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Graphing a Heaviside unit function

  1. Apr 13, 2008 #1
    [SOLVED] Graphing a Heaviside unit function

    1. The problem statement, all variables and given/known data
    find solution of the following differential eq and graph it:

    [tex] y'' + 4y = \delta(t-2\pi)[/tex]
    [tex]y(0)=0[/tex]
    [tex]y'(0)=0[/tex]

    2. Relevant equations
    [tex] \delta[/tex]
    is the Dirac delta function
    [tex] u_{c} [/tex]
    is the Heaviside unit step function


    3. The attempt at a solution

    I used the laplace transform and found the solution to be:

    [tex]\frac{1}{2}u_{\pi}(sin(2(t-\pi)))-\frac{1}{2}u_{2\pi}(sin(2(t-2\pi)))[/tex]

    which i checked and it is right. However, I'm not sure how to graph this. The following is what I have:

    before t = pi and after t = 2pi, y = 0.
    But, in between what will it be? If anyone could please help me out I would greatly appreciate it.
     
  2. jcsd
  3. Apr 13, 2008 #2
    As the sine function is periodic, you can write the expression as
    [tex]
    \frac{1}{2}sin2t(u_{\pi}-u_{2\pi})
    [/tex]
    Now can you proceed?
     
  4. Apr 13, 2008 #3
    Thank you for the help! So, then from pi to 2 pi I will just have that sine wave? Also, if you could please help me figure this out:

    so then is this:
    [tex]\frac{1}{2}u_{\pi}(sin(2(t-\pi)))-\frac{1}{2}u_{2\pi}(sin(2(t-2\pi)))[/tex]
    not the same as this?:
    [tex]\frac{1}{2}u_{\pi}(sin(2(t-\pi)))+\frac{1}{2}u_{2\pi}(-sin(2(t-2\pi)))[/tex]
     
  5. Apr 13, 2008 #4
    Correct :)

    About the second question, yes you can take the minus sign with the sine term, that's basic multiplication. sin(2t-2pi) = sin(2t-4pi) = sin(2t)
     
  6. Apr 13, 2008 #5
    Thanks again, but i'm not sure that i stated the second question right. I mean that if i bring the minus sign inside of the parenthesis it implies to me (maybe incorrectly) that the second sine wave will start at 2pi and propagate to infinity.
     
  7. Apr 13, 2008 #6
    Okay let me clear things up a bit,
    You follow how
    [tex]\frac{1}{2}u_{\pi}(sin(2(t-\pi)))-\frac{1}{2}u_{2\pi}(sin(2(t-2\pi)))[/tex]
    becomes
    [tex]\frac{1}{2}sin(2t)u_{\pi}-\frac{1}{2}sin(2t)u_{2\pi}[/tex]
    correct?
    Now the first term is the sine function from pi to infinity, the second term is the sine function from 2pi to infinity. Taking difference of the two graphs gives the combined graph.
     
    Last edited: Apr 13, 2008
  8. Apr 13, 2008 #7
    Ahh, I for some reason pictured this differently. But i definitely see now. Thank you for the help arunbg!
     
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