Laplace_06 - Laplace Transforms Extending the Fourier...

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Laplace Transforms Extending the Fourier Transform There are many signals for which the Fourier transform does not exist. By modifying the Fourier transform, it will be possible to significantly expand the class of signals which can be transformed. Modifying the Fourier transform Begin by considering a signal x(t) which is absolutely integrable and therefore has a Fourier transform, i.e., ( ) ( ) xt X j ω ℑ =  Now form a new signal , where ( ) gt σ ( ) ( ) t gt x te ± Here σ is a real quantity. Now determine the conditions under which this signal will have a Fourier transform. It will have a transform if it is absolutely integrable, i.e., () g td t −∞ <∞ But t t x te d t ∞∞ −∞ −∞ = ∫∫ Denote ( ) ( ) Gj g t σσ ± Now consider the conditions under which both have Fourier transforms. ( ) ( ) and g t Example : Let . This signal has a FT because ( ) ( ) 2 t xt te ut = 22 2 00 1 4 tt t te u t dt te dt te dt −− −∞ == =
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2 Then () 2 2 0 2 2 0 1 2 2 t tt t gtd t t eeu td t t e d t t e σ ∞∞ −+ −− −∞ −∞ ==  =−  + +  ∫∫ Evaluating this expression at the upper limit shows that it will become infinite if ( σ +2) < 0. In that case it will not have a Fourier transform. If, however, ( σ +2) > 0, then its value will be zero at the upper limit, and 2 2 0 1 for 2 2 t g t dt te dt −∞ > + From this example it is clear that can both have Fourier transforms. ( ) ( ) and xt g t Now observe that ( ) ( ) () () () and jt t Xj x te d t G j g t e d tx t e e d t e d t ω σω ωσ σσ −∞ −∞ −∞ −∞ = = From these two expressions it is clear that ( ) ( ) G j ωω += Now make a simple change of variable. Let s j + ± Then () () st X sx t e d t −∞ =
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3 Here X(s) is just the Fourier transform of . However, the above expression can ( ) gt σ also be thought of as a new transform of x(t) . This transform is called the Laplace Transform . Notice that if this Laplace transform of x(t) is evaluated at σ = 0, it is just the Fourier transform of x(t) . Inverting the Laplace Transform For a given value of σ , ( ) ( ) Xs G j ω = But () ( ) () ( ) 1 1 2 j tt Gj Gjed x t e σσ ωω π −− −∞ =ℑ  = =  Therefore 1 22 1 2 t jt e x t Gjed Gje d Xj ed σω ππ ∞∞ + −∞ −∞ + −∞ == =+ ∫∫ Now change variable in this expression. Let s = σ + j ω . For a fixed value of σ , and ds jd = 1 2 j st j xt X seds j +∞ −∞ = This is a rather strange looking integral that needs further explanation. However, assuming it can be evaluated, this suggests that () () 1 2 and Xs j st j st j xte d t −∞ = =
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4 These two expressions are called the Laplace transform pair . As with the Fourier transform, the Laplace transform can be thought of as another way of representing the signal. Moreover, it must contain all of the information about the signal, because it is possible to obtain the signal from it.
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This note was uploaded on 10/24/2009 for the course ECE 108 taught by Professor Li during the Spring '08 term at Lehigh University .

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Laplace_06 - Laplace Transforms Extending the Fourier...

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