Heaviside Function: Clarifying H(x-a) & H(a-x)

[Sudhir Regmi]

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.

 

[Replusz]

Since H(x)=-H(-x), it is true.
Hope this helps.

 

[Sudhir Regmi]

Since H(x)=-H(-x), it is true.
Hope this helps.

Thanks, it means H(a-x) = -H(x-a) is true.

 

[Replusz]

Exactly :)

 

[jasonRF]

Since H(x)=-H(-x), it is true.
Hope this helps.

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

 

[Replusz]

No. The Heaviside function is H(0)=0 and H(x<0)= -1 and H(x>1)=+1

 

[Replusz]

https://en.wikipedia.org/wiki/Heaviside_step_function

 

[Mark44]

No. The Heaviside function is H(0)=0 and H(x<0)= -1 and H(x>1)=+1

https://en.wikipedia.org/wiki/Heaviside_step_function

Take another look at that graph near the top of the page that you linked to. If x < 0, H(x) = 0, not -1.

 

[Sudhir Regmi]

Take another look at that graph near the top of the page that you linked to. If x < 0, H(x) = 0, not -1.

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.

 

[Mark44]

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.

$H(x - a) = \begin{cases} 1 & x > a \\ 0 & x < a \\ \end{cases}$
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.

 

[Sudhir Regmi]

Thank you Mark44, Replusz and jasonRF for responding to my question.