Explanation Problem design: Consider the following functions: (a) G ( s ) = 20 ( s + 2 ) s ( s 2 + 6 s + 25 ) (b) F ( s ) = 6 s 2 + 36 s + 20 ( s + 1 ) ( s + 2 ) ( s + 3 ) Find the inverse Laplace transform for the given functions. Formula used: Write the general expression for the inverse Laplace transform. g ( t ) = L − 1 [ G ( s ) ] (1) f ( t ) = L − 1 [ F ( s ) ] (2) Write the general expressions to find the inverse Laplace transform function. L − 1 [ 1 s + a ] = e − a t (3) L − 1 [ s + a ( s + a ) 2 + ω 2 ] = e − a t cos ω t (4) L − 1 [ ω ( s + a ) 2 + ω 2 ] = e − a t sin ω t (5) L − 1 [ 1 s ] = u ( t ) (6) Here, s + a is the frequency shift or frequency translation, and s is a complex variable. Write the general expression for the unit step function. u ( t ) = { 0 , t < 0 1 , t > 0 (7) Calculation: Part (a): The given Laplace transform, G ( s ) = 20 ( s + 2 ) s ( s 2 + 6 s + 25 ) (8) Expand G ( s ) using partial fraction. G ( s ) = 20 ( s + 2 ) s ( s 2 + 6 s + 25 ) = A s + B s + C s 2 + 6 s + 25 (9) Here, A , B , and C are the constants. Now, to find the constants by using algebraic method. Consider the partial fraction, 20 ( s + 2 ) s ( s 2 + 6 s + 25 ) = A s + B s + C s 2 + 6 s + 25 20 ( s + 2 ) s ( s 2 + 6 s + 25 ) = A ( s 2 + 6 s + 25 ) + ( B s + C ) s s ( s 2 + 6 s + 25 ) 20 ( s + 2 ) = A ( s 2 + 6 s + 25 ) + ( B s + C ) s 20 s + 40 = A s 2 + 6 A s + 25 A + B s 2 + C s Reduce the equation as follows, 20 s + 40 = ( A + B ) s 2 + ( 6 A + C ) s + 25 A (10) Compare the coefficients of s 2 in equation (10) on the both sides. A + B = 0 B = − A (11) Compare the coefficients of s in equation (10) on the both sides.. 6 A + C = 20 (12) Compare the coefficients of constant term in equation (10) to find the constant A . 25 A = 40 A = 40 25 A = 8 5 (13) Substitute equation (13) in equation (11) to find the constant B . B = − 8 5 Substitute equation (13) in equation (12) to find the constant C . 6 ( 8 5 ) + C = 20 C = 20 − 48 5 C = 52 5 Substitute 8 5 for A , − 8 5 for B and 52 5 for C in equation (9) to find G ( s ) . G ( s ) = ( 8 5 ) s + ( − 8 5 ) s − ( 24 5 ) + ( 24 5 ) + ( 52 5 ) s 2 + 6 s + 25 = ( 8 5 ) s + ( − 8 5 ) ( s + 3 ) + ( 24 5 ) + ( 52 5 ) s 2 + 6 s + 9 + 16 = ( 8 5 ) s + ( − 8 5 ) ( s + 3 ) ( s + 3 ) 2 + 16 + ( 76 5 ) ( s + 3 ) 2 + 16 { ∵ ( a + b ) 2 = a 2 + 2 a b + b 2 } = ( 8 5 ) s + ( − 8 5 ) ( s + 3 ) ( s + 3 ) 2 + 4 2 + ( 19 5 ) 4 ( s + 3 ) 2 + 4 2 Reduce the equation as follows, G ( s ) = 8 5 1 s − 8 5 s + 3 ( s + 3 ) 2 + 4 2 + 19 5 4 ( s + 3 ) 2 + 4 2 Substitute 8 5 1 s − 8 5 s + 3 ( s + 3 ) 2 + 4 2 + 19 5 4 ( s + 3 ) 2 + 4 2 for G ( s ) in equation (1). g ( t ) = L − 1 [ 8 5 1 s − 8 5 s + 3 ( s + 3 ) 2 + 4 2 + 19 5 4 ( s + 3 ) 2 + 4 2 ] = L − 1 [ 8 5 1 s ] − L − 1 [ 8 5 s + 3 ( s + 3 ) 2 + 4 2 ] + L − 1 [ 19 5 4 ( s + 3 ) 2 + 4 2 ] g ( t ) = 8 5 L − 1 [ 1 s ] − 8 5 L − 1 [ s + 3 ( s + 3 ) 2 + 4 2 ] + 19 5 L − 1 [ 4 ( s + 3 ) 2 + 4 2 ] (14) Apply inverse Laplace transform function given in equation (4), (5) and (6) to equation (14)

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ISBN: 8220102801448

Author: Alexander

Publisher: YUZU

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EBK FUNDAMENTALS OF ELECTRIC CIRCUITS

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Design a problem to make the better understand of the inverse Laplace transform.

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