plot curves in matlab

3 visualizaciones (últimos 30 días)
saman ahmadi
saman ahmadi el 15 de Ag. de 2020
Editada: John D'Errico el 15 de Ag. de 2020
Hi. How can i plot below equation(qa vs w)?
thank you
(9*w^4*cos(qa))/28000000 - (33*w^2*cos(qa))/32000 - (105*cos(qa))/64 + (603*w^2)/2560000 - (4891*w^4)/4480000000 + (93*w^6)/560000000000 + 105/64=0

Respuestas (2)

John D'Errico
John D'Errico el 15 de Ag. de 2020
Editada: John D'Errico el 15 de Ag. de 2020
fun = @(w,qa) (9*w.^4.*cos(qa))/28000000 - (33*w.^2.*cos(qa))/32000 - (105*cos(qa))/64 + (603*w.^2)/2560000 - (4891*w.^4)/4480000000 + (93*w.^6)/560000000000 + 105/64;
fimplicit(fun)
You can tell fimplicit how far out to go of course.
fimplicit(fun,[-100 100 -100 100])
It does appear as if much is happening when you broaden the view.
  2 comentarios
saman ahmadi
saman ahmadi el 15 de Ag. de 2020
thanks. from plotting sended equation must plot a curve like image attachmented.
John D'Errico
John D'Errico el 15 de Ag. de 2020
Editada: John D'Errico el 15 de Ag. de 2020
Sigh. It would have helped if you asked that question in the first place, instead of just asking how to plot qa as a function of w.
fun = @(qa,w) (9*w.^4.*cos(qa))/28000000 - (33*w.^2.*cos(qa))/32000 - (105*cos(qa))/64 + (603*w.^2)/2560000 - (4891*w.^4)/4480000000 + (93*w.^6)/560000000000 + 105/64;
fimplicit(fun,[-pi pi 0 100])
If you are looking for a specific result, then you need to tell people. That seems to come pretty close, though the scaling of the variables seems to be different from the plot you show. That is more a question of the specific function you gave us, which is surely different from that which generated your plot.

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Alan Stevens
Alan Stevens el 15 de Ag. de 2020
Somethig like this:
qafn = @(w) acos(( 4891*w.^4/4480000000 -93*w.^6/560000000000 -603*w.^2/2560000 -105/64 )...
./(9*w.^4/28000000 - 33*w.^2/32000 - 105/64 ));
w = -50:50;
qa = qafn(w);
plot(w,qa)
  6 comentarios
Alan Stevens
Alan Stevens el 15 de Ag. de 2020
Editada: Alan Stevens el 15 de Ag. de 2020
Looks like the implicit function is the way to go, though the w axis is scaled differently.
qawfn = @(qa,w) (9*w.^4.*cos(qa))/28000000 - (33*w.^2.*cos(qa))/32000 - (105*cos(qa))/64 ...
+ (603*w.^2)/2560000 - (4891*w.^4)/4480000000 + (93*w.^6)/560000000000 + 105/64;
fimplicit(qawfn,[-pi pi 0 90])
saman ahmadi
saman ahmadi el 15 de Ag. de 2020
thank you very much
dear Alan Stevens

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