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 ejx = cos(x) + j sin(x).Color_of_Cyan said:(-3e3tj/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?
No. It's (-3e-3t), not (-3e3t). Then, factor that out from the remaining expression.(-3e3tj/8)[(cos(4t) + sin(4t)j) - 1/(cos(4t) + sin(4t)j)]
Is this it for the -3 term now?
All wrong.Color_of_Cyan said: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?
No. Don't say "s1(t). Say "s1t" etc. So correct would beColor_of_Cyan said:The transform of F(s) = A/(s+s1) + B/(s+s2)
is Ae-s1(t) + Be-s2(t)
This is almost correct. Fix up your parentheses.Can I try from post 35 one more time though?:
f(t) = (-3j/[8(e-3t)(e-4jt)]) + (3j/[8(e-3t)(e4jt)])
How did you manage to get (1/e-3t)?(1/e-3t)[ (-j3/8(cos(-4t) + jsin(-4t))) + (j3/8(cos(4t) + jsin(4t))) ]
rude man said: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.
Hadn't noticed that. Right.LCKurtz said: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.
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)...rude man said: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.
You said take the square (s2 + 6s + 9 + 16) and do this?LCKurtz said: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.
This is "much better"? It's not. Not at all. It's a disaster!Color_of_Cyan said: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"?
Color_of_Cyan said: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!
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/e3 +- 4j out again?rude man said:This is "much better"? It's not. Not at all. It's a disaster!
Euler: ejx = cos(x) + jsin(x)
Please go back to post 41, last part.
I'm going to let LCKurtz guide you thru his approach.
Ah, so you use the table as much as possible for wherever it's L-1F(s)?LCKurtz said: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 said: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 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.
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
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?
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)
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?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.
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.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$$F(s) = \frac 1 {s^2+16}$$
It's probably much simpler than you seem to be thinking.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.
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)?
So you don't need to worry about the L-1 then, just as long as it's the same format?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: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 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.
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.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?
That's pretty evident.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.
What do you mean by 'that'? That I'm really bad at this or that it's really (1/4)sin(4t)? :(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.
Maybe some practice would help this stuff gel.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.
It would help us if you wrote complete mathematical statements. Is what really (1/4)sin(4t)?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?
Color_of_Cyan said:So what am I to do with the (1/4)e-3tsin(4t) now?
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.
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.
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.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.