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Chapter 4

Chapter 4 - 4 1 The Laplace Transform Exercises 4.1 1 1...

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4 The Laplace Transform Exercises 4.1 1. { f ( t ) } = Z 1 0 e st dt + Z 1 e st dt = 1 s e st ¯ ¯ ¯ ¯ 1 0 1 s e st ¯ ¯ ¯ ¯ 1 = 1 s e s 1 s µ 0 1 s e s = 2 s e s 1 s ,s > 0 2. { f ( t ) } = Z 2 0 4 e st dt = 4 s e st ¯ ¯ ¯ ¯ 2 0 = 4 s ( e 2 s 1) > 0 3. { f ( t ) } = Z 1 0 te st dt + Z 1 e st dt = µ 1 s te st 1 s 2 e st ¯ ¯ ¯ ¯ 1 0 1 s e st ¯ ¯ ¯ ¯ 1 = µ 1 s e s 1 s 2 e s µ 0 1 s 2 1 s (0 e s )= 1 s 2 (1 e s ) > 0 4. { f ( t ) } = Z 1 0 (2 t +1) e st dt = µ 2 s te st 2 s 2 e st 1 s e st ¯ ¯ ¯ ¯ 1 0 = µ 2 s e s 2 s 2 e s 1 s e s µ 0 2 s 2 1 s = 1 s (1 3 e s )+ 2 s 2 (1 e s ) > 0 5. { f ( t ) } = Z π 0 (sin t ) e st dt = µ s s 2 +1 e st sin t 1 s 2 e st cos t ¯ ¯ ¯ ¯ π 0 = µ 0+ 1 s 2 e πs µ 0 1 s 2 = 1 s 2 ( e > 0 6. { f ( t ) } = Z π/ 2 (cos t ) e st dt = µ s s 2 e st cos t + 1 s 2 e st sin t ¯ ¯ ¯ ¯ 2 =0 µ 1 s 2 e πs/ 2 = 1 s 2 e πs/ 2 > 0 7. f ( t ½ 0 , 0 <t< 1 t, t > 1 { f ( t ) } = Z 1 te st dt = µ 1 s te st 1 s 2 e st ¯ ¯ ¯ ¯ 1 = 1 s e s + 1 s 2 e s , s> 0 8. f ( t ½ 0 , 0 1 2 t 2 ,t> 1 { f ( t ) } =2 Z 1 ( t 1) e st dt µ 1 s ( t 1) e st 1 s 2 e st ¯ ¯ ¯ ¯ 1 = 2 s 2 e s , 0 9. f ( t ½ 1 t, 0 1 0 ,t > 0 { f ( t ) } = Z 1 0 (1 t ) e st dt = µ 1 s (1 t ) e st + 1 s 2 e st ¯ ¯ ¯ ¯ 1 0 = 1 s 2 e s + 1 s 1 s 2 , 0 10. f ( t 0 , 0 <t<a c, a<t<b 0 b ; { f ( t ) } = Z b a ce st dt = c s e st ¯ ¯ ¯ ¯ b a = c s ( e sa e sb ), 0 165

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Exercises 4.1 11. { f ( t ) } = Z 0 e t +7 e st dt = e 7 Z 0 e (1 s ) t dt = e 7 1 s e (1 s ) t ¯ ¯ ¯ ¯ 0 =0 e 7 1 s = e 7 s 1 ,s > 1 12. { f ( t ) } = Z 0 e 2 t 5 e st dt = e 5 Z 0 e ( s +2) t dt = e 5 s +2 e ( s +2) t ¯ ¯ ¯ ¯ 0 = e 5 s > 2 13. { f ( t ) } = Z 0 te 4 t e st dt = Z 0 te (4 s ) t dt = µ 1 4 s te (4 s ) t 1 (4 s ) 2 e (4 s ) t ¯ ¯ ¯ ¯ 0 = 1 (4 s ) 2 > 4 14. { f ( t ) } = Z 0 t 2 e 2 t e st dt = Z 0 t 2 e ( s +2) t dt = µ 1 s t 2 e ( s +2) t 2 ( s +2) 2 te ( s +2) t 2 ( s 3 e ( s +2) t ¯ ¯ ¯ ¯ 0 = 2 ( s 3 > 2 15. { f ( t ) } = Z 0 e t (sin t ) e st dt = Z 0 (sin t ) e ( s +1) t dt = µ ( s +1) ( s 2 +1 e ( s +1) t sin t 1 ( s 2 e ( s +1) t cos t ¯ ¯ ¯ ¯ 0 = 1 ( s 2 = 1 s 2 s > 1 16. { f ( t ) } = Z 0 e t (cos t ) e st dt = Z 0 (cos t ) e (1 s ) t dt = µ 1 s (1 s ) 2 e (1 s ) t cos t + 1 (1 s ) 2 e (1 s ) t sin t ¯ ¯ ¯ ¯ 0 = 1 s (1 s ) 2 = s 1 s 2 2 s > 1 17. { f ( t ) } = Z 0 t (cos t ) e st dt = ±Ã st s 2 s 2 1 ( s 2 2 ! (cos t ) e st + Ã t s 2 + 2 s ( s 2 2 ! (sin t ) e st ² 0 = s 2 1 ( s 2 2 > 0 18. { f ( t ) } = Z 0 t (sin t ) e st dt = ±Ã t s 2 2 s ( s 2 2 !
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Chapter 4 - 4 1 The Laplace Transform Exercises 4.1 1 1...

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