# Laplace Inverse using second shift theorem and heaviside unit function

I am trying to work out the Laplace inverse of $\frac{1}{s} e^{-s t_0-sb}$, I was trying to use the second shift theorem and rewrote it as $e^{-s t_0}\frac{1}{s}e^{-sb}$.

I ended up with $H(t-t_0)H(t-b-t_0)$, but have been told the solution is $H(t-b-T_0)$, what happens to the other heaviside unit function? Is there a general rule as to why this happens or have I just worked out the transform wrong?

If $b>0$ then: $$H(t-t_0)H(t-b-t_0)=H(t-b-t_0)$$ indeed:
• If $t<t_0$ then both $t-t_0$ and $t-b-t_0$ are negative so: $$H(t-t_0)H(t-b-t_0)=0 \times 0=0=H(t-b-t_0)$$
• If$t_0 \leq t<t_0+b$ then both $t-t_0 \geq 0$ and $t-b-t_<00$ : $$H(t-t_0)H(t-b-t_0)=1 \times 0=0=H(t-b-t_0)$$
• If $t\geq t_0 +b$ then both $t-t_0$ and $t-b-t_0$ are postive so: $$H(t-t_0)H(t-b-t_0)=1 \times 1=1=H(t-b-t_0)$$
The general rule is that, by denoting $\chi_A$ the characteristic function of a set $A$: $$\chi_A \cdot \chi_B =\chi_{A \cap B}$$ Here: $$H(t-\tau)=\chi_{[\tau,+\infty)}$$ and: $$[\tau_1,+\infty) \cap [\tau_2,+\infty)= [\max(\tau_1,\tau_2),+\infty)$$