Heaviside Function

[RobertPaulson]

I have a brief understanding of this topic but clearly not enough and I can't find any simple answers in my notes etc.

Basically the question asks to find the Laplace transform of g(x) where
\(\displaystyle g(x)=\left\{\begin{array}{lcl} 0, \;\ x<0\\ 1,\;\ 0\leq x<1\\0, \;\ x\geq 1\end{array}\right\)

I understand this function but they've said in the answers that this simply is \(\displaystyle g(x) = H(x) - H(x-1)\)

\(\displaystyle H(x)=\left\{\begin{array}{rcl}0, \;\ x<0\\1, \;\ x>0\end{array}\right\)

I don't know how they got this and I don't know how to represent H(x-1)

Please help me because I know this is probably really really simple :s

 

[Deleted member 4993]

I have a brief understanding of this topic but clearly not enough and I can't find any simple answers in my notes etc.

Basically the question asks to find the Laplace transform of g(x) where
\(\displaystyle g(x)=\left\{\begin{array}{lcl} 0, \;\ x<0\\ 1,\;\ 0\leq x<1\\0, \;\ x\geq 1\end{array}\right\)

I understand this function but they've said in the answers that this simply is \(\displaystyle g(x) = H(x) - H(x-1)\)

\(\displaystyle H(x)=\left\{\begin{array}{rcl}0, \;\ x<0\\1, \;\ x>0\end{array}\right\)

I don't know how they got this and I don't know how to represent H(x-1)

Please help me because I know this is probably really really simple :s

H(x-1) is same as H(x) - shifted 1 unit to the right (on x-axis). You have done this type of translation of functions in pre-calc.

 

[RobertPaulson]

H(x-1) is same as H(x) - shifted 1 unit to the right (on x-axis). You have done this type of translation of functions in pre-calc.

So the Heaviside H(x-1) is given as 0, x < 1; 1, x > 1 ?

If this is the case, I still don't understand how g(x) = H(x) - H(x-1)

 

[Deleted member 4993]

H(x-1) is same as H(x) - shifted 1 unit to the right (on x-axis). You have done this type of translation of functions in pre-calc.

So the Heaviside H(x-1) is given as 0, x < 1; 1, x > 1 ? <<< Correct

If this is the case, I still don't understand how g(x) = H(x) - H(x-1)

Sketch H(x) and H(x-1) - then ponder [H(x) - H(x-1)].... It should be clear....