# Laplace transform: Wikis

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In mathematics, the Laplace transform is a widely used integral transform. It has many important applications in mathematics, physics, engineering, and probability theory.

The Laplace transform is related to the Fourier transform, but whereas the Fourier transform resolves a function or signal into its modes of vibration, the Laplace transform resolves a function into its moments. Like the Fourier transform, the Laplace transform is used for solving differential and integral equations. In physics and engineering, it is used for analysis of linear time-invariant systems such as electrical circuits, harmonic oscillators, optical devices, and mechanical systems. In this analysis, the Laplace transform is often interpreted as a transformation from the time-domain, in which inputs and outputs are functions of time, to the frequency-domain, where the same inputs and outputs are functions of complex angular frequency, in radians per unit time. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.

Denoted $\displaystyle\mathcal{L} \left\{f(t)\right\}$, it is a linear operator on a function f(t) (original) with a real argument t (t ≥ 0) that transforms it to a function F(s) (image) with a complex argument s. This transformation is essentially bijective for the majority of practical uses; the respective pairs of f(t) and F(s) are matched in tables. The Laplace transform has the useful property that many relationships and operations over the originals f(t) correspond to simpler relationships and operations over the images F(s).[1]

## History

The Laplace transform is named in honor of mathematician and astronomer Pierre-Simon Laplace, who used the transform in his work on probability theory. From 1744, Leonhard Euler investigated integrals of the form

$z = \int X(x) e^{ax}\, dx\text{ and }z = \int X(x) x^A \, dx,$

— as solutions of differential equations but did not pursue the matter very far.[2] Joseph Louis Lagrange was an admirer of Euler and, in his work on integrating probability density functions, investigated expressions of the form

$\int X(x) e^{- a x } a^x\, dx,$

— which some modern historians have interpreted within modern Laplace transform theory.[3][4] These types of integrals seem first to have attracted Laplace's attention in 1782 where he was following in the spirit of Euler in using the integrals themselves as solutions of equations.[5] However, in 1785, Laplace took the critical step forward when, rather than just looking for a solution in the form of an integral, he started to apply the transforms in the sense that was later to become popular. He used an integral of the form:

$\int x^s \phi (s)\, dx,$

— akin to a Mellin transform, to transform the whole of a difference equation, in order to look for solutions of the transformed equation. He then went on to apply the Laplace transform in the same way and started to derive some of its properties, beginning to appreciate its potential power.[6]

Laplace also recognised that Joseph Fourier's method of Fourier series for solving the diffusion equation could only apply to a limited region of space as the solutions were periodic. In 1809, Laplace applied his transform to find solutions that diffused indefinitely in space.[7]

## Formal definition

The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by:

$F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt.$

The parameter s is a complex number:

$s = \sigma + i \omega, \,$ with real numbers σ and ω.

The meaning of the integral depends on types of functions of interest. For functions that decay at infinity or are of exponential type, it can be understood as a (proper) Lebesgue integral. However, for many applications it is necessary to regard it as a conditionally convergent improper integral at ∞. Still more generally, the integral can be understood in a weak sense, and this is dealt with below.

One can define the Laplace transform of a finite Borel measure μ by the Lebesgue integral[8]

$(\mathcal{L}\mu)(s) = \int_{[0,\infty)} e^{-st}d\mu(t).$

An important special case is where μ is a probability measure or, even more specifically, the Dirac delta function. In operational calculus, the Laplace transform of a measure is often treated as though the measure came from a distribution function ƒ. In that case, to avoid potential confusion, one often writes

$(\mathcal{L}f)(s) = \int_{0^-}^\infty e^{-st}f(t)\,dt$

where the lower limit of 0 is short notation to mean

$\lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\infty.$

This limit emphasizes that any point mass located at 0 is entirely captured by the Laplace transform. Although with the Lebesgue integral, it is not necessary to take such a limit, it does appear more naturally in connection with the Laplace–Stieltjes transform.

### Probability theory

In pure and applied probability, the Laplace transform is defined by means of an expectation value. If X is a random variable with probability density function ƒ, then the Laplace transform of ƒ is given by the expectation

$(\mathcal{L}f)(s) = E\left[ e^{-sX} \right]. \,$

By abuse of language, one often refers instead to this as the Laplace transform of the random variable X. Replacing s by −t gives the moment generating function of X. The Laplace transform has applications throughout probability theory, including first passage times of stochastic processes such as Markov chains, and renewal theory.

### Bilateral Laplace transform

When one says "the Laplace transform" without qualification, the unilateral or one-sided transform is normally intended. The Laplace transform can be alternatively defined as the bilateral Laplace transform or two-sided Laplace transform by extending the limits of integration to be the entire real axis. If that is done the common unilateral transform simply becomes a special case of the bilateral transform where the definition of the function being transformed is multiplied by the Heaviside step function.

The bilateral Laplace transform is defined as follows:

$F(s) = \mathcal{L}\left\{f(t)\right\} =\int_{-\infty}^{+\infty} e^{-st} f(t)\,dt.$

### Inverse Laplace transform

The inverse Laplace transform is given by the following complex integral, which is known by various names (the Bromwich integral, the Fourier-Mellin integral, and Mellin's inverse formula):

$f(t) = \mathcal{L}^{-1} \{F(s)\} = \frac{1}{2 \pi i} \lim_{T\to\infty}\int_{ \gamma - i T}^{ \gamma + i T} e^{st} F(s)\,ds,$

where γ is a real number so that the contour path of integration is in the region of convergence of F(s) normally requiring γ > Re(sp) for every singularity sp of F(s) and i2 = −1. If all singularities are in the left half-plane, that is Re(sp) < 0 for every sp, then γ can be set to zero and the above inverse integral formula becomes identical to the inverse Fourier transform.

An alternative formula for the inverse Laplace transform is given by Post's inversion formula.

## Region of convergence

If ƒ is a locally integrable function (or more generally a Borel measure locally of bounded variation), then the Laplace transform F(s) of ƒ converges provided that the limit

$\lim_{R\to\infty}\int_0^R f(t)e^{-ts}\,dt$

exists. The Laplace transform converges absolutely if the integral

$\int_0^\infty |f(t)e^{-ts}|\,dt$

exists (as proper Lebesgue integral). The Laplace transform is usually understood as conditionally convergent, meaning that it converges in the former instead of the latter sense.

The set of values for which F(s) converges absolutely is either of the form Re{s} > a or else Re{s} ≥ a, where a is an extended real constant, −∞ ≤ a ≤ ∞. (This follows from the dominated convergence theorem.) The constant a is known as the abscissa of absolute convergence, and depends on the growth behavior of ƒ(t).[9] Analogously, the two-sided transform converges absolutely in a strip of the form a < Re{s} < b, and possibly including the lines Re{s} = a or Re{s} = b.[10] The subset of values of s for which the Laplace transform converges absolutely is called the region of absolute convergence or the domain of absolute convergence. In the two-sided case, it is sometimes called the strip of absolute convergence. The Laplace transform is analytic in the region of absolute convergence.

Similarly, the set of values for which F(s) converges (conditionally or absolutely) is known as the region of conditional convergence, or simply the region of convergence (ROC). If the Laplace transform converges (conditionally) at s = s0, then it automatically converges for all s with Re{s} > Re{s0}. Therefore the region of convergence is a half-plane of the form Re{s} > a, possibly including some points of the boundary line Re{s} = a. In the region of convergence Re{s} > Re{s0}, the Laplace transform of ƒ can be expressed by integrating by parts as the integral

$F(s) = (s-s_0)\int_0^\infty e^{-(s-s_0)t}\beta(t)\,dt,\quad \beta(u)=\int_0^u e^{-s_0t}f(t)\,dt.$

That is, in the region of convergence F(s) can effectively be expressed as the absolutely convergent Laplace transform of some other function. In particular, it is analytic.

A variety of theorems, in the form of Paley–Wiener theorems, exist concerning the relationship between the decay properties of ƒ and the properties of the Laplace transform within the region of convergence.

In engineering applications, a function corresponding to a linear time-invariant (LTI) system is stable if every bounded input produces a bounded output. This is equivalent to the absolute convergence of the Laplace transform of the impulse response function in the region Re{s} ≥ 0. As a result, LTI systems are stable provided the poles of the Laplace transform of the impulse response function have negative real part.

## Properties and theorems

The Laplace transform has a number of properties that make it useful for analyzing linear dynamical systems. The most significant advantage is that differentiation and integration become multiplication and division, respectively, by s. (This is similar to the way that logarithms change an operation of multiplication of numbers to addition of their logarithms.) This changes integral equations and differential equations to polynomial equations, which are much easier to solve. Once solved, use of the inverse Laplace transform reverts back to the time domain.

Given the functions f(t) and g(t), and their respective Laplace transforms F(s) and G(s):

$f(t) = \mathcal{L}^{-1} \{ F(s) \}$
$g(t) = \mathcal{L}^{-1} \{ G(s) \}$

the following table is a list of properties of unilateral Laplace transform:

Properties of the unilateral Laplace transform
Time domain 's' domain Comment
Linearity $a f(t) + b g(t) \$ $a F(s) + b G(s) \$ Can be proved using basic rules of integration.
Frequency differentiation $t f(t) \$ $-F'(s) \$ $F'\,$ is the first derivative of $F\,$.
Frequency differentiation $t^{n} f(t) \$ $(-1)^{n} F^{(n)}(s) \$ More general form, (n)th derivative of F(s).
Differentiation $f'(t) \$ $s F(s) - f(0) \$ ƒ is assumed to be a differentiable function, and its derivative is assumed to be of exponential type. This can then be obtained by integration by parts
Second Differentiation $f''(t) \$ $s^2 F(s) - s f(0) - f'(0) \$ ƒ is assumed twice differentiable and the second derivative to be of exponential type. Follows by applying the Differentiation property to $f'(t)\,$.
General Differentiation $f^{(n)}(t) \$ $s^n F(s) - s^{n - 1} f(0) - \cdots - f^{(n - 1)}(0) \$ ƒ is assumed to be n-times differentiable, with nth derivative of exponential type. Follow by mathematical induction.
Frequency integration $\frac{f(t)}{t} \$ $\int_s^\infty F(\sigma)\, d\sigma \$
Integration $\int_0^t f(\tau)\, d\tau = (u * f)(t)$ ${1 \over s} F(s)$ u(t) is the Heaviside step function. Note (u * f)(t) is the convolution of u(t) and f(t).
Scaling $f(at) \$ ${1 \over |a|} F \left ( {s \over a} \right )$
Frequency shifting $e^{at} f(t) \$ $F(s - a) \$
Time shifting $f(t - a) u(t - a) \$ $e^{-as} F(s) \$ u(t) is the Heaviside step function
Convolution $(f * g)(t) \$ $F(s) \cdot G(s) \$ ƒ(t) and g(t) are extended by zero for t < 0 in the definition of the convolution.
Periodic Function $f(t) \$ ${1 \over 1 - e^{-Ts}} \int_0^T e^{-st} f(t)\,dt$ f(t) is a periodic function of period T so that $f(t) = f(t + T), \; \forall t\ge 0$. This is the result of the time shifting property and the geometric series.
• Initial value theorem:
$f(0^+)=\lim_{s\to \infty}{sF(s)}.$
• Final value theorem:
$f(\infty)=\lim_{s\to 0}{sF(s)}$, if all poles of sF(s) are in the left-hand plane.
The final value theorem is useful because it gives the long-term behaviour without having to perform partial fraction decompositions or other difficult algebra. If a function's poles are in the right-hand plane (e.g. et or sin(t)) the behaviour of this formula is undefined.

### Proof of the Laplace transform of a function's derivative

It is often convenient to use the differentiation property of the Laplace transform to find the transform of a function's derivative. This can be derived from the basic expression for a Laplace transform as follows:

$\mathcal{L} \left\{f(t)\right\} =\int_{0^-}^{+\infty} e^{-st} f(t)\,dt$
$~~ = \left[ \frac{f(t)e^{-st}}{-s} \right]_{0^-}^{+\infty} - \int_{0^-}^{+\infty} \frac{e^{-st}}{-s} f'(t)dt$ (by parts)
$~~ = \left[-\frac{f(0)}{-s}\right] + \frac{1}{s}\mathcal{L}\left\{f'(t)\right\},$

yielding

$\mathcal{L}\left\{\frac{df}{dt}\right\} = s\cdot\mathcal{L} \left\{ f(t) \right\}-f(0),$

and in the bilateral case, we have

$\mathcal{L}\left\{ { df \over dt } \right\} = s \int_{-\infty}^{+\infty} e^{-st} f(t)\,dt = s \cdot \mathcal{L} \{ f(t) \}.$

### Relationship to other transforms

#### Laplace–Stieltjes transform

The (unilateral) Laplace–Stieltjes transform of a function g : R → R is defined by the Lebesgue–Stieltjes integral

$\{\mathcal{L}^*g\}(s) = \int_0^\infty e^{-st}dg(t).$

The function g is assumed to be of bounded variation. If g is the antiderivative of ƒ:

$g(x) = \int_0^x f(t)\,dt$

then the Laplace–Stieltjes transform of g and the Laplace transform of ƒ coincide. In general, the Laplace–Stieltjes transform is the Laplace transform of the Stieltjes measure associated to g. So in practice, the only distinction between the two transforms is that the Laplace transform is thought of as operating on the density function of the measure, whereas the Laplace–Stieltjes transform is thought of as operating on its cumulative distribution function.[11]

#### Fourier transform

The continuous Fourier transform is equivalent to evaluating the bilateral Laplace transform with complex argument s = iω or s = 2πfi:

$\begin{array}{rcl} \hat{f}(\omega) & = & \mathcal{F}\left\{f(t)\right\} \\[1em] & = & \mathcal{L}\left\{f(t)\right\}|_{s = i\omega} = F(s)|_{s = i \omega}\\[1em] & = & \int_{-\infty}^{+\infty} e^{-\imath \omega t} f(t)\,\mathrm{d}t.\ \end{array}$

This expression excludes the scaling factor $1/\sqrt{2 \pi}$, which is often included in definitions of the Fourier transform. This relationship between the Laplace and Fourier transforms is often used to determine the frequency spectrum of a signal or dynamical system.

The above relation is valid as stated if and only if the region of convergence (ROC) of F(s) contains the imaginary axis, σ = 0. For example, the function f(t) = cos(ω0t)u(t) has a Laplace transform F(s) = s/(s2 + ω02) whose ROC is Re(s) > 0. Therefore, substituting s = iω in F(s) does not yield the Fourier transform of f(t) = cos(ω0t).

However, a relation of the form

$\lim_{\sigma\to 0^+} F(\sigma+i\omega) = \hat{f}(\omega)$

holds under much weaker conditions. For instance, this holds for the above example provided that the limit is understood as a weak limit of measures (see vague topology). General conditions relating the limit of the Laplace transform of a function on the boundary to the Fourier transform take the form of Paley-Wiener theorems.

#### Mellin transform

The Mellin transform and its inverse are related to the two-sided Laplace transform by a simple change of variables. If in the Mellin transform

$G(s) = \mathcal{M}\left\{g(\theta)\right\} = \int_0^\infty \theta^s g(\theta) \frac{d\theta}{\theta}$

we set θ = e-t we get a two-sided Laplace transform.

#### Z-transform

The unilateral or one-sided Z-transform is simply the Laplace transform of an ideally sampled signal with the substitution of

$z \ \stackrel{\mathrm{def}}{=}\ e^{s T} \$
where $T = 1/f_s \$ is the sampling period (in units of time e.g., seconds) and $f_s \$ is the sampling rate (in samples per second or hertz)

Let

$\Delta_T(t) \ \stackrel{\mathrm{def}}{=}\ \sum_{n=0}^{\infty} \delta(t - n T)$

be a sampling impulse train (also called a Dirac comb) and

$x_q(t) \ \stackrel{\mathrm{def}}{=}\ x(t) \Delta_T(t) = x(t) \sum_{n=0}^{\infty} \delta(t - n T)$
$= \sum_{n=0}^{\infty} x(n T) \delta(t - n T) = \sum_{n=0}^{\infty} x[n] \delta(t - n T)$

be the continuous-time representation of the sampled $x(t) \$

$x[n] \ \stackrel{\mathrm{def}}{=}\ x(nT) \$ are the discrete samples of $x(t) \$.

The Laplace transform of the sampled signal $x_q(t) \$ is

$X_q(s) = \int_{0^-}^{\infty} x_q(t) e^{-s t} \,dt$
$\ = \int_{0^-}^{\infty} \sum_{n=0}^{\infty} x[n] \delta(t - n T) e^{-s t} \, dt$
$\ = \sum_{n=0}^{\infty} x[n] \int_{0^-}^{\infty} \delta(t - n T) e^{-s t} \, dt$
$\ = \sum_{n=0}^{\infty} x[n] e^{-n s T}.$

This is precisely the definition of the unilateral Z-transform of the discrete function $x[n] \$

$X(z) = \sum_{n=0}^{\infty} x[n] z^{-n}$

with the substitution of $z \leftarrow e^{s T} \$.

Comparing the last two equations, we find the relationship between the unilateral Z-transform and the Laplace transform of the sampled signal:

$X_q(s) = X(z) \Big|_{z=e^{sT}}.$

The similarity between the Z and Laplace transforms is expanded upon in the theory of time scale calculus.

#### Borel transform

The integral form of the Borel transform

$F(s) = \int_0^\infty f(z)e^{-sz}\,dz$

is a special case of the Laplace transform for ƒ an entire function of exponential type, meaning that

$|f(z)|\le Ae^{B|z|}$

for some constants A and B. The generalized Borel transform allows a different weighting function to be used, rather than the exponential function, to transform functions not of exponential type. Nachbin's theorem gives necessary and sufficient conditions for the Borel transform to be well defined.

#### Fundamental relationships

Since an ordinary Laplace transform can be written as a special case of a two-sided transform, and since the two-sided transform can be written as the sum of two one-sided transforms, the theory of the Laplace-, Fourier-, Mellin-, and Z-transforms are at bottom the same subject. However, a different point of view and different characteristic problems are associated with each of these four major integral transforms.

## Table of selected Laplace transforms

The following table provides Laplace transforms for many common functions of a single variable. For definitions and explanations, see the Explanatory Notes at the end of the table.

Because the Laplace transform is a linear operator:

• The Laplace transform of a sum is the sum of Laplace transforms of each term.
$\mathcal{L}\left\{f(t) + g(t) \right\} = \mathcal{L}\left\{f(t)\right\} + \mathcal{L}\left\{ g(t) \right\}$
• The Laplace transform of a multiple of a function, is that multiple times the Laplace transformation of that function.
$\mathcal{L}\left\{a f(t)\right\} = a \mathcal{L}\left\{ f(t)\right\}$

The unilateral Laplace transform takes as input a function whose time domain is the non-negative reals, which is why all of the time domain functions in the table below are multiples of the Heaviside step function, u(t). The entries of the table that involve a time delay τ are required to be causal (meaning that τ > 0). A causal system is a system where the impulse response h(t) is zero for all time t prior to t = 0. In general, the region of convergence for causal systems is not the same as that of anticausal systems.

ID Function Time domain
$f(t) = \mathcal{L}^{-1} \left\{ F(s) \right\}$
Laplace s-domain
$F(s) = \mathcal{L}\left\{ f(t) \right\}$
Region of convergence
1 ideal delay $\delta(t-\tau) \$ $e^{-\tau s} \$
1a unit impulse $\delta(t) \$ 1 $\mathrm{all} \ s \,$
2 delayed nth power
with frequency shift
$\frac{(t-\tau)^n}{n!} e^{-\alpha (t-\tau)} \cdot u(t-\tau)$ $\frac{e^{-\tau s}}{(s+\alpha)^{n+1}}$ $\textrm{Re} \{ s \} > - \alpha \,$
2a nth power
( for integer n )
${ t^n \over n! } \cdot u(t)$ ${ 1 \over s^{n+1} }$ $\textrm{Re} \{ s \} > 0 \,$
$(n > -1) \,$
2a.1 qth power
( for complex q )
${ t^q \over \Gamma(q+1) } \cdot u(t)$ ${ 1 \over s^{q+1} }$ $\textrm{Re} \{ s \} > 0 \,$
$(\textrm{Re}\{q\} > -1) \,$
2a.2 unit step $u(t) \$ ${ 1 \over s }$ $\textrm{Re} \{ s \} > 0 \,$
2b delayed unit step $u(t-\tau) \$ ${ e^{-\tau s} \over s }$ $\textrm{Re} \{ s \} > 0 \,$
2c ramp $t \cdot u(t)\$ $\frac{1}{s^2}$ $\textrm{Re} \{ s \} > 0 \,$
2d nth power with frequency shift $\frac{t^{n}}{n!}e^{-\alpha t} \cdot u(t)$ $\frac{1}{(s+\alpha)^{n+1}}$ $\textrm{Re} \{ s \} > - \alpha \,$
2d.1 exponential decay $e^{-\alpha t} \cdot u(t) \$ ${ 1 \over s+\alpha }$ $\textrm{Re} \{ s \} > - \alpha \$
3 exponential approach $( 1-e^{-\alpha t}) \cdot u(t) \$ $\frac{\alpha}{s(s+\alpha)}$ $\textrm{Re} \{ s \} > 0\$
4 sine $\sin(\omega t) \cdot u(t) \$ ${ \omega \over s^2 + \omega^2 }$ $\textrm{Re} \{ s \} > 0 \$
5 cosine $\cos(\omega t) \cdot u(t) \$ ${ s \over s^2 + \omega^2 }$ $\textrm{Re} \{ s \} > 0 \$
6 hyperbolic sine $\sinh(\alpha t) \cdot u(t) \$ ${ \alpha \over s^2 - \alpha^2 }$ $\textrm{Re} \{ s \} > | \alpha | \$
7 hyperbolic cosine $\cosh(\alpha t) \cdot u(t) \$ ${ s \over s^2 - \alpha^2 }$ $\textrm{Re} \{ s \} > | \alpha | \$
8 Exponentially-decaying
sine wave
$e^{-\alpha t} \sin(\omega t) \cdot u(t) \$ ${ \omega \over (s+\alpha )^2 + \omega^2 }$ $\textrm{Re} \{ s \} > -\alpha \$
9 Exponentially-decaying
cosine wave
$e^{-\alpha t} \cos(\omega t) \cdot u(t) \$ ${ s+\alpha \over (s+\alpha )^2 + \omega^2 }$ $\textrm{Re} \{ s \} > -\alpha \$
10 nth root $\sqrt[n]{t} \cdot u(t)$ $s^{-(n+1)/n} \cdot \Gamma\left(1+\frac{1}{n}\right)$ $\textrm{Re} \{ s \} > 0 \,$
11 natural logarithm $\ln \left ( { t \over t_0 } \right ) \cdot u(t)$ $- { t_0 \over s} \ [ \ \ln(t_0 s)+\gamma \ ]$ $\textrm{Re} \{ s \} > 0 \,$
12 Bessel function
of the first kind,
of order n
$J_n( \omega t) \cdot u(t)$ $\frac{ \omega^n \left(s+\sqrt{s^2+ \omega^2}\right)^{-n}}{\sqrt{s^2 + \omega^2}}$ $\textrm{Re} \{ s \} > 0 \,$
$(n > -1) \,$
13 Modified Bessel function
of the first kind,
of order n
$I_n(\omega t) \cdot u(t)$ $\frac{ \omega^n \left(s+\sqrt{s^2-\omega^2}\right)^{-n}}{\sqrt{s^2-\omega^2}}$ $\textrm{Re} \{ s \} > | \omega | \,$
14 Bessel function
of the second kind,
of order 0
$Y_0(\alpha t) \cdot u(t)$ $-{2 \sinh^{-1}(s/\alpha) \over \pi \sqrt{s^2+\alpha^2}}$ $\textrm{Re} \{ s \} > 0 \,$
15 Modified Bessel function
of the second kind,
of order 0
$K_0(\alpha t) \cdot u(t)$
16 Error function $\mathrm{erf}(t) \cdot u(t)$ ${e^{s^2/4} \left(1 - \operatorname{erf} \left(s/2\right)\right) \over s}$ $\textrm{Re} \{ s \} > 0 \,$
Explanatory notes:
 $u(t) \,$ represents the Heaviside step function. $\delta(t) \,$ represents the Dirac delta function. $\Gamma (z) \,$ represents the Gamma function. $\gamma \,$ is the Euler-Mascheroni constant. $t \,$, a real number, typically represents time, although it can represent any independent dimension. $s \,$ is the complex angular frequency, and Re{s} is its real part. $\alpha \,$, $\beta \,$, $\tau \,$, and $\omega \,$ are real numbers. $n \,$, is an integer.

## s-Domain equivalent circuits and impedances

The Laplace transform is often used in circuit analysis, and simple conversions to the s-Domain of circuit elements can be made. Circuit elements can be transformed into impedances, very similar to phasor impedances.

Here is a summary of equivalents:

Note that the resistor is exactly the same in the time domain and the s-Domain. The sources are put in if there are initial conditions on the circuit elements. For example, if a capacitor has an initial voltage across it, or if the inductor has an initial current through it, the sources inserted in the s-Domain account for that.

The equivalents for current and voltage sources are simply derived from the transformations in the table above.

## Examples: How to apply the properties and theorems

The Laplace transform is used frequently in engineering and physics; the output of a linear time invariant system can be calculated by convolving its unit impulse response with the input signal. Performing this calculation in Laplace space turns the convolution into a multiplication; the latter being easier to solve because of its algebraic form. For more information, see control theory.

The Laplace transform can also be used to solve differential equations and is used extensively in electrical engineering. The Laplace transform reduces a linear differential equation to an algebraic equation, which can then be solved by the formal rules of algebra. The original differential equation can then be solved by applying the inverse Laplace transform. The English electrical engineer Oliver Heaviside first proposed a similar scheme, although without using the Laplace transform; and the resulting operational calculus is credited as the Heaviside calculus.

The following examples, derived from applications in physics and engineering, will use SI units of measure. SI is based on meters for distance, kilograms for mass, seconds for time, and amperes for electric current.

### Example #1: Solving a differential equation

The following example is based on concepts from nuclear physics.

Consider the following first-order, linear differential equation:

$\frac{dN}{dt} = -\lambda N.$

This equation is the fundamental relationship describing radioactive decay, where

$N \ = \ N(t)$

represents the number of undecayed atoms remaining in a sample of a radioactive isotope at time t (in seconds), and $\ \lambda$ is the decay constant.

We can use the Laplace transform to solve this equation.

Rearranging the equation to one side, we have

$\frac{dN}{dt} + \lambda N = 0.$

Next, we take the Laplace transform of both sides of the equation:

$\left( s \tilde{N}(s) - N_o \right) + \lambda \tilde{N}(s) \ = \ 0$

where

$\tilde{N}(s) = \mathcal{L}\{N(t)\}$

and

$N_o \ = \ N(0).$

Solving, we find

$\tilde{N}(s) = { N_o \over s + \lambda }.$

Finally, we take the inverse Laplace transform to find the general solution

$N(t) \ = \mathcal{L}^{-1} \{\tilde{N}(s)\} = \mathcal{L}^{-1} \left\{ \frac{N_o}{s + \lambda} \right\}$
$= \ N_o e^{-\lambda t},$

which is indeed the correct form for radioactive decay.

### Example #2: Deriving the complex impedance for a capacitor

This example is based on the principles of electrical circuit theory.

The constitutive relation governing the dynamic behavior of a capacitor is the following differential equation:

$i = C { dv \over dt}$

where C is the capacitance (in farads) of the capacitor, i = i(t) is the electric current (in amperes) through the capacitor as a function of time, and v = v(t) is the voltage (in volts) across the terminals of the capacitor, also as a function of time.

Taking the Laplace transform of this equation, we obtain

$I(s) = C \left( s V(s) - V_o \right)$

where

$I(s) = \mathcal{L} \{ i(t) \},$
$V(s) = \mathcal{L} \{ v(t) \},$ and
$V_o \ = \ v(t)|_{t=0}.$

Solving for V(s) we have

$V(s) = { I(s) \over sC } + { V_o \over s }.$

The definition of the complex impedance Z (in ohms) is the ratio of the complex voltage V divided by the complex current I while holding the initial state Vo at zero:

$Z(s) = { V(s) \over I(s) } \bigg|_{V_o = 0}.$

Using this definition and the previous equation, we find:

$Z(s) = \frac{1}{sC},$

which is the correct expression for the complex impedance of a capacitor.

### Example #3: Method of partial fraction expansion

Consider a linear time-invariant system with transfer function

$H(s) = \frac{1}{(s+\alpha)(s+\beta)}.$

The impulse response is simply the inverse Laplace transform of this transfer function:

$h(t) = \mathcal{L}^{-1}\{H(s)\}.$

To evaluate this inverse transform, we begin by expanding H(s) using the method of partial fraction expansion:

$\frac{1}{(s+\alpha)(s+\beta)} = { P \over s+\alpha } + { R \over s+\beta }.$

The unknown constants P and R are the residues located at the corresponding poles of the transfer function. Each residue represents the relative contribution of that singularity to the transfer function's overall shape. By the residue theorem, the inverse Laplace transform depends only upon the poles and their residues. To find the residue P, we multiply both sides of the equation by s + α to get

$\frac{1}{s+\beta} = P + { R (s+\alpha) \over s+\beta }.$

Then by letting s = −α, the contribution from R vanishes and all that is left is

$P = \left.{1 \over s+\beta}\right|_{s=-\alpha} = {1 \over \beta - \alpha}.$

Similarly, the residue R is given by

$R = \left.{1 \over s+\alpha}\right|_{s=-\beta} = {1 \over \alpha - \beta}.$

Note that

$R = {-1 \over \beta - \alpha} = - P$

and so the substitution of R and P into the expanded expression for H(s) gives

$H(s) = \left( \frac{1}{\beta-\alpha} \right) \cdot \left( { 1 \over s+\alpha } - { 1 \over s+\beta } \right).$

Finally, using the linearity property and the known transform for exponential decay (see Item #3 in the Table of Laplace Transforms, above), we can take the inverse Laplace transform of H(s) to obtain:

$h(t) = \mathcal{L}^{-1}\{H(s)\} = \frac{1}{\beta-\alpha}\left(e^{-\alpha t}-e^{-\beta t}\right),$

which is the impulse response of the system.

### Example #4: Mixing sines, cosines, and exponentials

Time function Laplace transform
$e^{-\alpha t}\left[ \cos{(\omega t)}+\left(\frac{\beta-\alpha}{\omega}\right)\sin{(\omega t)}\right]u(t)$ $\frac{s+\beta}{(s+\alpha)^2+\omega^2}$

Starting with the Laplace transform

$X(s) = \frac{s+\beta}{(s+\alpha)^2+\omega^2},$

we find the inverse transform by first adding and subtracting the same constant α to the numerator:

$X(s) = \frac{s+\alpha } { (s+\alpha)^2+\omega^2} + \frac{\beta - \alpha }{(s+\alpha)^2+\omega^2}.$

By the shift-in-frequency property, we have

$x(t) = e^{-\alpha t} \mathcal{L}^{-1} \left\{ {s \over s^2 + \omega^2} + { \beta - \alpha \over s^2 + \omega^2 } \right\}$
$= e^{-\alpha t} \mathcal{L}^{-1} \left\{ {s \over s^2 + \omega^2} + \left( { \beta - \alpha \over \omega } \right) \left( { \omega \over s^2 + \omega^2 } \right) \right\}$
$= e^{-\alpha t} \left[ \mathcal{L}^{-1} \left\{ {s \over s^2 + \omega^2} \right\} + \left( { \beta - \alpha \over \omega } \right) \mathcal{L}^{-1} \left\{ { \omega \over s^2 + \omega^2 } \right\} \right].$

Finally, using the Laplace transforms for sine and cosine (see the table, above), we have

$x(t) = e^{-\alpha t} \left[ \cos{(\omega t)}u(t)+\left(\frac{\beta-\alpha}{\omega}\right)\sin{(\omega t)}u(t)\right].$
$x(t) = e^{-\alpha t} \left[ \cos{(\omega t)}+\left(\frac{\beta-\alpha}{\omega}\right)\sin{(\omega t)}\right]u(t).$

### Example #5: Phase delay

Time function Laplace transform
$\sin{(\omega t+\phi)} \$ $\frac{s\sin\phi+\omega \cos\phi}{s^2+\omega^2} \$
$\cos{(\omega t+\phi)} \$ $\frac{s\cos\phi - \omega \sin\phi}{s^2+\omega^2} \$

Starting with the Laplace transform,

$X(s) = \frac{s\sin\phi+\omega \cos\phi}{s^2+\omega^2}$

we find the inverse by first rearranging terms in the fraction:

 $X(s) \,\!$ ${} = \frac{s \sin \phi}{s^2 + \omega^2} + \frac{\omega \cos \phi}{s^2 + \omega^2}$ ${} = (\sin \phi) \left(\frac{s}{s^2 + \omega^2} \right) + (\cos \phi) \left(\frac{\omega}{s^2 + \omega^2} \right).$

We are now able to take the inverse Laplace transform of our terms:

 $x(t) \,\!$ ${} = (\sin \phi) \mathcal{L}^{-1}\left\{\frac{s}{s^2 + \omega^2} \right\} + (\cos \phi) \mathcal{L}^{-1}\left\{\frac{\omega}{s^2 + \omega^2} \right\}$ ${}=(\sin \phi)(\cos \omega t) + (\sin \omega t)(cos \phi). \,\!$

To simplify this answer, we must recall the trigonometric identity that

$a \sin \omega t + b \cos \omega t = \sqrt{a^2+b^2} \cdot \sin \left(\omega t + \arctan (b/a) \right)$

and apply it to our value for x(t):

 $x(t) \,\!$ ${} = \sqrt{\cos^2 \phi + \sin^2 \phi} \cdot \sin \left( \omega t + \arctan \left(\frac{\sin \phi}{\cos \phi} \right) \right)$ ${}= \sin (\omega t + \phi) . \,\!$

We can apply similar logic to find that

$\mathcal{L}^{-1} \left\{ \frac{s\cos\phi - \omega \sin\phi}{s^2+\omega^2} \right\} = \cos{(\omega t+\phi)}. \$

## Notes

1. ^ Korn & Korn 1967, §8.1
2. ^ Euler 1744, (1753) and (1769)
3. ^ Lagrange 1773
4. ^ Grattan-Guinness 1997, p. 260
5. ^ Grattan-Guinness 1997, p. 261
6. ^ Grattan-Guinness 1997, pp. 261–262
7. ^ Grattan-Guinness 1997, pp. 262–266
8. ^ Feller 1971, §XIII.1
9. ^ Widder 1941, Chapter II, §1
10. ^ Widder 1941, Chapter VI, §2
11. ^ Feller 1971, p. 432

## References

### Modern

• Arendt, Wolfgang; Batty, Charles J.K.; Hieber, Matthias; Neubrander, Frank (2002), Vector-Valued Laplace Transforms and Cauchy Problems, Birkhäuser Basel, ISBN 3764365498 .
• Davies, Brian (2002), Integral transforms and their applications (Third ed.), New York: Springer, ISBN 0-387-95314-0 .
• Feller, William (1971), An introduction to probability theory and its applications. Vol. II., Second edition, New York: John Wiley & Sons, MR0270403 .
• Korn, G.A.; Korn, T.M. (1967), Mathematical Handbook for Scientists and Engineers (2nd ed.), McGraw-Hill Companies, ISBN 0-0703-5370-0 .
• Polyanin, A. D.; Manzhirov, A. V. (1998), Handbook of Integral Equations, Boca Raton: CRC Press, ISBN 0-8493-2876-4 .
• Schwartz, Laurent (1952), "Transformation de Laplace des distributions", Comm. Sém. Math. Univ. Lund [Medd. Lunds Univ. Mat. Sem.] 1952: 196–206, MR0052555 .
• Siebert, William McC. (1986), Circuits, Signals, and Systems, Cambridge, Massachusetts: MIT Press, ISBN 0-262-19229-2 .
• Widder, David Vernon (1941), The Laplace Transform, Princeton Mathematical Series, v. 6, Princeton University Press, MR0005923 .
• Widder, David Vernon (1945), "What is the Laplace transform?", The American Mathematical Monthly 52: 419–425, MR0013447, ISSN 0002-9890 .

### Historical

• Deakin, M. A. B. (1981), "The development of the Laplace transform", Archive for the History of the Exact Sciences 25: 343–390, doi:10.1007/BF01395660
• Deakin, M. A. B. (1982), "The development of the Laplace transform", Archive for the History of the Exact Sciences 26: 351–381
• Euler, L. (1744), "De constructione aequationum", Opera omnia, 1st series 22: 150–161 .
• Euler, L. (1753), "Methodus aequationes differentiales", Opera omnia, 1st series 22: 181–213 .
• Euler, L. (1769), "Institutiones calculi integralis, Volume 2", Opera omnia, 1st series 12 , Chapters 3-5.
• Grattan-Guinness, I (1997), "Laplace's integral solutions to partial differential equations", in Gillispie, C. C., Pierre Simon Laplace 1749-1827: A Life in Exact Science, Princeton: Princeton University Press, ISBN 0-691-01185-0 .
• Lagrange, J. L. (1773), Mémoire sur l'utilité de la méthode, Œuvres de Lagrange, 2, pp. 171–234 .

# Study guide

Up to date as of January 14, 2010

### From Wikiversity

In mathematics, the Laplace transform is a technique for analyzing linear time-invariant systems such as electrical circuits, harmonic oscillators, optical devices, and mechanical systems. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.

The Laplace transform is an important concept from the branch of mathematics called functional analysis.

In actual physical systems the Laplace transform is often interpreted as a transformation from the time-domain point of view, in which inputs and outputs are understood as functions of time, to the frequency-domain point of view, where the same inputs and outputs are seen as functions of complex angular frequency, or radians per unit time. This transformation not only provides a fundamentally different way to understand the behavior of the system, but it also drastically reduces the complexity of the mathematical calculations required to analyze the system.

The Laplace transform has many important applications in physics, optics, electrical engineering, control engineering, signal processing, and probability theory.

The Laplace transform is named in honor of mathematician and astronomer Pierre-Simon Laplace, who used the transform in his work on probability theory.

## Formal definition

The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by:

$F(s) = \mathcal{L} \left\{f(t)\right\}=\int_{0^-}^\infty e^{-st} f(t) \,dt.$

The lower limit of 0 is short notation to mean $\lim_{\epsilon \rightarrow +0} -\epsilon \$ and assures the inclusion of the entire Dirac delta function δ(t) at 0 if there is such an impulse in f(t) at 0.

The parameter s is in general complex:

$s = \sigma + i \omega \,$

This integral transform has a number of properties that make it useful for analyzing linear dynamical systems. The most significant advantage is that differentiation and integration become multiplication and division, respectively, with s. (This is similar to the way that logarithms change an operation of multiplication of numbers to addition of their logarithms.) This changes integral equations and differential equations to polynomial equations, which are much easier to solve.

### Bilateral Laplace transform

Main article: Two-sided Laplace transform

When one says "the Laplace transform" without qualification, the unilateral or one-sided transform is normally intended. The Laplace transform can be alternatively defined as the bilateral Laplace transform or two-sided Laplace transform by extending the limits of integration to be the entire real axis. If that is done the common unilateral transform simply becomes a special case of the bilateral transform where the definition of the function being transformed is multiplied by the Heaviside step function.

The bilateral Laplace transform is defined as follows:

$F(s) = \mathcal{L}\left\{f(t)\right\} =\int_{-\infty}^{+\infty} e^{-st} f(t)\,dt.$

### Inverse Laplace transform

The inverse Laplace transform is the Bromwich integral, which is a complex integral given by:

$f(t) = \mathcal{L}^{-1} \{F(s)\} = \frac{1}{2 \pi i} \int_{ \gamma - i \cdot \infty}^{ \gamma + i \cdot \infty} e^{st} F(s)\,ds,$

where γ is a real number so that the contour path of integration is in the region of convergence of F(s) normally requiring γ > Re(sp) for every singularity sp of F(s) and i2=-1. If all singularities are in the left half-plane, that is Re(sp) < 0 for every sp, then γ can be set to zero and the above inverse integral formula above becomes identical to the inverse Fourier transform.

An alternative formula for the inverse Laplace transform is given by Post's inversion formula.

## Region of convergence

The Laplace transform F(s) typically exists for all complex numbers such that Re{s} > a, where a is a real constant which depends on the growth behavior of f(t), whereas the two-sided transform is defined in a range a < Re{s} < b. The subset of values of s for which the Laplace transform exists is called the region of convergence (ROC) or the domain of convergence. In the two-sided case, it is sometimes called the strip of convergence.

There are no specific conditions that one can check a function against to know in all cases if its Laplace transform can be taken, other than to say the defining integral converges. It is however easy to give theorems on cases where it may or may not be taken.

## Properties and theorems

Given the functions f(t) and g(t), and their respective Laplace transforms F(s) and G(s):

$f(t) = \mathcal{L}^{-1} \{ F(s) \}$
$g(t) = \mathcal{L}^{-1} \{ G(s) \}$

the following table is a list of properties of unilateral Laplace transform:

Properties of the unilateral Laplace transform
Time domain Frequency domain Comment
Linearity $a f(t) + b g(t) \$ $a F(s) + b G(s) \$
Frequency differentiation $t f(t) \$ $-F'(s) \$
Frequency differentiation $t^{n} f(t) \$ $(-1)^{n} F^{(n)}(s) \$ more general
Differentiation $f' \$ $s F(s) - f(0^-) \$
Second Differentiation $f'' \$ $s^2 F(s) - s f(0^-) - f'(0^-) \$
General Differentiation $f^{(n)} \$ $s^n F(s) - s^{n - 1} f(0^-) - \cdots - f^{(n - 1)}(0^-) \$
Frequency integration $\frac{f(t)}{t} \$ $\int_s^\infty F(\sigma)\, d\sigma \$
Integration $\int_0^t f(\tau)\, d\tau = u(t) * f(t)$ ${1 \over s} F(s)$ u(t) is the Heaviside step function
Scaling $f(at) \$ ${1 \over |a|} F \left ( {s \over a} \right )$
Frequency shifting $e^{at} f(t) \$ $F(s - a) \$
Time shifting $f(t - a) u(t - a) \$ $e^{-as} F(s) \$ u(t) is the Heaviside step function
Convolution $f(t) * g(t) \$ $F(s) \cdot G(s) \$
Periodic Function $f(t) \$ ${1 \over 1 - e^{-Ts}} \int_0^T e^{-st} f(t)\,dt$ f(t) is a periodic function of period T so that $f(t) = f(t + T), \; \forall t$
• Initial value theorem:
$f(0^+)=\lim_{s\to \infty}{sF(s)}$
• Final value theorem:
$f(\infty)=\lim_{s\to 0}{sF(s)}$, all poles in left-hand plane.
The final value theorem is useful because it gives the long-term behaviour without having to perform partial fraction decompositions or other difficult algebra. If a function's poles are in the right hand plane (e.g. et or sin(t)) the behaviour of this formula is undefined.

### Proof of the Laplace transform of a function's derivative

It is often convenient to use the differentiation property of the Laplace transform to find the transform of a function's derivative. This can be derived from the basic expression for a Laplace transform as follows:

$\mathcal{L} \left\{f(t)\right\} =\int_{0^-}^{+\infty} e^{-st} f(t)\,dt$
$~~ = \left[ \frac{f(t)e^{-st}}{-s} \right]_{0^-}^{+\infty} - \int_{0^-}^{+\infty} \frac{e^{-st}}{-s} f'(t)dt$ (by parts)
$~~ = \left[-\frac{f(0)}{-s}\right] + \frac{1}{s}\mathcal{L}\left\{f'(t)\right\},$

yielding

$\mathcal{L}\left\{\frac{df}{dt}\right\} = s\cdot\mathcal{L} \left\{ f(t) \right\}-f(0),$

and in the bilateral case, we have

$\mathcal{L}\left\{ { df \over dt } \right\} = s \int_{-\infty}^{+\infty} e^{-st} f(t)\,dt = s \cdot \mathcal{L} \{ f(t) \}.$

### Relationship to other transforms

#### Fourier transform

The continuous Fourier transform is equivalent to evaluating the bilateral Laplace transform with complex argument s = iω:

$\begin{array}{rcl} F(\omega) & = & \mathcal{F}\left\{f(t)\right\} \\[1em] & = & \mathcal{L}\left\{f(t)\right\}|_{s = i \omega} = F(s)|_{s = i \omega}\\[1em] & = & \int_{-\infty}^{+\infty} e^{-\imath \omega t} f(t)\,\mathrm{d}t.\ \end{array}$

Note that this expression excludes the scaling factor $\frac{1}{\sqrt{2 \pi}}$, which is often included in definitions of the Fourier transform.

This relationship between the Laplace and Fourier transforms is often used to determine the frequency spectrum of a signal or dynamical system.

#### Mellin transform

The Mellin transform and its inverse are related to the two-sided Laplace transform by a simple change of variables. If in the Mellin transform

$G(s) = \mathcal{M}\left\{g(\theta)\right\} = \int_0^\infty \theta^s g(\theta) \frac{d\theta}{\theta}$

we set θ = e-t we get a two-sided Laplace transform.

#### Z-transform

The Z-transform is simply the Laplace transform of an ideally sampled signal with the substitution of

$z \ \stackrel{\mathrm{def}}{=}\ e^{s T} \$
where $T = 1/f_s \$ is the sampling period (in units of time e.g. seconds) and $f_s \$ is the sampling rate (in samples per second or hertz)

Let

$\Delta_T(t) \ \stackrel{\mathrm{def}}{=}\ \sum_{n=0}^{\infty} \delta(t - n T)$

be a sampling impulse train (also called a Dirac comb) and

$x_q(t) \ \stackrel{\mathrm{def}}{=}\ x(t) \Delta_T(t) = x(t) \sum_{n=0}^{\infty} \delta(t - n T)$
$= \sum_{n=0}^{\infty} x(n T) \delta(t - n T) = \sum_{n=0}^{\infty} x[n] \delta(t - n T)$

be the continuous-time representation of the sampled $x(t) \$.

$x[n] \ \stackrel{\mathrm{def}}{=}\ x(nT) \$ are the discrete samples of $x(t) \$.

The Laplace transform of the sampled signal $x_q(t) \$ is

$X_q(s) = \int_{0^-}^{\infty} x_q(t) e^{-s t} \,dt$
$\ = \int_{0^-}^{\infty} \sum_{n=0}^{\infty} x[n] \delta(t - n T) e^{-s t} \, dt$
$\ = \sum_{n=0}^{\infty} x[n] \int_{0^-}^{\infty} \delta(t - n T) e^{-s t} \, dt$
$\ = \sum_{n=0}^{\infty} x[n] e^{-n s T}.$

This is precisely the definition of the Z-transform of the discrete function $x[n] \$

$X(z) = \sum_{n=0}^{\infty} x[n] z^{-n}$

with the substitution of $z \leftarrow e^{s T} \$.

Comparing the last two equations, we find the relationship between the Z-transform and the Laplace transform of the sampled signal:

$X_q(s) = X(z) \Big|_{z=e^{sT}}.$

#### Borel transform

The integral form of the Borel transform is identical to the Laplace transform; indeed, these are sometimes mistakenly assumed to be synonyms. The generalized Borel transform generalizes the Laplace transform for functions not of exponential type.

#### Fundamental relationships

Since an ordinary Laplace transform can be written as a special case of a two-sided transform, and since the two-sided transform can be written as the sum of two one-sided transforms, the theory of the Laplace-, Fourier-, Mellin-, and Z-transforms are at bottom the same subject. However, a different point of view and different characteristic problems are associated with each of these four major integral transforms.

## Table of selected Laplace transforms

The following table provides Laplace transforms for many common functions of a single variable. For definitions and explanations, see the Explanatory Notes at the end of the table.

Because the Laplace transform is a linear operator:

• The Laplace transform of a sum is the sum of Laplace transforms of each term.
$\mathcal{L}\left\{f(t) + g(t) \right\} = \mathcal{L}\left\{f(t)\right\} + \mathcal{L}\left\{ g(t) \right\}$
• The Laplace transform of a multiple of a function, is that multiple times the Laplace transformation of that function.
$\mathcal{L}\left\{a f(t)\right\} = a \mathcal{L}\left\{ f(t)\right\}$

The unilateral Laplace transform is only valid when t is non-negative, which is why all of the time domain functions in the table below are multiples of the Heaviside step function, u(t).

ID Function Time Domain
$x(t) = \mathcal{L}^{-1} \left\{ X(s) \right\}$
Frequency Domain
$X(s) = \mathcal{L}\left\{ x(t) \right\}$
Region of convergence
for causal systems
1 ideal delay $\delta(t-\tau) \$ $e^{-\tau s} \$
1a unit impulse $\delta(t) \$ $1 \$ $\mathrm{all} \ s \,$
2 delayed nth power
with frequency shift
$\frac{(t-\tau)^n}{n!} e^{-\alpha (t-\tau)} \cdot u(t-\tau)$ $\frac{e^{-\tau s}}{(s+\alpha)^{n+1}}$ $s > 0 \,$
2a nth power
( for integer n )
${ t^n \over n! } \cdot u(t)$ ${ 1 \over s^{n+1} }$ $s > 0 \,$
2a.1 qth power
( for real q )
${ t^q \over \Gamma(q+1) } \cdot u(t)$ ${ 1 \over s^{q+1} }$ $s > 0 \,$
2a.2 unit step $u(t) \$ ${ 1 \over s }$ $s > 0 \,$
2b delayed unit step $u(t-\tau) \$ ${ e^{-\tau s} \over s }$ $s > 0 \,$
2c ramp $t \cdot u(t)\$ $\frac{1}{s^2}$ $s > 0 \,$
2d nth power with frequency shift $\frac{t^{n}}{n!}e^{-\alpha t} \cdot u(t)$ $\frac{1}{(s+\alpha)^{n+1}}$ $s > - \alpha \,$
2d.1 exponential decay $e^{-\alpha t} \cdot u(t) \$ ${ 1 \over s+\alpha }$ $s > - \alpha \$
3 exponential approach $( 1-e^{-\alpha t}) \cdot u(t) \$ $\frac{\alpha}{s(s+\alpha)}$ $s > 0\$
4 sine $\sin(\omega t) \cdot u(t) \$ ${ \omega \over s^2 + \omega^2 }$ $s > 0 \$
5 cosine $\cos(\omega t) \cdot u(t) \$ ${ s \over s^2 + \omega^2 }$ $s > 0 \$
6 hyperbolic sine $\sinh(\alpha t) \cdot u(t) \$ ${ \alpha \over s^2 - \alpha^2 }$ $s > | \alpha | \$
7 hyperbolic cosine $\cosh(\alpha t) \cdot u(t) \$ ${ s \over s^2 - \alpha^2 }$ $s > | \alpha | \$
8 Exponentially-decaying
sine wave
$e^{\alpha t} \sin(\omega t) \cdot u(t) \$ ${ \omega \over (s-\alpha )^2 + \omega^2 }$ $s > \alpha \$
9 Exponentially-decaying
cosine wave
$e^{\alpha t} \cos(\omega t) \cdot u(t) \$ ${ s-\alpha \over (s-\alpha )^2 + \omega^2 }$ $s > \alpha \$
10 nth root $\sqrt[n]{t} \cdot u(t)$ $s^{-(n+1)/n} \cdot \Gamma\left(1+\frac{1}{n}\right)$ $s > 0 \,$
11 natural logarithm $\ln \left ( { t \over t_0 } \right ) \cdot u(t)$ $- { t_0 \over s} \ [ \ \ln(t_0 s)+\gamma \ ]$ $s > 0 \,$
12 Bessel function
of the first kind,
of order n
$J_n( \omega t) \cdot u(t)$ $\frac{ \omega^n \left(s+\sqrt{s^2+ \omega^2}\right)^{-n}}{\sqrt{s^2 + \omega^2}}$ $s > 0 \,$
$(n > -1) \,$
13 Modified Bessel function
of the first kind,
of order n
$I_n(\omega t) \cdot u(t)$ $\frac{ \omega^n \left(s+\sqrt{s^2-\omega^2}\right)^{-n}}{\sqrt{s^2-\omega^2}}$ $s > | \omega | \,$
14 Bessel function
of the second kind,
of order 0
$Y_0(\alpha t) \cdot u(t)$ $-{2 \sinh^{-1}(s/\alpha) \over \pi \sqrt{s^2+\alpha^2}}$ $s > 0 \,$
15 Modified Bessel function
of the second kind,
of order 0
$K_0(\alpha t) \cdot u(t)$
16 Error function $\mathrm{erf}(t) \cdot u(t)$ ${e^{s^2/4} \left(1 - \operatorname{erf} \left(s/2\right)\right) \over s}$ $s > 0 \,$
Explanatory notes:
 $u(t) \,$ represents the Heaviside step function. $\delta(t) \,$ represents the Dirac delta function. $\Gamma (z) \,$ represents the Gamma function. $\gamma \,$ is the Euler-Mascheroni constant. $t \,$, a real number, typically represents time, although it can represent any independent dimension. $s \,$ is the complex angular frequency. $\alpha \,$, $\beta \,$, $\tau \,$, and $\omega \,$ are real numbers. $n \,$is an integer.
• A causal system is a system where the impulse response h(t) is zero for all time t prior to t = 0. In general, the ROC for causal systems is not the same as the ROC for anticausal systems. See also causality.

## s-Domain equivalent circuits and impedances

The Laplace transform is often used in circuit analysis, and simple conversions to the s-Domain of circuit elements can be made. Circuit elements can be transformed into impedances, very similar to phasor impedances.

Here is a summary of equivalents:

Note that the resistor is exactly the same in the time domain and the s-Domain. The sources are put in if there are initial conditions on the circuit elements. For example, if a capacitor has an initial voltage across it, or if the inductor has an initial current through it, the sources inserted in the s-Domain account for that.

The equivalents for current and voltage sources are simply derived from the transformations in the table above.

## Examples: How to apply the properties and theorems

The Laplace transform is used frequently in engineering and physics; the output of a linear dynamic system can be calculated by convolving its unit impulse response with the input signal. Performing this calculation in Laplace space turns the convolution into a multiplication; the latter being easier to solve because of its algebraic form. For more information, see control theory.

The Laplace transform can also be used to solve differential equations and is used extensively in electrical engineering. The method of using the Laplace Transform to solve differential equations was developed by the English electrical engineer Oliver Heaviside.

The following examples, derived from applications in physics and engineering, will use SI units of measure. SI is based on meters for distance, kilograms for mass, seconds for time, and amperes for electric current.

### Example #1: Solving a differential equation

The following example is based on concepts from nuclear physics.

Consider the following first-order, linear differential equation:

$\frac{dN}{dt} = -\lambda N.$

This equation is the fundamental relationship describing radioactive decay, where

$N \ = \ N(t)$

represents the number of undecayed atoms remaining in a sample of a radioactive isotope at time t (in seconds), and $\ \lambda$ is the decay constant.

We can use the Laplace transform to solve this equation.

Rearranging the equation to one side, we have

$\frac{dN}{dt} + \lambda N = 0$

Next, we take the Laplace transform of both sides of the equation:

$\left( s \tilde{N}(s) - N_o \right) + \lambda \tilde{N}(s) \ = \ 0$

where

$\tilde{N}(s) = \mathcal{L}\{N(t)\}$

and

$N_o \ = \ N(0).$

Solving, we find

$\tilde{N}(s) = { N_o \over s + \lambda }.$

Finally, we take the inverse Laplace transform to find the general solution

$N(t) \ = \mathcal{L}^{-1} \{\tilde{N}(s)\} = \mathcal{L}^{-1} \left\{ \frac{N_o}{s + \lambda} \right\}$
$= \ N_o e^{-\lambda t}$,

which is indeed the correct form for radioactive decay.

### Example #2: Deriving the complex impedance for a capacitor

This example is based on the principles of electrical circuit theory.

The constitutive relation governing the dynamic behavior of a capacitor is the following differential equation:

$i = C { dv \over dt}$

where C is the capacitance (in farads) of the capacitor, i = i(t) is the electrical current (in amperes) flowing through the capacitor as a function of time, and v = v(t) is the voltage (in volts) across the terminals of the capacitor, also as a function of time.

Taking the Laplace transform of this equation, we obtain

$I(s) = C \left( s V(s) - V_o \right)$

where

$I(s) = \mathcal{L} \{ i(t) \}$,
$V(s) = \mathcal{L} \{ v(t) \}$, and
$V_o \ = \ v(t)|_{t=0}.$

Solving for V(s) we have

$V(s) = { I(s) \over Cs } + { V_o \over s }.$

The definition of the complex impedance Z (in ohms) is the ratio of the complex voltage V divided by the complex current I while holding the initial state Vo at zero:

$Z(s) = { V(s) \over I(s) } \bigg|_{V_o = 0}.$

Using this definition and the previous equation, we find:

$Z(s) = \frac{1}{sC}$

which is the correct expression for the complex impedance of a capacitor.

### Example #3: Finding the transfer function from the impulse response

Relationship between the time domain and the frequency domain. Note the * in the time domain, denoting convolution.
This example is based on concepts from signal processing, and describes the dynamic behavior of a damped harmonic oscillator. See also RLC circuit.

Consider a linear time-invariant system with impulse response

$h(t) = A e^{- \alpha t} \cos(\omega_d t - \phi_d) \,$

such that

$\omega_d t - \phi_d \ge 0 \,$

where t is the time (in seconds), and

$0 \le \phi_d \le 2 \pi$

is the phase delay (in radians).

Suppose that we want to find the transfer function of the system. We begin by noting that

$h(t) = A e^{- \alpha t} \cos \left[ \omega_d (t - t_d) \right] \cdot u(t - t_d) \,$

where

$t_d = { \phi_d \over \omega_d }$

is the time delay of the system (in seconds), and $\ u(t) \$ is the Heaviside step function.

The transfer function is simply the Laplace transform of the impulse response:

$H(s) \ = \ \mathcal{L} \{ h(t) \} = A e^{-s t_d} {(s + \alpha) \over (s + \alpha)^2 + \omega_d^2 }$
$= \ A e^{-s t_d} {(s + \alpha) \over (s^2 + 2 \alpha s + \alpha^2) + \omega_d^2 }$
$= \ A e^{-s t_d} {(s + \alpha) \over (s^2 + 2 \alpha s + \omega_0^2 ) }$

where

$\omega_0 = \sqrt{\alpha^2 + \omega_d^2}$

is the (undamped) natural frequency or resonance of the system (in radians per second).

### Example #4: Method of partial fraction expansion

Consider a linear time-invariant system with transfer function

$H(s) = \frac{1}{(s+\alpha)(s+\beta)}.$

The impulse response is simply the inverse Laplace transform of this transfer function:

$h(t) = \mathcal{L}^{-1}\{H(s)\}.$

To evaluate this inverse transform, we begin by expanding H(s) using the method of partial fraction expansion:

$H(s) = \frac{1}{(s+\alpha)(s+\beta)} = { P \over (s+\alpha) } + { R \over (s+\beta) }$

for unknown constants P and R. To find these constants, we evaluate

$P = \left.{1 \over (s+\beta)}\right|_{s=-\alpha} = {1 \over (\beta - \alpha)}$

and

$R = \left.{1 \over (s+\alpha)}\right|_{s=-\beta} = {1 \over (\alpha - \beta)} = {-1 \over (\beta - \alpha)} = - P .$

Substituting these values into the expression for H(s), we find

$H(s) = \left( \frac{1}{\beta-\alpha} \right) \cdot \left( { 1 \over (s+\alpha) } - { 1 \over (s+\beta) } \right).$

Finally, using the linearity property and the known transform for exponential decay (see Item #3 in the Table of Laplace Transforms, above), we can take the inverse Laplace transform of H(s) to obtain:

$h(t) = \mathcal{L}^{-1}\{H(s)\} = \frac{1}{\beta-\alpha}\left(e^{-\alpha t}-e^{-\beta t}\right),$

which is the impulse response of the system.

### Example #5: Mixing sines, cosines, and exponentials

Time function Laplace transform
$e^{-\alpha t}\left[ \cos{(\omega t)}+\left(\frac{\beta-\alpha}{\omega}\right)\sin{(\omega t)}\right]u(t)$ $\frac{s+\beta}{(s+\alpha)^2+\omega^2}$

Starting with the Laplace transform

$X(s) = \frac{s+\beta}{(s+\alpha)^2+\omega^2},$

we find the inverse transform by first adding and subtracting the same constant α to the numerator:

$X(s) = \frac{s+\alpha } { (s+\alpha)^2+\omega^2} + \frac{\beta - \alpha }{(s+\alpha)^2+\omega^2}.$

By the shift-in-frequency property, we have

$x(t) = e^{-\alpha t} \mathcal{L}^{-1} \left\{ {s \over s^2 + \omega^2} + { \beta - \alpha \over s^2 + \omega^2 } \right\}$
$= e^{-\alpha t} \mathcal{L}^{-1} \left\{ {s \over s^2 + \omega^2} + \left( { \beta - \alpha \over \omega } \right) \left( { \omega \over s^2 + \omega^2 } \right) \right\}$
$= e^{-\alpha t} \left[ \mathcal{L}^{-1} \left\{ {s \over s^2 + \omega^2} \right\} + \left( { \beta - \alpha \over \omega } \right) \mathcal{L}^{-1} \left\{ { \omega \over s^2 + \omega^2 } \right\} \right].$

Finally, using the Laplace transforms for sine and cosine (see the table, above), we have

$x(t) = e^{-\alpha t} \left[ \cos{(\omega t)}u(t)+\left(\frac{\beta-\alpha}{\omega}\right)\sin{(\omega t)}u(t)\right].$
$x(t) = e^{-\alpha t} \left[ \cos{(\omega t)}+\left(\frac{\beta-\alpha}{\omega}\right)\sin{(\omega t)}\right]u(t).$

### Example #6: Phase delay

Time function Laplace transform
$\sin{(\omega t+\phi)} \$ $\frac{s\sin\phi+\omega \cos\phi}{s^2+\omega^2} \$
$\cos{(\omega t+\phi)} \$ $\frac{s\cos\phi - \omega \sin\phi}{s^2+\omega^2} \$

Starting with the Laplace transform,

$X(s) = \frac{s\sin\phi+\omega \cos\phi}{s^2+\omega^2}$

we find the inverse by first rearranging terms in the fraction:

 $X(s) \,\!$ ${} = \frac{s \sin \phi}{s^2 + \omega^2} + \frac{\omega \cos \phi}{s^2 + \omega^2}$ ${} = (\sin \phi) \left(\frac{s}{s^2 + \omega^2} \right) + (\cos \phi) \left(\frac{\omega}{s^2 + \omega^2} \right).$

We are now able to take the inverse Laplace transform of our terms:

 $x(t) \,\!$ ${} = (\sin \phi) \mathcal{L}^{-1}\left\{\frac{s}{s^2 + \omega^2} \right\} + (\cos \phi) \mathcal{L}^{-1}\left\{\frac{\omega}{s^2 + \omega^2} \right\}$ ${}=(\sin \phi)(\cos \omega t) + (\sin \omega t)(cos \phi). \,\!$

To simplify this answer, we must recall the trigonometric identity that

$a \sin \omega t + b \cos \omega t = \sqrt{a^2+b^2} \cdot \sin \left(\omega t + \arctan (b/a) \right)$

and apply it to our value for x(t):

 $x(t) \,\!$ ${} = \sqrt{\cos^2 \phi + \sin^2 \phi} \cdot \sin \left( \omega t + \arctan \left(\frac{\sin \phi}{\cos \phi} \right) \right)$ ${}= \sin (\omega t + \phi) . \,\!$

We can apply similar logic to find that

$\mathcal{L}^{-1} \left\{ \frac{s\cos\phi - \omega \sin\phi}{s^2+\omega^2} \right\} = \cos{(\omega t+\phi)}. \$

## References

• A. D. Polyanin and A. V. Manzhirov, Handbook of Integral Equations, CRC Press, Boca Raton, 1998. ISBN 0-8493-2876-4
• William McC. Siebert, Circuits, Signals, and Systems, MIT Press, Cambridge, Massachusetts, 1986. ISBN 0-262-19229-2
• Davies, Brian, Integral transforms and their applications (Springer, New York, 1978). ISBN 0-387-90313-5

Pierre-Simon Laplace

# Wiktionary

Up to date as of January 14, 2010

## English

### Noun

Wikipedia has an article on:

Wikipedia

Laplace transform

1. (mathematics) a function on positive real numbers such that differentiation and integration are reduced to multiplication and division

# Simple English

A Laplace transform is a way to turn functions into other functions in order to do certain calculations more easily. This way of turning functions to other functions is very similar to U Substitution. The aim of this change is to be able to turn the hard work of integration into a simple algebraic addition and subtraction, just as logarithms allow us to add and subtract instead of multiplying and dividing.