Evaluating the Laplace Transform of sin3(t)

Introduction

Evaluating the Laplace transform of mathematical functions is a fundamental aspect of signal processing and engineering. The Laplace transform is a powerful tool used to convert a time-domain signal into a frequency-domain representation, making analysis and problem-solving much more manageable. In this article, we will explore how to evaluate the Laplace transform of the function sin3(t). We will follow a detailed, step-by-step process that leverages trigonometric identities and the linearity of the Laplace transform.

Overview of the Laplace Transform

The Laplace transform of a function f(t), denoted as L{f(t)}, is defined as the integral:

L{f(t)} int_0^{infty} e^{-st} f(t) , dt

In this article, we will apply this definition to evaluate the Laplace transform of sin3(t).

Utilizing Trigonometric Identities

To improve the computational process, we can simplify the function using the following trigonometric identity:

sin^3(t) frac{3sin(t) - sin(3t)}{4}

Applying the Laplace Transform

Given the function sin^3(t), we can find its Laplace transform using the linearity property of the Laplace transform:

L{sin^3(t)} L{frac{3sin(t) - sin(3t)}{4}} frac{1}{4} left[3L{sin(t)} - L{sin(3t)}right]

Next, we need to recall the known Laplace transform formulas for sin(at), where a is a constant:

L{sin(at)} frac{a}{s^2 a^2}

Substituting Known Values

For sin(t) (where a 1):L{sin(t)} frac{1}{s^2 1}

For sin(3t) (where a 3):L{sin(3t)} frac{3}{s^2 9}

Substituting these values into our expression, we get:

L{sin^3(t)} frac{1}{4} left[3 cdot frac{1}{s^2 1} - frac{3}{s^2 9}right] frac{3}{4} left[frac{1}{s^2 1} - frac{1}{s^2 9}right]

Factoring out the 3 and finding a common denominator, we achieve:

frac{3}{4} left[frac{s^2 9 - (s^2 1)}{(s^2 1)(s^2 9)}right] frac{3}{4} left[frac{8}{(s^2 1)(s^2 9)}right] frac{6}{(s^2 1)(s^2 9)}

Thus, the Laplace transform of sin3(t) is:

mathcal{L}{sin^3(t)} frac{6}{s^2 1(s^2 9)}

Alternative Method Using Complex Substitution

A different method to find the Laplace transform of sin3(t) involves complex substitution. By definition, we have:

mathcal{L}_t[sin^3(t)](s) int_0^{infty} sin^3(t) e^{-st} dt int_0^{infty} left(frac{e^{it} - e^{-it}}{2i}right)^3 e^{-st} dt frac{1}{8i^3} int_0^{infty} e^{2it} e^{-2it} - 2e^{it} - e^{-it} e^{-st} dt

Further simplification leads to:

frac{1}{8i^3} int_0^{infty} e^{3it} - e^{-3it} - 3e^{it} e^{-st} - 3e^{-it} e^{-st} dt frac{1}{8i^3} left[-frac{e^{-s-3it}}{s-3i} - frac{e^{-s3it}}{s3i} - frac{3e^{-s-it}}{s-i} - frac{3e^{-sit}}{si} right]_0^{infty}

Finally, applying the limits and simplifying, we obtain:

frac{1}{8i^3} left[frac{-48i}{s^29 - s^21}right] frac{6}{s^29 - s^21}

This confirms the final result using complex substitution:

mathcal{L}{sin^3(t)} frac{6}{(s^2 1)(s^2 9)}

Conclusion

The Laplace transform of sin3(t) can be evaluated using either trigonometric identities or complex substitution. Both methods lead to the same result, providing a robust approach to understanding and solving problems involving Laplace transforms. This method is particularly useful in various fields of engineering and mathematics, including signal processing, control systems, and differential equations.