# 4.1 Properties of the laplace transform  (Page 2/2)

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## Differentiation

The Laplace transform of the derivative of a signal will be used widely. Consider

$\mathfrak{L}\left\{\frac{d}{dt},x,\left(t\right)\right\}={\int }_{{0}^{-}}^{\infty }{x}^{\text{'}}\left(t\right){e}^{-st}dt$

this can be integrated by parts:

$\begin{array}{cc}u={e}^{-st}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\hfill & {v}^{\text{'}}={x}^{\text{'}}\left(t\right)\hfill \\ {u}^{\text{'}}=-s{e}^{-st}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\hfill & v=x\left(t\right)\hfill \end{array}$

which gives

$\begin{array}{ccc}\hfill \mathfrak{L}\left\{\frac{d}{dt},x,\left(t\right)\right\}& =& {\left(u,v|}_{{0}^{-}}^{\infty }-{\int }_{{0}^{-}}^{\infty }{u}^{\text{'}}vdt\hfill \\ \hfill & =& {\left({e}^{-st},x,\left(t\right)|}_{{0}^{-}}^{\infty }+{\int }_{{0}^{-}}^{\infty }sx\left(t\right){e}^{-st}dt\hfill \\ \hfill & =& -x\left({0}^{-}\right)+sX\left(s\right)\hfill \\ \hfill \end{array}$

therefore we have,

$\frac{d}{dt}x\left(t\right)↔sX\left(s\right)-x\left({0}^{-}\right)$

## Higher order derivatives

The previous derivation can be extended to higher order derivatives. Consider

$y\left(t\right)=\frac{dx\left(t\right)}{dt}↔sX\left(s\right)-x\left({0}^{-}\right)$

it follows that

$\frac{dy\left(t\right)}{dt}↔sY\left(s\right)-y\left({0}^{-}\right)$

$\frac{{d}^{2}}{d{t}^{2}}x\left(t\right)↔{s}^{2}X\left(s\right)-sx\left({0}^{-}\right)-\frac{dx\left({0}^{-}\right)}{dt}$

This process can be iterated to get the Laplace transform of arbitrary higher order derivatives, giving

$\begin{array}{ccc}\hfill \frac{{d}^{n}x\left(t\right)}{d{t}^{n}}& ↔& {s}^{n}X\left(s\right)-{s}^{n-1}x\left({0}^{-}\right)-\sum _{k=2}^{n}{s}^{n-k}\frac{{d}^{k-1}x\left({0}^{-}\right)}{d{t}^{k-1}}\hfill \end{array}$

where it should be understood that

$\frac{{d}^{m}x\left({0}^{-}\right)}{d{t}^{m}}\equiv {\left(\frac{{d}^{m}x\left(t\right)}{d{t}^{m}}|}_{t={0}^{-}},m=1,...,n-1$

## Integration

Let

$g\left(t\right)={\int }_{{0}^{-}}^{t}x\left(\tau \right)d\tau$

it follows that

$\frac{dg\left(t\right)}{dt}=x\left(t\right)$

and $g\left({0}^{-}\right)=0$ . Moreover, we have

$\begin{array}{ccc}\hfill X\left(s\right)& =& \mathfrak{L}\left\{\frac{dg\left(t\right)}{dt}\right\}\hfill \\ \hfill & =& sG\left(s\right)-g\left({0}^{-}\right)\hfill \\ \hfill & =& sG\left(s\right)\hfill \end{array}$

therefore

$G\left(s\right)=\frac{X\left(s\right)}{s}$

but since

$G\left(s\right)=\mathfrak{L}\left\{{\int }_{{0}^{-}}^{t},x,\left(\tau \right),d,\tau \right\}$

we have

${\int }_{{0}^{-}}^{t}x\left(\tau \right)d\tau ↔\frac{X\left(s\right)}{s}$

Now suppose $x\left(t\right)$ has a non-zero integral over negative values of $t$ . We have

${\int }_{\infty }^{t}x\left(\tau \right)d\tau ={\int }_{-\infty }^{{0}^{-}}x\left(\tau \right)d\tau +{\int }_{{0}^{-}}^{t}x\left(\tau \right)d\tau$

The quantity ${\int }_{-\infty }^{{0}^{-}}x\left(\tau \right)d\tau$ is a constant for positive values of $t$ , and can be expressed as

$u\left(t\right){\int }_{-\infty }^{{0}^{-}}x\left(\tau \right)d\tau$

it follows that

${\int }_{\infty }^{t}x\left(\tau \right)d\tau ↔\frac{{\int }_{-\infty }^{{0}^{-}}x\left(\tau \right)d\tau }{s}+\frac{X\left(s\right)}{s}$

where we have used the fact that $u\left(t\right)↔\frac{1}{s}.$

## The initial value theorem

The initial value theorem makes it possible to determine $x\left(t\right)$ at $t={0}^{+}$ from $X\left(s\right)$ . From the derivative property of the Laplace transform, we can write

$\mathfrak{L}\left\{\frac{dx\left(t\right)}{dt}\right\}=sX\left(s\right)-x\left({0}^{-}\right)$

Taking the limit $s\to \infty$

$\begin{array}{cc}\hfill \underset{s\to \infty }{lim}{\int }_{{0}^{-}}^{\infty }\frac{dx\left(t\right)}{dt}{e}^{-st}dt& =\underset{s\to \infty }{lim}\left[s,X,\left(s\right),-,x,\left({0}^{-}\right)\right]\hfill \\ \hfill {\int }_{{0}^{-}}^{\infty }\underset{s\to \infty }{lim}\frac{dx\left(t\right)}{dt}{e}^{-st}dt& =\underset{s\to \infty }{lim}\left[s,X,\left(s\right),-,x,\left({0}^{-}\right)\right]\hfill \end{array}$

There are two cases, the first is when $x\left(t\right)$ is continuous at $t=0$ . In this case it is clear that $\frac{dx\left(t\right)}{dt}{e}^{-st}\to 0$ as $s\to \infty$ , so [link] can be written as

$0=\underset{s\to \infty }{lim}\left[s,X,\left(s\right),-,x,\left({0}^{-}\right)\right]$

Since $x\left(t\right)$ is continuous at $t=0$ , $x\left({0}^{-}\right)=x\left({0}^{+}\right)$ , the Initial Value Theorem follows,

$x\left({0}^{+}\right)=\underset{s\to \infty }{lim}sX\left(s\right)$

The second case is when $x\left(t\right)$ is discontinuous at $t=0$ . In this case, we use the fact that

${\left(\frac{dx\left(t\right)}{dt}|}_{t=0}=\left[x,\left({0}^{+}\right),-,x,\left({0}^{-}\right)\right]\delta \left(t\right)$

For example, if we integrate the right-hand side of [link] with $x\left({0}^{-}\right)=0$ and $x\left({0}^{+}\right)=1$ , we get the unit step function, $u\left(t\right)$ . Proceeding as before, we have

$\underset{s\to \infty }{lim}{\int }_{{0}^{-}}^{\infty }\frac{dx\left(t\right)}{dt}{e}^{-st}dt=\underset{s\to \infty }{lim}\left[s,X,\left(s\right),-,x,\left({0}^{-}\right)\right]$

The left-hand side of [link] can be written as

$\underset{s\to \infty }{lim}{\int }_{{0}^{+}}^{{0}^{-}}\left[x,\left({0}^{+}\right),-,x,\left({0}^{-}\right)\right]\delta \left(t\right){e}^{-st}dt+\underset{s\to \infty }{lim}{\int }_{{0}^{+}}^{\infty }\frac{dx\left(t\right)}{dt}{e}^{-st}dt$

From the sifting property of the unit impulse, the first term in [link] is

$\left[x,\left({0}^{+}\right),-,x,\left({0}^{-}\right)\right]$

while the second term is zero since in the limit, the real part of $s$ goes to infinity. Substituting these results into the left-hand side of [link] again leads to the initial value theorem, in [link] .

## The final value theorem

The Final Value Theorem allows us to determine

$\underset{t\to \infty }{lim}x\left(t\right)$

from $X\left(s\right)$ . Taking the limit as $s$ approaches zero in the derivative property gives

$\underset{s\to 0}{lim}{\int }_{{0}^{-}}^{\infty }\frac{dx\left(t\right)}{dt}{e}^{-st}dt=\underset{s\to 0}{lim}\left[s,X,\left(s\right),-,x,\left({0}^{-}\right)\right]$

The left-hand-side of [link] can be written as

${\int }_{{0}^{-}}^{\infty }\underset{s\to 0}{lim}\frac{dx\left(t\right)}{dt}{e}^{-st}dt={\int }_{{0}^{-}}^{\infty }\frac{dx\left(t\right)}{dt}dt=x\left(\infty \right)-x\left({0}^{-}\right)$

Substituting this result back into [link] leads to the Final Value Theorem

$x\left(\infty \right)=\underset{s\to 0}{lim}sX\left(s\right)$

which is only valid as long as the limit $x\left(\infty \right)$ exists.

 Property $y\left(t\right)$ $Y\left(s\right)$ Linearity $\alpha {x}_{1}\left(t\right)+\beta {x}_{2}\left(t\right)$ $\alpha {X}_{1}\left(s\right)+\beta {X}_{2}\left(s\right)$ Time Delay $x\left(t-\tau \right)$ $X\left(s\right){e}^{-s\tau }$ s-Shift $x\left(t\right){e}^{-at}$ $X\left(s+a\right)\right)$ Multiplication by $t$ $tx\left(t\right)$ $-\frac{dX\left(s\right)}{ds}$ Multiplication by ${t}^{n}$ ${t}^{n}x\left(t\right)$ ${\left(-1\right)}^{n}\frac{{d}^{n}X\left(s\right)}{d{s}^{n}}$ Convolution $x\left(t\right)*h\left(t\right)$ $X\left(s\right)H\left(s\right)$ Differentiation $\frac{dx\left(t\right)}{dt}$ $sX\left(s\right)-x\left({0}^{-}\right)$ $\frac{{d}^{2}x\left(t\right)}{d{t}^{2}}$ ${s}^{2}X\left(s\right)-sx\left({0}^{-}\right)-\frac{dx\left({0}^{-}\right)}{dt}$ $\frac{{d}^{n}x\left(t\right)}{d{t}^{n}}$ ${s}^{n}X\left(s\right)-{s}^{n-1}x\left({0}^{-}\right)-\sum _{k=2}^{n}{s}^{n-k}\frac{{d}^{k-1}x\left({0}^{-}\right)}{d{t}^{k-1}}$ Integration ${\int }_{\infty }^{t}x\left(\tau \right)d\tau$ $\frac{{\int }_{-\infty }^{{0}^{-}}x\left(\tau \right)d\tau }{s}+\frac{X\left(s\right)}{s}$ Initial Value Theorem $x\left({0}^{+}\right)=\underset{s\to \infty }{lim}sX\left(s\right)$ Final Value Theorem $x\left(\infty \right)=\underset{s\to 0}{lim}sX\left(s\right)$

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