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Unformatted text preview: CHAPTER 6 Laplace Transform Major Changes This chapter underwent some major changes regarding the order of the material and the emphasis placed on the various topics. In particular, Diracs delta was placed in a separate section. Partial fractions are now considered earlier. Their importance in this chapter has been diminished, and they are explained in terms of practical problems when they are needed, rather than in terms of impractical theoretical formulas. Convolution and nonhomogeneous linear ODEs are now considered earlier, whereas differentiation and integration of transforms (not of functions !) appears later and in a lesser role. SECTION 6.1. Laplace Transform. Inverse Transform. Linearity. sShifting, page 221 Purpose. To explain the basic concepts, to present a short list of basic transforms, and to show how these are derived from the definition. Main Content, Important Concepts Transform, inverse transform, linearity First shifting theorem Table 6.1 Existence and its practical significance Comment on Table 6.1 After working for a while in this chapter, the student should be able to memorize these transforms. Further transforms in Sec. 6.9 are derived as we go along, many of them from Table 6.1. SOLUTIONS TO PROBLEM SET 6.1, page 226 2. ( t 2 2 3) 2 5 t 4 2 6 t 2 1 9; transform 2 1 4. sin 2 4 t 5 2 cos 8 t ; transform 2 5 6. e 2 t sinh 5 t 5 ( e 4 t 2 e 2 6 t ); transform ( 2 ) 5 . This can be checked by the first shifting theorem. 8. sin (3 t 2 1 _ 2 ) 5 cos 1 _ 2 sin 3 t 2 sin 1 _ 2 cos 3 t ; transform 10. 12. ( t 1 1) 3 5 t 3 1 3 t 2 1 3 t 1 1; transform 1 1 1 1 } s 3 } s 2 6 } s 3 6 } s 4 2 1.6 } s 2 1 0.04 3 cos 1 _ 2 2 s sin 1 _ 2 } s 2 1 9 5 } ( s 1 1) 2 2 25 1 } s 1 6 1 } s 2 4 1 } 2 1 } 2 32 } s ( s 2 1 64) s } 2( s 2 1 64) 1 } 2 s 1 } 2 1 } 2 9 } s 12 } s 3 24 } s 5 117 im06.qxd 9/21/05 12:05 PM Page 117 14. k E b a e 2 st dt 5 ( e 2 as 2 e 2 bs ) 16. k (1 2 t / b ); transform k E b e 2 st ( 1 2 ) dt 5 2 e 2 st ( 1 2 )j b 1 E b ( 2 ) e 2 st dt 5 ( bs 1 e 2 bs 2 1) 18. E b e 2 st t dt 5 ( 2 ) 20. E 1 e 2 st t dt 1 E 2 1 e 2 st (2 2 t ) dt . Integration by parts gives 2 j 1 1 E 1 e 2 st dt 2 j 2 1 2 E 2 1 e 2 st dt 5 2 2 ( e 2 s 2 1) 1 1 ( e 2 2 s 2 e 2 s ) 5 ( 2 e 2 s 1 1 1 e 2 2 s 2 e 2 s ) 5 . 22. Let st 5 t , t 5 t / s , dt 5 d t / s . Then + (1/ t w ) 5 E ` e 2 st t 2 1/2 dt 5 E ` e 2 t ( t / s ) 2 1/2 d t 5 s 2 1/2 E ` e 2 t t 2 1/2 d t 5 s 2 1/2 G ( 1 _ 2 ) 5 p / s w . 24. No matter how large we choose M and k , we have e t 2 . Me kt for all t greater than some t because t 2 . ln M 1 kt for all sufficiently large t (and fixed positive M and k ). 26. Use e at 5 cosh at 1 sinh at . 28. Let 5 + 2 1 ( F ), g 5 + 2 1 ( G ). Since the transform is linear, we obtain aF 1 bG 5 a + () 1 b + ( g ) 5 + ( a 1 bg )....
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This note was uploaded on 11/26/2009 for the course ECE 212321 taught by Professor Simpson during the Spring '09 term at Adelphi.
 Spring '09
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