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Laplace Transforms: Transfer Functions and Impulse

  1. Sep 28, 2015 #1
    1. The problem statement, all variables and given/known data
    I uploaded the question as a picture and attached it.

    2. Relevant equations
    Unit step function -
    [itex]u_c (t) =
    \begin{cases}
    1 & \text{if } t \geq c \\
    0 & \text{if } t < c
    \end{cases}[/itex]

    Impulse function -
    [itex] δ(t) = \displaystyle\lim_{Δ\rightarrow 0} δ_Δ (t)[/itex]

    Multiplication Property for Impulse function -
    [itex] f(t)⋅δ(t - t_d) = f(t_d)⋅δ(t - t_d)[/itex]
    *A function [itex]f(t)[/itex] becomes a value [itex]f(t_d)[/itex]*

    3. The attempt at a solution

    (a and b)

    I have determined that both of the transfer functions are the same,

    [itex] H(s) = V(s)/F_{p / w}(s) = {\frac{1}{75s + 0.0046}} [/itex]

    (c)

    The Laplace transform of the impulse function is 1 so,

    [itex] V(s) = {\frac{1}{75s + 0.0046}} [/itex]

    [itex] v(t) = {\frac{1}{75}} e^{{\frac{-0.0046}{75}}t} [/itex]

    (d)

    The Laplace transform of the unit step function is 1/s so,

    [itex] V(s) = {\frac{1}{s(75s + 0.0046)}} [/itex]

    [itex] v(t) = 217.391 - 217.391 e^{{\frac{-0.0046}{75}}t} [/itex]

    ***Am I right up to this point?***

    (e)

    [itex]f_{wind} (t) =
    \begin{cases}
    4.5 & \text{if }1 \geq t < 10 \\
    0 & \text{otherwise,}
    \end{cases}[/itex]

    Does that mean that from 1 -> 10 there is a constant force of only 4.5N? That just seems negligible compared to the force applied by the skaters pushes.

    [itex] f(t) = f_{wind}(t) + 1160δ(t-4) + 935δ(t-6) + 708δ(t-7.8) [/itex]

    Not quite sure how to model this!
     

    Attached Files:

  2. jcsd
  3. Sep 28, 2015 #2

    Mark44

    Staff: Mentor

    Assuming units of m/sec, this seems way too large. Can you show what you did in going from V(s) to v(t)? There's a partial fractions decomposition involved.
    You have a typo: The first case restriction is ##1 \le t < 10##
     
  4. Sep 28, 2015 #3
    I can't write them out right now since I'm on my phone. I just checked them on wolfram alpha and it gave the same answer.
     
  5. Sep 29, 2015 #4
    Ok I have fixed the function, [itex]f_{wind}(t)[/itex].

    For,

    [itex] V(s) = {\frac{1}{s(75s + 0.0046)}} [/itex]

    Partial Fraction Decomp.

    [itex] V(s) = A/s + B/(75s + 0.0046) [/itex]

    [itex] 1 = A(75s + 0.0046) + B(s) [/itex]

    subbing in s = 0,

    [itex] A = 1/0.0046 = 217.391 [/itex]

    subbing in s = -0.0046/75

    [itex] 1 = B(-0.0046/75) [/itex]

    [itex] B = -75/0.0046 [/itex]

    [itex] V(s) = 217.391/s + (-75/0.0046)/(75s + 0.0046) [/itex]

    [itex] v(t) = 217.391 - 217.391e^{-0.0046/75} [/itex]

    Could someone help me with part 5e? I'm not sure how to plot this.
     
  6. Sep 29, 2015 #5
    Here is what I have so far,

    [itex]f(t) = 1160dirac(t-4) + 935dirac(t-6) + 708dirac(t-7.8) + 4.5(u(t-1) - u(t-10))[/itex]
     
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