## 1. What is Dirac delta function?

It is very common in mechanical systems the action of an external force which acts with a huge intensity in a very short period of time. For example, the moment when a player kicks violently a ball. The intensity of this force may be modeled by the function

\displaystyle\delta_{\epsilon}(t-t_{0})=\left\{\begin{array}{ll} 0, & 0\leq t<t_{0}-\epsilon \\\frac{1}{2\epsilon}, & t_{0}-\epsilon\leq t<t_{0}+\epsilon \\ 0, & t\geq t_{0}+\epsilon \end{array}\right.

We call this function unit impulse and it basically represents a constant application in a very short period of time around t_{0}.

It is clear the function above is not, actually, a fuction. Also, the Dirac delta function is characterized by the following properties

\delta(t-t_{0})=\left\{\begin{array}{ll} \infty, & t=t_{0} \\ 0, & t\neq t_{0} \end{array}\right.

and

\displaystyle\int_{-\infty}^{\infty}\delta(t-t_{0})dt=1

## 2. Laplace Transform of the Dirac delta function

### Proof:

To prove the equality above, note that

\displaystyle \delta_{\epsilon}(t-t_{0})=\frac{1}{2\epsilon}\left[u(t-(t_{0}-\epsilon))-u(t-(t_{0}+\epsilon))\right]

and since

\displaystyle \mathscr{L} \{u(t-a)\}=\frac{e^{-sa}}{s}

then, by the linearity of Laplace transform, we have

\begin{aligned} \mathscr{L} \{\delta_{\epsilon}(t-t_{0})\} & = \mathscr{L} \left\{\frac{1}{2\epsilon}\left[u(t-(t_{0}-\epsilon))-u(t-(t_{0}+\epsilon))\right]\right\} \\ &= \frac{1}{2\epsilon}\left( \mathscr{L} \{u(t-(t_{0}-\epsilon))\}- \mathscr{L} \{u(t-(t_{0}+\epsilon))\}\right) \\ &= \frac{1}{2\epsilon}\left(\frac{e^{-s(t_{0}-\epsilon)}}{s}-\frac{e^{-s(t_{0}+\epsilon)}}{s}\right) \end{aligned}

As

\displaystyle\begin{aligned} e^{-s(t_{0}-\epsilon)}-e^{-s(t_{0}+\epsilon)} & = e^{-st_{0}+s\epsilon}-e^{-st_{0}-s\epsilon} \\ & = e^{-st_{0}}\cdot e^{s\epsilon}-e^{-st_{0}}\cdot e^{-s\epsilon} \\ & = e^{-st_{0}}(e^{s\epsilon}-e^{-s\epsilon}) \end{aligned}

so

\displaystyle \mathscr{L} \{\delta_{\epsilon}(t-t_{0})\}=e^{-st_{0}}\left(\frac{e^{s\epsilon}-e^{-s\epsilon}}{2\epsilon}\right)

An by the definition of \delta:

\displaystyle\mathscr{L} \{\delta(t-t_{0})\}=\lim_{\epsilon\to0} \mathscr{L} \{\delta_{\epsilon}(t-t_{0})\}=\lim_{\epsilon\to0}e^{-st_{0}}\left(\frac{e^{s\epsilon}-e^{-s\epsilon}}{2s\epsilon}\right)

Hence,

\displaystyle\mathscr{L} \{\delta(t-t_{0})\}=e^{-st_{0}}\lim_{\epsilon\to0}\left(\frac{e^{s\epsilon}-e^{-s\epsilon}}{2s\epsilon}\right)

When we evaluate the limit above, we get \dfrac{0}{0}; then we apply L’Hôpital’s Rule:

\displaystyle\begin{aligned} \mathscr{L} \{\delta(t-t_{0})\}&=e^{-st_{0}}\lim_{\epsilon\to0}\left(\dfrac{\dfrac{d}{ds}(e^{s\epsilon}-e^{-s\epsilon})}{{\dfrac{d}{ds}(2s\epsilon)}}\right) \\ & = e^{-st_{0}}\lim_{\epsilon\to0}\left(\frac{\epsilon e^{s\epsilon}+\epsilon e^{-s\epsilon}}{2\epsilon}\right) \\ & = e^{-st_{0}}\lim_{\epsilon\to0}\left(\frac{e^{s\epsilon}+e^{-s\epsilon}}{2}\right) \\ & = e^{-st_{0}}\frac{1+1}{2} \\ & = e^{-st_{0}} \end{aligned}

Observe that for t_{0}=0, we may concluded

\displaystyle\mathscr{L} \{\delta(t)\}=1

## Example 1: shifted Dirac function

To evaluate the Laplace transform of \delta(t-2\pi), we use the time shift property. Note that t_{0}=2\pi, we have:

\displaystyle\mathscr{L} \{\delta(t-2\pi)\}=1\cdot e^{-s2\pi}=e^{-2\pi s}

## Example 2: Differential equation with Dirac function

Applying the Laplace transform, we get:

\begin{aligned} & \mathscr{L} \{y''+16y\}= \mathscr{L} \{\delta(t-2\pi)\} \\ & \mathscr{L} \{y''\}+16 \mathscr{L} \{y\}= \mathscr{L} \{\delta(t-2\pi)\} \end{aligned}

From the table of common Laplace transforms and from the previous example of shifted Dirac function, we have

\begin{aligned} & s^{2}Y(s)-sy(0)-y'(0)+16Y(s)=e^{-2\pi s} \\ & s^{2}Y(s)-s\cdot0-0+16Y(s)=e^{-2\pi s} \\ & (s^{2}+16)Y(s)=e^{-2\pi s} \\ & Y(s)=\frac{e^{-2\pi s}}{s^{2}+16} \end{aligned}

We may rewrite Y(s) as follows

\displaystyle Y(s)=e^{-2\pi s}\cdot\frac{1}{s^{2}+16}

Let

\displaystyle F(s)=\frac{1}{s^{2}+16}

then

Y(s)=e^{-2\pi s}F(s)

i.e.,

y(t)= \mathscr{L} ^{-1}\left\{e^{-2\pi s}F(s)\right\}

From the line 9 of the Laplace transform table, we know:

\displaystyle \mathscr{L} ^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a)

where f(t)= \mathscr{L} ^{-1}\{F(s)\}. So, since

\displaystyle \begin{aligned} F(s) & = \frac{1}{s^{2}+16} \\ & = \frac{1}{s^{2}+4^{2}} \\ & = \frac{1}{4}\cdot\frac{4}{s^{2}+4^{2}} \end{aligned}

thus

\displaystyle \begin{aligned} f(t) & = \mathscr{L} ^{-1}\{F(s)\} \\ f(t) & = \mathscr{L} ^{-1}\left\{\frac{1}{4}\cdot\frac{4}{s^{2}+4^{2}}\right\} \\ f(t) & =\frac{1}{4} \mathscr{L} ^{-1}\left\{\frac{4}{s^{2}+4^{2}}\right\} \\ f(t) & =\frac{1}{4}\sin(4t) \end{aligned}

where we considered b=4 in the third line of the table. Therefore, for a=2\pi

\displaystyle y(t)=f(t-2\pi)u(t-2\pi)

which gives us

\displaystyle y(t)=\frac{1}{4}\sin(4(t-2\pi))u(t-2\pi)

However,

\displaystyle \sin(4(t-2pi))=\sin(4t)

so

\displaystyle y(t)=\frac{1}{4}\sin(4t)u(t-2\pi)

That’s all!

We reached the end of this exercise about calculating the Laplace transform of the Dirac function using the definition.