Laplace transform ( "find x(t)" though ? )

[Mark44]

Given property: e^-at (in t domain) = 1/(s+a) (in s-domain).

NOT given: ke^-at (in t domain) = k/(s+a) (in s-domain), where k is a constant.

First off, ke^-at ≠ k/(s + a).
You're completely ignoring the idea that you're taking a Laplace transform here.

As a correct mathematical equation, you should say this:
$\mathcal{L} [e^{at}] = \frac{1}{s - a}$
To see the LaTeX I used, right click on the equation above, and "Show math as" ... "TeX commands."

L¹[e^-at] = 1/(s+a)

Closer, but you shouldn't have the exponent.

f(t) = e^-2t

k = 1, a = -2, --> f(s) = 1/(s - 2)

Put everything together in a coherent thought.
$\mathcal{L}[e^{-2t}] = \frac{1}{s - (-2)} = \frac{1}{s + 2}$
On the right is F(s), the Laplace transform of f(t). Note the difference in capitalization.

Given property: sin(bt) (in t domain) = b/(s²+b²) (in s-domain).

Again, these aren't equal. You're ignoring the fact that this time you're taking the inverse Laplace transform.

NOT given: ksin(bt) (in t domain) = kb/(s²+b²) (in s-domain)., where k is a constant.

f(s) = 8/(s² + 4);

a = 4, k = (1/2)

---> f(t) = (1/2)sin(4t)

No.
What you want is more like this:
$\mathcal{L}^{-1}[\frac{8}{s^2 + 4}] = \mathcal{L}^{-1}[\frac{4 * 2}{s^2 + 2^2}] = 4 * \mathcal{L}^{-1}[\frac{2}{s^2 + 2^2}] = 4sin(2t)$
The last expression is f(t).

After the 2nd equals, I had things set up in the form a/(s² + a²). Notice that the $\mathcal{L}^{-1}$ symbol stayed around for a while, until I was ready to actually take the inverse LP transform.

I also used the property that I mentioned a few posts ago about the linearity of the transform and inverse.

You couldn't just start off with the entire numerator first? I was thinking about it, so how should I start with that? Cya though, thanks.

 

[Color_of_Cyan]

Noted, thanks. And I forgot for that property the 4 was squared, ughh.

So the real property of it all is this?:

L[f(t)] = F(s)

L^-1 = f(t)

 

[Mark44]

Noted, thanks. And I forgot for that property the 4 was squared, ughh.

So the real property of it all is this?:

L[f(t)] = F(s)

L^-1 = f(t)

No, this is just notation that shows the relationship between a function and its Laplace transform, and vice versa. BTW, the 2nd line above is missing F(s). IOW, it should say
$\mathcal{L}^{-1}[F(s)] = f(t)$

The real property (that is, the definition) is this:
$\mathcal{L}[f(t)] = \int_0^{\infty}e^{-st}f(t)dt$
The limits of integration are s values, so the integral on the right is a function of s (i.e., F(s)).

 

[Color_of_Cyan]

No, this is just notation that shows the relationship between a function and its Laplace transform, and vice versa. BTW, the 2nd line above is missing F(s). IOW, it should say
$\mathcal{L}^{-1}[F(s)] = f(t)$

The real property (that is, the definition) is this:
$\mathcal{L}[f(t)] = \int_0^{\infty}e^{-st}f(t)dt$
The limits of integration are s values, so the integral on the right is a function of s (i.e., F(s)).

But it still correctly describes it though (with the table and all), right? Descriptions are more what I'm looking for, at least :)

So for the original numerator, 4e^-4s -3, I can just put it as (-3/4)sin(4t) for the '-3'. But not for the 'e^-4s', because of the 's'?

 

[LCKurtz]

So for the original numerator, 4e^-4s -3, I can just put it as (-3/4)sin(4t) for the '-3'.

Yes.

But not for the 'e^-4s', because of the 's'?

Correct. Look in your tables for$$
\mathcal L(f(t-a)u(t-a))$$where $u$ is the unit step function and see if you can use that.

 

[Color_of_Cyan]

Yes.
Correct. Look in your tables for$$
\mathcal L(f(t-a)u(t-a))$$where $u$ is the unit step function and see if you can use that.

Yes.
Correct. Look in your tables for$$
\mathcal L(f(t-a)u(t-a))$$where $u$ is the unit step function and see if you can use that.

Okay. There's one that says this:

L[f(t - t₀)u(t - t₀)] = e^-t₀sF(s)

so is 4e^-4s just this?:

4(t - 4)u(t - 4) ?

 

[LCKurtz]

Okay. There's one that says this:

L[f(t - t₀)u(t - t₀)] = e^-t₀sF(s)

so is 4e^-4s just this?:

4(t - 4)u(t - 4) ?

You don't have just $4e^{-4s}$. You have$$
\frac{4e^{-4s}}{s^2+6s+25}$$to inverse.

 

[Color_of_Cyan]

You don't have just $4e^{-4s}$. You have$$
\frac{4e^{-4s}}{s^2+6s+25}$$to inverse.

Do I use a different property for that then?

I'm thinking I might have to do use -dF(s)/ds = L[tf(t)]

If so, it would be this:

$$ \frac { 4e^{-4s}(2s + 6) + (16e^{-4s})(s^{2} + 6s + 25) } {(s^{2} + 6s + 25)^{2}} $$

 

[LCKurtz]

Apply the formula you found in post #76 to the expression in post #77. I'm not going to do it for you because it is the last step in working your problem. If you still can't get it after all these posts, it is time to sit down with your teacher or a tutor.

 

[Color_of_Cyan]

Trying to figure out the notation is too much :( Do you only use the property in post 76 for

$$ \frac {4e^{-4s}} {s^{2} + 6s + 25} $$ ? I think it is this though:

$$ (\frac {1} {4}) (e^{-3t})sin4(t - 4)u(t - 4) $$I think I got it now. Thank you for all the help (and extreme patience)