Question

Find the inverse Laplace transform of the given function. $$ \frac{2 s-3}{s^{2}+2 s+10} $$

Short answer

Answer: The inverse Laplace transform of the function \( \frac{2 s-3}{s^{2}+2 s+10} \) is \( f(t) = 2e^{-t}\cos(3t) - \frac{5}{3}e^{-t}\sin(3t) \).

Step-by-step solution

Identify the form of the given function

The given function can be written as a combination of standard Laplace transforms. To do that, we need to use partial fraction decomposition and rewrite the given function in the form of a sum of simpler fractions.

Partial fraction decomposition

We'll first complete the square in the denominator, and then perform partial fraction decomposition: $$ \frac{2s-3}{s^{2}+2 s+10} = \frac{2s - 3}{(s + 1)^{2} + 3^{2}} $$ Now we rewrite the given function in the form of a sum of simpler fractions: $$ \frac{2s - 3}{(s + 1)^{2} + 3^{2}} = \frac{A(s + 1) + B}{(s + 1)^{2} + 3^{2}} $$ Equating the numerators, we have: $$ 2s - 3 = A(s + 1) + B $$ By comparing coefficients, we get: \(A = 2\), and \(B = -5\) So, the given function can be written as: $$ \frac{2s - 3}{(s + 1)^{2} + 3^{2}} = \frac{2(s + 1) - 5}{(s + 1)^{2} + 3^{2}} $$

Inverse Laplace Transform

Now, we can identify the standard Laplace transform and find its inverse using inverse Laplace transform formulas: The inverse Laplace transform of a given function is given by: $$ \mathcal{L}^{-1}\left\{\frac{2s - 3}{(s + 1)^{2} + 3^{2}}\right\} = \mathcal{L}^{-1}\left\{\frac{2(s + 1) - 5}{(s + 1)^{2} + 3^{2}}\right\} $$ Inverse Laplace transform of a standard function in the form \(\frac{s - a}{(s - a)^{2} + b^{2}}\) is given by: $$ \mathcal{L}^{-1}\left\{\frac{s-a}{(s-a)^{2} + b^{2}}\right\} = e^{at}\cos(bt) $$ and the inverse Laplace transform of a standard function in the form \(\frac{1}{(s - a)^{2} + b^{2}}\) is given by: $$ \mathcal{L}^{-1}\left\{\frac{1}{(s - a)^{2} + b^{2}}\right\} = e^{at}\frac{\sin(bt)}{b} $$ Using the above formulas, we get: $$ \mathcal{L}^{-1}\left\{\frac{2(s + 1) - 5}{(s + 1)^{2} + 3^{2}}\right\} = \mathcal{L}^{-1}\left\{2\frac{(s + 1)}{(s + 1)^{2} + 3^{2}}\right\} - \mathcal{L}^{-1}\left\{5\frac{1}{(s + 1)^{2} + 3^{2}}\right\} $$ $$ = 2e^{-t}\cos(3t) - 5e^{-t}\frac{\sin(3t)}{3} $$ So, the inverse Laplace transform of the given function is: $$ f(t) = 2e^{-t}\cos(3t) - \frac{5}{3}e^{-t}\sin(3t) $$

Key concepts

partial fraction decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions. This technique is valuable for finding inverse Laplace transforms, as it simplifies the process of matching terms with known Laplace transform pairs.
In the given exercise, we started with the fraction \( \frac{2s - 3}{s^{2} + 2s + 10} \). To facilitate decomposition, we first completed the square on the denominator, rewriting it as \((s + 1)^{2} + 3^{2}\).
Then, we represented the numerator as a linear combination aligned with the denominator, \( A(s + 1) + B \). Solving for \( A \) and \( B \) by comparing coefficients gave us simple terms that are easier to invert using known Laplace transform formulas.

standard Laplace transforms

Standard Laplace transforms are fundamental transformations used to convert time-domain functions into the complex frequency domain, where they are easier to manipulate and solve.
The inverse of these standard transforms allows us to revert back to the time domain. In solving the exercise, the function was rewritten using standard forms with \( s \) shifted by "+1" due to the completed square in the denominator.
This manipulation is crucial since it aligns the problem with standard inverse Laplace transform formulas, allowing a straightforward calculation. Remember, this typically involves either cosine or sine formulas.

inverse Laplace transform formulas

Inverse Laplace transform formulas are essential for translating functions from the complex frequency domain back to the time domain. These formulas match complex expressions to time-domain functions, such as exponentials and trigonometric functions.
For the exercise function \( \frac{2s - 3}{(s + 1)^{2} + 3^{2}} \), we used the following inverse formulas:

• \( \mathcal{L}^{-1}\{ \frac{s-a}{(s-a)^{2} + b^{2}} \} = e^{at}\cos(bt) \)
• \( \mathcal{L}^{-1}\{ \frac{1}{(s-a)^{2} + b^{2}} \} = e^{at}\frac{\sin(bt)}{b} \)

These formulas relate the time shift and the oscillating behavior (either cosine or sine) based on shifts in the complex frequency "s".
They were applied directly to the terms of the partial fraction decomposition, resulting in a combination of exponential decay, cosine, and sine terms.

cosine and sine transforms

Cosine and sine transforms form the backbone of inverse Laplace transforms for harmonic oscillators in engineering and physics. They describe oscillatory behaviors seen in mechanical and electrical systems.
In the exercise, the inverse transform involved two main components: an exponential decay function multiplied by either the cosine or sine functions.

• The formula \( e^{at}\cos(bt) \) corresponds to the inverse Laplace transform of shifting along \( s \) and relates to damped cosine oscillations.
• The formula \( e^{at}\frac{\sin(bt)}{b} \) deals with similar sine oscillations.

These components combined to describe the overall time-domain behavior of the original complex expression, denoted by \( f(t) = 2e^{-t}\cos(3t) - \frac{5}{3}e^{-t}\sin(3t) \). Thus, by utilizing cosine and sine transforms, we can represent how a system evolves over time with oscillatory decay.