9 Laplace Transform

The Fourier Transform is only possible for well-behaved functions that are Lebesgue integrable, i.e. \(f(x) \in L^1[(-\infty, \infty)]\). This excludes functions like the exponential \(\mathrm{e}^{\lambda t}\) or the Heaviside function.

In order to tackle these badly behaved functions we can use the Laplace transform, which is basically a one sided Fourier-like transform.

Definition 9.1 (The Laplace Transform) A function \(f\,:\, [0, \infty) \to \mathbb{R}\) is called Laplace-transformable if \[ F(s) = \mathcal{L}\{f(t)\} := \int_{0}^{\infty} f(t)\mathrm{e}^{-s t}\, \mathrm{d}t \] exists for all \(s\in\mathrm{R}\). In this case we call \(F(s) \equiv \mathcal{L}\{f(t)\}\) the Laplace transform of \(f(t)\).

The inverse Laplace transform is defined as \[ f(t) = \mathcal{L}^{-1}\{F(s)\} = \frac{1}{2 \pi \mathrm{i}} \int_{\gamma - \mathrm{i} \infty}^{\gamma + \mathrm{i} \infty} F(s)\mathrm{e}^{s t}\, \mathrm{d}s \] for the complex valued \(s=\gamma + \mathrm{i} \omega\).

It is quite common to use the capitalized function name for the Laplace transform, as we did here.

To give a better understanding of theses formula we follow the derivation from (Brunton and Kutz 2022, sec. 2.5).

Let us consider the function \(f(t)=\mathrm{e}^{\lambda t}\). As mentioned before we can not directly use the Fourier transform as the function is not bounded. In order to trap the function for \(t\to\infty\) we multiply by \(\mathrm{e}^{-\gamma t}\) where \(\gamma\) is damping more that \(f\) grows. In order to handle \(t\to-\infty\) we multiply by the Heaviside function \[ H(t) = \begin{cases} 0, &t \leq 0,\\ 1, &t > 0, \end{cases} \] and transform it into a one-sided function as in definition Definition 9.1 required. We end up with the function \[ \underline{f}(t) = f(t) \mathrm{e}^{-\gamma t}H(t) = \begin{cases} 0, &t \leq 0,\\ f(t)\mathrm{e}^{-\gamma t}, &t > 0. \end{cases} \] Taking the Fourier transform we get \[ \begin{aligned} \hat{\underline{f}}(\omega) = \mathcal{F}\{\underline{f}(t)\} &= \int_{-\infty}^\infty \underline{f}(t)\mathrm{e}^{-\mathrm{i}\omega t}\, \mathrm{d}t = \int_{0}^\infty f(t)\mathrm{e}^{-\gamma t}\mathrm{e}^{-\mathrm{i}\omega t}\, \mathrm{d}t = \\ &= \int_{0}^\infty f(t)\mathrm{e}^{-(\gamma + \omega) t}\, \mathrm{d}t = \int_{0}^\infty f(t)\mathrm{e}^{-s t}\, \mathrm{d}t = \\ &= \mathcal{L}\{f(t)\} = F(s) \end{aligned} \] giving rise to the Laplace transform.

To get the inverse we start with the inverse Fourier transform of \(\hat{\underline{f}}(\omega)\) \[ \underline{f}(t) = \mathcal{F}^{-1}\{\hat{\underline{f}}(\omega)\} = \frac{1}{2\pi} \int_{-\infty}^\infty \hat{\underline{f}}(\omega) \mathrm{e}^{\mathrm{i}\omega t}\, \mathrm{d}\omega. \] Multiply both sides with \(\mathrm{e}^{\gamma t}\) we get \[ \begin{aligned} f(t)H(t) = \underline{f}(t)\mathrm{e}^{\gamma t} &= \frac{1}{2\pi} \int_{-\infty}^\infty \mathrm{e}^{\gamma t} \hat{\underline{f}}(\omega) \mathrm{e}^{\mathrm{i}\omega t}\, \mathrm{d}\omega = \\ &= \frac{1}{2\pi} \int_{-\infty}^\infty \hat{\underline{f}}(\omega) \mathrm{e}^{(\gamma + \mathrm{i}\omega) t}\, \mathrm{d}\omega \end{aligned} \] By a change of variables \(s=\gamma + \mathrm{i}\omega\) we get \(\mathrm{d}\omega = \tfrac{1}{\mathrm{i}} \mathrm{d}s\) and \[ f(t)H(t) = \frac{1}{2 \pi \mathrm{i}} \int_{\gamma - \mathrm{i} \infty}^{\gamma + \mathrm{i} \infty} F(x)\mathrm{e}^{s t}\, \mathrm{d}s = \mathcal{L}^{-1}\{F(s)\}. \]

Note

Form this derivation we see that the Laplace transform is a generalized Fourier transform for a broader spectrum of functions.

Sometimes the Laplace transform is simpler than the Fourier transform. In particular this is the case for the Dirac delta function that has infinitely many frequencies in Fourier domain but is constant \(1\) in Laplace domain. We will use this to compute the impulse response of systems with forcing.

A lot of the properties we saw for the Fourier transform carry over to the Laplace transform.

Linearity \[ \mathcal{L}\{\alpha f(t) + \beta g(t)\} = \alpha \mathcal{L}\{f(t)\} + \beta \mathcal{L}\{g(t)\} = \alpha F(s)+ \beta G(s). \]
Conjugation \[ \mathcal{L}\{\overline{f(t)}\} = \overline{\hat{f}(\overline{s})}. \]
Scaling, for \(\alpha \geq 0\) \[ \mathcal{L}\{f(\alpha t)\} = \frac{1}{\alpha}F\left(\frac{s}{\alpha}\right). \]
Drift in time, for \(a\in\mathbb{R}\) \[ \mathcal{L}\{f(t - a)H(t - a)\} = \mathrm{e}^{-a s}F(s) \] and \[ \mathcal{L}\{f(t)H(t - a)\} = \mathrm{e}^{-a s}\mathcal{L}\{f(t+a)\}. \]
Drift in frequency, for \(a\in\mathbb{R}\) \[ \mathrm{e}^{a t} \mathcal{L}\{f(t)\} = F(s - a). \]
Derivative in time \[ \mathcal{L}\{\partial_t f(t)\} = s F(s) - f(0) \] We are going to prove this by going through the lines \[ \begin{aligned} \mathcal{L}\left\{\frac{d}{d\,t}f(t)\right\} &= \int_{0}^\infty f'(t)\mathrm{e}^{-s t}\, \mathrm{d}t \\ &= \left[f(t)\mathrm{e}^{-s t}\right]_{0}^\infty - \int_{0}^\infty f(t) (- s\mathrm{e}^{-s t})\, \mathrm{d}t \\ &= -f(0) + s \mathcal{L}\{f(t)\} \end{aligned} \] For higher derivatives we get \[ \mathcal{L}\{\partial_t^n f(t)\} = s^n F(s) - \sum_{k=1}^n s^{n-k} f^{(k-1)}(0). \]
Integral in time \[ \mathcal{L}\left\{\int_0^t f(\tau)\, \mathrm{d}\tau\right\} = \frac{1}{s} F(s). \]
Derivative in frequency \[ \mathcal{L}\{t^n f(t)\} = (-1)^n \partial_\omega^n F(s) \]
Integral in frequency \[ \mathcal{L}\left\{\frac{1}{t} f(t)\right\} = \int_s^\infty F(u)\, \mathrm{d}u. \]
The convolution of two functions is defined as \[ (f \ast g)(t) = \int_{0}^{\infty}f(\tau) g(t - \tau)\, \mathrm{d}\tau, \] and for the Laplace transform \[ \mathcal{L}\{(f \ast g)(t)\} = F(s) \cdot G(s). \]

Important

The following examples are mostly from Meyberg and Vachenauer (1992).

Example 9.1 (Laplace transform of some basic functions)

For the constant \(f(t) = 1\) function we get \[ \mathcal{L}\{f(t)\} = \frac{1}{s},\quad s>0. \]
For the exponential \(f(t) = \mathrm{e}^{\lambda t}\) we get \[ \mathcal{L}\{f(t)\} = \frac{1}{s- \lambda},\quad s>\lambda\in\mathbb{R}. \]
For the cosine \(f(t) = \cos{\omega t}\) we get \[ \mathcal{L}\{f(t)\} = \frac{s}{s^2 + \omega^2},\quad s>0. \]
For the sine \(f(t) = \sin{\omega t}\) we get \[ \mathcal{L}\{f(t)\} = \frac{\omega}{s^2 + \omega^2},\quad s>0. \]

Example 9.2 (Laplace transform of a step function) For a rectangle with width \(a\geq 0\) and height \(1\) we get \[ f(t) = H(t) - H(t-a) \] we get \[ \mathcal{L}\{f(t)\} = \frac{1 - \mathrm{e}^{-a s}}{s} \]

We can extend this to a step function \[ g(t) = \sum_{n=0}^\infty H(t-na) \] and we get \[ \mathcal{L}\{g(t)\} = \frac{1}{s (1- \mathrm{e}^{-a s})} \]

Example 9.3 (Cam drive) A cam drive has an elastic and lightly damped punch \(S\) with eigenfrequency \(\omega = 2\).

We excite the system with a half-wave sine of half the eigenfrequency. \[ \begin{aligned} x_A(t) &= \frac{1}{2}(\sin t + |\sin t|) \\ &= \begin{cases} \sin(t), & 2 n \pi \leq t \le (2n+1)\pi,\\ 0 , & (2n+1)\pi \leq t \le (2n + 2)\pi, \end{cases} \quad \text{for}\quad n = 0, 1, 2, \ldots \\ &= \sum_{n=0}^\infty H(t - n \pi) \sin(t-n\pi). \end{aligned} \]

For the excitement of \(S\) the initial value problem \[ \ddot{x} + 4 x = x_A(t), \quad x(0) = \dot{x}(0) = 0, \] is given.

Via the Laplace transform we can solve for \(X(s)\) in \[ s^2 X(s) + 4 X(s) = \sum_{n=0}^\infty \frac{1}{1+s^2} \mathrm{e}^{-n \pi s} \] and obtain \[ X(s) = \frac{1}{(1+s^2)(s^2 + 4)} \sum_{n=0}^\infty \mathrm{e}^{-n \pi s}. \]

Via the inverse Laplace transform we can get back to \(x(t)\) \[ x(t) = \frac{1}{6}(\sin(t) + |\sin(t)|)- \frac{\sin(2t)}{6}\sum_0^\infty H(t-n\pi) \] and without the heavy side function \[ x(t) = \begin{cases} \frac{1}{3} \sin (t) - \frac{2n+1}{6} \sin(2t), & 2 n \pi \leq t \le (2n+1)\pi, \\ - \frac{n+1}{3} \sin(2t), &(2n+1)\pi \leq t \le (2n + 2)\pi. \end{cases} \]

Show the code for the figure

import numpy as np
import matplotlib.pyplot as plt
%config InlineBackend.figure_formats = ["svg"]

heavy_side = lambda t: np.ones_like(t) * (t>=0)
g = lambda n, t: (np.heaviside(t - 2 * n * np.pi, 1) - \
                  np.heaviside(t - (2 * n + 1) * np.pi, 1))
h = lambda n, t: (np.heaviside(t - (2 * n + 1) * np.pi, 1) - \
                  np.heaviside(t - (2 * n + 2) * np.pi, 1))

xa = lambda t: 1 / 2 * (np.sin(t) + np.abs(np.sin(t)))

fun1 = lambda n, t: (1 / 3 * np.sin(t) - (2 * n + 1) / 6 * \
                     np.sin(2 * t)) * g(n, t)
fun2 = lambda n, t:-1 * (n + 1) / 3 * np.sin(2 * t) * h(n, t)
funn = lambda n, t: fun1(n, t) + fun2(n, t) 
fun = lambda t: funn(0, t) + funn(1, t) + funn(2, t) + funn(3, t)

t = np.linspace(0, 8 * np.pi, 1024)
i = np.argmax(funn(0, t))
j = np.argmax(funn(3, t))
k = (funn(3, t[j]) - funn(0, t[i])) / (t[j] - t[i])
line = lambda t: k * (t - t[i]) + funn(0, t[i])

plt.plot(t, xa(t), "--", label=r"$x_A$")
plt.plot(t, fun(t), label=r"$x$")
plt.plot(t, line(t), "k:")
plt.xlabel(r"t")
plt.ylabel(r"x")
plt.legend()
plt.gca().set_aspect(8*np.pi / (3 * 3.0))
plt.ylim([-1.5, 1.5])

Figure 9.2: Exciting function and response, we indicate the growth rate as a dotted line.

Example 9.4 (Electric Circuit) For the electric components \(R\) (resistor), \(C\) capacitor, and \(L\) coil we can use the Laplace transform to show how voltage \(u(t)\) and current \(i(t)\) is transformed.

We need to assume that \(i(0)=0\) than we get

Table 9.1: Laplace transform for common electrical components

component	\(u(t)\) and \(i(t)\)	Laplace transform
R	\(u_R(t) = R i(t)\)	\(U_r(s) = R I(s)\)
C	\(u_C(t) = \frac{1}{x} \int_0^t i(\tau)\, \mathrm{d}\tau\)	\(U_C(s) = \frac{1}{sC}I(s)\)
L	\(u_L(t) = L \frac{\mathrm{d} i}{\mathrm{d} t}\)	\(U_L(s) = sL I(s)\)

An Ohm’s law applies for \[ U(s) = Z(s) I(s), \quad Z(s) \in \{R, \frac{1}{Cs}, Ls\} \]

For a RCL electric circuit

we have \[ u(t) = u_R(t) + u_c(t) + u_L(t) \] and we get for the Laplace transformed current \(I(s)\) \[ I(s) = \frac{U(s)}{R+\frac{1}{sC}+ sL} = \frac{s C U(s)}{s^2 LC + s RC + 1} = Z(s) U(s) \] and \[ i(t) = \mathcal{L}^{-1}\{z(t)\} \ast u(t). \]

Example 9.5 (The Linear Time Invariant transfer function of LTI Systems) In signal and control theory the transfer function \(H(s)\) is used the describe a system. If we assume that at \(t=0\) we have an initial state (zero state) we get the relation between input signal \(x(t)\) and output signal \(y(t)\) in the Laplace space as \[ Y(s) = H(s) X(s) \] and consequently \[ y(t) = h(t) \ast x(t) \] where \(h(t)= \mathcal{L}^{-1}\{H(s)\}\) is called the impulse response.

Figure 9.4: Linear Time invariant System

The relation between \(x\) and \(y\) can often be described via a linear differential equation with constant coefficients. In these cases \[ \alpha_n y^{(n)} + \alpha_{n-1} y^{(n-1)} + \cdots + \alpha_1 \dot{y} + \alpha_0 y = x(t) \] and for \(y(0) = \dot{y}(0) = \cdots = y^{(n)}(0) = 0\) we get \[ H(s) = \frac{1}{\alpha_n s^n + \alpha_{n-1}s^{n-1} + \cdots \alpha_1 s + \alpha_0}. \]

We can rewrite this as \[ \alpha_n \left( s^n H(s) - \frac{1}{\alpha_n} \right) + \alpha_{n-1}s^{n-1} H(s) + \cdots + \alpha_1 s H(s) + \alpha_0 H(s) = 0 \] and with the properties of the Laplace transform we get \[ \alpha_n h^{(n)} + \alpha_{n-1} h^{(n-1)} + \cdots + \alpha_1 \dot{h} + \alpha_0 h = 0 \] with \(h(0) = \dot{h}(0) = \cdots = h^{(n-2)}(0) = 0\), and \(h^{(n-1)}(0) = \tfrac{1}{\alpha_n}\).

This tells us the answer to the zero input signal \(x(t) = 0\) is the solution of above initial value problem with initial state as an impulse of size \(\tfrac{1}{\alpha_n}\).

In terms of the Dirac-delta function \(\delta(t)\) the function \(h(t)\) is the answer to the initial value problem \[ \alpha_n h^{(n-1)} + \alpha_{n-1} h^{(n-1)} + \cdots + \alpha_1 \dot{h} + \alpha_0 h = \delta(t). \]

Example 9.6 (The RC low pass filter) The RC low pass filter

has the transfer function \[ H(s) = \frac{1}{sRC + 1} \] and the impulse response \[ h(t) = \frac{1}{RC}\mathrm{e}^{-\frac{t}{RC}}. \]

Show the code for the figure

import numpy as np
import matplotlib.pyplot as plt
%config InlineBackend.figure_formats = ["svg"]

RC = 1 
fun = lambda t: 1 / RC * np.exp(-t / RC) * np.heaviside(t, 1)
t = np.linspace(-0.5, 8.5, 1024)

plt.plot(t, fun(t))
plt.text(-0.4, 0.94, r"$\frac{1}{RC}$")
plt.xlabel(r"t")
plt.ylabel(r"y")
plt.gca().set_aspect(9 / (3 * 1.0))

Figure 9.6: Impulse response for the first order RC low pass filter

If we extend this to a second order low pass filter Figure 9.5 (b) we get \[ H(s) = \frac{1}{(sRC + 1)^2} \] and the impulse response \[ h(t) = \frac{1}{(RC)^2}t\mathrm{e}^{-\frac{t}{RC}}. \]

Show the code for the figure

import numpy as np
import matplotlib.pyplot as plt
%config InlineBackend.figure_formats = ["svg"]

RC = 1 
fun = lambda t: 1 / RC**2 * t * np.exp(-t / RC) * np.heaviside(t, 1)
t = np.linspace(-0.5, 8.5, 1024)

plt.plot(t, fun(t))
plt.text(-0.5, 0.94/(np.exp(1)), r"$\frac{1}{RC \mathrm{e}}$")
plt.text(0.8, -0.05, r"$RC$")
plt.plot([-0.1, 1, 1], [1/np.exp(1), 1/np.exp(1), 0], "k:")
plt.xlabel(r"t")
plt.ylabel(r"y")
plt.gca().set_aspect(9 / (3 * 0.37))

Figure 9.7: Impulse response for the first order RC low pass filter