Chapter 6: Problem 5
Find the inverse Laplace transform of the given function. $$ \frac{2 s+2}{s^{2}+2 s+5} $$
Short Answer
Expert verified
Answer: The inverse Laplace transform of the given function is \(e^{-t}\left(\frac{2}{5}\cos(2t)+\frac{1}{4}\sin(2t)\right)\).
Step by step solution
01
Perform Partial Fraction Decomposition
For this step, we begin by performing partial fraction decomposition on the given function.
$$
\frac{2s+2}{s^2+2s+5} = \frac{A}{s+\alpha} + \frac{Bs + C}{s^2+\beta s+\gamma}
$$
However, we can see that the denominator is already in a quadratic form that cannot be further factorized. Thus, partial fraction decomposition will be in this form:
$$
\frac{2s + 2}{s^2 + 2s + 5} = \frac{As + B}{s^2 + 2s + 5}
$$
Now we need to solve for A and B.
02
Solve for A and B
Multiply both sides of the equation by the denominator \((s^2 + 2s + 5)\):
$$
2s + 2 = (As + B)(s^2 + 2s + 5).
$$
Expanding the equation, we get:
$$
2s + 2 = As^2 + 2As + 5As + 5B
$$
By comparing coefficients of similar terms, we get:
$$
A + 2A + 5A = 2 \Rightarrow A = \frac{2}{8} = \frac{1}{4}
$$
Furthermore, we have:
$$
5B = 2 \Rightarrow B = \frac{2}{5}
$$
Now that we have A and B, we can rewrite our function:
03
Rewrite the function with A and B
With the values of A and B, rewrite the function as:
$$
\frac{2s + 2}{s^2 + 2s + 5} = \frac{\frac{1}{4}s + \frac{2}{5}}{s^2 + 2s + 5}
$$
04
Find the Inverse Laplace Transform
Check Laplace transform table to find inverse Laplace transforms for this type of function. We have the form:
$$
\mathcal{L}^{-1}\{ \frac{as+b}{s^2+2asw+ws^2} \} = e^{-at} (b\cos(wt)+a\sin(wt))
$$
So in our case, \(a=1\), \(b=\frac{2}{5}\) and \(w=2\). Using the table result, apply inverse Laplace transform to get the result:
$$
\mathcal{L}^{-1}\{ \frac{\frac{1}{4}s + \frac{2}{5}}{s^2 + 2s + 5} \} = e^{-t}\left(\frac{2}{5}\cos(2t)+\frac{1}{4}\sin(2t)\right)
$$
05
Final Answer
The final answer for the inverse Laplace transform of the given function is:
$$
e^{-t}\left(\frac{2}{5}\cos(2t)+\frac{1}{4}\sin(2t)\right)
$$
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to simplify complex rational expressions, especially when dealing with Laplace transforms. The goal is to break down a fraction with a polynomial in the denominator into a sum of simpler fractions, where the denominators are factors of the original denominator. This can make it easier to perform operations like finding the inverse Laplace transform.
Application in Inverse Laplace Transforms
When you have a rational expression where the degree of the numerator is less than the degree of the denominator, partial fraction decomposition can be particularly useful. Once decomposed, each term of the broken down expression often has a straightforward inverse Laplace transform, based on standard forms found in Laplace transform tables. Remember, it is critical to first determine whether the denominator can be factorized further. As shown in the step-by-step solution for our exercise, when it cannot, like with an irreducible quadratic, we express the numerator as a linear combination of terms.Laplace Transform Table
A Laplace transform table is an invaluable tool for engineers and mathematicians working with differential equations and complex systems. It lists many common functions along with their Laplace transforms, enabling quick lookup and application of these transformations to various problems, including inverse Laplace transforms.
Using the Table for Inverse Transforms
To use a Laplace transform table for inverse transforms, first, ensure that the function you're working with matches a form in the table. For inverse transforms, you're essentially reversing the Laplacian process to find the time-domain function from a given s-domain function. In our textbook solution, after employing partial fraction decomposition, we turned to the table to identify the corresponding time-domain functions that matched the s-domain functions we obtained.Comparing Coefficients
Comparing coefficients is a technique used in algebra to find values of unknown constants by matching the coefficients of corresponding powers of variables on both sides of an equation. It's based on the principle that if two polynomials are equal, then their corresponding coefficients for the same powers must also be equal.