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

[Color_of_Cyan]

It's probably much simpler than you seem to be thinking.

If F(s) = $\frac{s}{s^2 + 1}$ then F(s + 2) = $\frac{s + 2}{(s + 2)^2 + 1}$
What you need to do is to recognize the basic underlying function F, and what the formulas look like with a variety of horizontal translations. I.e., the difference between F(s) and F(s + a), for example. A lot of the work in working with the tables is to put what you're starting with in a form that matches a row in your table.

So you don't need to worry about the L^-1 then, just as long as it's the same format?

From what Kurtz gave, would I now have 4e^-3tsin(4t)?

 

[LCKurtz]

Is it only because the 16 was from completing the square (and so it becomes part of the function)? And a is only what you have grouped inside with 's'? It was very confusing.

The only close property I had to the one you gave for that was this

$$ L[e^{-at}f(t)] = F(s + a) $$

Is it really the same thing then? This is important; I do not have any property given that exclusively says 'L^-1'

Every transform formula has a corresponding inverse transform, just like this one.

Moving on though. You get the result, but you still have ANOTHER function F(s) to transform? And it seems I still need more than the table (or maybe just more help understanding the notation of it which has been very alien to me); I can't just do: b/(s² + b²) ---> sin(bt) yet?

Can I say that L^-11/(s² + 16) transforms to 4sin(4t)?

You can answer that yourself. Is the transform of $4\sin(4t)$ equal to $\frac 1 {s^2+16}$?

 

[Color_of_Cyan]

So it's (1/4)(sin(4t)) then? I don't think I can be 100% sure of the way the transform notation (or algebra) works just yet.

 

[Mark44]

It's probably much simpler than you seem to be thinking.
If F(s) = $\frac{s}{s^2 + 1}$ then F(s + 2) = $\frac{s + 2}{(s + 2)^2 + 1}$
What you need to do is to recognize the basic underlying function F, and what the formulas look like with a variety of horizontal translations. I.e., the difference between F(s) and F(s + a), for example. A lot of the work in working with the tables is to put what you're starting with in a form that matches a row in your table.

So you don't need to worry about the L^-1 then, just as long as it's the same format?

I wasn't doing anything with Laplace transforms or inverses. I was trying to help you see the algebra of what's going on, apart from the business of taking a Laplace transform.

 

[Mark44]

So it's (1/4)(sin(4t)) then? I don't think I can be 100% sure of the way the transform notation (or algebra) works just yet.

That's pretty evident.

Both the Laplace transform and its inverse are linear in the sense that $\mathcal{L}[k f(t)] = k\mathcal{L}[f(t)]$, where k is a constant. In other words, you can bring constants in or out.

 

[Color_of_Cyan]

That's pretty evident.

Both the Laplace transform and its inverse are linear in the sense that $\mathcal{L}[k f(t)] = k\mathcal{L}[f(t)]$, where k is a constant. In other words, you can bring constants in or out.

What do you mean by 'that'? That I'm really bad at this or that it's really (1/4)sin(4t)? :(

So what am I to do with the (1/4)e^-3tsin(4t) now?

 

[Mark44]

So it's (1/4)(sin(4t)) then? I don't think I can be 100% sure of the way the transform notation (or algebra) works just yet.

Maybe some practice would help this stuff gel.

Exercise 1. Here's a function in the time domain: $f(t) = e^{-2t}$
Write a mathematical equation that shows the work for finding the Laplace transform of this function. I.e., find F(s).

Exercise 2. Here's a function in the s domain (AKA the frequency domain, I believe it's called): $F(s) = \frac{8}{s^2 + 4}$
Write a mathematical equation that shows the work for finding the inverse Laplace transform of this function. I.e., find f(t).

Be sure to include the symbols $\mathcal{L}$ and $\mathcal{L}^{-1}$ where appropriate.

 

[Mark44]

What do you mean by 'that'? That I'm really bad at this or that it's really (1/4)sin(4t)? :(

So what am I to do with the (1/4)e^-3tsin(4t) now?

It would help us if you wrote complete mathematical statements. Is what really (1/4)sin(4t)?

 

[LCKurtz]

So what am I to do with the (1/4)e^-3tsin(4t) now?

You started with the problem$$
\mathcal L^{-1}\frac{4e^{-4s}-3}{s^2+6s+25}$$You have finally got the inverse of the$$
\frac 1 {s^2+6s+25}$$Now you have to figure out how the $4e^{-4s}$ and the $-3$ affect the answer. This is getting confusing so I'm going to let Mark take it from here.

 

[Color_of_Cyan]

Maybe some practice would help this stuff gel.

Exercise 1. Here's a function in the time domain: $f(t) = e^{-2t}$
Write a mathematical equation that shows the work for finding the Laplace transform of this function. I.e., find F(s).

Be sure to include the symbols $\mathcal{L}$ and $\mathcal{L}^{-1}$ where appropriate.

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.

L¹[ke^-at] = 1/(s+a)

f(t) = e^-2t

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

Exercise 2. Here's a function in the s domain (AKA the frequency domain, I believe it's called): $F(s) = \frac{8}{s^2 + 4}$
Write a mathematical equation that shows the work for finding the inverse Laplace transform of this function. I.e., find f(t).

Be sure to include the symbols $\mathcal{L}$ and $\mathcal{L}^{-1}$ where appropriate.

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

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

f(s) = 8/(s² + 4);

L^-1[kb/(s²+b²)] = ksin(bt)

a = 4, k = (1/2)

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

You started with the problem$$
\mathcal L^{-1}\frac{4e^{-4s}-3}{s^2+6s+25}$$You have finally got the inverse of the$$
\frac 1 {s^2+6s+25}$$Now you have to figure out how the $4e^{-4s}$ and the $-3$ affect the answer. This is getting confusing so I'm going to let Mark take it from here.

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.

 

[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)