HigherLin4

# The laplace transform also turns a translation of t

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The Laplace transform also turns a translation of t into multiplication by an expo- nential in s . Notice that L [ f ]( s ) only depends on the values of f ( t ) over [0 . ). Therefore before we translate f we multiply it by the unit step function u ( t ) defined by u ( t ) = braceleftbigg 1 for t 0 , 0 for t < 0 . (8.4) Because the functions uf and f agree over [0 , ), it is clear that L [ uf ]( s ) = L [ f ]( s ). We now consider the Laplace transform of the translation u ( t c ) f ( t c ) for every c > 0. Theorem. If L [ f ]( s ) exists for every s > α and c > 0 then L [ u ( t c ) f ( t c )]( s ) exists for every s > α with L [ u ( t c ) f ( t c )]( s ) = e cs L [ f ]( s ) for s > α . Proof. For every T > c one has integraldisplay T 0 e st u ( t c ) f ( t c ) d t = integraldisplay T c e st f ( t c ) d t = e cs integraldisplay T c e s ( t c ) f ( t c ) d t = e cs integraldisplay T c 0 e st f ( t ) d t . Therefore L [ u ( t c ) f ( t c )]( s ) = lim T →∞ integraldisplay T 0 e st u ( t c ) f ( tc ) d t = e cs lim T →∞ integraldisplay T c 0 e st f ( t ) d t = e cs L [ f ]( s ) for s > α . square

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6 8.3: Existence and Differentiablity of the Transform. In each of the above examples the definite integrals over [0 , T ] that appear in the limit (8.2) were proper. Indeed, we were able to evaluate the definite integrals analytically and determine the limit (8.2) for every real s . More generally, from calculus we know that a definite integral over [0 , T ] is proper whenever its integrand is: bounded over [0 , T ], continuous at all but a finite number of points in [0 , T ]. Such an integrand is said to be piecewise continuous over [0 , T ]. Because for e st is continuous (and therefore bounded) function of t over every [0 , T ] for each real s , the definite integrals over [0 , T ] that appear in the limit (8.2) will be proper whenever f ( t ) is piecewise continuous over every [0 , T ]. Example. The function f ( t ) = braceleftbigg 0 for 0 t < π , cos( t ) for t π , is piecewise continuous over every [0 , T ] because it is clearly bounded over [0 , ) and its only discontinuity is at the point t = π . Example. The so-called sawtooth function f ( t ) = t k for k t < k + 1 where k = 0 , 1 , 2 , 3 , · · · , is piecewise continuous over every [0 , T ] because it is clearly bounded over [0 , ) and has discontinuities at the points t = 1 , 2 , 3 , · · · , only a finite number of which lie in each [0 , T ]. If we assume that f ( t ) is piecewise continuous over every [0 , T ], we still have to give a condition under which the limit (8.2) will exist for certain s . Such a condition is provided by the following definition. Definition. A function f ( t ) defined over [0 , ) is said to be of exponential order α as t → ∞ provided that for every σ > α there exist K σ and T σ such that | f ( t ) | ≤ K σ e σt for every t T σ . (8.5) Roughly speaking, a function is of exponential order α as t → ∞ if its absolute value does not grow faster than e σt as t → ∞ for every σ > α .
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