# Heaviside Step Function?

• Aug 15th 2008, 11:58 AM
scott_237
Heaviside Step Function?
This seems like an easy enough question to do but could somebody explain how to do question 8)b) HERE using the heaviside step functions? Thanks a lot
• Aug 15th 2008, 01:40 PM
CaptainBlack
Quote:

Originally Posted by scott_237
This seems like an easy enough question to do but could somebody explain how to do question 8)b) HERE using the heaviside step functions? Thanks a lot

First there is an ambiguity in how the Heaviside step is defined at the jump. if you are working with the $\mathcal{H}(0)=1/2$ definition you can't match the function at the jumps. I'm not geing to worry about the definition at the jumps and assume the definition of $\mathcal{H}$ that you have is OK for this function.

The jumps are at $x=-1$ and $x=+1$, so:

$f(x)=(-1-x)+\mathcal{H}(x+1)[(1+x)+2]+\mathcal{H}(x-1)[-2+(x-1)^2]$

That is the first term on the right is the function definition on $x<-1$ extended to the whole of $\mathbb{R}$ , the second term is is zero for $x<-1$ and for $x \ge 1$ subtracts the first term to get rid of it where we dont need it and adds in the function on the diddle segment of the domain.

The third term removes the middle function and adds in the right most part of the function for $x \ge 1.$

RonL