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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 8 25 Copyright © Orchard Publications Derivation of the Fourier Transform from the Laplace Transform 8.4.9 The Function Pair (8.75) Proof: We express the sine function as From (8.73), and Using we obtain (8.75). 8.5 Derivation of the Fourier Transform from the Laplace Transform If a time function is zero for , we can obtain the Fourier transform of from the one- sided Laplace transform of by substitution of with . Example 8.1 It is known that . Compute Solution: Thus, we have obtained the following Fourier transform pair. (8.76) ω 0 sin t () u 0 t ω 0 sin t u 0 t π j2 ---- δω ω 0 δ ω ω 0 + + [] ω 2 ω 0 2 ω 2 ----------------- + ω 0 sin t 1 ---- e j ω 0 t e j ω 0 t = e j ω 0 t u 0 t πδω ω 0 1 j ωω 0 ---------------------- + e j ω 0 t u 0 t 0 + 1 j 0 + ----------------------- + u 0 t πδω 1 j ω ----- + ft t0 sj ω L e α t u 0 t 1 s α + ----------- = F e α t u 0 t {} F e α t u 0 t L e α t u 0 t ω = 1 s α + ω = 1 j ωα + --------------- == = e α t u 0 t 1 j +

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Chapter 8 The Fourier Transform 8 26 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications Example 8.2 It is known that Compute Solution: Thus, we have obtained the following Fourier transform pair. (8.77) We can also find the Fourier transform of a time function that has non zero values for , and it is zero for all . But because the one sided Laplace transform does not exist for , we must first express the negative time function in the domain, and compute the one sided Laplace transform. Then, the Fourier transform of can be found by substituting with . In other words, when for , and for , we use the substitution (8.78) Example 8.3 Compute the Fourier transform of a. using the Fourier transform definition b. by substitution into the Laplace transform equivalent Solution: a. Using the Fourier transform definition, we obtain L e α t ω 0 t cos () u 0 t [] s α + s α + 2 ω 0 2 + -------------------------------- = F e α t ω 0 t cos u 0 t {} F e α t ω 0 t cos u 0 t L e α t ω 0 t cos u 0 t sj ω = = s α + s α + 2 ω 0 2 + ω = j ωα + j + 2 ω 0 2 + ----------------------------------- = = e α t ω 0 t cos u 0 t j + j + 2 ω 0 2 + ft t0 < > < > ω 0 = 0 < F L ω = = e at =
Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 8 27 Copyright © Orchard Publications Fourier Transforms of Common Waveforms and thus we have the transform pair (8.79) b.By substitution into the Laplace transform equivalent, we obtain and this result is the same as (8.79). We observe that since is real and even, is also real and even.

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