Question

Use the Convolution Theorem to compute the inverse transform of the following: a. \(F(s)=\frac{2}{s^{2}\left(s^{2}+1\right)} .\) b. \(F(s)=\frac{e^{-3 s}}{s^{2}}\) c. \(F(s)=\frac{1}{s\left(s^{2}+2 s+5\right)}\)

Short answer

The respective inverse Laplace transforms are: (a) \(2(t - sin(t))\), (b) \(\mathcal{u}(t-3)t\) where \(\mathcal{u}(t-3)t\) is the unit step function with shift 3, and (c) \( 1 + e^{-t} cos(2t) \)

Step-by-step solution

Part (a) - Inverse Laplace transform of \(F(s)=\frac{2}{s^{2}\left(s^{2}+1\right)}\)

Break down the division as a product of two inverse Laplace transforms: \(L^{-1}\{2/s^{2}\}*L^{-1}\{1/(s^{2}+1)\}\), which represent \(2t\) and \(sin(t)\) respectively.

Convolution of \(2t\) and \(sin(t)\)

Obtain the convolution of the two separate functions, given as \(\int_0^t 2\tau sin(t-\tau) d\tau\). By integrating this function, you get the result of \(2(t - sin(t))\).

Part (b) - Inverse Laplace transform of \(F(s)=\frac{e^{-3 s}}{s^{2}}\)

Start by using the time shift property and rewrite as \(L^{-1}\{e^{-3s}/ s^{2}\}\), which represents \(\mathcal{u}(t-3)t\).

Part (c) - Inverse Laplace transform of \(F(s)=\frac{1}{s\left(s^{2}+2 s+5\right)}\)

Start with partial fraction decomposition to get \(F(s) = \frac{1}{s} - \frac{s + 1}{s^{2}+2s+5}\), and then determine the inverse for each term separately. Inverse Laplace of \(1/s\) is \(1\) and the inverse Laplace transform of \( \frac{s+1}{s^{2}+2s+5} \) is \( e^{-t} cos(2t) \). Adding these inverse transforms, you get the final result.

Key concepts

Convolution Theorem

The Convolution Theorem is a mathematical principle which states that under certain conditions, the Laplace transform of a convolution of two functions is the product of their individual Laplace transforms. In simpler terms, if you have two time-domain functions whose behavior you understand, you can predict how they will interact with each other over time.

For example, if you have the inverse Laplace transform of two functions, say f(t) and g(t), and you are looking to find the inverse transform of their product F(s)G(s), the result is the convolution of f(t) and g(t). Mathematically, the convolution is represented by an integral:
\[(f * g)(t) = t_0^t f(\tau) g(t-\tau) d\tau]\]
When applied to solving differential equations, this theorem can simplify complex problems by breaking them into more manageable parts.

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used to simplify complex rational expressions, particularly when finding the inverse Laplace Transform. This method involves breaking down a fraction into a sum of simpler fractions, with the aim of making the inverse transformation more straightforward.

To apply this technique, you first express the complex fraction as a sum of simpler fractions, each with its own distinct denominator. These simpler fractions can then be inverted individually, which is vastly easier than inverting the original complex expression. This method is akin to decomposing a difficult task into smaller, more manageable pieces. A classic example is decomposing a fraction with a quadratic denominator, allowing for simple inverse Laplace transforms of linear terms.

Time Shift Property

The Time Shift Property of the Laplace Transform describes how a shift in the time domain affects the transform in the s-domain. If a function f(t) has a corresponding Laplace Transform F(s), then a shifted version of the function, f(t - a) where a is a constant, will have a Laplace Transform that includes a factor of e^{-as}.

Mathematically, it's shown as:
\[L\big\{u(t-a)f(t-a)\big\} = e^{-as}F(s)\]
Here, u(t - a) is the Heaviside step function that essentially 'turns on' the function f(t) at t = a. This property is particularly useful when solving differential equations that involve delayed or shifted inputs.

Laplace Transform

The Laplace Transform is a powerful integral transform used to make solving differential equations easier by converting functions of time into functions of a complex variable s. The transform takes a time-domain function, typically representing a physical system's output, to the s-domain where algebraic techniques can be applied.

Given a function f(t), its Laplace Transform F(s) is defined as:
\[F(s) = 0^{fty} e^{-st} f(t) dt\]
The inverse Laplace Transform, denoted as L^-1{F(s)}, enables you to revert back to the time-domain expression. It's widely used in engineering and physics, as it conveniently handles operations like differentiation and integration by converting them into algebraic operations.