Question

Recall that cos bt $=(e^{ibt}+e^{tbt})/2$ and $\sin bt=(e^{jbt}-$ $e^{-tbt})/2i.$ Assuming that the necessary elementary integration formulas extend to this case, find the Laplace transform of the given function; $a$ and $b$ are real constants. $$\sin bt$$

Short answer

Answer: The Laplace transform of the function $\sin bt$ is $\frac{2s}{s^2+b^2}$.

Step-by-step solution

Convert sine function using given property

We are given that $\sin bt=\frac{e^{jbt}-e^{-jbt}}{2i}$. So our function becomes: $$f(t)=\frac{e^{jbt}-e^{-jbt}}{2i}$$

Applying the Laplace transform formula

Using the Laplace transform definition, we need to evaluate the integral: $$F(s)=\mathcal{L}\{f(t)\}={\int }_0^{\mathrm{\infty }}{e}^{-st}\frac{e^{jbt}-e^{-jbt}}{2i}dt$$

Separate the integral into two integrals

We can separate the integral into two parts, one for each term inside the brackets: $$F(s)=\frac{1}{2i}[{\int }_0^{\mathrm{\infty }}{e}^{-st}{e}^{jbt}dt-{\int }_0^{\mathrm{\infty }}{e}^{-st}{e}^{-jbt}dt]$$

Combine exponential terms in each integral

In both integrals, we can combine the exponential terms to simplify calculations: $$F(s)=\frac{1}{2i}[{\int }_0^{\mathrm{\infty }}{e}^{(jbt-st)}dt-{\int }_0^{\mathrm{\infty }}{e}^{(-jbt-st)}dt]$$

Integrate both terms

Now, we integrate each term with respect to $t$: $$F(s)=\frac{1}{2i}[\frac{e^{(jbt-st)}}{(jb-s)}{|}_0^{\mathrm{\infty }}-\frac{e^{(-jbt-st)}}{(-jb-s)}{|}_0^{\mathrm{\infty }}]$$

Evaluate the limits of integration

Next, we evaluate the limits for both terms: $$F(s)=\frac{1}{2i}[(\frac{e^{(jbt-st)}}{(jb-s)}{|}_{\mathrm{\infty }}-\frac{e^{(jbt-st)}}{(jb-s)}{|}_0)-(\frac{e^{(-jbt-st)}}{(-jb-s)}{|}_{\mathrm{\infty }}-\frac{e^{(-jbt-st)}}{(-jb-s)}{|}_0)]$$ Since $e^{-st}$ approaches 0 as $t\to \mathrm{\infty }$, we have: $$F(s)=\frac{1}{2i}[(-\frac{1}{(jb-s)})-(-\frac{1}{(-jb-s)})]$$

Simplify and find the Laplace transform of the given function

After simplification and removing the constant $\frac{1}{2i}$ from both terms, we arrive at the result: $$F(s)=\frac{1}{(s-jb)}+\frac{1}{(s+jb)}=\frac{2s}{s^2+b^2}$$ Thus, the Laplace transform of the given function $\sin bt$ is: $$\mathcal{L}\{\sin bt\}=\frac{2s}{s^2+b^2}$$

Key concepts

Laplace Transform

Understanding the Laplace transform is essential for solving complex problems in engineering and physics. The Laplace transform is a powerful mathematical tool used to switch a function from the time domain to the s-domain (complex frequency domain). This transformation turns differential equations into algebraic ones, which are generally easier to solve.

The basic definition of the Laplace transform of a function f(t), where t represents time, is given by the integral: $$\mathcal{L}\{f(t)\}={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)dt,$$where s is a complex number, and e^{-st} is the kernel of the transform. When applied to the sine function, the transform helps in solving differential equations involving the sine function without directly integrating the function over time.

Complex Exponential Representation

The complex exponential representation is an elegant way of expressing trigonometric functions like sine and cosine. Euler's formula states that $$e^{i\theta }=\cos (\theta )+i\sin (\theta ),$$and this is pivotal for understanding the behavior of sinusoidal functions in the complex plane.

In the context of the Laplace transform, we use this representation to convert the sine function into a form that is more conducive for the transformation process. The sine function can be expressed as $$\sin (\theta )=\frac{e^{i\theta }-e^{-i\theta }}{2i}.$$This approach not only simplifies the integration process but also provides deeper insight into the nature of the function in the frequency domain.

Integration Techniques in Laplace Transform

Integration techniques are critical when performing the Laplace transform. The ability to skillfully manipulate integrals dictates the ease of transforming functions. In the case of the Laplace transform of the sine function, integration involves complex exponentials, requiring you to integrate terms like $$e^{(jbt-st)}$$and $$e^{(-jbt-st)}.$$Utilizing algebraic properties and breaking the integral into smaller, more manageable parts simplifies the process. After separating the exponentials, you combine like terms before integrating. The result is a transformed function that reflects the behavior of the original function in the s-domain. Understanding these techniques ensures accurate transformation and simplifies the process involved.

Elementary Differential Equations

Elementary differential equations form the bedrock of dynamic systems and describe how physical quantities change over time. Solutions to these equations reveal the system's behavior and are often essential in the fields of science and engineering. The Laplace transform is a pivotal tool used to solve linear differential equations with constant coefficients by transforming them into algebraic equations.

By applying the Laplace transform to both sides of a differential equation, you can eliminate the derivatives and obtain an equation in the Laplace domain. After finding the solution in this domain, you can then apply the inverse Laplace transform to revert back to the time domain. In the exercise at hand, the process starts by converting the sine function to its exponential form and then, integrating to find the Laplace transform helps in solving an equation that would otherwise involve more complex integration steps directly in the time domain.