4.1.1. The Laplace Transform¶

4.1.1.1. Definition¶

Let \(x(t)\) be a continous time signal. Its one-sided Laplace transform \(X(s)\) is defined with:

\[X(s) = \int_{0}^{\infty} x(t) e^{-st} dt\]

The two-sided or bilateral Laplace transform integrates from \(-\infty\) to \(\infty\). Using the one-sided Laplace transform is equivalent with transforming causal signals and systems, i.e. a signal such that \(x(t)=0\) for \(x<0\). And most often we assume that \(x(0)=0\).

Just like we have done for the Z-transform we have to specify the region of convergence (ROC) for the Laplace transform. Existence of the integral defining the Laplace transform in general is not be guaranteed for all points in the complex plane.

A signal \(x(t)\) and its Laplace transform \(X(s)\) are denoted as the transform pair:

\[x(t) \LTright X(s)\]

For completeness we also give the inverse Laplace transform:

\[x(t) = \frac{1}{2\pi j} \lim_{T\rightarrow\infty}\int_{\gamma-jT}^{\gamma+jT} e^{st} X(s) ds\]

where \(\gamma\) is a real number so that the contour path of integration is in the region of convergence of \(X(s)\).

The inverse transform is not much use in practice. Most often the inverse transform is deduced from known transform pairs and the properties of the Laplace transform.

The Laplace transform is important because:

Convolution in the continuous time domain corresponds with multiplication in the S-domain.

\[x(t) \ast y(t) = X(s) Y(s)\]

The Laplace transform shares this property with the Fourier transform.
Any linear partial differential equation with constant coefficients is transformed into an algebraic equation using the Laplace transform. These PDE’s are often occurring in physics to model time dependent systems. Control theory for these type of systems is highly dependent on the Laplace transform. Another example, more close to home, is modeling frequency dependent components in electronics and building analog filters with these components.

Table 4.1 **Laplace Transform**¶
Synthesis Equation	Analysis Equation
\[x(t) = \frac{1}{2\pi j} \lim_{T\rightarrow\infty}\int_{\gamma-jT}^{\gamma+jT} e^{st} X(s) ds\]	\[X(s) = \int_{0}^{\infty} x(t) e^{-st} dt\]

4.1.1.2. Eigenfunctions of an LTI system¶

The function \(x(t)=e^{st}\) is an eigenfunction of an LTI system. When we feed this as input to an LTI system with impulse response \(h(t)\) we get the response \(y(t)\) as the convolution

\[\begin{split}y(t) &= x(t) \ast h(t)\\ &= \int_{-\infty}^{\infty} x(t-u)h(u)du\\ &= \int_{-\infty}^{\infty} e^{s(t-u)}h(u)du\\ &= e^{st} \int_{-\infty}^{\infty} e^{-su}h(u)du\\ &= H(s) e^{st}\end{split}\]

The complex variable \(s\) can be written as \(s=\sigma + j\w\) (both \(\sigma\) and \(\w\) real) leading to eigenfunctions of the form \(e^{st} = e^{(\sigma+j\w)t} = e^{\sigma t} \left(\cos(\w t+ j\sin(\w t)\right)\). So the complex exponentials from Fourier analysis are replaced with decaying (\(\sigma<0\)) or increasing (\(\sigma>0\)) complex exponentials.

4.1.1.3. The Laplace Transform and the Fourier Transform¶

Remember the continuous time Fourier transform (CTFT):

\[X(\w) = \int_{-\infty}^{\infty} x(t) e^{-j\w t}\]

Comparing this with the definition of the Laplace transform \(X(s)\) we see that

\[X(s)\Bigr|_{s=j\w} = X(\w)\]

Note that we don’t write \(X(j\w)=X(\w)\) because there is no context that makes it clear that the first \(X\) is the Laplace transform and the second \(X\) is the Fourier transform. This mathematical rather sloppy way of distinguishing functions by their argument is common practice in computer science but not as common in math.

Furthermore let’s look at the Laplace transform in more detail. The definition can be rewritten as:

\[\begin{split}\LT\{x(t)\} = X(s) = X(\sigma + j\w) &= \int_{-\infty}^{\infty} x(t) e^{-st}\\ &= \int_{-\infty}^{\infty} x(t) e^{-(\sigma + j\w) t}\\ &= \int_{-\infty}^{\infty} e^{-\sigma t} x(t) e^{-j\w t}\\ &= \FT\left\{ e^{-\sigma t} x(t) \right\}\end{split}\]

So along the vertical imaginaire axis \(\sigma=0\) we find the Fourier transform in the s-domain, and along the vertical lines with real value \(\sigma\) we find the Fourier transform of the original function after multiplication with \(e^{-\sigma t}\).