Loop Trough Time t = t+dt

3 visualizaciones (últimos 30 días)
Obaja Triputera Wijaya
Obaja Triputera Wijaya el 13 de Nov. de 2019
Comentada: darova el 18 de Nov. de 2019
I have a simple code like this
t = 1380;
dt = 0.1
for i = 1:1000
t = t+dt;
disp(t)
end
I believe the answer should be obvious that the final answer should be
t = 1480. I dont know why Matlab shows the answer a little bit different which is 1479.99999999991.
Anyone know why?

Respuestas (1)

Star Strider
Star Strider el 13 de Nov. de 2019
You have encountered floating-point approximation error.
See the documentation section on Floating-Point Numbers, and the colon operator.
  3 comentarios
Star Strider
Star Strider el 18 de Nov. de 2019
Yes.
Change it to:
if abs(t - 1400) < 0.05
a = 2
end
Since the code counts up by ‘dt’, this will introduce a tolerance in the calculation, so the floating-point approximation error are taken into account.
To see this graphically:
t = linspace(1399, 1401);
figure
plot(t, (abs(t - 1400) < 0.05))
grid
That will show the effect of using the inequality to test for a range of values for ‘t’ near 1400.
Experiment to get the result you want.
darova
darova el 18 de Nov. de 2019
Don't use equal sign for float numbers
if abs(t-1400) < 1e-6 % tolerance
a = 2;
end

Iniciar sesión para comentar.

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by