Universidade Federal do Rio Grande do Sul (UFRGS)
Programa de Pós-Graduação em Engenharia Civil (PPGEC)

PEC00025: Introduction to Vibration Theory¶

Class 03 - Laplace Transform¶

1. Introduction
2. Formal definition
3. Laplace transform of some basic functions
3.1. Unit step (Heaviside's) function
3.2. Unit impulse (Dirac's Delta) function
3.3. Sine and cosine functions
4. Translation theorems
5. Transform of derivatives
6. Transform of integrals
7. Convolution theorem
8. Assignment

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Prof. Marcelo M. Rocha, Dr.techn. (ORCID)
Porto Alegre, RS, Brazil

1. Introduction ¶

Whenever a function $f(t)$ is defined by means of a differential equation and a suitable set of boundary conditions, it may be convenient to transform the problem of finding $f(t)$, a function of an independent variable $t$ (time), into a new problem of finding $\bar{f}(s)$, function of a transformed independent variable $s$ (complex frequency). This linear operation may be generically expressed as:

$$\bar{f}(s) = \int_{a}^{b} K(s,t) \; f(t) \; dt$$

where the function $K(s,t)$ is called transform kernel. If the integral limits $a$ and $b$ are finites, the function $\bar{f}(s)$ is a finite transformation. Among many possibilities, we shall study the two most important kernels in our context of vibration analysis, namely the Laplace and the Fourier transforms. These transforms feature the possibility of being reversed in a unique way and can be used to find solutions of linear systems (differential equations) both algebraically or numerically.

In the particular case of Laplace Transform, we are concerned with the time domain solutions of the dynamic equilibrium equation for some special loading functions, such as the unit impulse and the unit step functions, which are of fundamental importance for understanding the dynamic behavior of structural systems.

2. Formal definition ¶

Let $f(t)$ be a function defined for all $t > 0$. The Laplace Transform of $f(t)$ is defined as:

$$\mathscr{L}\left\{ f(t) \right\} = \bar{f}(s) = \int_{0}^{\infty} e^{-st}\; f(t) \; dt$$

The transform kernel in this case is

$$K(s,t) = e^{-st}$$

with $s$ being a complex frequency $s = \sigma + i\omega$, with units in radians per second for $t$ in seconds. Observing that one of the integral bounds is infinite, the function $f(t)$ must fulfill some special conditions such that the integral convergence can be ensured. For instance, the function $f(t)$ must have a finite number of finite descontinuities and no infinite descontinuity (be piecewise continuous) along the integration interval. We will not give further details about these conditions, for there will be no issues regarding the functions we will be addressing here.

Although the Laplace transform can be reversed in a unique way, its inverse form cannot be expressed in a closed form as the direct transform. The transform of a given function and its inverse constitutes a so-called transform pair, usually represented as:

$$f(t) \Longleftrightarrow \bar{f}(s)$$

The most straightforward way of using Laplace transforms is by means of lookup tables or CAS (Computer Algebra Systems). However, some fundamental functions and theorems must receive special attention as follows.

3. Laplace transform of some basic functions ¶

3.1. Unit step (Heaviside's) function ¶

The unit step, also known as Heaviside's function, is defined as

$$H(t-\tau) = \left\{ \begin{array}{lll} 0 & \mbox{if} &t < \tau \\ 1 & \mbox{if} &t \geq \tau \end{array} \right .$$

and is depicted in the figure below (for $\tau = 0$):

By applying the Laplace transform to this function and adjusting integral limits gives:

$$\mathscr{L}\left\{ H(t-\tau) \right\} = \bar{H}(s) = \int_{\tau}^{\infty} e^{-st} \cdot 1 \cdot dt$$

and hence:

$$\bar{H}(s) = \frac {e^{-s\tau }}{s}$$

For the particular case where the discontinuity occurs for $t = \tau = 0$, it results:

$$\bar{H}(s) = \frac{1}{s}$$

The Heaviside's function usually represents a sudden change in a otherwise static load, which may cause some significant dynamic response.

3.2. Unit impulse (Dirac's Delta) function ¶

The unit impulse, also known as Dirac's Delta function, is defined as

$$\delta(t-\tau ) = \left\{ \begin{array}{lll} 0 & \mbox{if} &t \neq \tau \\ \infty & \mbox{if} &t = \tau \end{array} \right .$$

and is depicted in the figure below (for $\tau = 0$):

The main point about Dirac's function is that, although the function goes to infinity at $t = \tau$, the function has a finite integral:

$$\int_{-\infty}^{\infty} \delta(t-\tau) \; dt = 1$$

and this implies that for any function $f(t)$:

$$\int_{-\infty}^{\infty} f(t) \; \delta(t-\tau) \; dt = f(\tau)$$

It can be observed that the Dirac's Delta is not a function in the classical sense, but rather a distribution, like the Gaussian probability density function. In fact, Dirac's Delta can be understood as a Gaussian density function with standard deviation equal to zero and mean value equal to $\tau$ or, in order words, a deterministic distribution. From this viewpoint, the integral above corresponds to the definition of the expected value of $f(t)$, calculated as the function $f(\tau)$ at the deterministic time instant $t=\tau$.

By applying the Laplace transform to this function it results:

$$\mathscr{L}\left\{ \delta(t-\tau) \right\} = \bar{\delta}(s) = \int_{0}^{\infty} e^{-st} \; \delta(t-\tau) \; dt$$

and hence:

$$\bar{\delta}(s) = e^{-s\tau}$$

For the particular case where the discontinuity occurs for $t = \tau = 0$:

$$\bar{\delta}(s) = 1$$

The Dirac's Delta function usually represents an impulsive load with very short duration, which also may cause some significant dynamic response. By comparing the above plots for both the Heaviside's and the Dirac's Delta functions, it can be readily observed that they are related by integration or derivation as:

$$\begin{align*} H(t-\tau) &= \int \delta(t-\tau) \; dt \\ \delta(t-\tau) &= \frac{d H(t-\tau)}{dt} \end{align*}$$

We will come back later on this relation.

3.3. Sine and cosine functions ¶

The well known functions $\sin(\omega t)$ and $\cos(\omega t)$ can also undergo a Laplace transform, as shown in the following. The parameter $\omega$ is a frequency, usually in radians per second with time $t$ in seconds, as depicted below for $x = \omega t$.

Applying the Laplace transform to the sine function gives:

$$\mathscr{L} \left\{ \sin (\omega t) \right\} = \int_{0}^{\infty} e^{-st} \; \sin(\omega t) \; dt = \frac{\omega}{s^2 + \omega^2}$$

and similarly for the cosine function:

$$\mathscr{L} \left\{ \cos (\omega t) \right\} = \int_{0}^{\infty} e^{-st} \; \cos(\omega t) \; dt = \frac{s}{s^2 + \omega^2}$$

In vibration analysis, the sine and cosine functions are used to represent both loadings and system response.

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Note: Laplace transforms, both direct and inverse, are available at the HP Prime calculator operating in CAS mode. For instance, for cosine transform by typing laplace(cos(s*w),s) will result in s/(s^2 + w^2), while by typing invlaplace(s/(s^2 + w^2),s) will result in cos(s*|w|). Hence, this calculator is a nice replacement for Laplace transform lookup tables.

4. Translation theorems ¶

Translation theorems are available both in time and in frequency domain. Let us start by calculating the Laplace transform of a function $e^{\omega t} f(t)$:

$$\mathscr{L} \left\{ e^{\omega t} f(t) \right\} = \int_{0}^{\infty} e^{-st} \; e^{\omega t} f(t) \; dt = \int_{0}^{\infty} e^{-(s - \omega)t} \; f(t) \; dt$$

Hence:

$$\mathscr{L} \left\{ e^{\omega t} f(t) \right\} = \bar{f}(s - \omega)$$

which is the translation theorem in frequency domain. It means that a shift of $s$ by $\omega$ corresponds, in time domain, to multiplying the function $f(t)$ by the exponential function $e^{\omega t} f(t)$.

The other way around, let us replace the dummy variable $t$ by $\xi = t - \tau$ in the definition of Laplace transform. Recognizing that $d\xi = dt$ and that for $\xi = 0$ one has $t = \tau$ gives:

$$\int_{\tau}^{\infty} e^{-s(t - \tau)} \; f(t - \tau) \; dt = e^{s \tau} \; \int_{0}^{\infty} e^{-st} \; H(t - \tau) \; f(t - \tau) \; dt$$

and consequently:

$$\mathscr{L} \left\{ H(t - \tau) f(t - \tau) \right\} = e^{-s\tau} \; \bar{f}(s)$$

which is the translation theorem in time domain. It means that by multiplying the transformed function $\bar{f}(s)$ by $e^{-s\tau}$ corresponds to shifting the time domain by $\tau$ and zeroing all function values before the instant $\tau$.

It will be seen that the translation theorem in frequency domain is used to solve the equilibrium equation of linear damped systems.

5. Transform of derivatives ¶

The Laplace transform applied to derivatives is essential for solving differential equations. It is expressed as:

$$\mathscr{L} \left\{ \dot{f}(t) \right\} = \int_{0}^{\infty} e^{-st} \; \dot{f}(t) \; dt$$

where the dot notation is used for time derivative:

$$\dot{f}(t) = \frac{df}{dt}$$

This can be solved through integration by parts by defining:

$$\begin{array}{ll} u &= e^{-st} \hspace{2cm} &dv = \dot{f}(t) \; dt \\ du &= -s e^{-st} dt &v = f(t) \end{array}$$

and replacing in:

$$\int u \; dv = uv - \int v \; du$$

to get:

$$\int_{0}^{\infty} e^{-st} \; \dot{f}(t) \; dt = -f(0) + s \int_{0}^{\infty} e^{-st} \; f(t) \; dt$$

and hence:

$$\mathscr{L} \left\{ \dot{f}(t) \right\} = -f(0) + s \bar{f}(s)$$

where $f(0)$ is the initial value of $f(t)$ in time domain. For solving the dynamic equilibrium equation of linear systems, the second time derivative of $f(t)$ will also be necessary. Applying again the derivation rule results:

$$\mathscr{L} \left\{ \ddot{f}(t) \right\} = -s f(0) - \dot{f}(0) + s^2 \bar{f}(s)$$

Time derivatives of higher order can be calculated but will not be necessary in the present context.

6. Transform of integrals ¶

Although it is not necessary for solving the basic form of the dynamic equilibrium equation, the same technique used in the previous section can be used for transforming integrals as:

$$\mathscr{L} \left\{ \int_0^t f(\tau) \; d\tau \right\} = \int_{0}^{\infty} e^{-st} \; \int_0^t f(\tau) \; d\tau \; dt$$

This can be also be solved through integration by parts by defining:

$$\begin{array}{ll} u &= \int_0^t f(\tau) \; d\tau \hspace{2cm} &dv = e^{-st} \; dt \\ du &= f(t) \; dt &v = -e^{-st} / s \end{array}$$

and replacing in:

$$\int u \; dv = uv - \int v \; du$$

to get:

$$\int_{0}^{\infty} e^{-st} \; \int_{0}^{t} f(\tau) \; d\tau \; dt = 0 + \frac{1}{s} \int_{0}^{\infty} e^{-st} \; f(t) \; dt$$

and hence:

$$\mathscr{L} \left\{ \int_0^t f(\tau) \; d\tau \right\} = \frac{\bar{f}(s)}{s}$$

It can be observed that integration or derivation in frequency domain is simply a matter of dividing or multiplying by complex frequency $s$, respectivelly. This can be exemplified by the Heaviside's and Dirac's Delta function by observing that:

$$\bar{H}(s) = \frac{\bar{\delta}(s)}{s}$$

in agreement to the fact that:

$$H(t-\tau) = \int \delta(t-\tau) \; dt$$

as stated at the end of section 3.2.

7. Convolution theorem ¶

The convolution between two functions $f(t)$ e $g(t)$ is defined as the integral:

$$f(t) * g(t) \hspace{3mm} = \int_{0}^{t} f(\tau) g(t - \tau) \; d\tau$$

It can be easily verified that this operation is both commutative and associative:

$$\begin{align*} f(t) * g(t) &= g(t) * f(t) \\ \left[ f(t) * g(t) \right] * h(t) &= f(t) * \left[ g(t) * h(t) \right] \end{align*}$$

The convolution operation is graphically depicted in the figure below.

Applying the Laplace transform to the convolution gives:

$$\mathscr{L} \left\{ f(t) * g(t) \right\} = \int_{0}^{\infty} e^{-st} \int_{0}^{t} f(\tau) g(t - \tau) \; d\tau \; dt$$

The upper limit for the inner integral can be changed by multiplying the integrand by the Heaviside's function:

$$g(t - \tau) H(t - \tau) = \left\{ \begin{array}{lll} g(t - \tau) & \mbox{if} &\tau \leq t \\ 0 & \mbox{if} &\tau > t \end{array} \right .$$

what gives:

$$\mathscr{L} \left\{ f(t) * g(t) \right\} = \int_{0}^{\infty} e^{-st} \int_{0}^{\infty} f(\tau) g(t - \tau) H(t - \tau) \; d\tau \; dt$$

By re-ordering the terms and introducing a variable change $\xi = t - \tau$ gives:

$$\mathscr{L} \left\{ f(t) * g(t) \right\} = \int_{0}^{\infty} f(\tau) \int_{0}^{\infty} e^{-s(\tau + \xi)} g(\xi) H(\xi) \; d\xi \; d\tau$$

Recognizing now that $H(\xi) = 1$ for all domain allows the integration variables to be separated:

$$\mathscr{L} \left\{ f(t) * g(t) \right\} = \int_{0}^{\infty} e^{-s\tau} f(\tau) \; d\tau \cdot \int_{0}^{\infty} e^{-s\xi} g(\xi) \; d\xi$$

what finally implies that:

$$\mathscr{L} \left\{ f(t) * g(t) \right\} = \bar{f}(s) \; \bar{g}(s)$$

This is the Convolution Theorem for Laplace transform. It implies that whenever a Laplace transform can be expressed as the product of two functions in the complex frequency $s$, the inverse Laplace of this product is the convolution between the inverse transform of each single function. This theorem will be used to solve the differential equilibrium equations of dynamic systems subjected to forced vibration.

8. Assignment ¶