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

In summary, you are allowed to do the laplace transform, but you should not write ##\frac{\omega _{1}}{s^{2}+\omega _{1}^{2}}=\frac{\omega _{1}}{(s-4)^{2}+\omega _{1}^{2}}## because those are not equal.
  • #1
iRaid
559
8

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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  • #2
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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  • #3
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.
 
  • #4
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!
 

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

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 commonly used in engineering and physics to solve differential equations.

2. How do you find the Laplace Transform of a function?

To find the Laplace Transform of a function, you need to use the integral formula:
F(s) = ∫0 f(t)e-st dt
where F(s) is the transformed function and f(t) is the original function.

3. What is the significance of the Laplace Transform?

The Laplace Transform is important because it allows us to solve differential equations, which are used to model many real-world phenomena. It also simplifies calculations by converting a differential equation into a simple algebraic equation.

4. Can the Laplace Transform be used for all functions?

No, the Laplace Transform can only be used for functions that satisfy certain conditions, such as being continuous and having finite limits as t approaches infinity. It also cannot be used for functions with discontinuities or infinite discontinuities.

5. How is the inverse Laplace Transform calculated?

The inverse Laplace Transform is calculated using a table of Laplace Transform pairs or by using partial fraction decomposition and the formula:
f(t) = (1/2πi)∫c-i∞c+i∞ F(s)est ds
where F(s) is the transformed function and c is any complex number to the right of all singularities of F(s).

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