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plot curves in matlab

조회 수: 3 (최근 30일)
saman ahmadi
saman ahmadi 2020년 8월 15일
편집: John D'Errico 2020년 8월 15일
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

답변 (2개)

John D'Errico
John D'Errico 2020년 8월 15일
편집: John D'Errico 2020년 8월 15일
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.
  댓글 수: 2
saman ahmadi
saman ahmadi 2020년 8월 15일
thanks. from plotting sended equation must plot a curve like image attachmented.
John D'Errico
John D'Errico 2020년 8월 15일
편집: John D'Errico 2020년 8월 15일
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 2020년 8월 15일
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)
  댓글 수: 6
Alan Stevens
Alan Stevens 2020년 8월 15일
편집: Alan Stevens 2020년 8월 15일
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 2020년 8월 15일
thank you very much
dear Alan Stevens

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