Section 8.1: Basic Table of Laplace Transforms.pdf

Laplace transform Given a function, f (t), the Laplace Transform, denoted L{f (t)}(s), is defined by Z ∞ e −ts f (t) dt, L{f }(s) = 0 where the domain is all values of s for which the integral exists. We will almost always use the standard notation L{f (t)}(s) = F (s) to denote the Laplace transform of f (t). So lower-case letters will correspond to functions of t and upper-case functions will be the corresponding laplace transform. Example Find the Laplace transform of the following functions. 1. 1 2. e αt 3. t 4. t 2 5. t 3 6. sin(βt) 7. cos(βt) Existence of the Laplace Transform A function f (t) is said to be piecewise continuous on a finite interval [a, b] if f (t) is continuous at every point of [a, b], except possibly for a finite number of points at which f (t) has a jump discontinuity. A function f (t) is said to be piecewise continuous on [0, ∞) if f (t) is piecewise continuous on [0, N] for all N > 0. A function f (t) is said to be of exponential order α if there exist positive constants T and M such that |f (t)| ≤ Me αt for all t ≥ T . Conditions for Existence of the Laplace Transform If f (t) is piecewise continuous on [0, ∞) and of exponential order α, then L{f }(s) exists for s > α. Example Discuss whether the following functions are of exponential order. 1. t 3 sin(t) 2. 100e 49t 2 3. sin(e t ) + e sin(t) 4. e t 3 Brief Table of Laplace Transforms f (t) F (s) 1 1 s, s > 0 1 s−α , s > α n! ,s>0 s n+1 n! ,s>α (s−α)n+1 β ,s>α (s−α)2 +β 2 s−α ,s>α (s−α)2 +β 2 e αt t n , n = 1, 2, 3, . . . e αt t n , n = 1, 2, 3, . . . e αt sin(βt) e αt cos(βt) Property 1: Linearity of the Transform Let f (t), f1 (t), and f2 (t) be functions whose Laplace transforms exist for s > α and let c be a constant. Then, for s > α, 1. L{f1 + f2 } = L{f1 } + L{f2 } 2. L{cf } = cL{f } Example Let f (t) = 11 + 5e 4t − 6 sin(t). Find L{f }(s). Property 2: Limits of Laplace Transforms Let f (t) be a function whose Laplace Transform exists for s > α. Then, Z ∞ e −st f (t) dt lim lim L{f (t)}(s) = s→∞ 0 s→∞ Z ∞ ≤ lim e −st f (t) dt s→∞ 0 Z ∞ ≤ lim Me −(s−α)t dt s→∞ 0 M s→∞ s − α = 0, = and, thus, lim L{f (t)}(s) = 0. s→∞ lim