how to plot curve in matlab

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saman ahmadi
saman ahmadi le 5 Sep 2020
Modifié(e) : KSSV le 6 Sep 2020
Hi. How can i plot below equation?
thank you very much
f=(31*f^6)/49000000000 - (667*f^2)/39200 - (657*f^4)/98000000 - (375*cos(qa))/28 - (f^2*cos(qa))/400 + (3*f^4*cos(qa))/7000000 + 375/28;
  3 commentaires
saman ahmadi
saman ahmadi le 5 Sep 2020
Excuse me, it is below, I want to plot curve(qa vs f). thank you
(31*f^6)/49000000000 - (667*f^2)/39200 - (657*f^4)/98000000 - (375*cos(qa))/28 - (f^2*cos(qa))/400 + (3*f^4*cos(qa))/7000000 + 375/28=0;
David Hill
David Hill le 5 Sep 2020
For each qa there potentailly could be six real solutions for f. How do you want to plot that? You could plot the equation's value for different values of qa. For example:
qa=pi/4;%change qa to desired values.
y=@(f)(31*f.^6)/49000000000 - (667*f.^2)/39200 - (657*f.^4)/98000000 - (375*cos(qa))/28 - (f.^2*cos(qa))/400 + (3*f.^4*cos(qa))/7000000 + 375/28;
f=-115:.01:115;
plot(f,y(f));

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Réponses (1)

KSSV
KSSV le 6 Sep 2020
Modifié(e) : KSSV le 6 Sep 2020
Are you looking for something like this?
qa = linspace(-pi,+pi,200) ; % give your ranges
f= linspace(-115,115,200) ; % give your ranges
[f,qa] = meshgrid(f,qa) ; % for a mesh
y=@(f,qa)(31*f.^6)/49000000000 - (667*f.^2)/39200 - (657*f.^4)/98000000 - (375*cos(qa))/28 - (f.^2.*cos(qa))/400 + (3*f.^4.*cos(qa))/7000000 + 375/28;
y = y(f,qa) ;
contour(f,qa,y,[0 0])

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