201.ev1.12.Lec14

# True however for simple y s s it is less efficient

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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

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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 s 1 s 2 + s /2+3 bracketrightbigg = 1 2 bracketleftBigg s 1 ( s +1/4) 2 + 47 / 16 bracketrightBigg = 1 2 s +1/4 ( s +1/4) 2 + 47 / 16 1 2 5/4 (
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