# Laplace Transform - Differentiation in Time property demonstration

I have a signal

x[t_]=t/3 Exp[-3t] UnitStep[t]


and the laplace transform as

X[s_]=Integrate[x[t]Exp[-s t],{t,-Infinity,Infinity}]


I am trying to take the derivative of x[t] and applying a Laplace transform on that to show that it equals sX[s]

This is how I take the derivative: dx[t_]=D[x[t],t] and this is how I apply the transform to it: X0[s_] = Integrate[x'[t] Exp[-s t], {t, -Infinity, Infinity}] Something keeps going wrong with this last line of code and integration bit.

I also want to show that the ROC for sX(s) is AT LEAST R. I believe I might've just picked the wrong signal for it so if any of you have any suggestions on how to make this demonstration work, I am all ears!

Thank you sooo much in advance!

You can use the Laplace Transform function:

x[t_] := t/3 Exp[-3 t] UnitStep[t];
eqn1 = LaplaceTransform[x[t], t, s]
eqn2 = LaplaceTransform[D[x[t], t], t, s]


You can verify the derivative property using:

FullSimplify[s eqn1] == FullSimplify[eqn2]
True


Of course it's also true in general:

LaplaceTransform[D[z[t], t], t, s]


which returns

s LaplaceTransform[z[t], t, s] - z[0]

• Thanks! This works great! I was just wondering, how would I go about plotting eqn1 and eqn2 and the Laplace transforms in general? Dec 16, 2020 at 12:11
• UnitStep is superfluous since LaplaceTransform[t/3 Exp[-3 t], t, s] produces the same result as LaplaceTransform[x[t], t, s]. Jan 15, 2021 at 7:19