1 Transforms and causal functions

Without perhaps realising it, we are used to employing transformations in mathematics. For example, we often transform problems in algebra to an equivalent problem in geometry in which our natural intuition and experience can be brought to bear. Thus, for example, if we ask:

q What are those values of $x$ for which $x\left(x-1\right)\left(x+2\right)>0$ ’ then perhaps the simplest way to solve this problem is to sketch the curve $y=x\left(x-1\right)\left(x+2\right)$ and then, by inspection, find for what values of $x$ it is positive. We obtain the following figure.

Figure 1

We have transformed a problem in algebra into an equivalent geometrical problem.

Clearly, by inspection of the curve, this inequality is satisfied if

$-2<x<0\text{ or if}x>1$

and we have transformed back again to algebraic form.

The Laplace transform is a more complicated transformation than the simple geometric transformation considered above. What is done is to transform a function $f\left(t\right)$ of a single variable $t$ into another function $F\left(s\right)$ of a single variable $s$ through the relation:

$F\left(s\right)={\int \; <!--nolimits\; -->}_0^{\infty }{\text{e}}^{-st}f\left(t\right)dt.$

The procedure is to produce, for each $f\left(t\right)$ of interest, the corresponding expression $F\left(s\right)$ . As a simple example, if $f\left(t\right)={\text{e}}^{-2t}$ then

(We remind the reader that ${\text{e}}^{-kt}\to 0$ as $t\to \infty$ if $k>0$ .)

Task!

Find $F\left(s\right)$ if $f\left(t\right)=t$ using $F\left(s\right)={\int \; <!--nolimits\; -->}_0^{\infty }{\text{e}}^{-st}tdt$

Answer

The integral ${\int \; <!--nolimits\; -->}_0^{\infty }{\text{e}}^{-st}f\left(t\right)dt$ is called the Laplace transform of $f\left(t\right)$ and is denoted by $\mathcal{L}\left\{f\left(t\right)\right\}$ .

1.1 Causal functions

As we have seen above, the Laplace transform involves an integral with limits $t=0$ and $t=\infty$ . Because of this, the nature of the function being transformed, $f\left(t\right)$ , when $t$ is negative is of no importance. In order to emphasize this we shall only consider so-called causal functions all of which take the value $0$ when $t<0$ .

The simplest causal function is the Heaviside or step function denoted by $u\left(t\right)$ and defined by:

$u\left(t\right)=\left\{\begin{array}{cc}1\hfill & \text{if}t\ge 0\hfill \\ 0\hfill & \text{if}t<0\hfill \end{array}\right.$

with graph as in Figure 2.

Figure 2

Similarly we can consider other ‘step-functions’. For example, from the above definition we deduce

$u\left(t-3\right)=\left\{\begin{array}{cc}1\hfill & \text{if}t-3\ge 0\hfill \\ 0\hfill & \text{if}t-3<0\hfill \end{array}\right.\text{or, rearranging the inequalities:}u\left(t-3\right)=\left\{\begin{array}{cc}1\hfill & \text{if}t\ge 3\hfill \\ 0\hfill & \text{if}t<3\hfill \end{array}\right.$

with graph as in Figure 3:

Figure 3

The step function has a useful property: multiplying an ordinary function $f\left(t\right)$ by the step function $u\left(t\right)$ changes it into a causal function; e.g. if $f\left(t\right)=sint$ then $sint.u\left(t\right)$ is causal. This is illustrated in the change from Figure 4 to Figure 5:

Figure 4

Figure 5

The step function can be used to ‘switch on’ functions at other values of $t$ (which we will normally interpret as time). For example $u\left(t-1\right)$ has the value 1 if $t\ge 1$ and $0$ otherwise so that $sint.u\left(t-1\right)$ is described by the (solid) curve in Figure 6:

Figure 6

The step function can also be used to ‘switch-off’ signals. For example, the step function $u\left(t-1\right)-u\left(t-3\right)$ in Figure 7 has the effect on $f\left(t\right)$ such that $f\left(t\right)\left[u\left(t-1\right)-u\left(t-3\right)\right]$ (described by the solid curve in Figure 8) switches on at $t=1$ (because then $u\left(t-1\right)-u\left(t-3\right)$ takes the value 1), remains ‘on’ for $1\le t\le 3$ , and then switches ‘off’ when $t>3$ (because then $u\left(t-1\right)-u\left(t-3\right)=1-1=0$ ).

Figure 7

Figure 8

If we have an expression $f\left(t-a\right)u\left(t-a\right)$ then this is the function $f\left(t\right)$ translated along the $t$ -axis through a time $a$ . For example $sin\left(t-2\right).u\left(t-2\right)$ is simply the causal sine curve $sint.u\left(t\right)$ shifted to the right by two units as described in the following Figure 9.

Figure 9

Task!

Sketch the curve $f\left(t\right)={\text{e}}^t\left(u\left(t-1\right)-u\left(t-2\right)\right).$

Answer