To determine The value of f ( t ) using the theorem (3) if the value of L ( f ) = 1 s ( s 2 + ω 2 ) . Explanation Theorem used: Laplace transform of integral (Theorem 3) Let F ( s ) denote the transform of a function f ( t ) which is piecewise continuous for t ≥ 0 and satisfies a growth restriction Then, for s > 0 , s > k , and t > 0 , L ( ∫ 0 t f ( τ ) d τ ) = 1 s F ( s ) , thus ∫ 0 t f ( τ ) d τ = L − 1 ( 1 s F ( s ) ) ( 1 ) . Calculation: It is given that F ( s ) = 1 s ( s 2 + ω 2 ) . Rewrite the given transform of the function as follows. L ( f ) = 1 s ( s 2 + ω 2 ) = ω ω s ( s 2 + ω 2 ) Let f ( t ) be the inverse transform of F ( s ) . Now, using the above mentioned theorem observe that the inverse transform of F ( s ) can be obtained using the formula f ( t ) = ∫ 0 t g ( τ ) d τ where g ( t ) = 1 ω L − 1 ( ω s 2 + ω 2 )

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Author: Kreyszig

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Chapter 6.2, Problem 25P

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The value of f(t) using the theorem (3) if the value of L(f)=1s(s2+ω2).

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Chapter 6.2, Problem 24P

Chapter 6.2, Problem 26P

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