### 3 The first shift theorem

The first and second shift theorems enable an even wider range of Laplace transforms to be easily obtained than the transforms we have already found. They also enable a significantly wider range of inverse transforms to be found. Here we introduce the first shift theorem. If $f\left(t\right)$ is a causal function with Laplace transform $F\left(s\right)$ , i.e. $\mathsc{L}\left\{f\left(t\right)\right\}=F\left(s\right)$ , then as we shall see, the Laplace transform of ${\text{e}}^{-at}f\left(t\right)$ , where $a$ is a given constant, can easily be found in terms of $F\left(s\right)$ .

Using the definition of the Laplace transform:

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

But if

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

then simply replacing ‘ $s$ ’ by ‘ $s+a$ ’ on both sides gives:

$\phantom{\rule{2em}{0ex}}F\left(s+a\right)={\int }_{0}^{\infty }{\text{e}}^{-\left(s+a\right)t}f\left(t\right)\phantom{\rule{0.3em}{0ex}}dt$

That is, the parameter $s$ is shifted to the value $s+a$ .

We have then the statement of the first shift theorem :

##### Key Point 7

First Shift Theorem

If $\mathsc{L}\left\{f\left(t\right)\right\}=F\left(s\right)$ then $\mathsc{L}\left\{{\text{e}}^{-at}f\left(t\right)\right\}=F\left(s+a\right).$

For example, we already know (from Table 1) that

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{{t}^{3}u\left(t\right)\right\}=\frac{6}{{s}^{4}}$

and so, by the first shift theorem:

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{{\text{e}}^{-2t}{t}^{3}u\left(t\right)\right\}=\frac{6}{{\left(s+2\right)}^{4}}$

Use the first shift theorem to determine $\mathsc{L}\left\{{\text{e}}^{2t}cos3t.u\left(t\right)\right\}$ .

You should obtain $\frac{s-2}{{\left(s-2\right)}^{2}+9}$ since $\mathsc{L}\left\{cos3t.u\left(t\right)\right\}=\frac{s}{{s}^{2}+9}$

and so by the first shift theorem (with $a=-2$ )

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{{\text{e}}^{2t}cos3t.u\left(t\right)\right\}=\frac{s-2}{{\left(s-2\right)}^{2}+9}$

obtained by simply replacing ‘ $s$ ’ by ‘ $s-2$ ’.

We can also employ the first shift theorem to determine some inverse Laplace transforms.

Find the inverse Laplace transform of $F\left(s\right)=\frac{3}{{s}^{2}-2s-8}$ .

Begin by completing the square in the denominator:

$\frac{3}{{s}^{2}-2s-8}=\frac{3}{{\left(s-1\right)}^{2}-9}$ Recalling that $\mathsc{L}\left\{sinh3t\phantom{\rule{1em}{0ex}}u\left(t\right)\right\}=\frac{3}{{s}^{2}-9}$ (from the Task on page 15) complete the inversion using the first shift theorem:

You should obtain

$\phantom{\rule{2em}{0ex}}{\mathsc{L}}^{-1}\left\{\frac{3}{{\left(s-1\right)}^{2}-9}\right\}={\text{e}}^{t}sinh3t\phantom{\rule{1em}{0ex}}u\left(t\right)$

Here, in the notation of the shift theorem:

#### 3.1 Inverting using completion of the square

The function:

$\phantom{\rule{2em}{0ex}}F\left(s\right)=\frac{4s}{{s}^{2}+2s+5}$

does not appear in the table of transforms and, again, needs amending before we can find its inverse transform. In this case, because ${s}^{2}+2s+5$ does not have nice factors, we complete the square in the denominator:

$\phantom{\rule{2em}{0ex}}{s}^{2}+2s+5\equiv {\left(s+1\right)}^{2}+4$

and so

$\phantom{\rule{2em}{0ex}}F\left(s\right)=\frac{4s}{{s}^{2}+2s+5}=\frac{4s}{{\left(s+1\right)}^{2}+4}$

Now the numerator needs amending slightly to enable us to use the appropriate rule in the table of transforms (Table 1, Rule 8):

$\begin{array}{rcll}F\left(s\right)=\frac{4s}{{\left(s+1\right)}^{2}+4}& =& 4\left\{\frac{s+1-1}{{\left(s+1\right)}^{2}+4}\right\}& \text{}\\ & =& 4\left\{\frac{s+1}{{\left(s+1\right)}^{2}+4}-\frac{1}{{\left(s+1\right)}^{2}+4}\right\}& \text{}\\ & =& \frac{4\left(s+1\right)}{{\left(s+1\right)}^{2}+4}-2\left[\frac{2}{{\left(s+1\right)}^{2}+4}\right]\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Hence

$\begin{array}{rcll}{\mathsc{L}}^{-1}\left\{F\left(s\right)\right\}& =& 4{\mathsc{L}}^{-1}\left\{\frac{s+1}{{\left(s+1\right)}^{2}+4}\right\}-2{\mathsc{L}}^{-1}\left\{\frac{2}{{\left(s+1\right)}^{2}+4}\right\}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}& \text{}\\ & =& 4{\text{e}}^{-t}cos2t\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{0.3em}{0ex}}u\left(t\right)-2{\text{e}}^{-t}sin2t\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{0.3em}{0ex}}u\left(t\right)& \text{}\\ & =& {\text{e}}^{-t}\left[4cos2t-2sin2t\right]u\left(t\right)& \text{}\end{array}$

Find the inverse Laplace transform of $\frac{3}{{s}^{2}-4s+6}$ .

Begin by completing the square in the denominator of this expression:

$\phantom{\rule{2em}{0ex}}\frac{3}{{s}^{2}-4s+6}=\frac{3}{{\left(s-2\right)}^{2}+2}$

Now obtain the inverse:

You should obtain:

${\mathsc{L}}^{-1}\left\{\frac{3}{{\left(s-2\right)}^{2}+2}\right\}={\mathsc{L}}^{-1}\left\{\frac{3}{\sqrt{2}}\left[\frac{\sqrt{2}}{{\left(s-2\right)}^{2}+2}\right]\right\}=\frac{3}{\sqrt{2}}{\text{e}}^{2t}sin\sqrt{2}t.u\left(t\right)$ (Table 1, Rule 7)

##### Exercise

Determine the inverse Laplace transforms of the following functions.

1. $\frac{10}{{s}^{4}}$
2. $\frac{s-1}{{s}^{2}+8s+17}$
3. $\frac{3s-7}{{s}^{2}+9}$
4. $\frac{3s+3}{\left(s-1\right)\left(s+2\right)}$
5. $\frac{s+3}{{s}^{2}+4s}$
6. $\frac{2}{\left(s+1\right)\left({s}^{2}+1\right)}$
1. $\frac{10}{6}{t}^{3}$
2. ${\text{e}}^{-4t}cost-5{\text{e}}^{-4t}sint$
3. $3cos3t-\frac{7}{3}sin3t$
4. $2{\text{e}}^{t}+{\text{e}}^{-2t}$
5. $\frac{3}{4}u\left(t\right)+\frac{1}{4}{\text{e}}^{-4t}u\left(t\right)$
6. $\left({\text{e}}^{-t}-cost+sint\right)u\left(t\right)$