# Quick Math Inverse Laplace Transform

## Quick Math Inverse Laplace Transform

1
##### Enter expression in this form

Example

Enter $$F(s) = \frac{s}{s^2 + 1}$$ as s/(s^2 + 1)

### Inverse Laplace Transform

The Inverse Laplace Transform is the process of finding the original time-domain function $$f(t)$$ from its Laplace transform $$F(s)$$. This operation is denoted as $$\mathcal{L}^{-1}\{F(s)\}$$. The inverse transform allows us to revert a function from the complex frequency domain back to the time domain, which is particularly useful for solving differential equations and analyzing systems in engineering and physics.

#### Definition

Given a function $$F(s)$$ which is the Laplace transform of $$f(t)$$, the inverse Laplace transform is defined as:
$\mathcal{L}^{-1}\{F(s)\} = f(t)$

#### Properties

– Linearity: $$\mathcal{L}^{-1}\{a F(s) + b G(s)\} = a \mathcal{L}^{-1}\{F(s)\} + b \mathcal{L}^{-1}\{G(s)\}$$
– First Shifting Theorem: If $$F(s) = \mathcal{L}\{f(t)\}$$, then $$\mathcal{L}^{-1}\{F(s-a)\} = e^{at} f(t)$$
– Second Shifting Theorem: If $$F(s) = \mathcal{L}\{f(t)\}$$, then $$\mathcal{L}^{-1}\{e^{-as} F(s)\} = u(t-a) f(t-a)$$, where $$u(t-a)$$ is the Heaviside step function.

#### Example

Example: Inverse Laplace Transform of $$\frac{s}{s^2 + 1}$$

Problem: Find the inverse Laplace transform of $$F(s) = \frac{s}{s^2 + 1}$$.

#### Solution:

1. Identify the form: Recognize that $$\frac{s}{s^2 + 1}$$ matches the standard Laplace transform of the derivative of the sine function.
2. Reference Standard Transform: The Laplace transform of $$\cos(t)$$ is $$\mathcal{L}\{\cos(t)\} = \frac{s}{s^2 + 1}$$.

Thus:
$\mathcal{L}^{-1}\left\{\frac{s}{s^2 + 1}\right\} = \cos(t)$

Answer: The inverse Laplace transform of $$\frac{s}{s^2 + 1}$$ is $$\cos(t)$$.

The inverse Laplace transform is a crucial tool for translating functions from the frequency domain back to the time domain. By utilizing standard Laplace transform pairs and properties, we can effectively find the original time-domain functions. In the given example, the inverse Laplace transform of $$\frac{s}{s^2 + 1}$$ was found to be $$\cos(t)$$, demonstrating the process of identifying and applying known transform pairs.