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SUMMARY OF LAPLACE TRANSFORMS LANCE D. DRAGER Definition of Laplace Transform: If f (t) is defined for 0 ≤ t < ∞, L {f (t)} = F (s) where Z ∞ F (s) = e−st f (t) dt. 0 Transforms of Derivatives: L {f 0 (t)} = sF (s) − f (0) L {f 00 (t)} = s2 F (s) − sf (0) − f 0 (0) n o L f (n) (t) = sn F (s) − sn−1 f (0) − sn−2 f 0 (0) − · · · − f (n−1) (0). Translation in s: L eat f (t) = F (s − a). Unit Step Function: The unit step function (a.k.a., the Heaviside function) is ( 0, −∞ < t < 0, U (t) = 1, 0 ≤ t < ∞. For a > 0, we have ( 0, −∞ < t < a U (t − a) = 1, a ≤ t < ∞. Indicator functions: The indicator function of the interval (a, b) is ( 1, t ∈ (a, b) I(a,b) (t) = 0, t ∈ / (a, b). Assuming that 0 ≤ a < b < ∞ and that t is restricted to 0 ≤ t < ∞, indicator functions can be expressed in terms of unit step functions for the purposes of Laplace transforms by I(a,b) (t) = U (t − a) − U (t − b) I(0,b) (t) = 1 − U (t − b) I(a,∞) = U (t − a). (Given the book’s definition of U (t − a),as above, these are the indicator functions of [a, b), but that doesn’t matter for the Laplace transform.) Translation in t: For a > 0, L {U (t − a)f (t − a)} = e−as F (s) L {U (t − a)g(t)} = e−as L {g(t + a)} . Time-stamp: ”2013-10-29 14:07:30 drager”. 1 2 LANCE D. DRAGER Derivatives of Transforms: dn F (s). dsn Transform of a Periodic Function: If f (t) has period T , then Z T 1 L {f (t)} = e−st f (t) dt. 1 − e−sT 0 L {tn f (t)} = (−1)n Definition of Convolution: If f (t) and g(t) are defined for 0 ≤ t < ∞, then f ∗ g is defined on the same t range by Z t Z t g(τ )f (t − τ ) dτ, f (τ )g(t − τ ) dτ = (f ∗ g)(t) = 0 0 so f ∗ g = g ∗ f . Convolution Theorem: L {(f ∗ g)(t)} = F (s)G(s). Transform of Integrals: Z t F (s) L f (τ ) dτ = . s 0 Transform of the Delta Function: L {δ(t − a)} = e−as . Impulse response function: Let P (s) be a polynomial of degree n. The solution of the “formal” initial value problem P (D)y = δ(t), y(0) = 0, y 0 (0) = 0, . . . , y (n−1) (0) = 0 is 1 h(t) = L , P (s) which is called the impulse response function of the operator P (D). The solution of the initial value problem −1 P (D)y = f (x), y(0) = 0, y 0 (0) = 0, . . . , y (n−1) (0) = 0 (∗) is y(t) = L −1 Z {F (s)/P (s)} = (h ∗ f )(t) = t h(t − τ )f (τ ) dτ. 0 The solution of the homogeneous IVP P (D)y = 0, y(0) = A0 , y 0 (0) = A1 , . . . , y (n−1) (0) = An−1 , (∗∗) is g(t) = L −1 {Q(s)/P (s)} for some polynomial Q(s) of degree n − 1 that you’d have to calculate. The solution of the IVP P (D)y = f (t), y(0) = A0 , y 0 (0) = A1 , . . . , y (n−1) (0) = An−1 is the sum of the solutions to (∗) and (∗∗), Z t y(t) = h(t − τ )f (τ ) dτ + g(t), 0 SUMMARY OF LAPLACE TRANSFORMS 3 where (of course) g(t) does not depend of f (t). Short Table of Laplace Transforms Ref. f (t) (1) 1 (2) t (3) tn (4) eat (5) tn eat (6) cos(ωt) (7) sin(ωt) (8) eat cos(ωt) (9) eat sin(ωt) (10) t cos(ωt) (11) t sin(ωt) (12) sin(ωt) − ωt cos(ωt) F (s) 1 s 1 s2 n! sn+1 1 s−a n! (s − a)n+1 s s2 + ω 2 ω s2 + ω 2 s−a (s − a)2 + ω 2 ω (s − a)2 + ω 2 s2 − ω 2 (s2 + ω 2 )2 2ωs 2 (s + ω 2 )2 (s2 2ω 3 + ω 2 )2 Department of Mathematics and Statistics, Texas Tech University, Lubbock, TX 79409-1042 E-mail address: [email protected]