Why I got dirac

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john
john on 25 Jan 2014
Commented: john on 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?

Accepted Answer

Mischa Kim
Mischa Kim on 25 Jan 2014
Edited: Mischa Kim on 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 Comments
Mischa Kim
Mischa Kim on 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 on 29 Jan 2014
OK, thank you

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