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

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SUMMARY

The discussion focuses on finding the Laplace transform of the function $$z(t)=e^{-6t}\sin(\omega_{1}t)+e^{4t}\cos(\omega_{2}t)$$ for $$t \geq 0$$. The correct approach involves using the property of Laplace transforms for exponential functions, specifically $$\mathcal{L}(e^{at}f(t))=\mathcal{L}(f(t))(s-a)$$. The participants clarify that $$\mathcal{L}(e^{4t}\sin(\omega_1t))$$ should be expressed as $$\frac{\omega _1}{(s-4)^2+\omega _1^2}$$ instead of incorrectly equating it to $$\frac{\omega _1}{s^{2}+\omega _1^{2}}$$. The discussion emphasizes the importance of proper notation and understanding the transformation properties.

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  • Familiarity with exponential functions and trigonometric functions
  • Knowledge of the notation used in calculus and differential equations
  • Ability to manipulate algebraic expressions involving complex variables
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  • Study the properties of Laplace transforms, particularly the shifting theorem
  • Learn about the Laplace transform of common functions, including sine and cosine
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Students studying differential equations, engineers working with control systems, and mathematicians interested in transform techniques will benefit from this discussion.

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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