Question

Find the inverse Laplace transform of the given function by using the convolution theorem. \(F(s)=\frac{G(s)}{s^{2}+1}\)

Short answer

Answer: The inverse Laplace transform of the given function is \(f(t) = \int_0^t g(\tau) \cdot \sin(t - \tau) \,d\tau\), where g(t) is the inverse Laplace transform of \(G(s)\).

Step-by-step solution

Express the function in a suitable form for the convolution theorem

First, let's rewrite the given function \(F(s) = \frac{G(s)}{s^2+1}\) as a product of two functions \(F(s) = G(s) \cdot H(s)\), where \(H(s) = \frac{1}{s^2+1}\). This form is suitable for the convolution theorem.

Find the inverse Laplace transform of each individual function

Now, we need to find the inverse Laplace transforms of the functions \(G(s)\) and \(H(s)\). We are not given the form of \(G(s)\), so we will denote its inverse Laplace transform as \(g(t)\). For \(H(s)\), we recognize it as the Laplace transform of \(\sin(t)\), so: $$\mathcal{L}^{-1}\{H(s)\} = \sin(t)$$

Apply the convolution theorem

The convolution theorem states that if \(F(s) = G(s) \cdot H(s)\), then the inverse Laplace transform of \(F(s)\) is obtained by convoluting the inverse Laplace transforms of \(G(s)\) and \(H(s)\). In this case: $$f(t) = g(t) * \sin(t)$$ The convolution of two functions is defined as: $$f(t) = \int_0^t g(\tau) \cdot \sin(t - \tau) \,d\tau$$ So, the inverse Laplace transform of the given function is: $$f(t) = \int_0^t g(\tau) \cdot \sin(t - \tau) \,d\tau$$

Key concepts

Convolution Theorem

The convolution theorem is pivotal in the study of Laplace transforms. It provides a method to find the inverse Laplace transform of a product of two functions. Suppose you have two functions, say \(G(s)\) and \(H(s)\). Instead of finding their inverse Laplace transforms separately and multiplying them, you can use the convolution theorem. Specifically, the inverse Laplace transform of the product \(G(s) \cdot H(s)\) is the convolution of the inverse transforms of \(G(s)\) and \(H(s)\).

Essentially, convolution in this context is integrating the product of the two inverse transforms over a variable. The formula for convolution is given by:

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

Here, \(g(t)\) and \(h(t)\) are the inverse Laplace transforms of \(G(s)\) and \(H(s)\), respectively. This approach simplifies complex products, allowing direct computation of their time-domain counterparts. It is especially useful when dealing with functions expressed as rational expressions in the Laplace domain.

Laplace Transform

The Laplace transform is a powerful tool used in mathematics and engineering to transform a function from the time domain to the complex frequency domain. This transformation is exceptionally beneficial in analyzing systems, solving differential equations, and simplifying complex mathematical models.

The process works by converting a time-dependent function, \( f(t) \), into a complex function, \( F(s) \). This is represented by the integral:

• \( \mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) \, dt \)

The variable \(s\) often represents a complex number and encompasses both the damping aspects (the real part) and the oscillatory aspects (the imaginary part) of a function.

The Laplace transform can simplify the process of handling linear time-invariant systems. By transferring the function into the \(s\)-domain, complex differential equations become more manageable algebraic equations, making it easier to apply techniques like convolution.

Sin Function

The sine function, represented as \( \sin(t) \), is a fundamental building block in trigonometry and arises naturally when dealing with periodic phenomena. Its importance is magnified in the world of Laplace transforms, especially due to its simple representation in the \(s\)-domain.

In terms of Laplace transforms, the function \( \sin(t) \) can be represented by the transform:

• \( \mathcal{L}\{\sin(t)\} = \frac{1}{s^2 + 1} \)

This transformation shows that the sine function corresponds to a simple rational expression in the \(s\)-domain, making it an integral part of solving differential equations or analyzing systems subject to sinusoidal inputs.

Understanding the relationship between functions like \(\sin(t)\) and their respective Laplace transforms is crucial when employing methods such as the convolution theorem to solve equations and retrieve time-domain characteristics of a system.