Properties of the Laplace transform

Existence

As we have already seen, the Laplace transform exists for more functions than does the Fourier transform. Let \(f:\mathbb{R}_+ \rightarrow \mathbb{R}\) have a finite number of finite-magnitude discontinuites. If there can be found \(M, \alpha \in \mathbb{R}\) such that \[\begin{aligned} |f(t)| \le M e^{\alpha t}\quad \forall t \in \mathbb{R}_+ \end{aligned}\] then the transform exists (converges) for \(\sigma > \alpha\).

Note that this is a sufficient condition, not necessary. That is, there may be (and are) functions for which a transform exists that do not meet the condition above.

Linearity

The Laplace transform is a linear map. Let \(a,b \in \mathbb{R}\); \(f,g \in T\) where \(T\) is a set of functions of nonnegative time \(t\); and \(F, G\) the Laplace transform images of \(f, g\). The following identity holds: \[\begin{aligned} \mathcal{L}(a f(t) + b g(t)) = a F(s) + b G(s). \end{aligned}\]

Time-shifting

Shifting the time-domain function \(f(t)\) in time corresponds to a simple product in the \(s\)-domain Laplace transform image. Let the Laplace transform image of \(f(t)\) be \(F(s)\) and \(\tau\in\mathbb{R}\). The following identity holds: \[\begin{aligned} \mathcal{L}(f(t+\tau)) = e^{s\tau}F(s). \end{aligned}\]

Time-differentiation

Differentiating the time-domain function \(f(t)\) with respect to time yields a simple relation in the \(s\)-domain. Let \(F(s)\) be the Laplace transform image of \(f(t)\) and \(f(0)\) the value of \(f\) at \(t=0\). The following identity holds:¹ \[\begin{aligned} \mathcal{L}\frac{\diff f} {\diff t} = s F(s) - f(0). \end{aligned}\]

Time-integration

Similarly, integrating the time-domain function \(f(t)\) with respect to time yields a simple relation in the \(s\)-domain. Let \(F(s)\) be the Laplace transform image of \(f(t)\). The following identity holds:² \[\begin{aligned} \mathcal{L}\int_0^t f(\tau)\diff \tau = \frac{1} {s} F(s). \end{aligned}\]

Convolution

The convolution operator \(*\) is defined for real functions of time \(f, g\) by \[\begin{aligned} \label{eq:lap_conv} (f*g)(t) &\equiv \int_{-\infty}^\infty f(\tau) g(t-\tau)\, d\tau. \end{aligned}\] This too has a simple Laplace transform. Let \(F, G\) be the Laplace transforms of \(f, g\). The following identity holds: \[\begin{aligned} \mathcal{L}(f*g)(t) = F(s) G(s). \end{aligned}\]

Final value theorem

The final value theorem is a property of the Laplace transform. This theorem allows the computation of constant time-domain steady-state values from the frequency domain, which can be quite convenient when the inverse Laplace transform is elusive. Let \(f(t)\) have tranform \(F(s)\) and its time-derivative have an existing transform. If the limit in time exists, \[\begin{aligned} \lim_{t\rightarrow\infty} f(t) = \lim_{s\rightarrow 0} s F(s). \end{aligned}\] Note that if the steady-state of \(f(t)\) is not a constant (e.g. it is sinusoidal), the limit does not exist.