Question

Find the inverse transforms of the functions \(F(p)\). \(\frac{2 p-1}{p^{2}-2 p+10} \quad\) Hint: You can use \(L 7\) and \(L 8\) with complex \(a\)

Short answer

The inverse transform is \( f(t) = 2e^{t} \sin(3t) + \frac{1}{3}e^{t} \cos(3t) \).

Step-by-step solution

Identify the Laplace Transform Formula

Given: \( F(p) = \frac{2p - 1}{p^2 - 2p + 10} \).We need to use the inverse Laplace transform to find the original function. Notice that the denominator can be manipulated into a recognizable form.

Complete the Square for the Denominator

Rewrite the denominator to complete the square:\[ p^2 - 2p + 10 = (p - 1)^2 + 9 \].So, \( F(p) = \frac{2p - 1}{(p - 1)^2 + 9} \).

Split the Numerator

Express the numerator in terms of \( p-1 \):\[ 2p - 1 = 2(p - 1) + 1 \].Thus, \( F(p) = \frac{2(p - 1) + 1}{(p - 1)^2 + 9} \).

Write as Sum of Two Fractions

Split the function into two simpler fractions:\( F(p) = \frac{2(p - 1)}{(p - 1)^2 + 9} + \frac{1}{(p - 1)^2 + 9} \).

Recognize Inverse Laplace Transforms

Using standard Laplace transform tables:\( \mathcal{L}^{-1} \left\{ \frac{a}{(p - b)^2 + a^2} \right\} = e^{bt} \sin(at) \) and \( \mathcal{L}^{-1} \left\{ \frac{a^2}{(p - b)^2 + a^2} \right\} = e^{bt} \cos(at) \).

Apply Formulas

\( \mathcal{L}^{-1} \left\{ \frac{2(p - 1)}{(p - 1)^2 + 9} \right\} = 2e^{t} \sin(3t) \) and \( \mathcal{L}^{-1} \left\{ \frac{1}{(p - 1)^2 + 9} \right\} = \frac{1}{3}e^{t} \cos(3t) \).

Combine Results

Combine the inverse transforms:\( f(t) = 2e^{t} \sin(3t) + \frac{1}{3}e^{t} \cos(3t) \).

Key concepts

Laplace transform

The Laplace transform is a powerful mathematical tool used predominantly in engineering and physics. It converts a time-domain function, usually denoted as f(t), into a complex frequency-domain representation known as F(p). This technique simplifies the analysis and solution of linear differential equations.
To calculate the Laplace transform, we apply the integral formula:
\[ \mathcal{L}\{ f(t) \} = \int_{0}^{\infty} e^{-pt} f(t) dt \]
This process can make solving complex differential equations much more manageable. For instance, instead of solving the differential equation directly, one can transform the problem into an algebraic equation in the p-domain, solve it, and then perform an inverse Laplace transform to get back to the time domain.

Partial fraction decomposition

Partial fraction decomposition is a method used to break down complex rational functions into simpler fractions. This is particularly useful in the context of inverse Laplace transforms, making it easier to work with standard Laplace transform tables.
To perform partial fraction decomposition, follow these steps:

• Ensure the degree of the numerator is less than that of the denominator. If not, perform polynomial division.
• Factor the denominator into its simplest form.
• Express the function as a sum of fractions with unknown coefficients.
• Solve for the coefficients by multiplying through by the common denominator and equating coefficients.

In our example, we used partial fraction decomposition to rewrite the given function into simpler forms which were easier to handle with Laplace transform tables:
\[ F(p) = \frac{2(p - 1)}{(p - 1)^2 + 9} + \frac{1}{(p - 1)^2 + 9} \]

Inverse transforms

Applying inverse Laplace transforms allows us to convert back from the frequency domain to the time domain. This step reveals the original time-dependent function. To do this efficiently, one typically utilizes standard Laplace transform tables, which map various kinds of functions between the time and frequency domains.
Typical formulas you might encounter include:

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

Through these transformations, complex expressions in the frequency domain are converted back into recognizable time-domain functions. By applying these known inverse Laplace transforms to our partial fractions, we reconstructed the original function.

Trigonometric functions

Trigonometric functions frequently appear in the results of inverse Laplace transforms. These include sine and cosine functions, which describe oscillatory behavior in the time domain.
Key trigonometric formulas used in inverse Laplace transforms include:

• \( \sin(at) \): Often comes from terms of the form \( \frac{a}{(p - b)^2 + a^2} \)
• \( \cos(at) \): Often comes from terms of the form \( \frac{a^2}{(p - b)^2 + a^2} \)

In our specific problem, applying these transforms allowed us to recover components involving sines and cosines, summarizing as:
\[ f(t) = 2e^{t} \sin(3t) + \frac{1}{3}e^{t} \cos(3t) \]
These expressions represent oscillatory behavior modulated by an exponential term, indicating a combination of sine and cosine waves scaled by an exponentially growing or decaying factor.