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Why I got dirac

8 vues (au cours des 30 derniers jours)
john
john le 25 Jan 2014
Commenté : john le 29 Jan 2014
Hi,
why I got dirac in result:
a=ilaplace((134.66666666666666666666666666667*s - 134666.66666666666666666666666667)/(s^2 + 1000000.0) + 0.09046252753979897587299774386338);
a=(404*cos(1000*t))/3 - (4627111433557333*sin(1000*t))/34359738368000 + (3259256042553855*dirac(t))/36028797018963968;
What does it mean form me?

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Mischa Kim
Mischa Kim le 25 Jan 2014
Modifié(e) : Mischa Kim le 28 Jan 2014
Hello John, the "why" is pretty easy: in the transfer function you have a contstant term (0.0904..). This constant term produces the dirac when you are taking the inverse Laplace. What it means for you depends on the particular problem you are trying to solve. I am just guessing right now, but is the constant term really supposed to be there?
  3 commentaires
Mischa Kim
Mischa Kim le 28 Jan 2014
John, since I do not know how the circuit looks like and what you are calculating the inverse Laplace of it is impossible to say. As an example, if you look at the voltage-current relationship for an ideal inductor, a current step results in a dirac voltage. In other words, if your computations are correct, I recommend looking at the physics of the problem to try to explain why you get the dirac.
john
john le 29 Jan 2014
OK, thank you

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