Determine [tex]$e^{-1}{F}$[/tex].
[tex]$F(s)=\frac{4 s^3-9 s^2-s+9}{s^3(s-9)}$[/tex]
Answer (2)
Decompose F ( s ) into partial fractions: F ( s ) = s 1 − s 3 1 + s − 9 3 .
Find the inverse Laplace transform of each term: L − 1 { s 1 } = 1 , L − 1 { s 3 1 } = 2 t 2 , and L − 1 { s − 9 1 } = e 9 t .
Combine the inverse Laplace transforms: f ( t ) = 1 − 2 t 2 + 3 e 9 t .
The inverse Laplace transform of F ( s ) is 1 − 2 t 2 + 3 e 9 t .
Explanation
Problem Analysis We are asked to find the inverse Laplace transform of the function F ( s ) = s 3 ( s − 9 ) 4 s 3 − 9 s 2 − s + 9 . This involves finding a function of time, f ( t ) , such that when we take its Laplace transform, we obtain F ( s ) .
Partial Fraction Decomposition Setup To find the inverse Laplace transform, we first need to decompose F ( s ) into partial fractions. This will allow us to express F ( s ) as a sum of simpler terms, each of which has a known inverse Laplace transform. We assume the partial fraction decomposition has the form:
s 3 ( s − 9 ) 4 s 3 − 9 s 2 − s + 9 = s A + s 2 B + s 3 C + s − 9 D
Clearing Denominators Next, we multiply both sides of the equation by s 3 ( s − 9 ) to clear the denominators:
4 s 3 − 9 s 2 − s + 9 = A s 2 ( s − 9 ) + B s ( s − 9 ) + C ( s − 9 ) + D s 3
Solving for C and D Now, we solve for the constants A , B , C , and D . We can do this by substituting convenient values of s :
If s = 0 , we have:
4 ( 0 ) 3 − 9 ( 0 ) 2 − 0 + 9 = A ( 0 ) + B ( 0 ) + C ( 0 − 9 ) + D ( 0 )
9 = − 9 C
C = − 1
If s = 9 , we have:
4 ( 9 ) 3 − 9 ( 9 ) 2 − 9 + 9 = A ( 0 ) + B ( 0 ) + C ( 0 ) + D ( 9 ) 3
4 ( 729 ) − 9 ( 81 ) = 729 D
2916 − 729 = 729 D
2187 = 729 D
D = 3
Solving for A and B Now that we have C = − 1 and D = 3 , we can substitute these values back into the equation:
4 s 3 − 9 s 2 − s + 9 = A s 2 ( s − 9 ) + B s ( s − 9 ) − 1 ( s − 9 ) + 3 s 3
To find A and B , we can compare coefficients of the powers of s . Comparing the coefficients of s 3 , we have:
4 = A + 3
A = 1
Comparing the coefficients of s 2 , we have:
− 9 = − 9 A + B
− 9 = − 9 ( 1 ) + B
B = 0
Partial Fraction Decomposition Result Now we have A = 1 , B = 0 , C = − 1 , and D = 3 . So the partial fraction decomposition is:
F ( s ) = s 1 + s 2 0 + s 3 − 1 + s − 9 3 = s 1 − s 3 1 + s − 9 3
Inverse Laplace Transforms of Individual Terms Now we find the inverse Laplace transform of each term:
L − 1 { s 1 } = 1
L − 1 { s 3 1 } = 2 ! t 2
L − 1 { s − 9 1 } = e 9 t
Final Inverse Laplace Transform Therefore, the inverse Laplace transform of F ( s ) is:
f ( t ) = L − 1 { F ( s ) } = 1 − 2 t 2 + 3 e 9 t
Conclusion Thus, the inverse Laplace transform of F ( s ) is 1 − 2 t 2 + 3 e 9 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 the analysis much simpler. For example, you can determine the voltage or current response of the circuit to a specific input signal, like a step function or a sinusoidal wave. This helps in designing and troubleshooting electrical systems.
To find e − 1 F , we first perform partial fraction decomposition of F ( s ) and find the inverse Laplace transform f ( t ) = 1 − 2 t 2 + 3 e 9 t . Multiplying this by e − 1 gives the final expression for e − 1 F ( s ) .
;
