Determine [tex]L ^{-1}{F}[/tex].
[tex]s^2 F(s)+s F(s)-20 F(s)=\frac{s^2+5}{s^2+6 s}[/tex]
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[tex]L ^{-1}{F}=\square[/tex]
Answer (2)
First, isolate F ( s ) by factoring and dividing: F ( s ) = s ( s + 6 ) ( s + 5 ) ( s − 4 ) s 2 + 5 .
Perform partial fraction decomposition: s ( s + 6 ) ( s + 5 ) ( s − 4 ) s 2 + 5 = s A + s + 6 B + s + 5 C + s − 4 D , and solve for A, B, C, and D.
Apply the inverse Laplace transform to each term using the table: L − 1 s 1 = 1 and L − 1 s + a 1 = e − a t .
Combine the results to obtain the final answer: − 24 1 − 60 41 e − 6 t + 3 2 e − 5 t + 120 7 e 4 t .
Explanation
Problem Setup We are given the equation s 2 F ( s ) + s F ( s ) − 20 F ( s ) = s 2 + 6 s s 2 + 5 and asked to find the inverse Laplace transform of F ( s ) , denoted as L − 1 F ( s ) .
Isolating F(s) First, we solve for F ( s ) by factoring out F ( s ) on the left side of the equation: F ( s ) ( s 2 + s − 20 ) = s 2 + 6 s s 2 + 5
F(s) Isolated Then, we divide both sides by ( s 2 + s − 20 ) to isolate F ( s ) :
F ( s ) = ( s 2 + 6 s ) ( s 2 + s − 20 ) s 2 + 5
Factoring Denominators Next, we factor the denominators: F ( s ) = s ( s + 6 ) ( s + 5 ) ( s − 4 ) s 2 + 5
Partial Fraction Decomposition Now, we perform partial fraction decomposition on F ( s ) :
s ( s + 6 ) ( s + 5 ) ( s − 4 ) s 2 + 5 = s A + s + 6 B + s + 5 C + s − 4 D
Clearing Denominators To find A, B, C, and D, we multiply both sides by s ( s + 6 ) ( s + 5 ) ( s − 4 ) :
s 2 + 5 = A ( s + 6 ) ( s + 5 ) ( s − 4 ) + B ( s ) ( s + 5 ) ( s − 4 ) + C ( s ) ( s + 6 ) ( s − 4 ) + D ( s ) ( s + 6 ) ( s + 5 )
Solving for Constants Solving for the constants, we find: A = -1/24 B = -41/60 C = 2/3 D = 7/120
Rewriting F(s) So, we can rewrite F ( s ) as: F ( s ) = s − 1/24 + s + 6 − 41/60 + s + 5 2/3 + s − 4 7/120
Applying Inverse Laplace Transform Now, we apply the inverse Laplace transform to each term: L − 1 F ( s ) = L − 1 s − 1/24 + L − 1 s + 6 − 41/60 + L − 1 s + 5 2/3 + L − 1 s − 4 7/120 L − 1 F ( s ) = − 24 1 L − 1 s 1 − 60 41 L − 1 s + 6 1 + 3 2 L − 1 s + 5 1 + 120 7 L − 1 s − 4 1
Using Laplace Transform Table Using the Laplace transform table, we know: L − 1 s 1 = 1 L − 1 s + a 1 = e − a t Therefore: L − 1 F ( s ) = − 24 1 − 60 41 e − 6 t + 3 2 e − 5 t + 120 7 e 4 t
Final Answer Combining the terms, we get: L − 1 F ( s ) = − 24 1 − 60 41 e − 6 t + 3 2 e − 5 t + 120 7 e 4 t
Conclusion Therefore, the inverse Laplace transform of F ( s ) is: L − 1 F ( s ) = − 24 1 − 60 41 e − 6 t + 3 2 e − 5 t + 120 7 e 4 t
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with resistors, capacitors, and inductors. By using Laplace transforms, you can convert differential equations that describe the circuit's behavior into algebraic equations, making them much easier to solve. Once you find the solution in the Laplace domain, you can use the inverse Laplace transform to convert it back to the time domain, giving you the actual voltage or current as a function of time. This allows engineers to predict how a circuit will behave under different conditions and design it accordingly.
To find L − 1 F ( s ) , we isolate F ( s ) , perform partial fraction decomposition, and apply the inverse Laplace transform. The final result is L − 1 F ( s ) = − 24 1 − 60 41 e − 6 t + 3 2 e − 5 t + 120 7 e 4 t .
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