In this section we look at the problem of ﬁnding inverse Laplace transforms. 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 … Determine L 1fFgfor (a) F(s) = 2 s3, (b) F(s) = 3 s 2+ 9, (c) F(s) = s 1 s 2s+ 5. Example 1. Solution. nding inverse Laplace transforms is a critical step in solving initial value problems. It can be shown that the Laplace transform of a causal signal is unique; hence, the inverse Laplace transform is uniquely deﬁned as well. 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 of Laplace and Z-transforms X(s) x(t) x(kT) or x(k) X(z) 1. Table of Laplace Transforms Definition of Laplace transform 0 L{f (t)} e st f (t)dt f (t) L 1{F(s)} F(s) L{f (t)} Laplace transforms of elementary functions 1 s 1 tn 1! The text has a more detailed table. An abbreviated table of Laplace transforms was given in the previous lecture. TABLE OF LAPLACE TRANSFORM FORMULAS L[tn] = n! In other … Linearity: Lfc1f(t)+c2g(t)g = c1Lff(t)g+c2Lfg(t)g. 2. – – δ0(n-k) 1 n = k 0 n ≠ k z-k 3. s 1 1(t) 1(k) 1 1 1 −z− 4. s +a 1 e-at e-akT 1 1 1 −e−aT z− 5. 2 1 s t kT ()2 1 1 1 − −z Tz 6. 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) The ﬁnal stage in that solution procedure involves calulating inverse Laplace transforms. 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)]. 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. The following table are useful for applying this technique. 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. A List of Laplace and Inverse Laplace Transforms Related to Fractional Order Calculus 3 F(s) f(t) k s2+k2 coth ˇs 2k jsinkt 1 s e k=s J 0(2 p kt) p1 s e k=s p1 ˇt cos2 p kt p1 s … s n+1 L−1 1 s = 1 (n−1)! 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. 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! 6.3 Inverse Laplace Transforms Recall the solution procedure outlined in Figure 6.1. First derivative: Lff0(t)g = sLff(t)g¡f(0). 3 2 s t2 (kT)2 ()1 3 2 1 1 Signals & Systems - Reference Tables 1 Table of Fourier Transform Pairs Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform f t F( )ej td 2 1 ( ) Definition of Fourier Transform F() f (t)e j tdt f (t t0) F( )e j t0 f (t)ej 0t F 0 f ( t) ( ) 1 F F(t) 2 f n n dt In this course we shall use lookup tables to evaluate the inverse Laplace transform. – – Kronecker delta δ0(k) 1 k = 0 0 k ≠ 0 1 2. Properties of Laplace transform: 1.

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