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First order Laplace transform

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    A process can be represented by the first order equation

    (4δy(t)/δt) + y(t) = 3u(t)

    Assume the initial state is steady (y = 0 at t = –0).

    (a) Determine the transfer function of this process in the s domain.
    (b) If the input is a ramp change in u(t) = 4t, determine the value of y(t)
    when t = 10 s.

    2. Relevant equations
    Ramp change 1/s2

    3. The attempt at a solution
    (a)
    Transfer function

    {sY(s) + y(-0)} + Y(s)/4 = 3U(s)/4

    sY(s) + Y(s)/4 = 3U(s)/4

    G(s) = Y(s)/U(s) = ¾/(s + ¼) = 4/4(s + ¼)

    (b)
    This is where i am stuck i realise this is where the ramp change 1/s2 comes in.

    Y(s) =¾/(s + ¼) × 1/s2 = ¾/s(s + ¼)
    I think this is correct but I don't know where to go from here.

    any help would be appreciated.
     
  2. jcsd
  3. Mar 27, 2016 #2

    Mark44

    Staff: Mentor

    Check your algebra. ##\frac{3/4}{s + 1/4} \times \frac 1 {s^2} \ne \frac{3/4}{s(s + 1/4)}##

    I haven't verified the rest of your work...
     
  4. Mar 27, 2016 #3
    sorry copied it in wrong
    Y(s) =¾/(s + ¼) × 1/s2 = 3/s(s + ¼)
     
  5. Mar 27, 2016 #4

    Mark44

    Staff: Mentor

    That is still incorrect.

    Also, when you write 3/s(s + 1/4), the usual interpretation is ##\frac 3 s (s + 1/4)##, which is not what you meant.
     
  6. Mar 28, 2016 #5
    [tex]Y(s) =\frac{\frac{3}{4}}{(s+\frac{1}{4})}\frac{1}{s^2} = \frac{\frac{3}{4}}{s^2(s+\frac{1}{4})}[/tex]

    Is this better?
     
  7. Mar 28, 2016 #6

    Mark44

    Staff: Mentor

    Yes. To find y(t), split up the fraction using partial fraction decomposition, and then take the inverse Laplace transform of each of the resulting terms.
     
  8. Mar 28, 2016 #7
    OK so partial fraction. [tex]\frac{a}{s^2 (s + a)} = \frac{(As + B)}{s^2} + \frac{C}{(s + a)}[/tex]
    Multiply through by [tex] s^2 (s + a) [/tex]
    Giving [tex]a = (As + B) (s + a) + Cs^2[/tex]
    Removing the brackets[tex]As^2 + Bs + Aas + Ba + C^2[/tex]
    [tex] s^0..... a = Ba.... B = 1[/tex] [tex]s^1....0 = B + Aa[/tex] substituting in for B [tex]A = \frac{-1}{a}[/tex] [tex]s^2.... 0 = A + C[/tex] substituting in for A[tex]C = \frac{1}{a}[/tex]
    gives [tex]= \frac{(-\frac{s}{a} + 1)}{s^2} + \frac{\frac{1}{a}}{(s + a)}[/tex]
    simplifying [tex]= \frac{(-\frac{s}{a} + 1)}{s^2}[/tex] to [tex] -\frac{\frac{s}{a}}{s^2} + \frac{1}{s^2} = \frac{\frac{1}{a}}{s} + \frac{1}{s^2}[/tex]
    Final result[tex]\frac{a}{s^2(s + a)} = -\frac{\frac{1}{a}}{s} + \frac{1}{s^2} + \frac{\frac{1}{a}}{(s + a)}[/tex]
    [tex]-\frac{\frac{1}{a}}{s}\text { inverse to } -\frac{1}{a}[/tex]
    [tex]\frac{1}{s^2}\text{ inverse to } t [/tex]
    [tex]\frac{\frac{1}{a}}{(s + a)}\text{ inverse to }\frac{1}{a}e^{-at}[/tex]
    [tex]\left[-\frac{1}{a} + t + \frac{1}{a}e^{-at}\right][/tex] as a = ¼ ad t = 10
    [tex]3\left[-\frac{1}{\frac{1}{4}} + 10 + \frac{1}{\frac{1}{4}}e^{\frac{1}{4}}\right][/tex]
    So [tex] y(t) = 3\left[-4 + 10 +4e^{-\frac{1}{4}}\right] = 27.35[/tex]

    Is that any where near?
     
  9. Mar 29, 2016 #8

    Mark44

    Staff: Mentor

    By my calculation, you're low by about a factor of 3.

    What you show as y(t) above is actually y(10). What do you have for y(t)? You need to check that it is a solution of the diff. equation y' + (1/4)y = 3t, and that y(0) = 0. The solution I found satisifies this diff. equation and initial condition, and I get y(10) ≈ 75.94.
     
  10. Mar 29, 2016 #9

    Mark44

    Staff: Mentor

    I think you have a mistake in the last line above.
    3u(t) = 12t, so after dividing by 4 you should get 3t. The Laplace transform of this is ##\frac 3 {s^2}##, not ##\frac {3/4}{s^2}##
     
  11. Mar 29, 2016 #10
    Yes I see the mistake there is also another.
    [tex]Y(s) = \frac{\frac{12}{4}}{s^2(s + \frac{1}{4})}[/tex]
    also the second error.
    [tex]\frac{1}{a}e^{at} = \frac{1}{\frac{1}{4}}e^{-\frac{10}{4}}[/tex]
    giving [tex]y(10) =12 \left[ -4 + 10 + 4e^{-\frac{10}{4}}\right ] = 75.94[/tex]

    Is this better?
     
    Last edited by a moderator: Mar 29, 2016
  12. Mar 29, 2016 #11

    Mark44

    Staff: Mentor

    Your answer agrees with mine.
    One thing though that you're a little sloppy on: You wrote ##\frac{1}{a}e^{at} = \frac{1}{\frac{1}{4}}e^{-\frac{10}{4}}##. The left side is a function of t, but the right side is a number. You should have ##y(t) = \frac{1}{a}e^{-at} = 4e^{-\frac{t}{4}}##. From this you get y(10) = 75.94 (approx.).

    I also don't see much advantage of bringing a into things as you did in your work.
     
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