Laplace Transform

Posted By on January 3, 2015

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The Laplace transform is an integral transform perhaps second only to the Fourier transform in its utility in solving physical problems. The Laplace transform is particularly useful in solving linear ordinary differential equations such as those arising in the analysis of electronic circuits.

The (unilateral) Laplace transform L (not to be confused with the Lie derivative, also commonly denoted L) is defined by


where f(t) is defined for t>=0 (Abramowitz and Stegun 1972). The unilateral Laplace transform is almost always what is meant by “the” Laplace transform, although a bilateral Laplace transform is sometimes also defined as


(Oppenheim et al. 1997). The unilateral Laplace transform L_t[f(t)](s) is implemented in Mathematica as LaplaceTransform[f[t], t, s].

The inverse Laplace transform is known as the Bromwich integral, sometimes known as the Fourier-Mellin integral (see also the related Duhamel’s convolution principle).

A table of several important one-sided Laplace transforms is given below.

f L_t[f(t)](s) conditions
1 1/s
t 1/(s^2)
t^n (n!)/(s^(n+1)) n in Z>=0
t^a (Gamma(a+1))/(s^(a+1)) R[a]>-1
e^(at) 1/(s-a)
cos(omegat) s/(s^2+omega^2) omega in R
sin(omegat) omega/(s^2+omega^2) s>|I[omega]|
cosh(omegat) s/(s^2-omega^2) s>|R[omega]|
sinh(omegat) omega/(s^2-omega^2) s>|I[omega]|
e^(at)sin(bt) b/((s-a)^2+b^2) s>a+|I[b]|
e^(at)cos(bt) (s-a)/((s-a)^2+b^2) b in R
delta(t-c) e^(-cs)
H_c(t) {1/s   for c<=0; (e^(-cs))/s   for c>0
J_0(t) 1/(sqrt(s^2+1))
J_n(at) ((sqrt(s^2+a^2)-s)^n)/(a^nsqrt(s^2+a^2)) n in Z>=0

In the above table, J_0(t) is the zeroth-order Bessel function of the first kind, delta(t) is the delta function, and H_c(t) is the Heaviside step function.

The Laplace transform has many important properties. The Laplace transform existence theorem states that, if f(t) is piecewise continuous on every finite interval in [0,infty) satisfying


for all t in [0,infty), then L_t[f(t)](s) exists for all s>a. The Laplace transform is also unique, in the sense that, given two functions F_1(t) and F_2(t) with the same transform so that


then Lerch’s theorem guarantees that the integral


vanishes for all a>0 for a null function defined by


The Laplace transform is linear since

L_t[af(t)+bg(t)] = int_0^infty[af(t)+bg(t)]e^(-st)dt
= aint_0^inftyfe^(-st)dt+bint_0^inftyge^(-st)dt
= aL_t[f(t)]+bL_t[g(t)].

The Laplace transform of a convolution is given by


Now consider differentiation. Let f(t) be continuously differentiable n-1 times in [0,infty). If |f(t)|<=Me^(at), then


This can be proved by integration by parts,

L_t[f^'(t)](s) = lim_(a->infty)int_0^ae^(-st)f^'(t)dt
= lim_(a->infty){[e^(-st)f(t)]_0^a+sint_0^ae^(-st)f(t)dt}
= lim_(a->infty)[e^(-sa)f(a)-f(0)+sint_0^ae^(-st)f(t)dt]
= sL_t[f(t)]-f(0).

Continuing for higher-order derivatives then gives


This property can be used to transform differential equations into algebraic equations, a procedure known as the Heaviside calculus, which can then be inverse transformed to obtain the solution. For example, applying the Laplace transform to the equation




which can be rearranged to


If this equation can be inverse Laplace transformed, then the original differential equation is solved.

The Laplace transform satisfied a number of useful properties. Consider exponentiation. If L_t[f(t)](s)=F(s) for s>alpha (i.e., F(s) is the Laplace transform of f), then L_t[e^(at)f](s)=F(s-a) for s>a+alpha. This follows from

F(s-a) = int_0^inftyfe^(-(s-a)t)dt
= int_0^infty[f(t)e^(at)]e^(-st)dt
= L_t[e^(at)f(t)](s).

The Laplace transform also has nice properties when applied to integrals of functions. If f(t) is piecewise continuous and |f(t)|<=Me^(at), then


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Posted by Akash Kurup

Founder and C.E.O, World4Engineers Educationist and Entrepreneur by passion. Orator and blogger by hobby