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Unformatted text preview: Chapter 5 Section 5.1 1. 1 2 3 4 0.5 1 1.5 2 2.5 3 t The function is piecewise continuous. 2. 1 2 3 4 5 0.5 1 1.5 2 2.5 3 t The function is neither continuous nor piecewise continuous. 3. 0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 2.5 3 t 1 The function is continuous. 4. 0.5 1 1.5 2 0.5 1 1.5 2 2.5 3 t The function is piecewise continuous. 5. The function is of exponential order. We can take K = 3, a = 5 and M = 0. 6. Since f ( t ) = e 3 t =  f ( t )  = e 3 t , the function is of exponential order. We can take K = 1, a = 3 and M = 0. 7. Since  f ( t )  =  e 2 t sin(3 t )  e 2 t , the function is of exponential order. We can take K = 1, a = 2 and M = 0. 8. Since t 10 = e ln t 10 = e 10ln t e 10 t for all t 0, the function is of exponential order. We can take K = 1, a = 10 and M = 0. 9. cosh( t 2 ) = e t 2 + e t 2 2 is not of exponential order since for all a &gt; 0, there exists an M such that e t 2 e at for all t M . 10. Since  sin( t 4 )  1 = e t is of exponential order. We can take K = 1, a = 0 and M = 0. 11. Since 1 1 + t 1 for all t 0, the function is of exponential order. We can take K = 1, a = 0 and M = 0. 12. e t 3 is not of exponential order since for all a &gt; 0, there exists an M &gt; 0 such that e t 3 e at for all t M . 13. 2 (a) For f ( t ) = t , the Laplace transform is F ( s ) = Z e st tdt = e st t e st s 2 fl fl fl fl t =0 = 1 s 2 . (b) For f ( t ) = t 2 , the Laplace transform is F ( s ) = Z e st t 2 dt = e st s 2 t 2 + 2 e st st + 2 e st s 3 fl fl fl fl t =0 = 2 s 3 . (c) For f ( t ) = t n , the Laplace transform is F ( s ) = Z e st t n dt = n ! s n +1 . 14. For f ( t ) = t t 1 1 1 &lt; t the Laplace transform F ( s ) = Z e st f ( t ) dt = Z 1 e st tdt + Z 1 e st dt. We integrate the first integral on the righthand side by parts. In particular, we have Z t e st tdt = te st s fl fl fl fl 1 Z 1 e st s dt = e s s + 1 s Z 1 e st dt = e s s + 1 s e st s fl fl fl fl 1 = e s s e s s 2 + 1 s 2 . The second integral above can be evaluated as follows: Z 1 e st ,dt = e st s fl fl fl fl 1 = e s s . Combining these two integrals, we have F ( s ) = e s s e s s 2 + 1 s 2 + e s s = 1 s 2 e s s 2 . 15. For f ( t ) = t 1 1 1 &lt; t 2 2 &lt; t, 3 the Laplace transform is given by F ( s ) = Z 1 e st dt + Z 2 1 e st dt + Z 2 e st dt = Z 2 1 e st dt = e st s fl fl fl fl 2 1 = e 2 s s e s s = e s e 2 s s . 16. For f ( t ) = t 1 e t 1 &lt; t the Laplace transform is given by F ( s ) = Z 1 e st e t dt = Z 1 e ( s +1) t dt = e ( s +1) t ( s + 1) fl fl fl fl 1 = 0 e ( s +1) ( s + 1) = e ( s +1) s + 1 ....
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 Spring '08
 comech
 Equations, Laplace, dt, e−st dt

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