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

[rude man]

B = (-3/8j)

B = (3j/8)

Right!
So now, given F(s) = (-j3/8)/(s+s1) + j3/8/(s+s2), where s1 = a + jb, s2 = a - jb, a = 3, b = 4,
what is f(t)?

 

[Color_of_Cyan]

That's what I was going to ask. Any hints on that, or just look at the Laplace table? What about the 4e^-4s term from the beginning?
Well just for the -3 term from the beginning (and looking back at the Laplace table) I'd say this:
f(t) ::::::::: F(s)
e^-at ::::::::: 1/(s + a)
__________________
F(s) = (-j3/8)/(s+s1) + j3/8/(s+s2),
and a = s1 and s2? What happens to the -+3j/8 ?
I think this? (Just for the -3 at the beginning though):
=-3j/8e^{(-3 - 4j)t} + 3j/8e^{(-3 + 4j)t}

 

[rude man]

That's what I was going to ask. Any hints on that, or just look at the Laplace table? What about the 4e^-4s term from the beginning?

For the n^th time, ignore that term until you've inverse-transformed the -3 term!

Well just for the -3 term from the beginning (and looking back at the Laplace table) I'd say this:
f(t) ::::::::: F(s)
e^-at ::::::::: 1/(s + a)
__________________
F(s) = (-j3/8)/(s+s1) + j3/8/(s+s2),

Perfect!

I think this? (Just for the -3 at the beginning though):
f(t) = -3j/8e^{(-3 - 4j)t} + 3j/8e^{(-3 + 4j)t}

This is 100% right!
Now, you know f(t) must be real, so what are you going to do with the above?
Hint: Euler relation! And remember from high school, e^{(x + y)} = e^xe^y.

 

[Color_of_Cyan]

For the n^th time, ignore that term until you've inverse-transformed the -3 term!
Hint: Euler relation! And remember from high school, e^{(x + y)} = e^xe^y.

Haha, okay sorry, I thought the '-3' was already done.
f(t) = -3j/8e^{(-3 - 4j)t} + 3j/8e^{(-3 + 4j)t}
= f(t) = (-3j/[8(e^-3t)(e^-4jt] + (3j/[8(e^-3t)(e^4jt)
Euler's relation:
e^xj = (cos x) + (sin x)j
(e^-4jt) = cos(-4)t + sin (-4)jt
e^4jt = cos(4)t + sin(4)jt
Do you just plug these back in and then that's -3 done?

 

[rude man]

Haha, okay sorry, I thought the '-3' was already done.
f(t) = -3j/8e^{(-3 - 4j)t} + 3j/8e^{(-3 + 4j)t}
= f(t) = (-3j/[8(e^-3t)(e^-4jt] + (3j/[8(e^-3t)(e^4jt)

At this point, factor out the e^-3t in both terms, then work with what's left using Euler.

Euler's relation:
(e^-4jt) = cos(-4)t + sin (-4)jt
e^4jt = cos(4)t + sin(4)jt

Rewrite as e^j4t = cos(4t) + j sin(4t) etc.

 

[Color_of_Cyan]

(-3e^3tj/8)[(cos(4t) + sin(4t)j) - 1/(cos(4t) + sin(4t)j)]

I feel more comfortable putting the 'j' after the terms not before them is that bad? Is this it for the -3 term now?

 

[rude man]

(-3e^3tj/8)[(cos(4t) + sin(4t)j) - 1/(cos(4t) + sin(4t)j)]

I feel more comfortable putting the 'j' after the terms not before them is that bad?

Yes, sin(4t)j = sin(4tj) which is obviously not the same as j sin(4t). The j should really always come first, as in j100 instead of 100j. For another example, Euler is always written as e^jx = cos(x) + j sin(x).

(-3e3tj/8)[(cos(4t) + sin(4t)j) - 1/(cos(4t) + sin(4t)j)]
Is this it for the -3 term now?

No. It's (-3e^-3t), not (-3e^3t). Then, factor that out from the remaining expression.
Then, what's with the division sign all of a sudden?
You are not translating your -3t term from exponential form to sines and cosines correctly.

 

[Color_of_Cyan]

I thought I could do that because it was 1/e^-3 (for BOTH denominators), and I thought I could factor both out as just e^3, And then I divided out only one of the Euler relations.. so I can't do that or I did it wrong?

 

[rude man]

I thought I could do that because it was 1/e^-3 (for BOTH denominators), and I thought I could factor both out as just e^3, And then I divided out only one of the Euler relations.. so I can't do that or I did it wrong?

All wrong.
Look, you have F(s) = A/(s+s1) + B/(s+s2).
What is the inverse transform of F(s) for this?

 

[Color_of_Cyan]

The transform of F(s) = A/(s+s1) + B/(s+s2)

is Ae^-s1(t) + Be^-s2(t)

Can I try from post 35 one more time though?:

f(t) = (-3j/[8(e^-3t)(e^-4jt)]) + (3j/[8(e^-3t)(e^4jt)])

(1/e^-3t)[ (-j3/8(cos(-4t) + jsin(-4t))) + (j3/8(cos(4t) + jsin(4t))) ]

 

[rude man]

The transform of F(s) = A/(s+s1) + B/(s+s2)

is Ae^-s1(t) + Be^-s2(t)

No. Don't say "s₁(t). Say "s₁t" etc. So correct would be
"The inverse transform of F(s) = A/(s+s₁) + B/(s+s₂)
is Ae^-s₁t + Be^-s₂t ... (1)

Can I try from post 35 one more time though?:

f(t) = (-3j/[8(e^-3t)(e^-4jt)]) + (3j/[8(e^-3t)(e^4jt)])

This is almost correct. Fix up your parentheses.

(1/e^-3t)[ (-j3/8(cos(-4t) + jsin(-4t))) + (j3/8(cos(4t) + jsin(4t))) ]

How did you manage to get (1/e^-3t)?
Also, your parentheses don't look right. Too many of them!
Other than that this looks OK now.
So now combine (-j3/8)(cos(-4t) + jsin(-4t)) + (j3/8)(cos(4t) + jsin(4t)) using Euler to come up with a real function of time. In other words, all the j's have to disappear.

 

[rude man]

Since you are seemingly having a lot of trouble with this complex math maybe you should look at posts 2 and 16 again. It would avoid all the complex math if you do it that way. Let us know if you want this explained to you.

 

[LCKurtz]

Since you are seemingly having a lot of trouble with this complex math maybe you should look at posts 2 and 16 again. It would avoid all the complex math if you do it that way. Let us know if you want this explained to you.

I have been waiting for the OP to get done with this method to suggest that. But note that I had a typo in post #16 which I can't edit. It should have said complete the square on$$\frac 1 {s^2+6s+25}$$and use that shifting formula.

 

[rude man]

I have been waiting for the OP to get done with this method to suggest that. But note that I had a typo in post #16 which I can't edit. It should have said complete the square on$$\frac 1 {s^2+6s+25}$$and use that shifting formula.

Hadn't noticed that. Right.

EDIT: If the OP wants to investigate it that way, is it OK with you if I turn him over to you?

 

[Color_of_Cyan]

Since you are seemingly having a lot of trouble with this complex math maybe you should look at posts 2 and 16 again. It would avoid all the complex math if you do it that way. Let us know if you want this explained to you.

I'd be able to make it look better here on the forums if only I knew the way Kurtz posted his fractions (like on post 43 on here)...

$$\frac {1} {e^{-3t}} ( \frac {-j3} {8(cos(-4t) + jsin(-4t))} + \frac {j3} {8(cos(4t) + jsin(4t))})$$

Much better :) And it still looks like a challenge to get rid of all the 'j's (is that even possible?). What did you mean use "Euler"?

I have been waiting for the OP to get done with this method to suggest that. But note that I had a typo in post #16 which I can't edit. It should have said complete the square on$$\frac 1 {s^2+6s+25}$$and use that shifting formula.

You said take the square (s² + 6s + 9 + 16) and do this?

L^-1F(S + a) = e^-atL^-1F(s)

So is s = (s+3)² and a = 16?

What do you have for L^-1F(s) then?

Don't go anywhere Rude Man!

 

[rude man]

I'd be able to make it look better here on the forums if only I knew the way Kurtz posted his fractions (like on post 43 on here)...

$$\frac {1} {e^{-3t}} ( \frac {-j3} {8(cos(-4t) + jsin(-4t))} + \frac {j3} {8(cos(4t) + jsin(4t))})$$

Much better :) And it still looks like a challenge to get rid of all the 'j's (is that even possible?). What did you mean use "Euler"?

This is "much better"? It's not. Not at all. It's a disaster!
Euler: e^jx = cos(x) + jsin(x)
Please go back to post 41, last part.
I'm going to let LCKurtz guide you thru his approach.

 

[LCKurtz]

You said take the square (s² + 6s + 9 + 16) and do this?

L^-1F(S + a) = e^-atL^-1F(s)

So is s = (s+3)² and a = 16?

What do you have for L^-1F(s) then?

Don't go anywhere Rude Man!

You have $\mathcal L^{-1}\frac 1 {(s+3)^2 + 4^2}$ so that fraction is $F(s+3)$. So what does the equation I highlighted in red tell you for this problem?

 

[Color_of_Cyan]

This is "much better"? It's not. Not at all. It's a disaster!
Euler: e^jx = cos(x) + jsin(x)
Please go back to post 41, last part.
I'm going to let LCKurtz guide you thru his approach.

The code works fine for me in Chrome, and I simplified it the way you said before, didn't I? Or was I supposed to simplify the 1/e^{3 +- 4j} out again?

You have $\mathcal L^{-1}\frac 1 {(s+3)^2 + 4^2}$ so that fraction is $F(s+3)$. So what does the equation I highlighted in red tell you for this problem?

Ah, so you use the table as much as possible for wherever it's L^-1F(s)?

Because looking back at it, F(s) with b/(s² + a²) = sin bt

so sin(4t) = L^-1 of that ?

Then it's sin(4t)e^-4t?

 

[LCKurtz]

Ah, so you use the table as much as possible for wherever it's L^-1F(s)?

Because looking back at it, F(s) with b/(s² + a²) = sin bt

Those aren't equal. You have $s$ on the left and $t$ on the right.

so sin(4t) = L^-1 of that ?

Then it's sin(4t)e^-4t?

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.

 

[NascentOxygen]

CoC, it's often helpful when you can check your working as you proceed to learn from your mistakes, and wolfram alpha is of incalculable value here. A couple of examples apposite to the mathematics you have been using:-

e.g., http://m.wolframalpha.com/input/?i=partial fractions -3/((s+3+i4)(s+3-i4))&x=0&y=0

e.g., http://m.wolframalpha.com/input/?i=inverse Laplace Transform 1/((s+3)^2+4^2)&x=0&y=0
maybe scroll down to alternate forms (for some reason if I use a lower-case "t" in spelling "transform" it doesn't list the explicit "sin" solution.)

Good luck with your studies!

 

[Color_of_Cyan]

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)
e^at :::::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s - a)
te^-at ::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s+a)²

 

[LCKurtz]

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)
e^at :::::::::::::::::::::::::::::::::::::::::::::::::::::::: 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.

 

[Color_of_Cyan]

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?

 

[LCKurtz]

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?

 

[Color_of_Cyan]

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)

 

[LCKurtz]

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)

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.

 

[Color_of_Cyan]

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.

 

[LCKurtz]

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.

 

[Color_of_Cyan]

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)?

 

[Mark44]

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.

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)?