How to draw the tangent line on a curve

109 visualizaciones (últimos 30 días)
Inou
Inou el 29 de Jun. de 2021
Comentada: Star Strider el 23 de En. de 2022
I would like to plot the tangent of the curve I (V) and find the slope and then the intersection of the tangent with the x axis (V0).
I am attaching the curve data (I, U) and a summary image of what I am looking for. Thank you in advance for your help

Respuesta aceptada

Star Strider
Star Strider el 29 de Jun. de 2021
Try this —
T1 = readtable('https://www.mathworks.com/matlabcentral/answers/uploaded_files/668953/Tan.xlsx')
T1 = 42×2 table
I V ______ ________ 0 0 0 0.064516 0 0.14337 0 0.20072 0 0.36559 0 0.45878 4.6681 0.55197 19.593 0.63799 40.963 0.73118 58.96 0.7957 79.604 0.86022 114.14 0.9534 138.19 1.0108 164.38 1.0681 181.86 1.1039 207.75 1.1541
I = T1.I;
V = T1.V;
dI = gradient(I);
dV = gradient(V);
Vi = 1.3; % Choose Voltage Value To Calculate Slope
Ii = interp1(V, I, Vi);
dVi = interp1(V, dV, Vi)
dVi = 0.0299
dIi = interp1(V, dI, Vi)
dIi = 18.8851
Slope = dIi/dVi
Slope = 631.9586
YIntercept = Ii - Slope * Vi
YIntercept = -528.9232
XIntercept = -YIntercept / Slope
XIntercept = 0.8370
figure
plot(V,I)
hold on
plot(Vi, Ii, 'p')
plot(xlim, xlim*Slope+YIntercept, '-r')
plot(XIntercept, 0, 'sr')
hold off
grid
xlabel('V')
ylabel('I')
text(Vi, Ii, sprintf('$I = %.3f \\times V %+.3f \\rightarrow \\ $', Slope, YIntercept), 'Horiz','right', 'Vert','middle', 'Interpreter','latex')
text(XIntercept, 0, sprintf('$%.4f V \\downarrow $', XIntercept), 'Horiz','right', 'Vert','bottom', 'Interpreter','latex')
ylim([0 max(ylim)])
.
  6 comentarios
Sharatkumar Kondikoppa
Sharatkumar Kondikoppa el 23 de En. de 2022
In your code you have taken Vi =1.3 , did you choose randomly or took a mean value ? Can you explain me.
Or can I choose any value ?
Star Strider
Star Strider el 23 de En. de 2022
I chose that value for ‘Vi’ since although the exact point was not stated, it was certainly implied in the ‘image001.png’ plot.
So ‘Vi’ can be any point in the curve.

Iniciar sesión para comentar.

Más respuestas (1)

Asmit Singh
Asmit Singh el 29 de Jun. de 2021
This answer might be useful.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by