Inverse laplace and limits

7 vues (au cours des 30 derniers jours)
Ali Almakhmari
Ali Almakhmari le 8 Déc 2020
How can I find the limit as t goes to infinity of the inverse laplace function (A)? Code Below
syms s
G = (14*s^2+32)/(s*(s^2+4)*(2*s^2+3*s+8));
A = ilaplace(G)
  2 commentaires
David Goodmanson
David Goodmanson le 9 Déc 2020
Hello Ali,
A =
sin(2*t) - exp(-(3*t)/4)*(cos((55^(1/2)*t)/4) + (55^(1/2)*sin((55^(1/2)*t)/4))/5) + 1
t-->infinity makes the term involving exp go away, but that leaves sin(2*t) +1 which does not go to a limit as t-->infinity. Life is like that sometimes.
The Final Value Theorem gives an answer of 1 which gets the constant part right, but unfortunately it only applies if there actually is a limit.
The culprit is the (s^2+4) factor, which has roots s = +-2i with no negative real part.
Bjorn Gustavsson
Bjorn Gustavsson le 9 Déc 2020
From my use of the Laplace-transform (as a physicist) I'd take that sin(2*t)+1 as a limit-behaviour of the ODE-solution as time becomes large - and be happy with that.

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