Laplace Transform: Find $$z(t)$$

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Homework Help Overview

The discussion revolves around finding the Laplace transform of the function $$z(t)=e^{-6t}\sin(\omega_{1}t)+e^{4t}\cos(\omega_{2}t)$$ for $$t \geq 0$$. Participants are exploring the application of Laplace transform properties and questioning the correctness of their approaches.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the Laplace transform to the sine function and questions whether their manipulation of the transform is valid. They express uncertainty about the equality of two expressions involving the Laplace transform.

Discussion Status

Some participants provide feedback on the original poster's approach, suggesting a clarification on the equality of the expressions. There is an ongoing exploration of the correct representation of the Laplace transform for the given function.

Contextual Notes

Participants are navigating potential confusion regarding the properties of Laplace transforms and the rendering of mathematical expressions in the forum's format.

iRaid
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Homework Statement


Wondering if I did this correctly..

Find the laplace transform:
$$z(t)=e^{-6t}sin(\omega_{1}t)+e^{4t}cos(\omega_{2}t)$$ for ##t\geq 0##

Homework Equations


The Attempt at a Solution



For the first part, I assume I can do this, but I'm not too sure. This is my main question, am I allowed to do this?
$$\mathcal{L}(sin(\omega_{1}t))=F(s+6)$$
Which gives me:
$$\frac{\omega _{1}}{s^{2}+\omega _{1}^{2}}=\frac{\omega _{1}}{(s-4)^{2}+\omega _{1}^{2}}$$

I figure since:
$$F(s+a)=\int_{0}^{\infty}f(t)e^{-(s+a)t}dt$$
I can do the above?Sorry if this question is stupid, I haven't done laplace transforms in a long time.
 
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iRaid said:

Homework Statement


Wondering if I did this correctly..

Find the laplace transform:
$$z(t)=e^{-6t}sin(\omega_{1}t)+e^{4t}cos(\omega_{2}t)$$ for ##t\geq 0##

Homework Equations


The Attempt at a Solution



For the first part, I assume I can do this, but I'm not too sure. This is my main question, am I allowed to do this?
$$\mathcal{L}(sin(\omega_{1}t))=F(s+6)$$
Which gives me:
$$\frac{\omega _{1}}{s^{2}+\omega _{1}^{2}}=\frac{\omega _{1}}{(s-4)^{2}+\omega _{1}^{2}}$$

I figure since:
$$F(s+a)=\int_{0}^{\infty}f(t)e^{-(s+a)t}dt$$
I can do the above?

Yes. But I wouldn't write ##\frac{\omega _{1}}{s^{2}+\omega _{1}^{2}}=\frac{\omega _{1}}{(s-4)^{2}+\omega _{1}^{2}}## because those aren't equal. Instead write
$$\mathcal{L}(e^{4t}\sin(\omega_1t))=\frac{\omega _1}{(s-4)^2+\omega _1^2}$$
[Edit] Not sure why that won't render. Or why it is boldface.
 
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LCKurtz said:
Yes. But I wouldn't write ##\frac{\omega _{1}}{s^{2}+\omega _{1}^{2}}=\frac{\omega _{1}}{(s-4)^{2}+\omega _{1}^{2}}## because those aren't equal. Instead write
$$\mathcal{L}(e^{4t}\sin(\omega_1t))=\frac{\omega _1}{(s-4)^2+\omega _1^2}$$
[Edit] Not sure why that won't render. Or why it is boldface.

Try editing using the BBCode editor instead of the default rich text editor - there should be a little icon at the top right that let's you switch to that mode, and you'll see some bogus boldface tags mixed up in your latex.

It is, aside from being in boldface, rendering properly. You just have to refresh the page to get it to render.
 
LCKurtz said:
Yes. But I wouldn't write ##\frac{\omega _{1}}{s^{2}+\omega _{1}^{2}}=\frac{\omega _{1}}{(s-4)^{2}+\omega _{1}^{2}}## because those aren't equal. Instead write
$$\mathcal{L}(e^{4t}\sin(\omega_1t))=\frac{\omega _1}{(s-4)^2+\omega _1^2}$$
[Edit] Not sure why that won't render. Or why it is boldface.

Right, I didn't mean to do that. That's all I needed.

Thanks!
 

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