Inverse Laplace transform of $\frac{r_1e^{-t_0s}}{s + r_2 + r_3}$

Question

I have the Laplace transform of $x$ as follows:

$$ x_L=\frac{r_1e^{-t_0s}}{s + r_2 + r_3}, $$

where $x$ is a function of $t$, and $x_L$ is a function of $s$.

I know the inverse Laplace transform of

$$ \frac{e^{-cs}}{s} $$

is

$$ H(t-c), $$

but I'm not sure how to deal with the $r_y$ factors in my equation for $x_L$.

What is the inverse Laplace transform of $x_L$ as defined above?

Answer 1

Note that

$$ x_L=\frac{r_1e^{-t_0s}}{s + r_2 + r_3} $$

can be split into

$$ x_L=r_1e^{-t_0s}\frac{1}{s + r_2 + r_3}. $$

Define $f_L=\frac{1}{s + r_2 + r_3}$.

According to my table of commonly encountered Laplace transforms, I have

$$ H(t-c)f(t-c)=e^{-cs}f_L(s). $$

Hence, at this point, in order to solve the overall problem, I just need to know how to compute the inverse Laplace transform of my $f_L$ to get $f(t)$.

Another entry in my table is

$$ e^{at}=\frac{1}{s-a}, s>a. $$

If we define $a=-r_2-r_3$, we have

$$ e^{(-r_2-r_3)t}=f(t). $$

Defining $c=t_0$, we can solve the overall problem:

\begin{align} x&=r_1[H(t-t_0)f(t-c)]\\ x&=r_1\bigg[H(t-t_0)e^{(-r_2-r_3)(t-t_0)}\bigg]. \end{align}