Problem 497 - Advanced Calculus

Question

(a) Define the convolution of two functions \(\mathrm{f}(\mathrm{t})\) and \(\mathrm{g}(\mathrm{t})\). (b) State the convolution theorem for Laplace transforms. (c) Find the inverse Laplace transform $$ \mathrm{f}(\mathrm{t})=\mathrm{L}^{-1}\\{\mathrm{~F}(\mathrm{~S})\\}=\mathrm{L}^{-1}\left\\{1 /\left(\mathrm{s}^{2}+\mathrm{c}^{2}\right)^{2}\right\\} $$ \((\mathrm{c}=\) constant \() .\)

Step-by-Step Solution

Answer

The inverse Laplace transform of the given function \(F(s) = \frac{1}{(s^2 + c^2)^2}\) is \[f(t) = \frac{1}{c^2} \int_{0}^{\infty} \sin(c\tau) \sin(c(t - \tau)) d\tau\]

Step 1: Part (a) - Define convolution of two functions

The convolution of two functions \(f(t)\) and \(g(t)\), denoted as \((f*g)(t)\), is a function derived by "smearing" one function over the other and defined mathematically as the integral of the product of the functions evaluated at the difference of their arguments. The convolution is given by: \[(f*g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t - \tau) d\tau\]

Step 2: Part (b) - Convolution theorem for Laplace transforms

The convolution theorem for Laplace transforms states that the Laplace transform of the convolution of two functions in the time domain is equal to the product of the Laplace transforms of the individual functions in the frequency domain. Mathematically, this can be expressed as: \[\mathcal{L}\{(f*g)(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\}\] where \(\mathcal{L}\) denotes the Laplace transform operation.

Step 3: Part (c) - Find the inverse Laplace transform

We are given the function \(F(s) = \frac{1}{(s^2 + c^2)^2}\), and we need to find its inverse Laplace transform \(f(t) = \mathcal{L}^{-1}\{F(s)\}\). Noticing that \(F(s)\) can be expressed as the product of two Laplace transforms: \[\frac{1}{(s^2 + c^2)^2} = \frac{1}{s^2 + c^2} \cdot \frac{1}{s^2 + c^2}\] Now, apply the convolution theorem in reverse, which states that if \(F(s) = F_1(s) \cdot F_2(s)\), then the inverse Laplace transform is: \[\mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\{F_1(s)\} * \mathcal{L}^{-1}\{F_2(s)\}\] Since both \(F_1(s)\) and \(F_2(s)\) have the same form, we only need to find the inverse Laplace transform of one, multiply it by itself through convolution, and we'll have the inverse Laplace transform of the given function. We know that the inverse Laplace transform of \(\frac{1}{s^2 + c^2}\) is \(f_1(t) = \frac{\sin(ct)}{c}\) (refer to the Laplace transform table for this result). Therefore, we have: \[\mathcal{L}^{-1}\{F(s)\} = f_1(t) * f_1(t) = \left(\frac{\sin(ct)}{c}\right) * \left(\frac{\sin(ct)}{c}\right)\] Now, we can calculate the convolution integral: \fbox{\begin{minipage}{15em} \[\begin{aligned} f(t) &= \int_{-\infty}^{\infty} \frac{\sin(c\tau)}{c} \cdot \frac{\sin(c(t - \tau))}{c} d\tau \\&= \frac{1}{c^2} \int_{0}^{\infty} \sin(c\tau) \sin(c(t - \tau)) d\tau \end{aligned}\] \end{minipage}} _this_is_the_result_of_the_exercise_ Thus, we have found the inverse Laplace transform of the given function \(F(s)\), which is expressed as: \[f(t) = \frac{1}{c^2} \int_{0}^{\infty} \sin(c\tau) \sin(c(t - \tau)) d\tau\]

Key Concepts

Laplace Transforms

Laplace transforms are a powerful mathematical tool used in engineering and physics to transform complex time-domain functions into simpler s-domain representations. This transformation aids in solving differential equations and analyzing systems. The Laplace transform of a function \( f(t) \) is denoted by \( \mathcal{L}\{f(t)\} \) and defined as:
\[\mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt\]where \( s \) is a complex number. Some key points about the Laplace transform:

• It helps in converting differential equations into algebraic problems.
• This method is particularly useful for functions that are piecewise, periodic, or exponential in nature.
• By converting to the frequency domain, it simplifies the analysis of systems and allows us to apply easier mathematical operations.

Understanding Laplace transforms enables seamless transitions between time and frequency domains, proving essential for control system analysis and signal processing.

Convolution Theorem

The convolution theorem is a fundamental theorem that links the convolution of two functions in the time domain to the multiplication of their Laplace transforms in the frequency domain.
This connection can drastically simplify calculations. If \( f(t) \) and \( g(t) \) are two time-domain functions with Laplace transforms \( F(s) \) and \( G(s) \) respectively, then the convolution \( (f*g)(t) \) is given by
\[(f*g)(t) = \int_{0}^{t} f(\tau) g(t-\tau) \, d\tau\]
With the convolution theorem, we have:

• \( \mathcal{L}\{(f*g)(t)\} = F(s) \cdot G(s) \)

This theorem is particularly helpful because transforming a convolution in the time domain often results in just a simple product in the frequency domain.

• It eases solving integral equations and differential systems.
• By using the multiplication property, it simplifies finding inverse Laplace transforms as well.

Inverse Laplace Transform

The inverse Laplace transform is a process used to convert functions from the s-domain back to the time domain. This process is essential when one needs to analyze the behavior of time-dependent systems. For a given function \( F(s) \), the inverse Laplace transform is represented by \( \mathcal{L}^{-1}\{F(s)\} \).
The mathematical process involves complex contour integration, but in practice:

• It is often performed using transform tables or applying techniques like partial fraction decomposition.
• Understanding inverse transforms is essential for system response analysis.

Using the convolution theorem, as shown in the problem, can also simplify finding the inverse. For example, a product of two Laplace transforms can be reverted to time domain through convolution.

• For time functional analysis, inverse Laplace is crucial in solving differential equations.
• It enables engineers and scientists to predict how physical systems respond over time.