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Laplace Transforms Involving: Unit-Step, and Ramp Functions

  1. Sep 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Here is an imgur link to my assignment: http://imgur.com/N0l2Buk
    I also uploaded it as a picture and attached it to this post.

    2. Relevant equations

    [itex]u_c (t) =
    \begin{cases}
    1 & \text{if } t \geq c \\
    0 & \text{if } t < c
    \end{cases}[/itex]

    3. The attempt at a solution

    Question 1.1 -

    [itex]L[tu(t)] = \int_0^∞ tu(t)e^{-st} \,dt[/itex]

    Using the definition of the step function, [itex] t \geq 0, u(t) = 1[/itex]
    *Is it right to assume that [itex] c = 0 [/itex]?*

    [itex]L[tu(t)] = \int_0^∞ t(1)e^{-st} \,dt[/itex]

    [itex]L[tu(t)] = \int_0^∞ te^{-st} \,dt[/itex]

    [itex]L[tu(t)] = 1/s^2 [/itex]

    I'm not sure if this is correct. Should it be solved using the rule, [itex] L[tf(t)] = -F'(s)[/itex]

    Question 1.2 -

    Let [itex] r_1 (t), r_2 (t) [/itex] be the two ramp functions

    Let [itex] u_1 (t), u_2 (t) [/itex] be the two unit-step functions

    [itex]r_1 (t) =
    \begin{cases}
    t & \text{if } 0 \leq t < 1
    \end{cases}[/itex]

    [itex]r_2 (t) =
    \begin{cases}
    t+1 & \text{if } 1 \leq t < 2
    \end{cases}[/itex]

    [itex]u_2 (t) =
    \begin{cases}
    3 & \text{if } 2 \leq t < 4
    \end{cases}[/itex]

    I'm not quite sure what to do for the unit-step functions. Could someone help me figure out what they should be?
     

    Attached Files:

    Last edited: Sep 19, 2015
  2. jcsd
  3. Sep 20, 2015 #2

    Orodruin

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    Your transformation is correct. For the second part, I do not think you have done what was intended. You are supposed to write the function as a linear combination, not as different expressions in different regions.
     
  4. Sep 20, 2015 #3
    So would it be something like,

    Let [itex]y(t)[/itex] be the function on the graph that I am trying to recreate.
    ** Is this true, [itex]r(t) = tu(t)[/itex]? **

    So,

    [itex]y(t) = tu(t) - (t+1)u(t-1) - 3u(t-2)[/itex]

    Although I still don't know what the second unit-step function should be?
     
    Last edited: Sep 20, 2015
  5. Sep 20, 2015 #4

    Orodruin

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    Have you tried plotting the function you gave? How does it differ from the one you should find?
     
  6. Sep 20, 2015 #5

    vela

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    Try plotting the function g(t) = u(t-1)-u(t-2). You should see it's a pulse. Now consider what you'll get if you multiply g(t) by some function. Try plotting the function f(t) alone and the product f(t)g(t). For instance, try it with f(t) = t2. Do you understand the effect of multiplying by g(t) on the graph?
     
  7. Sep 20, 2015 #6
    I tried plotting it and from what I see it's not similar at all.
     
  8. Sep 20, 2015 #7
    So multiplying f(t) by g(t) gives me a point on that function?
     
  9. Sep 20, 2015 #8

    vela

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    Which function? Which point? Can you elaborate?
     
  10. Sep 20, 2015 #9
    So for g(t) that you suggested it gave a pulse at (1,1), then when I plotted [itex]f(t) = t^2[/itex] I got a parabola. When I multiplied f(t) by g(t), I got a pulse again at (1,1).
     
  11. Sep 20, 2015 #10

    Orodruin

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    I suggest that you think along the following lines:
    • What happens when you pass the step of a step function? What changes? Where do these changes occur for the sought function?
    • What happens when you pass the base of a function of the form ##(t-a) u(t-a)##? Where do these changes occur for the sought function?
     
  12. Sep 20, 2015 #11

    vela

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    I don't think you plotted g(t) correctly then. I'm not sure what you mean by a pulse at (1,1). I've attached a plot of what you should've gotten.
     

    Attached Files:

  13. Sep 20, 2015 #12
    So I did some work with graphing the equations and I think this is the right one,

    [itex] y(t) = tu(t) + u(t-1) - (t-2)u(t-2) - 3u(t-4) [/itex]

    Let me know if this latex is working or not, I'm on my phone and can't see if it's showing up correctly!
     
  14. Sep 20, 2015 #13

    Orodruin

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    Looks reasonable.
     
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