HW11 - MA341 Homework 11 Solutions 7.6 5. g ( t ) = ,...

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Unformatted text preview: MA341 Homework 11 Solutions 7.6 5. g ( t ) = , < t < 1; 2 , 1 < t < 2; 1 , 2 < t < 3; 3 , 3 < t . g ( t ) = 2 u ( t- 1)- u ( t- 2) + 2 u ( t- 3). The Laplace transform of u ( t- a ) is L{ u ( t- a ) } ( s ) = e- as s . Thus the Laplace transform of g ( t ) is L{ g ( t ) } ( s ) = 2 e- s s- e- 2 s s + 2 e- 3 s s = 2 e- s- e- 2 s + 2 e- 3 s s . 9. From the graph of g ( t ) we find that the function of g ( t ) can be written as g ( t ) = , < t 1; t- 1 , 1 < t 2;- t + 3 , 2 < t 3; , 3 < t . Hence, the function g ( t ) in terms of the unit step function is g ( t ) = ( t- 1) u ( t- 1) + (- 2 t + 4) u ( t- 2) + ( t- 3) u ( t- 3) = ( t- 1) u ( t- 1)- 2( t- 2) u ( t- 2) + ( t- 3) u ( t- 3) . Then, by formula (5) of section 7.6, we get: L{ g ( t ) } ( s ) = e- s 1 s 2- 2 e- 2 s 1 s 2 + e- 3 s 1 s 2 = e- s- 2 e- 2 s + e- 3 s s 2 . 13. L- 1 { e- 2 s- 3 e- 4 s s + 2 } = L- 1 { e- 2 s s + 2 } - 3 L- 1 { e- 4 s s + 2 } To use the translation in t property given by formula (6), we first express e- 2 s s +2 as e- as F ( s ). For that purpose, we put e- as = e- 2 s and F ( s ) = 1 s +2 . Thus, a = 2 and f ( t ) = L- 1 { 1 s +2 } ( t ) = e- 2 t . It now follows from the translation property that L- 1 { e- 2 s s + 2 } ( t ) = f ( t- 2) u ( t- 2) = e- 2( t- 2) u ( t- 2) . Similarly, L- 1 { e- 4 s s + 2 } ( t ) = f ( t- 4) u ( t- 4) = e- 2( t- 4) u ( t- 4) . 1 MA341 Homework 11 Solutions Consequently, L- 1 { e- 2 s- 3 e- 4 s s + 2 } = e- 2( t- 2) u ( t- 2)- 3 e- 2( t- 4) u ( t- 4) . 15. To use the translation in t property given by formula (6), we first express se- 3 s s 2 +4 s +5 as e- as F ( s ). For that purpose, we put e- as = e- 3 s and F ( s ) = s s 2 +4 s +5 . Thus, a = 3 and f ( t ) = L- 1 { s s 2 + 4 s + 5 } ( t ) = L- 1 { s + 2 ( s + 2) 2 + 1 } ( t )- 2 L- 1 { 1 ( s + 2) 2 + 1...
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HW11 - MA341 Homework 11 Solutions 7.6 5. g ( t ) = ,...

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