### 1 The Laplace transform

If $f\left(t\right)$ is a causal function then the Laplace transform of $f\left(t\right)$ is written $\mathsc{L}\left\{f\left(t\right)\right\}$ and defined by:

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{f\left(t\right)\right\}={\int }_{0}^{\infty }{\text{e}}^{-st}f\left(t\right)\phantom{\rule{0.3em}{0ex}}dt.$

Clearly, once the integral is performed and the limits substituted the resulting expression will involve the $s$ parameter alone since the dependence upon $t$ is removed in the integration process. This resulting expression in $s$ is denoted by $F\left(s\right)$ ; its precise form is dependent upon the form taken by $f\left(t\right)$ . We now refine Key Point 1 (page 4).

##### Key Point 3

The Laplace Transform of a Causal Function

$\mathsc{L}\left\{f\left(t\right)u\left(t\right)\right\}\equiv {\int }_{0}^{\infty }{\text{e}}^{-st}f\left(t\right)u\left(t\right)\phantom{\rule{0.3em}{0ex}}dt\equiv F\left(s\right)$

To begin, we determine the Laplace transform of some simple causal functions. For example, if we consider the ramp function $f\left(t\right)=t.u\left(t\right)$ with graph

Figure 11

we find:

Now we have the difficulty of substituting in the limits of integration. The only problem arises with the upper limit ( $t=\infty$ ). We shall always assume that the parameter $s$ is so chosen that no contribution ever arises from the upper limit ( $t=\infty$ ). In this particular case we need only demand that $s$ is real and positive. Using this ‘rule of thumb’:

$\begin{array}{rcll}\mathsc{L}\left\{t\phantom{\rule{1em}{0ex}}u\left(t\right)\right\}& =& \left[0-0\right]-\left[0-\left(\frac{1}{{\left(-s\right)}^{2}}\right)\right]\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& \frac{1}{{s}^{2}}& \text{}\end{array}$

Thus, if $f\left(t\right)=t\phantom{\rule{1em}{0ex}}u\left(t\right)$ then $F\left(s\right)=1∕{s}^{2}$ .

A similar, but more tedious, calculation yields the result that if $f\left(t\right)={t}^{n}u\left(t\right)$ in which $n$ is a positive integer then:

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{{t}^{n}u\left(t\right)\right\}=\frac{n!}{{s}^{n+1}}$

[We remember $n!\equiv n\left(n-1\right)\left(n-2\right)\dots \left(3\right)\left(2\right)\left(1\right).$ ]

Find the Laplace transform of the step function $u\left(t\right)$ .

Begin by obtaining the Laplace integral:

You should obtain ${\int }_{0}^{\infty }{\text{e}}^{-st}\phantom{\rule{0.3em}{0ex}}dt$ since in the range of integration, $t>0$ and so $u\left(t\right)=1$ leading to

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{u\left(t\right)\right\}={\int }_{0}^{\infty }{\text{e}}^{-st}u\left(t\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{0.3em}{0ex}}dt={\int }_{0}^{\infty }{\text{e}}^{-st}\phantom{\rule{0.3em}{0ex}}dt$

Now complete the integration:

You should have obtained:

$\begin{array}{rcll}\mathsc{L}\left\{u\left(t\right)\right\}& =& {\int }_{0}^{\infty }{\text{e}}^{-st}\phantom{\rule{0.3em}{0ex}}dt& \text{}\\ & =& {\left[\frac{{\text{e}}^{-st}}{\left(-s\right)}\right]}_{0}^{\infty }=0-\left[\frac{1}{\left(-s\right)}\right]=\frac{1}{s}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

where, again, we have assumed the contribution from the upper limit is zero.

As a second example, we consider the decaying exponential $f\left(t\right)={\text{e}}^{-at}u\left(t\right)$ where $a$ is a positive constant. This function has graph:

Figure 12

In this case,

Therefore, if $f\left(t\right)={\text{e}}^{-at}u\left(t\right)$ then $F\left(s\right)=\frac{1}{s+a}$ .

Following this approach we can develop a table of Laplace transforms which records, for each causal function $f\left(t\right)$ listed, its corresponding transform function $F\left(s\right)$ . Table 1 gives a limited table of transforms.

Table 1: Table of Laplace Transforms
 Rule Causal function Laplace transform 1 $f\left(t\right)$ $F\left(s\right)$ 2 $u\left(t\right)$ $\frac{1}{s}$ 3 ${t}^{n}u\left(t\right)$ $\frac{n!}{{s}^{n+1}}$ 4 ${\text{e}}^{-at}u\left(t\right)$ $\frac{1}{s+a}$ 5 $sinat\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{0.3em}{0ex}}u\left(t\right)$ $\frac{a}{{s}^{2}+{a}^{2}}$ 6 $cosat\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{0.3em}{0ex}}u\left(t\right)$ $\frac{s}{{s}^{2}+{a}^{2}}$ 7 ${\text{e}}^{-at}sinbt\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{0.3em}{0ex}}u\left(t\right)$ $\frac{b}{{\left(s+a\right)}^{2}+{b}^{2}}$ 8 ${\text{e}}^{-at}cosbt\phantom{\rule{0.3em}{0ex}}u\left(t\right)$ $\frac{s+a}{{\left(s+a\right)}^{2}+{b}^{2}}$

Note: For convenience, this table is repeated at the end of the Workbook.

#### 1.1 The linearity property of the Laplace transformation

If $f\left(t\right)$ and $g\left(t\right)$ are causal functions and ${c}_{1}$ , ${c}_{2}$ are constants then

$\begin{array}{rcll}\mathsc{L}\left\{{c}_{1}f\left(t\right)+{c}_{2}g\left(t\right)\right\}& =& {\int }_{0}^{\infty }{\text{e}}^{-st}\left[{c}_{1}f\left(t\right)+{c}_{2}g\left(t\right)\right]\phantom{\rule{0.3em}{0ex}}dt& \text{}\\ & =& {c}_{1}{\int }_{0}^{\infty }{\text{e}}^{-st}f\left(t\right)\phantom{\rule{0.3em}{0ex}}dt+{c}_{2}{\int }_{0}^{\infty }{\text{e}}^{-st}g\left(t\right)\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& {c}_{1}\mathsc{L}\left\{f\left(t\right)\right\}+{c}_{2}\mathsc{L}\left\{g\left(t\right)\right\}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$
##### Key Point 4

Linearity Property of the Laplace Transform

$\mathsc{L}\left\{{c}_{1}f\left(t\right)+{c}_{2}g\left(t\right)\right\}={c}_{1}\mathsc{L}\left\{f\left(t\right)\right\}+{c}_{2}\mathsc{L}\left\{g\left(t\right)\right\}$

That is, the Laplace transform of a linear sum of causal functions is a linear sum of Laplace transforms. For example,

$\begin{array}{rcll}\mathsc{L}\left\{2cost\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{0.3em}{0ex}}u\left(t\right)-3{t}^{2}u\left(t\right)\right\}& =& 2\mathsc{L}\left\{cost\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{0.3em}{0ex}}u\left(t\right)\right\}-3\mathsc{L}\left\{{t}^{2}u\left(t\right)\right\}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& 2\left(\frac{s}{{s}^{2}+1}\right)-3\left(\frac{2}{{s}^{3}}\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Obtain the Laplace transform of the hyperbolic function $sinhat$ .

Begin by expressing $sinhat$ in terms of exponential functions:

$sinhat=\frac{1}{2}\left({\text{e}}^{at}-{\text{e}}^{-at}\right)$ Now use the linearity property (Key Point 4) to obtain the Laplace transform of the causal function $sinhat.u\left(t\right)$ :

You should obtain $a∕\left({s}^{2}-{a}^{2}\right)$ since

Obtain the Laplace transform of the hyperbolic function $coshat$ .

You should obtain $\frac{s}{{s}^{2}-{a}^{2}}$ since

Find the Laplace transform of the delayed step-function $u\left(t-a\right)$ , $a>0$ .

Write the delayed step-function here in terms of an integral:

You should obtain $\mathsc{L}\left\{u\left(t-a\right)\right\}={\int }_{a}^{\infty }{\text{e}}^{-st}\phantom{\rule{0.3em}{0ex}}dt$ (note the lower limit is $a$ ) since:

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{u\left(t-a\right)\right\}={\int }_{0}^{\infty }{\text{e}}^{-st}u\left(t-a\right)\phantom{\rule{0.3em}{0ex}}dt={\int }_{0}^{a}{\text{e}}^{-st}u\left(t-a\right)\phantom{\rule{0.3em}{0ex}}dt+{\int }_{a}^{\infty }{\text{e}}^{-st}u\left(t-a\right)\phantom{\rule{0.3em}{0ex}}dt$

In the first integral $0 and so $\left(t-a\right)<0$ , therefore $u\left(t-a\right)=0$ .

In the second integral $a and so $\left(t-a\right)>0$ , therefore $u\left(t-a\right)=1$ . Hence

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{u\left(t-a\right)\right\}=0+{\int }_{a}^{\infty }{\text{e}}^{-st}\phantom{\rule{0.3em}{0ex}}dt.$

Now complete the integration:

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{u\left(t-a\right)\right\}={\int }_{a}^{\infty }{\text{e}}^{-st}\phantom{\rule{0.3em}{0ex}}dt={\left[\frac{{\text{e}}^{-st}}{\left(-s\right)}\right]}_{a}^{\infty }=\frac{{\text{e}}^{-sa}}{s}$

##### Exercise

Determine the Laplace transform of the following functions.

1. ${\text{e}}^{-3t}u\left(t\right)$
2. $u\left(t-3\right)$
3. ${\text{e}}^{-t}sin3t.u\left(t\right)$
4. $\left(5cos3t-6{t}^{3}\right).u\left(t\right)$
1. $\frac{1}{s+3}$
2. $\frac{{\text{e}}^{-3s}}{s}$
3. $\frac{3}{{\left(s+1\right)}^{2}+9}$
4. $\frac{5s}{{s}^{2}+9}-\frac{36}{{s}^{4}}$