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Partial fractions before Inverse Laplace

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data
    I have this lowpass circuit which I have transformed to the S-domain.
    The circuit is to be exposed to a unit step, and then I shall convert the transient response to the time domain.

    Here's the transfer function of the lowpass circuit:

    [tex]H(s) = \frac{\frac{1}{LC}}{s^2 + s \frac{1}{RC} + \frac{1}{LC}}[/tex]

    [itex]\frac{1}{LC} = 1000000[/itex]
    [itex]\frac{1}{RC} = 0,001415[/itex]


    The function of the unit step is
    [tex]x(t)=1 --> X(s) = \frac{1}{s}[/tex]


    2. Relevant equations

    [tex]
    Y(s) = H(s) * X(s)
    [/tex]


    [tex]
    Y(s) = \frac{\frac{1}{LC}}{s^2 + s \frac{1}{RC} + \frac{1}{LC}} * \frac{1}{s}
    [/tex]

    3. The attempt at a solution

    Now, my problem is that I have great difficulties "arranging" the equation before converting it back to the time domain.
    I know that it involves some partial fractions and some unknows (A, B, C and so forth), but even though I have studied the relevant subject in my text book, I cant f*cking do it.

    I'll show you what I got so far (wrong as it may be)

    [tex]
    1: \frac{A}{s} * \frac{B}{s^2 + s \frac{1}{RC} + \frac{1}{LC}} = \frac{\frac{1}{LC}}{s*(s^2 + s \frac{1}{RC} + \frac{1}{LC}})
    [/tex]

    [tex]
    2: A*(s^2 + s \frac{1}{RC} + \frac{1}{LC}) + Bs = 1
    [/tex]


    I feel that I'am wondering in the dark, so if someone could point me in the right direction or even shed some light over what I am doing and how I am suppose to do it I would be very very happy :)
     
  2. jcsd
  3. Apr 26, 2012 #2

    Mark44

    Staff: Mentor

    Let's simplify things a bit by writing what you have as
    $$Y(s) = \frac{b}{s(s^2 + as + b)}$$

    Here a = 1/(RC) and b = 1/(LC)

    The decomposition you need to use depends on whether s2 + as + b has real factors or is an irreducible quadratic.

    If s2 + as + b can be factored into (s - r1)(s - r2), then Y(s) can be decomposed like this:
    $$Y(s) = \frac{A}{s} + \frac{B}{s - r_1} + \frac{C}{s - r_2}$$

    A, B, and C are constants; C is unrelated to the capacitance C of the problem.

    If s2 + as + b is an irreducible quadratic, then the decomposition goes like this:
    $$Y(s) = \frac{A}{s} + \frac{Bs + C}{s^2 + as + b} $$

    Whichever form is applicable, the idea is to solve for the constants A, B, and C, so that you can more easily find the inverse Laplace transform, resulting in y(t).
     
  4. Apr 26, 2012 #3

    Mark44

    Staff: Mentor

    I should mention that a and b are different from A and B.
     
  5. Apr 26, 2012 #4
    Ok, thanks.

    The demoninator is irreducible quadratic.

    I am working may way further now and I'll get back when/if it stops
     
  6. Apr 26, 2012 #5
    OK, I am officially stuck - again.

    I am looking at the below example, but I cant figure out where the numbers in the red squares comes from (10, 13, 60).

    https://docs.google.com/open?id=0B-sl9wXn3g43b3FadGg1VUU1X2s [Broken]


    Edit: Ok, I guess the number 10 is the numerator in F(s) and 13 is the last part of the equation in the denominator. That leaves "60". Where the hell comes that from?
     
    Last edited by a moderator: May 5, 2017
  7. Apr 26, 2012 #6

    Mark44

    Staff: Mentor

    The numbers in the red squares are actually 10/13, -10/13, and -60/13.

    Starting with
    $$\frac{10}{s(s^2 + 6s + 13)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 6s + 13}$$

    multiply both sides by s(s2 + 6s + 13).

    That gives you
    10 = A(s2 + 6s + 13) + s(Bs + C)

    This equation is actually an identity that must hold for all values of s, other than the three that make the denominators zero in the original equation.

    Solve for A, B, and C, and you should get 10/13, -10/13, and -60/13, respectively, assuming the work in the photo you sent is correct.
     
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