# How to Solve Laplace Transforms with a Fractional Term?

• MHB
• rannasquaer
In summary: Right. I meant the $\cosh$ and $\sinh$ versions.Also note that $\mathscr{L}^{-1} F(s-\alpha)=e^{\alpha t}f(t)$.So we can do:$\mathscr{L}^{-1} \frac{a(s+\lambda)}{(s+ \lambda)^2- \omega^2} + \mathscr{L}^{-1} \frac{b + \lambda a}{(s+ \lambda)^2- \omega^2} =a e^{-\lambda t}\sinh(\omega t) \cdot u(t rannasquaer How to solve the transforms below \[ \mathscr{L}^{-1} \frac{a(s+2 \lambda)+b}{(s+ \lambda)^2- \omega^2}$

rannasquaer said:
How to solve the transforms below

$\mathscr{L}^{-1} \frac{a(s+2 \lambda)+b}{(s+ \lambda)^2- \omega^2}$

The table of Laplace transforms lists that $\mathscr{L}^{-1} \frac{s+\alpha}{(s+\alpha)^2+\omega^2} = e^{-\alpha t}\cos(\omega t)\cdot u(t)$ and $\mathscr{L}^{-1} \frac{\omega}{(s+\alpha)^2+\omega^2} = e^{-\alpha t}\sin(\omega t)\cdot u(t)$.

Can we use those to find the requested transform?

Klaas van Aarsen said:
The table of Laplace transforms lists that $\mathscr{L}^{-1} \frac{s+\alpha}{(s+\alpha)^2+\omega^2} = e^{-\alpha t}\cos(\omega t)\cdot u(t)$ and $\mathscr{L}^{-1} \frac{\omega}{(s+\alpha)^2+\omega^2} = e^{-\alpha t}\sin(\omega t)\cdot u(t)$.

Can we use those to find the requested transform?

I think yes, if I rewrite like

$\mathscr{L}^{-1} \frac{a(s+\lambda)}{(s+ \lambda)^2- \omega^2} + \mathscr{L}^{-1} \frac{b + \lambda a}{(s+ \lambda)^2- \omega^2}$

but I have $(s+\lambda)^2-\omega^2$ and not $(s+\lambda)^2+\omega^2$

The table of Laplace transforms lists that $\mathscr{L}^{-1} \frac{\alpha}{s^2- \alpha^2} = \sin h(\alpha t).u(t)$ and $\mathscr{L}^{-1} \frac{s}{s^2- \alpha^2} = \cos h(\alpha t).u(t)$

I do not know what to do now

rannasquaer said:
I think yes, if I rewrite like

$\mathscr{L}^{-1} \frac{a(s+\lambda)}{(s+ \lambda)^2- \omega^2} + \mathscr{L}^{-1} \frac{b + \lambda a}{(s+ \lambda)^2- \omega^2}$

but I have $(s+\lambda)^2-\omega^2$ and not $(s+\lambda)^2+\omega^2$

The table of Laplace transforms lists that $\mathscr{L}^{-1} \frac{\alpha}{s^2- \alpha^2} = \sin h(\alpha t).u(t)$ and $\mathscr{L}^{-1} \frac{s}{s^2- \alpha^2} = \cos h(\alpha t).u(t)$

I do not know what to do now

Right. I meant the $\cosh$ and $\sinh$ versions.

Also note that $\mathscr{L}^{-1} F(s-\alpha)=e^{\alpha t}f(t)$.

So we can do:
$\mathscr{L}^{-1} \frac{a(s+\lambda)}{(s+ \lambda)^2- \omega^2} + \mathscr{L}^{-1} \frac{b + \lambda a}{(s+ \lambda)^2- \omega^2} =a e^{-\lambda t}\sinh(\omega t) \cdot u(t)+ \frac{b+\lambda a}{\omega}e^{-\lambda t}\cosh(\omega t) \cdot u(t)$
And if we want to, we can rewrite it using $\sinh x= \frac 12(e^x-e^{-x})$ and $\cosh x=\frac 12(e^x + e^{-x})$.

Great, I understood how to continue to do the math. Thank you!

## 1. What is the Laplace Transform?

The Laplace Transform is a mathematical tool used to convert a function of time into a function of complex frequency. It is often used in engineering and physics to solve differential equations and analyze systems.

## 2. How do you find the Laplace Transform?

The Laplace Transform can be found by taking the integral of a function multiplied by a decaying exponential. This integral can be solved using tables, formulas, or by using software such as MATLAB.

## 3. What are the applications of the Laplace Transform?

The Laplace Transform has many applications, including solving differential equations, analyzing systems in control theory, and studying signals and systems in electrical engineering. It is also used in the field of image processing and in the study of fluid dynamics.

## 4. What are the advantages of using the Laplace Transform?

The Laplace Transform has many advantages, including simplifying complex differential equations, allowing for easy analysis of systems, and providing a way to solve problems in the frequency domain. It also has applications in circuit analysis and signal processing.

## 5. Are there any limitations to using the Laplace Transform?

While the Laplace Transform has many useful applications, it does have some limitations. It can only be applied to functions that are defined for all positive time values, and it is not suitable for solving certain types of differential equations, such as those with discontinuous or multivariable solutions.

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