drawing a point in the graph
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回答 (2 件)
x = 1:5;
y = 20:24;
plot(x,y,x(3),y(3),'*') % specify a marker symbol * in index 3 of y
please follow this for more information about plot
3 件のコメント
Ismita
2022 年 12 月 6 日
try this approach. I am not sure about his efficiency.
t1 = 0:0.5:2;
x1 = t1/2-1;
t2 = 2:0.5:20;
x2 = t2 - sqrt(2*t2);
t3 = 0:0.5:1;
x3 = t3+1;
t4 = 1:0.5:11.6;
x4 = 2*sqrt(t4);
figure(1)
[x1equality ia]=find(ismember(t1,t2));
[x3equality ib]=find(ismember(t3,t4));
plot(x1, t1, 'b', x2, t2, 'r', x3, t3, 'k', x4, t4, 'g',x1(ia),t1(ia),'*',x3(ib),t3(ib),'*')
Ismita
2022 年 12 月 7 日
x = 0:0.5:10;
t = (x/2).^2;
abs(x-2*sqrt(t)) %cross-check weether x == 2*sqrt(t)
plot(x, t, '*-')
5 件のコメント
Walter Roberson
2022 年 12 月 6 日
You need to calculate the point of interesection between (x2,t2) and (x4,t4) . Then you can plot() that point, specifying a marker. For example,
plot(interesection_x, intersection_y, 'b*')
Ismita
2022 年 12 月 7 日
Walter Roberson
2022 年 12 月 7 日
plot(6.78233, 11.66, 'r*')
It does not matter if there is no exact match for this in the curves, if it is the correct intersection point.
Unless, that is, what you want to do is find the closest point on each of the two curves and mark those close points rather than the intersection point?
Ismita
2022 年 12 月 7 日
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