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

Differentiation

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

this can be integrated by parts:

which gives

therefore we have,

Higher order derivatives

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

it follows that

which leads to

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

where it should be understood that

Integration

Let

it follows that

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

therefore

but since

we have

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

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

it follows that

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

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

Taking the limit $s\to \infty$

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

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

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

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

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

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

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

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

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

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

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