THE LAPLACE TRANSFORM METHOD
Kirchhoﬀ’s voltage law, applied to the circuit for t > 0, yields the familiar diﬀerential equation (i). This equation is transformed, term by term, into the s-domain equation (ii). The unknown current iðtÞ becomes IðsÞ, while the known voltage v ¼ 50uðtÞ is transformed to 50/s. Also, di=dt is transformed into ið0þ Þ þ sIðsÞ, in which ið0þ Þ is 2 A. Equation (iii) is solved for IðsÞ, 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 IðsÞ, which is iðtÞ. 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, Lið0þ Þ. The s-domain current establishes the voltage terms RIðsÞ and sLIðsÞ 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 ¼ þ j! to a part of the complex plane called the convergence region. As an example, the transform of xðtÞ ¼ eat uðtÞ is 1=ðs þ aÞ, provided Re ½s > a, which deﬁnes its region of convergence. EXAMPLE 16.2
Find the Laplace transform of xðtÞ ¼ 3e2t uðtÞ and show the region of convergence. ð1 ð1 3 3 ½eðs2Þt 1 ; Re ½s > 2 XðsÞ ¼ 3e2t est dt ¼ 3eðs2Þt dt ¼ 0 ¼ s 2 s 2 0 0
The region of convergence of XðsÞ is the right half plane > 2, shown hatched in Fig. 16-3.
INITIAL-VALUE AND FINAL-VALUE THEOREMS
Taking the limit as s ! 1 (through real values) of the direct Laplace transform of the derivative, df ðtÞ=dt, lim l
ð1 df ðtÞ df ðtÞ st e dt ¼ lim fsFðsÞ f ð0þ Þg ¼ lim s!1 0 s!1 dt dt
Published on May 10, 2013