# Laplace transform of the Dirac Delta function does not match the Laplace transform of the Unit Step function

If $u_s(t)$ is the unit step function, $\delta(t)$ the dirac delta function, and $f(t) = u_s(t)$, then $\frac {d} {dx}f(t) = \delta(t)$.

The laplace transform of $\frac {d} {dx}f(t)$ is $s(F(t))-f(0)$, meaning that $\frac {d} {dx}u_s(t) = s(\frac 1 s) - 1$, which is $0$.

The laplace transform of $\delta(t)$ is $1$.

Why do these values not match?

• You have two choices : using $f(0) = 1$ and $f'(t) = 0$, or using $f(0) = 0$ and $f'(t) = \delta(t)$. – reuns Jul 1 '17 at 21:58
• You have used $f(0+)$ so you miss the step. Use $f(0-) = 0$ instead. – md2perpe Jul 2 '17 at 0:31