Problem 25 - Differential Equations and Linear Algebra

Question

Determine a function \(f(t)\) that has the given Laplace transform \(F(s)\). $$F(s)=\frac{2 s+5}{s\left(s^{2}+4 s+20\right)}$$

Step-by-Step Solution

Answer

The function \(f(t)\) that has the given Laplace transform \(F(s)\) is $$f(t) = e^{-2t}\left(2\cos(4t)+\frac{5}{2}\sin(4t)\right)$$.

Step 1: Perform Partial Fraction Decomposition

First, perform partial fraction decomposition to express \(F(s)\) as the sum of simpler fractions. $$\frac{2s+5}{s(s^2 + 4s + 20)} = \frac{A}{s} + \frac{Bs+C}{s^2+4s+20}$$ Now, we need to find the constants A, B, and C.

Step 2: Find the Constants A, B, and C

To determine A, B, and C, we can clear the denominators and match the numerators: $$(2s+5) = A(s^2+4s+20) + (Bs+C)s$$ We can equate coefficients of s-powers: 1. Coefficient of \(s^2\): \(0 = A\) 2. Coefficient of \(s\): \(2 = 4A+B\) → \(B=2\) 3. Constant term: \(5 = 20A+C\) → \(C=5\) Thus, the constants A, B, and C are 0, 2, and 5 respectively. Now we can rewrite the decomposed function: $$F(s) = \frac{0}{s} + \frac{2s+5}{s^{2}+4s+20}$$

Step 3: Inverse Laplace Transform

Now, we can take the inverse Laplace transform of each term to find the time-domain function f(t). For the first term, since A=0, it will not contribute to f(t). For the second term, completing the square of the quadratic part in the denominator, we have $$s^2 + 4s + 20 = (s+2)^2 + 16 = (s+2)^2 + (4i)^2$$. We observe that the second term resembles the inverse Laplace transform of a sinusoidal function. Using the inverse transform formula, we know $$\mathcal{L}^{-1}\left\{\frac{s-a}{(s-a)^{2}+b^{2}}\right\}=e^{at}\cos(bt)$$. In our case, a=-2 and b=4. Therefore, we have: $$\mathcal{L}^{-1}\left\{\frac{2s+5}{s^2 + 4s + 20}\right\} = e^{-2t}\left(2\cos(4t)+\frac{5}{2}\sin(4t)\right)$$ So, the function f(t) will be:

Step 4: Final Answer

$$f(t) = e^{-2t}\left(2\cos(4t)+\frac{5}{2}\sin(4t)\right)$$

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a crucial step when working with Laplace transforms, especially when the transform's expression includes rational functions. In simple terms, it involves breaking down a complex fraction into several simpler fractions that are easier to work with. This decomposition helps simplify calculations, particularly when evaluating inverse Laplace transforms.

In the given exercise, the function to decompose is \( \frac{2s+5}{s(s^2 + 4s + 20)} \). We set up the decomposition as follows:

• We express the fraction as the sum: \( \frac{A}{s} + \frac{Bs+C}{s^2 + 4s + 20} \).
• The goal is to determine the constants \( A, B, \) and \( C \) by clearing the denominators and equating coefficients of corresponding powers of \( s \).
• From the solution: \( A=0 \), \( B=2 \), and \( C=5 \).

With this result, the complex rational expression is simplified, making it ready for the next steps, including the inverse Laplace transformation.

Inverse Laplace Transform

The inverse Laplace transform is the process that converts a Laplace transform back into a time-domain function. This is essential when determining the original function from its transform for practical applications in engineering and physics.

In the exercise, after decomposing the Laplace transform function, **the focus moves towards finding what time-domain function corresponds to the given transforms.**

We see that one term vanished because it was linked to \( A = 0 \). What remains is to find the inverse of \( \frac{2s+5}{s^2 + 4s + 20} \).

By completing the square in the denominator,

• we reformulate \( s^2 + 4s + 20 \) into \( (s+2)^2 + (4i)^2 \).

This reformulation allows us to apply known inverse Laplace transform formulas for sinusoidal functions:

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

By applying these formulas, **the time-dependent function** is uncovered, \( f(t) = e^{-2t}(2\cos(4t) + \frac{5}{2}\sin(4t)) \), representing a mix of exponential decay and oscillations.

Sinusoidal Function

A sinusoidal function is a mathematical curve that describes a smooth periodic oscillation. Sinusoidal functions are prevalent in wave-related phenomena, describing sound waves, light waves, and other periodic functions.

In the context of Laplace transforms, sinusoidal functions often appear when solving differential equations involving oscillations. During the inverse Laplace transform step of the exercise, the decomposed term aligns perfectly with forms that produce sinusoidal functions in the time domain:
The transformation results in
**\( e^{-2t}(2\cos(4t) + \frac{5}{2}\sin(4t)) \).**

• The part \( e^{-2t} \) indicates exponential decay, affecting the amplitude of the sinusoidal components over time.
• The \( \cos(4t) \) and \( \sin(4t) \) functions introduce the oscillatory behavior.

The mix of exponential and sinusoidal functions captures the dynamics of systems that experience both decay and periodic oscillations.

The resulting function, \( f(t) \), combines these elements to model real-world phenomena effectively, showcasing the power of sinusoidal functions in engineering and physics applications.