Laplace Transform and Inverse

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The Laplace transform was first introduced by Pierre-Simon Laplace in his work on probability theory. The transform turns out to have many applications in science and engineering, especially in solving linear differential equations.

The Laplace Transform

Let f(t)f(t) be defined for 0t<0\leq t<\infty. The Laplace transform of f(t)f(t), which is denoted by F(s)F(s), or L{f(t)}\mathcal{L}\left\{f(t)\right\} is given by

F(s)=L{f(t)}=0estf(t)dt\begin{equation} F\left(s\right)=\mathcal{L}\left\{f\left(t\right)\right\}=\int_{0}^{\infty}e^{-st}f\left(t\right)\mathrm{d}t \end{equation}

where the improper integral is understood as

0estf(t)dt=limA0Aestf(t)dt\int_{0}^{\infty}e^{-st}f\left(t\right)\mathrm{d}t=\lim_{A\to \infty}\int_{0}^{A}e^{-st}f\left(t\right)\mathrm{d}t

Remarks:

Two conditions that ff should be satisfied to have a Laplace transform (for sufficiently large ss) are

  1. The function f(t)f(t) is of exponential order, i.e., there exist constants MM and cc such that for all 0t<0\le t<\infty,
f(t)Mect\left|f\left(t\right)\right|\le Me^{ct}
  1. The function f(t)f(t) is piecewise continuous, i.e., f(t)f(t) has at most a finite number of discontinuities on any interval 0tA0\le t\le A, and both the limit from the right and the limit from the left of ff exist at every point of discontinuity.

The Laplace transform doesn’t care how the function f(t)f(t) behaves when t<0t<0, this means that different functions can have the same Laplace transform as long as they agree on t0t\ge 0.

Properties

In the following, we will assume L{f(t)}=F(s)\mathcal{L}\left\{f(t)\right\}=F(s), L{g(t)}=G(s)\mathcal{L}\left\{g(t)\right\}=G(s), or denotes as f(t)F(s)f(t)\rightsquigarrow F(s), g(t)G(s)g(t)\rightsquigarrow G(s). We have the following important properties.

  1. Linearity
L{αf(t)+βg(t)}=αF(s)+βG(s)\mathcal{L}\left\{\alpha f(t)+\beta g(t)\right\}=\alpha F(s)+\beta G(s)
  1. Frequency domain derivatives
tf(t)F(s) tf\left(t\right)\rightsquigarrow-F'\left(s\right)
tnf(t)(1)nF(n)(s)t^{n}f\left(t\right) \rightsquigarrow \left(-1\right)^{n}F^{\left(n\right)}\left(s\right)
  1. Time domain derivatives
f(t)sF(s)f(0)f'\left(t\right)\rightsquigarrow sF(s)-f(0)
f(t)s2F(s)sf(0)f(0)f''(t)\rightsquigarrow s^2F(s)-sf(0)-f'(0)
f(n)(t)snF(s)k=1nsnkf(k1)(0)f^{\left(n\right)}(t)\rightsquigarrow s^{n}F(s)-\sum_{k=1}^{n}s^{n-k}f^{\left(k-1\right)}\left(0\right)
  1. Frequency shifting
eatf(t)F(sa)e^{at}f(t)\rightsquigarrow F\left(s-a\right)
  1. Time shifting
u(ta)f(ta)easF(s)u(t-a)f(t-a)\rightsquigarrow e^{-as}F(s)
f(t)u(ta)easL(f(t+a))f(t)u(t-a)\rightsquigarrow e^{-as}\mathcal{L}\left(f(t+a)\right)

where u(t)u(t) is the Heaviside step function (unit step function), and a>0a>0.

  1. Time scaling
f(at)1aF(sa)f(at)\rightsquigarrow \frac{1}{a}F\left(\frac{s}{a}\right)

where a>0a>0.

  1. Convolution

The convolution between two functions is defined as

(fg)(t)=0tf(τ)g(tτ)dτ(f*g)\left(t\right)=\int_{0}^{t}f(\tau)g(t-\tau)\mathrm{d}\tau

Then, the Laplace transform of the convolution is the ordinary product of the Laplace transforms,

(fg)(t)F(s)G(s)(f*g)\left(t\right)\rightsquigarrow F(s)G(s)

Laplace Transform of Common Functions

The following lists the Laplace transform of some common functions:

  1. Dirac Delta function (unit impulse)
δ(t)1\delta\left(t\right)\rightsquigarrow 1
δ(ta)eas\delta\left(t-a\right)\rightsquigarrow e^{-as}
  1. Heaviside step function (unit step)
u(t)1su(t)\rightsquigarrow \frac{1}{s}
u(ta)1seasu(t-a)\rightsquigarrow \frac{1}{s}e^{-as}
  1. Polynomial
t1s2t\rightsquigarrow \frac{1}{s^2}

For integersnn, the following two formulas hold

tnn!sn+1t^{n}\rightsquigarrow \frac{n!}{s^{n+1}}
tn1s1n+1Γ(1n+1)\sqrt[n]{t}\rightsquigarrow \frac{1}{s^{\frac{1}{n}+1}}\Gamma\left(\frac{1}{n}+1\right)

where

Γ(z)=0tz1etdt\Gamma\left(z\right)=\int_{0}^{\infty}t^{z-1}e^{-t}\mathrm{d}t

is the Euler’s Gamma function, with Γ(1/2)=π\Gamma(1/2)=\pi.

  1. Exponential decay
eat1s+ae^{-at}\rightsquigarrow \frac{1}{s+a}
  1. Sine and Cosine
sin(ωt)ωs2+ω2\sin\left(\omega t\right)\rightsquigarrow \frac{\omega}{s^2+\omega^2}
cos(ωt)ss2+ω2\cos\left(\omega t\right)\rightsquigarrow \frac{s}{s^2+\omega^2}
  1. Hyperbolic Sine and Cosine
sinh(ωt)ωs2ω2\sinh\left(\omega t\right)\rightsquigarrow \frac{\omega}{s^2-\omega^2}
cosh(ωt)ss2ω2\cosh\left(\omega t\right)\rightsquigarrow \frac{s}{s^2-\omega^2}

Inverse Laplace Transform

The inversion of the Laplace Transform is given by the Mellin inversion formula. In particular, if ff is continuous on [0,)\left[0,\infty\right), continuously differentiable on (0,)\left(0,\infty\right), and satisfies the condition for the Laplace transform to exist, then

f(s)=[M(Lf)](s)f\left(s\right)=\left[\mathcal{M}\left(\mathcal{L}f\right)\right]\left(s\right)

for all s[0,)s\in\left[0,\infty\right), where

(MF)(t)=12πiCeptF(p)dp=12πiβiβ+ieptF(p)dp\left(\mathcal{M}F\right)\left(t\right)=\frac{1}{2\pi i}\int_{\mathcal{C}^*}e^{pt}F\left(p\right)\mathrm{d}p=\frac{1}{2\pi i}\int_{\beta-i\infty}^{\beta+i\infty}e^{pt}F\left(p\right)\mathrm{d}p

is called the Bromwich integral of FF, and the (positively oriented) contour is defined as

C={zC:Rez=β}\mathcal{C}=\left\{z\in\mathbb{C}: \mathrm{Re}z=\beta\right\}

where βR\beta\in\R is any number such that FF is analytic for all zCz\in\mathbb{C} with Rezβ\mathrm{Re}z\ge\beta.

To calculate the Bromwich integral, we should consider the case when t0t\ge0 and t<0t<0 separately.

For t0t\ge 0, the integration contour closed on the left.

Let Γ\Gamma be the straight line and Cβ,R\mathcal{C}_{\beta,R} be the semicircle (orientation indicated by the arrows). Then, by the residue theorem

ΓeptF(p)dp+Cβ,ReptF(p)dp=2πik=1NreszkF\int_{\Gamma}e^{pt}F\left(p\right)\mathrm{d}p+\int_{\mathcal{C}_{\beta,R}}e^{pt}F\left(p\right)\mathrm{d}p=2\pi i\sum_{k=1}^{N}\mathrm{res}_{z_k}F

Letting RR\to \infty, the first integral be the Bromwich integral we want. For the second integral, by applying the substitution rule,

Cβ,ReptF(p)dp=ieβtCReitpF(β+ip)dp\int_{\mathcal{C}_{\beta,R}}e^{pt}F\left(p\right)\mathrm{d}p=ie^{\beta t}\int_{\mathcal{C}_R}e^{itp}F\left(\beta+ip\right)\mathrm{d}p

where CR\mathcal{C}_R is a semi-circle of radius RR in the upper half-plane. In most cases, we can use Jordan’s Lemma to show that the right-hand side goes to 0 as RR\to \infty.

Then, the Bromwich integral of FF is just the sum of all the residues of FF,

12πiβiβ+ieptF(p)dp=k=1NreszkF\frac{1}{2\pi i}\int_{\beta-i\infty}^{\beta+i\infty}e^{pt}F\left(p\right)\mathrm{d}p=\sum_{k=1}^{N}\mathrm{res}_{z_k}F

For t<0t<0, the integration contour is closed on the left,

Let Γ\Gamma be the straight line and Cβ,R(2)\mathcal{C}_{\beta,R}^{(2)} be the semi-circle (orientation indicated by the arrows). Using a similar argument, we see that

Cβ,R(2)eptF(p)dp=ieβtCReitpF(βip)dp\int_{\mathcal{C}_{\beta,R}^{(2)}}e^{pt}F\left(p\right)\mathrm{d}p=ie^{\beta t}\int_{\mathcal{C}_R}e^{-itp}F\left(\beta-ip\right)\mathrm{d}p

Since the integral over the whole closed contour is zero, and the orientation of the straight line is in the opposite direction, the Bromwich integral is

12πiβiβ+ieptF(p)dp=12πilimRCβ,R(2)eptF(p)dp\frac{1}{2\pi i}\int_{\beta-i\infty}^{\beta+i\infty}e^{pt}F\left(p\right)\mathrm{d}p=\frac{1}{2\pi i}\lim_{R\to\infty}\int_{\mathcal{C}_{\beta,R}^{(2)}}e^{pt}F\left(p\right)\mathrm{d}p

which in most cases vanishes according to Jordan’s Lemma.

Reference

[1]. VV286 Lecture Slides, Horst Hohberger, UM-SJTU JI.

Categories: Mathematics