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.


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

| improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Jacob Schwartz Dec 6 '18 at 14:41
  • 1
    $\begingroup$ If you use inexact numbers like 6.5 in your input, you will have an inexact result. $\endgroup$ – John Doty Dec 6 '18 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];

enter image description here

| improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Jacob Schwartz Dec 7 '18 at 0:18

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