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

  1. Oct 29, 2014 #1
    Mod note: Please don't tinker with SIZE tags. Things are perfectly readable without them.
    1. The problem statement, all variables and given/known data
    Find x(t) = L-1[(4e-4s - 3)/(s2 + 6s + 25)].

    2. Relevant equations
    L(x(t)) = x(t)e-stdt
    L-1(x(s)) = (1/2π)(σ - ∞j)(σ + ∞j)[x(s)est]ds, "But you want to avoid this integral."
    Laplace Table:
    f(t), t > 0 ::::::::::::::::::::::::::::::::::::::::::::::: F(s)
    _______________________________________
    σ(t) :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1
    u(t) :::::::::::::::::::::::::::::::::::::::::::::::::::::: (1/s)
    t :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: (1/s2)
    e-at :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s + a)
    eat :::::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s - a)
    te-at ::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s+a)2
    sin(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: B/(s2 + B2)
    cos(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: s/(s2 + B2)
    _______________________________________________________
    Properties:
    L[tf(t)] = -dF(s)/ds
    L[x(at)] = (1/|a|) X (s/a), (unsure what this means at the moment though)
    L[0tf(T)dT] = F(s)/s
    Other properties

    3. The attempt at a solution

    It seems this problem is asking for doing the inverse Laplace transform on the equation
    (4e-4s - 3)/(s2 + 6s + 25) but it seems I can't just do the integral. From what I hear "Laplace tables" are pretty normal to use, but I seem to be missing how they are supposed to help here, do they apply to Inverse Laplace transforms too?
     
    Last edited by a moderator: Oct 29, 2014
  2. jcsd
  3. Oct 29, 2014 #2

    Mark44

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    I don't think they expect you to use the definition (evaluate the integral).
    Yes. If one column has the function, f(t), the other column has ##\mathcal{L}(f(t)) = F(s)##, the Laplace transform of f. To find the inverse, match a function in the transform column with its counterpart on the same row.

    Both the Laplace transform and inverse Laplace transform are linear, which means that the Laplace transform of a sum is the sum of the Laplace transforms. The same holds for the inverse Laplace transform. Break up your function into the sum of two functions, and work on those. You will also need to complete the square in the denominator to get it in the form of (s + <something>)2 + <something>.
     
  4. Oct 30, 2014 #3

    rude man

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    Break up the denominator into (s + a)(s + b), then use partial fraction expansion.
    If the poles a and b turn out to be complex conjugates you also need to remember Euler's equation.
    By no means try to do the integral. That's advanced math.
    You know how to handle L-1F(s)(e-sT) given L-1F(s) I hope ...
     
  5. Nov 2, 2014 #4
    Sorry for getting back to this so late.

    Welp, that's even more math review for me then (please bear with me again):

    I'm not even sure how to do partial fractions here. It seems I can't divide out (4e-4s - 3)/(s2 + 6s + 25) or factor out (s2 + 6s + 25) easily so I use the quadratic formula, and with that I got this:

    s = [-6 +- (36 - 100)1/2]/2

    s = [-6 +- (-64)1/2]/2

    s = [-6 +- (-64)1/2]/2

    s = [-6 +- (8i)]/2

    s = -3 + 4i and s = -3 - 4i (and I got the same thing completing the square).

    Is partial fractions with it now this?:

    4e-4s-3 = A(-4i - 3) + B(4i - 3)
     
  6. Nov 2, 2014 #5

    rude man

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    You derived the partial fraction denominators correctly.
    You next rewrite 3/(s2 + 6s + 25) in partial fractions and invert those into the time domain.
     
  7. Nov 2, 2014 #6
    Where did you get the 3 from? Ah, do you mean change -3 +- 4i to time domain as in convert to polar & phasor form (because it's rectangular form)?

    θ = tan-1(j/x)

    mag = (j2 + x2)1/2

    It would be this:

    -3 - 4i = (5 ∠ 53.1)

    3 + 4i = (5 ∠ -53.1)
     
  8. Nov 3, 2014 #7

    rude man

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    No.
    Your expression to invert is
    (4e-4s - 3)/(s2 + 6s + 25)
    = (-3)/(s2 + 6s + 25) + 4e-4s/(s2 + 6s + 25).
    The "3" is the numerator of the first term above with a minus sign. Deal with this term first, then do the e-4s part afterwards.

    In general: if you have a complex-conjugate pole pair as you have in this instance, if
    F(s) = A/(s + a + jb) + A/(s + a - jb)
    then f(t) = Ae-(a + jb)t + Ae-(a - jb)t
    and then use the Euler relation ejx = cos(x) + j sin(x)
    to get a real f(t).

    This is just a simple and obvious expansion of 1/(s + a) → e-at for complex-conjugate pole pairs.
    You have already found the poles in your post 4, to wit, s = -3 + 4i and s = -3 - 4j.
     
  9. Nov 3, 2014 #8
    I take it you would end up putting the "pole expression" where it is "a + jb" then? Not really sure what goes in place of a or jb. Unless you meant to take the terms from the expression

    What would happen to the 4 in the second term?

    Would it then be something like this?:

    f(t) = -3e-(-3 +4j)t - 4e-4se-(-3 - 4j)t.

    What would the x be for the Euler equation?
     
  10. Nov 3, 2014 #9

    rude man

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    a = -3
    b = 4

    x = b or -b.
    Study my post 7 some more, then get going with inverse-transforming A/(s+a+jb) + A/(s+a-jb).
     
  11. Nov 5, 2014 #10
    Ok so it's this:

    f(t) = -3e-(-3 +4j)t - 4e-4se-(-3 - 4j)t.

    But at the same time "s = -3 + 4i and s = -3 - 4j"? Or is this just for the imaginary (s) domain ?

    So F(s) is likely this then?:

    3/(s - 3 + 4j) + 4e-4s/(s - 3 - 4j).

    You have to do the inverse Laplace transform in F(s) first?

    Afraid I may need another hint with this, I'll get on this again soon though. Noted Mark's post that F(s) in the table is also the Laplace transform of f(t)
     
  12. Nov 5, 2014 #11

    rude man

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    Wrong.

    As I said before, forget about the 4 e-s term until you've finished with the -3 term.
    You have not followed what I stated in my post 7.
     
  13. Nov 5, 2014 #12
    It got confusing because you have A = -3 and also -3 as one of the pole terms.

    F(s) = -3/(s - 3 + 4j) - 3/(s - 3 - 4j) then

    f(t) = -3e-(-3 + 4j)t - 3e-(-3 - 4j)t

    f(t) = -3e(3 - 4j)t - 3e(3 + 4j)t

    f(t) = -3e(3t)e-4jt - 3e(3t)e4jt

    I assume you're supposed to only use the Euler relation to swap out the ejx term in f(t) then? Would it still be done with the t you have in the F(t) formula? Because it would be e-4jt and e4jt instead, wouldn't it?

    Otherwise it would be this..

    eJj = cos(J) + sin(J)j; so...

    e-4j = cos(-4) + sin(-4)j = -0.65364 + 0.7568j

    e4j = cos(4) + sin(4)j = -0.65364 - 0.7568j

    Is this correct so far?
     
  14. Nov 6, 2014 #13

    rude man

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    You have not yet correctly done the partial fraction expansion. In other words, we have

    -3/(s^2 + 6s + 25) = A/(s + a + jb) + B/(s + a - jb). (sorry, I said only "A" before but it's this) & you need to solve for A and B. If you do it right you will find B = -A.

    Also, I meant to say a = 3, not -3, in post 9. The two denominator factors are
    (s + s1) = (s + a + jb)
    (s + s2) = (s + a - jb)
    a = 3, b = 4. You're right, it's a coincidence that "3" appears in 2 places.

    You need to solve for A and B before proceeding to the Euler relation thing.
    Try to stick to a and b, not 3 and 4, until the end please. I get muddled myself with all those numbers floating around.
     
  15. Nov 6, 2014 #14
    No worries.

    -3/(s2 + 6s + 25) = A/(s + a +jb) + B/(s + a - jb)

    Since you have this:

    (s + s1) = (s + a + jb)
    (s + s2) = (s + a - jb)

    can I write this instead?

    -3/(s2 + 6s + 25) = A/(s + s1) + B/(s + s2)

    Also for partial fraction expansion the numerator degree is lower than the denominator, but how do I solve for A and B now then? Don't you set A or B equal to 0 first then solve for each of them? What is 's' (while assuming s1 and s2 are the poles I solved for earlier)?
     
  16. Nov 6, 2014 #15

    rude man

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    Yes. You should. Makes things easier down the road.
    Yes. Look this up somewhere if you have to. It's standard partial fraction expansion.
    You know what s1 and s2 are. Look above.
    What is "s"? I can't believe you asked that question.
     
  17. Nov 6, 2014 #16

    LCKurtz

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    Since using partial fractions doesn't seem to be getting anywhere with the OP, I have another suggestion. Your basic problem is to calculate$$
    {\mathcal L}^{-1}\frac 1 {s^2+6s + 9}$$If you know the formula$$
    {\mathcal L}^{-1} F(s+a) = e^{-at}{\mathcal L}^{-1}F(s)$$all you have to do is complete the square on the denominator, use that formula, and look at your sine and cosine transforms. You don't need partial fractions.
     
  18. Nov 7, 2014 #17

    rude man

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    Sure. Very elegant! But that requires availability of the sine and cosine transforms. Partial fractions is more fundamental, requiring only the pair 1/(s+a) ↔ exp(-at). (Higher-order numerators can be dealt with by f'(t) ↔ sF(s) etc.)

    You might as well then just go one step further and look up the inverse transform of the entire original function in one fell swoop, which as a matter of fact I did to double-check my p.f. derivation. Which, in fact, is what a practicing EE would do automatically of course, since he/she would have an extensive table to work with. Mine has 157 transform pairs.

    But IMO the beginning student is I think better advised to familiarize him/herself with the partial fraction expansion method, including the realization that the method covers complex-conjugate poles as well as real ones.
     
  19. Nov 7, 2014 #18

    LCKurtz

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    Of course, I understand your point. But given that the OP in fact gave table entries, my assumption would be that the exercise in question was for practice in using the tables.
     
  20. Nov 7, 2014 #19

    rude man

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    You're right, I missed seeing those on the OP's 1st post. But it still requires use of your special relation which I doubt he/she has encountered (anyway, not on his/her list.). Your approach is really clever - I hadn't thought of it - but perhaps a bit advanced for the OP's level of expertise. Thanks for your inputs!
     
  21. Nov 7, 2014 #20
    I need to find a way to put more time in to this, sorry. I'm a desperate guy by the way (if that matters, heh).

    s = -3 + 4j and s = -3 - 4j, and that's added to 's1' and 's2' in the denominator (which is it supposed to be)?

    Actually wait, I think I see you just keep 's' as the variable PLUS either of the previous pole terms (s1 and s2)? It's really been some time seen I've seen any algebra II equation like this (if at all, simplifying something with partial fractions involving imaginary terms). How do you solve for A easily then?

    But let me see..

    -3/(s2 + 6s + 25) = A/(s + s1) + B/(s + s2)

    The way I think it goes, (s - 3 + 4i)(s - 3 - 4i) composes (s2 + 6s + 25)?

    B = 0, then maybe s1 = -3 + 4i (while keeping +s in the denominator)

    -3/(s2 + 6s + 25) = A/(s - 3 + 4i)

    A = -3(s - 3 + 4i)/[(s2 + 6s + 25)]

    A = -3(s - 3 + 4i)/[(s - 3 + 4i)(s - 3 - 4i)]

    A = -3/(s - 3 - 4i) ?

    Unless there's a mistake somewhere.
     
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