Page 410

CHAP. 16]

THE LAPLACE TRANSFORM METHOD

399

domain, a process called the inverse Laplace transform, l1 ½FðsÞ ¼ f ðtÞ. The inverse Laplace transform can also be expressed as an integral, the complex inversion integral: ð 1 0 þj1 l1 ½FðsÞ ¼ f ðtÞ ¼ FðsÞest ds ð3Þ 2j 0 j1 In (3) the path of integration is a straight line parallel to the j!-axis, such that all the poles of FðsÞ lie to the left of the line. Here again, the integration need not actually be performed unless it is a question of adding to existing tables of transform pairs. It should be remarked that taking the direct Laplace transform of a physical quantity introduces an extra time unit in the result. For instance, if iðtÞ is a current in A, then IðsÞ has the units A  s (or C). Because the extra unit s will be removed in taking the inverse Laplace transform, we shall generally omit to cite units in the s-domain, shall still call IðsÞ a ‘‘current,’’ indicate it by an arrow, and so on.

16.3

SELECTED LAPLACE TRANSFORMS The Laplace transform of the unit step function is easily obtained: ð1 1 1 l½uðtÞ ¼ ð1Þest dt ¼  ½est 1 0 ¼ s s 0

From the linearity of the Laplace transform, it follows that vðtÞ ¼ VuðtÞ in the time domain has the sdomain image VðsÞ ¼ V=s. The exponential decay function, which appeared often in the transients of Chapter 7, is another time function which is readily transformed. ð1 A ðaþsÞt 1 A ½e l½Aeat  ¼ Aeat est dt ¼ 0 ¼ Aþs sþa 0 or, inversely, l

1



 A ¼ Aeat sþa

The transform of a sine function is also easily obtained.  1 ð1 sðsin !tÞest  est ! cos !t ! ðsin !tÞest dt ¼ ¼ 2 l½sin !t ¼ 2 2 þ ! þ !2 s s 0 0 It will be useful now to obtain the transform of a derivative, df ðtÞ=dt.   ð1 df ðtÞ df ðtÞ st e dt l ¼ dt dt 0 Integrating by parts,   ð1 ð1 df ðtÞ st 1 st þ f ðtÞðse Þ dt ¼ f ð0 Þ þ s f ðtÞest dt ¼ f ð0þ Þ þ sFðsÞ l ¼ ½e f ðtÞ0þ  dt 0 0 A small collection of transform pairs, including those obtained above, is given in Table 16-1. The last five lines of the table present some general properties of the Laplace transform. EXAMPLE 16.1 Consider a series RL circuit, with R ¼ 5  and L ¼ 2:5 mH. At t ¼ 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.

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An