Laplace Transform

Intro

The $\mathcal{L}aplace$ transform is able to convert a real-valued function into a function with a complex variable $s$ . Such a real-valued function can for example be an input $x(t)$ . The complex variable $s$ can be written by $s = \sigma + i\omega$ . Which essentially stands for the complex frequency, it can also be called the s-domain.

The $\mathcal{L}aplace$ transform is often useful in parts of control engineering where computations become more easier in the s-domain. Namely, the $\mathcal{L}aplace$ transform is able to turn an ordinary differential equation into an algebraic equation. Secondly, it is convenient to transform a convolution into a multiplication. Both are shown in the article about transfer functions.

The official way to compute a $\mathcal{L}aplace$ transform is by the use of the following integral;

\begin{equation} \mathcal{L}\{f(t)\} = \int^{\infty}_{0}f(t)e^{-st}dt \end{equation}

However, for using the $\mathcal{L}aplace$ transform it often suffices to use the lookup table below. The reason for this is that the computation of the integral takes in general quite some time, which results in the lookup table being much faster. A more in-depth explanation of the use of this integral is shown in Laplace Integral. For now, we will simply use the lookup table. In this table, the most encountered functions, not all, are shown with their time-domain function and the $\mathcal{L}aplace$ transform of it.

Time Domain $f(t)$	Frequency Domain $\mathcal{L}\{f(t)\}$
$1$	$\frac{1}{s}$
$\delta(t)$	$1$
$\delta(t-\tau)$	$e^{-\tau s}$
$t^n$	$\frac{n!}{s^{n+1}}$
$t^ne^{-\alpha t}$	$\frac{n!}{(s+\alpha)^{n+1}}$
$e^{\alpha t}$	$\frac{1}{s-\alpha}$
$sin(\omega t)$	$\frac{\omega}{s^2 + \omega^2}$
$e^{\alpha t}sin(\omega t)$	$\frac{\omega}{(s-\alpha)^2 + \omega^2}$
$cos(\omega t)$	$\frac{s}{s^2 + \omega^2}$
$e^{\alpha t}cos(\omega t)$	$\frac{s-\alpha}{(s-\alpha)^2 + \omega^2}$
$...$	$...$

Examples Lookup Table

Let's say we want to apply the transform to the following function in time domain;

\begin{equation} \begin{aligned} f(t) &= e^{-2t}\\ F(s) &= \frac{1}{s+2} \end{aligned} \end{equation}

A combination of different components from the table is also possible. Let's take the same exponent as in the previous example and multiply it with a sinusoid.

\begin{equation} \begin{aligned} f(t) &= e^{-2t}\cdot sin(5t)\\ F(s) &= \frac{1}{s+2} \cdot \frac{5}{s^2 + 5^2} \\ &= \frac{5}{(s+2)(s^2 + 5^2)} \end{aligned} \end{equation}

In this article, we introduced the $\mathcal{L}aplace$ transform, explained why it is useful, and showed how to compute it via a lookup table. As there can be noticed, by using the lookup table the transform can be executed rather quickly. In further articles, we will elaborate on the inverse $\mathcal{L}aplace$ transform in order to go from the s-domain back to the time-domain.

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Intro

Examples Lookup Table

More about Laplace

[1]

Laplace Transform

[2]

Inverse Laplace Transform

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