Intersection of two curves
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    Chinmayraj Doddarajappa
 el 14 de Ag. de 2022
  
    
    
    
    
    Comentada: Chinmayraj Doddarajappa
 el 14 de Ag. de 2022
            Hi,
I am trying to find the intersection point of two curves of different functions plotted in a same graph. Can someone help me to find the [x] value of the intersection point.
Thanks
N = [1 2 3 4 5 6]
P = [1.48982012272225e-09	1.20413388047633e-08	4.17435214011468e-08	1.03854604485768e-07	2.17722472607192e-07	4.11611145129198e-07]
f = [1254.80279734744	316.932254638916	144.685314017159	85.4215488617609	58.6806449463084	44.5833312773839]
The script is,
figure
yyaxis left
plot(N,Pi_I)
yyaxis right
plot(N,f)

1 comentario
  Star Strider
      
      
 el 14 de Ag. de 2022
				The curves do not intersect in the region-of-interest (defined by the ‘N’ vector).  
It may be possible to force an intersection by extrapolating both of them, however I would not trust that result because you have no idea what the curves do outside the known values.  They might never intersect, or they could have an infinite number of intersections if one or both are oscillatory and have appropriate magnitudes.  
Respuesta aceptada
  Bruno Luong
      
      
 el 14 de Ag. de 2022
        
      Editada: Torsten
      
      
 el 14 de Ag. de 2022
  
      Intersection of to things that are not comparable? why not? just avoid using it to conclude anything that matters.
N = [1 2 3 4 5 6]
P = [1.48982012272225e-09	1.20413388047633e-08	4.17435214011468e-08	1.03854604485768e-07	2.17722472607192e-07	4.11611145129198e-07]
f = [1254.80279734744	316.932254638916	144.685314017159	85.4215488617609	58.6806449463084	44.5833312773839]
figure
yyaxis left
plot(N,P)
yyaxis right
plot(N,f)
yyaxis right
ylimr = ylim;
yyaxis left
yliml = ylim;
Ps = (P-yliml(1))./diff(yliml);
fs = (f-ylimr(1))./diff(ylimr);
ds = Ps-fs;
i=find(ds(1:end-1).*ds(2:end)<0,1);
Nx = interp1(ds(i:i+1),N(i:i+1),0);
yyaxis left
hold on
plot(Nx,interp1(N,P,Nx),'ob')
yyaxis right
hold on
plot(Nx,interp1(N,f,Nx),'xr')
Más respuestas (2)
  Torsten
      
      
 el 14 de Ag. de 2022
        Do you see a point of intersection ? I don't.
N = [1 2 3 4 5 6];
P = [1.48982012272225e-09	1.20413388047633e-08	4.17435214011468e-08	1.03854604485768e-07	2.17722472607192e-07	4.11611145129198e-07];
f = [1254.80279734744	316.932254638916	144.685314017159	85.4215488617609	58.6806449463084	44.5833312773839];
plot(N,P)
hold on
plot(N,f)
2 comentarios
  Torsten
      
      
 el 14 de Ag. de 2022
				Don't you see the different scales of the axes ? 
The curves don't intersect (at least not within the region where data are available), as can be seen in the "real" plot above.
  dpb
      
      
 el 14 de Ag. de 2022
        One approach might be something like
R_PF=mean(P./f);                    % the average scaling between the two
fn1=@(x)interp1(N,f*R_PF,x);        % interpolate the scaled values
fn2=@(x)interp1(N,P,x);             % and the other
fnD=@(x)fn2(x)-fn1(x);              % evaluate the difference between the two, for
x=interp1(fnD(N),N,0);              % finding the zero point reverse lookup/interpolation for "X"
The above produces
>> x 
x =
    4.5738
>> fnD(x)
ans =
   5.2940e-23
>> 
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