Linearity: Lfc1f(t)+c2g(t)g = c1Lff(t)g+c2Lfg(t)g. 2. Determine L 1fFgfor (a) F(s) = 2 s3, (b) F(s) = 3 s 2+ 9, (c) F(s) = s 1 s 2s+ 5. First derivative: Lff0(t)g = sLff(t)g¡f(0). The following table are useful for applying this technique. We get the solution y(t) by taking the inverse Laplace transform. 18.031 Laplace Transform Table Properties and Rules Function Transform f(t) F(s) = Z 1 0 f(t)e st dt (De nition) af(t) + bg(t) aF(s) + bG(s) (Linearity) eatf(t) F(s a) (s-shift) f0(t) sF(s) f(0 ) f00(t) s2F(s) sf(0 ) f0(0 ) f(n)(t) snF(s) sn 1f(0 ) f(n 1)(0 ) tf(t) F0(s) t nf(t) ( 1)nF( )(s) u(t a)f(t a) e asF(s) (t-translation or t-shift) u(t a)f(t) e asL(f(t+ a)) (t-translation) In other … Properties of Laplace transform: 1. The text has a more detailed table. Table 1: Table of Laplace Transforms Number f(t) F(s) 1 δ(t)1 2 us(t) 1 s 3 t 1 s2 4 tn n! Solution. 6.2: Solution of initial value problems (4) Topics: † Properties of Laplace transform, with proofs and examples † Inverse Laplace transform, with examples, review of partial fraction, † Solution of initial value problems, with examples covering various cases. Table 1: A List of Laplace and Inverse Laplace Transforms Related to Fractional Order Calculus. INVERSE LAPLACE TRANSFORMS In this appendix, we provide additional unilateral Laplace transform pairs in Table B.1 and B.2, giving the s-domain expression first. – – Kronecker delta δ0(k) 1 k = 0 0 k ≠ 0 1 2. The inverse Laplace transform We can also define the inverse Laplace transform: given a function X(s) in the s-domain, its inverse Laplace transform L−1[X(s)] is a function x(t) such that X(s) = L[x(t)]. nding inverse Laplace transforms is a critical step in solving initial value problems. Example 1. It can be shown that the Laplace transform of a causal signal is unique; hence, the inverse Laplace transform is uniquely defined as well. To determine the inverse Laplace transform of a function, we try to match it with the form of an entry in the right-hand column of a Laplace table. tedious to deal with, one usually uses the Cauchy theorem to evaluate the inverse transform using f(t) = Σ enclosed residues of F (s)e st. tn−1 L eat = 1 s−a L−1 1 s−a = eat L[sinat] = a s 2+a L−1 1 s +a2 = 1 a sinat L[cosat] = s s 2+a L−1 s s 2+a = cosat Differentiation and integration L d dt f(t) = sL[f(t)]−f(0) L d2t dt2 f(t) = s2L[f(t)]−sf(0)−f0(0) L dn … The final stage in that solution procedure involves calulating inverse Laplace transforms. TABLE OF LAPLACE TRANSFORM FORMULAS L[tn] = n! 3 2 s t2 (kT)2 ()1 3 2 1 1 s n+1 L−1 1 s = 1 (n−1)! 2 1 s t kT ()2 1 1 1 − −z Tz 6. 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