undefined Solution The function f ( t ) and to sketch f ( t ) if L ( f ) = ( e − 3 s ) 4 . The function f ( t ) is f ( t ) = ( t − 3 ) 3 6 u ( t − 3 ) _ . Formula used: 1. The Laplace transform of the function 1 s n is 1 s n = t n − 1 ( n − 1 ) ! . Theorem used: First shifting theorem: If f ( t ) has the transform F ( s ) , then e a t f ( t ) has the transform F ( s − a ) . That is, L ( e a t f ( t ) ) = F ( s − a ) . Second shifting theorem: If f ( t ) has the transform F ( s ) , then the shifted function f ˜ ( t ) = f ( t − a ) u ( t − a ) = { 0 if t < a f ( t − a ) if t > a } has the transform e − a s F ( s ) . That is, if L ( f ( t ) ) = F ( s ) then, L ( f ( t − a ) u ( t − a ) ) = e − a s F ( s ) ⋯ ( 1 ) Calculation: The given function is L ( f ) = ( e − 3 s ) s 4 . Rewrite the given function as follows. L ( f ) = ( e − 3 s ) s 4 f = L − 1 ( ( e − 3 s ) s 4 ) From the above mentioned formula, it can be noted that the inverse Laplace transform of the function 1 s 4 is t 3 6 . Note that the term ( e − 3 s ) s 4 is of the form e − a s F ( s ) . Therefore, obtain the inverse Laplace transform of ( e − 3 s ) s 4 using the second shifting theorem as shown below. L − 1 ( ( e − 3 s ) s 4 ) = f ( t − 3 ) u ( t − 3 ) ( ∵ f ( t ) = t 3 6 ) = ( t − 3 ) 3 6 u ( t − 3 ) Therefore, the function f ( t ) is f ( t ) = ( t − 3 ) 3 6 u ( t − 3 ) _ . Note that the function u ( t − 3 ) can be written as follows. u ( t − 3 ) = { 0 , x < 3 1 , x > 3 } Now, use the online graphing calculator and sketch the given function as shown below. Thus, the required graph of the function ( t − 3 ) 3 6 u ( t − 3 ) is as shown in Figure 1.

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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