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Conceptual Misunderstanding in dealing with Heaviside Function

  1. Oct 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi guys, so I'm struggling with understanding the heaviside function clearly. In particular, I am struggling to understand how changing f(t) to f(t-3) ''shifts'' everything to the right.

    2. Relevant equations

    Say we take the following heaviside function

    f(t-3) = 7 for t >= 3
    f(t-3) = 0 for t < 0

    We get a graph that loops something like this:

    http://img41.imageshack.us/img41/2830/q8e8.png [Broken]


    3. The attempt at a solution

    However, this doesn't make much intuitive sense to me. Say we are to plug in the value of t=3, then we should get at t=3, the value of f(0), but there has not been anything told to us about how the function behaves at f(0). Only that for values t>3, the value will be 7. So I don't really understand how this function is working.

    Thanks for your time.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 29, 2013 #2

    Mark44

    Staff: Mentor

    This is a basic property of functions in general, not just the Heaviside function. If you know what the graph of y = f(t) looks like, then the graph of y = f(t - 3) will be a shift (translation) to the right by 3 units of the graph of y = f(t).

    Suppose y = g(t) = t2. The graph is a parabola with vertex at (0, 0) and opening upward. What does the graph of the new function y = g(t - 2) = (t - 2)2 look like?
    You've been told that if t ≥ 3, then t - 3 ≥ 0, so the output of your function is 7.
    The function definition says t ≥ 3, not t > 3.
     
    Last edited by a moderator: May 6, 2017
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