_Signal_FAQ12_Fourier_&_Laplace_Transform

_Signal_FAQ12_Fourier_&_Laplace_Transform - Is it...

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Is it true that by simply replacing s with j ϖ , we will get Fourier transform from Laplace transform? A simple answer is sometimes yes and sometimes no. The Laplace transform of ( 29 t x a is ( 29 ( 29 - = 0 dt e t x s X st a a Since the integration is from 0 to infinite, we will have the same ( 29 s X a regardless of what ( 29 t x a is like for negative t . One way to make sure that ( 29 s X a is unique and has a one-to-one relationship with ( 29 t x a is therefore to require ( 29 t x a to be 0 for negative t . In other words, only those ( 29 t x a that is 0 for negative t has Laplace transform. As an example, there is no Laplace transform for ( 29 2 1 = t x , which has a value 2 for all time. There is also no Laplace transform for ( 29 ( 29 t t x 4 cos 3 2 = , which is non-zero for negative time. However, the Laplace transform for ( 29 ( 29 t u t x a 2 3 = , where ( 29 t u a is the analog unit step function, does exist and is s 2 . Similarly, the Laplace transform for the one-sided signal
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_Signal_FAQ12_Fourier_&_Laplace_Transform - Is it...

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