tailieunhanh - Electric Circuits, 9th Edition P48

Electric Circuits, 9th Edition P48. Designed for use in a one or two-semester Introductory Circuit Analysis or Circuit Theory Course taught in Electrical or Computer Engineering Departments. Electric Circuits 9/e is the most widely used introductory circuits textbook of the past 25 years. As this book has evolved over the years to meet the changing learning styles of students, importantly, the underlying teaching approaches and philosophies remain unchanged. | 446 Introduction to the Laplace Transform or 96 -3 4 -8 -2 K2 72. Then is 96 5 5 .y 12 5 5 8 K3 48. s -6 From Eq. and the K values obtained 96 5 5 s 12 120 48 72 --------- ----------- ------1-----------------. 5 5 8 s 6 5 5 6 5 8 At this point testing the result to protect against computational errors is a good idea. As we already mentioned a partial fraction expansion creates an identity thus both sides of Eq. must be the same for all 5 values. The choice of test values is completely open hence we choose values that are easy to verify. For example in Eq. testing at either -5 or -12 is attractive because in both cases the left-hand side reduces to zero. Choosing -5 yields 120 48 72 y - -24 48 - 24 0 whereas testing -12 gives 120 48 72 -10-8 18 0. Now confident that the numerical values of the various Ks are correct we proceed to find the inverse transform . 96 5 5 5 12 5 7----------f 120 48e-6 - 72e S M r . I 5 5 8 5 6 J v v assessment problems Objective 2 Be able to calculate the inverse Laplace transform using partial fraction expansion and the Laplace transform table Find Z if _ Ô52 265 26 s l s 2 s 3 Answer f t - 3c- 2e 21 e 3t u t . Find r if 7s2 635 134 5 3 5 4 5 5 Answer f f 4e-3 6e 4 - 3e yl u f . NOTE Also try Chapter Problems a and b . Inverse Transforms 447 Partial Fraction Expansion Distinct Complex Roots of D s The only difference between finding the coefficients associated with distinct complex roots and finding those associated with distinct real roots is that the algebra in the former involves complex numbers. We illustrate by expanding the rational function r s 100 5 3 F s ----------------------. 5 6 .y2 65 25 We begin by noting that F s is a proper rational function. Next we must find the roots of the quadratic term s2 6s 25 s2 6s 25 5 3 - J4 5 3 4 . With the denominator in factored form we proceed as before 100 5 3 _ 5 6 s2 6s 25 Ky Kt L I __ I _J_ 6 5

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