Determine [tex]L ^{-1}{F}[/tex].
[tex]s F(s)-F(s)=\frac{4 s+5}{s^2+2 s+1}[/tex]
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[tex]L ^{-1}{F}=\square[/tex]
Answer (2)
Factor out F ( s ) and express it as F ( s ) = ( s − 1 ) ( s + 1 ) 2 4 s + 5 .
Perform partial fraction decomposition: F ( s ) = s − 1 A + s + 1 B + ( s + 1 ) 2 C , solving for A = 4 9 , B = − 4 9 , and C = − 2 1 .
Apply the inverse Laplace transform to each term using known transforms.
Obtain the final result: L − 1 { F ( s ) } = 4 9 e t − 4 9 e − t − 2 1 t e − t .
Explanation
Problem Analysis We are given the equation s F ( s ) − F ( s ) = s 2 + 2 s + 1 4 s + 5 and asked to find the inverse Laplace transform of F ( s ) , denoted as L − 1 { F ( s ) } .
Factoring First, factor out F ( s ) on the left side of the equation: ( s − 1 ) F ( s ) = s 2 + 2 s + 1 4 s + 5 .
Solving for F(s) Next, solve for F ( s ) : F ( s ) = ( s − 1 ) ( s 2 + 2 s + 1 ) 4 s + 5 = ( s − 1 ) ( s + 1 ) 2 4 s + 5 .
Partial Fraction Decomposition Now, perform partial fraction decomposition on F ( s ) : ( s − 1 ) ( s + 1 ) 2 4 s + 5 = s − 1 A + s + 1 B + ( s + 1 ) 2 C .
Clearing Denominators Multiply both sides by ( s − 1 ) ( s + 1 ) 2 to clear the denominators: 4 s + 5 = A ( s + 1 ) 2 + B ( s − 1 ) ( s + 1 ) + C ( s − 1 ) .
Solving for Coefficients Solve for the coefficients A , B , and C by substituting suitable values for s or by equating coefficients of powers of s .
Solving for A If s = 1 , then 4 ( 1 ) + 5 = A ( 1 + 1 ) 2 + B ( 0 ) + C ( 0 ) , so 9 = 4 A , which means A = 4 9 .
Solving for C If s = − 1 , then 4 ( − 1 ) + 5 = A ( 0 ) + B ( 0 ) + C ( − 1 − 1 ) , so 1 = − 2 C , which means C = − 2 1 .
Solving for B Expanding the equation, we have 4 s + 5 = A ( s 2 + 2 s + 1 ) + B ( s 2 − 1 ) + C ( s − 1 ) = A s 2 + 2 A s + A + B s 2 − B + C s − C . Then ( A + B ) s 2 + ( 2 A + C ) s + ( A − B − C ) = 4 s + 5 . So A + B = 0 , 2 A + C = 4 , and A − B − C = 5 . Since A = 4 9 , B = − 4 9 . Since C = − 2 1 , 2 A + C = 2 ( 4 9 ) − 2 1 = 2 9 − 2 1 = 2 8 = 4 , which is correct. Also, A − B − C = 4 9 − ( − 4 9 ) − ( − 2 1 ) = 4 9 + 4 9 + 2 1 = 4 18 + 4 2 = 4 20 = 5 , which is also correct.
F(s) with Coefficients Thus, F ( s ) = s − 1 9/4 − s + 1 9/4 − ( s + 1 ) 2 1/2 .
Inverse Laplace Transform Apply the inverse Laplace transform to each term: L − 1 { F ( s ) } = L − 1 { s − 1 9/4 } − L − 1 { s + 1 9/4 } − L − 1 { ( s + 1 ) 2 1/2 } .
Known Laplace Transforms Use the known Laplace transforms: L − 1 { s − a 1 } = e a t and L − 1 { ( s − a ) 2 1 } = t e a t .
Final Answer Therefore, L − 1 { F ( s ) } = 4 9 e t − 4 9 e − t − 2 1 t e − t .
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with resistors, capacitors, and inductors. The behavior of this circuit can be described by differential equations. By using Laplace transforms, you can convert these differential equations 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 behavior of the circuit over time. This is particularly helpful for analyzing transient responses, like what happens when you switch the circuit on or off.
To find the inverse Laplace transform of F(s), we first rearranged the given equation and solved for F(s). We then performed partial fraction decomposition and found the coefficients A, B, and C. Finally, we applied the inverse Laplace transforms to each term to arrive at the result: L^{-1}{F(s)} = \frac{9}{4} e^{t} - \frac{9}{4} e^{-t} - \frac{1}{2}te^{-t}.
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