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how to plot curve in matlab

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saman ahmadi
saman ahmadi 2020-9-5
编辑： KSSV 2020-9-6
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;
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saman ahmadi
saman ahmadi 2020-9-5
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 2020-9-5
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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回答（1 个）

KSSV
KSSV 2020-9-6
编辑：KSSV 2020-9-6
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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