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Heaviside help please?/

  1. Oct 11, 2013 #1

    If you have a Heaviside function such as (u(x-2)-u(x-6), then am I right in thinking that this would be = 1 or 'on' between x=2 and x=6? What happens if you put x in front of the equation? E.g. x(u(x-2)-u(x-6))

    Im trying to understand Heaviside but I am getting quite confused by it all at the moment????
  2. jcsd
  3. Oct 11, 2013 #2


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    you are right about that it is only in the interval 2<=x<6 that u(x-2)-u(x-6=1; otherwise, it equals zero.

    Thus, x*(u(x-2)-u(x-6))=x for 2<=x<6, while 0 otherwise
  4. Oct 11, 2013 #3
    Hi Thanks for the reply. If you replaced the x at the front with a number, does this mean that the function would then be true or 'on' between 2 and 6, but at the amplitude of the number that has replaced x?
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