University of Central FloridaSchool of Electrical Engineering and Computer ScienceEGN-3420 - Engineering Analysis.Fall 2009 - dcmLaplace Transform and its application for solving differentialequations. Fourier and Z TransformsMotivation. Transform methods are widely used in many areas of science and engineering. Forexample, transform methods are used in signal processing and circuit analysis, in applicationsof probability theory. The basic idea is to transform a function from its original domain intoa transform domain where certain operations can be carried out more efficiently, carrying outthe operation in the transform domain, and then carrying out an inverse transform of theresult (from the transform domain to the original domain).For example, the convolution operation of two functions of time t, f(t) and g(t) is definedas:f(t) ∗g(t) =Z+∞−∞f(τ ) · g(t −τ)dτ =Z+∞−∞f(t −τ) ·g(τ )dτwith τ a real number. The convolution in the time domain becomes multiplication in theLaplace or Fourier domain. As an application, consider a linear circuit with the impulseresponse h(t) and with the signal x(t) as input. Then the output of the linear circuit isy(t) = x(t)∗h(t). If H(s), X(s) and Y (s) are the Laplace transforms of the impulse response,the input, and the output, respectively, then Y (s) = H(s)·X(s), as seen in Figure 1. Once weknow Y (s) we can apply the inverse Laplace Transform to obtain the response of the circuit,y(t), function of time t.Transform methods provide a bridge between the commonly used method of separationof variables and numerical techniques for solving linear partial differential equations. Whilein some ways similar to separation of variables, transform methods can be effective for awider class of problems. Even when the inverse of the transform cannot be found analytically,numeric and asymptotic techniques now exist for their inversion.Laplace Transform. Let R be the field of real numbers and C the field of complex numbers.Consider a function f : R 7→ R such that f(t), t ∈ R, t ≥ 0. Then the Laplace Transform off(t) is denoted as L[f(t)] and it is defined as F (s) with s ∈ C:F (s) = L[f(t)] =R∞0e−stf(t)dt.The Laplace transform F (s) typically exists for all complex numbers s such that Re(s) > awhere a ∈ R is a constant which depends on the behavior of f(t).The Inverse Laplace Transform is given by the following complex integral:f(t) = L−1[F (s)] =12πilimT →∞Rγ+iTγ−iTestF (s)ds1Figure 1: A circuit with the impulse response h(t) and x(t) as input. Then the output isy(t) = x(t) ∗ h(t). If H(s), X(s) and Y (s) are respectively the Laplace transforms of theimpulse response, the input, and the output, then Y (s) = H(s) · X(s)where γ is a real number so that the contour path of integration is in the region of convergenceof F (s) normally requiring sρ> Re(sρ) for every singularity sρof F (s) and with i =√−1.If all singularities are in the left half-plane (in this case Re(sρ) < 0 for every sρ), then γ canbe set to zero and the above inverse integral formula becomes identical to the inverse Fouriertransform. This integral is known as the Fourier-Mellin integral.The Bilateral Laplace Transform is defined as:F (s) = L[f(t)] =Z∞−∞e−stf(t)dt.In probability theory, the Laplace transform is defined by means of an expectation value.If X is a random variable with probability density function fX, then the Laplace transformof fXis given by the expectation:(LfX)(s) = E[e−sX].By abuse of language, one often refers instead to this as the Laplace transform of therandom variable X. Replacing the variable s by −t gives the moment generating functionof X. The Laplace transform has applications throughout probability theory, including firstpassage times of stochastic processes such as Markov chains and renewal theory.Laplace Transforms of a few functions f(t). In each case we start from the definition. Forexample, f(t) = k:F (s) = L[k] =Z∞0e−stkdt = −ks£e−st¤t7→∞t=0=ks.2Now f(t) = eαtF (s) = L£eαt¤=Z∞0e−steαtdt = −1s − α£e−(s−α)t¤t7→∞t=0=1s − α.Properties of the Laplace Transform. The properties of the Laplace Transform summarizedin Table 1 can be derived easily starting from the definition.Table 1: Properties of the Laplace Transform. The function f(t) is assumed to be n-timesdifferentiable, with n-th derivative of exponential type. Notations: F (s) = L[f(t)], G(s) =L[g(t)], f(n)is the n-th derivative of f(t), F(n)(s) is the n-th derivative of F (s), u(t) is theHeaviside step function, u(t) =Rt−∞δ(τ)dτ with δ the Dirac delta function, (f ∗ g)(t) is theconvolution of f(t) and g(t), α, β ∈ R.Property Time-domain s-domainLinearity αf(t) + βg(t) αF (s) + βG(s)Scaling f(α t)1αF (sα)Frequency shifting eαtf(t) F (s − α)Time shifting f(t − α)u(t − α) e−αsF (s)Frequency differentiation tnf(t) (−1)nF(n)(s)Frequency integration f(t)/tR∞sF (r)drDifferentiation f(n)(t) snF (s) − sn−1f(0) − ... − f(n−1)(0))IntegrationRt0f(τ )dτ (u ∗ f)(t)1sF (s)Convolution (f ∗ g)(t) F (s) ∗ G(s)Periodic function f(t) = f(t + T ) f(t)11−e−T sRT0e−stf(t)dtFor example, to prove linearity consider two functions f (t) and g(t) and their LaplaceTransforms:F (s) = L[f(t)] =Z∞0e−stf(t)dt and G(s) = L[g(t)] =Z∞0e−stg(t)dt.From the definition of the Laplace Transform it follows thatL[f(t) + g(t)] =Z∞0e−st[f(t) + g(t)] dt =Z∞0e−stf(t)dt +Z∞0e−stg(t)dt = F (s) + G(s).It is also easy to see that F (0) represents the area under the curve f(t):F (s = 0)Z∞0f(t)dtThe Laplace Transform can be expressed as:L[f(t)] =f(0)s+f0(0)s2+f00(0)s3+f000(0)s4+ . . . .3Proof: This important property of the Laplace Transform is a consequence of the followingequality:Ze−αxf(x)dx = −e−αxα·f(x) +f0(x)α+f00(x)α2+f000(x)α3+ . . .¸This is easy to prove by applying the derivation operator of both sides; then the left handside becomes A = e−αxf(x). The right hand side is the sum of two terms B and C:B = αe−αxα·f(x) +f0(x)α+f00(x)α2+f000(x)α3+ . . .¸C = −e−αxα·f0(x) +f00(x)α+f000(x)α2+fiv(x)α3+ . . .¸ThenB + C = e−αxf(x).Thus A = B + C and this equality allows us to express the Laplace Transform as:L[f(t)] =½−e−sts·f(t) +f0(t)s+f00(t)s2+f000(t)s3+ . . .¸¾∞0.But:limt7→∞½−e−sts·f(t) +f0(t)s+f00(t)s2+f000(t)s3+ . . .¸¾= 0After we subtract the value of the expression for t = 0 we obtain the result