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The transform of a sine function is also easily obtained.  1 1 s sin !t e st e st ! cos !t ! sin !t e st dt 2 l sin !t s2 !2 s !2 0 0 It will be useful now to obtain the transform of a derivative, df t =dt.   1 df t df t st e dt l dt dt 0 Integrating by parts,   1 1 df t st 1 st f t se dt f 0 s f t e st dt f 0 sF s l e f t 0 dt 0 0 A small collection of transform pairs, including those obtained above, is given in Table 16-1. The last ve lines of the table present some general properties of the Laplace transform.
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EXAMPLE 16.1 Consider a series RL circuit, with R 5
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and L 2:5 mH. At t 0, when the current in the circuit is 2 A, a source of 50 V is applied. The time-domain circuit is shown in Fig. 16-1.
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THE LAPLACE TRANSFORM METHOD
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Time Domain (i) Ri + L di = L dt
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s-Domain (ii) RI(s) + L[_ i(0+) + sI(s)] = V(s) (iii) 5I(s) + (2.5 10 3)[_2 +sI(s)]=
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(classical methods) (iv) I(s)= (v) (vii) i(t) = 10 _ 8e
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_2000t
(vi)
_8 10 + s s+2000 _ 1 =10 10L 1 s _ _ 1 (_8)L 1 = _ 8e 2000t s+2000
Table 16-1 Laplace Transform Pairs f t 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. t 14. 15. 16. 17.
F s 1 s 1 s2 1 s a 1 s a 2 ! s2 ! 2 s s2 ! 2 s sin  ! cos  s2 ! 2 s cos  ! sin  s2 ! 2 ! s a 2 !2 s a s a 2 !2 ! s2 ! 2 s s2 ! 2 sF s f 0 F s s e t1 s F s c1 F1 s c2 F2 s F1 s F2 s
1 t e at te at sin !t cos !t sin !t  cos !t  e at sin !t e at cos !t sinh !t cosh !t df dt f  d f t t1 c1 f1 t c2 f2 t t f1  f2 t  d
CHAP. 16]
THE LAPLACE TRANSFORM METHOD
Fig. 16-1
Fig. 16-2
Kirchho s voltage law, applied to the circuit for t > 0, yields the familiar di erential equation (i). This equation is transformed, term by term, into the s-domain equation (ii). The unknown current i t becomes I s , while the known voltage v 50u t is transformed to 50/s. Also, di=dt is transformed into i 0 sI s , in which i 0 is 2 A. Equation (iii) is solved for I s , and the solution is put in the form (iv) by the techniques of Section 16.6. Then lines 1, 3, and 16 of Table 16-1 are applied to obtain the inverse Laplace transform of I s , which is i t . A circuit can be drawn in the s-domain, as shown in Fig. 16-2. The initial current appears in the circuit as a voltage source, Li 0 . The s-domain current establishes the voltage terms RI s and sLI s in (ii) just as a phasor current I and an impedance Z create a phasor voltage IZ.
CONVERGENCE OF THE INTEGRAL
For the Laplace transform to exist, the integral (2) should converge. This limits the variable s  j! to a part of the complex plane called the convergence region. As an example, the transform of x t e at u t is 1= s a , provided Re s > a, which de nes its region of convergence.
EXAMPLE 16.2 Find the Laplace transform of x t 3e2t u t and show the region of convergence. 1 1 3 3 e s 2 t 1 ; Re s > 2 X s 3e2t e st dt 3e s 2 t dt 0 s 2 s 2 0 0
The region of convergence of X s is the right half plane  > 2, shown hatched in Fig. 16-3.
Fig. 16-3
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