

but you do not use the value of rload anywhere in your code.

Note: that loop could be replaced by just

fk = abs(x(:,1));

[Mallouli Marwa]

I have chaged this for loop by this :

Please correct it

index = 1;

for i=0:20:50000

      xm =max(x(index,1))-min (x(index,1))

      xk(index) = xm;

      index = index + 1;

end

figure;

plot([0:20:50000],xk,'b')

xlabel('rload(Ohm)');

ylabel('magnitude of displacement(m))');

[Walter Roberson]

x(index,1) is a scalar, so min() and max() of it are the same values, so your xm is always going to be 0.

What is wrong with

plot([0:20:50000], abs(x(:,1)), 'b')

?