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◦ the only thing we haven’t discussed so far is

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Unformatted text preview: ◦ The only thing we haven’t discussed so far is the last step. Remark 4. There is indeed an inverse transform formula: L − 1 { Y ( s ) } = 1 2 π i lim T →∞ integraldisplay α − iT α + iT e st Y ( s ) d s (29) Here α is any real number satisfying: α + β i belonds to the domain of Y ( s ) for all β. (30) As we can see that integration takes place in the complex plane C . This formula is universally true, however for simple Y ( s ) ’s it is less efficient than the “simplification + table” approach below. Therefore we will not discuss further the above inversion formula. ◦ Simple inverse transforms. Reading the transform table backwards, we have L{ 1 } = 1 s L − 1 braceleftbigg 1 s bracerightbigg = 1 (31) L{ t n } = n ! s n +1 L − 1 braceleftbigg 1 s n +1 bracerightbigg = t n n ! (32) L{ e at } = 1 s − a L − 1 braceleftbigg 1 s − a bracerightbigg = e at (33) L{ sin b t } = b s 2 + b 2 L − 1 braceleftbigg b s 2 + b 2 bracerightbigg = sin b t, (34) L{ cos b t } = s s 2 + b 2 L − 1 braceleftBig s s 2 + b 2 bracerightBig = cos b t, (35) L{ e at f } = L{ f } ( s − a ) L − 1 { F ( s − a ) } = e at f , (36) L{ t n f } = ( − 1) n d n d s n L{ f } L − 1 braceleftbigg ( − 1) n d n d s n F bracerightbigg = t n f (37) L{ c 1 f + c 2 g } = c 1 L{ f } + c 2 L{ g } L − 1 { c 1 F + c 2 G } = c 1 L − 1 { F } + c 2 L − 1 { G } . (38) Feb. 8, 2012 3 Combining some of the above properties, we have the following very useful inverse transform formulas: L − 1 braceleftbigg 1 ( s − a ) n +1 bracerightbigg = e at t n n ! ; (39) L − 1 braceleftbigg s − α ( s − α ) 2 + β 2 bracerightbigg = e αt cos β t ; (40) L − 1 braceleftbigg β ( s − α ) 2 + β 2 bracerightbigg = e αt sin β t. (41) Example 5. (7.4 3; 7.4 3) Find L − 1 braceleftbigg s + 1 s 2 + 2 s + 10 bracerightbigg . (42) Solution. We notice s + 1 s 2 + 2 s + 10 = s + 1 ( s + 1) 2 + 9 = s + 1 ( s + 1) 2 + 3 2 . (43) Thus recalling the transform formulas for e at f and for cos b t , we have L − 1 braceleftbigg s + 1 s 2 + 2 s + 10 bracerightbigg = e − t L − 1 braceleftBig s s 2 + 3 2 bracerightBig = e − t cos 3 t. (44) Example 6. (7.4 6; 7.4 6) Find L − 1 braceleftBigg 3 (2 s + 5) 3 bracerightBigg . (45) Solution. We have L − 1 braceleftBigg 3 (2 s + 5) 3 bracerightBigg = L − 1 braceleftBigg 3 2 3 ( s + 5 2 ) 3 bracerightBigg = 3 8 L − 1 braceleftBigg 1 ( s + 5 2 ) 3 bracerightBigg = 3 8 e − 5 2 t L − 1 braceleftbigg 1 s 3 bracerightbigg = 3 16 e − 5 2 t t 2 . (46) Example 7. (7.4 10; 7.4 10) Find L − 1 braceleftbigg s − 1 2 s 2 + s + 6 bracerightbigg . (47) Solution. Note that the denominator is not of the form ( s − 1) 2 + b 2 . But we can write s − 1 2 s 2 + s + 6 = 1 2 bracketleftbigg...
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