Laplace Transform: Understanding the Inverse Transformation

The question asks us to find the function \(f(t)\) given its Laplace transform, \(L(s) = \frac{1}{{\left( {{s^2} + {\omega ^2}} \right)}}\). This process is known as finding the inverse Laplace transform.

Inverse Laplace Transform: Key Concepts

The Laplace transform is a powerful mathematical tool used to convert functions from the time domain (t-domain) to the complex frequency domain (s-domain). Conversely, the inverse Laplace transform converts functions from the s-domain back to the t-domain. To solve this problem, we need to recall some standard Laplace transform pairs.

• One of the most common Laplace transform pairs involves sinusoidal functions. We know that the Laplace transform of \(\sin(\omega t)\) is given by:
• $$L\left\{ {\sin \left( {\omega t} \right)} \right\} = \frac{\omega }{{{s^2} + {\omega ^2}}}$$
• Similarly, for \(\cos(\omega t)\):
• $$L\left\{ {\cos \left( {\omega t} \right)} \right\} = \frac{s}{{{{s^2} + {\omega ^2}}}}$$

Function f(t): Step-by-Step Derivation

We are given the Laplace transform of the function \(f(t)\) as:

$$L(s) = \frac{1}{{\left( {{s^2} + {\omega ^2}} \right)}}$$

Our goal is to find \(f(t) = L^{-1}\left\{L(s)\right\}\). Let's compare the given \(L(s)\) with the standard Laplace transform of \(\sin(\omega t)\). The standard form requires an \(\omega\) in the numerator.

To match the standard form \( \frac{\omega }{{{s^2} + {\omega ^2}}} \), we can multiply and divide the given expression by \(\omega\):

$$L(s) = \frac{1}{\omega } \cdot \frac{\omega }{{\left( {{s^2} + {\omega ^2}} \right)}}$$

Now, we can apply the inverse Laplace transform. Due to the linearity property of the Laplace transform (and its inverse), a constant multiplier can be taken outside the inverse Laplace transform operator:

$$f(t) = {L^{ - 1}}\left\{ {\frac{1}{\omega } \cdot \frac{\omega }{{\left( {{s^2} + {\omega ^2}} \right)}}} \right\}$$

$$f(t) = \frac{1}{\omega }{L^{ - 1}}\left\{ {\frac{\omega }{{\left( {{s^2} + {\omega ^2}} \right)}}} \right\}$$

From our knowledge of standard Laplace transform pairs, we know that \( {L^{ - 1}}\left\{ {\frac{\omega }{{\left( {{s^2} + {\omega ^2}} \right)}}} \right\} = \sin \left( {\omega t} \right) \).

Substituting this back into the equation for \(f(t)\):

$$f(t) = \frac{1}{\omega }\sin \left( {\omega t} \right)$$

Comparing with Options

Let's check this result against the provided options:

1. Option 1: \(f\left( t \right) = \frac{1}{{{\omega ^2}}}\left( {1 - \cos \omega t} \right)\)
2. Option 2: \(f\left( t \right) = \frac{1}{\omega }\cos \omega t\)
3. Option 3: \(f\left( t \right) = \frac{1}{\omega }\sin \omega t\;\)
4. Option 4: \(f\left( t \right) = \frac{1}{{{\omega ^2}}}\left( {1 - \sin \omega t} \right)\)

The derived function \(f(t) = \frac{1}{\omega }\sin \left( {\omega t} \right)\) perfectly matches Option 3.

Summary of Inverse Laplace Transform

To find the inverse Laplace transform of \(L(s) = \frac{1}{{\left( {{s^2} + {\omega ^2}} \right)}}\), we identified the standard Laplace transform pair for \(\sin(\omega t)\). By adjusting the constant multiplier, we successfully transformed the function from the s-domain back to the t-domain, yielding \(f(t) = \frac{1}{\omega }\sin \omega t\).