Finding the intersection points between two curves
이전 댓글 표시
I need to find the intersection points of this equation:
this is my code. I dont know what is the best way . Also when I plot in terms of 
I have some periods but when I plot in terms of β I have only one period. How I can plot it in terms of beta but with nulti perids?
I have some periods but when I plot in terms of β I have only one period. How I can plot it in terms of beta but with nulti perids?many thanks in advance
h=20;
K=13.6;
l=1148e-6;
g1=fplot(@(beta)tan(beta*l));
hold on
g2=fplot(@(beta)2*h*beta*K/(K^2*beta^2-h^2));
채택된 답변
Several options —
h=20;
K=13.6;
l=1148e-6;
f1 = @(beta)tan(beta*l);
f2 = @(beta)2*h*beta*K./(K^2*beta.^2-h^2);
figure
g1 = fplot(f1);
hold on
g2 = fplot(f2);
g3 = fplot(@(beta) f2(beta)-f1(beta), '--'); % Use 'fsolve' On 'g3' To Find The Intersections
Another option:
format long
Xint = interp1(g3.YData, g3.XData, 0)
.
댓글 수: 11
raha ahmadi
2021년 7월 30일
There do not appear to be any other intersections —
h=20;
K=13.6;
l=1148e-6;
f1 = @(beta)tan(beta*l);
f2 = @(beta)2*h*beta*K./(K^2*beta.^2-h^2);
figure
g1 = fplot(f1);
hold on
g2 = fplot(f2);
g3 = fplot(@(beta) f2(beta)-f1(beta), [-50 50], '--'); % Use 'fsolve' On 'g3' To Find The Intersections
grid
format long
Xint = interp1(g3.YData, g3.XData, 0)
If there sere other intersections, the easiest way to find them would be:
zxi = find(diff(sign(g3.YData)))
And to my surprise, it seems that there are more roots!
So to get more precise results:
for k = 1:numel(zxi)
idxrng = (-1:1)+zxi(k); % Index Range For Interpolation
rootval(k) = interp1(g3.YData(idxrng), g3.XData(idxrng), 0);
end
rootval
The slope of ‘g1’ prevents the roots from being symmetrical about the origin.
.
My pleasure!
If you want to calculate those roots —
h=20;
K=13.6;
l=1148e-6;
g1=fplot(@(betaL)tan(betaL));
hold on
g2=fplot(@(betaL)2*h.*betaL/l*K./(h^2-K^2.*betaL/l.^2));
hold on
f3 = @(betaL) (tan(betaL)) - (2*h.*betaL/l*K./(h^2-K^2.*betaL/l.^2));
g3 = fplot(f3, [-1 1]*20, ':');
zxi = find(diff(sign(g3.YData)));
for k = 1:numel(zxi)
idxrng = (-1:1)+zxi(k); % Index Range For Interpolation
xv(k,:) = interp1(g3.YData(idxrng), g3.XData(idxrng), 0);
yv(k,:) = interp1(g3.XData, g3.YData, xv(k));
end
plot(xv(abs(yv)<1), zeros(size(xv(abs(yv)<1))), 'xr')
hold off
Intersections = table(xv(abs(yv)<1),yv(abs(yv)<1), 'VariableNames',{'x','y'})
.
raha ahmadi
2021년 7월 31일
Star Strider
2021년 7월 31일
As always, my pleasure!
Thank you for your compliment!
I calculated the intersections. I do not understand.
.
raha ahmadi
2021년 8월 1일
편집: raha ahmadi
2021년 8월 1일
Hi Star Strider sorry about the delay in response, I did not see your comment. I think I know the reason. In the first picture which you and I plotted the blue graph is the tangent function (in spite of what we expect). In this picture the period is very long and if you extend the axes limits you can see the difference between scales. But in the last picture you sent, by changing the scale of argument of tangent made the period of tangent very short.(remember that in the first picture scaling factor is about 1e-6 and this make period very long) I attached the two pictures for convenience
O.K.
I do not understand the problem or how this relates to the tangent function.
figure
hfp = fplot(@(x) 548*x./(187.7*x.^2-400), [-1 1]*30);
hold on
rootx = interp1(hfp.YData, hfp.XData, 0)
plot(rootx, 0, 'xr', 'MarkerSize',7.5)
hold off
grid
legend('$f(x)=\frac{548\ x}{187.7\ x^2-400}$', 'Intercept', 'Interpreter','latex', 'Location','best')
The one root is easy enough to calculate.
.
raha ahmadi
2021년 8월 7일
편집: raha ahmadi
2021년 8월 7일
There is nothing to fix.
Look at the results:
x=[-1000,1000]
y=sin (x)
roots(y)
so, plotting ‘y’ using polyval:
xv = linspace(-10,10);
pv = polyval(y, xv);
figure
plot(xv, pv)
grid
The reason is readily apparent! The ‘y’ vector corresponds to:
This is a linear relationship with one zero-crossing (root) at 
.
. However,
x = -1000:1000;
y = sin (x);
roots(y)
Except for the first element (1.1126), the absolute values of all the others are uniformly 1.
This time, ‘y’ corresponds to:
syms x
yp = vpa(poly2sym(y,x),5)
figure
fplot(yp, [-1000, 1000])
grid
axis([-pi pi -5 5])
The vertical dashed lines indicate singularities 
.
. The sin function has an infiinty of roots, those being 
radians, where n is an integer. It is necessary to use a zero-finding algorithm (such as fzero or fsolve) or interpolation (interp1) to locate them, not the roots function.
radians, where n is an integer. It is necessary to use a zero-finding algorithm (such as fzero or fsolve) or interpolation (interp1) to locate them, not the roots function. .
raha ahmadi
2021년 8월 7일
Star Strider
2021년 8월 7일
Thank you!
As always, my pleasure!
.
추가 답변 (1개)
darova
2021년 7월 30일
I'd try fsolve for solving
Read help carefully:
fplot(f,[0 10])
댓글 수: 1
raha ahmadi
2021년 7월 30일
편집: raha ahmadi
2021년 7월 30일
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