Heaviside function in the function whose Laplace transformation is $e^{-(\gamma+s)}/[(s+\gamma)^2+b^2]$

Question

This is from a homework question 13.22 part (c) from "Mathematical Methods for Physic and Engineering" by Riley et. al on p. 464

I don't understand why the heaviside function is in the solution to this problem. I would be very grateful if someone could point me in the right direction.

The question states:

Find the function y(t) whose Laplace transforms is the following: $$e^{-(\gamma+s)}/[(s+\gamma)^2+b^2]$$

The correct solution uses the convolution formula for a Laplace transform to get:

$$ \begin{align} f(t) &= \frac{e^{-\gamma t_0}}{b} \int_0^{t} e^{-\gamma u}\sin(bu)H(u)\delta(t-u-t_0) \, du \\&= \frac{e^{-\gamma t_0}}{b} e^{-\gamma (t-t_0)} \sin[b(t-t_0)]H(t-t_0) \\ &= \frac{1}{b} e^{-\gamma t} sin[b(t-t_0)] H(t-t_0)\end{align}$$

where $\delta$ is the Dirac delta function and $H$ is the heavyside function.

To approach this problem I took $\bar g(s) = e^{-s t_0}$ and $\bar f(s) = \frac{b}{(s^2 + \gamma)^2 + b^2}$ as the Laplace transforms. From there I looked up the Laplace transforms in a table provided by the book and got $g(t) = \delta (t-t_0)$ and $f(t) = e^{-\gamma t} \sin(bt)$. I then applied the convolution theorem in the following way: $$\begin{align} \mathcal{L}^{-1} \left(\tfrac{1}{b} e^{-\gamma t_0 } \bar f(s) \bar g(s) \right) &= \frac{1}{b} e^{-\gamma t_0} \int_0^t e^{-\gamma u} \sin(bu) \delta(t-u-t_0) \, du \\ &= \frac{1}{b} e^{-\gamma t} \sin(b(t-t_0))\end{align}$$

where $\mathcal{L}$ is the Laplace transform. As you can see, my solution was similar to the correct solution, but has no heaviside function. I still can't figure out where the heaviside function comes from! Thank you in advance for your help.

Answer 1

By using the inverse Laplace transform, we should be able to clear up the confusion of where the Heaviside function comes into play. The inverse Laplace transform is $$ \mathcal{L}^{-1}\{F(s)\} = \frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma + i\infty}F(s)e^{st}ds = \sum\text{Res} $$ If we expand the denominator, we have that $$ (s+\gamma)^2 +b^2 = s^2 + 2\gamma s+\gamma^2 + b^2. $$ Then the poles occur when the denominator is zero. Using the quadratic equation, we obtain that the poles occur when $$ s = -\gamma\pm ib. $$ Now, let's set up the inverse Laplace transform. $$ \mathcal{L}^{-1}\{F(s)\} = \frac{e^{-\gamma}}{2\pi i}\int_{\sigma-i\infty}^{\sigma + i\infty}\frac{\exp[s(t-1)]}{(s+\gamma - ib)(s+\gamma+ib)}ds $$ The numerator is $\exp[s(t-1)]$. In order for convergence, $s(t-1)<0$ or $t<1$. The definition of the Heaviside function (Unit step function) is $$ \mathcal{H}(t) = \mathcal{U}(t) = \begin{cases} 1, & t>0\\ 0, & t<0 \end{cases} $$ In order for convergence of the exponential term, we need $t<1$ or $\mathcal{H}(t-1)$ which will zero out all the terms with $t<1$. \begin{align} \frac{e^{-\gamma}}{2\pi i}\int_{\sigma-i\infty}^{\sigma + i\infty}\frac{\exp[s(t-1)]}{(s+\gamma - ib)(s+\gamma+ib)}ds &= e^{-\gamma}\biggl[\lim_{s\to -\gamma+ib}(s+\gamma-ib)\frac{e^{s(t-1)}}{(s+\gamma - ib)(s+\gamma+ib)}+\lim_{s\to -\gamma-ib}(s+\gamma+ib)\frac{e^{s(t-1)}}{(s+\gamma - ib)(s+\gamma+ib)}\biggr]\\ &=\frac{e^{-\gamma-\gamma(t-1)}\mathcal{H}(t-1)}{b}\biggl[\frac{e^{ib(t-1)}-e^{-ib(t-1)}}{2i}\biggr]\\ &= \frac{e^{-\gamma-\gamma(t-1)}\mathcal{H}(t-1)}{b}\sin[b(t-1)] \end{align}

where I took $\gamma = b = 1$.