# Laplace transform

## 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##

## 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.

## Answers and Replies

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LCKurtz
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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##

## 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
$$\mathcal{L}(e^{4t}\sin(\omega_1t))=\frac{\omega _1}{(s-4)^2+\omega _1^2}$$
 Not sure why that won't render. Or why it is boldface.

Last edited by a moderator:
iRaid
Nugatory
Mentor
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
$$\mathcal{L}(e^{4t}\sin(\omega_1t))=\frac{\omega _1}{(s-4)^2+\omega _1^2}$$
 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 lets 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.

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
$$\mathcal{L}(e^{4t}\sin(\omega_1t))=\frac{\omega _1}{(s-4)^2+\omega _1^2}$$
 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!