### 1 The second shift theorem

The second shift theorem is similar to the first except that, in this case, it is the time-variable that is shifted not the $s$ -variable. Consider a causal function $f\left(t\right)u\left(t\right)$ which is shifted to the right by amount $a$ , that is, the function $f\left(t-a\right)u\left(t-a\right)$ where $a>0$ . Figure 13 illustrates the two causal functions.

Figure 13

The Laplace transform of the shifted function is easily obtained:

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

(Note the change in the lower limit from $0$ to $a$ resulting from the step function switching on at $t=a$ ). We can re-organise this integral by making the substitution $x=t-a$ . Then $\phantom{\rule{1em}{0ex}}dt=dx\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}$

Therefore

$\begin{array}{rcll}{\int }_{a}^{\infty }{\text{e}}^{-st}f\left(t-a\right)\phantom{\rule{0.3em}{0ex}}dt& =& {\int }_{0}^{\infty }{\text{e}}^{-s\left(x+a\right)}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\text{e}}^{-sa}{\int }_{0}^{\infty }{\text{e}}^{-sx}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

The final integral is simply the Laplace transform of $f\left(x\right)$ , which we know is $F\left(s\right)$ and so, finally, we have the statement of the second shift theorem:

##### Key Point 8

Second Shift Theorem

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

Obviously, this theorem has its uses in finding the Laplace transform of time-shifted causal functions but it is also of considerable use in finding inverse Laplace transforms since, using the inverse formulation of the theorem of Key Point 8 we get:

##### Key Point 9

Inverse Second Shift Theorem

If ${\mathsc{L}}^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)$ then ${\mathsc{L}}^{-1}\left\{{\text{e}}^{-sa}F\left(s\right)\right\}=f\left(t-a\right)u\left(t-a\right)$

Find the inverse Laplace transform of $\frac{{\text{e}}^{-3s}}{{s}^{2}}$ .

You should obtain $\left(t-3\right)u\left(t-3\right)$ for the following reasons. We know that the inverse Laplace transform of $1∕{s}^{2}$ is $t.u\left(t\right)$ (Table 1, Rule 3) and so, using the second shift theorem (with $a=3$ ), we have

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

This function is graphed in the following figure:

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

You should obtain ${\text{e}}^{t}\left(cost+sint\right)$ .

To obtain this, complete the square in the denominator: ${s}^{2}-2s+2={\left(s-1\right)}^{2}+1$ and so

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

Now, using the first shift theorem

and

Thus

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