Problem 24 - Differential Equations and Linear Algebra

Question

Express \(L^{-1}[F(s) G(s)]\) in terms of a convolution integral. $$F(s)=\frac{s+4}{s^{2}+8 s+25}, \quad G(s)=\frac{s e^{-\pi / 2}}{s^{2}+16}$$

Step-by-Step Solution

Answer

The convolution integral representing the inverse Laplace transform of \(F(s)G(s)\) is given by: $$ L^{-1}[F(s)G(s)](t) = \int_0^t e^{-4\tau}\cos(3\tau)e^{-\pi (t-\tau)/2}\sin(4(t-\tau))d\tau $$

Step 1: Identify Inverse Laplace Transforms

To express the inverse Laplace transform of F(s)G(s), we need to find the inverse Laplace transforms of both F(s) and G(s) separately: $$ F(s) = \frac{s+4}{s^2 + 8s + 25} \quad \text{and} \quad G(s) = \frac{s e^{-\pi s/2}}{s^2 + 16} $$ We note that F(s) can be rewritten as the sum of two terms to facilitate the inverse Laplace transform: $$ F(s) = \frac{s+4}{(s+4)^2 + 3^2} = \frac{s+4}{s^2 + 8s + 25} = \frac{(s+4)-4+4}{(s+4)^2 + 3^2} = \frac{(s+4)+0}{(s+4)^2 + 3^2} = \frac{s+4}{(s+4)^2 + 3^2} $$ Let's find the inverse Laplace of F(s) and G(s) individually.

Step 2: Inverse Laplace Transform of F(s)

The inverse Laplace transform of F(s) resembles the inverse Laplace transform of a standard function: $$ L^{-1}\left[\frac{A}{(s-a)^{2}+b^{2}}\right] = e^{at}\cos(bt) $$ Comparing the standard function with our F(s), we see that \(A = 1\), \(a=-4\), and \(b=3\). Therefore, the inverse Laplace transform of F(s) is given by: $$ f(t) = L^{-1}[F(s)](t) = e^{-4t}\cos(3t) $$

Step 3: Inverse Laplace Transform of G(s)

The inverse Laplace transform of G(s) can be computed directly using the properties of the convolution integral. Note that G(s) is the Laplace transform of \(g(t)e^{-\pi t/2}\), where g(t) is the inverse Laplace transform of: $$ \frac{s}{s^2 + 16} = \frac{s}{(0s)^2 + 4^2} $$ Following the same process as Step 2 and using the standard function: $$ L^{-1}\left[\frac{As}{s^{2}+b^{2}}\right] = e^{at}\sin(bt) $$ Comparing the components: \(A=1-a=0\), \(b=4\). Hence, the inverse Laplace transform is given by: $$ g(t) = L^{-1}\left[\frac{s e^{-\pi / 2}}{s^{2}+16}\right](t) = e^{-\pi t/2}\sin(4t) $$

Step 4: Compute the Convolution Integral

Now that we have the inverse Laplace transforms of F(s) and G(s), we can write the convolution integral as: $$ L^{-1}[F(s)G(s)](t) = (f*g)(t) = \int_0^t f(\tau)g(t-\tau)d\tau $$ Substituting our previously obtained values for f(τ) and g(t-τ), we get: $$ L^{-1}[F(s)G(s)](t) = \int_0^t e^{-4\tau}\cos(3\tau)e^{-\pi (t-\tau)/2}\sin(4(t-\tau))d\tau $$ This is the convolution integral that represents the inverse Laplace transform of F(s)G(s).

Key Concepts

Laplace Transform

The Laplace Transform is a powerful tool in mathematics and engineering for transforming a time-domain function into a complex frequency-domain representation. It takes a function of time, often denoted as \( f(t) \), and converts it into a function of a complex variable \( s \), denoted as \( F(s) \). This transformation is extremely useful for solving differential equations, analyzing linear time-invariant systems, and performing operational calculus.

The Laplace Transform is defined by the integral:

• \( F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \)

This integral transforms the original function from the time domain to the s-domain, smoothing out and highlighting different aspects of its behavior and making differential equations easier to handle. For example, step functions turn into algebraic equations when transformed, simplifying the processes of solving complex differential equations.

Inverse Laplace Transform

The Inverse Laplace Transform is the process of converting back from the frequency domain to the time domain. It is used to find the original function \( f(t) \) given its Laplace transform \( F(s) \). Mathematically, it is represented as \( L^{-1}[F(s)](t) = f(t) \).

Finding the Inverse Laplace Transform can be done using tables of standard transforms or techniques such as partial fraction decomposition, complex contour integration (for more advanced applications), and looking for patterns in known transforms.

In practical applications, the Inverse Laplace Transform is crucial for interpreting results in the time domain after problems have been solved in the frequency domain using Laplace Transforms.

Differential Equations

Differential equations are equations that involve the rates of change of quantities and are widely used to model various natural phenomena. They appear in fields such as physics, biology, economics, and engineering. A differential equation relates a function to its derivatives.

There are various types of differential equations, including ordinary differential equations (ODEs) and partial differential equations (PDEs), each with its own methods for finding solutions.

Using the Laplace Transform to solve differential equations is a common technique. The transform converts a differential equation in the time domain into an algebraic equation in the s-domain, which is typically simpler to solve. Once solved, the Inverse Laplace Transform is used to convert back to the time domain, providing the solution to the original differential problem.

Convolution Theorem

The Convolution Theorem is a central concept in the context of Laplace Transforms and is particularly useful for solving complex problems involving differential equations. It states that under suitable conditions, the Laplace transform of the convolution of two functions is the product of their individual Laplace transforms.

In mathematical terms, if \( f(t) \) and \( g(t) \) are two functions, their convolution is defined as:

• \((f * g)(t) = \int_0^t f(\tau) g(t - \tau) \, d\tau\)

The Convolution Theorem then provides a way to handle the inverse transform of products of Laplace transforms, allowing for efficient circular solutions to complex systems. It simplifies the inverse operation by connecting integral transformations and encourages a more straightforward approach to system analysis. This theorem is particularly helpful because it reduces the computation from a complex product back to an integral, which is generally easier to solve.