I tried this:

InverseLaplaceTransform[-((-6.5 - 3.25 s - 5.5 s^2 + 0.5 s^3 + 
      s^4)/((2 + s) (1 + 0.5 s + s^2) (1.5 + s + s^2))), s, t] // 
   Simplify // ComplexExpand // Simplify

In this way I get the following:

0.714286 E^(-2. t) - 1.71429 E^(-0.5 t) Cos[(1.11803 + 0. I) t] + 
 3.19438 E^(-0.5 t) Sin[(1.11803 + 0. I) t]

which is similar to the answer I want:

$ [0.7143 e^{-2t}-1.7145 e^{-0.5t}cos(1.25t)+3.194 e^{-0.5t}sin(1.25t)]u(t) $

I have tried with FullSimplify, Apart, TrigReduce, ExpToTrig and I have not got the answer in terms of the UnitStep function.

I will appreciate any suggestion to get the answer I want.

up vote 2 down vote accepted

It seems to be that The unit step is implied in the result, so it is not needed unless there is an ambiguity. You can see this from

LaplaceTransform[UnitStep[t], t, s]

Mathematica graphics

InverseLaplaceTransform[1/s, s, t]

Mathematica graphics

So Mathematica returned 1 and not unitstep. If you want a unit step in the result, just multiply the result by unit step. But it is assumed that the result of the inverse Laplace transform is for t>=0 (because Laplace transform works from t>=0 by definition.

Mathematica graphics

  • But still the answer does not match because there are complex numbers. If I add // Chop the complex numbers are removed but it differs a bit. – Jacob Schwartz Dec 6 at 14:41
  • 1
    If you use inexact numbers like 6.5 in your input, you will have an inexact result. – John Doty Dec 6 at 20:15

It can be computed as the impulse response. (I also rationalized it to get the exact result.)

tfm=TransferFunctionModel[Rationalize[-((-6.5-3.25 s-5.5 s^2+0.5 
     s^3+s^4)/((2+s) (1+0.5 s+s^2) (1.5+s+s^2)))],s];
or=OutputResponse[tfm,DiracDelta[t],t][[1]]//Simplify

enter image description here

  • InverseLaplaceTransform[ Rationalize[-((-6.5 - 3.25 s - 5.5 s^2 + 0.5 s^3 + s^4)/((2 + s) (1 + 0.5 s + s^2) (1.5 + s + s^2)))], s, t] // Expand. I used Rationalize that I saw in your answer, thanks. – Jacob Schwartz Dec 7 at 0:18

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.