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I Heaviside function

  1. Dec 14, 2016 #1
    Hi everyone,
    I could not figure out whether H(x-a) and H(a-x) are same or different. Please help me to understand this.
     
  2. jcsd
  3. Dec 14, 2016 #2
    • Member warned to be careful of posting incorrect information
    Since H(x)=-H(-x), it is true.
    Hope this helps.
     
  4. Dec 14, 2016 #3
    Thanks, it means H(a-x) = -H(x-a) is true.
     
  5. Dec 14, 2016 #4
    Exactly :)
     
  6. Dec 14, 2016 #5

    jasonRF

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    Science Advisor
    Gold Member

    The Heaviside step function is usually define: H(x) = 0 for x<0, and H(x) = 1 for x>0. So it is not true that H(x) = -H(-x). For example, H(1) = 1 but H(-1) = 0.

    Jason
     
  7. Dec 14, 2016 #6
    No. The Heaviside function is H(0)=0 and H(x<0)= -1 and H(x>1)=+1
     
  8. Dec 14, 2016 #7
  9. Dec 14, 2016 #8

    Mark44

    Staff: Mentor

  10. Dec 14, 2016 #9
    Hello Mark44,
    What do you think about my original question, is H( x-a) equal to H( a-x)? Where x is a variable and a is a constant.
     
  11. Dec 14, 2016 #10

    Mark44

    Staff: Mentor

    ##H(x - a) = \begin{cases} 1 & x > a \\ 0 & x < a \\ \end{cases}##
    H(a - x) = H(-(x - a)) -- this is the reflection of the graph of H(x - a) across the line x = a, so the graphs of these two functions are not the same.
     
  12. Dec 14, 2016 #11
    Thank you Mark44, Replusz and jasonRF for responding to my question.
     
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