1
$\begingroup$

This question already has an answer here:

Is it possible to Plot the output of LaplaceTransform with Mathematica?

Obviously when I apply LaplaceTransform to a function I obtain a function of parameter s... but is implicit defined as sigma+j*omega, but I can't see this....(like the following example shown).

LaplaceTransform[Sin[t], t, s]
1/(1 + s^2)

Can I plot this function? Can I plot the Abs[] of the Laplace transformed function?

I wish obtain a graphic like this

enter image description here

Thx :)

$\endgroup$

marked as duplicate by Artes, happy fish, Community May 16 '17 at 12:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Artes I don't see the Re+Im part, because appear only parameter " s " , so i think that my question isn't a duplicate $\endgroup$ – plus91 May 16 '17 at 12:00
  • $\begingroup$ Plot3D[Abs[1/(1 + s^2) /. s -> x + I y], {x, -2, 2}, {y, -2, 2}, PlotRange -> {0, 2}] $\endgroup$ – Artes May 16 '17 at 12:01
  • $\begingroup$ @Artes thx, now it's clear :D $\endgroup$ – plus91 May 16 '17 at 12:21
3
$\begingroup$

The answer is YES, it is possible to Plot the output of LaplaceTransform with Mathematica.

To understand how to plot functions in general read the documentation for Plot.

Plot[
 1/(1 + s^2)
 , {s, -4, 4}
 ]

Mathematica graphics

Your would get the same if you use

Plot[
 Evaluate@LaplaceTransform[Sin[t], t, s]
 , {s, -4, 4}
 ]

or if you assign a new function name to the the transformed function

out[s_] = LaplaceTransform[Sin[t], t, s]

Plot[
 out[s]
 , {s, -4, 4}
 ]

For complex s you could use

Plot3D[
 Abs[1/(1 + Complex[x, y]^2)]
 , {x, -1, 1}
 , {y, -1, 1}
 , MaxRecursion -> 5
 ]

Mathematica graphics

Or

Plot3D[
 Evaluate@ReIm[1/(1 + Complex[x, y]^2)]
 , {x, -1, 1}
 , {y, 0, 2}
 , MaxRecursion -> 5
 , PlotStyle -> {
   Directive[Red, Opacity[0.3]]
   , Directive[Blue, Opacity[0.3]]
   }
 ]

Mathematica graphics

$\endgroup$

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