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

AI Thread Summary
The discussion revolves around finding the inverse Laplace transform of the function (4e-4s - 3)/(s^2 + 6s + 25). Participants emphasize the importance of using Laplace tables for both transforms and suggest breaking the function into simpler parts for easier handling. Completing the square in the denominator is recommended to identify complex conjugate poles, which can then be addressed using partial fraction expansion. The conversation highlights the need to avoid direct integration for the inverse transform, as it is deemed too complex for this context. Overall, the approach focuses on leveraging known properties and methods of Laplace transforms to derive the solution efficiently.
  • #51
LCKurtz said:
Those aren't equal. You have ##s## on the left and ##t## on the right.It's close but you are being very careless. What is ##a##? What is ##b##? You have to match up the table entries exactly with your problem.

I rushed that too much and then got too excited, sorry. It seems I can't really use anything else given to me then on the table here yet; on the table only these seem to come close:

sin(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: B/(s2 + B2)
cos(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: s/(s2 + B2)
e-at :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s + a)
eat :::::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s - a)
te-at ::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s+a)2
 
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  • #52
Color_of_Cyan said:
I rushed that too much and then got too excited, sorry. It seems I can't really use anything else given to me then on the table here yet; on the table only these seem to come close:

sin(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: B/(s2 + B2)
cos(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: s/(s2 + B2)
e-at :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s + a)
eat :::::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s - a)
te-at ::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s+a)2

It doesn't facilitate the discussion for you to just copy the table. I already know what's in the table. Answer the question what is ##a## and what is ##b##. Write down what you are taking the inverse of. Like$$\mathcal L^{-1}(...) =~ ?$$I have to know what you are inversing at any step to show you what your mistakes are.
 
  • #53
In the one above a = 4 and b = 1. In other words I can't really figure out how to find that inverse laplace:

L-1[1/((s+3)2 + 42)]

Does this mean s isn't really equal to s+3 in this case then?
 
  • #54
Your basic problem is to calculate$$
{\mathcal L}^{-1}\frac 1 {s^2+6s + 25} ={\mathcal L}^{-1} \frac 1 {(s+3)^2+4^2}$$and you have the general formula you can use on your problem$$
{\mathcal L}^{-1} F(s+a) = e^{-at}{\mathcal L}^{-1}F(s)$$
Please quote just this post and answer the following two questions to get started:
1. What is ##a## in your problem?
2. What is ##F(s)## in your problem?
 
Last edited:
  • #55
LCKurtz said:
Your basic problem is to calculate$$
{\mathcal L}^{-1}\frac 1 {s^2+6s + 9} ={\mathcal L}^{-1} \frac 1 {(s+3)^2+4^2}$$and you have the general formula you can use on your problem$$
{\mathcal L}^{-1} F(s+a) = e^{-at}{\mathcal L}^{-1}F(s)$$
Please quote just this post and answer the following two questions to get started:
1. What is ##a## in your problem?
2. What is ##F(s)## in your problem?

The closest given property I have that looks like that is L[e-atf(t)] = F(s + a) otherwise I don't have it.

I would say a is either 16 or 4 and F(s) = 1/(s2 + 6s + 9)
 
  • #56
LCKurtz said:
Please quote just this post and answer the following two questions to get started:
1. What is ##a## in your problem?
2. What is ##F(s)## in your problem?
Color_of_Cyan said:
The closest given property I have that looks like that is L[e-atf(t)] = F(s + a) otherwise I don't have it.

I would say a is either 16 or 4 and F(s) = 1/(s2 + 6s + 9)

You have$$
{\mathcal L}^{-1} \frac 1 {(s+3)^2+4^2}$$that you are trying to calculate and you have the formula$$
{\mathcal L}^{-1} F(s+a) = e^{-at}{\mathcal L}^{-1}F(s)$$and you think ##a=4## or ##a=16##? If you are going to use that formua don't you have to have$$
F(s+a) = \frac 1 {(s+3)^2+4^2}$$Think again about what ##a## is and what ##F(s)## is in your problem. You have to figure that out to use the right side of the formula.
 
  • #57
LCKurtz said:
You have$$
{\mathcal L}^{-1} \frac 1 {(s+3)^2+4^2}$$that you are trying to calculate and you have the formula$$
{\mathcal L}^{-1} F(s+a) = e^{-at}{\mathcal L}^{-1}F(s)$$and you think ##a=4## or ##a=16##? If you are going to use that formua don't you have to have$$
F(s+a) = \frac 1 {(s+3)^2+4^2}$$Think again about what ##a## is and what ##F(s)## is in your problem. You have to figure that out to use the right side of the formula.
Well okay, to review, F(s+a) means whatever F(s+a) is equal to is a function of 2 variables then? (It's like, I need a time machine when reviewing all of this. Or maybe just a better moment to think more about it). All I am doing first is finding the function F(s) then?

I think it is this:

$$ F(s) = \frac {1} {(s^{2} + a^{2})} $$

with a = 4,

That formula was not on my list of properties, but thanks for telling me.
 
  • #58
You still don't have it. You have$$F(s+a) = \frac 1 {(s+3)^2 + 16}$$It should be obvious to you that in this setting ##a=3## and$$F(s) = \frac 1 {s^2+16}$$ So the formula gives you$$
\mathcal L^{-1}\frac 1 {(s+3)^2 + 16} = e^{-3t}\mathcal L^{-1}\frac 1 {s^2 + 16}$$Hopefully you can finish it now with the table. I'm coming very close to violating forum rules by working the whole problem for you.
 
  • #59
LCKurtz said:
You still don't have it. You have$$F(s+a) = \frac 1 {(s+3)^2 + 16}$$It should be obvious to you that in this setting ##a=3## and$$F(s) = \frac 1 {s^2+16}$$
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'

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)?
 
  • #60
Color_of_Cyan said:
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.
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.
Color_of_Cyan said:
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'

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)?
 
  • #61
Mark44 said:
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)?
 
  • #62
Color_of_Cyan said:
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}##?
 
  • #63
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.
 
  • #64
Mark44 said:
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.
Color_of_Cyan said:
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.
 
  • #65
Color_of_Cyan said:
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.
 
  • #66
Mark44 said:
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?
 
  • #67
Color_of_Cyan said:
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.
 
  • #68
Color_of_Cyan said:
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)?
 
  • #69
Color_of_Cyan said:
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.
 
  • #70
Mark44 said:
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)

Mark44 said:
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)
LCKurtz said:
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.
 
  • #71
Color_of_Cyan said:
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."
Color_of_Cyan said:
L1[e-at] = 1/(s+a)
Closer, but you shouldn't have the exponent.
Color_of_Cyan said:
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.
Color_of_Cyan said:
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.
Color_of_Cyan said:
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.
Color_of_Cyan said:
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.
 
  • #72
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)
 
  • #73
Color_of_Cyan said:
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)).
 
  • #74
Mark44 said:
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'?
 
  • #75
Color_of_Cyan said:
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.
 
  • #76
LCKurtz said:
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.
LCKurtz said:
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) ?
 
  • #77
Color_of_Cyan said:
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.
 
  • #78
LCKurtz said:
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:

$$ \frac { 4e^{-4s}(2s + 6) + (16e^{-4s})(s^{2} + 6s + 25) } {(s^{2} + 6s + 25)^{2}} $$
 
  • #79
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.
 
  • #80
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) :H
 
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