How does the step function work?
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Arcadius
el 10 de Abr. de 2022
Comentada: Star Strider
el 10 de Abr. de 2022
Hello, I am trying to figure out how the 'step' function work. I have a transfer function:
num = [-5.21];
den = [1 6 73];
G = tf(num, den);
Getting the inverse laplace of this transfer function manually, I get:
y(t)=((-5.21)/8)*exp(-3t)*sin(8t)
When I use the 'step' function the final steady state in the graph is -0.0714, when the final steady state of the inverse laplace approaches 0. There are great differences too in the graphs as the one worked manually it shows that the oscillation first peaks at -0.39 while matlab shows the first oscillation first peaks at -0.09. Where did I go wrong? I would greatly appreciate an answer. Thanks!
Matlab:

Desmos:

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Star Strider
el 10 de Abr. de 2022
You forgot to actually use the Heaviside unit step function as an input!
num = [-5.21];
den = [1 6 73];
G = tf(num, den)
syms s t
H = num / poly2sym(den,s) * laplace(heaviside(t))
h = ilaplace(H)
figure
hfp = fplot(h, [0 2]);
grid
EndVal = hfp.YData(end)
So the step result is correct!
.
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