Determine [tex]L^{-1}\left\{F\right\}[/tex].
[tex]F(s)=\frac{4 s^2+84 s+336}{(s-2)\left(s^2+12 s+37\right)}[/tex]
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[tex]L^{-1}\left\{F\right\}=\square[/tex]
Answer (2)
Perform partial fraction decomposition on F ( s ) to express it as a sum of simpler fractions: s − 2 A + s 2 + 12 s + 37 B s + C .
Solve for the constants A , B , and C , obtaining A = 8 , B = − 4 , and C = − 20 .
Find the inverse Laplace transform of each term separately, using the properties of Laplace transforms and completing the square where necessary.
Combine the inverse Laplace transforms to obtain the final result: 8 e 2 t − 4 e − 6 t cos ( t ) + 4 e − 6 t sin ( t ) .
Explanation
Problem Analysis We are given the function F(s)=\frac{4 s^2+84 s+336}{(s-2)\(s^2+12 s+37\)} . Our goal is to find the inverse Laplace transform of F ( s ) , denoted as L − 1 { F ( s ) } .
Partial Fraction Decomposition To find the inverse Laplace transform, we first perform partial fraction decomposition on F ( s ) . We express F ( s ) as ( s − 2 ) ( s 2 + 12 s + 37 ) 4 s 2 + 84 s + 336 = s − 2 A + s 2 + 12 s + 37 B s + C .
Clearing Denominators Multiplying both sides by ( s − 2 ) ( s 2 + 12 s + 37 ) , we get 4 s 2 + 84 s + 336 = A ( s 2 + 12 s + 37 ) + ( B s + C ) ( s − 2 ) .
Expanding and Collecting Terms Expanding and collecting like terms, we have 4 s 2 + 84 s + 336 = ( A + B ) s 2 + ( 12 A − 2 B + C ) s + ( 37 A − 2 C ) .
Equating Coefficients Equating coefficients, we obtain the following system of equations:
A + B = 4
12 A − 2 B + C = 84
37 A − 2 C = 336
Solving for Coefficients Solving this system of equations, we find A = 8 , B = − 4 , and C = − 20 .
Rewriting F(s) Thus, we can rewrite F ( s ) as F ( s ) = s − 2 8 + s 2 + 12 s + 37 − 4 s − 20 .
Inverse Laplace Transform of First Term Now, we find the inverse Laplace transform of each term. We have L − 1 { s − 2 8 } = 8 e 2 t .
Completing the Square For the second term, we complete the square in the denominator: s 2 + 12 s + 37 = ( s + 6 ) 2 + 1. Then we rewrite the second term as ( s + 6 ) 2 + 1 − 4 s − 20 = ( s + 6 ) 2 + 1 − 4 ( s + 6 ) + 24 − 20 = ( s + 6 ) 2 + 1 − 4 ( s + 6 ) + 4 = − 4 ( s + 6 ) 2 + 1 s + 6 + 4 ( s + 6 ) 2 + 1 1 .
Inverse Laplace Transform of Second Term The inverse Laplace transform of the second term is L − 1 { s 2 + 12 s + 37 − 4 s − 20 } = − 4 L − 1 { ( s + 6 ) 2 + 1 s + 6 } + 4 L − 1 { ( s + 6 ) 2 + 1 1 } = − 4 e − 6 t cos ( t ) + 4 e − 6 t sin ( t ) .
Combining Inverse Laplace Transforms Combining the inverse Laplace transforms, we get L − 1 { F ( s ) } = 8 e 2 t − 4 e − 6 t cos ( t ) + 4 e − 6 t sin ( t ) .
Final Answer Therefore, the inverse Laplace transform of F ( s ) is L − 1 { F ( s ) } = 8 e 2 t − 4 e − 6 t cos ( t ) + 4 e − 6 t sin ( t ) .
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with a voltage source, resistors, and capacitors. 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 (the 's' domain), you can use the inverse Laplace transform to convert it back to the time domain, giving you the actual voltage and current as functions of time. This helps engineers design and troubleshoot circuits more efficiently.
To find the inverse Laplace transform of F ( s ) = ( s − 2 ) ( s 2 + 12 s + 37 ) 4 s 2 + 84 s + 336 , we performed partial fraction decomposition, solved for the constants, and found the individual inverse transforms for each term. The final answer is L − 1 { F ( s )} = 8 e 2 t − 4 e − 6 t cos ( t ) + 4 e − 6 t sin ( t ) .
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