Explanation Given data: The approximation function f ( t ) is, f ( t ) = e − ∈ | t | cos ω 0 t (1) Where, ∈ is a positive real constant. Calculation: Applying Laplace transform to equation (1) as follows, F ( s ) = s + ∈ ( s + ∈ ) 2 + ω 0 2 = s + ∈ ( s + ∈ ) 2 − ( j ω 0 ) 2 F ( s ) = s + ∈ ( s + ∈ + j ω 0 ) ( s + ∈ − j ω 0 ) { ∵ ( a 2 − b 2 ) = ( a + b ) ( a − b ) } (2) Simplifying equation (2) by partial fraction method as follows, F ( s ) = s + ∈ ( s + ∈ − j ω 0 ) ( s + ∈ + j ω 0 ) = A ( s + ∈ − j ω 0 ) A * ( s + ∈ + j ω 0 ) (3) Therefore, A = s + ∈ s + ∈ + j ω 0 = − ∈ + j ω 0 + ∈ ( − ∈ + j ω 0 + ∈ + j ω 0 ) { ∵ s = − ∈ + j ω 0 } = j ω 0 2 j ω 0 = 0.5 Since A = 0.5 , therefore, A * = 0.5 Substituting the values of A and A * in equation (3) as follows, F ( s ) = 0.5 ( s + ∈ − j ω 0 ) + 0.5 ( s + ∈ + j ω 0 ) Therefore, the above function is written as follows, F ( ω ) = F ( s ) | s = j ω + F ( s ) | s = − j ω = 0.5 ( j ω + ∈ − j ω 0 ) + 0.5 ( j ω + ∈ + j ω 0 ) + 0.5 ( − j ω + ∈ − j ω 0 ) + 0.5 ( − j ω + ∈ + j ω 0 ) = 0.5 ( ∈ + j ( ω − ω 0 ) ) + 0.5 ( ∈ + j ( ω + j ω 0 ) ) + 0.5 ( ∈ − j ( ω + ω 0 ) ) + 0.5 ( ∈ − j ( ω − ω 0 ) ) = 0.5 [ ∈ − j ( ω − ω 0 ) ] + 0.5 [ ∈ + j ( ω − ω 0 ) ] ( ∈ 2 + ( ω − ω 0 ) 2 ) + 0.5 [ ∈ − j ( ω + ω 0 ) ] + 0.5 [ ∈ + j ( ω + ω 0 ) ] ( ∈ 2 + ( ω + ω 0 ) 2 ) Therefore, the above equation will be, F ( ω ) = ∈ ∈ 2 + ( ω − ω 0 ) 2 + ∈ ∈ 2 + ( ω + ω 0 ) 2 (4) As ∈ → 0 , then F ( ω ) → 0 everywhere except at ω = ± ω 0

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ISBN: 9780134746968

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Chapter 17, Problem 9P

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Calculate the Fourier transform of F{cosω0t} by using the approximation function f(t)=e−∈|t|cosω0t.

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Chapter 17, Problem 8P

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