Antworten (2)

Star Strider
Star Strider am 28 Jan. 2017

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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 Kommentare

qpei9f dkaslfjl
qpei9f dkaslfjl am 28 Jan. 2017
Thank you. How do I reduce the scale?
Star Strider
Star Strider am 28 Jan. 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
qpei9f dkaslfjl am 30 Jan. 2017
Again, thanks a lot.
Star Strider
Star Strider am 30 Jan. 2017
My pleasure.

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Image Analyst
Image Analyst am 28 Jan. 2017

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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 Kommentar

qpei9f dkaslfjl
qpei9f dkaslfjl am 28 Jan. 2017
Thanks a lot. Seems it is periodic as I hoped.

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