Bilateral Laplace transform

When one says "the Laplace transform" without qualification, the unilateral or one-sided transform is usually 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 F(s) is defined as follows:

${\displaystyle F(s)=\int _{-\infty }^{\infty }e^{-st}f(t)\,dt}$

(Eq.2)

An alternate notation for the bilateral Laplace transform is ${\displaystyle {\mathcal {B}}\{f\}}$, instead of F.

Inverse Laplace transform

Two integrable functions have the same Laplace transform only if they differ on a set of Lebesgue measure zero. This means that, on the range of the transform, there is an inverse transform. In fact, besides integrable functions, the Laplace transform is a one-to-one mapping from one function space into another in many other function spaces as well, although there is usually no easy characterization of the range.

Typical function spaces in which this is true include the spaces of bounded continuous functions, the space L^∞(0, ∞), or more generally tempered distributions on (0, ∞). The Laplace transform is also defined and injective for suitable spaces of tempered distributions.

In these cases, the image of the Laplace transform lives in a space of analytic functions in the region of convergence. 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):

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

(Eq.3)

where γ is a real number so that the contour path of integration is in the region of convergence of F(s). In most applications, the contour can be closed, allowing the use of the residue theorem. An alternative formula for the inverse Laplace transform is given by Post's inversion formula. The limit here is interpreted in the weak-* topology.

In practice, it is typically more convenient to decompose a Laplace transform into known transforms of functions obtained from a table, and construct the inverse by inspection.

Probability theory

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

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

By convention, this is referred to as the Laplace transform of the random variable X itself. Here, 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.

Of particular use is the ability to recover the cumulative distribution function of a continuous random variable X, by means of the Laplace transform as follows:^[19]

$${\displaystyle F_{X}(x)={\mathcal {L}}^{-1}\!\left\{{\frac {1}{s}}\operatorname {E} \left[e^{-sX}\right]\right\}\!(x)={\mathcal {L}}^{-1}\!\left\{{\frac {1}{s}}{\mathcal {L}}\{f\}(s)\right\}\!(x).}$$

Algebraic construction

The Laplace transform can be alternatively defined in a purely algebraic manner by applying a field of fractions construction to the convolution ring of functions on the positive half-line. The resulting space of abstract operators is exactly equivalent to Laplace space, but in this construction the forward and reverse transforms never need to be explicitly defined (avoiding the related difficulties with proving convergence).^[20]

Relation to power series

The Laplace transform can be viewed as a continuous analogue of a power series.^[25] If a(n) is a discrete function of a positive integer n, then the power series associated to a(n) is the series

$${\displaystyle \sum _{n=0}^{\infty }a(n)x^{n}}$$

where x is a real variable (see Z-transform). Replacing summation over n with integration over t, a continuous version of the power series becomes

$${\displaystyle \int _{0}^{\infty }f(t)x^{t}\,dt}$$

where the discrete function a(n) is replaced by the continuous one f(t).

Changing the base of the power from x to e gives

$${\displaystyle \int _{0}^{\infty }f(t)\left(e^{\ln {x}}\right)^{t}\,dt}$$

For this to converge for, say, all bounded functions f, it is necessary to require that ln x < 0. Making the substitution −s = ln x gives just the Laplace transform:

$${\displaystyle \int _{0}^{\infty }f(t)e^{-st}\,dt}$$

In other words, the Laplace transform is a continuous analog of a power series, in which the discrete parameter n is replaced by the continuous parameter t, and x is replaced by e^−s.

Relation to moments

The quantities

$${\displaystyle \mu _{n}=\int _{0}^{\infty }t^{n}f(t)\,dt}$$

are the moments of the function f. If the first n moments of f converge absolutely, then by repeated differentiation under the integral,

$${\displaystyle (-1)^{n}({\mathcal {L}}f)^{(n)}(0)=\mu _{n}.}$$

This is of special significance in probability theory, where the moments of a random variable X are given by the expectation values ${\displaystyle \mu _{n}=\operatorname {E} [X^{n}]}$. Then, the relation holds

$${\displaystyle \mu _{n}=(-1)^{n}{\frac {d^{n}}{ds^{n}}}\operatorname {E} \left[e^{-sX}\right](0).}$$

Computation 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:

$${\displaystyle {\begin{aligned}{\mathcal {L}}\left\{f(t)\right\}&=\int _{0^{-}}^{\infty }e^{-st}f(t)\,dt\\[6pt]&=\left[{\frac {f(t)e^{-st}}{-s}}\right]_{0^{-}}^{\infty }-\int _{0^{-}}^{\infty }{\frac {e^{-st}}{-s}}f'(t)\,dt\quad {\text{(by parts)}}\\[6pt]&=\left[-{\frac {f(0^{-})}{-s}}\right]+{\frac {1}{s}}{\mathcal {L}}\left\{f'(t)\right\},\end{aligned}}}$$

yielding

$${\displaystyle {\mathcal {L}}\{f'(t)\}=s\cdot {\mathcal {L}}\{f(t)\}-f(0^{-}),}$$

and in the bilateral case,

$${\displaystyle {\mathcal {L}}\{f'(t)\}=s\int _{-\infty }^{\infty }e^{-st}f(t)\,dt=s\cdot {\mathcal {L}}\{f(t)\}.}$$

The general result

$${\displaystyle {\mathcal {L}}\left\{f^{(n)}(t)\right\}=s^{n}\cdot {\mathcal {L}}\{f(t)\}-s^{n-1}f(0^{-})-\cdots -f^{(n-1)}(0^{-}),}$$

where ${\displaystyle f^{(n)}}$ denotes the nth derivative of f, can then be established with an inductive argument.

Evaluating integrals over the positive real axis

A useful property of the Laplace transform is the following:

$${\displaystyle \int _{0}^{\infty }f(x)g(x)\,dx=\int _{0}^{\infty }({\mathcal {L}}f)(s)\cdot ({\mathcal {L}}^{-1}g)(s)\,ds}$$

under suitable assumptions on the behaviour of ${\displaystyle f,g}$ in a right neighbourhood of ${\displaystyle 0}$ and on the decay rate of ${\displaystyle f,g}$ in a left neighbourhood of ${\displaystyle \infty }$. The above formula is a variation of integration by parts, with the operators ${\displaystyle {\frac {d}{dx}}}$ and ${\displaystyle \int \,dx}$ being replaced by ${\displaystyle {\mathcal {L}}}$ and ${\displaystyle {\mathcal {L}}^{-1}}$. Let us prove the equivalent formulation:

$${\displaystyle \int _{0}^{\infty }({\mathcal {L}}f)(x)g(x)\,dx=\int _{0}^{\infty }f(s)({\mathcal {L}}g)(s)\,ds.}$$

By plugging in ${\displaystyle ({\mathcal {L}}f)(x)=\int _{0}^{\infty }f(s)e^{-sx}\,ds}$ the left-hand side turns into:

$${\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }f(s)g(x)e^{-sx}\,ds\,dx,}$$

but assuming Fubini's theorem holds, by reversing the order of integration we get the wanted right-hand side.

This method can be used to compute integrals that would otherwise be difficult to compute using elementary methods of real calculus. For example,

$${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x}}dx=\int _{0}^{\infty }{\mathcal {L}}(1)(x)\sin xdx=\int _{0}^{\infty }1\cdot {\mathcal {L}}(\sin )(x)dx=\int _{0}^{\infty }{\frac {dx}{x^{2}+1}}={\frac {\pi }{2}}.}$$

Laplace–Stieltjes transform

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

$${\displaystyle \{{\mathcal {L}}^{*}g\}(s)=\int _{0}^{\infty }e^{-st}\,d\,g(t)~.}$$

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

$${\displaystyle g(x)=\int _{0}^{x}f(t)\,d\,t}$$

then the Laplace–Stieltjes transform of g and the Laplace transform of f 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.^[26]

Fourier transform

The Fourier transform is a special case (under certain conditions) of the bilateral Laplace transform. While the Fourier transform of a function is a complex function of a real variable (frequency), the Laplace transform of a function is a complex function of a complex variable. The Laplace transform is usually restricted to transformation of functions of t with t ≥ 0. A consequence of this restriction is that the Laplace transform of a function is a holomorphic function of the variable s. Unlike the Fourier transform, the Laplace transform of a distribution is generally a well-behaved function. Techniques of complex variables can also be used to directly study Laplace transforms. As a holomorphic function, the Laplace transform has a power series representation. This power series expresses a function as a linear superposition of moments of the function. This perspective has applications in probability theory.

The Fourier transform is equivalent to evaluating the bilateral Laplace transform with imaginary argument s = iω or s = 2πiξ^[27] when the condition explained below is fulfilled,

$${\displaystyle {\begin{aligned}{\hat {f}}(\omega )&={\mathcal {F}}\{f(t)\}\\[4pt]&={\mathcal {L}}\{f(t)\}|_{s=i\omega }=F(s)|_{s=i\omega }\\[4pt]&=\int _{-\infty }^{\infty }e^{-i\omega t}f(t)\,dt~.\end{aligned}}}$$

This convention of the Fourier transform (${\displaystyle {\hat {f}}_{3}(\omega )}$ in Fourier transform § Other conventions) requires a factor of 1/2π on the inverse 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) has a Laplace transform F(s) = s/(s² + ω₀²) whose ROC is Re(s) > 0. As s = iω₀ is a pole of F(s), substituting s = iω in F(s) does not yield the Fourier transform of f(t)u(t), which is proportional to the Dirac delta function δ(ω − ω₀).

However, a relation of the form

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

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

$${\displaystyle G(s)={\mathcal {M}}\{g(\theta )\}=\int _{0}^{\infty }\theta ^{s}g(\theta )\,{\frac {d\theta }{\theta }}}$$

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

$${\displaystyle z{\stackrel {\mathrm {def} }{{}={}}}e^{sT},}$$

where T = 1/f_s is the sampling interval (in units of time e.g., seconds) and f_s is the sampling rate (in samples per second or hertz).

Let

$${\displaystyle \Delta _{T}(t)\ {\stackrel {\mathrm {def} }{=}}\ \sum _{n=0}^{\infty }\delta (t-nT)}$$

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

$${\displaystyle {\begin{aligned}x_{q}(t)&{\stackrel {\mathrm {def} }{{}={}}}x(t)\Delta _{T}(t)=x(t)\sum _{n=0}^{\infty }\delta (t-nT)\\&=\sum _{n=0}^{\infty }x(nT)\delta (t-nT)=\sum _{n=0}^{\infty }x[n]\delta (t-nT)\end{aligned}}}$$

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

$${\displaystyle x[n]{\stackrel {\mathrm {def} }{{}={}}}x(nT)~.}$$

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

$${\displaystyle {\begin{aligned}X_{q}(s)&=\int _{0^{-}}^{\infty }x_{q}(t)e^{-st}\,dt\\&=\int _{0^{-}}^{\infty }\sum _{n=0}^{\infty }x[n]\delta (t-nT)e^{-st}\,dt\\&=\sum _{n=0}^{\infty }x[n]\int _{0^{-}}^{\infty }\delta (t-nT)e^{-st}\,dt\\&=\sum _{n=0}^{\infty }x[n]e^{-nsT}~.\end{aligned}}}$$

This is the precise definition of the unilateral Z-transform of the discrete function x[n]

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

with the substitution of z → e^sT.

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

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

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

$${\displaystyle F(s)=\int _{0}^{\infty }f(z)e^{-sz}\,dz}$$

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

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

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.

Evaluating improper integrals

Let ${\displaystyle {\mathcal {L}}\left\{f(t)\right\}=F(s)}$. Then (see the table above)

$${\displaystyle \partial _{s}{\mathcal {L}}\left\{{\frac {f(t)}{t}}\right\}=\partial _{s}\int _{0}^{\infty }{\frac {f(t)}{t}}e^{-st}\,dt=-\int _{0}^{\infty }f(t)e^{-st}=-F(s)}$$

From which one gets:

$${\displaystyle {\mathcal {L}}\left\{{\frac {f(t)}{t}}\right\}=\int _{s}^{\infty }F(p)\,dp.}$$

In the limit ${\displaystyle s\rightarrow 0}$, one gets

$${\displaystyle \int _{0}^{\infty }{\frac {f(t)}{t}}\,dt=\int _{0}^{\infty }F(p)\,dp,}$$

provided that the interchange of limits can be justified. This is often possible as a consequence of the final value theorem. Even when the interchange cannot be justified the calculation can be suggestive. For example, with a ≠ 0 ≠ b, proceeding formally one has

$${\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {\cos(at)-\cos(bt)}{t}}\,dt&=\int _{0}^{\infty }\left({\frac {p}{p^{2}+a^{2}}}-{\frac {p}{p^{2}+b^{2}}}\right)\,dp\\[6pt]&=\left[{\frac {1}{2}}\ln {\frac {p^{2}+a^{2}}{p^{2}+b^{2}}}\right]_{0}^{\infty }={\frac {1}{2}}\ln {\frac {b^{2}}{a^{2}}}=\ln \left|{\frac {b}{a}}\right|.\end{aligned}}}$$

The validity of this identity can be proved by other means. It is an example of a Frullani integral.

Another example is Dirichlet integral.

Complex impedance of a capacitor

In the theory of electrical circuits, the current flow in a capacitor is proportional to the capacitance and rate of change in the electrical potential (with equations as for the SI unit system). Symbolically, this is expressed by the differential equation

$${\displaystyle i=C{dv \over dt},}$$

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

Taking the Laplace transform of this equation, we obtain

$${\displaystyle I(s)=C(sV(s)-V_{0}),}$$

where

$${\displaystyle {\begin{aligned}I(s)&={\mathcal {L}}\{i(t)\},\\V(s)&={\mathcal {L}}\{v(t)\},\end{aligned}}}$$

and

$${\displaystyle V_{0}=v(0).}$$

Solving for V(s) we have

$${\displaystyle V(s)={I(s) \over sC}+{V_{0} \over s}.}$$

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 V₀ at zero:

$${\displaystyle Z(s)=\left.{V(s) \over I(s)}\right|_{V_{0}=0}.}$$

Using this definition and the previous equation, we find:

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

which is the correct expression for the complex impedance of a capacitor. In addition, the Laplace transform has large applications in control theory.

Impulse response

Consider a linear time-invariant system with transfer function

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

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

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

Partial fraction expansion

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

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

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

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

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

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

Similarly, the residue R is given by

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

Note that

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

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

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

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

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

which is the impulse response of the system.

Convolution

The same result can be achieved using the convolution property as if the system is a series of filters with transfer functions 1/(s + α) and 1/(s + β). That is, the inverse of

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

is

$${\displaystyle {\mathcal {L}}^{-1}\!\left\{{\frac {1}{s+\alpha }}\right\}*{\mathcal {L}}^{-1}\!\left\{{\frac {1}{s+\beta }}\right\}=e^{-\alpha t}*e^{-\beta t}=\int _{0}^{t}e^{-\alpha x}e^{-\beta (t-x)}\,dx={\frac {e^{-\alpha t}-e^{-\beta t}}{\beta -\alpha }}.}$$

Phase delay

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

Starting with the Laplace transform,

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

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

$${\displaystyle {\begin{aligned}X(s)&={\frac {s\sin(\varphi )}{s^{2}+\omega ^{2}}}+{\frac {\omega \cos(\varphi )}{s^{2}+\omega ^{2}}}\\&=\sin(\varphi )\left({\frac {s}{s^{2}+\omega ^{2}}}\right)+\cos(\varphi )\left({\frac {\omega }{s^{2}+\omega ^{2}}}\right).\end{aligned}}}$$

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

$${\displaystyle {\begin{aligned}x(t)&=\sin(\varphi ){\mathcal {L}}^{-1}\left\{{\frac {s}{s^{2}+\omega ^{2}}}\right\}+\cos(\varphi ){\mathcal {L}}^{-1}\left\{{\frac {\omega }{s^{2}+\omega ^{2}}}\right\}\\&=\sin(\varphi )\cos(\omega t)+\cos(\varphi )\sin(\omega t).\end{aligned}}}$$

This is just the sine of the sum of the arguments, yielding:

$${\displaystyle x(t)=\sin(\omega t+\varphi ).}$$

We can apply similar logic to find that

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

Statistical mechanics

In statistical mechanics, the Laplace transform of the density of states ${\displaystyle g(E)}$ defines the partition function.^[36] That is, the canonical partition function ${\displaystyle Z(\beta )}$ is given by

$${\displaystyle Z(\beta )=\int _{0}^{\infty }e^{-\beta E}g(E)\,dE}$$

and the inverse is given by