How to plot this function?
Mostrar comentarios más antiguos
(exp(-t)*(exp(-1)^floor(-t) - 1))/(exp(1) - 1)
Respuestas (2)
Star Strider
el 28 de En. de 2017
It’s straightforward:
t = linspace(0, 99);
f = @(t) (exp(-t).*(exp(-floor(-t)) - 1))/(exp(1) - 1);
figure(1)
plot(t, f(t), '-pg')
grid
See the documentation for the plot function, and Function Basics (link) to understand the anonymous function.
4 comentarios
qpei9f dkaslfjl
el 28 de En. de 2017
Star Strider
el 28 de En. de 2017
My pleasure.
You can adjust the range of ‘t’ by changing the first two arguments of the linspace function, and the number of elements in the vector it returns by specifying that in the third argument. So to set ‘t’ to be a vector of 500 elements going from -5 to +10, you would define it as:
t = linspace(-5, 10, 500);
To adjust the range of the axes in the plot, see the documentation for the axis (link) function. It will allow you to set the limits for the axes. So to set the x-axis to go from 0 to 10 and the y-axis to go from 0.5 to 0.6, after the plot call, you would set the axis limits as:
axis([0 10 0.5 0.6])
qpei9f dkaslfjl
el 30 de En. de 2017
Star Strider
el 30 de En. de 2017
My pleasure.
Image Analyst
el 28 de En. de 2017
Try this:
t = linspace(-4, 15, 1000);
y = (exp(-t) .* (exp(-1) .^ floor(-t) - 1)) ./ (exp(1) - 1);
plot(t, y, 'LineWidth', 2);
grid on;
fontSize = 20;
xlabel('t', 'FontSize', fontSize);
ylabel('y', 'FontSize', fontSize);
ax = gca
ax.YAxisLocation = 'origin'
ax.XAxisLocation = 'origin'

1 comentario
qpei9f dkaslfjl
el 28 de En. de 2017
Categorías
Más información sobre Annotations en Centro de ayuda y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!