Therefore l t e at s is only defined for s a with l t

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Therefore L [ t e at ]( s ) is only defined for s > a with L [ t e at ]( s ) = 1 ( s a ) 2 for s > a . Example. Use definition (8.2) to compute L [ e ibt ]( s ) for any real b . For b negationslash = 0 you see from (8.2) that for any real s we have L [ e ibt ]( s ) = lim T →∞ integraldisplay T 0 e st e ibt d t = lim T →∞ integraldisplay T 0 e ( s ib ) t d t = lim T →∞ parenleftbigg e ( s ib ) t s ib parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle T t =0 = lim T →∞ bracketleftbigg 1 s ib e ( s ib ) T s ib bracketrightbigg = 1 s ib for s > 0 , undefined for s 0 . The case b = 0 is identical to our first example with a = 0. In every case L [ e ibt ]( s ) is only defined for s > 0 with L [ e ibt ]( s ) = 1 s ib for s > 0 . 8.2: Properties of the Transform. If we always had to return to the definition of the Laplace transform everytime we wanted to apply it, it would not be easy to use. Rather, we will use the definition to compute the Laplace transform for a few basic functions and to establish some general properties that will allow us to build formulas for more complicated functions. The most important such property is the fact that the Laplace transform L is a linear operator.
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4 Theorem. If L [ f ]( s ) and L [ g ]( s ) exist for some s then so does L [ f + g ]( s ) and L [ cf ]( s ) for every constant c with L [ f + g ]( s ) = L [ f ]( s ) + L [ g ]( s ) , L [ cf ]( s ) = c L [ f ]( s ) . (8.3) Proof. This follows directly from definition (8.2) and the facts that definite integrals and limits depend linearly on their arguments. Specifically, one sees that L [ f + g ]( s ) = lim T →∞ integraldisplay T 0 e st ( f ( t ) + g ( t ) ) d t = lim T →∞ integraldisplay T 0 e st f ( t ) d t + lim T →∞ integraldisplay T 0 e st g ( t ) d t = L [ f ]( s ) + L [ g ]( s ) , L [ cf ]( s ) = lim T →∞ integraldisplay T 0 e st cf ( t ) d t = c lim T →∞ integraldisplay T 0 e st f ( t ) d t = c L [ f ]( s ) . square Example. Compute L [cos( bt )]( s ) and L [sin( bt )]( s ) for any real b negationslash = 0. This can be done by using the Euler identity e ibt = cos( bt ) + i sin( bt ) and the linearity of L . Then L [cos( bt )]( s ) + i L [sin( bt )]( s ) = L [ e ibt ]( s ) = 1 s ib = s + ib s 2 + b 2 for s > 0 . By equating the real and imaginary parts above, we see that L [cos( bt )]( s ) = s s 2 + b 2 for s > 0 , L [sin( bt )]( s ) = b s 2 + b 2 for s > 0 . Another general property of the Laplace transform is that it turns multiplication by an exponential in t into a translation of s . Theorem. If L [ f ]( s ) exists for every s > α and a is any real number then L [ e at f ( t )]( s ) exists for every s > α + a with L [ e at f ( t )]( s ) = L [ f ]( s a ) . Proof. This follows directly from definition (8.2). Specifically, one sees that L [ e at f ]( s ) = lim T →∞ integraldisplay T 0 e st e at f ( t ) d t = lim T →∞ integraldisplay T 0 e ( s a ) t f ( t ) d t = L [ f ]( s a ) .
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5 square Examples. From our previous examples and the above theorem we see that L [ e ( a + ib ) t ]( s ) = 1 s a ib for s > a , L [ e at cos( bt )]( s ) = s a ( s a ) 2 + b 2 for s > a , L [ e at sin( bt )]( s ) = b ( s a ) 2 + b 2 for s > a .
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  • Spring '10
  • LEVERMORE
  • Laplace

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