Integration by parts gives Z te s ia t dt te s ia t s ia e s ia t s ia 2 1 s ia

Integration by parts gives z te s ia t dt te s ia t s

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Integration by parts gives Z 0 te - ( s - ia ) t dt = te - ( s - ia ) t s - ia + e - ( s - ia ) t ( s - ia ) 2 0 = 1 ( s - ia ) 2 , s > 0 . Similarly, Z 0 te - ( s + ia ) t dt = 1 ( s + ia ) 2 , s > 0 . Thus, L [ t cos( at )] = 1 2 1 ( s - ia ) 2 + 1 ( s + ia ) 2 = s 2 - a 2 ( s 2 + a 2 ) 2 , s > 0 . Joseph M. Mahaffy, h [email protected]u i Lecture Notes – Laplace Transforms: Part A — (11/26) Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Piecewise Continuous Functions Definition (Piecewise Continuous) A function f is said to be a piecewise continuous on an interval α t β if the interval can be partitioned by a finite number of points α = t 0 < t 1 < ... < t n = β so that: 1 f is continuous on each subinterval t i - 1 < t < t i , and 2 f approaches a finite limit as the endpoints of each subinterval are approached from within the subinterval. α β t 1 t 2 t y The figure to the right shows the graph of a piecewise continuous function defined for t [ α, β ) with jump discontinuities at t = t 1 and t 2 . It is continuous on the subintervals ( α, t 1 ), ( t 1 , t 2 ), and ( t 2 , β ). Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (12/26)
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Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Examples: Laplace Transform Example 7: Define the piecewise continuous function f ( t ) = e 2 t , 0 t < 1 , 4 , 1 t. The Laplace transform satisfies: F ( s ) = Z 0 e - st f ( t ) dt = Z 1 0 e - st e 2 t dt + Z 1 e - st · 4 dt = Z 1 0 e - ( s - 2) t dt + 4 lim A →∞ Z A 1 e - st dt = - e - ( s - 2) t s - 2 1 t =0 - 4 lim A →∞ e - st s A t =1 = 1 s - 2 - e - ( s - 2) s - 2 + 4 e - s s , s > 0 , s 6 = 2 . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (13/26) Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Existence of Laplace Transform Definition (Exponential Order) A function f ( t ) is of exponential order (as t + ) if there exist real constants M 0, K > 0, and a , such that | f ( t ) | ≤ Ke at , when t M . Examples: f ( t ) = cos( αt ) satisfies being of exponential order with M = 0, K = 1, and a = 0 f ( t ) = t 2 satisfies being of exponential order with a = 1, K = 1, and M = 1. By L’Hˆ opital’s Rule (twice) lim t →∞ t 2 e t = lim t →∞ 2 e t = 0 . f ( t ) = e t 2 is NOT of exponential order Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (14/26) Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Existence of Laplace Transform Theorem (Existence of Laplace Transform) Suppose 1 f is piecewise continuous on the interval 0 t A for any positive A 2 f is of exponential order , i.e., there exist real constants M 0 , K > 0 , and a , such that | f ( t ) | ≤ Ke at , when t M . Then the Laplace transform given by L [ f ( t )] = F ( s ) = Z 0 e - st f ( t ) dt, exists for s > a .
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