Heaviside help please?/

  • #1
Hi,

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????
 

Answers and Replies

  • #2
arildno
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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
 
  • #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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